TurbomachineryDesign and Theory Rama S. R. Gorla Cleveland State University Cleveland, Ohio, U.S.A. Aijaz A. Khan N.E.D. University of Engineering and Technology Karachi, Pakistan M A R C E L MARCEL DEKKER, INC. D E K K E R NEW YORK . BASEL Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Although great care has been taken to provide accurate and current information, neither the author(s) nor the publisher, nor anyone else associated with this publication, shall be liable for any loss, damage, or liability directly or indirectly caused or alleged to be caused by this book. The material contained herein is not intended to provide specific advice or recommendations for any specific situation. Trademark notice: Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress. ISBN: 0-8247-0980-2 This book is printed on acid-free paper. Headquarters Marcel Dekker, Inc., 270 Madison Avenue, New York, NY 10016, U.S.A. tel: 212-696-9000; fax: 212-685-4540 Distribution and Customer Service Marcel Dekker, Inc., Cimarron Road, Monticello, New York 12701, U.S.A. tel: 800-228-1160; fax: 845-796-1772 Eastern Hemisphere Distribution Marcel Dekker AG, Hutgasse 4, Postfach 812, CH-4001 Basel, Switzerland tel: 41-61-260-6300; fax: 41-61-260-6333 World Wide Web http://www.dekker.com The publisher offers discounts on this book when ordered in bulk quantities. For more information, write to Special Sales/Professional Marketing at the headquarters address above. Copyright q 2003 by Marcel Dekker, Inc. All Rights Reserved. Neither this book nor any part may be reproduced or transmitted in any formor by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage and retrieval system, without permission in writing from the publisher. Current printing (last digit): 10 9 8 7 6 5 4 3 2 1 PRINTED IN THE UNITED STATES OF AMERICA Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved MECHANICAL ENGINEERING A Series of Textbooks and Reference Books Founding Editor L. L. Faulkner Columbus Division, Battelle Memorial Institute and Department of Mechanical Engineering The Ohio State University Columbus, Ohio 1. Spring Designer's Handbook, Harold Carlson 2. Computer-Aided Graphics and Design, Daniel L. Ryan 3. Lubrication Fundamentals, J. George Wills 4. Solar Engineering for Domestic Buildings, William A. Himmelmant 5. Applied Engineering Mechanics: Statics and Dynamics, G. Booithroyd and C. Poli 6. Centrifugal Pump Clinic, lgor J. Karassik 7. Computer-Aided Kinetics for Machine Design, Daniel L. Ryan 8. Plastics Products Design Handbook, Part A: Materials and Components; Part B: Processes and Design for Processes, edited by Edward Miller 9. Turbomachinery: Basic Theory and Applications, Earl Logan, Jr. 10. Vibrations of Shells and Plates, Werner Soedel 11. Flat and Corrugated Diaphragm Design Handbook, Mario Di Giovanni 12. Practical Stress Analysis in Engineering Design, Alexander Blake 13. An lntroduction to the Design and Behavior of Bolted Joints, John H. Bickford 14. Optimal Engineering Design: Principles and Applications, James N, Siddall 15. Spring Manufacturing Handbook, Harold Carlson 16. lndustrial Noise Control: Fundamentals and Applications, edited by Lewis H. Bell 17. Gears and Their Vibration: A Basic Approach to Understanding Gear Noise, J. Derek Smith 18. Chains for Power Transmission and Material Handling: Design and Appli- cations Handbook, American Chain Association 19. Corrosion and Corrosion Protection Handbook, edited by Philip A. Schweitzer 20. Gear Drive Systems: Design and Application, Peter Lynwander 2 1 . Controlling In-Plant Airborne Contaminants: Systems Design and Cal- culations, John D. Constance 22. CAD/CAM Systems Planning and Implementation, Charles S. Knox 23. Probabilistic Engineering Design: Principles and Applications, James N . Siddall 24. Traction Drives: Selection and Application, Frederick W. Heilich Ill and Eugene E. Shube 25. Finite Element Methods: An lntroduction, Ronald L. Huston anld Chris E. Passerello Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 26. Mechanical Fastening of Plastics: An Engineering Handbook, Brayton Lincoln, Kenneth J. Gomes, and James F. Braden 27. Lubrication in Practice: Second Edition, edited by W. S. Robertson 28. Principles of Automated Drafting, Daniel L. Ryan 29. Practical Seal Design, edited by Leonard J. Martini 30. Engineering Documentation for CAD/CAM Applications, Charles S. Knox 31. Design Dimensioning with Computer Graphics Applications, Jerome C. Lange 32. Mechanism Analysis: Simplified Graphical and Analytical Techniques, Lyndon 0. Barton 33. CAD/CAM Systems: Justification, Implementation, Productivity Measurement, Edward J. Preston, George W. Crawford, and Mark E. Coticchia 34. Steam Plant Calculations Manual, V. Ganapathy 35. Design Assurance for Engineers and Managers, John A. Burgess 36. Heat Transfer Fluids and Systems for Process and Energy Applications, Jasbir Singh 37. Potential Flows: Computer Graphic Solutions, Robert H. Kirchhoff 38. Computer-Aided Graphics and Design: Second Edition, Daniel L. Ryan 39. Electronically Controlled Proportional Valves: Selection and Application, Michael J. Tonyan, edited by Tobi Goldoftas 40. Pressure Gauge Handbook, AMETEK, U.S. Gauge Division, edited by Philip W. Harland 41. Fabric Filtration for Combustion Sources: Fundamentals and Basic Tech- nology, R. P. Donovan 42. Design of Mechanical Joints, Alexander Blake 43. CAD/CAM Dictionary, Edward J. Preston, George W. Crawford, and Mark E. Coticchia 44. Machinery Adhesives for Locking, Retaining, and Sealing, Girard S. Haviland 45. Couplings and Joints: Design, Selection, and Application, Jon R. Mancuso 46. Shaft Alignment Handbook, John Piotrowski 47. BASIC Programs for Steam Plant Engineers: Boilers, Combustion, Fluid Flow, and Heat Transfer, V. Ganapathy 48. Solving Mechanical Design Problems with Computer Graphics, Jerome C. Lange 49. Plastics Gearing: Selection and Application, Clifford E. Adams 50. Clutches and Brakes: Design and Selection, William C. Orthwein 51. Transducers in Mechanical and Electronic Design, Harry L. Trietley 52. Metallurgical Applications of Shock- Wave and High-Strain-Rate Phenom- ena, edited by Lawrence E. Murr, Karl P. Staudhammer, and Marc A. Meyers 53. Magnesium Products Design, Robert S. Busk 54. How to Integrate CAD/CAM Systems: Management and Technology, William D. Engelke 55. Cam Design and Manufacture: Second Edition; with cam design software for the IBM PC and compatibles, disk included, Preben W. Jensen 56. Solid-state AC Motor Controls: Selection and Application, Sylvester Campbell 57. Fundamentals of Robotics, David D. Ardayfio 58. Belt Selection and Application for Engineers, edited by Wallace D. Erickson 59. Developing Three-Dimensional CAD Software with the ISM PC, C. Stan Wei 60. Organizing Data for CIM Applications, Charles S. Knox, with contributions by Thomas C. Boos, Ross S. Culverhouse, and Paul F. Muchnicki Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 61. Computer-Aided Simulation in Railway Dynamics, by Rao V. Dukkipati and Joseph R. Amyot 62. Fiber-Reinforced Composites: Materials, Manufacturing, and Design, P. K. Mallick 63. Photoelectric Sensors and Controls: Selection and Application, Scott M. Juds 64. Finite Element Analysis with Personal Computers, Edward R. Champion, Jr., and J. Michael Ensminger 65. Ultrasonics: Fundamentals, Technology, Applications: Second Edition, Revised and Expanded, Dale Ensminger 66. Applied Finite Element Modeling: Practical Problem Solving for Engineers, Jeffrey M. Steele 67. Measurement and Instrumentation in Engineering: Principles and Basic Laboratory Experiments, Francis S. Tse and Ivan E. Morse 68. Centrifugal Pump Clinic: Second Edition, Revised and Expanded, lgor J. Karassi k 69. Practical Stress Analysis in Engineering Design: Second Edition, Revised and Expanded, Alexander Blake 70. An Introduction to the Design and Behavior of Bolted Joints: Second Edition, Revised and Expanded, John H. Bickford 71. High Vacuum Technology: A Practical Guide, Marsbed H. Hablanian 72. Pressure Sensors: Selection and Application, Duane Tandeske 73. Zinc Handbook: Properties, Processing, and Use in Design, Frank Porter 74. Thermal Fatigue of Metals, Andrzej Weronski and Tadeusz Hejwowski 75. Classical and Modern Mechanisms for Engineers and Inventors, Preben W. Jensen 76. Handbook of Electronic Package Design, edited by Michael Pecht 77. Shock- Wave and High-Strain-Rate Phenomena in Materials, edited by Marc A. Meyers, Lawrence E. Murr, and Karl P. Staudhammer 78. Industrial Refrigeration: Principles, Design and Applications, P. C. Koelet 79. Applied Combustion, Eugene L. Keating 80. Engine Oils and Automotive Lubrication, edited by Wilfried J. Barb 8 1 . Mechanism Analysis: Simplified and Graphical Techniques, Second Edition, Revised and Expanded, Lyndon 0. Barton 82. Fundamental Fluid Mechanics for the Practicing Engineer, James W. Murdock 83. Fiber-Reinforced Composites: Materials, Manufacturing, and Design, Second Edition, Revised and Expanded, P. K. Mallick 84. Numerical Methods for Engineering Applications, Edward R. Champion, Jr. 85. Turbomachinery: Basic Theory and Applications, Second Edition, Revised and Expanded, Earl Logan, Jr. 86. Vibrations of Shells and Plates: Second Edition, Revised and Expanded, Werner Soedel 87. Steam Plant Calculations Manual: Second Edition, Revised and Expanded, V. Ganapathy 88. Industrial Noise Control: Fundamentals and Applications, Second Edition, Revised and Expanded, Lewis H. Bell and Douglas H. Bell 89. Finite Elements: Their Design and Performance, Richard H. MacNeal 90. Mechanical Properties of Polymers and Composites: Second Edition, Re- vised and Expanded, Lawrence E. Nielsen and Robert F. Landel 91. Mechanical Wear Prediction and Prevention, Raymond G. Bayer Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 92. Mechanical Power Transmission Components, edited by David W. South and Jon R. Mancuso 93. Handbook of Turbomachinery, edited by Earl Logan, Jr. 94. Engineering Documentation Control Practices and Procedures, Ray E. Monahan 95. Refractory Linings Thermomechanical Design and Applications, Charles A. Schacht 96. Geometric Dimensioning and Tolerancing: Applications and Techniques for Use in Design, Manufacturing, and Inspection, James D. Meadows 97. An lntroduction to the Design and Behavior of Bolted Joints: Third Edition, Revised and Expanded, John H. Bickford 98. Shaft Alignment Handbook: Second Edition, Revised and Expanded, John Piotrowski 99. Computer-Aided Design of Polymer-Matrix Composite Structures, edited by Suong Van Hoa 100. Friction Science and Technology, Peter J. Blau 101, lntroduction to Plastics and Composites: Mechanical Properties and Engi- neering Applications, Edward Mi I le r 102. Practical Fracture Mechanics in Design, Alexander Blake 103. Pump Characteristics and Applications, Michael W. Volk 104. Optical Principles and Technology for Engineers, James E. Stewart 105. Optimizing the Shape of Mechanical Elements and Structures, A. A. Seireg and Jorge Rodriguez 106. Kinematics and Dynamics of Machinery, Vladimir Stejskal and Michael ValaSek 107. Shaft Seals for Dynamic Applications, Les Home 108. Reliability-Based Mechanical Design, edited by Thomas A. Cruse 109. Mechanical Fastening, Joining, and Assembly, James A. Speck 1 10. Turbomachinery Fluid Dynamics and Heat Transfer, edited by Chunill Hah 1 1 1. High-Vacuum Technology: A Practical Guide, Second Edition, Revised and Expanded, Marsbed H. Hablanian 1 12. Geometric Dimensioning and Tolerancing: Workbook and Answerbook, James D. Meadows 1 13. Handbook of Materials Selection for Engineering Applications, edited by G. T. Murray 114. Handbook of Thermoplastic Piping System Design, Thomas Sixsmith and Reinhard Hanselka 1 15. Practical Guide to Finite Elements: A Solid Mechanics Approach, Steven M. Lepi 1 16. Applied Computational Fluid Dynamics, edited by Vijay K. Garg 117. Fluid Sealing Technology, Heinz K. Muller and Bernard S. Nau 1 18. Friction and Lubrication in Mechanical Design, A. A. Seireg 1 19. lnfluence Functions and Matrices, Yuri A. Melnikov 120. Mechanical Analysis of Electronic Packaging Systems, Stephen A. McKeown 121 . Couplings and Joints: Design, Selection, and Application, Second Edition, Revised and Expanded, Jon R. Mancuso 122. Thermodynamics: Processes and Applications, Earl Logan, Jr. 123. Gear Noise and Vibration, J. Derek Smith 124. Practical Fluid Mechanics for Engineering Applications, John J. Bloomer 125. Handbook of Hydraulic Fluid Technology, edited by George E. Totten 126. Heat Exchanger Design Handbook, T. Kuppan Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 127. Designing for Product Sound Quality, Richard H. Lyon 128. Probability Applications in Mechanical Design, Franklin E. Fisher and Joy R. Fisher 129. Nickel Alloys, edited by Ulrich Heubner 1 30. Rotating Machinery Vibration: Problem Analysis and Troubleshooting, Maurice L. Adams, Jr. 131. Formulas for Dynamic Analysis, Ronald L. Huston and C. Q. Liu 132. Handbook of Machinery Dynamics, Lynn L. Faulkner and Earl Logan, Jr. 1 33. Rapid Prototyping Technology: Selection and Application, Kenneth G. Cooper 1 34. Reciprocating Machinery Dynamics: Design and Analysis, Abdulla S. Rangwala 1 35. Maintenance Excellence: Optimizing Equipment Life-Cycle Decisions, edi- ted by John D. Campbell and Andrew K. S. Jardine 136. Practical Guide to Industrial Boiler Systems, Ralph L. Vandagriff 137. Lubrication Fundamentals: Second Edition, Revised and Expanded, D. M. Pirro and A. A. Wessol 138. Mechanical Life Cycle Handbook: Good Environmental Design and Manu- facturing, edited by Mahendra S. Hundal 139. Micromachining of Engineering Materials, edited by Joseph McGeough 140. Control Strategies for Dynamic Systems: Design and Implementation, John H. Lumkes, Jr. 141 . Practical Guide to Pressure Vessel Manufacturing, Sunil Pullarcot 142. Nondestructive Evaluation: Theory, Techniques, and Applications, edited by Peter J. Shull 1 43. Diesel Engine Engineering: Thermodynamics, Dynamics, Design, and Control, Andrei Makartchouk 144. Handbook of Machine Tool Analysis, loan D. Marinescu, Constantin Ispas, and Dan Boboc 145. Implementing Concurrent Engineering in Small Companies, Susan Carlson Skalak 146. Practical Guide to the Packaging of Electronics: Thermal and Mechanical Design and Analysis, Ali Jamnia 147. Bearing Design in Machinery: Engineering Tribology and Lubrication, Avraham Harnoy 148. Mechanical Reliability Improvement: Probability and Statistics for Experi- mental Testing, R. E. Little 149. Industrial Boilers and Heat Recovery Steam Generators: Design, Ap- plications, and Calculations, V. Ganapathy 150. The CAD Guidebook: A Basic Manual for Understanding and Improving Computer-Aided Design, Stephen J. Schoonmaker 151. Industrial Noise Control and Acoustics, Randall F. Barron 1 52. Mechanical Properties of Engineered Materials, Wole Soboyejo 153. Reliability Verification, Testing, and Analysis in Engineering Design, Gary S. Wasserman 154. Fundamental Mechanics of Fluids: Third Edition, I. G. Currie 1 55. Intermediate Heat Transfer, Kau-Fui Vincent W ong 1 56. HVAC Water Chillers and Cooling Towers: Fundamentals, Application, and Operation, Herbert W. Stanford Ill 157. Gear Noise and Vibration: Second Edition, Revised and Expanded, J. Derek Smith Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 1 58. Handbook of Turbomachinery: Second Edition, Revised and Expanded, Earl Logan, Jr., and Ramendra Roy 1 59. Piping and Pipeline Engineering: Design, Construction, Maintenance, Integ- rity, and Repair, George A. Antaki 160. Turbomachinery: Design and Theory, Rama S. R. Gorla and Aijaz Ahmed Khan Additional Volumes in Preparation Target Costing: Market-Driven Product Design, M. Bradford Clifton, VVesley P. Townsend, Henry M. B. Bird, and Robert E. Albano Theory of Dimensioning: An Introduction to Parameterizing Geometric Models, Vijay Srinivasan Fluidized Bed Combustion, Simeon N. Oka Structural Analysis of Polymeric Composite Materials, Mark E. Tuttle Handbook of Pneumatic Conveying Engineering, David Mills, Mark G. Jones, and Vijay K. Agarwal Handbook of Mechanical Design Based on Material Composition, Geolrge E. Totten, Lin Xie, and Kiyoshi Funatani Mechanical Wear Fundamentals and Testing: Second Edition, Revised and Expanded, Raymond G. Bayer Engineering Design for Wear: Second Edition, Revised and Expanded, Raymond G. Bayer Clutches and Brakes: Design and Selection, Second Edition, William C. Orthwein Progressing Cavity Pumps, Downhole Pumps, and Mudmotors, Lev Nelik Mechanical Engineering Sofmare Spring Design with an ISM PC, A1Dietrich Mechanical Design Failure Analysis: With Failure Analysis System SolyWare for the IBM PC, David G. Ullman Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved To my parents, Tirupelamma and Subba Reddy Gorla, who encouraged me to strive for excellence in education —R. S. R. G. To my wife, Tahseen Ara, and to my daughters, Shumaila, Sheema, and Afifa —A. A. K. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Preface Turbomachinery: Design and Theory offers an introduction to the subject of turbomachinery and is intended to be a text for a single-semester course for senior undergraduate and beginning graduate students in mechanical engineering, aerospace engineering, chemical engineering, design engineering, and manu- facturing engineering. This book is also a valuable reference to practicing engineers in the fields of propulsion and turbomachinery. A basic knowledge of thermodynamics, fluid dynamics, and heat transfer is assumed. We have introduced the relevant concepts from these topics and reviewed them as applied to turbomachines in more detail. An introduction to dimensional analysis is included. We applied the basic principles to the study of hydraulic pumps, hydraulic turbines, centrifugal compressors and fans, axial flow compressors and fans, steam turbines, and axial flow and radial flow gas turbines. A brief discussion of cavitation in hydraulic machinery is presented. Each chapter includes a large number of solved illustrative and design example problems. An intuitive and systematic approach is used in the solution of these example problems, with particular attention to the proper use of units, which will help students understand the subject matter easily. In addition, we have provided several exercise problems at the end of each chapter, which will allow students to gain more experience. We urge students to take these exercise problems seriously: they are designed to help students fully grasp each topic Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved and to lead them toward a more concrete understanding and mastery of the techniques presented. This book has been written in a straightforward and systematic manner, without including irrelevant details. Our goal is to offer an engineering textbook on turbomachinery that will be read by students with enthusiasm and interest— we have made special efforts to touch students’ minds and assist them in exploring the exciting subject matter. R.S.R.G. would like to express thanks to his wife, Vijaya Lakshmi, for her support and understanding during the preparation of this book. A.A.K. would like to extend special recognition to his daughter, Shumaila, a practicing computer engineer, for her patience and perfect skills in the preparation of figures; to Sheema Aijaz, a civil engineer who provided numerous suggestions for enhancement of the material on hydraulic turbomachines; and to M. Sadiq, who typed some portions of the manuscript. A.A.K. is also indebted to Aftab Ahmed, Associate Professor of Mechanical Engineering at N.E.D. University of Engineering and Technology, for his many helpful discussions during the writing of this book. We would like to thank Shirley Love for her assistance in typing portions of the manuscript. We also thank the reviewers for their helpful comments, and we are grateful to John Corrigan, editor at Marcel Dekker, Inc., for encouragement and assistance. Rama S. R. Gorla Aijaz A. Khan Preface vi Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Contents Preface 1. Introduction: Dimensional Analysis—Basic Thermodynamics and Fluid Mechanics 1.1 Introduction to Turbomachinery 1.2 Types of Turbomachines 1.3 Dimensional Analysis 1.4 Dimensions and Equations 1.5 The Buckingham P Theorem 1.6 Hydraulic Machines 1.7 The Reynolds Number 1.8 Model Testing 1.9 Geometric Similarity 1.10 Kinematic Similarity 1.11 Dynamic Similarity 1.12 Prototype and Model Efficiency 1.13 Properties Involving the Mass or Weight of the Fluid 1.14 Compressible Flow Machines 1.15 Basic Thermodynamics, Fluid Mechanics, and Definitions of Efficiency Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 1.16 Continuity Equation 1.17 The First Law of Thermodynamics 1.18 Newton’s Second Law of Motion 1.19 The Second Law of Thermodynamics: Entropy 1.20 Efficiency and Losses 1.21 Steam and Gas Turbines 1.22 Efficiency of Compressors 1.23 Polytropic or Small-Stage Efficiency 1.24 Nozzle Efficiency 1.25 Diffuser Efficiency 1.26 Energy Transfer in Turbomachinery 1.27 The Euler Turbine Equation 1.28 Components of Energy Transfer Examples Problems Notation 2. Hydraulic Pumps 2.1 Introduction 2.2 Centrifugal Pumps 2.3 Slip Factor 2.4 Pump Losses 2.5 The Effect of Impeller Blade Shape on Performance 2.6 Volute or Scroll Collector 2.7 Vaneless Diffuser 2.8 Vaned Diffuser 2.9 Cavitation in Pumps 2.10 Suction Specific Speed 2.11 Axial Flow Pump 2.12 Pumping System Design 2.13 Life Cycle Analysis 2.14 Changing Pump Speed 2.15 Multiple Pump Operation Examples Problems Notation 3. Hydraulic Turbines 3.1 Introduction 3.2 Pelton Wheel 3.3 Velocity Triangles 3.4 Pelton Wheel (Losses and Efficiencies) Examples 3.5 Reaction Turbine Contents viii Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 3.6 Turbine Losses 3.7 Turbine Characteristics 3.8 Axial Flow Turbine 3.9 Cavitation Examples Problems Notation 4. Centrifugal Compressors and Fans 4.1 Introduction 4.2 Centrifugal Compressor 4.3 The Effect of Blade Shape on Performance 4.4 Velocity Diagrams 4.5 Slip Factor 4.6 Work Done 4.7 Diffuser 4.8 Compressibility Effects 4.9 Mach Number in the Diffuser 4.10 Centrifugal Compressor Characteristics 4.11 Stall 4.12 Surging 4.13 Choking Examples Problems Notation 5. Axial Flow Compressors and Fans 5.1 Introduction 5.2 Velocity Diagram 5.3 Degree of Reaction 5.4 Stage Loading 5.5 Lift-and-Drag Coefficients 5.6 Cascade Nomenclature and Terminology 5.7 3-D Consideration 5.8 Multi-Stage Performance 5.9 Axial Flow Compressor Characteristics Examples Problems Notation 6. Steam Turbines 6.1 Introduction 6.2 Steam Nozzles 6.3 Nozzle Efficiency Contents ix Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 6.4 The Reheat Factor 6.5 Metastable Equilibrium Examples 6.6 Stage Design 6.7 Impulse Stage 6.8 The Impulse Steam Turbine 6.9 Pressure Compounding (The Rateau Turbine) 6.10 Velocity Compounding (The Curtis Turbine) 6.11 Axial Flow Steam Turbines 6.12 Degree of Reaction 6.13 Blade Height in Axial Flow Machines Examples Problems Notation 7. Axial Flow and Radial Flow Gas Turbines 7.1 Introduction to Axial Flow Turbines 7.2 Velocity Triangles and Work Output 7.3 Degree of Reaction (L 7.4 Blade-Loading Coefficient 7.5 Stator (Nozzle) and Rotor Losses 7.6 Free Vortex Design 7.7 Constant Nozzle Angle Design Examples 7.8 Radial Flow Turbine 7.9 Velocity Diagrams and Thermodynamic Analysis 7.10 Spouting Velocity 7.11 Turbine Efficiency 7.12 Application of Specific Speed Examples Problems Notation 8. Cavitation in Hydraulic Machinery 8.1 Introduction 8.2 Stages and Types of Cavitation 8.3 Effects and Importance of Cavitation 8.4 Cavitation Parameter for Dynamic Similarity 8.5 Physical Significance and Uses of the Cavitation Parameter Contents x Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 8.6 The Rayleigh Analysis of a Spherical Cavity in an Inviscid Incompressible Liquid at Rest at Infinity 8.7 Cavitation Effects on Performance of Hydraulic Machines 8.8 Thoma’s Sigma and Cavitation Tests Notation Appendix The International System of Units (SI) Thermodynamic Properties of Water Thermodynamic Properties of Liquids Thermodynamic Properties of Air Bibliography Contents xi Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 1 Introduction: Dimensional Analysis—Basic Thermodynamics and Fluid Mechanics 1.1 INTRODUCTION TO TURBOMACHINERY Aturbomachine is a device in which energy transfer occurs between a flowing fluid and a rotating element due to dynamic action, and results in a change in pressure and momentumof the fluid. Mechanical energy transfer occurs inside or outside of the turbomachine, usually in a steady-flow process. Turbomachines include all those machines that produce power, such as turbines, as well as those types that produce a head or pressure, such as centrifugal pumps and compressors. The turbomachine extracts energy from or imparts energy to a continuously moving stream of fluid. However in a positive displacement machine, it is intermittent. The turbomachine as described above covers a wide range of machines, such as gas turbines, steam turbines, centrifugal pumps, centrifugal and axial flow compressors, windmills, water wheels, and hydraulic turbines. In this text, we shall deal with incompressible and compressible fluid flow machines. 1.2 TYPES OF TURBOMACHINES There are different types of turbomachines. They can be classified as: 1. Turbomachines in which (i) work is done by the fluid and (ii) work is done on the fluid. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 1.1 Types and shapes of turbomachines. Chapter 1 2 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 2. Turbomachines in which fluid moves through the rotating member in axial direction with no radial movement of the streamlines. Such machines are called axial flow machines whereas if the flow is essentially radial, it is called a radial flow or centrifugal flow machine. Some of these machines are shown in Fig. 1.1, and photographs of actual machines are shown in Figs. 1.2–1.6. Two primary points will be observed: first, that the main element is a rotor or runner carrying blades or vanes; and secondly, that the path of the fluid in the rotor may be substantially axial, substantially radial, or in some cases a combination of both. Turbomachines can further be classified as follows: Turbines: Machines that produce power by expansion of a continuously flowing fluid to a lower pressure or head. Pumps: Machines that increase the pressure or head of flowing fluid. Fans: Machines that impart only a small pressure-rise to a continuously flowing gas; usually the gas may be considered to be incompressible. Figure 1.2 Radial flow fan rotor. (Courtesy of the Buffalo Forge Corp.) Basic Thermodynamics and Fluid Mechanics 3 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 1.3 Centrifugal compressor rotor (the large double-sided impellar on the right is the main compressor and the small single-sided impellar is an auxiliary for cooling purposes). (Courtesy of Rolls-Royce, Ltd.) Figure 1.4 Centrifugal pump rotor (open type impeller). (Courtesy of the Ingersoll- Rand Co.) Chapter 1 4 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 1.5 Multi-stage axial flow compressor rotor. (Courtesy of the Westinghouse Electric Corp.) Figure 1.6 Axial flow pump rotor. (Courtesy of the Worthington Corp.) Basic Thermodynamics and Fluid Mechanics 5 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Compressors: Machines that impart kinetic energy to a gas by compressing it and then allowing it to rapidly expand. Compressors can be axial flow, centrifugal, or a combination of both types, in order to produce the highly compressed air. In a dynamic compressor, this is achieved by imparting kinetic energy to the air in the impeller and then this kinetic energy is converted into pressure energy in the diffuser. 1.3 DIMENSIONAL ANALYSIS To study the performance characteristics of turbomachines, a large number of variables are involved. The use of dimensional analysis reduces the variables to a number of manageable dimensional groups. Usually, the properties of interest in regard to turbomachine are the power output, the efficiency, and the head. The performance of turbomachines depends on one or more of several variables. A summary of the physical properties and dimensions is given in Table 1.1 for reference. Dimensional analysis applied to turbomachines has two more important uses: (1) prediction of a prototype’s performance from tests conducted on a scale Table 1.1 Physical Properties and Dimensions Property Dimension Surface L 2 Volume L 3 Density M/L 3 Velocity L/T Acceleration L/T 2 Momentum ML/T Force ML/T 2 Energy and work ML 2 /T 2 Power ML 2 /T 3 Moment of inertia ML 2 Angular velocity I/T Angular acceleration I/T 2 Angular momentum ML 2 /T Torque ML 2 /T 2 Modules of elasticity M/LT 2 Surface tension M/T 2 Viscosity (absolute) M/LT Viscosity (kinematic) L 2 /T Chapter 1 6 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved model (similitude), and (2) determination of the most suitable type of machine, on the basis of maximum efficiency, for a specified range of head, speed, and flow rate. It is assumed here that the student has acquired the basic techniques of forming nondimensional groups. 1.4 DIMENSIONS AND EQUATIONS The variables involved in engineering are expressed in terms of a limited number of basic dimensions. For most engineering problems, the basic dimensions are: 1. SI system: mass, length, temperature and time. 2. English system: mass, length, temperature, time and force. The dimensions of pressure can be designated as follows P ¼ F L 2 ð1:1Þ Equation (1.1) reads as follows: “The dimension of P equals force per length squared.” In this case, L 2 represents the dimensional characteristics of area. The left hand side of Eq. (1.1) must have the same dimensions as the right hand side. 1.5 THE BUCKINGHAM P THEOREM In 1915, Buckingham showed that the number of independent dimensionless group of variables (dimensionless parameters) needed to correlate the unknown variables in a given process is equal to n 2 m, where n is the number of variables involved and m is the number of dimensionless parameters included in the variables. Suppose, for example, the drag force F of a flowing fluid past a sphere is known to be a function of the velocity (v) mass density (r) viscosity (m) and diameter (D). Then we have five variables (F, v, r, m, and D) and three basic dimensions (L, F, and T) involved. Then, there are 5 2 3 ¼ 2 basic grouping of variables that can be used to correlate experimental results. 1.6 HYDRAULIC MACHINES Consider a control volume around the pump through which an incompressible fluid of density r flows at a volume flow rate of Q. Since the flow enters at one point and leaves at another point the volume flow rate Q can be independently adjusted by means of a throttle valve. The discharge Q of a pump is given by Q ¼ f ðN; D; g; H; m; rÞ ð1:2Þ Basic Thermodynamics and Fluid Mechanics 7 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved where H is the head, D is the diameter of impeller, g is the acceleration due to gravity, r is the density of fluid, N is the revolution, and m is the viscosity of fluid. In Eq. (1.2), primary dimensions are only four. Taking N, D, and r as repeating variables, we get P 1 ¼ ðNÞ a ðDÞ b r _ _ c ðQÞ M 0 L 0 T 0 ¼ ðT 21 Þ a ðLÞ b ðML 23 Þ c ðL 3 T 21 Þ For dimensional homogeneity, equating the powers of M, L, and T on both sides of the equation: for M, 0 ¼ c or c ¼ 0; for T, 0 ¼ 2 a 21 or a ¼ 21; for L, 0 ¼ b 2 3c þ 3 or b ¼ 23. Therefore, P 1 ¼ N 21 D 23 r 0 Q ¼ Q ND 3 ð1:3Þ Similarly, P 2 ¼ ðNÞ d ðDÞ e r _ _ f ðgÞ M 0 L 0 T 0 ¼ ðT 21 Þ d ðLÞ e ðML 23 Þ f ðLT 22 Þ Now, equating the exponents: for M, 0 ¼ f or f ¼ 0; for T, 0 ¼ 2 d 2 2 or d ¼ 22; for L, 0 ¼ e 2 3f þ 1 or e ¼ 21. Thus, P 2 ¼ N 22 D 21 r 0 g ¼ g N 2 D ð1:4Þ Similarly, P 3 ¼ ðNÞ g ðDÞ h r _ _ i ðHÞ M 0 L 0 T 0 ¼ ðT 21 Þ g ðLÞ h ðML 23 Þ i ðLÞ Equating the exponents: for M, 0 ¼ i or i ¼ 0; for T, 0 ¼ 2g or g ¼ 0; for L, 0 ¼ h 2 3i þ 1 or h ¼ 21. Thus, P 3 ¼ N 0 D 21 r 0 H ¼ H D ð1:5Þ and, P 4 ¼ ðNÞ j ðDÞ k r _ _ l ðmÞ M 0 L 0 T 0 ¼ ðT 21 Þ j ðLÞ k ðML 23 Þ l ðML 21 T 21 Þ Equating the exponents: for M, 0 ¼ l þ 1 or l ¼ 21; for T, 0 ¼ 2j 2 1 or j ¼ 21; for L, 0 ¼ k-3l 2 1 or k ¼ 22. Chapter 1 8 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Thus, P 4 ¼ N 21 D 22 r 21 m ¼ m ND 2 r ð1:6Þ The functional relationship may be written as f Q ND 3 ; g N 2 D ; H D ; m ND 2 r _ _ ¼ 0 Since the product of two Pterms is dimensionless, therefore replace the terms P 2 and P 3 by gh/N 2 D 2 f Q ND 3 ; gH N 2 D 2 ; m ND 2 r _ _ ¼ 0 or Q ¼ ND 3 f gH N 2 D 2 ; m ND 2 r _ _ ¼ 0 ð1:7Þ A dimensionless term of extremely great importance that may be obtained by manipulating the discharge and head coefficients is the specific speed, defined by the equation N s ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Flow coefficient Head coefficient _ ¼ N ffiffiffiffi Q _ _ gH _ _ 3/4 ð1:8Þ The following few dimensionless terms are useful in the analysis of incompressible fluid flow machines: 1. The flow coefficient and speed ratio: The term Q/(ND 3 ) is called the flow coefficient or specific capacity and indicates the volume flow rate of fluid through a turbomachine of unit diameter runner, operating at unit speed. It is constant for similar rotors. 2. The head coefficient: The term gH/N 2 D 2 is called the specific head. It is the kinetic energy of the fluid spouting under the head H divided by the kinetic energy of the fluid running at the rotor tangential speed. It is constant for similar impellers. c ¼ H/ U 2 /g _ _ ¼ gH/ p 2 N 2 D 2 _ _ ð1:9Þ 3. Power coefficient or specific power: The dimensionless quantity P/(rN 2 D 2 ) is called the power coefficient or the specific power. It shows the relation between power, fluid density, speed and wheel diameter. 4. Specific speed: The most important parameter of incompressible fluid flow machinery is specific speed. It is the non-dimensional term. All turbomachineries operating under the same conditions of flow and head Basic Thermodynamics and Fluid Mechanics 9 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved having the same specific speed, irrespective of the actual physical size of the machines. Specific speed can be expressed in this form N s ¼ N ffiffiffiffi Q _ /ðgHÞ 3/4 ¼ N ffiffiffi P p / b r 1/2 gH _ _ 5/4 c ð1:10Þ The specific speed parameter expressing the variation of all the variables N, Q and H or N,P and H, which cause similar flows in turbomachines that are geometrically similar. The specific speed represented by Eq. (1.10) is a nondimensional quantity. It can also be expressed in alternate forms. These are N s ¼ N ffiffiffiffi Q _ /H 3/4 ð1:11Þ and N s ¼ N ffiffiffi P p /H 5/4 ð1:12Þ Equation (1.11) is used for specifying the specific speeds of pumps and Eq. (1.12) is used for the specific speeds of turbines. The turbine specific speed may be defined as the speed of a geometrically similar turbine, which develops 1 hp under a head of 1 meter of water. It is clear that N s is a dimensional quantity. In metric units, it varies between 4 (for very high head Pelton wheel) and 1000 (for the low-head propeller on Kaplan turbines). 1.7 THE REYNOLDS NUMBER Reynolds number is represented by Re ¼ D 2 N/n where y is the kinematic viscosity of the fluid. Since the quantity D 2 N is proportional to DV for similar machines that have the same speed ratio. In flow through turbomachines, however, the dimensionless parameter D 2 N/n is not as important since the viscous resistance alone does not determine the machine losses. Various other losses such as those due to shock at entry, impact, turbulence, and leakage affect the machine characteristics along with various friction losses. Consider a control volume around a hydraulic turbine through which an incompressible fluid of density r flows at a volume flow rate of Q, which is controlled by a valve. The head difference across the control volume is H, and if the control volume represents a turbine of diameter D, the turbine develops a shaft power P at a speed of rotation N. The functional equation may be written as P ¼ f ðr; N; m; D; Q; gHÞ ð1:13Þ Chapter 1 10 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Equation (1.13) may be written as the product of all the variables raised to a power and a constant, such that P ¼ const: r a N b m c D d Q e gH _ _ f _ _ ð1:14Þ Substituting the respective dimensions in the above Eq. (1.14), ML 2 /T 3 _ _ ¼ const:ðM/L 3 Þ a ð1/TÞ b ðM/LTÞ c ðLÞ d ðL 3 /TÞ e ðL 2 /T 2 Þ f ð1:15Þ Equating the powers of M, L, and T on both sides of the equation: for M, 1 ¼ a þ c; for L, 2 ¼ 23a 2 c þ d þ3e þ2f; for T, 23 ¼ 2b 2 c 2 e 2 2f. There are six variables and only three equations. It is therefore necessary to solve for three of the indices in terms of the remaining three. Solving for a, b, and d in terms of c, e, and f we have: a ¼ 1 2c b ¼ 3 2c 2e 22f d ¼ 5 22c 23e 22f Substituting the values of a, b, and d in Eq. (1.13), and collecting like indices into separate brackets, P ¼ const: rN 3 D 5 _ _ ; m rND 2 _ _ c ; Q ND 3 _ _ e ; gH N 2 D 2 _ _ f _ _ ð1:16Þ In Eq. (1.16), the second term in the brackets is the inverse of the Reynolds number. Since the value of c is unknown, this term can be inverted and Eq. (1.16) may be written as P/rN 3 D 5 ¼ const: rND 2 m _ _ c ; Q ND 3 _ _ e ; gH N 2 D 2 _ _ f _ _ ð1:17Þ In Eq. (1.17) each group of variables is dimensionless and all are used in hydraulic turbomachinery practice, and are known by the following names: the power coefficient P/rN 3 D 5 ¼ P _ _ ; the flow coefficient _ Q/ND 3 ¼ f _ ; and the head coefficient gH/N 2 D 2 ¼ c _ _ . Eqution (1.17) can be expressed in the following form: P ¼ f Re; f; c _ _ ð1:18Þ Equation (1.18) indicates that the power coefficient of a hydraulic machine is a function of Reynolds number, flow coefficient and head coefficient. In flow through hydraulic turbomachinery, Reynolds number is usually very high. Therefore the viscous action of the fluid has very little effect on the power output of the machine and the power coefficient remains only a function of f and c. Basic Thermodynamics and Fluid Mechanics 11 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Typical dimensionless characteristic curves for a hydraulic turbine and pump are shown in Fig. 1.7 (a) and (b), respectively. These characteristic curves are also the curves of any other combination of P, N, Q, and H for a given machine or for any other geometrically similar machine. 1.8 MODEL TESTING Some very large hydraulic machines are tested in a model form before making the full-sized machine. After the result is obtained from the model, one may transpose the results from the model to the full-sized machine. Therefore if the curves shown in Fig 1.7 have been obtained for a completely similar model, these same curves would apply to the full-sized prototype machine. 1.9 GEOMETRIC SIMILARITY For geometric similarity to exist between the model and prototype, both of them should be identical in shape but differ only in size. Or, in other words, for geometric similarity between the model and the prototype, the ratios of all the corresponding linear dimensions should be equal. Let L p be the length of the prototype, B p, the breadth of the prototype, D p, the depth of the prototype, and L m , B m, and D m the corresponding dimensions of Figure 1.7 Performance characteristics of hydraulic machines: (a) hydraulic turbine, (b) hydraulic pump. Chapter 1 12 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved the model. For geometric similarity, linear ratio (or scale ratio) is given by L r ¼ L p L m ¼ B p B m ¼ D p D m ð1:19Þ Similarly, the area ratio between prototype and model is given by A r ¼ L p L m _ _ 2 ¼ B p B m _ _ 2 ¼ D p D m _ _ 2 ð1:20Þ and the volume ratio V r ¼ L p L m _ _ 3 ¼ B p B m _ _ 3 ¼ D p D m _ _ 3 ð1:21Þ 1.10 KINEMATIC SIMILARITY For kinematic similarity, both model and prototype have identical motions or velocities. If the ratio of the corresponding points is equal, then the velocity ratio of the prototype to the model is V r ¼ V 1 v 1 ¼ V 2 v 2 ð1:22Þ where V 1 is the velocity of liquid in the prototype at point 1, V 2, the velocity of liquid in the prototype at point 2, v 1 , the velocity of liquid in the model at point 1, and v 2 is the velocity of liquid in the model at point 2. 1.11 DYNAMIC SIMILARITY If model and prototype have identical forces acting on them, then dynamic similarity will exist. Let F 1 be the forces acting on the prototype at point 1, and F 2 be the forces acting on the prototype at point 2. Then the force ratio to establish dynamic similarity between the prototype and the model is given by F r ¼ F p1 F m1 ¼ F p2 F m2 ð1:23Þ 1.12 PROTOTYPE AND MODEL EFFICIENCY Let us suppose that the similarity laws are satisfied, h p and h m are the prototype and model efficiencies, respectively. Now from similarity laws, representing the model and prototype by subscripts m and p respectively, H p N p D p _ _ 2 ¼ H m N m D m ð Þ 2 or H p H m ¼ N p N m _ _ 2 D p D m _ _ 2 Basic Thermodynamics and Fluid Mechanics 13 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Q p N p D 3 p ¼ Q m N m D 3 m or Q p Q m ¼ N p N m _ _ D p D m _ _ 3 P p N 3 p D 5 p ¼ P m N 3 m D 5 m or P p P m ¼ N p N m _ _ 3 D p D m _ _ 5 Turbine efficiency is given by h t ¼ Power transferred from fluid Fluid power available: ¼ P rgQH Hence; h m h p ¼ P m P p _ _ Q p Q m _ _ H p H m _ _ ¼ 1: Thus, the efficiencies of the model and prototype are the same providing the similarity laws are satisfied. 1.13 PROPERTIES INVOLVING THE MASS OR WEIGHT OF THE FLUID 1.13.1 Specific Weight (g) The weight per unit volume is defined as specific weight and it is given the symbol g (gamma). For the purpose of all calculations relating to hydraulics, fluid machines, the specific weight of water is taken as 1000 l/m 3 . In S.I. units, the specific weight of water is taken as 9.80 kN/m 3 . 1.13.2 Mass Density (r) The mass per unit volume is mass density. In S.I. systems, the units are kilograms per cubic meter or NS 2 /m 4 . Mass density, often simply called density, is given the greek symbol r (rho). The mass density of water at 15.58 is 1000 kg/m 3 . 1.13.3 Specific Gravity (sp.gr.) The ratio of the specific weight of a given liquid to the specific weight of water at a standard reference temperature is defined as specific gravity. The standard reference temperature for water is often taken as 48C Because specific gravity is a ratio of specific weights, it is dimensionless and, of course, independent of system of units used. 1.13.4 Viscosity (m) We define viscosity as the property of a fluid, which offers resistance to the relative motion of fluid molecules. The energy loss due to friction in a flowing Chapter 1 14 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved fluid is due to the viscosity. When a fluid moves, a shearing stress develops in it. The magnitude of the shearing stress depends on the viscosity of the fluid. Shearing stress, denoted by the symbol t (tau) can be defined as the force required to slide on unit area layers of a substance over another. Thus t is a force divided by an area and can be measured in units N/m 2 or Pa. In a fluid such as water, oil, alcohol, or other common liquids, we find that the magnitude of the shearing stress is directly proportional to the change of velocity between different positions in the fluid. This fact can be stated mathematically as t ¼ m Dv Dy _ _ ð1:24Þ where Dv Dy is the velocity gradient and the constant of proportionality m is called the dynamic viscosity of fluid. Units for Dynamic Viscosity Solving for m gives m ¼ t Dv/Dy ¼ t Dy Dv _ _ Substituting the units only into this equation gives m ¼ N m 2 £ m m/s ¼ N £ s m 2 Since Pa is a shorter symbol representing N/m 2 , we can also express m as m ¼ Pa · s 1.13.5 Kinematic Viscosity (n) The ratio of the dynamic viscosity to the density of the fluid is called the kinematic viscosity y (nu). It is defined as n ¼ m r ¼ mð1/rÞ ¼ kg ms £ m 3 kg ¼ m 2 s ð1:25Þ Any fluid that behaves in accordance with Eq. (1.25) is called a Newtonian fluid. 1.14 COMPRESSIBLE FLOW MACHINES Compressible fluids are working substances in gas turbines, centrifugal and axial flow compressors. To include the compressibility of these types of fluids (gases), some new variables must be added to those already discussed in the case of hydraulic machines and changes must be made in some of the definitions used. The important parameters in compressible flow machines are pressure and temperature. Basic Thermodynamics and Fluid Mechanics 15 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved In Fig. 1.8 T-s charts for compression and expansion processes are shown. Isentropic compression and expansion processes are represented by s and the subscript 0 refers to stagnation or total conditions. 1 and 2 refer to the inlet and outlet states of the gas, respectively. The pressure at the outlet, P 02 , can be expressed as follows P 02 ¼ f D; N; m; P 01 ; T 01 ; T 02 ; r 01 ; r 02 ; m _ _ ð1:26Þ The pressure ratio P 02 /P 01 replaces the head H, while the mass flow rate m (kg/s) replaces Q. Using the perfect gas equation, density may be written as r ¼ P/RT. Now, deleting density and combining R with T, the functional relationship can be written as P 02 ¼ f ðP 01 ; RT 01 ; RT 02 ; m; N; D; mÞ ð1:27Þ Substituting the basic dimensions and equating the indices, the following fundamental relationship may be obtained P 02 P 01 ¼ f RT 02 RT 01 _ _ ; m RT 01 _ _ 1/2 P 01 D 2 _ _ _ _ _ _; ND RT 01 ð Þ 1/2 _ _ ; Re _ _ _ _ _ _ ð1:28Þ In Eq. (1.28), R is constant and may be eliminated. The Reynolds number in most cases is very high and the flow is turbulent and therefore changes in this parameter over the usual operating range may be neglected. However, due to Figure 1.8 Compressionandexpansionincompressible flowmachines: (a) compression, (b) expansion. Chapter 1 16 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved large changes of density, a significant reduction in Re can occur which must be taken into consideration. For a constant diameter machine, the diameter D may be ignored, and hence Eq. (1.28) becomes P 02 P 01 ¼ f T 02 T 01 _ _ ; mT 1/2 01 P 01 _ _ ; N T 1/2 01 _ _ _ _ ð1:29Þ In Eq. (1.29) some of the terms are new and no longer dimensionless. For a particular machine, it is typical to plot P 02 /P 01 and T 02 /T 01 against the mass flow Figure 1.9 Axial flow compressor characteristics: (a) pressure ratio, (b) efficiency. Figure 1.10 Axial flow gas turbine characteristics: (a) pressure ratio, (b) efficiency. Basic Thermodynamics and Fluid Mechanics 17 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved rate parameter mT 1/2 01 /P 01 for different values of the speed parameter N/T 1/2 01 . Equation (1.28) must be used if it is required to change the size of the machine. The term ND/ðRT 01 Þ 1/2 indicates the Mach number effect. This occurs because the impeller velocity v / ND and the acoustic velocity a 01 / RT 01 , while the Mach number M ¼ V/a 01 ð1:30Þ The performance curves for an axial flow compressor and turbine are shown in Figs. 1.9 and 1.10. 1.15 BASIC THERMODYNAMICS, FLUID MECHANICS, AND DEFINITIONS OF EFFICIENCY In this section, the basic physical laws of fluid mechanics and thermodynamics will be discussed. These laws are: 1. The continuity equation. 2. The First Law of Thermodynamics. 3. Newton’s Second Law of Motion. 4. The Second Law of Thermodynamics. The above items are comprehensively dealt with in books on thermo- dynamics with engineering applications, so that much of the elementary discussion and analysis of these laws need not be repeated here. 1.16 CONTINUITY EQUATION For steady flow through a turbomachine, m remains constant. If A 1 and A 2 are the flow areas at Secs. 1 and 2 along a passage respectively, then _ m ¼ r 1 A 1 C 1 ¼ r 2 A 2 C 2 ¼ constant ð1:31Þ where r 1 , is the density at section 1, r 2, the density at section 2, C 1 , the velocity at section 1, and C 2 , is the velocity at section 2. 1.17 THE FIRST LAW OF THERMODYNAMICS According to the First Law of Thermodynamics, if a system is taken through a complete cycle during which heat is supplied and work is done, then _ dQ 2dW ð Þ ¼ 0 ð1:32Þ where _ dQ represents the heat supplied to the system during this cycle and _ dW Chapter 1 18 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved the work done by the system during the cycle. The units of heat and work are taken to be the same. During a change of state from 1 to 2, there is a change in the internal energy of the system U 2 2U 1 ¼ _ 2 1 dQ 2dW ð Þ ð1:33Þ For an infinitesimal change of state dU ¼ dQ 2dW ð1:34Þ 1.17.1 The Steady Flow Energy Equation The First Law of Thermodynamics can be applied to a system to find the change in the energy of the system when it undergoes a change of state. The total energy of a system, E may be written as: E ¼ Internal Energy þKinetic Energy þPotential Energy E ¼ U þK:E: þP:E: ð1:35Þ where U is the internal energy. Since the terms comprising E are point functions, we can write Eq. (1.35) in the following form dE ¼ dU þdðK:E:Þ þdðP:E:Þ ð1:36Þ The First Law of Thermodynamics for a change of state of a system may therefore be written as follows dQ ¼ dU þdðKEÞ þdðPEÞ þdW ð1:37Þ Let subscript 1 represents the system in its initial state and 2 represents the system in its final state, the energy equation at the inlet and outlet of any device may be written Q 122 ¼ U 2 2U 1 þ mðC 2 2 2C 2 1 Þ 2 þmgðZ 2 2Z 1 Þ þW 1–2 ð1:38Þ Equation (1.38) indicates that there are differences between, or changes in, similar forms of energy entering or leaving the unit. In many applications, these differences are insignificant and can be ignored. Most closed systems encountered in practice are stationary; i.e. they do not involve any changes in their velocity or the elevation of their centers of gravity during a process. Thus, for stationary closed systems, the changes in kinetic and potential energies are negligible (i.e. K(K.E.) ¼ K(P.E.) ¼ 0), and the first law relation Basic Thermodynamics and Fluid Mechanics 19 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved reduces to Q 2W ¼ DE ð1:39Þ If the initial and final states are specified the internal energies 1 and 2 can easily be determined from property tables or some thermodynamic relations. 1.17.2 Other Forms of the First Law Relation The first law can be written in various forms. For example, the first law relation on a unit-mass basis is q 2w ¼ DeðkJ/kgÞ ð1:40Þ Dividing Eq. (1.39) by the time interval Dt and taking the limit as Dt ! 0 yields the rate form of the first law _ Q 2 _ W ¼ dE dt ð1:41Þ where Q ˙ is the rate of net heat transfer, W ˙ the power, and dE dt is the rate of change of total energy. Equations. (1.40) and (1.41) can be expressed in differential form dQ 2dW ¼ dEðkJÞ ð1:42Þ dq 2dw ¼ deðkJ/kgÞ ð1:43Þ For a cyclic process, the initial and final states are identical; therefore, DE ¼ E 2 2E 1 . Then the first law relation for a cycle simplifies to Q 2W ¼ 0ðkJÞ ð1:44Þ That is, the net heat transfer and the net work done during a cycle must be equal. Defining the stagnation enthalpy by: h 0 ¼ h þ 1 2 c 2 and assuming g (Z 2 2 Z 1 ) is negligible, the steady flow energy equation becomes _ Q 2 _ W ¼ _ mðh 02 2h 01 Þ ð1:45Þ Most turbomachinery flow processes are adiabatic, and so Q ˙ ¼ 0. For work producing machines, W ˙ . 0; so that _ W ¼ _ mðh 01 2h 02 Þ ð1:46Þ For work absorbing machines (compressors) W , 0; so that _ W !2 _ W ¼ _ mðh 02 2h 01 Þ ð1:47Þ 1.18 NEWTON’S SECOND LAW OF MOTION Newton’s Second Law states that the sum of all the forces acting on a control volume in a particular direction is equal to the rate of change of momentum of the fluid across the control volume. For a control volume with fluid entering with Chapter 1 20 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved uniform velocity C 1 and leaving with uniform velocity C 2 , then F ¼ _ mðC 2 2C 1 Þ ð1:48Þ Equation (1.48) is the one-dimensional form of the steady flow momentum equation, and applies for linear momentum. However, turbomachines have impellers that rotate, and the power output is expressed as the product of torque and angular velocity. Therefore, angular momentumis the most descriptive parameter for this system. 1.19 THE SECOND LAW OF THERMODYNAMICS: ENTROPY This law states that for a fluid passing through a cycle involving heat exchanges _ dQ T # 0 ð1:49Þ where dQis an element of heat transferred to the systemat an absolute temperature T. If all the processes in the cycle are reversible, so that dQ ¼ dQ R , then _ dQ R T ¼ 0 ð1:50Þ The property called entropy, for a finite change of state, is then given by S 2 2S 1 ¼ _ 2 1 dQ R T ð1:51Þ For an incremental change of state dS ¼ mds ¼ dQ R T ð1:52Þ where m is the mass of the fluid. For steady flow through a control volume in which the fluid experiences a change of state from inlet 1 to outlet 2, _ 2 1 d _ Q T # _ m s 2 2s 1 ð Þ ð1:53Þ For adiabatic process, dQ ¼ 0 so that s 2 $ s 1 ð1:54Þ For reversible process s 2 ¼ s 1 ð1:55Þ Basic Thermodynamics and Fluid Mechanics 21 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved In the absence of motion, gravity and other effects, the first law of thermodynamics, Eq. (1.34) becomes Tds ¼ du þpdv ð1:56Þ Putting h ¼ u þ pv and dh ¼ du þ pdv þ vdp in Eq. (1.56) gives Tds ¼ dh 2vdp ð1:57Þ 1.20 EFFICIENCY AND LOSSES Let H be the head parameter (m), Q discharge (m 3 /s) The waterpower supplied to the machine is given by P ¼ rQgHðin wattsÞ ð1:58Þ and letting r ¼ 1000 kg/m 3 , ¼ QgHðin kWÞ Now, let DQ be the amount of water leaking from the tail race. This is the amount of water, which is not providing useful work. Then: Power wasted ¼ DQðgHÞðkWÞ For volumetric efficiency, we have h n ¼ Q 2DQ Q ð1:59Þ Net power supplied to turbine ¼ ðQ 2DQÞgHðkWÞ ð1:60Þ If H r is the runner head, then the hydraulic power generated by the runner is given by P h ¼ ðQ 2DQÞgH r ðkWÞ ð1:61Þ The hydraulic efficiency, h h is given by h h ¼ Hydraulic output power Hydraulic input power ¼ ðQ 2DQÞgH r ðQ 2DQÞgH ¼ H r H ð1:62Þ Chapter 1 22 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved If P m represents the power loss due to mechanical friction at the bearing, then the available shaft power is given by P s ¼ P h 2P m ð1:63Þ Mechanical efficiency is given by h m ¼ P s P h ¼ P s P m 2P s ð1:64Þ The combined effect of all these losses may be expressed in the form of overall efficiency. Thus h 0 ¼ P s WP ¼ h m P h WP ¼ h m WPðQ 2DQÞ WPQDH ¼ h m h v h h ð1:65Þ 1.21 STEAM AND GAS TURBINES Figure 1.11 shows an enthalpy–entropy or Mollier diagram. The process is represented by line 1–2 and shows the expansion from pressure P 1 to a lower pressure P 2 . The line 1–2s represents isentropic expansion. The actual Figure 1.11 Enthalpy–entropy diagrams for turbines and compressors: (a) turbine expansion process, (b) compression process. Basic Thermodynamics and Fluid Mechanics 23 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved turbine-specific work is given by W t ¼ h 01 2h 02 ¼ ðh 1 2h 2 Þ þ 1 2 ðC 2 1 2C 2 2 Þ ð1:66Þ Similarly, the isentropic turbine rotor specific work between the same two pressures is W 0 t ¼ h 01 2h 02s ¼ ðh 1 2h 2s Þ þ 1 2 C 2 1 2C 2 2s _ _ ð1:67Þ Efficiency can be expressed in several ways. The choice of definitions depends largely upon whether the kinetic energy at the exit is usefully utilized or wasted. In multistage gas turbines, the kinetic energy leaving one stage is utilized in the next stage. Similarly, in turbojet engines, the energy in the gas exhausting through the nozzle is used for propulsion. For the above two cases, the turbine isentropic efficiency h tt is defined as h tt ¼ W t W 0 t ¼ h 01 2h 02 h 01 2h 02s ð1:68Þ When the exhaust kinetic energy is not totally used but not totally wasted either, the total-to-static efficiency, h ts , is used. In this case, the ideal or isentropic turbine work is that obtained between static points 01 and 2s. Thus h ts ¼ h 01 2h 02 h 01 2h 02s þ 1 2 C 2 2s ¼ h 01 2h 02 h 01 2h 2s ð1:69Þ If the difference between inlet and outlet kinetic energies is small, Eq. (1.69) becomes h ts ¼ h 1 2h 2 h 1 2h 2s þ 1 2 C 2 1s An example where the outlet kinetic energy is wasted is a turbine exhausting directly to the atmosphere rather than exiting through a diffuser. 1.22 EFFICIENCY OF COMPRESSORS The isentropic efficiency of the compressor is defined as h c ¼ Isentropic work Actual work ¼ h 02s 2h 01 h 02 2h 01 ð1:70Þ If the difference between inlet and outlet kinetic energies is small, 1 2 C 2 1 ¼ 1 2 C 2 2 and h c ¼ h 2s 2h 1 h 2 2h 1 ð1:71Þ Chapter 1 24 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 1.23 POLYTROPIC OR SMALL-STAGE EFFICIENCY Isentropic efficiency as described above can be misleading if used for compression and expansion processes in several stages. Turbomachines may be used in large numbers of very small stages irrespective of the actual number of stages in the machine. If each small stage has the same efficiency, then the isentropic efficiency of the whole machine will be different from the small stage efficiency, and this difference is dependent upon the pressure ratio of the machine. Isentropic efficiency of compressors tends to decrease and isentropic efficiency of turbines tends to increase as the pressure ratios for which the machines are designed are increased. This is made more apparent in the following argument. Consider an axial flow compressor, which is made up of several stages, each stage having equal values of h c , as shown in Fig. 1.12. Then the overall temperature rise can be expressed by DT ¼ DT 0 s h s ¼ 1 h s DT 0 s Figure 1.12 Compression process in stages. Basic Thermodynamics and Fluid Mechanics 25 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved (Prime symbol is used for isentropic temperature rise, and subscript s is for stage temperature). Also, DT ¼ D T 0 /h c by definition of h c , and thus: h s /h c ¼ DT s 0 /DT 0 . It is clear from Fig. 1.12 that DT 0 s . DT 0 . Hence, h c , h s and the difference will increase with increasing pressure ratio. The opposite effect is obtained in a turbine where h s (i.e., small stage efficiency) is less than the overall efficiency of the turbine. The above discussions have led to the concept of polytropic efficiency, h 1 , which is defined as the isentropic efficiency of an elemental stage in the process such that it is constant throughout the entire process. The relationship between a polytropic efficiency, which is constant through the compressor, and the overall efficiency h c may be obtained for a gas of constant specific heat. For compression, h 1c ¼ dT 0 dT ¼ constant But, T p ðg21Þ/g ¼ constant for an isentropic process, which in differential form is dT 0 dT ¼ g 21 g dP P Now, substituting dT 0 from the previous equation, we have h 1c dT 0 dT ¼ g 21 g dP P Integrating the above equation between the inlet 1 and outlet 2, we get h 1c ¼ lnðP 2 /P 1 Þ g21 g lnðT 2 /T 1 Þ ð1:72Þ Equation (1.72) can also be written in the form T 2 T 1 ¼ P 2 P 1 _ _g21 gh1c ð1:73Þ The relation between h 1c and h c is given by h c ¼ ðT 0 2 /T 1 Þ 21 ðT 2 /T 1 Þ 21 ¼ ðP 2 /P 1 Þ g21 g 21 ðP 2 /P 1 Þ g21 gh1c 21 ð1:74Þ From Eq. (1.74), if we write g21 gh 1c as n21 n , Eq. (1.73) is the functional relation between P and T for a polytropic process, and thus it is clear that the non isentropic process is polytropic. Chapter 1 26 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Similarly, for an isentropic expansion and polytropic expansion, the following relations can be developed between the inlet 1 and outlet 2: T 1 T 2 ¼ P 1 P 2 _ _ h 1t g21 ð Þ g and h t ¼ 1 2 1 P 1 /P 2 _ _ h 1t g21 ð Þ g 1 2 1 P 1 /P 2 _ _ g21 ð Þ g ð1:75Þ where h 1t is the small-stage or polytropic efficiency for the turbine. Figure 1.13 shows the overall efficiency related to the polytropic efficiency for a constant value of g ¼ 1.4, for varying polytropic efficiencies and for varying pressure ratios. As mentioned earlier, the isentropic efficiency for an expansion process exceeds the small-stage efficiency. Overall isentropic efficiencies have been Figure 1.13 Relationships among overall efficiency, polytropic efficiency, and pressure ratio. Basic Thermodynamics and Fluid Mechanics 27 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved calculated for a range of pressure ratios and different polytropic efficiencies. These relationships are shown in Fig. 1.14. 1.24 NOZZLE EFFICIENCY The function of the nozzle is to transform the high-pressure temperature energy (enthalpy) of the gasses at the inlet position into kinetic energy. This is achieved by decreasing the pressure and temperature of the gasses in the nozzle. From Fig. 1.15, it is clear that the maximum amount of transformation will result when we have an isentropic process between the pressures at the entrance and exit of the nozzle. Such a process is illustrated as the path 1–2s. Now, when nozzle flow is accompanied by friction, the entropy will increase. As a result, the path is curved as illustrated by line 1–2. The difference in the enthalpy change between the actual process and the ideal process is due to friction. This ratio is known as the nozzle adiabatic efficiency and is called nozzle efficiency (h n ) or jet Figure 1.14 Turbine isentropic efficiency against pressure ratio for various polytropic efficiencies (g ¼ 1.4). Chapter 1 28 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved pipe efficiency (h j ). This efficiency is given by: h j ¼ Dh D h0 ¼ h 01 2h 02 h 01 2h 02 0 ¼ c p T 01 2T 02 ð Þ c p T 01 2T 02 0 ð Þ ð1:76Þ 1.25 DIFFUSER EFFICIENCY The diffuser efficiency h d is defined in a similar manner to compressor efficiency (see Fig. 1.16): h d ¼ Isentropic enthalpy rise Actual enthalpy rise ¼ h 2s 2h 1 h 2 2h 1 ð1:77Þ The purpose of diffusion or deceleration is to convert the maximum possible kinetic energy into pressure energy. The diffusion is difficult to achieve and is rightly regarded as one of the main problems of turbomachinery design. This problem is due to the growth of boundary layers and the separation of the fluid molecules from the diverging part of the diffuser. If the rate of diffusion is too rapid, large losses in stagnation pressure are inevitable. On the other hand, if Figure 1.15 Comparison of ideal and actual nozzle expansion on a T-s or h–s plane. Basic Thermodynamics and Fluid Mechanics 29 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved the rate of diffusion is very low, the fluid is exposed to an excessive length of wall and friction losses become predominant. To minimize these two effects, there must be an optimum rate of diffusion. 1.26 ENERGY TRANSFER IN TURBOMACHINERY This section deals with the kinematics and dynamics of turbomachines by means of definitions, diagrams, and dimensionless parameters. The kinematics and dynamic factors depend on the velocities of fluid flow in the machine as well as the rotor velocity itself and the forces of interaction due to velocity changes. 1.27 THE EULER TURBINE EQUATION The fluid flows through the turbomachine rotor are assumed to be steady over a long period of time. Turbulence and other losses may then be neglected, and the mass flow rate m is constant. As shown in Fig. 1.17, let v (omega) be the angular velocity about the axis A–A. Fluid enters the rotor at point 1 and leaves at point 2. In turbomachine flow analysis, the most important variable is the fluid velocity and its variation in the different coordinate directions. In the designing of blade shapes, velocity vector diagrams are very useful. The flow in and across Figure 1.16 Mollier diagram for the diffusion process. Chapter 1 30 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved the stators, the absolute velocities are of interest (i.e., C). The flowvelocities across the rotor relative to the rotating blade must be considered. The fluid enters with velocity C 1 , which is at a radial distance r 1 from the axis A–A. At point 2 the fluid leaves with absolute velocity (that velocity relative to an outside observer). The point 2 is at a radial distance r 2 fromthe axis A–A. The rotating disc may be either a turbine or a compressor. It is necessary to restrict the flowto a steady flow, i.e., the mass flowrate is constant (no accumulation of fluid in the rotor). The velocity C 1 at the inlet to the rotor can be resolved into three components; viz.; C a1 — Axial velocity in a direction parallel to the axis of the rotating shaft. C r1 — Radial velocity in the direction normal to the axis of the rotating shaft. C w1 — whirl or tangential velocity in the direction normal to a radius. Similarly, exit velocity C 2 can be resolved into three components; that is, C a2 , C r2 , and C w2 . The change in magnitude of the axial velocity components through the rotor gives rise to an axial force, which must be taken by a thrust bearing to the stationary rotor casing. The change in magnitude of the radial velocity components produces radial force. Neither has any effect on the angular motion of the rotor. The whirl or tangential components C w produce the rotational effect. This may be expressed in general as follows: The unit mass of fluid entering at section 1 and leaving in any unit of time produces: The angular momentum at the inlet: C w1 r 1 The angular momentum at the outlet: C w2 r 2 And therefore the rate of change of angular momentum ¼ C w1 r 1 – C w2 r 2 Figure 1.17 Velocity components for a generalized rotor. Basic Thermodynamics and Fluid Mechanics 31 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved By Newton’s laws of motion, this is equal to the summation of all the applied forces on the rotor; i.e., the net torque of the rotor t (tau). Under steady flow conditions, using mass flow rate m, the torque exerted by or acting on the rotor will be: t ¼ m C w1 r 1 2C w2 r 2 ð Þ Therefore the rate of energy transfer, W, is the product of the torque and the angular velocity of the rotor v (omega), so: W ¼ tv ¼ mv C w1 r 1 2C w2 r 2 ð Þ For unit mass flow, energy will be given by: W ¼ vðC w1 r 1 2C w2 r 2 Þ ¼ C w1 r 1 v 2C w2 r 2 v ð Þ But, v r 1 ¼ U 1 and v r 2 ¼ U 2. Hence; W ¼ C w1 U 1 2C w2 U 2 ð Þ; ð1:78Þ where, W is the energy transferred per unit mass, and U 1 and U 2 are the rotor speeds at the inlet and the exit respectively. Equation (1.78) is referred to as Euler’s turbine equation. The standard thermodynamic sign convention is that work done by a fluid is positive, and work done on a fluid is negative. This means the work produced by the turbine is positive and the work absorbed by the compressors and pumps is negative. Therefore, the energy transfer equations can be written separately as W ¼ C w1 U 1 2C w2 U 2 ð Þ for turbine and W ¼ C w2 U 2 2C w1 U 1 ð Þ for compressor and pump: The Euler turbine equation is very useful for evaluating the flow of fluids that have very small viscosities, like water, steam, air, and combustion products. To calculate torque from the Euler turbine equation, it is necessary to know the velocity components C w1 , C w2 , and the rotor speeds U 1 and U 2 or the velocities V 1 , V 2 , C r1 , C r2 as well as U 1 and U 2 . These quantities can be determined easily by drawing the velocity triangles at the rotor inlet and outlet, as shown in Fig. 1.18. The velocity triangles are key to the analysis of turbo- machinery problems, and are usually combined into one diagram. These triangles are usually drawn as a vector triangle: Since these are vector triangles, the two velocities U and V are relative to one another, so that the tail of V is at the head of U. Thus the vector sum of U and V is equal to the vector C. The flow through a turbomachine rotor, the absolute velocities C 1 and C 2 as well as the relative velocities V 1 and V 2 can have three Chapter 1 32 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved components as mentioned earlier. However, the two velocity components, one tangential to the rotor (C w ) and another perpendicular to it are sufficient. The component C r is called the meridional component, which passes through the point under consideration and the turbomachine axis. The velocity components C r1 and C r2 are the flow velocity components, which may be axial or radial depending on the type of machine. 1.28 COMPONENTS OF ENERGY TRANSFER The Euler equation is useful because it can be transformed into other forms, which are not only convenient to certain aspects of design, but also useful in Figure 1.18 Velocity triangles for a rotor. Basic Thermodynamics and Fluid Mechanics 33 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved understanding the basic physical principles of energy transfer. Consider the fluid velocities at the inlet and outlet of the turbomachine, again designated by the subscripts 1 and 2, respectively. By simple geometry, C 2 r2 ¼ C 2 2 2C 2 w2 and C 2 r2 ¼ V 2 2 2 U 2 2C w2 ð Þ 2 Equating the values of C 2 r2 and expanding, C 2 2 2C 2 w2 ¼ V 2 2 2U 2 2 þ2U 2 C w2 2C 2 w2 and U 2 C w2 ¼ 1 2 C 2 2 þU 2 2 2V 2 2 _ _ Similarly, U 1 C w1 ¼ 1 2 ðC 2 1 þU 2 1 2V 2 1 Þ Inserting these values in the Euler equation, E ¼ 1 2 ðC 2 1 2C 2 2 Þ þðU 2 1 2U 2 2 Þ þðV 2 1 2V 2 2 Þ _ ¸ ð1:79Þ The first term, 1 2 ðC 2 1 2C 2 2 Þ, represents the energy transfer due to change of absolute kinetic energy of the fluid during its passage between the entrance and exit sections. In a pump or compressor, the discharge kinetic energy from the rotor, 1 2 C 2 2 , may be considerable. Normally, it is static head or pressure that is required as useful energy. Usually the kinetic energy at the rotor outlet is converted into a static pressure head by passing the fluid through a diffuser. In a turbine, the change in absolute kinetic energy represents the power transmitted from the fluid to the rotor due to an impulse effect. As this absolute kinetic energy change can be used to accomplish rise in pressure, it can be called a “virtual pressure rise” or “a pressure rise” which is possible to attain. The amount of pressure rise in the diffuser depends, of course, on the efficiency of the diffuser. Since this pressure rise comes from the diffuser, which is external to the rotor, this term, i.e., 1 2 ðC 2 1 2C 2 2 Þ, is sometimes called an “external effect.” The other two terms of Eq. (1.79) are factors that produce pressure rise within the rotor itself, and hence they are called “internal diffusion.” The centrifugal effect, 1 2 ðU 2 1 2U 2 2 Þ, is due to the centrifugal forces that are developed as the fluid particles move outwards towards the rim of the machine. This effect is produced if the fluid changes radius as it flows from the entrance to the exit section. The third term, 1 2 ðV 2 1 2V 2 2 Þ, represents the energy transfer due to the change of the relative kinetic energy of the fluid. If V 2 . V 1 , the passage acts like a nozzle and if V 2 , V 1 , it acts like a diffuser. From the above discussions, it is Chapter 1 34 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved apparent that in a turbocompresser, pressure rise occurs due to both external effects and internal diffusion effect. However, in axial flow compressors, the centrifugal effects are not utilized at all. This is why the pressure rise per stage is less than in a machine that utilizes all the kinetic energy effects available. It should be noted that the turbine derives power from the same effects. Illustrative Example 1.1: A radial flow hydraulic turbine produces 32 kW under a head of 16mand runningat 100 rpm. Ageometrically similar model producing 42 kWanda headof 6mis tobe testedunder geometricallysimilar conditions. If model efficiency is assumed to be 92%, find the diameter ratio between the model and prototype, the volume flow rate through the model, and speed of the model. Solution: Assuming constant fluid density, equating head, flow, and power coefficients, using subscripts 1 for the prototype and 2 for the model, we have from Eq. (1.19), P 1 r 1 N 3 1 D 5 1 _ _ ¼ P 2 r 2 N 3 2 D 5 2 _ _ ; where r 1 ¼ r 2 : Then, D 2 D 1 ¼ P 2 P 1 _ _1 5 N 1 N 2 _ _3 5 or D 2 D 1 ¼ 0:032 42 _ _1 5 N 1 N 2 _ _3 5 ¼ 0:238 N 1 N 2 _ _3 5 Also, we know from Eq. (1.19) that gH 1 N 1 D 1 ð Þ 2 ¼ gH 2 N 2 D 2 ð Þ 2 ðgravity remains constantÞ Then D 2 D 1 ¼ H 2 H 1 _ _1 2 N 1 N 2 _ _ ¼ 6 16 _ _1 2 N 1 N 2 _ _ Equating the diameter ratios, we get 0:238 N 1 N 2 _ _3 5 ¼ 6 16 _ _1 2 N 1 N 2 _ _ or N 2 N 1 _ _2 5 ¼ 0:612 0:238 ¼ 2:57 Therefore the model speed is N 2 ¼ 100 £ 2:57 ð Þ 5 2 ¼ 1059 rpm Basic Thermodynamics and Fluid Mechanics 35 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Model scale ratio is given by D 2 D 1 ¼ 0:238 ð Þ 100 1059 _ _3 5 ¼ 0:238ð0:094Þ 0:6 ¼ 0:058: Model efficiency is h m ¼ Power output Water power input or, 0:92 ¼ 42 £ 10 3 rgQH ; or, Q ¼ 42 £ 10 3 0:92 £ 10 3 £ 9:81 £ 6 ¼ 0:776 m 3 /s Illustrative Example 1.2: A centrifugal pump delivers 2.5 m 3 /s under a head of 14 m and running at a speed of 2010 rpm. The impeller diameter of the pump is 125 mm. If a 104 mm diameter impeller is fitted and the pump runs at a speed of 2210 rpm, what is the volume rate? Determine also the new pump head. Solution: First of all, let us assume that dynamic similarity exists between the two pumps. Equating the flow coefficients, we get [Eq. (1.3)] Q 1 N 1 D 3 1 ¼ Q 2 N 2 D 3 2 or 2:5 2010 £ ð0:125Þ 3 ¼ Q 2 2210 £ ð0:104Þ 3 Solving the above equation, the volume flow rate of the second pump is Q 2 ¼ 2:5 £ 2210 £ ð0:104Þ 3 2010 £ ð0:125Þ 3 ¼ 1:58 m 3 /s Now, equating head coefficients for both cases gives [Eq. (1.9)] gH 1 /N 2 1 D 2 1 ¼ gH 2 /N 2 2 D 2 2 Substituting the given values, 9:81 £ 14 ð2010 £ 125Þ 2 ¼ 9:81 £ H 2 ð2210 £ 104Þ 2 Therefore, H 2 ¼ 11.72 m of water. Chapter 1 36 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved IllustrativeExample 1.3: Anaxial flowcompressor handlingair anddesigned to run at 5000rpm at ambient temperature and pressure of 188C and 1.013 bar, respectively. The performance characteristic of the compressor is obtained at the atmosphere temperature of 258C. What is the correct speed at which the compressor must run? If an entry pressure of 65kPa is obtained at the point where the mass flow rate would be 64kg/s, calculate the expected mass flow rate obtained in the test. Solution: Since the machine is the same in both cases, the gas constant R and diameter can be cancelled from the operating equations. Using first the speed parameter, N 1 ffiffiffiffiffiffiffi T 01 p ¼ N 2 ffiffiffiffiffiffiffi T 02 p Therefore, N 2 ¼ 5000 273 þ25 273 þ18 _ _1 2 ¼ 5000 298 291 _ _ 0:5 ¼ 5060 rpm Hence, the correct speed is 5060 rpm. Now, considering the mass flow parameter, m 1 ffiffiffiffiffiffiffi T 01 p p 01 ¼ m 2 ffiffiffiffiffiffiffi T 02 p p 02 Therefore, m 2 ¼ 64 £ 65 101:3 _ _ 291 298 _ _ 0:5 ¼ 40:58 kg/s Illustrative Example 1.4: A pump discharges liquid at the rate of Q against a head of H. If specific weight of the liquid is w, find the expression for the pumping power. Solution: Let Power P be given by: P ¼ f ðw; Q; HÞ ¼ kw a Q b H c where k, a, b, and c are constants. Substituting the respective dimensions in the above equation, ML 2 T 23 ¼ kðML 22 T 22 Þ a ðL 3 T 21 Þ b ðLÞ c Equating corresponding indices, for M, 1 ¼ a or a ¼ 1; for L, 2 ¼ 22a þ 3b þ c; and for T, 23 ¼ 22a 2 b or b ¼ 1, so c ¼ 1. Basic Thermodynamics and Fluid Mechanics 37 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Therefore, P ¼ kwQH Illustrative Example 1.5: Prove that the drag force F on a partially submerged body is given by: F ¼ V 2 l 2 r f k l ; lg V 2 _ _ where V is the velocity of the body, l is the linear dimension, r, the fluid density, k is the rms height of surface roughness, and g is the gravitational acceleration. Solution: Let the functional relation be: F ¼ f ðV; l; k; r; gÞ Or in the general form: F ¼ f ðF; V; l; k; r; gÞ ¼ 0 In the above equation, there are only two primary dimensions. Thus, m ¼ 2. Taking V, l, and r as repeating variables, we get: P 1 ¼ ðVÞ a ðlÞ b r _ _ c F M o L o T o ¼ ðLT 21 Þ a ðLÞ b ðML 23 Þ c ðMLT 22 Þ Equating the powers of M, L, and T on both sides of the equation, for M, 0 ¼ c þ 1 or c ¼ 21; for T, 0 ¼ 2a 2 2 or a ¼ 22; and for L, 0 ¼ a þ b2 3c þ 1 or b ¼ 22. Therefore, P 1 ¼ ðVÞ 22 ðlÞ 22 ðrÞ 21 F ¼ F V 2 l 2 r Similarly, P 2 ¼ ðVÞ d ðlÞ e r _ _ f ðkÞ Therefore, M 0 L 0 T 0 ¼ ðLT 21 Þ d ðLÞ e ðML 23 Þ f ðLÞ for M, 0 ¼ f or f ¼ 0; for T, 0 ¼ 2d or d ¼ 0; and for L, 0 ¼ d þ e 2 3f þ 1 or e ¼ 21. Thus, P 2 ¼ ðVÞ 0 ðlÞ 21 ðrÞ 0 k ¼ k l Chapter 1 38 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved and P 3 ¼ ðVÞ g ðlÞ h r _ _ i ðgÞ M 0 L 0 T 0 ¼ ðLT 21 Þ g ðLÞ h ðML 23 Þ i ðLT 22 Þ Equating the exponents gives, for M, 0 ¼ i or i ¼ 0; for T, 0 ¼ 2g–2 or g ¼ 2 2; for L, 0 ¼ g þ h 2 3i þ 1 or h ¼ 1. Therefore; P 3 ¼ V 22 l 1 r 0 g ¼ lg V 2 Now the functional relationship may be written as: f F V 2 l 2 r ; k l ; lg V 2 _ _ ¼ 0 Therefore, F ¼ V 2 l 2 r f k l ; lg V 2 _ _ Illustrative Example 1.6: Consider an axial flow pump, which has rotor diameter of 32 cm that discharges liquid water at the rate of 2.5 m 3 /min while running at 1450 rpm. The corresponding energy input is 120 J/kg, and the total efficiency is 78%. If a second geometrically similar pump with diameter of 22 cm operates at 2900 rpm, what are its (1) flow rate, (2) change in total pressure, and (3) input power? Solution: Using the geometric and dynamic similarity equations, Q 1 N 1 D 2 1 ¼ Q 2 N 2 D 2 2 Therefore, Q 2 ¼ Q 1 N 2 D 2 2 N 1 D 2 1 ¼ ð2:5Þð2900Þð0:22Þ 2 ð1450Þð0:32Þ 2 ¼ 2:363 m 3 /min As the head coefficient is constant, W 2 ¼ W 1 N 2 2 D 2 2 N 2 1 D 2 1 ¼ ð120Þð2900Þ 2 ð0:22Þ 2 ð1450Þ 2 ð0:32Þ 2 ¼ 226:88 J/kg The change in total pressure is: DP ¼ W 2 h tt r ¼ ð226:88Þð0:78Þð1000Þ N/m 2 ¼ ð226:88Þð0:78Þð1000Þ10 25 ¼ 1:77 bar Basic Thermodynamics and Fluid Mechanics 39 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Input power is given by P ¼ _ mW 2 ¼ ð1000Þð2:363Þð0:22688Þ 60 ¼ 8:94 kW Illustrative Example 1.7: Consider an axial flow gas turbine in which air enters at the stagnation temperature of 1050 K. The turbine operates with a total pressure ratio of 4:1. The rotor turns at 15500 rpm and the overall diameter of the rotor is 30 cm. If the total-to-total efficiency is 0.85, find the power output per kg per second of airflow if the rotor diameter is reduced to 20 cm and the rotational speed is 12,500 rpm. Take g ¼ 1.4. Solution: Using the isentropic P–T relation: T 0 02 ¼ T 01 P 02 P 01 _ _ g21 ð Þ 2 ¼ ð1050Þ 1 4 _ _ 0:286 ¼ 706:32K Using total-to-total efficiency, T 01 2T 02 ¼ T 01 2T 0 02 _ _ h tt ¼ ð343:68Þð0:85Þ ¼ 292:13 K and W 1 ¼ c p DT 0 ¼ ð1:005Þð292:13Þ ¼ 293:59 kJ/kg W 2 ¼ W 1 N 2 2 D 2 2 N 2 1 D 2 1 ¼ ð293:59 £ 10 3 Þð12; 500Þ 2 ð0:20Þ 2 ð15; 500Þ 2 ð0:30Þ 2 ¼ 84; 862 J/kg [ Power output ¼ 84.86 kJ/kg Illustrative Example 1.8: At what velocity should tests be run in a wind tunnel on a model of an airplane wing of 160 mm chord in order that the Reynolds number should be the same as that of the prototype of 1000 mm chord moving at 40.5 m/s. Air is under atmospheric pressure in the wind tunnel. Solution: Let Velocity of the model: V m Length of the model : L m ¼ 160 mm Length of the prototype: L p ¼ 1000 mm Velocity of the prototype: V p ¼ 40:5 m/s Chapter 1 40 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved According to the given conditions: ðReÞ m ¼ ðReÞ p V m L m n m ¼ V p L p n p ; Therefore; v m ¼ v p ¼ v air Hence V m L m ¼ V p L p ; or V m ¼ L p V p /L m ¼ 40:5 £ 1000/160 ¼ 253:13 m/s Illustrative Example 1.9: Show that the kinetic energy of a body equals kmV 2 using the method of dimensional analysis. Solution: Since the kinetic energy of a body depends on its mass and velocity, K:E: ¼ f ðV; mÞ; or K:E: ¼ kV a m b : Dimensionally, FLT 0 ¼ ðLT ÿ1 Þ a ðFT 2 L ÿ1 Þ b Equating the exponents of F, L, and T, we get: F: 1 ¼ b; L: 1 ¼ a 2b; T: 0 ¼2a þ2b This gives b ¼ 1 and a ¼ 2. So, K.E. ¼ kV 2 m, where k is a constant. Illustrative Example 1.10: Consider a radial inward flow machine, the radial and tangential velocity components are 340 m/s and 50 m/s, respectively, and the inlet and the outlet radii are 14 cm and 7 cm, respectively. Find the torque per unit mass flow rate. Solution: Here, r 1 ¼ 0:14 m C w1 ¼ 340 m/s; r 2 ¼ 0:07 m C w2 ¼ 50 m/s Basic Thermodynamics and Fluid Mechanics 41 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Torque is given by: T ¼ r 1 C w1 2r 2 C w2 ¼ ð0:14 £ 340 20:07 £ 50Þ ¼ ð47:6 23:5Þ ¼ 44:1 N-m per kg/s PROBLEMS 1.1 Show that the power developed by a pump is given by P ¼ kwQH where k ¼ constant, w ¼ specific weight of liquid, Q ¼ rate of discharge, and H ¼ head dimension. 1.2 Develop an expression for the drag force on a smooth sphere of diameter D immersed in a liquid (of density r and dynamic viscosity m) moving with velocity V. 1.3 The resisting force F of a supersonic plane in flight is given by: F ¼ f ðL; V; r; m; kÞ where L ¼ the length of the aircraft, V ¼ velocity, r ¼ air density, m ¼ air viscosity, and k ¼ the bulk modulus of air. 1.4 Show that the resisting force is a function of Reynolds number and Mach number. 1.5 The torque of a turbine is a function of the rate of flow Q, head H, angular velocity v, specific weight w of water, and efficiency. Determine the torque equation. 1.6 The efficiency of a fan depends on density r, dynamic viscosity m of the fluid, angular velocity v, diameter D of the rotor and discharge Q. Express efficiency in terms of dimensionless parameters. 1.7 The specific speed of a Kaplan turbine is 450 when working under a head of 12 m at 150 rpm. If under this head, 30,000 kW of energy is generated, estimate how many turbines should be used. (7 turbines). 1.8 By using Buckingham’s P theorem, show that dimensionless expression KP is given by: DP ¼ 4 f V 2 rl 2D Chapter 1 42 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved where KP ¼ pressure drop in a pipe, V ¼ mean velocity of the flow, l ¼ length of the pipe, D ¼ diameter of the pipe, m ¼ viscosity of the fluid, k ¼ average roughness of the pipe, and r ¼ density of the fluid. 1.9 If H f is the head loss due to friction (KP/w) and w is the specific weight of the fluid, show that H f ¼ 4 f V 2 l 2gD (other symbols have their usual meaning). 1.10 Determine the dimensions of the following in M.L.T and F.L.T systems: (1) mass, (2) dynamic viscosity, and (3) shear stress. M; FT 2 L 21 ; ML 21 T 21 ; FTL 22 ; ML 21 T 22 ; FL 23 _ _ NOTATION A r area ratio a sonic velocity B r breadth of prototype C velocity of gas, absolute velocity of turbo machinery D diameter of pipe, turbine runner, or pump D p depth of the prototype E energy transfer by a rotor or absorbed by the rotor F force F r force ratio g local acceleration due to gravity H head h specific enthalpy h 0 stagnation enthalpy K.E. kinetic energy L length L p length of prototype L r scale ratio M Mach number m mass rate of flow N speed N s specific speed P power P h hydraulic power P m power loss due to mechanical friction at the bearing P s shaft power P.E. potential energy Basic Thermodynamics and Fluid Mechanics 43 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved p fluid pressure p 0 stagnation pressure Q volume rate of flow, heat transfer R gas constant Re Reynolds number r radius of rotor s specific entropy sp . gr specific gravity of fluid T temperature, time T 0 stagnation temperature t time U rotor speed V relative velocity, mean velocity W work V r volume ratio, velocity ratio W t actual turbine work output W t 0 isentropic turbine work output a absolute air angle b relative air angle g specific weight, specific heat ratio h efficiency h /c polytropic efficiency of compressor h /t polytropic efficiency of turbine h c compressor efficiency h d diffuser efficiency h h hydraulic efficiency h j jet pipe or nozzle efficiency h m mechanical efficiency h o overall efficiency h p prototype efficiency h s isentropic efficiency h t turbine efficiency h ts total-to-static efficiency h tt total-to-total efficiency h v volumetric efficiency m absolute or dynamic viscosity n kinematic viscosity P dimensionless parameter r mass density t shear stress, torque exerted by or acting on the rotor v angular velocity Chapter 1 44 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved SUFFIXES 0 stagnation conditions 1 inlet to rotor 2 outlet from the rotor 3 outlet from the diffuser a axial h hub r radial t tip w whirl or tangential Basic Thermodynamics and Fluid Mechanics 45 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 2 Hydraulic Pumps 2.1 INTRODUCTION Hydraulics is defined as the science of the conveyance of liquids through pipes. The pump is often used to raise water from a low level to a high level where it can be stored in a tank. Most of the theory applicable to hydraulic pumps has been derived using water as the working fluid, but other liquids can also be used. In this chapter, we will assume that liquids are totally incompressible unless otherwise specified. This means that the density of liquids will be considered constant no matter how much pressure is applied. Unless the change in pressure in a particular situation is very great, this assumption will not cause a significant error in calculations. Centrifugal and axial flow pumps are very common hydraulic pumps. Both work on the principle that the energy of the liquid is increased by imparting kinetic energy to it as it flows through the pump. This energy is supplied by the impeller, which is driven by an electric motor or some other drive. The centrifugal and axial flow pumps will be discussed separately in the following sections. 2.2 CENTRIFUGAL PUMPS The three important parts of centrifugal pumps are (1) the impeller, (2) the volute casing, and (3) the diffuser. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 2.2.1 Impeller The centrifugal pump is used to raise liquids from a lower to a higher level by creating the required pressure with the help of centrifugal action. Whirling motion is imparted to the liquid by means of backward curved blades mounted on a wheel known as the impeller. As the impeller rotates, the fluid that is drawn into the blade passages at the impeller inlet or eye is accelerated as it is forced radially outwards. In this way, the static pressure at the outer radius is much higher than at the eye inlet radius. The water coming out of the impeller is then lead through the pump casing under high pressure. The fluid has a very high velocity at the outer radius of the impeller, and, to recover this kinetic energy by changing it into pressure energy, diffuser blades mounted on a diffuser ring may be used. The stationary blade passages have an increasing cross-sectional area. As the fluid moves through them, diffusion action takes place and hence the kinetic energy is converted into pressure energy. Vaneless diffuser passages may also be used. The fluid moves from the diffuser blades into the volute casing. The functions of a volute casing can be summarized as follows: It collects water and conveys it to the pump outlet. The shape of the casing is such that its area of cross-section gradually increases towards the outlet of the pump. As the flowing water progresses towards the delivery pipe, more and more water is added from the outlet periphery of the impeller. Figure 2.1 shows a centrifugal pump impeller with the velocity triangles at inlet and outlet. For the best efficiency of the pump, it is assumed that water enters the impeller radially, i.e., a 1 = 908 and C w1 = 0. Using Euler’s pump equation, the work done per second on the water per unit mass of fluid flowing E = W m = U 2 C w2 2U 1 C w1 ( ) (2:1) Where C w is the component of absolute velocity in the tangential direction. E is referred to as the Euler head and represents the ideal or theoretical head developed by the impeller only. The flow rate is Q = 2pr 1 C r1 b 1 = 2pr 2 C r2 b 2 (2:2) Where C r is the radial component of absolute velocity and is perpendicular to the tangent at the inlet and outlet and b is the width of the blade. For shockless entry and exit to the vanes, water enters and leaves the vane tips in a direction parallel to their relative velocities at the two tips. As discussed in Chapter 1, the work done on the water by the pump consists of the following three parts: 1. The part (C 2 2 – C 1 2 )/2 represents the change in kinetic energy of the liquid. 2. The part (U 2 2 – U 1 2 )/2 represents the effect of the centrifugal head or energy produced by the impeller. Chapter 2 48 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 3. The part (V 2 2 2 V 1 2 )/2 represents the change in static pressure of the liquid, if the losses in the impeller are neglected. 2.3 SLIP FACTOR From the preceding section, it may be seen that there is no assurance that the actual fluid will follow the blade shape and leave the impeller in a radial direction. There is usually a slight slippage of the fluid with respect to the blade rotation. Figure 2.2 shows the velocity triangles at impeller tip. In Fig. 2.2, b 2 / is the angle at which the fluid leaves the impeller, and b 2 is the actual blade angle, and C w2 and C w2 / are the tangential components of absolute velocity corresponding to the angles b 2 and b 2 / , respectively. Thus, C w2 is reduced to C w2 / and the difference DC w is defined as the slip. The slip factor is defined as Slip factor; s = C w2 / C w2 According to Stodola’s theory, slip in centrifugal pumps and compressors is due to relative rotation of fluid in a direction opposite to that of impeller with the same Figure 2.1 Velocity triangles for centrifugal pump impeller. Hydraulic Pumps 49 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved angular velocity as that of an impeller. Figure 2.3 shows the leading side of a blade, where there is a high-pressure region while on the trailing side of the blade there is a low-pressure region. Due to the lower pressure on the trailing face, there will be a higher velocity and a velocity gradient across the passage. This pressure distribution is associated with the existence of circulation around the blade, so that lowvelocity on the high- pressure side and high velocity on the low-pressure side and velocity distribution is not uniform at any radius. Due to this fact, the flow may separate from the suction surface of the blade. Thus, C w2 is less than C w2 / and the difference is defined as the slip. Another way of looking at this effect, as given by Stodola, is shown in Fig. 2.4, the impeller itself has an angular velocity v so that, relative to the impeller, the fluid must have an angular velocity of 2v; the result of this being a circulatory motion relative to the channel or relative eddy. The net result of the previous discussion is that the fluid is discharged from the impeller at an angle relative to the impeller, which is less than the vane angle as mentioned earlier. Figure 2.2 Velocity triangle at impeller outlet with slip. Figure 2.3 Pressure distribution on impeller vane. LP = low pressure, HP = high pressure. Chapter 2 50 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Hence, the slip factor s is defined as s = C / w2 C w2 (2:3) For purely radial blades, which are often used in centrifugal compressors, b 2 will be 908 and the Stodola slip factor becomes s = 1 2 p n (2:4) where n is the number of vanes. The Stanitz slip factor is given by s = 1 2 0:63p n (2:5) When applying a slip factor, the Euler pump equation becomes W m = sU 2 C w2 2U 1 C w1 (2:6) Typically, the slip factor lies in the region of 0.9, while the slip occurs even if the fluid is ideal. 2.4 PUMP LOSSES The following are the various losses occurring during the operation of a centrifugal pump. 1. Eddy losses at entrance and exit of impeller, friction losses in the impeller, frictional and eddy losses in the diffuser, if provided. 2. Losses in the suction and delivery pipe. The above losses are known as hydraulic losses. 3. Mechanical losses are losses due to friction of the main bearings, and stuffing boxes. Thus, the energy supplied by the prime mover to Figure 2.4 Relative eddy in impeller channel. Hydraulic Pumps 51 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved impeller is equal to the energy produced by impeller plus mechanical losses. A number of efficiencies are associated with these losses. Let r be the density of liquid; Q, flow rate; H, total head developed by the pump; P s , shaft power input; H i , total head across the impeller; and h i , head loss in the impeller. Then, the overall efficiency h o is given by: h o = Fluid power developed by pump Shaft power input = rgQH P s (2:7) Casing efficiency h c is given by: h c = Fluid power at casing outlet/fluid power at casing inlet = Fluid power at casing outlet/(fluid power developed by impeller 2leakage loss) = rgQH/rgQH i = H/H i (2:8) Impeller efficiency h i is given by: h i = Fluid power at impeller exit/fluid power supplied to impeller = Fluid power at impeller exit/(fluid power developed by impeller - impeller loss) = rgQ i H i / rgQ i H i - h i ( )  à = H i /(H i - h i ) (2:9) Volumetric efficiency h v is given by: h v = Flow rate through pump/flow rate through impeller = Q/(Q - q) (2:10) Mechanical efficiency h m is given by: h m = Fluid power supplied to the impeller/power input to the shaft = rgQ i (h i - H i )/P s (2:11) Therefore, h o = h c h i h v h m (2:12) Chapter 2 52 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved A hydraulic efficiency may be defined as h h = Actual head developed by pump Theoretical head developed by impeller = H (H i - h i ) (2:13) The head H is also known as manometric head. 2.5 THE EFFECT OF IMPELLER BLADE SHAPE ON PERFORMANCE The various blade shapes utilized in impellers of centrifugal pumps/compressors are shown in Fig. 2.5. The blade shapes can be classified as: 1. Backward-curved blades (b 2 , 908) 2. Radial blades (b 2 = 908) 3. Forward-curved blades (b 2 . 908) As shown in Fig. 2.5, for backward-curved vanes, the value of C w2 (whirl component at outlet) is much reduced, and thus, such rotors have a low energy transfer for a given impeller tip speed, while forward-curved vanes have a high value of energy transfer. Therefore, it is desirable to design for high values of b 2 (over 908), but the velocity diagrams show that this also leads to a very high value of C 2 . High kinetic energy is seldom required, and its reduction to static pressure by diffusion in a fixed casing is difficult to perform in a reasonable sized casing. However, radial vanes (b 2 = 908) have some particular advantages for very high- speed compressors where the highest possible pressure is required. Radial vanes are relatively easy to manufacture and introduce no complex bending stresses (Fig. 2.6). Figure 2.5 Centrifugal pump outlet velocity triangles for varying blade outlet angle. Hydraulic Pumps 53 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 2.6 VOLUTE OR SCROLL COLLECTOR A volute or scroll collector consists of a circular passage of increasing cross- sectional area (Fig. 2.7). The advantage of volute is its simplicity and low cost. The cross-sectional area increases as the increment of discharge increases around the periphery of the impeller, and, if the velocity is constant in the volute, Figure 2.6 Characteristics for varying outlet blade angle. Figure 2.7 Volute or scroll collector. Chapter 2 54 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved then the static pressure is likewise constant and the radial thrust will be zero. Any deviation in capacity (i.e., flow rate) from the design condition will result in a radial thrust which if allowed to persist could result in shaft bending. The cross-sectional shape of the volute is generally similar to that shown in Fig. 2.8, with the sidewalls diverging from the impeller tip and joined by a semicircular outer wall. The circular section is used to reduce the losses due to friction and impact when the fluid hits the casing walls on exiting from the impeller. 2.7 VANELESS DIFFUSER For the diffusion process, the vaneless diffuser is reasonably efficient and is best suited for a wide range of operations. It consists simply of an annular passage without vanes surrounding the impeller. A vaneless diffuser passage is shown in Fig. 2.9. The size of the diffuser can be determined by using the continuity equation. The mass flow rate in any radius is given by m = rAC r = 2prbrC r (2:14) where b is the width of the diffuser passage, C r = r 2 b 2 r 2 C r2 rbr (2:15) where subscripted variables represent conditions at the impeller outlet and the unsubscripted variables represent conditions at any radius r in the vaneless diffuser. Assuming the flow is frictionless in the diffuser, angular momentum Figure 2.8 Cross-section of volute casing. Hydraulic Pumps 55 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 2.9 Vaneless diffuser. Figure 2.10 Logarithmic spiral flow in vaneless space. Chapter 2 56 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved is constant and C w = C w2 r 2 ( )/r (2:16) But the tangential velocity component (C w ) is usually very much larger than the radial velocity component C r, and, therefore, the ratio of the inlet to outlet diffuser velocities C 2 C 3 = r 3 r 2 . It means that for a large reduction in the outlet kinetic energy, a diffuser with a large radius is required. For an incompressible flow, rC r is constant, and, therefore, tan a = C w /C r = constant. Thus, the flow maintains a constant inclination to radial lines, the flow path traces a logarithmic spiral. As shown in Fig. 2.10, for an incremental radius dr, the fluid moves through angle du, then rdu = dr tan a. Integrating we have u 2u 2 = tan a log r/r 2 ( ) (2:17) Substituting a = 788 and (r/r 2 ) = 2, the change in angle of the diffuser is equal to 1808. Because of the long flow path with this type of diffuser, friction effects are high and the efficiency is low. 2.8 VANED DIFFUSER The vaned diffuser is advantageous where small size is important. In this type of diffuser, vanes are used to diffuse the outlet kinetic energy of the fluid at a much higher rate than is possible by a simple increase in radius, and hence it is possible to reduce the length of flow path and diameter. The vane number, the angle of divergence is smaller, and the diffuser becomes more efficient, but greater is the friction. The cross section of the diffuser passage should be square to give a maximum hydraulic radius. However, the number of diffuser vanes should have no common factor with the number of impeller vanes. The collector and diffuser operate at their maximum efficiency at the design point only. Any deviation from the design discharge will change the outlet velocity triangle and the subsequent flow in the casing. 2.9 CAVITATION IN PUMPS Cavitation is caused by local vaporization of the fluid, when the local static pressure of a liquid falls below the vapor pressure of the liquid. Small bubbles or cavities filled with vapor are formed, which suddenly collapse on moving forward with the flow into regions of high pressure. These bubbles collapse with tremendous force, giving rise to pressure as high as 3500 atm. In a centrifugal pump, these low-pressure zones are generally at the impeller inlet, where the fluid is locally accelerated over the vane surfaces. In turbines, cavitation is most likely Hydraulic Pumps 57 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved to occur at the downstream outlet end of a blade on the low-pressure leading face. When cavitation occurs, it causes the following undesirable effects: 1. Local pitting of the impeller and erosion of the metal surface. 2. Serious damage can occur from this prolonged cavitation erosion. 3. Vibration of machine and noise is also generated in the form of sharp cracking sounds when cavitation takes place. 4. A drop in efficiency due to vapor formation, which reduces the effective flow areas. The avoidance of cavitation in conventionally designed machines can be regarded as one of the essential tasks of both pump and turbine designers. This cavitation imposes limitations on the rate of discharge and speed of rotation of the pump. A cavitation parameter is defined as s c = pump total inlet head above vapor pressure/head developed by the pump or at the inlet flange s c = p 1 rg - V 2 1 2g 2 p v rg = H (2:18) The numerator of Eq. (2.18) is a suction head and is called the net positive suction Figure 2.11 Cavitation limits for radial flow pumps. Chapter 2 58 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved head (NPSH) of the pump. It is a measure of the energy available on the suction side of the pump, and H is the manometric head. The cavitation parameter is a function of specific speed, efficiency of the pump, and number of vanes. Figure 2.11 shows the relationship between s c and N s . It may be necessary in the selection of pumps that the value of s c does not fall below the given value by the plots in Fig. 2.11 for any condition of operation. 2.10 SUCTION SPECIFIC SPEED The efficiency of the pump is a function of flow coefficient and suction specific speed, which is defined as N suc = NQ 1/2 g NPSH ( )  à 23/4 Thus, h = f Q; N suc À Á The cavitation parameter may also be determined by the following equation N s /N suc = (NPSH) 3/4 /H 3/4 = s 3/4 c (2:19) 2.11 AXIAL FLOW PUMP In an axial flow pump, pressure is developed by flow of liquid over blades of airfoil section. It consists of a propeller-type of impeller running in a casing. The advantage of an axial flow pump is its compact construction as well as its ability to run at extremely high speeds. The flow area is the same at inlet and outlet and the minimum head for this type of pump is the order of 20 m. 2.12 PUMPING SYSTEM DESIGN Proper pumping system design is the most important single element in minimizing the life cycle cost. All pumping systems are comprised of a pump, a driver, pipe installation, and operating controls. Proper design considers the interaction between the pump and the rest of the system and the calculation of the operating duty point(s) (Fig. 2.12). The characteristics of the piping system must be calculated in order to determine required pump performance. This applies to both simple systems as well as to more complex (branched) systems. Both procurement costs and the operational costs make up the total cost of an installation during its lifetime. A number of installation and operational costs are directly dependent on the piping diameter and the components in the piping system. Hydraulic Pumps 59 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved A considerable amount of the pressure losses in the system are caused by valves, in particular, control valves in throttle-regulated installations. In systems with several pumps, the pump workload is divided between the pumps, which together, and in conjunction with the piping system, deliver the required flow. The piping diameter is selected based on the following factors: . Economy of the whole installation (pumps and system) . Required lowest flow velocity for the application (e.g., avoid sedimentation) . Required minimum internal diameter for the application (e.g., solid handling) . Maximum flow velocity to minimize erosion in piping and fittings . Plant standard pipe diameters. Decreasing the pipeline diameter has the following effects: . Piping and component procurement and installation costs will decrease. . Pump installation procurement costs will increase as a result of increased flow losses with consequent requirements for higher head pumps and larger motors. Costs for electrical supply systems will therefore increase. . Operating costs will increase as a result of higher energy usage due to increased friction losses. Some costs increase with increasing pipeline size and some decrease. Because of this, an optimum pipeline size may be found, based on minimizing costs over the life of the system. A pump application might need to cover several Figure 2.12 The duty point of the pump is determined by the intersection of the system curve and the pump curve as shown above. Chapter 2 60 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved duty points, of which the largest flow and/or head will determine the rated duty for the pump. The pump user must carefully consider the duration of operation at the individual duty points to properly select the number of pumps in the installation and to select output control. 2.12.1 Methods for Analyzing Existing Pumping Systems The following steps provide an overall guideline to improve an existing pumping system. . Assemble a complete document inventory of the items in the pumping system. . Determine the flow rates required for each load in the system. . Balance the system to meet the required flow rates of each load. . Minimize system losses needed to balance the flow rates. . Affect changes to the pump to minimize excessive pump head in the balanced system. . Identify pumps with high maintenance cost. One of two methods can be used to analyze existing pumping systems. One consists of observing the operation of the actual piping system, and the second consists of performing detailed calculations using fluid analysis techniques. The first method relies on observation of the operating piping system (pressures, differential pressures, and flow rates), the second deals with creating an accurate mathematical model of the piping system and then calculating the pressure and flow rates with the model. The following is a checklist of some useful means to reduce the life cycle cost of a pumping system. . Consider all relevant costs to determine the life cycle cost. . Procure pumps and systems using life cycle cost considerations. . Optimize total cost by considering operational costs and procurement costs. . Consider the duration of the individual pump duty points. . Match the equipment to the system needs for maximum benefit. . Match the pump type to the intended duty. . Do not oversize the pump. . Match the driver type to the intended duty. . Specify motors to have high efficiency. . Monitor and sustain the pump and the system to maximize benefit. . Consider the energy wasted using control valves. Hydraulic Pumps 61 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 2.12.2 Pump System Interaction The actual operating point on the pump systemcharacteristic curve is defined by its interaction with the operating characteristics of the installed system (Fig. 2.13). The system characteristics will consist of: . The total static head, being the difference in elevation between the upstream and downstream controls (generally represented by reservoir levels), . The energy losses in the system (generally pipe friction), which are normally expressed as a function of velocity head. . The interaction point of these curves represents the actual operating point (as shown later), defining the Head supplied by the pump and the Discharge of the system. The efficiency of the pump under these conditions will also be defined. Note that the efficiency of the pump at this operating point is the critical parameter inpumpselectionandsuitabilityfor a particular system(Figs. 2.14 and 2.15). 2.13 LIFE CYCLE ANALYSIS Over the economic life of a pumped supply system, a number of design parameter will change. System behavior for all possible operating environments is needed (Fig. 2.16). Parameters that will change include: Figure 2.13 Typical pump characteristics. Chapter 2 62 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 2.15 Selection of pump type. Figure 2.14 Pump–system interaction point and pump efficiency. Hydraulic Pumps 63 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved . Seasonal variations in demand. . Water demand increases as the system population expands. . Increasing pipe friction as the system ages. For all operating conditions, it is necessary to maintain pump operation close to peak efficiency. This can be achieved using multiple pumps and timed pumping periods. Figure 2.16 Variations in demand and operating characteristics. Chapter 2 64 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 2.14 CHANGING PUMP SPEED The most common pump–motor combination employed in water supply operations is a close coupled system powered by an electric motor. These units can only operate at a speed related to the frequency of the A.C. supply (50 cycles/s or 3000 cycles/min), with the number of pairs of poles in the motor (N) reducing the pump speed to 3000/N revolutions per minute. Pumps driven through belt drives or powered by petrol or diesel motors are more flexible allowing the pump speed to be adjusted to suit the operational requirements. Analysis of system operation will require the head–discharge- efficiency characteristic for the particular operating speed. Given the head–discharge-efficiency characteristics for speed N (in tabular form), the corresponding characteristics for any speed N / can be established as follows: Q / = N / N Q flow points (2:20) H / = N / N 2 H head points (2:21) h / = h efficiency points (2:22) The data set for the new pump speed can then be matched to the system characteristics. 2.15 MULTIPLE PUMP OPERATION The most common type of pumping station has two or more pumps operating in parallel. This provides flexibility of operation in coping with a range of flow conditions and allows maintenance of individual units while other units are in operation. Occasionally situations will be encountered where pumps are operated in series to boost outputs. This is generally a temporary measure as any failure of one unit will severely affect system operation. Composite characteristics (head–discharge curves) are obtained by combining the individual curves. The composite curve is employed in the same manner (i.e., intersection with system curve) s an individual curve (Fig. 2.17). Where pumps operate in parallel, the composite curve is obtained by adding the flow rates for a given head. Where pumps operate in series, the composite is obtained by adding heads for a given flow rate. Hydraulic Pumps 65 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 2.15.1 Net Positive Suction Head Simply, NPSH is the minimum suction condition (pressure) required to prevent pump cavitation. Conceptually, NPSH can be imagined as the pressure drop between the pump inlet flange and the point inside the pump where the fluid dynamic action, as it leaves the impeller, causes a pressure rise. Sufficient NPSH allows for pumping without liquid vaporizing in the pump first-stage impeller eye as the fluid pressure drops due to pump suction losses (Fig. 2.18). The NPSH required is reported in head of fluid (being pumped) required at the pump inlet. As such, NPSH required has units of length (height). Usually, the datum line for pump NPSH is the centerline of the inlet. This is satisfactory for small pumps. For larger pumps, the NPSH requirements reported by the manufacturer should be verified for the datum line being discussed. The NPSH Figure 2.17 Composite pump characteristics. Chapter 2 66 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved available differs from NPSH required. The NPSH required determined during the manufacturers test and shown on the vendor’s pump curve is based upon a 3%head pump differential loss. The NPSH available must be large enough to eliminate head loss. The NPSH available is the excess of pressure over the liquid’s vapor pressure at the pump suction flange. Except in rare circumstances, centrifugal pumps require the NPSH available to be greater than NPSH required to avoid cavitation during operation. Determining the NPSH available is the process engineer’s job. Determining the NPSH required is the job of the pump vendor. Our concern here is the process system side of determining what NPSH is available. Pressure balance and NPSH available derive from Bernoulli’s equation for a stationary conduit where the total energy (head) in a system is the same for any point along a streamline (assuming no friction losses). Incorporating friction losses and restating the formula in a form familiar to process engineers, the NPSH available in a system can be expressed as: NPSH a = 2:31 P - P a 2P v ( ) g - S 2B 2L - V 2 2g (2:23) where NPSH a is the net positive suction head available (ft); P, pressure above liquid (psi gage); P a , atmospheric pressure (psi); P v , vapor pressure of liquid at pumping conditions (psia); g, specific gravity of liquid at pumping conditions; Figure 2.18 The elements of Eq. (2.24) are illustrated with a pump taking suction from a tower. Hydraulic Pumps 67 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved S, static height of liquid from grade (ft); B, distance of pump centerline (suction nozzle centerline for vertical pumps); L, suction system friction losses (ft of pumping liquid); and V is the average liquid velocity at pump suction nozzle (ft/s). Converting to absolute pressures, fluid density and resetting the datum line to the pump centerline results in: NPSH a = 144 P abs 2P v ( ) r - H 2L - V 2 2g (2:24) where P abs is the pressure above liquids (psia); r, fluid density (lb/ft 3 ); and H is the static height of liquid between liquid level and pump suction centerline (datum line), ft. Illustrative Example 2.1: A centrifugal pump runs at a tip speed of 12 m/s and a flow velocity of 1.5 m/s. The impeller diameter is 1.2 m and delivers 3.8 m 3 /min of water. The outlet blade angle is 288 to the tangent at the impeller periphery. Assuming that the fluid enters in the axial direction and zero slip, calculate the torque delivered by the impeller. Solution: From Fig. 2.2, for zero slip b 2 = b 2 1 . Using Eq. (2.1), the Euler head H = E = (U 2 C w2 2 U 1 C w1 )/g. Since C w1 = 0, as there is no inlet whirl component, head H is given by H = U 2 C w2 g = U 2 g U 2 2 1:5 tan 28 8 = 12 9:81 12 2 1:5 tan 28 8 = 11:23 m Power delivered = rgQH J s = 1000(9:81)(3:8)(11:23) 60(1000) = 6:98 kW Torque delivered = Power/angular velocity = 6980(0.6)/12 = 349 Nm. Illustrative Example 2.2: A fluid passes through an impeller of 0.22 m outlet diameter and 0.1 m inlet diameter. The impeller is rotating at 1250 rpm, and the outlet vane angle is set back at an angle of 228 to the tangent. Assuming that the fluid enters radially with velocity of flow as 3.5 m/s, calculate the head imparted to a fluid. Solution: Since fluid enters in the radial direction, C w1 = 0, a 1 = 908, b 2 = 228, C a1 = 3.5 m/s = C a2 Chapter 2 68 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Head developed H = C w2 U 2 /g Impeller tip speed, U 2 = pDN 60 = p(0:22)(1250) 60 = 14:40 m/s Whirl velocity at impeller outlet, from velocity diagram, C w2 = U 2 2(C a2 /tan b 2 ) = 14:40 2(3:5/tan 22 8 ) = 5:74 m/s Therefore, the head imparted is given by H = 5:74(14:40)/9:81 = 8:43 m Design Example 2.3: A centrifugal pump impeller runs at 1400 rpm, and vanes angle at exit is 258. The impeller has an external diameter of 0.4 m and an internal diameter of 0.2 m. Assuming a constant radial flow through the impeller at 2.6 m/s, calculate (1) the angle made by the absolute velocity of water at exit with the tangent, (2) the inlet vane angle, and (3) the work done per kg of water (Fig. 2.19). Solution: 1. Impeller tip speed is given by U 2 = pD 2 N 60 = p(0:4)(1400) 60 = 29:33 m/s Whirl velocity at impeller tip C w2 = U 2 2 C r2 tan b 2 = 29:33 2 2:6 tan 25 8 = 23:75 m/s Figure 2.19 Velocity triangle at outlet. Hydraulic Pumps 69 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Now, from the velocity triangle at impeller tip, tan a 2 = C r2 C w2 = 2:6 23:75 = 0:1095 Therefore, a 2 = 6.258. 2. Impeller velocity at inlet U 1 = pD 1 N 60 = p(0:2)1400 60 = 14:67 m/s tan b 1 = C r1 U 1 = 2:6 14:67 = 0:177 Therefore, b 1 = 10.058. 3. Work done per kg of water is given by C w2 U 2 = 23:75(29:33) = 696:59 Nm = 696:59 J: Design Example 2.4: A centrifugal pump impeller has a diameter of 1.2 m; rpm 210; area at the outer periphery 0.65 m 2 ; angle of vane at outlet 258, and ratio of external to internal diameter 2:1. Calculate (1) the hydraulic efficiency, (2) power, and (3) minimum speed to lift water against a head of 6.2 m. Assume that the pump discharges 1550 l/s (Fig. 2.20). Solution: 1. Here, Q = 1550 l/s, b 2 = 258, H = 6:2 m; D 2 /D 1 = 2; D 2 = 1:2 m, N = 210 rpm; A = 0:65 m 2 . Velocity of flow at impeller tip C r2 = Q A = 1550 1000(0:65) = 2:385 m/s Figure 2.20 Velocity triangle at impeller outlet. Chapter 2 70 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Impeller tip speed, U 2 = p(1:2)(210) 60 = 13:2 m/s C w2 = U 2 2 C r2 tan b 2 = 13:2 2 2:385 tan 258 = 8:09 m/s Assuming radial entry, theoretical head is given by H = C w2 U 2 9:81 = 8:09(13:2) 9:81 = 10:89 m Assuming slip factor, s = 1, hydraulic efficiency is given by h h = 6:2(100) 10:89 = 56:9%: 2. Power P = (1550)(10.89)(9.81)/1000 = 165.59 kW. 3. Centrifugal head is the minimum head. Therefore, U 2 2 2U 2 1 2g = 6:2 It is given that U 1 = U 2 /2. Therefore, U 2 2 20:25U 2 2 = 2(9:81)(6:2) i.e., U 2 = 12.74 m/s Hence, minimum speed is = 12.74(60)/p(1.2) = 203 rpm. Illustrative Example 2.5: A centrifugal pump is required to pump water against a total head of 35 m at the rate of 45 l/s. Find the horsepower of the pump, if the overall efficiency is 60%. Solution: Total head; H = 35m Discharge; Q = 45l/s = 0:045 m 3 /s Overall efficiency;h o = 60% = 0:60 Power; P = rgQH h o J s = rgQH 1000h o kW = 9:81(0:045)(35) 0:746(0:60) = 34:5 hp Illustrative Example 2.6: A centrifugal pump impeller has 0.3 m inlet diameter and 0.6 m external diameters. The pump runs at 950 rpm, and the entry Hydraulic Pumps 71 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved of the pump is radial. The velocity of flow through the impeller is constant at 3.5 m/s. The impeller vanes are set back at angle of 468 to the outer rim. Calculate (1) the vane angle at inlet of a pump, (2) the velocity direction of water at outlet, and (3) the work done by the water per kg of water (Fig. 2.21). Solution: 1. Velocity of flow, C r1 = C r2 = 3.5 m/s. Let a 1 be the vane angle at inlet. Tangential velocity of impeller at inlet U 1 = pD 1 N 60 = p(0:3)(950) 60 = 14:93 m/s From inlet velocity triangle tan a 1 = C r1 U 1 = 3:5 14:93 = 0:234 Therefore, a 1 = 13.198. 2. Tangential velocity of impeller at outlet U 2 = pD 2 N 60 = p(0:6)(950) 60 = 29:86 m/s For velocity of whirl at impeller outlet, using velocity triangle at outlet C w2 = U 2 2 C r2 tan 468 = 29:86 2 3:5 tan 468 = 26:48 m/s and C 2 2 = C 2 r2 - C 2 w2 = 3:5 2 - 26:48 2 ; C 2 = 26:71 m/s where C 2 is the velocity of water at outlet. Let a 2 be the direction of Figure 2.21 Velocity triangle at impeller outlet and inlet. Chapter 2 72 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved water at outlet, and, therefore, a 2 is given by tan a 2 = C r2 C w2 = 3:5 26:48 = 0:132 i.e., a 2 = 7.538. 3. Work done by the wheel per kg of water W = C w2 U 2 = 26:48(29:86) = 790:69 Nm Design Example 2.7: A centrifugal pump delivers water at the rate of 8.5 m 3 /min against a head of 10 m. It has an impeller of 50 cm outer diameter and 25 cm inner diameter. Vanes are set back at outlet at an angle of 458, and impeller is running at 500 rpm. The constant velocity of flow is 2 m/s. Determine (1) the manometric efficiency, (2) vane angle at inlet, and (3) minimum starting speed of the pump (Fig. 2.22). Solution: 1. The manometric efficiency is given by h man = H (C w2 U 2 /g) From outlet velocity triangle U 2 = pD 2 N 60 = p(0:5)(500) 60 = 13:0 m/s Figure 2.22 Velocity triangle (a) outlet, (b) inlet. Hydraulic Pumps 73 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Now, tan b 2 = C r2 /(U 2 2 C w2 ), or tan 458 = 2/(U 2 2 C w2 ), or 1 = 2/(13 2 C w2 ), C w2 = 11 m/s. Hence, h man = H (C w2 U 2 /g) = 10(9:81) 11(13) = 68:6% 2. Vane angle at inlet b 1 tan b 1 = C r1 U 1 and U 1 = 0:5 £ U 2 = 6:5 m/s [ tan b 1 = 2 6:5 = 0:308 i.e., b 1 = 178. 3. The minimum starting speed is (U 2 2 2U 2 1 )/2g = H or pD 2 N 60 À Á 2 2 pD 1 N 60 À Á 2 2(9:81) = 10 Therefore, N = 618 rpm. Illustrative Example 2.8: A centrifugal pump impeller has 0.6 m outside diameter and rotated at 550 rpm. Vanes are radial at exit and 8.2 cm wide. Velocity of flow through the impeller is 3.5 m/s, and velocity in the delivery pipe is 2.5 m/s. Neglecting other losses, calculate head and power of the pump (Fig. 2.23). Solution: 1. D 2 = 0.6 m, N = 550 rpm, C r2 = 3.5 m/s, C w2 = U 2 . Impeller speed at outlet U 2 = pD 2 N 60 = p(0:6)(550) 60 = 17:29 m/s: Figure 2.23 Velocity triangle for Example 2.8. Chapter 2 74 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Head, through which the water can be lifted, H = C w2 U 2 g 2 V 2 2g (neglecting all losses) = (17:29)(17:29) 9:81 2 2:5 2 2(9:81) = 30:47 20:319 = 30:2 m of water: 2. Power = rgQH 1000 kW where Q = pD 2 b 2 C r2 (where b 2 is width) = p(0:6)(0:082)(3:5) = 0:54 m 3 /s Therefore, power is given by P = rgQH 1000 kW = 1000(9:81)(0:54)(30:2) 1000 = 160 kW: Illustrative Example 2.9: A centrifugal pump impeller has a diameter of 1 m and speed of 11 m/s. Water enters radially and discharges with a velocity whose radial component is 2.5 m/s. Backward vanes make an angle of 328 at exit. If the discharge through the pump is 5.5 m 3 /min, determine (1) h.p. of the pump and (2) turning moment of the shaft (Fig. 2.24). Solution: 1. Data D 2 = 1 m, U 2 = 11 m/s, Figure 2.24 Velocity triangles for Example 2.9. Hydraulic Pumps 75 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved a 1 = 908, C r2 = 2.5 m/s, b 2 = 328, Q = 5.5 m 3 /min. First, consider outlet velocity triangle C w2 = U 2 2 C r2 tan b 2 = 11 2 2:5 tan 328 = 7 m/s Power of the pump is given by P = rQC w2 U 2 1000 kW P = 1000 ( ) 5:5 ( ) 7 ( ) 11 ( ) 60 ( ) 1000 ( ) = 7 kW: 2. Now, h:p: = 2pNT 60 where T is the torque of the shaft. Therefore, T = h:p: £ 60 2pN But U 2 = pD 2 N 60 or N = 60£U 2 pD 2 = 60(11) p(1) = 210 rpm Hence, T = (7)(1000)(60) 2p(210) = 318 Nm/s: Illustrative Example 2.10: A centrifugal pump running at 590 rpm and discharges 110 l/min against a head of 16 m. What head will be developed and quantity of water delivered when the pump runs at 390 rpm? Solution: N 1 = 590, Q 1 = 110 l/min = 1.83 l/s, H 1 = 16 m, N 2 = 390 rpm, H 2 = ? Chapter 2 76 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved As ffiffiffiffiffiffi H 1 p N 1 = ffiffiffiffiffiffi H 2 p N 2 Then, ffiffiffiffiffi 16 _ 590 = ffiffiffiffiffiffi H 2 p 390 Therefore, H 2 = 6.98 m Therefore, head developed by the pump at 390 rpm = 6.98 m. In order to find discharge through the pump at 390 rpm, using the Eq. (1.11) N 1 ffiffiffiffiffiffi Q 1 _ H 3/4 1 = N 2 ffiffiffiffiffiffi Q 2 _ H 3/4 2 590 ffiffiffiffiffiffiffiffiffi 1:83 _ 16 ( ) 3/4 = 390 ffiffiffiffiffiffi Q 2 _ 6:98 ( ) 3/4 or 798:14 8 = 390 ffiffiffiffiffiffi Q 2 _ 4:29 ffiffiffiffiffiffi Q 2 p = 1:097 i.e., Q = 1.203 l/s Illustrative Example 2.11: The impeller of a centrifugal pump has outlet diameter of 0.370 m, runs at 800 rpm, and delivers 30 l/s of water. The radial velocity at the impeller exit is 2.5 m/s. The difference between the water levels at the overhead tank and the pump is 14 m. The power required to drive the pump is 8 hp, its mechanical and volumetric effectiveness being 0.96 and 0.97, respectively. The impeller vanes are backward curved with an exit angle of 458. Calculate (1) ideal head developed with no slip and no hydraulic losses and (2) the hydraulic efficiency. Solution: 1. Impeller tip speed U 2 = pD 2 N 60 or U 2 = p(0:37)(800) 60 = 15:5 m/s: As the radial velocity at the impeller exit = 2.5 m/s. Therefore, C w2 = U 2 2 C r2 tan b 2 = 15:5 2 2:5 tan 458 = 13 m/s. When there is no slip, the head developed will be H = C w2 U 2 g = (13)(15:5) 9:81 = 20:54 m Hydraulic Pumps 77 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved If there are no hydraulic internal losses, the power utilized by the pump will be: P = (0:96)(8) = 7:68 hp Theoretical flow rate = Q h v = 0:03 0:97 = 0:031 m 3 /s Ideal head, H i , is given by H i = (7:68)(0:746) (9:81)(0:031) = 18:84 m: 2. The hydraulic efficiency is h h = H H i = 14 18:84 = 0:746 or 74:3%: Illustrative Example 2.12: The impeller of a centrifugal pump has outer diameter of 1.06 m and speed is 56 m/s. The blades are backward curved and they make an angle of 208 with the wheel tangent at the blade tip. If the radial velocity of the flow at the tip is 7.5 m/s and the slip factor is 0.88. Determine (1) the actual work input per kg of water flow and (2) the absolute velocity of fluid at the impeller. Solution: 1. Exit blade angle, b 2 = 208 [C w2 = U 2 2 C r2 tan b 2 = 56 2 7:5 tan 208 = 35:4 m/s Using slip factor, s = 0.88, the velocity whirl at exit is, C w2 = s £ 35.4 = 0.88 £ 35.4 = 31.2 m/s. Work input per kg of water flow W = C w2 U 2 1000 = (56)(31:2) 1000 = 1:75 kJ/kg: 2. Absolute velocity at impeller tip C 2 = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 2 r2 - C 2 W2 À Á q = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 7:5 2 - 31:2 2 À Á q C 2 = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 56:25 - 973:44 _ = 32:09 m/s Design Example 2.13: A centrifugal pump impeller of 260 mm diameter runs at 1400 rpm, and delivers 0.03 m 3 /s of water. The impeller has Chapter 2 78 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved a backward curved facing blades inclined at 308 to the tangent at outlet. The blades are 20 mm in depth at the outlet, and a slip factor of 0.78 may be assumed. Calculate the theoretical head developed by the impeller, and the number of impeller blades. Solution: Assuming the blades are of infinitesimal thickness, the flow area is given by A = impeller periphery £ blade depth = p £ 0:26 £ 0:02 = 0:0163 m 2 Flow velocity is given by C r2 = Q A = 0:03 0:0163 = 1:84 m/s Impeller tip speed, U 2 , is U 2 = pD 2 N 60 = p(0:26)(1400) 60 = 19:07 m/s Absolute whirl component, C w2 is given by C w2 = U 2 2 C r2 tan 308 = 19:07 2 1:84 tan 308 = 15:88 m/s Using Euler’s equation, and assuming C w1 = 0 (i.e., no whirl at inlet) H = U 2 C w2 g = (19:07)(15:88) 9:81 = 30:87 m Theoretical head with slip is H = 0.78 £ 30.87 = 24.08 m. To find numbers of impeller blades, using Stanitz formula Slip factor, s = 1 2 0:63p n or 0:78 = 1 2 0:63p n or 0:63p n = 1 2 0:78 = 0:22 [ n = 0:63p 0:22 = 9 Number of blades required = 9 Design Example 2.14: A centrifugal pump impeller runs at 1500 rpm, and has internal and external diameter of 0.20 m and 0.4 m, respectively, assuming constant radial velocity at 2.8 m/s and the vanes at the exit are set back at an angle of 308. Calculate (1) the angle absolute velocity of water at exit makes with the tangent, (2) inlet vane angle, and (3) the work done per kg of water. Hydraulic Pumps 79 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: 1. D 1 = 0.2 m, D 2 = 0.4 m, N = 1500 rpm, C r2 = 2.8 m/s, b 2 = 308. Impeller tip speed, U 2 , is U 2 = pD 2 N 60 = p(0:4)(1500) 60 = 31:43 m/s Whirl component of absolute velocity at impeller exit is C w2 = U 2 2 C r2 tan 308 = 31:43 2 2:8 tan 308 = 26:58 m/s tan a 2 = 2:8 26:58 = 0:1053 i.e., a 2 = 68. 2. Impeller speed at inlet U 1 = pD 1 N 60 = p(0:2)(1500) 60 = 15:7 m/s tan b 1 = 2:8 15:7 = 0:178 i.e., b 1 = 10.18. 3. Work done per kg of water C w2 U 2 = 26:58 £ 31:43 = 835:4 Nm: Design Example 2.15: An axial flow pump discharges water at the rate of 1.30 m 3 /s and runs at 550 rpm. The total head is 10 m. Assume blade velocity = 22 m/s, the flow velocity = 4.5 m/s, hydraulic efficiency = 0.87, and the overall pump efficiency = 0.83, find (1) the power delivered to the water, and power input, (2) the impeller hub diameter and tip diameter, and (3) the inlet and outlet blade angles for the rotor. Solution: 1. Power delivered to the water P = rgHQ/1000 kW = (9:81)(1:30)(10) = 127:53 kW Power input to the pump P = 127:53 0:83 = 153:65 kW: Chapter 2 80 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 2. Rotor tip diameter is given by D 2 = 60U 2 pN = (60)(22) p(550) = 0:764 m Rotor hub diameter D 2 1 = D 2 2 2 Q p/4 ( ) £ C a = 0:764 2 2 1:3 p/4 ( )(4:5) = 0:216 m i.e., D 1 = 0.465 m. 3. Rotor velocity at hub is given by U 1 = D 1 D 2 U 2 = (0:465)(22) 0:764 = 13:39 m/s Since, the axial velocity is constant, we have: rotor inlet angle at tip a 1t = tan 21 C a /U 1 ( ) = tan 21 4:5/13:39 ( ) = 18:588 Rotor outlet angle a 2t = tan 21 C a /U 2 ( ) = tan 21 4:5/22 ( ) = 11:568: Design Example 2.16: A single stage, radial flow, and double suction centrifugal pump having the following data: Discharge 72 l/s Inner diameter 90 mm Outer diameter 280 mm Revolution/minute 1650 Head 25 m Width at inlet 20 mm/side Width at outlet 18 mm/side Absolute velocity angle at inlet 908 Leakage losses 2 l/s Mechanical losses 1:41 kW Contraction factor due to vane thickness 0:85 Relative velocity angle measured from tangential direction 358 Overall efficiency of the pump 0:56 Hydraulic Pumps 81 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Determine (1) inlet vane angle, (2) the angle at which the water leaves the wheel, (3) the absolute velocity of water leaving impeller, (4) manometric efficiency, and (5) the volumetric and mechanical efficiencies. Solution: Total quantity of water to be handled by the pump Q t = Q del - Q leak = 72 - 2 = 74 Total quantity of water per side = 74/2 = 37 l/s 1. Impeller speed at inlet U 1 = pD 1 N 60 = p(0:09)(1650) 60 = 7:78 m/s Flow area at inlet = PD 1 b 1 £ contraction factor = (P)(0:09)(0:02)(0:85) = 0:0048 m 2 Therefore, the velocity of flow at inlet C r1 = Q Area of flow = 37 £ 10 23 0:0048 = 7:708 m/s From inlet velocity triangle tan b 1 = C r1 U 1 = 7:708 7:78 = 0:9907 b 1 = 44:738 2. Area of flow at outlet A 2 = P£ D 2 £ b 2 £ contraction factor Where b 2 = 18/2 = 9 mm for one side. So, A 2 = (P)(0.28)(0.009) (0.85) = 0.0067 m 2 . Therefore, the velocity of flow at outlet C r2 = Q Area of flow = 37 £ 10 23 0:0067 = 5:522 m/s The impeller speed at outlet U 2 = pD 2 N 60 = p(0:28)(1650) 60 = 24:2 m/s Chapter 2 82 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Now using velocity triangle at outlet tan b 2 = C r2 U 2 2C w2 = 5:522 24:2 2C w2 C w2 = 16:99 m/s Further, tan a 2 = C r2 C w2 = 5:522 16:99 = 0:325 a 2 = 188: 3. The absolute velocity of water leaving the impeller C 2 = C w2 Cos a 2 = 16:99 Cos 188 = 17:8 m/s: 4. The manometric efficiency h mano = (g)(H mano ) (U 2 )(C W2 ) = 9:81 £ 25 24:2 £ 16:99 = 0:596: 5. The volumetric efficiency h v = Q 2 Q Total = 72 74 = 0:973 Water power = rgQH = 1000 £ 9:81 £ 72 £ 25/1000 = 17:66 kW Shaft power = Water power h o = 17:66 0:56 = 31:54 kW Mechanical efficiency is h m = P s 2P loss P s = 31:54 21:41 31:54 = 0:955 or 95:5%: Illustrative Example 2.17: A single stage centrifugal pump is designed to give a discharge of Q when working against a manometric head of 20 m. On test, it was found that head actually generated was 21.5 m for the designed discharge, Q. If it is required to reduce the original diameter 32 cm without reducing the speed of the impeller, compute the required diameter to be reduced. Hydraulic Pumps 83 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: Head generated by the pump H = U 2 2g = (pDN/60) 2 2g or H « D 2 H H / = D D / 2 H = 21:5 m; H / = 20 m; D = 32 cm So, D / = D H / H 1/2 = 32 20 21:5 2 = 30:86 cm Design Example 2.18: A two stage centrifugal pump is designed to discharge 55 l/s at a head of 70 m. If the overall efficiency is 76% and specific speed per stage about 38, find (1) the running speed in rpm and (2) the power required to run pump. If the actual manometric head developed is 65% of the theoretical head, assuming no slip, the outlet angle of the blades 288, and radial velocity at exit 0.14 times the impeller tip speed at exit, find the diameter of impeller. Solution: 1. The specific speed is N s = N ffiffiffiffi Q _ H 3/4 N = N s H 3/4 ffiffiffiffi Q _ = 38(70/2) 3/4 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 55 £ 10 23 _ = 546:81 0:235 = 2327 rpm: 2. Q = 55 £ 10 23 m 3 /s Power required to drive pump = rgQH 0:76 = 1000 £ 9:81 £ 55 £ 10 23 £ 70 0:76 £ 1000 = 49:7 kW Chapter 2 84 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved H mano = 0.65 H. Here b 2 = 288 and C r2 = 0:14U 2 . From velocity triangle at outlet tan b 2 = C r2 U 2 2C W2 or tan 288 = 0:14U 2 U 2 2C W2 U 2 U 2 2C W2 = 0:5317 0:14 = 3:798 (A) As the flow at entrance is radial and a 1 = 908, the fundamental equation of pump would be H mano h mano = U 2 C W2 g Where h mano manometric efficiency of pump which is 65%. Therefore, 35 0:65 = U 2 C W2 g U 2 C W2 = 35 £ 9:81 0:65 C W2 = 528:23 U 2 (B) Substituting for C W2 in Eq. (A) and solving U 2 U 2 2 528:23 U 2 = 3:798 U 2 = 26:78 m/s: Also, U 2 = pD 2 N 60 or 26:78 = p£D 2 £2327 60 D 2 = 0:2197 m or 21:97 cm: Design Example 2.19: Two multistage centrifugal pumps are used in series to handle water flow rate of 0.0352m 3 /s, and total head required is 845 m. Each pump is required to produce a head of half the total and run at 1445rpm. If the impeller in all the stages is identical and specific speed is 14, determine (1) head developed per stage and the required number of stages in each pump, (2) The required impeller diameters assuming the speed ratio based on the outer Hydraulic Pumps 85 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved tip diameter to be 0.96 and the shaft power input, if the overall efficiency of each pump is 0.75. Solution: Head developed in each stage is H 3/4 = N ffiffiffiffi Q _ N s = 1445 ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:0352 _ 14 H = 51:93 m Total head required = 845 m (of water) Number of stages needed = 845 51:93 = 16 Number of stages in each pump = 8 Impeller speed at tip is U 2 = 0:96(2gH) 0:5 = 0:96[2 £ 9:81 £ 51:93] 0:5 = 30:6 m/s Impeller diameter at tip, D 2 = p £ 60 £ 30:6 £ 1445. But U 2 = pD 2 N 60 or D 2 = U 2 £ 60 p £ 1445 = 30:6 £ 60 p £ 1445 = 0:4043 m or 40:43 cm: Design Example 2.20: A centrifugal pump is required to be made to lift water through 105 m heights from a well. Number of identical pumps having their designed speed 900 rpm and specific speed 700 rpm with a rated discharge of 5500 l/min are available. Determine the number of pumps required and how they should be connected? Solution: Specific speed for a single impeller is given by N s = N ffiffiffiffi Q _ H 3/4 Given; N s = 700; H = 105 N = 900; and Q = 5500 60 = 91:67 l/s Substituting, 700 = 900 ffiffiffiffiffiffiffiffiffiffiffi 91:67 _ H 3/4 ; H = 28 m Chapter 2 86 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Hence number of stages required = Total head to be developed Head per stage = 105 28 = 4 stages in series Design Example 2.21: The specific speed of an axial flow pump impeller is 1150 and velocity of flow is 2.5 m/s. The outer and inner diameters of the impeller are 0.90 m and 0.45 m, respectively. Calculate the suitable speed of the pump to give a head of 5.5 m. Also, calculate vane angle at the entry of the pump. Solution: Given, D 2 = 0.9 m, D 1 = 0.45 m, N s = 1150, C r = 2.5 m/s, H = 5.5 m. As discharge; Q = area of flow £ velocity of flow = p 4 (0:9 2 20:45 2 ) £ 2:5 = 1:193 m 3 /s = 1193 l/s Also, N s = N ffiffiffiffi Q _ H 3/4 or 1150 = N ffiffiffiffiffiffiffiffiffiffi 1193 _ (5:5) 3/4 N = (5:5) 3/4 £ 1150 ffiffiffiffiffiffiffiffiffiffi 1193 _ = 120 rpm In order to find vane angle at entry, using velocity triangle at inlet, U 1 = pD 1 N 60 = p £ 0:45 £ 120 60 = 2:82 m/s tan a 1 C r1 U 1 = 2:5 2:82 = 0:8865 i.e., a = 41:568: Hydraulic Pumps 87 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved PROBLEMS 2.1 A centrifugal pump of 25 cm impeller diameter running at 1450 rpm, develops a head of 15 m. If the outlet flow area is 480 cm 2 , and discharging water 0.12 m 3 /s, and loss of head in the pump can be taken as 0.003C 2 1 , find the outlet blade angle. (148) 2.2 A centrifugal pump having vane angles at inlet and outlet are 258 and 308, respectively. If internal and external diameters of impeller are 0.15 and 0.30 m, respectively, calculate the work done per kg of water. Assume velocity of flow constant. (197.18 Nm) 2.3 A centrifugal pump discharges 50 liters/second of water against a total head of 40 m. Find the horsepower of the pump, if the overall efficiency is 62%. (42 hp) 2.4 A centrifugal pump delivers 26 l/s against a total head of 16 m at 1450 rpm. The impeller diameter is 0.5 m. A geometrically similar pump of 30 cm diameter is running at 2900 rpm. Calculate head and discharge required assuming equal efficiencies between the two pumps. (11.52 m, 11.23 l/s) 2.5 A centrifugal pump is built to work against a head of 20 m. A model of this pump built to one-fourth its size is found to generate a head of 7 m when running at its best speed of 450 rpm and requires 13.5 hp to run it. Find the speed of the prototype. (190 rpm) 2.6 Derive the expression for power required for a pump when it discharges a liquid of specific weight w at the rate of Q against a head of H. 2.7 Show that the pressure rise in an impeller of a centrifugal pump is given by C 2 r1 -U 2 2 2C 2 r2 cosec 2 b 2 2g (where C r 1 = velocity of flow at inlet, U 2 = blade velocity at outlet, C r 2 = velocity of flow at outlet, and b 2 = blade angle at outlet). Assuming that friction and other losses are neglected. 2.8 Derive an expression for static head developed by a centrifugal pump having radial flow at inlet. 2.9 A centrifugal pump discharges 0.15 m 3 /s of water against a head of 15 m. The impeller has outer and inner diameter of 35 and 15 cm, respectively. The outlet vanes are set back at an angle 408. The area of flow is constant from inlet to outlet and is 0.06 m 2 . Calculate the manometric efficiency Chapter 2 88 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved and vane angle at inlet if the speed of the pump is 960 rpm. Take slip factor = 1. (57.3%, 188) 2.10 A centrifugal pump of 35 cm diameter running at 1000 rpm develops a head of 18 m. The vanes are curved back at an angle of 308 to the tangent at outlet. If velocity flow is constant at 2.4 m/s, find the manometric efficiency of the pump. (76.4%) 2.11 An axial flow pump is required to deliver 1 m 3 /s at 7 m head while running at 960 rpm. Its outer diameter is 50 and hub diameter is 25 cm. Find (1) flow velocity, which is assumed to be constant from hub to tip and (2) power required to drive the pump if overall efficiency is 84%. (6.791 m/s, 81.75 kW) 2.12 An axial flow pump has the following data: Rotational speed 750 rpm Discharge of water 1:75 m 3 /s Head 7:5 m Hub to runner diameter ratio 0:45 Through flow velocity is 0.35 times the peripheral velocity. Find the diameter and minimum speed ratio. (0.59 m, 0.83) 2.13 In an axial flow pump, the rotor has an outer diameter of 75 cm and an inner diameter of 40 cm; it rotates at 500 rpm. At the mean blade radius, the inlet blade angle is 128 and the outlet blade angle is 158. Sketch the corresponding velocity diagrams at inlet and outlet, and estimate from them (1) the head the pump will generate, (2) the discharge or rate of flow in l/s, (3) the shaft h.p. input required to drive the pump, and (4) the specific speed of the pump. Assume a manometric or hydraulic efficiency of 88% and a gross or overall efficiency of 81%. (19.8 m; 705 l/s; 230 hp; 45) 2.14 If an axial flow pump delivers a discharge Q against a head H when running at a speed N, deduce an expression for the speed of a geometrically similar pump of such a size that when working against unit head, it will transmit unit power to the water flowing through it. Show that this value is proportional to the specific speed of the pump. Hydraulic Pumps 89 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved NOTATION b width of the diffuser passage C w2 tangential components of absolute velocity corresponding to the angle b 2 E Euler head H total head developed by the pump H i total head across the impeller N suc Suction specific speed m mass flow rate n number of vanes P s shaft power input Q flow rate r radius U impeller speed V relative velocity a absolute velocity angle b relative velocity angle h c casing efficiency h R hydraulic efficiency h i impeller efficiency h m mechanical efficiency h o overall efficiency h v volumetric efficiency r density of liquid s slip factor v angular velocity SUFFIXES 1 inlet to impeller 2 outlet from the impeller 3 outlet from the diffuser a axial r radial w whirl Chapter 2 90 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 3 Hydraulic Turbines 3.1 INTRODUCTION In a hydraulic turbine, water is used as the source of energy. Water or hydraulic turbines convert kinetic and potential energies of the water into mechanical power. The main types of turbines are (1) impulse and (2) reaction turbines. The predominant type of impulse machine is the Pelton wheel, which is suitable for a range of heads of about 150–2,000 m. The reaction turbine is further subdivided into the Francis type, which is characterized by a radial flow impeller, and the Kaplan or propeller type, which is an axial-flow machine. In the sections that follow, each type of hydraulic turbine will be studied separately in terms of the velocity triangles, efficiencies, reaction, and method of operation. 3.2 PELTON WHEEL An American Engineer Lester A. Pelton discovered this (Fig. 3.1) turbine in 1880. It operates under very high heads (up to 1800 m.) and requires comparatively less quantity of water. It is a pure impulse turbine in which a jet of fluid delivered is by the nozzle at a high velocity on the buckets. These buckets are fixed on the periphery of a circular wheel (also known as runner), which is generally mounted on a horizontal shaft. The primary feature of the impulse Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved turbine with respect to fluid mechanics is the power production as the jet is deflected by the moving vane(s). The impact of water on the buckets causes the runner to rotate and thus develops mechanical energy. The buckets deflect the jet through an angle of about 160 and 1658 in the same plane as the jet. After doing work on the buckets water is discharged in the tailrace, and the whole energy transfer from nozzle outlet to tailrace takes place at constant pressure. The buckets are so shaped that water enters tangentially in the middle and discharges backward and flows again tangentially in both the directions to avoid thrust on the wheel. The casing of a Pelton wheel does not perform any hydraulic function. But it is necessary to safeguard the runner against accident and also to prevent the splashing water and lead the water to the tailrace. 3.3 VELOCITY TRIANGLES The velocity diagrams for the Pelton wheel are shown in Fig. 3.2. Since the angle of entry of the jet is nearly zero, the inlet velocity triangle is a straight line, as shown in Fig. 3.2. If the bucket is brought to rest, then the relative fluid velocity, V 1 , is given by V 1 = jet velocity 2 bucket speed = C 1 2 U 1 The angle turned through by the jet in the horizontal plane during its passage over the bucket surface is a and the relative velocity at exit is V 2 . The absolute Figure 3.1 Single-jet, horizontal shaft Pelton turbine. Chapter 3 92 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved velocity, C 2 , at exit can be obtained by adding bucket speed vector U 2 and relative velocity, V 2 , at exit. Now using Euler’s turbine Eq. (1.78) W = U 1 C W1 2 U 2 C W2 Since in this case C W2 is in the negative x direction, W = U (U - V 1 ) - V 1 cos(180 2a) 2 U [ ] ¦ ¦ Neglecting loss due to friction across the bucket surface, that is, V 1 = V 2 , then W = U(V 1 2 V 1 cos a) Therefore E = U(C 1 2 U)(1 2 cos a)/g (3:1) the units of E being Watts per Newton per second weight of flow. Eq. (3.1) can be optimized by differentiating with respect to U, and equating it to zero. Therefore dE dU = (1 2 cos a)(C 1 2 2U)/g = 0 Then C 1 = 2U or U = C 1 /2 (3:2) Figure 3.2 Velocity triangles for a Pelton wheel. Hydraulic Turbines 93 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Substituting Eq. (3.2) into Eq. (3.1) we get E max = C 2 1 (1 2 cos a)/4g In practice, surface friction is always present and V 1 – V 2 , then Eq. (3.1) becomes E = U(C 1 2 U)(1 2 k cos a)/g (3:3) where k = V 2 V 1 Introducing hydraulic efficiency as h h = Energy Transferred Energy Available in jet i:e: h h = E (C 2 1 /2g) (3:4) if a = 1808, the maximum hydraulic efficiency is 100%. In practice, deflection angle is in the order of 160–1658. 3.4 PELTON WHEEL (LOSSES AND EFFICIENCIES) Head losses occur in the pipelines conveying the water to the nozzle due to friction and bend. Losses also occur in the nozzle and are expressed by the velocity coefficient, C v . The jet efficiency (h j ) takes care of losses in the nozzle and the mechanical efficiency (h m ) is meant for the bearing friction and windage losses. The overall efficiency (h o ) for large Pelton turbine is about 85–90%. Following efficiency is usually used for Pelton wheel. Pipeline transmission efficiency = Energy at end of the pipe Energy available at reservoir Figure 3.3 shows the total headline, where the water supply is from a reservoir at a head H 1 above the nozzle. The frictional head loss, h f , is the loss as the water flows through the pressure tunnel and penstock up to entry to the nozzle. Then the transmission efficiency is h trans = (H 1 2 h f )/H 1 = H/H 1 (3:5) The nozzle efficiency or jet efficiency is h j = Energy at nozzle outlet Energy at nozzle inlet = C 2 1 /2gH (3:6) Chapter 3 94 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Nozzle velocity coefficient C v = Actual jet velocity Theoretical jet velocity = C 1 = ffiffiffiffiffiffiffiffiffi 2gH _ Therefore the nozzle efficiency becomes h j = C 2 1 =2gH = C 2 v (3:7) The characteristics of an impulse turbine are shown in Fig. 3.4. Figure 3.4 shows the curves for constant head and indicates that the peak efficiency occurs at about the same speed ratio for any gate opening and that Figure 3.3 Schematic layout of hydro plant. Figure 3.4 Efficiency vs. speed at various nozzle settings. Hydraulic Turbines 95 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved the peak values of efficiency do not vary much. This happens as the nozzle velocity remaining constant in magnitude and direction as the flow rate changes, gives an optimum value of U/C 1 at a fixed speed. Due to losses, such as windage, mechanical, and friction cause the small variation. Fig. 3.5 shows the curves for power vs. speed. Fixed speed condition is important because generators are usually run at constant speed. Illustrative Example 3.1: A generator is to be driven by a Pelton wheel with a head of 220 m and discharge rate of 145 L/s. The mean peripheral velocity of wheel is 14 m/s. If the outlet tip angle of the bucket is 1608, find out the power developed. Solution: Dischargerate; Q = 145 L/s Head; H = 220 m U 1 = U 2 = 14 m/s b 2 = 180 2 1608 = 208 Refer to Fig. 3.6 Using Euler’s equation, work done per weight mass of water per sec. = (C w1 U 1 2 C w2 U 2 ) But for Pelton wheel C w2 is negative Figure 3.5 Power vs. speed of various nozzle setting. Chapter 3 96 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Therefore Work done / s = (C w1 U 1 - C w2 U 2 ) Nm / s From inlet velocity triangle C w1 = C 1 and C 2 1 2g = H Hence, C 1 = ffiffiffiffiffiffiffiffiffi 2gH _ = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 £ 9:81 £ 220 _ = 65:7 m/s Relative velocity at inlet is V 1 = C 1 2 U 1 = 65:7 2 14 = 51:7 m/s From outlet velocity triangle V 1 = V 2 = 51:7 m/s(neglecting friction) and cos b 2 = U 2 -C w2 V 2 or cos(20) = 14 - C w2 51:7 Therefore C w2 = 34:58 m/s Hence, work done per unit mass of water per sec. = (65:7)(14) - (34:58)(14) = 1403:92 Nm Power developed = (1403:92)(145) 1000 = 203:57 kW Figure 3.6 Inlet and outlet velocity triangles. Hydraulic Turbines 97 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Design Example 3.2: A Pelton wheel is supplied with 0.035 m 3 /s of water under a head of 92 m. The wheel rotates at 725 rpm and the velocity coefficient of the nozzle is 0.95. The efficiency of the wheel is 82% and the ratio of bucket speed to jet speed is 0.45. Determine the following: 1. Speed of the wheel 2. Wheel to jet diameter ratio 3. Dimensionless power specific speed of the wheel Solution: Overall efficiency h o = Power developed Power available [ P = rgQHh o J/s = rgQHh o 1000 kW = 9:81(0:035)(92)(0:82) = 25:9 kW Velocity coefficient C v = C 1 ffiffiffiffiffiffiffiffiffi 2gH _ or C 1 = C v ffiffiffiffiffiffiffiffiffi 2gH _ = 0:95[(2)(9:81)(92)] 1/2 = 40:36 m/s 1. Speed of the wheel is given by U = 0:45(40:36) = 18:16 m/s 2. If D is the wheel diameter, then U = vD 2 or D = 2U v = (2)(18:16)(60) 725(2p) = 0:478 m Jet area A = Q C 1 = 0:035 40:36 = 0:867 £ 10 23 m 2 and Jet diameter, d, is given by d = 4A p _ _ 1/2 = (4)(0:867 £ 10 23 ) p _ _ 1/2 = 0:033 m Diameter ratio D d = 0:478 0:033 = 14:48 Chapter 3 98 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 3. Dimensionless specific speed is given by Eq. (1.10) N sp = NP 1/2 r 1/2 (gH) 5/4 = 725 60 _ _ £ (25:9)(1000) 10 3 _ _ 1/2 £ 1 (9:81) £ (92) _ _ 5/4 = (12:08)(5:09)(0:0002) = 0:0123 rev = (0:0123)(2p) rad = 0:077 rad Illustrative Example 3.3: The speed of Pelton turbine is 14 m/s. The water is supplied at the rate of 820 L/s against a head of 45 m. If the jet is deflected by the buckets at an angle of 1608, find the hP and the efficiency of the turbine. Solution: Refer to Fig. 3.7 U 1 = U 2 = 14 m/s Q = 820 L/s = 0.82 m 3 /s H = 45 m b 2 = 180 2 1608 = 208 Velocity of jet C 1 = C v ffiffiffiffiffiffiffiffiffi 2gH _ , assuming C v = 0:98 = 0:98 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2)(9:81)(45) _ = 29:12 m/s Figure 3.7 Velocity triangle for Example 3.3. Hydraulic Turbines 99 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Assuming b 1 = 1808 b 2 = 180 2 1608 = 208 C w1 = C 1 = 29:12 m/s V 1 = C 1 2 U 1 = 29:12 2 14 = 15:12 m/s From outlet velocity triangle, U 1 = U 2 (neglecting losses on buckets) V 2 = 15:12 m/ s and U 2 = 14 m/s C w2 = V 2 cosa 2 2 U 2 = 15:12 cos 208 2 14 = 0:208 m/s Work done per weight mass of water per sec = (C w1 - C w2 )U = (29:12 - 0:208) £ (14) = 410:6 Nm/ s [ Power developed = (410:6)(0:82 £ 10 3 ) 1000 = 336:7 kW = 451 hP Efficiencyh 1 = Power developed Available Power = (1000)(336:7) (1000)(9:81)(0:82)(45) = 0:930 or 93:0% Illustrative Example 3.4: A Pelton wheel develops 12,900 kW at 425 rpm under a head of 505 m. The efficiency of the machine is 84%. Find (1) discharge of the turbine, (2) diameter of the wheel, and (3) diameter of the nozzle. Assume C v = 0.98, and ratio of bucket speed to jet speed = 0.46. Solution: Head, H = 505 m. Power, P = 12,900 kW Speed, N = 425 rpm Efficiency, h o = 84% 1. Let Q be the discharge of the turbine Using the relation h o = P 9:81QH Chapter 3 100 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved or 0:84 = 12; 900 (9:81)(505)Q = 2:60 Q or Q = 3:1 m 3 /s 2. Velocity of jet C = C v ffiffiffiffiffiffiffiffiffi 2gH _ (assume C v = 0:98) or C = 0:98 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2)(9:81)(505) _ = 97:55 m/s Tangential velocity of the wheel is given by U = 0:46C = (0:46)(97:55) = 44:87 m/s and U = pDN 60 ; hence wheel diameter is D = 60U pN = (60)(44:87) (p)(425) = 2:016 m 3. Let d be the diameter of the nozzle The discharge through the nozzle must be equal to the discharge of the turbine. Therefore Q = p 4 £ d 2 £ C 3:1 = ( p 4 )(d 2 )(97:55) = 76:65 d 2 [ d = ffiffiffiffiffiffiffiffiffiffiffi 3:1 76:65 _ = 0:20 m Illustrative Example 3.5: A double Overhung Pelton wheel unit is to operate at 12,000 kW generator. Find the power developed by each runner if the generator is 95%. Solution: Output power = 12,000 kW Efficiency, h = 95% Therefore, power generated by the runner = 12; 000 0:95 = 12; 632 kW Hydraulic Turbines 101 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Since there are two runners, power developed by each runner = 12; 632 2 = 6316 kW Design Example 3.6: At the power station, a Pelton wheel produces 1260 kW under a head of 610 m. The loss of head due to pipe friction between the reservoir and nozzle is 46 m. The buckets of the Pelton wheel deflect the jet through an angle of 1658, while relative velocity of the water is reduced by 10% due to bucket friction. The bucket/jet speed ratio is 0.46. The bucket circle diameter of the wheel is 890 mm and there are two jets. Find the theoretical hydraulic efficiency, speed of rotation of the wheel, and diameter of the nozzle if the actual hydraulic efficiency is 0.9 times that calculated above. Assume nozzle velocity coefficient, C v = 0.98. Solution: Refer to Fig. 3.8. Hydraulic efficiency h h = Power output Energy available in the jet = P 0:5mC 2 1 At entry to nozzle H = 610 2 46 = 564 m Using nozzle velocity coefficient C 1 = C v ffiffiffiffiffiffiffiffiffi 2gH _ = 0:98 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2)(9:81)(564) _ = 103:1 m/s Now W m = U 1 C w1 2 U 2 C w2 = U U - V 1 ( ) 2 U 2 V 2 cos 1808 2a ( ) [ ] ¦ ¦ = U C 1 2 U ( ) 1 2 kcos a ( ) [ ] where V 2 = kV 1 Therefore, W/m = 0.46C 1 (C 1 2 0.46C 1 )(1 2 0.9 cos 1658) Substitute the value of C 1 W/m = 5180:95 Theoretical hydraulic efficiency = Power output Energy available in the jet = 5180:95 0:5 £ 103 2 = 98% Chapter 3 102 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Actual hydraulic efficiency = (0:9)(0:98) = 0:882 Wheel bucket speed = (0:46)(103) = 47:38 m/s Wheel rotational speed = N = (47:38)(60) (0:445)(2p) = 1016 rpm Actual hydraulic efficiency = Actual power energy in the jet = (1260 £ 10 3 ) 0:5 mC 2 1 Therefore, m = (1260 £ 10 3 ) (0:882)(0:5)(103 2 ) = 269 kg/s For one nozzle, m = 134.5 kg/s For nozzle diameter, using continuity equation, m = rC 1 A = rC 1 pd 2 4 Hence, d = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (134:5)(4) (p)(103 £ 10 3 ) _ = 0:041 m = 41 mm Illustrative Example 3.7: A Pelton wheel has a head of 90 m and head lost due to friction in the penstock is 30 m. The main bucket speed is 12 m/s and the nozzle discharge is 1.0 m 3 /s. If the bucket has an angle of 158 at the outlet and C v = 0.98, find the power of Pelton wheel and hydraulic efficiency. Figure 3.8 Velocity triangle for Example 3.6. Hydraulic Turbines 103 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: (Fig. 3.9) Head = 90 m Head lost due to friction = 30 m Head available at the nozzle = 90 2 30 = 60 m Q = 1 m 3 /s From inlet diagram C 1 = C v ffiffiffiffiffiffiffiffiffi 2gH _ = 0:98 £ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2)(9:81)(60 _ ) = 33:62 m/s Therefore, V 1 = C 1 2 U 1 = 33.62 2 12 = 21.62 m/s From outlet velocity triangle V 2 = V 1 = 21:16 m/s (neglecting losses) U 2 = U 1 = 12 m/s C w2 = V 2 cos a 2 U 2 = 21:62 cos 158 2 12 = 8:88 m/s Figure 3.9 Velocity triangle for Example 3.7. Chapter 3 104 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved and Cr 2 = V 2 sin a = 21:62 sin 158 = 5:6 m/s Therefore, C 2 = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 2 w2 - Cr 2 2 _ = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (8:88) 2 - (5:6) 2 _ = 10:5 m/s [ Work done = C 2 1 2 C 2 2 2 = (33:62) 2 2 (10:5) 2 2 = 510 kJ/kg Note Work done can also be found by using Euler’s equation (C w1 U 1 - C w2 U 2 ) Power = 510 kW Hydraulic efficiency h h = work done kinetic energy = (510)(2) (33:62) 2 = 90:24% Design Example 3.8: A single jet Pelton wheel turbine runs at 305 rpm against a head of 515 m. The jet diameter is 200 mm, its deflection inside the bucket is 1658 and its relative velocity is reduced by 12% due to friction. Find (1) the waterpower, (2) resultant force on the bucket, (3) shaft power if the mechanical losses are 4% of power supplied, and (4) overall efficiency. Assume necessary data. Solution: (Fig. 3.10) Velocity of jet, C 1 = C v ffiffiffiffiffiffiffiffiffi 2gH _ = 0:98 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2)(9:81)(515 _ ) = 98:5 m/s Discharge, Q is given by Q = Area of jet £ Velocity = p 4 £ (0:2) 2 (98:5) = 3:096 m 3 /s 1. Water power is given by P = rgQH = (9:81)(3:096)(515) = 15641:5 kW 2. Bucket velocity, U 1 , is given by U 1 = C v ffiffiffiffiffiffiffiffiffi 2gH _ = 0:46 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2)(9:81)(515) _ = 46 m/s (assuming C v = 0:46) Relative velocity, V 1 , at inlet is given by V 1 = C 1 2 U 1 = 98:5 2 46 = 52:5 m/s Hydraulic Turbines 105 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved and V 2 = 0:88 £ 52:5 = 46:2 m/s From the velocity diagram C w2 = U 2 2 V 2 cos 15 = 46 2 46:2 £ 0:966 = 1:37 m/s Therefore force on the bucket = rQ(C w1 2 C w2 ) = 1000 £ 3:096(98:5 2 1:37) = 300714 N 3. Power produced by the Pelton wheel = (300714)(46) 1000 = 13832:8 kW Taking mechanical loss = 4% Therefore, shaft power produced = 0.96 £ 13832.8 = 13279.5 kW 4. Overall efficiency h o = 13279:5 15641:5 = 0:849 or 84:9% Figure 3.10 Velocity triangles for Example 3.8. Chapter 3 106 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 3.5 REACTION TURBINE The radial flow or Francis turbine is a reaction machine. In a reaction turbine, the runner is enclosed in a casing and therefore, the water is always at a pressure other than atmosphere. As the water flows over the curved blades, the pressure head is transformed into velocity head. Thus, water leaving the blade has a large relative velocity but small absolute velocity. Therefore, most of the initial energy of water is given to the runner. In reaction turbines, water leaves the runner at atmospheric pressure. The pressure difference between entrance and exit points of the runner is known as reaction pressure. The essential difference between the reaction rotor and impulse rotor is that in the former, the water, under a high station head, has its pressure energy converted into kinetic energy in a nozzle. Therefore, part of the work done by the fluid on the rotor is due to reaction from the pressure drop, and part is due to a change in kinetic energy, which represents an impulse function. Fig. 3.11 shows a cross-section through a Francis turbine and Fig. 3.12 shows an energy distribution through a hydraulic reaction turbine. In reaction turbine, water from the reservoir enters the turbine casing through penstocks. Hence, the total head is equal to pressure head plus velocity head. Thus, the water enters the runner or passes through the stationary vanes, which are fixed around the periphery of runners. The water then passes immediately into the rotor where it moves radially through the rotor vanes and exits from the rotor blades at a smaller diameter, after which it turns through 908 into the draft tube. The draft tube is a gradually increasing cross-sectional area passage. It helps in increasing the work done by the turbine by reducing pressure at the exit. The penstock is a waterway, which carries water from the reservoir to the turbine casing. The inlet and outlet velocity triangles for the runner are shown in Fig. 3.13. Figure 3.11 Outlines of a Francis turbine. Hydraulic Turbines 107 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 3.12 Reaction turbine installation. Chapter 3 108 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 3.13 (a) Francis turbine runner and (b) velocity triangles for inward flow reaction turbine. Hydraulic Turbines 109 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Let C 1 = Absolute velocity of water at inlet D 1 = Outer diameter of the runner N = Revolution of the wheel per minute U 1 = Tangential velocity of wheel at inlet V 1 = Relative velocity at inlet C r1 = radial velocity at inlet a 1 = Angle with absolute velocity to the direction of motion b 1 = Angle with relative velocity to the direction of motion H = Total head of water under which turbine is working C 2 ; D 2 ; U 2 ; V 2 ; C r2 = Corresponding values at outlet Euler’s turbine equation Eq. (1.78) and E is maximum when C w2 (whirl velocity at outlet) is zero that is when the absolute and flow velocities are equal at the outlet. 3.6 TURBINE LOSSES Let P s = Shaft power output P m = Mechanical power loss P r = Runner power loss P c = Casing and draft tube loss P l = Leakage loss P = Water power available P h = P r - P c - P l = Hydraulic power loss Runner power loss is due to friction, shock at impeller entry, and flow separation. If h f is the head loss associated with a flow rate through the runner of Q r , then P s = rgQ r h f (Nm/s) (3:8) Leakage power loss is due to leakage in flow rate, q, past the runner and therefore not being handled by the runner. Thus Q = Q r - q (3:9) Chapter 3 110 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved If H r is the head across the runner, the leakage power loss becomes P l = rgH r q (Nm / s) (3:10) Casing power loss, P c , is due to friction, eddy, and flow separation losses in the casing and draft tube. If h c is the head loss in casing then P c = rgQh c (Nm / s) (3:11) From total energy balance we have rgQH = P m -r g (h f Q r - h c Q - H r q - P s ) Then overall efficiency, h o , is given by h o = Shaft power output Fluid power available at inlet or h o = P s rgQH (3:12) Hydraulic efficiency, h h , is given by h h = Power available at runner Fluid power available at inlet or h h = (P s - P m ) rgQH (3:13) Eq. (3.13) is the theoretical energy transfer per unit weight of fluid. Therefore the maximum efficiency is h h = U 1 C w1 /gH (3:14) 3.7 TURBINE CHARACTERISTICS Part and overload characteristics of Francis turbines for specific speeds of 225 and 360 rpm are shown in Fig. 3.14 Figure 3.14 shows that machines of low specific speeds have a slightly higher efficiency. It has been experienced that the Francis turbine has unstable characteristics for gate openings between 30 to 60%, causing pulsations in output and pressure surge in penstocks. Both these problems were solved by Paul Deriaz by designing a runner similar to Francis runner but with adjustable blades. The part-load performance of the various types are compared in Fig. 3.15 showing that the Kaplan and Pelton types are best adopted for a wide range of load but are followed fairly closely by Francis turbines of low specific speed. Hydraulic Turbines 111 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 3.14 Variation of efficiency with load for Francis turbines. Figure 3.15 Comparison of part-load efficiencies of various types of hydraulic turbine. Chapter 3 112 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 3.8 AXIAL FLOW TURBINE In an axial flow reaction turbine, also known as Kaplan turbine, the flow of water is parallel to the shaft. A Kaplan turbine is used where a large quantity of water is available at low heads and hence the blades must be long and have large chords so that they are strong enough to transmit the very high torque that arises. Fig. 3.16 and 3.17 shows the outlines of the Kaplan turbine. The water from the scroll flows over the guide blades and then over the vanes. The inlet guide vanes are fixed and are situated at a plane higher than the runner blades such that fluid must turn through 908 to enter the runner in the axial direction. The function of the guide vanes is to impart whirl to the fluid so that the radial distribution of velocity is the same as in a free vortex. Fig. 3.18 shows the velocity triangles and are usually drawn at the mean radius, since conditions change from hub to tip. The flow velocity is axial at inlet and outlet, hence C r1 = C r2 = C a C 1 is the absolute velocity vector at anglea 1 toU 1 , andV 1 is the relative velocity at an angle b 1 . For maximum efficiency, the whirl component C w2 = 0, in which case the absolute velocity at exit is axial and then C 2 = C r2 Using Euler’s equation E = U(C w1 2 C w2 )/g and for zero whirl (C w2 = 0) at exit E = UC w1 /g 3.9 CAVITATION In the design of hydraulic turbine, cavitation is an important factor. As the outlet velocity V 2 increases, then p 2 decreases and has its lowest value when the vapor pressure is reached. At this pressure, cavitation begins. The Thoma parameter s = NPSH H and Fig. 3.19 give the permissible value of s c in terms of specific speed. The turbines of high specific speed have a high critical value of s, and must therefore be set lower than those of smaller specific speed (N s ). Illustrative Example 3.9: Consider an inward flow reaction turbine in which velocity of flow at inlet is 3.8 m/s. The 1 m diameter wheel rotates at 240 rpm and absolute velocity makes an angle of 168 with wheel tangent. Determine (1) velocity of whirl at inlet, (2) absolute velocity of water at inlet, (3) vane angle at inlet, and (4) relative velocity of water at entrance. Hydraulic Turbines 113 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 3.16 Kaplan turbine of water is available at low heads. Chapter 3 114 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: From Fig. 3.13b 1. From inlet velocity triangle (subscript 1) tana 1 = C r1 C w1 or C w1 = C r1 tana 1 = 3:8 tan168 = 13:3 m/s 2. Absolute velocity of water at inlet, C 1 , is sina 1 = C r1 C 1 or C 1 = C r1 sina 1 = 3:8 sin168 = 13:79 m/s 3. U 1 = (pD 1 )(N) 60 = (p)(1)(240) 60 = 12:57 m/s Figure 3.17 Kaplan turbine runner. Hydraulic Turbines 115 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved and tan b 1 = C r1 (C w1 2 U 1 ) = 3:8 (13:3 2 12:57) = 3:8 0:73 = 5:21 [ b 1 = 798 nearby 4. Relative velocity of water at entrance sin b 1 = C r1 V 1 or V 1 = C r1 sin b 1 = 3:8 sin 798 = 3:87m/s Illustrative Example 3.10: The runner of an axial flow turbine has mean diameter of 1.5 m, and works under the head of 35 m. The guide blades make an angle of 308 with direction of motion and outlet blade angle is 228. Assuming axial discharge, calculate the speed and hydraulic efficiency of the turbine. Figure 3.18 Velocity triangles for an axial flow hydraulic turbine. Chapter 3 116 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 3.19 Cavitation limits for reaction turbines. Figure 3.20 Velocity triangles (a) inlet and (b) outlet. Hydraulic Turbines 117 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: Since this is an impulse turbine, assume coefficient of velocity = 0.98 Therefore the absolute velocity at inlet is C 1 = 0:98 ffiffiffiffiffiffiffiffiffi 2gH _ = 0:98 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2)(9:81)(35 _ ) = 25:68 m/s The velocity of whirl at inlet C w1 = C 1 cos a 1 = 25:68 cos 308 = 22:24 m/s Since U 1 = U 2 = U Using outlet velocity triangle C 2 = U 2 tan b 2 = U tan b 2 = U tan 228 Hydraulic efficiency of turbine (neglecting losses) h h = C w1 U 1 gH = H 2 C 2 2 /2g H 22:24U g = H 2 (U tan 228) 2 2g or 22:24U g - (U tan 22) 2 2g = H or 22:24U - 0:082U 2 2 9:81H = 0 or 0:082U 2 - 22:24U 2 9:81H = 0 or U = 222:24 ^ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (22:24) 2 - (4)(0:082)(9:81)(35) _ (2)(0:082) As U is positive, U = 222:24 - ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 494:62 - 112:62 _ 0:164 = 222:24 - 24:64 0:164 = 14:63 m/s Now using relation U = pDN 60 Chapter 3 118 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved or N = 60U pD = (60)(14:63) (p)(1:5) = 186 rpm Hydraulic efficiency h h = C w1 U gH = (22:24)(14:63) (9:81)(35) = 0:948 or 94:8% Illustrative Example 3.11: A Kaplan runner develops 9000 kW under a head of 5.5 m. Assume a speed ratio of 2.08, flow ratio 0.68, and mechanical efficiency 85%. The hub diameter is 1/3 the diameter of runner. Find the diameter of the runner, and its speed and specific speed. Solution: U 1 = 2:08 ffiffiffiffiffiffiffiffiffi 2gH _ = 2:08 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2)(9:81)(5:5) _ = 21:61 m/s C r1 = 0:68 ffiffiffiffiffiffiffiffiffi 2gH _ = 0:68 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2)(9:81)(5:5) _ = 7:06 m/s Now power is given by 9000 = (9:81)(5:5)(0:85)Q Therefore, Q = 196:24 m 3 /s If D is the runner diameter and, d, the hub diameter Q = p 4 (D 2 2 d 2 )C r1 or p 4 D 2 2 1 9 D 2 _ _ 7:06 = 196:24 Solving D = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (196:24)(4)(9) (p)(7:06)(8) _ = 6:31 m N s = N ffiffiffi P _ H 5/4 = 65 ffiffiffiffiffiffiffiffiffiffi 9000 _ 5:5 5/4 = 732 rpm Design Example 3.12: A propeller turbine develops 12,000 hp, and rotates at 145 rpm under a head of 20 m. The outer and hub diameters are 4 m and 1.75 m, Hydraulic Turbines 119 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved respectively. Calculate the inlet and outlet blade angles measured at mean radius if overall and hydraulic efficiencies are 85% and 93%, respectively. Solution: Mean diameter = 4 - 1:75 2 = 2:875 m U 1 = pDN 60 = (p)(2:875)(145) 60 = 21:84 m/s Using hydraulic efficiency h h = C w1 U 1 gH = (C w1 )(21:84) (9:81)(20) = 0:93C w1 or C w1 = 8:35 m/s Power = (12; 000)(0:746) = 8952 kW Power = rgQHh o or 8952 = 9:81 £ Q £ 20 £ 0:85 Therefore, Q = 8952 (9:81)(20)(0:85) = 53:68 m 3 /s Discharge, Q = 53:68 = p 4 (4 2 2 1:75 2 )C r1 [ C r1 = 5:28 m/s tan b 1 = C r1 U 1 2 C w1 = 5:28 21:84 2 8:35 = 5:28 13:49 = 0:3914 b 1 = 21:388 and tan b 2 = C r2 U 2 = 5:28 21:84 = 0:2418 b 2 = 13:598 Illustrative Example 3.13: An inward flow reaction turbine wheel has outer and inner diameter are 1.4 m and 0.7 m respectively. The wheel has radial vanes and discharge is radial at outlet and the water enters the vanes at an angle of 128. Assuming velocity of flow to be constant, and equal to 2.8 m/s, find Chapter 3 120 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 1. The speed of the wheel, and 2. The vane angle at outlet. Solution: Outer diameter, D 2 = 1.4 m Inner diameter, D 1 = 0.7 m Angle at which the water enters the vanes, a 1 = 128 Velocity of flow at inlet, C r1 = C r2 = 2:8 m/s As the vanes are radial at inlet and outlet end, the velocity of whirl at inlet and outlet will be zero, as shown in Fig. 3.21. Tangential velocity of wheel at inlet, U 1 = C r1 tan 128 = 2:8 0:213 = 13:15 m/s Also, U 1 = pD 2 N 60 or N = 60U 1 pD 2 = (60)(13:15) (p)(1:4) = 179 rpm Figure 3.21 Velocity triangles at inlet and outlet for Example 3.13. Hydraulic Turbines 121 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Let b 2 is the vane angle at outlet U 2 = pD 1 N 60 = (p)(0:7)(179) 60 = 6:56 m/s From Outlet triangle, tan b 2 = C r2 U 2 = 2:8 6:56 = 0:4268 i:e: b 2 = 23:118 Illustrative Example 3.14: Consider an inward flow reaction turbine in which water is supplied at the rate of 500 L/s with a velocity of flow of 1.5 m/s. The velocity periphery at inlet is 20 m/s and velocity of whirl at inlet is 15 m/s. Assuming radial discharge, and velocity of flow to be constant, find 1. Vane angle at inlet, and 2. Head of water on the wheel. Solution: Discharge, Q = 500 L/s = 0.5 m 3 /s Velocity of flow at inlet, C r1 = 1.5 m/s Velocity of periphery at inlet, U 1 = 20 m/s Velocity of whirl at inlet, C w1 = 15 m/s As the velocity of flow is constant, C r1 = C r2 = 1.5 m/s Let b 1 = vane angle at inlet From inlet velocity triangle tan (180 2b 1 ) = C r1 U 1 2 C w1 = 1:5 20 2 15 = 0:3 [ (180 2b 1 ) = 16841 / or b 1 = 1808 2 16841 / = 163819 / Since the discharge is radial at outlet, ad so the velocity of whirl at outlet is zero Therefore, C w1 U 1 g = H 2 C 2 1 2g = H 2 C 2 r1 2g Chapter 3 122 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved or (15)(20) 9:81 = H 2 1:5 2 (2)(9:81) [ H = 30:58 2 0:1147 = 30:47 m DesignExample 3.15: Inner and outer diameters of anoutwardflowreaction turbine wheel are 1 m and 2 m respectively. The water enters the vane at angle of 208 and leaves the vane radially. Assuming the velocity of flowremains constant at 12 m/s and wheel rotates at 290 rpm, find the vane angles at inlet and outlet. Solution: Inner diameter of wheel, D 1 = 1 m Outer diameter of wheel, D 2 = 2 m a 1 = 208 Velocity of flow is constant That is, C r1 = C r2 = 12 m/s Speed of wheel, N = 290 rpm Vane angle at inlet = b 1 U 1 is the velocity of periphery at inlet. Therefore, U 1 = pD 1 N 60 = (p)(1)(290) 60 = 15:19 m/s From inlet triangle, velocity of whirl is given by C w1 = 12 tan 20 = 12 0:364 = 32:97 m/s Hence, tan b 1 = C r1 C w1 2 U 1 = 12 32:97 2 15:19 = 12 17:78 = 0:675 i.e. b 1 = 348 Let b 2 = vane angle at outlet U 2 = velocity of periphery at outlet Therefore U 2 = pD 2 N 60 = (p)(2)(290) 60 = 30:38 m/s From the outlet triangle tan b 2 = C r2 U 2 = 12 30:38 = 0:395 Hydraulic Turbines 123 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved i.e., b 2 = 21833 / Illustrative Example 3.16: An inward flow turbine is supplied with 245 L of water per second and works under a total head of 30 m. The velocity of wheel periphery at inlet is 16 m/s. The outlet pipe of the turbine is 28 cm in diameter. The radial velocity is constant. Neglecting friction, calculate 1. The vane angle at inlet 2. The guide blade angle 3. Power. Solution: If D 1 is the diameter of pipe, then discharge is Q = p 4 D 2 1 C 2 or C 2 = (4)(0:245) (p)(0:28) 2 = 3:98 m/s But C 2 = C r1 = C r2 Neglecting losses, we have C w1 U 1 gH = H 2 C 2 2 /2g H or C w1 U 1 = gH 2 C 2 2 /2 = [(9:81)(30)] 2 (3:98) 2 2 = 294:3 2 7:92 = 286:38 Power developed P = (286:38)(0:245) kW = 70:16 kW and C w1 = 286:38 16 = 17:9 m/s tan a 1 = 3:98 17:9 = 0:222 i.e. a 1 = 12831 / tan b 1 = C r1 C w1 2 U 1 = 3:98 17:9 2 16 = 3:98 1:9 = 2:095 i.e. b 1 = 64.43 or b 1 = 64825 / Chapter 3 124 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Design Example 3.17: A reaction turbine is to be selected from the following data: Discharge = 7:8 m 3 /s Shaft power = 12; 400 kW Pressure head in scroll casing at the entrance to turbine = 164 m of water Elevation of turbine casing above tail water level = 5:4 m Diameter of turbine casing = 1 m Velocity in tail race = 1:6 m/s Calculate the effective head on the turbine and the overall efficiency of the unit. Solution: Velocity in casing at inlet to turbine C c = Discharge Cross 2 sectional area of casing = 7:8 (p/4)(1) 2 = 9:93 m/s The net head on turbine = Pressure head - Head due to turbine position - C 2 c 2 C 2 1 2g = 164 - 5:4 - (9:93) 2 2 (1:6) 2 2g = 164 - 5:4 - 98:6 2 2:56 19:62 = 174:3 m of water Waterpower supplied to turbine = QgH kW = (7:8)(9:81)(174:3) = 13; 337 kW Hence overall efficiency, h o = Shaft Power Water Power = 12; 400 13; 337 = 0:93 or 93% Design Example 3.18: A Francis turbine wheel rotates at 1250 rpm and net head across the turbine is 125 m. The volume flow rate is 0.45 m 3 /s, radius of the runner is 0.5 m. The height of the runner vanes at inlet is 0.035 m. and the angle of Hydraulic Turbines 125 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved the inlet guide vanes is set at 708 from the radial direction. Assume that the absolute flow velocity is radial at exit, find the torque and power exerted by the water. Also calculate the hydraulic efficiency. Solution: For torque, using angular momentum equation T = m(C w2 r 2 2 C w1 r 1 ) As the flow is radial at outlet, C w2 = 0 and therefore T = 2mC w1 r 1 = 2rQC w1 r 1 = 2(10 3 )(0:45)(0:5C w1 ) = 2225C w1 Nm If h 1 is the inlet runner height, then inlet area, A, is A = 2pr 1 h 1 = (2)(p)(0:5)(0:035) = 0:11m 2 C r1 = Q/A = 0:45 0:11 = 4:1 m/s From velocity triangle, velocity of whirl C w1 = C r1 tan708 = (4:1)(2:75) = 11:26m/s Substituting C w1 , torque is given by T = 2(225)(11:26) = 22534 Nm Negative sign indicates that torque is exerted on the fluid. The torque exerted by the fluid is -2534 Nm Power exerted P = Tv = (2534)(2)(p)(1250) (60)(1000) = 331:83 kW Chapter 3 126 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Hydraulic efficiency is given by h h = Power exerted Power available = (331:83)(10 3 ) rgQH = 331:83 £ 10 3 (10 3 )(9:81)(0:45)(125) = 0:6013 = 60:13% Design Example 3.19: An inward radial flow turbine develops 130 kW under a head of 5 m. The flowvelocity is 4 m/s and the runner tangential velocity at inlet is 9.6 m/s. The runner rotates at 230 rpmwhile hydraulic losses accounting for 20% of the energy available. Calculate the inlet guide vane exit angle, the inlet angle to the runner vane, the runner diameter at the inlet, and the height of the runner at inlet. Assume radial discharge, and overall efficiency equal to 72%. Solution: Hydraulic efficiency is h h = Power deleloped Power available = m(C w1 U 1 2 C w2 U) rgQH Since flow is radial at outlet, then C w2 = 0 and m = rQ, therefore h h = C w1 U 1 gH 0:80 = (C w1 )(9:6) (9:81)(5) C w1 = (0:80)(9:81)(5) 9:6 = 4:09 m/s Radial velocity C r1 = 4 m/s tan a 1 = C r1 /C w1 (from velocity triangle) = 4 4:09 = 0:978 Hydraulic Turbines 127 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved i.e., inlet guide vane angle a 1 = 44821 / tan b 1 = C r1 C w1 2 U 1 ( ) = 4 4:09 2 9:6 ( ) = 4 25:51 = 20:726 i.e., b 1 = 235.988 or 1808 2 35.98 = 144.028 Runner speed is U 1 = pD 1 N 60 or D 1 = 60U 1 pN = (60)(9:6) (p)(230) D 1 = 0:797 m Overall efficiency h o = Power output Power available or rgQH = (130)(10 3 ) 0:72 or Q = (130)(10 3 ) (0:72)(10 3 )(9:81)(5) = 3:68 m 3 /s But Q = pD 1 h 1 C r1 (where h 1 is the height of runner) Therefore, h 1 = 3:68 (p)(0:797)(4) = 0:367 m Illustrative Example 3.20: The blade tip and hub diameters of an axial hydraulic turbine are 4.50 m and 2 m respectively. The turbine has a net head of 22 m across it and develops 22 MW at a speed of 150 rpm. If the hydraulic efficiency is 92% and the overall efficiency 84%, calculate the inlet and outlet blade angles at the mean radius assuming axial flow at outlet. Chapter 3 128 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: Mean diameter, D m , is given by D m = D h - D t 2 = 2 - 4:50 2 = 3:25 m Overall efficiency, h o , is given by h o = Power develpoed Power available [ Power available = 22 0:84 = 26:2 MW Also, available power = rgQH (26:2)(10 6 ) = (10 3 )(9:81)(22)Q Hence flow rate, Q, is given by Q = (26:2)(10 6 ) (10 3 )(9:81)(22) = 121:4 m 3 /s Now rotor speed at mean diameter U m = pD m N 60 = (p)(3:25)(150) 60 = 25:54 m/s Power given to runner = Power available £ h h = 26:2 £ 10 6 £ 0:92 = 24:104 MW Theoretical power given to runner can be found by using P = rQU m C w1 (C w2 = 0) (24:104)(10 6 ) = (10 3 )(121:4)(25:54)(C w1 ) [ C w1 = (24:104)(10 6 ) (10 3 )(121:4)(25:54) = 7:77 m/s Axial velocity is given by C r = Q £ 4 p(D 2 t 2 D 2 h ) = (121:4)(4) p(4:50 2 2 2 2 ) = 9:51 m/s Using velocity triangle tan (180 2b 1 ) = C r U m 2 C w1 = 9:51 25:54 2 7:77 Hydraulic Turbines 129 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Inlet angle, b 1 = 151:858 At outlet tan b 2 = C r V cw 2 But V cw2 equals to U m since C w2 is zero. Hence tan b 2 = 9:51 25:54 = 0:3724 that is, b 2 = 20:438 Design Example 3.21: The following design data apply to an inward flow radial turbine: Overall efficiency 75% Net head across the turbine 6 m Power output 128 kW The runner tangential velocity 10:6 m/s Flow velocity 4 m/s Runner rotational speed 235 rpm Hydraulic losses 18% of energy available Calculate the inlet guide vane angle, the inlet angle of the runner vane, the runner diameter at inlet, and height of the runner at inlet. Assume that the discharge is radial. Solution: Hydraulic efficiency, h h , is given by h h = Power given to runner Water Power available = m U 1 C w1 2 U 2 C w2 ( ) rgQH Since flow is radial at exit, C w2 = 0 and m = rQ. Therefore h h = U 1 C w1 gH 0:82 = (10:6)(C w1 ) (9:81)(6) or C w1 = 4:6 m/s Chapter 3 130 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Now tan a 1 = C r1 /C w1 = 4 4:6 = 0:8695 that is; a 1 = 418 From Figs. 3.22 and 3.23 tan (180 2b 1 ) = C r1 U 1 2 C w1 = 4 10:6 2 4:6 = 0:667 that is; b 1 = 33:698 Hence blade angle, b 1 , is given by 1808 2 33:698 = 146:318 Runner speed at inlet U 1 = pD 1 N 60 Figure 3.22 Velocity triangles for Example 3.14. Hydraulic Turbines 131 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 3.23 Velocity triangles at inlet and outlet for Example 3.15. Figure 3.24 Inlet velocity triangle. Chapter 3 132 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved or D 1 = U 1 (60) pN = (10:6)(60) (p)(235) = 0:86 m Overall efficiency h o = Power output Power available rgQH = (128)(10 3 ) 0:75 From which flow rate Q = (128)(10 3 ) (0:75)(10 3 )(9:81)(6) = 2:9 m 3 /s Also, Q = pD 1 hC r1 where h 1 is the height of runner Therefore, h 1 = 2:9 (p)(0:86)(4) = 0:268 m Design Example 3.22: A Kaplan turbine develops 10,000 kW under an effective head 8 m. The overall efficiency is 0.86, the speed ratio 2.0, and flow ratio 0.60. The hub diameter of the wheel is 0.35 times the outside diameter of the wheel. Find the diameter and speed of the turbine. Solution: Head, H = 8 m, Power, P = 10,000 kW Overall efficiency, h o = 0.86 Speed ratio 2 = U 1 (2gH) 1=2 ; or U l = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 £ 9:81 £ 8 _ = 25:06 m/s Flow ratio C r1 (2gH) 1=2 = 0:60 or C r1 = 0:60 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 £ 9:81 £ 8 _ = 7:52 m/s Hub diameter, D 1 = 0.35 D 2 Hydraulic Turbines 133 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Overall efficiency, h o = P rgQH Or 0:86 = 10000 1000 £ 9:81 £ Q £ 8 [ Q = 148:16 m 3 /s Now using the relation Q = C r1 £ p 4 D 2 1 2 D 2 2 _ ¸ Or 148:16 = 7:52 £ p 4 D 2 1 2 0:35D 2 1 _ _ _ ¸ [ D 1 = 5:35 m The peripheral velocity of the turbine at inlet 25:06 = pD 1 N 60 = p £ 5:35 £ N 60 [ N = 60 £ 25:06 p £ 5:35 = 89 rpm Design Example 3.23: An inward flow reaction turbine, having an inner and outer diameter of 0.45 m and 0.90 m, respectively. The vanes are radial at inlet and the discharge is radial at outlet and the water enters the vanes at an angle of 128. Assuming the velocity of flow as constant and equal to 2.8 m/s, find the speed of the wheel and the vane angle at outlet. Solution: Inner Diameter, D 2 = 0.45 m Outer Diameter, D 1 = 0.9 m a 2 = 908(radial discharge) a 1 = 128; C r1 = C r2 = 2:8 m/s Chapter 3 134 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved From velocity triangle at inlet (see Fig. 3.11), The peripheral velocity of the wheel at inlet U 1 = C r1 tan a 1 = 2:8 tan 128 = 13:173 m/s Now, U 1 = pD 1 N 60 or N = 60U 1 pD 1 = 60 £ 13:173 p £ 0:9 = 279 rpm Considering velocity triangle at outlet peripheral velocity at outlet U 2 = pD 2 N 60 = p £ 0:45 £ 279 60 = 6:58 m/s tan b 2 = C r2 U 2 = 2:8 6:58 = 0:426 [ b 2 = 23:058 Design Example 3.24: An inward flow reaction turbine develops 70 kW at 370 rpm. The inner and outer diameters of the wheel are 40 and 80 cm, respectively. The velocity of the water at exit is 2.8 m/s. Assuming that the discharge is radial and that the width of the wheel is constant, find the actual and theoretical hydraulic efficiencies of the turbine and the inlet angles of the guide and wheel vanes. Turbine discharges 545 L/s under a head of 14 m. Solution: Q = 545 L/s = 0.545 m 3 /s D 1 = 80 cm, D 2 = 40 cm H = 14 m, a 2 = 908 (radial discharge) b 1 = b 2 Peripheral velocity of the wheel at inlet U 1 = pD 1 N 60 = p £ 0:80 £ 370 60 = 15:5 m/s Velocity of flow at the exit, C r2 = 2.8 m/s As a 2 = 908, C r2 = C 2 Work done/s by the turbine per kg of water = Cw£U 1 g Hydraulic Turbines 135 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved But this is equal to the head utilized by the turbine, i.e. C w1 U 1 g = H 2 C 2 2g (Assuming there is no loss of pressure at outlet) or C w1 £ 15:5 9:81 = 14 2 (2:8) 2 2 £ 9:81 = 13:6 m or C w1 = 13:6 £ 9:81 15:5 = 8:6 m/s Work done per second by turbine = rQ g C w1 U 1 = 1000 £ 0:545 £ 8:6 £ 15:5 1000 = 72:65kW Available power or water power = rgQH 1000 = 74:85 Actual available power = 70 kW Overall turbine efficiency ish t = 70 74:85 £ 100 h t = 93:52% This is the actual hydraulic efficiency as required in the problem. Hydraulic Efficiency is h h = 72:65 75:85 £ 100 = 97:06% This is the theoretical efficiency Q = pD 1 b 1 C r1 = pD 2 b 2 C r2 (Neglecting blade thickness) C r1 = C r2 D 2 D 1 = 2:8 £ 40 20 = 1:4 m/s Drawing inlet velocity triangle tan b 1 = C r1 U 1 2 C w1 = 1:4 15:5 2 8:6 = 1:4 6:9 = 0:203 Chapter 3 136 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved i.e., b 1 = 11.478 C 1 = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C w1 - C r1 _ = 8:6 2 - 1:4 2 _ _ 0:5 = 8:64 m/s and cosa 1 = C w1 C 1 = 8:6 8:64 = 0:995 i.e., a 1 = 5.58 Design Example 3.25: An inward flow Francis turbine, having an overall efficiency of 86%, hydraulic efficiency of 90%, and radial velocity of flow at inlet 0.28 ffiffiffiffiffiffiffiffiffi 2gH _ . The turbine is required to develop 5000 kW when operating under a net head of 30 m, specific speed is 270, assume guide vane angle 308, find 1. rpm of the wheel, 2. the diameter and the width of the runner at inlet, and 3. the theoretical inlet angle of the runner vanes. Solution: Power, P = 5000 kW; a 1 = 308; H = 30 m; C r1 = 0:28 ffiffiffiffiffiffiffiffiffi 2gH _ , N s = 270, h h = 0.90, h o = 0.86 1. Specific speed of the turbine is N s = N ffiffiffi P _ H 5=4 or N = N s H 5=4 ffiffiffi P _ = 270 £ (30) 1:25 ffiffiffiffiffiffiffiffiffiffi 5000 _ = 18957 71 = 267 rpm 2. Velocity of Flow: C r1 = 0:28 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 £ 9:81 £ 30 _ = 6:79m/s From inlet velocity triangle C r1 = C 1 sin a 1 or 6:79 = C 1 sin 308 or C 1 = 6:79 0:5 = 13:58 m/s C w1 = C 1 cos308 = 13.58 £ 0.866 = 11.76 m/s Hydraulic Turbines 137 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Work done per (sec) (kg) of water = C w1 £ U 1 g = h h £ H = 0:9 £ 30 = 27 mkg/s Peripheral Velocity, U 1 = 27 £ 9:81 11:76 = 22:5 m/s But U 1 = pD 1 N 60 or D 1 = 60U 1 pN = 60 £ 22:5 p £ 267 = 1:61 m Power, P = rgQHh o or 5000 = 1000 £ 9.81 £ Q £ 30 £ 0.86 or Q = 19.8 m 3 /s Also Q = kpD 1 b 1 C r1 (where k is the blade thickness coefficient and b 1 is the breath of the wheel at inlet) or b 1 = Q kpD 1 C r1 = 19:8 0:95 £ p £ 1:61 £ 6:79 = 0:61 m 3. From inlet velocity triangle tan b 1 = C r1 U 1 2 C w1 = 6:79 22:5 2 11:76 = 6:79 10:74 = 0:632 i.e. b 1 = 32.308 Design Example 3.26: A 35 MW generator is to operate by a double overhung Pelton wheel. The effective head is 350 m at the base of the nozzle. Find the size of jet, mean diameter of the runner and specific speed of wheel. Assume Pelton wheel efficiency 84%, velocity coefficient of nozzle 0.96, jet ratio 12, and speed ratio 0.45. Solution: In this case, the generator is fed by two Pelton turbines. Chapter 3 138 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Power developed by each turbine, P T = 35; 000 2 = 17; 500 kW Using Pelton wheel efficiency in order to find available power of each turbine P = 17; 500 0:84 = 20; 833 kW But, P = rgQH Q = P rgH = 20833 1000 £ 9:81 £ 350 = 6:07 m 3 /s Velocity of jet,C j = C v ffiffiffiffiffiffiffiffiffi 2gH _ = 0:96 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 £ 9:81 £ 350 _ C j = 79:6 m/s Area of jet, A = Q C j = 6:07 79:6 = 0:0763 m 2 [ Diameter of jet,d = 4A p _ _ 0:5 = 4 £ 0:0763 p _ _ 0:5 = 0:312 m d = 31:2 cm Diameter of wheel D = d £ jet ratio = 0.312 £ 12 = 3.744 m Peripheral velocity of the wheel U = speed ratio ffiffiffiffiffiffiffiffiffi 2gH _ = 0:45 £ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 £ 9:81 £ 350 _ = 37:29 m/s But U = pDN 60 or N = 60U pD = 60 £ 37:29 p £ 3:744 = 190 rpm Specific speed, N s = N ffiffiffiffiffiffi P T _ H 5=4 = 190 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 17; 500 _ (350) 1:25 = 16:6 PROBLEMS 3.1 A Pelton wheel produces 4600 hP under a head of 95 m, and with an overall efficiency of 84%. Find the diameter of the nozzle if the coefficient of velocity for the nozzle is 0.98. (0.36 m) Hydraulic Turbines 139 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 3.2 Pelton wheel develops 13,500 kW under a head of 500 m. The wheel rotates at 430 rpm. Find the size of the jet and the specific speed. Assume 85% efficiency. (0.21 m, 21) 3.3 A Pelton wheel develops 2800 bhP under a head of 300 m at 84% efficiency. The ratio of peripheral velocity of wheel to jet velocity is 0.45 and specific speed is 17. Assume any necessary data and find the jet diameter. (140 mm) 3.4 A Pelton wheel of power station develops 30,500 hP under a head of 1750 m while running at 760 rpm. Calculate (1) the mean diameter of the runner, (2) the jet diameter, and (3) the diameter ratio. (2.14 m, 0.104 m, 20.6) 3.5 Show that in an inward flow reaction turbine, when the velocity of flow is constant and wheel vane angle at entrance is 908, the best peripheral velocity is ffiffiffiffiffiffiffiffiffi 2gH _ / ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 - tan 2 a _ where H is the head and a the angle of guide vane. 3.6 A Pelton wheel develops 740 kW under a head of 310 m. Find the jet diameter if its efficiency is 86% and C v = 0:98: (0.069 m) 3.7 A reaction turbine runner diameter is 3.5 m at inlet and 2.5 m at outlet. The turbine discharge 102 m 3 per second of water under a head of 145 m. Its inlet vane angle is 1208. Assume radial discharge at 14 m/s, breadth of wheel constant and hydraulic efficiency of 88%, calculate the power developed and speed of machine. (128 MW, 356 rpm) 3.8 Show that in a Pelton wheel, where the buckets deflect the water through an angle of (1808 2 a) degrees, the hydraulic efficiency of the wheel is given by h h = 2U(C 2 U)(1 - cos a) C 2 where C is the velocity of jet and U is mean blade velocity. 3.9 A Kaplan turbine produces 16000 kW under a head of 20 m, while running at 166 rpm. The diameter of the runner is 4.2 m while the hub diameter is 2 m, the discharge being 120 m 3 /s. Calculate (1) the turbine efficiency, Chapter 3 140 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved (2) specific speed, (3) the speed ratio based on the tip diameter of the blade, and (4) the flow ratio. (78%, 497, 1.84, 0.48) 3.10 Evolve a formula for the specific speed of a Pelton wheel in the following form N s = k· ffiffiffiffiffiffiffi h· _ d D where N s = specific speed, h = overall efficiency, d = diameter of jet, D = diameter of bucket circle, and k = a constant. NOTATION C jet velocity, absolute C v nozzle velocity coefficient C w velocity of whirl D wheel diameter d diameter of nozzle E energy transfer by bucket H r head across the runner h f frictional head loss N s specific speed P water power available P c casing and draft tube losses P h hydraulic power loss P l leakage loss P m mechanical power loss P r runner power loss P s shaft power output U bucket speed W work done a angle of the blade tip at outlet b angle with relative velocity h i nozzle efficiency h trans transmission efficiency k relative velocity ratio Hydraulic Turbines 141 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 4 Centrifugal Compressors and Fans 4.1 INTRODUCTION This chapter will be concerned with power absorbing turbomachines, used to handle compressible fluids. There are three types of turbomachines: fans, blowers, and compressors. A fan causes only a small rise in stagnation pressure of the flowing fluid. A fan consists of a rotating wheel (called the impeller), which is surrounded by a stationary member known as the housing. Energy is transmitted to the air by the power-driven wheel and a pressure difference is created, providing airflow. The air feed into a fan is called induced draft, while the air exhausted from a fan is called forced draft. In blowers, air is compressed in a series of successive stages and is often led through a diffuser located near the exit. The overall pressure rise may range from 1.5 to 2.5 atm with shaft speeds up to 30,000 rpm or more. 4.2 CENTRIFUGAL COMPRESSOR The compressor, which can be axial flow, centrifugal flow, or a combination of the two, produces the highly compressed air needed for efficient combustion. In turbocompressors or dynamic compressors, high pressure is achieved by imparting kinetic energy to the air in the impeller, and then this kinetic energy Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved converts into pressure in the diffuser. Velocities of airflow are quite high and the Mach number of the flow may approach unity at many points in the air stream. Compressibility effects may have to be taken into account at every stage of the compressor. Pressure ratios of 4:1 are typical in a single stage, and ratios of 6:1 are possible if materials such as titanium are used. There is renewed interest in the centrifugal stage, used in conjunction with one or more axial stages, for small turbofan and turboprop aircraft engines. The centrifugal compressor is not suitable when the pressure ratio requires the use of more than one stage in series because of aerodynamic problems. Nevertheless, two-stage centrifugal compressors have been used successfully in turbofan engines. Figure 4.1 shows part of a centrifugal compressor. It consists of a stationary casing containing an impeller, which rotates and imparts kinetic energy to the air and a number of diverging passages in which the air decelerates. The deceleration converts kinetic energy into static pressure. This process is known as diffusion, and the part of the centrifugal compressor containing the diverging passages is known as the diffuser. Centrifugal compressors can be built with a double entry or a single entry impeller. Figure 4.2 shows a double entry centrifugal compressor. Air enters the impeller eye and is whirled around at high speed by the vanes on the impeller disc. After leaving the impeller, the air passes through a diffuser in which kinetic energy is exchanged with pressure. Energy is imparted to the air by the rotating blades, thereby increasing the static pressure as it moves from eye radius r 1 to tip radius r 2 . The remainder of the static pressure rise is achieved in the diffuser. The normal practice is to design the compressor so that about half the pressure rise occurs in the impeller and half in the diffuser. The air leaving the diffuser is collected and delivered to the outlet. Figure 4.1 Typical centrifugal compressor. Chapter 4 144 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 4.3 THE EFFECT OF BLADE SHAPE ON PERFORMANCE As discussed in Chapter 2, there are three types of vanes used in impellers. They are: forward-curved, backward-curved, and radial vanes, as shown in Fig. 4.3. The impellers tend to undergo high stress forces. Curved blades, such as those used in some fans and hydraulic pumps, tend to straighten out due to centrifugal force and bending stresses are set up in the vanes. The straight radial Figure 4.2 Double-entry main stage compressor with side-entry compressor for cooling air. (Courtesy of Rolls-Royce, Ltd.) Figure 4.3 Shapes of centrifugal impellar blades: (a) backward-curved blades, (b) radial blades, and (c) forward-curved blades. Centrifugal Compressors and Fans 145 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved blades are not only free from bending stresses, they may also be somewhat easier to manufacture than curved blades. Figure 4.3 shows the three types of impeller vanes schematically, along with the velocity triangles in the radial plane for the outlet of each type of vane. Figure 4.4 represents the relative performance of these types of blades. It is clear that increased mass flow decreases the pressure on the backward blade, exerts the same pressure on the radial blade, and increases the pressure on the forward blade. For a given tip speed, the forward-curved blade impeller transfers maximum energy, the radial blade less, and the least energy is transferred by the backward-curved blades. Hence with forward-blade impellers, a given pressure ratio can be achieved from a smaller-sized machine than those with radial or backward-curved blades. 4.4 VELOCITY DIAGRAMS Figure 4.5 shows the impeller and velocity diagrams at the inlet and outlet. Figure 4.5a represents the velocity triangle when the air enters the impeller in the axial direction. In this case, absolute velocity at the inlet, C 1 ¼ C a1 . Figure 4.5b represents the velocity triangle at the inlet to the impeller eye and air enters through the inlet guide vanes. Angle u is made by C 1 and C a1 and this angle is known as the angle of prewhirl. The absolute velocity C 1 has a whirl component C w1 . In the ideal case, air comes out from the impeller tip after making an angle of 908 (i.e., in the radial direction), so C w2 ¼ U 2 . That is, the whirl component is exactly equal to the impeller tip velocity. Figure 4.5c shows the ideal velocity Figure 4.4 Pressure ratio or head versus mass flow or volume flow, for the three blade shapes. Chapter 4 146 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved triangle. But there is some slip between the impeller and the fluid, and actual values of C w1 are somewhat less than U 2 . As we have already noted in the centrifugal pump, this results in a higher static pressure on the leading face of a vane than on the trailing face. Hence, the air is prevented from acquiring Figure 4.5 Centrifugal impellar and velocity diagrams. Centrifugal Compressors and Fans 147 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved a whirl velocity equal to the impeller tip speed. Figure 4.5d represents the actual velocity triangle. 4.5 SLIP FACTOR From the above discussion, it may be seen that there is no assurance that the actual fluid will follow the blade shape and leave the compressor in a radial direction. Thus, it is convenient to define a slip factor s as: s ¼ C w2 U 2 ð4:1Þ Figure 4.6 shows the phenomenon of fluid slip with respect to a radial blade. In this case, C w2 is not equal to U 2 ; consequently, by the above definition, the slip factor is less than unity. If radial exit velocities are to be achieved by the actual fluid, the exit blade angle must be curved forward about 10–14 degrees. The slip factor is nearly constant for any machine and is related to the number of vanes on the impeller. Various theoretical and empirical studies of the flow in an impeller channel have led to formulas for Figure 4.6 Centrifugal compressor impeller with radial vanes. Chapter 4 148 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved slip factors: For radial vaned impellers, the formula for s is given by Stanitz as follows: s ¼ 1 2 0:63p n ð4:2Þ where n is the number of vanes. The velocity diagram indicates that C w2 approaches U 2 as the slip factor is increased. Increasing the number of vanes may increase the slip factor but this will decrease the flow area at the inlet. A slip factor of about 0.9 is typical for a compressor with 19–21 vanes. 4.6 WORK DONE The theoretical torque will be equal to the rate of change of angular momentum experienced by the air. Considering a unit mass of air, this torque is given by theoretical torque, t ¼ C w2 r 2 ð4:3Þ where, C w2 is whirl component of C 2 and r 2 is impeller tip radius. Let v ¼ angular velocity. Then the theoretical work done on the air may be written as: Theoretical work done W c ¼ C w2 r 2 v ¼ C w2 U 2 . Using the slip factor, we have theoretical W c ¼ sU 2 2 (treating work done on the air as positive) In a real fluid, some of the power supplied by the impeller is used in overcoming losses that have a braking effect on the air carried round by the vanes. These include windage, disk friction, and casing friction. To take account of these losses, a power input factor can be introduced. This factor typically takes values between 1.035 and 1.04. Thus the actual work done on the air becomes: W c ¼ csU 2 2 ð4:4Þ (assuming C w1 ¼ 0, although this is not always the case.) Temperature equivalent of work done on the air is given by: T 02 2T 01 ¼ csU 2 2 C p where T 01 is stagnation temperature at the impeller entrance; T 02 is stagnation temperature at the impeller exit; and C p is mean specific heat over this temperature range. As no work is done on the air in the diffuser, T 03 ¼ T 02 , where T 03 is the stagnation temperature at the diffuser outlet. The compressor isentropic efficiency (h c ) may be defined as: h c ¼ T 03 0 2T 01 T 03 2T 01 Centrifugal Compressors and Fans 149 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved (where T 03 0 ¼ isentropic stagnation temperature at the diffuser outlet) or h c ¼ T 01 T 03 0 =T 01 2 1 _ _ T 03 2T 01 Let P 01 be stagnation pressure at the compressor inlet and; P 03 is stagnation pressure at the diffuser exit. Then, using the isentropic P–T relationship, we get: P 03 P 01 ¼ T 03 0 T 01 _ _ g/ðg21Þ ¼ 1 þ h c ðT 03 2T 01 Þ T 01 _ _ g /ðg21Þ ¼ 1 þ h c csU 2 2 C p T 01 _ _ g/ðg21Þ ð4:5Þ Equation (4.5) indicates that the pressure ratio also depends on the inlet temperature T 01 and impeller tip speed U 2 . Any lowering of the inlet temperature T 01 will clearly increase the pressure ratio of the compressor for a given work input, but it is not under the control of the designer. The centrifugal stresses in a rotating disc are proportional to the square of the rim. For single sided impellers of light alloy, U 2 is limited to about 460 m/s by the maximum allowable centrifugal stresses in the impeller. Such speeds produce pressure ratios of about 4:1. To avoid disc loading, lower speeds must be used for double-sided impellers. 4.7 DIFFUSER The designing of an efficient combustion system is easier if the velocity of the air entering the combustion chamber is as low as possible. Typical diffuser outlet velocities are in the region of 90m/s. The natural tendency of the air in a diffusion process is to break awayfromthe walls of the diverging passage, reverse its direction and flow back in the direction of the pressure gradient, as shown in Fig. 4.7. Eddy formation during air deceleration causes loss by reducing the maximum pressure rise. Therefore, the maximum permissible included angle of the vane diffuser passage is about 118. Any increase in this angle leads to a loss of efficiency due to Figure 4.7 Diffusing flow. Chapter 4 150 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved boundary layer separation on the passage walls. It should also be noted that any change from the design mass flow and pressure ratio would also result in a loss of efficiency. The use of variable-angle diffuser vanes can control the efficiency loss. The flow theory of diffusion, covered in Chapter 2, is applicable here. 4.8 COMPRESSIBILITY EFFECTS If the relative velocity of a compressible fluid reaches the speed of sound in the fluid, separation of flow causes excessive pressure losses. As mentioned earlier, diffusion is a very difficult process and there is always a tendency for the flow to break away fromthe surface, leading to eddy formation and reduced pressure rise. It is necessary to control the Mach number at certain points in the flow to mitigate this problem. The value of the Mach number cannot exceed the value at which shock waves occur. The relative Mach number at the impeller inlet must be less than unity. As shown in Fig. 4.8a, the air breakaway from the convex face of the curved part of the impeller, and hence the Mach number at this point, will be very important and a shock wave might occur. Now, consider the inlet velocity triangle again (Fig. 4.5b). The relative Mach number at the inlet will be given by: M 1 ¼ V 1 ffiffiffiffiffiffiffiffiffiffiffi gRT 1 p ð4:6Þ where T 1 is the static temperature at the inlet. It is possible to reduce the Mach number by introducing the prewhirl. The prewhirl is given by a set of fixed intake guide vanes preceding the impeller. As shown in Fig. 4.8b, relative velocity is reduced as indicated by the dotted triangle. One obvious disadvantage of prewhirl is that the work capacity of Figure 4.8 a) Breakaway commencing at the aft edge of the shock wave, and b) Compressibility effects. Centrifugal Compressors and Fans 151 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved the compressor is reduced by an amount U 1 C w1 . It is not necessary to introduce prewhirl down to the hub because the fluid velocity is low in this region due to lower blade speed. The prewhirl is therefore gradually reduced to zero by twisting the inlet guide vanes. 4.9 MACH NUMBER IN THE DIFFUSER The absolute velocity of the fluid becomes a maximum at the tip of the impeller and so the Mach number may well be in excess of unity. Assuming a perfect gas, the Mach number at the impeller exit M 2 can be written as: M 2 ¼ C 2 ffiffiffiffiffiffiffiffiffiffiffi gRT 2 p ð4:7Þ However, it has been found that as long as the radial velocity component (C r2 ) is subsonic, Mach number greater than unity can be used at the impeller tip without loss of efficiency. In addition, supersonic diffusion can occur without the formation of shock waves provided constant angular momentum is maintained with vortex motion in the vaneless space. High Mach numbers at the inlet to the diffuser vanes will also cause high pressure at the stagnation points on the diffuser vane tips, which leads to a variation of static pressure around the circumference of the diffuser. This pressure variation is transmitted upstream in a radial direction through the vaneless space and causes cyclic loading of the impeller. This may lead to early fatigue failure when the exciting frequency is of the same order as one of the natural frequencies of the impeller vanes. To overcome this concern, it is a common a practice to use prime numbers for the impeller vanes and an even number for the diffuser vanes. Figure 4.9 The theoretical centrifugal compressor characteristic. Chapter 4 152 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 4.10 CENTRIFUGAL COMPRESSOR CHARACTERISTICS The performance of compressible flow machines is usually described in terms of the groups of variables derived in dimensional analysis (Chapter 1). These characteristics are dependent on other variables such as the conditions of pressure and temperature at the compressor inlet and physical properties of the working fluid. To study the performance of a compressor completely, it is necessary to plot P 03 /P 01 against the mass flow parameter m ffiffiffiffiffi T 01 p P 01 for fixed speed intervals of N ffiffiffiffiffi T 01 p . Figure 4.9 shows an idealized fixed speed characteristic. Consider a valve placed in the delivery line of a compressor running at constant speed. First, suppose that the valve is fully closed. Then the pressure ratio will have some value as indicated by Point A. This pressure ratio is available from vanes moving the air about in the impeller. Now, suppose that the valve is opened and airflow begins. The diffuser contributes to the pressure rise, the pressure ratio increases, and at Point B, the maximum pressure occurs. But the compressor efficiency at this pressure ratio will be below the maximum efficiency. Point C indicates the further increase in mass flow, but the pressure has dropped slightly from the maximum possible value. This is the design mass flow rate pressure ratio. Further increases in mass flow will increase the slope of the curve until point D. Point D indicates that the pressure rise is zero. However, the above- described curve is not possible to obtain. 4.11 STALL Stalling of a stage will be defined as the aerodynamic stall, or the breakaway of the flow from the suction side of the blade airfoil. A multistage compressor may operate stably in the unsurged region with one or more of the stages stalled, and the rest of the stages unstalled. Stall, in general, is characterized by reverse flow near the blade tip, which disrupts the velocity distribution and hence adversely affects the performance of the succeeding stages. Referring to the cascade of Fig. 4.10, it is supposed that some nonuniformity in the approaching flow or in a blade profile causes blade B to stall. The air now flows onto blade A at an increased angle of incidence due to blockage of channel AB. The blade A then stalls, but the flow on blade C is now at a lower incidence, and blade C may unstall. Therefore the stall may pass along the cascade in the direction of lift on the blades. Rotating stall may lead to vibrations resulting in fatigue failure in other parts of the gas turbine. Centrifugal Compressors and Fans 153 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 4.12 SURGING Surging is marked by a complete breakdown of the continuous steady flow throughout the whole compressor, resulting in large fluctuations of flow with time and also in subsequent mechanical damage to the compressor. The phenomenon of surging should not be confused with the stalling of a compressor stage. Figure 4.11 shows typical overall pressure ratios and efficiencies h c of a centrifugal compressor stage. The pressure ratio for a given speed, unlike the temperature ratio, is strongly dependent on mass flow rate, since the machine is usually at its peak value for a narrow range of mass flows. When the compressor is running at a particular speed and the discharge is gradually reduced, the pressure ratio will first increase, peaks at a maximum value, and then decreased. The pressure ratio is maximized when the isentropic efficiency has the maximum value. When the discharge is further reduced, the pressure ratio drops due to fall in the isentropic efficiency. If the downstream pressure does not drop quickly there will be backflow accompanied by further decrease in mass flow. In the mean time, if the downstream pressure drops below the compressor outlet pressure, there will be increase in mass flow. This phenomenon of sudden drop in delivery pressure accompanied by pulsating flow is called surging. The point on the curve where surging starts is called the surge point. When the discharge pipe of the compressor is completely choked (mass flow is zero) the pressure ratio will have some value due to the centrifugal head produced by the impeller. Figure 4.10 Mechanism of stall propagation. Chapter 4 154 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Between the zero mass flow and the surge point mass flow, the operation of the compressor will be unstable. The line joining the surge points at different speeds gives the surge line. 4.13 CHOKING When the velocity of fluid in a passage reaches the speed of sound at any cross- section, the flow becomes choked (air ceases to flow). In the case of inlet flow passages, mass flow is constant. The choking behavior of rotating passages Figure 4.11 Centrifugal compressor characteristics. Centrifugal Compressors and Fans 155 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved differs from that of the stationary passages, and therefore it is necessary to make separate analysis for impeller and diffuser, assuming one dimensional, adiabatic flow, and that the fluid is a perfect gas. 4.13.1 Inlet When the flow is choked, C 2 ¼ a 2 ¼ gRT. Since h 0 ¼ h þ 1 2 C 2 , then C p T 0 ¼ C p T þ 1 2 g RT, and T T 0 ¼ 1 þ gR 2C p _ _ 21 ¼ 2 g þ 1 ð4:8Þ Assuming isentropic flow, we have: r r 0 _ _ ¼ P P 0 _ _ T 0 T _ _ ¼ 1 þ 1 2 g 2 1 _ _ M 2 _ _ 12g ð Þ= g21 ð Þ ð4:9Þ and when C ¼ a, M ¼ 1, so that: r r 0 _ _ ¼ 2 g þ 1 _ _ _ _ 1= g21 ð Þ ð4:10Þ Using the continuity equation, _ m A _ _ ¼ rC ¼ r gRT _ ¸ 1=2 , we have _ m A _ _ ¼ r 0 a 0 2 g þ 1 _ _ gþ1 ð Þ=2 g21 ð Þ ð4:11Þ where (r 0 and a 0 refer to inlet stagnation conditions, which remain unchanged. The mass flow rate at choking is constant. 4.13.2 Impeller When choking occurs in the impeller passages, the relative velocity equals the speed of sound at any section. The relative velocity is given by: V 2 ¼ a 2 ¼ gRT _ ¸ and T 01 ¼ T þ gRT 2C p _ _ 2 U 2 2C p Therefore, T T 01 _ _ ¼ 2 g þ 1 _ _ 1 þ U 2 2C p T 01 _ _ ð4:12Þ Chapter 4 156 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Using isentropic conditions, r r 01 _ _ ¼ T T 01 _ _ 1= g21 ð Þ and; from the continuity equation: _ m A _ _ ¼ r 0 a 01 T T 01 _ _ ðgþ1Þ=2ðg21Þ ¼ r 01 a 01 2 g þ 1 _ _ 1 þ U 2 2C p T 01 _ _ _ _ ðgþ1Þ=2ðg21Þ ¼ r 01 a 01 2 þ g 2 1 _ _ U 2 =a 2 01 g þ 1 _ _ ðgþ1Þ=2ðg21Þ ð4:13Þ Equation (4.13) indicates that for rotating passages, mass flow is dependent on the blade speed. 4.13.3 Diffuser For choking in the diffuser, we use the stagnation conditions for the diffuser and not the inlet. Thus: _ m A _ _ ¼ r 02 a 02 2 g þ 1 _ _ gþ1 ð Þ=2 g21 ð Þ ð4:14Þ It is clear that stagnation conditions at the diffuser inlet are dependent on the impeller process. Illustrative Example 4.1: Air leaving the impeller with radial velocity 110 m/s makes an angle of 25830 0 with the axial direction. The impeller tip speed is 475 m/s. The compressor efficiency is 0.80 and the mechanical efficiency is 0.96. Find the slip factor, overall pressure ratio, and power required to drive the compressor. Neglect power input factor and assume g ¼ 1.4, T 01 ¼ 298 K, and the mass flow rate is 3 kg/s. Solution: From the velocity triangle (Fig. 4.12), tan b 2 _ _ ¼ U 2 2C w2 C r2 tan 25:58 ð Þ ¼ 475 2C w2 110 Therefore, C w2 ¼ 422:54 m/s. Centrifugal Compressors and Fans 157 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Now, s ¼ C w2 U 2 ¼ 422:54 475 ¼ 0:89 The overall pressure ratio of the compressor: P 03 P 01 ¼ 1þ h c scU 2 2 C p T 01 _ _ g= g21 ð Þ ¼ 1þ ð0:80Þð0:89Þð475 2 Þ ð1005Þð298Þ _ _ 3:5 ¼ 4:5 The theoretical power required to drive the compressor: P¼ mscU 2 2 1000 _ _ kW¼ ð3Þð0:89Þð475 2 Þ 1000 _ _ ¼602:42kW Using mechanical efficiency, the actual power required to drive the compressor is: P ¼ 602.42/0.96 ¼ 627.52 kW. Illustrative Example 4.2: The impeller tip speed of a centrifugal compressor is 370 m/s, slip factor is 0.90, and the radial velocity component at the exit is 35 m/s. If the flow area at the exit is 0.18 m 2 and compressor efficiency is 0.88, determine the mass flow rate of air and the absolute Mach number at the impeller tip. Assume air density ¼ 1.57 kg/m 3 and inlet stagnation temperature is 290 K. Neglect the work input factor. Also, find the overall pressure ratio of the compressor. Solution: Slip factor: s ¼ C w2 U 2 Therefore: C w2 ¼ U 2 s ¼ (0.90)(370) ¼ 333 m/s The absolute velocity at the impeller exit: C 2 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 2 r2 þC 2 v2 _ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 333 2 þ 35 2 _ ¼ 334:8 m/s The mass flow rate of air: _ m ¼ r 2 A 2 C r2 ¼ 1:57*0:18*35 ¼ 9:89 kg/s Figure 4.12 Velocity triangle at the impeller tip. Chapter 4 158 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The temperature equivalent of work done (neglecting c): T 02 2T 01 ¼ sU 2 2 C p Therefore, T 02 ¼ T 01 þ sU 2 2 C p ¼ 290 þ ð0:90Þð370 2 Þ 1005 ¼ 412:6 K The static temperature at the impeller exit, T 2 ¼ T 02 2 C 2 2 2C p ¼ 412:6 2 334:8 2 ð2Þð1005Þ ¼ 356:83 K The Mach number at the impeller tip: M 2 ¼ C 2 ffiffiffiffiffiffiffiffiffiffiffi gRT 2 p ¼ 334:8 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1:4Þð287Þð356:83Þ p ¼ 0:884 The overall pressure ratio of the compressor (neglecting c): P 03 P 01 ¼ 1 þ h c scU 2 2 C p T 01 _ _ 3:5 ¼ 1 þ ð0:88Þð0:9Þð370 2 Þ ð1005Þð290Þ _ _ 3:5 ¼ 3:0 Illustrative Example 4.3: A centrifugal compressor is running at 16,000 rpm. The stagnation pressure ratio between the impeller inlet and outlet is 4.2. Air enters the compressor at stagnation temperature of 208C and 1 bar. If the impeller has radial blades at the exit such that the radial velocity at the exit is 136 m/s and the isentropic efficiency of the compressor is 0.82. Draw the velocity triangle at the exit (Fig. 4.13) of the impeller and calculate slip. Assume axial entrance and rotor diameter at the outlet is 58 cm. Figure 4.13 Velocity triangle at exit. Centrifugal Compressors and Fans 159 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: Impeller tip speed is given by: U 2 ¼ pDN 60 ¼ ðpÞð0:58Þð16000Þ 60 ¼ 486 m/s Assuming isentropic flow between impeller inlet and outlet, then T 02 0 ¼ T 01 4:2 ð Þ 0:286 ¼ 441:69 K Using compressor efficiency, the actual temperature rise T 02 2T 01 ¼ T 02 0 2T 01 _ _ h c ¼ 441:69 2 293 ð Þ 0:82 ¼ 181:33K Since the flow at the inlet is axial, C w1 ¼ 0 W ¼ U 2 C w2 ¼ C p T 02 2T 01 ð Þ ¼ 1005ð181:33Þ Therefore: C w2 ¼ 1005 181:33 ð Þ 486 ¼ 375 m/s Slip ¼ 486–375 ¼ 111 m/s Slip factor: s ¼ C w2 U 2 ¼ 375 486 ¼ 0:772 Illustrative Example 4.4: Determine the adiabatic efficiency, temperature of the air at the exit, and the power input of a centrifugal compressor from the following given data: Impeller tip diameter ¼ 1 m Speed ¼ 5945 rpm Mass flow rate of air ¼ 28 kg/s Static pressure ratio p 3 /p 1 ¼ 2:2 Atmospheric pressure ¼ 1 bar Atmospheric temperature ¼ 258C Slip factor ¼ 0:90 Neglect the power input factor. Solution: The impeller tip speed is given by: U 2 ¼ pDN 60 ¼ ðpÞð1Þð5945Þ 60 ¼ 311 m/s The work input: W ¼ sU 2 2 ¼ ð0:9Þð311 2 Þ 1000 ¼ 87 kJ/kg Chapter 4 160 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Using the isentropic P–T relation and denoting isentropic temperature by T 3 0 , we get: T 3 0 ¼ T 1 P 3 P 1 _ _ 0:286 ¼ ð298Þð2:2Þ 0:286 ¼ 373:38 K Hence the isentropic temperature rise: T 3 0 2T 1 ¼ 373:38 2 298 ¼ 75:38 K The temperature equivalent of work done: T 3 2T 1 ¼ W C p _ _ ¼ 87/1:005 ¼ 86:57 K The compressor adiabatic efficiency is given by: h c ¼ T 3 0 2T 1 _ _ T 3 2T 1 ð Þ ¼ 75:38 86:57 ¼ 0:871 or 87:1% The air temperature at the impeller exit is: T 3 ¼ T 1 þ 86:57 ¼ 384:57 K Power input: P ¼ _ mW ¼ ð28Þð87Þ ¼ 2436 kW Illustrative Example 4.5: A centrifugal compressor impeller rotates at 9000 rpm. If the impeller tip diameter is 0.914 m and a 2 ¼ 208, calculate the following for operation in standard sea level atmospheric conditions: (1) U 2 , (2) C w2 , (3) C r2 , (4) b 2 , and (5) C 2 . 1. Impeller tip speed is given by U 2 ¼ pDN 60 ¼ ðpÞð0:914Þð9000Þ 60 ¼ 431 m/s 2. Since the exit is radial and no slip, C w2 ¼ U 2 ¼ 431 m/s 3. From the velocity triangle, C r2 ¼ U 2 tan(a 2 ) ¼ (431) (0.364) ¼ 156.87 m/s 4. For radial exit, relative velocity is exactly perpendicular to rotational velocity U 2 . Thus the angle b 2 is 908 for radial exit. 5. Using the velocity triangle (Fig. 4.14), C 2 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi U 2 2 þC 2 r2 _ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 431 2 þ 156:87 2 _ ¼ 458:67 m/s Illustrative Example 4.6: A centrifugal compressor operates with no prewhirl is run with a rotor tip speed of 457 m/s. If C w2 is 95%of U 2 and h c ¼ 0.88, Centrifugal Compressors and Fans 161 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved calculate the following for operation in standard sea level air: (1) pressure ratio, (2) work input per kg of air, and (3) the power required for a flow of 29 k/s. Solution: 1. The pressure ratio is given by (assuming s ¼ c ¼ 1): P 03 P 01 ¼ 1 þ h c scU 2 2 C p T 01 _ _ g= g21 ð Þ ¼ 1 þ ð0:88Þð0:95Þð457 2 Þ ð1005Þð288Þ _ _ 3:5 ¼ 5:22 2. The work per kg of air W ¼ U 2 C w2 ¼ ð457Þð0:95Þð457Þ ¼ 198:4 kJ/kg 3. The power for 29 kg/s of air P ¼ _ mW ¼ ð29Þð198:4Þ ¼ 5753:6 kW Illustrative Example 4.7: A centrifugal compressor is running at 10,000 rpm and air enters in the axial direction. The inlet stagnation temperature of air is 290 K and at the exit from the impeller tip the stagnation temperature is 440 K. The isentropic efficiency of the compressor is 0.85, work input factor c ¼ 1.04, and the slip factor s ¼ 0.88. Calculate the impeller tip diameter, overall pressure ratio, and power required to drive the compressor per unit mass flow rate of air. Figure 4.14 Velocity triangle at impeller exit. Chapter 4 162 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: Temperature equivalent of work done: T 02 2T 01 ¼ scU 2 2 C p or ð0:88Þð1:04ÞðU 2 2 Þ 1005 : Therefore; U 2 ¼ 405:85 m/s and D ¼ 60U 2 pN ¼ ð60Þð405:85Þ ðpÞð10; 000Þ ¼ 0:775 m The overall pressure ratio is given by: P 03 P 01 ¼ 1 þ h c scU 2 2 C p T 01 _ _ g= g21 ð Þ ¼ 1 þ ð0:85Þð0:88Þð1:04Þð405:85 2 Þ ð1005Þð290Þ _ _ 3:5 ¼ 3:58 Power required to drive the compressor per unit mass flow: P ¼ mcsU 2 2 ¼ ð1Þð0:88Þð1:04Þð405:85 2 Þ 1000 ¼ 150:75 kW Design Example 4.8: Air enters axially in a centrifugal compressor at a stagnation temperature of 208C and is compressed from 1 to 4.5 bars. The impeller has 19 radial vanes and rotates at 17,000 rpm. Isentropic efficiency of the compressor is 0.84 and the work input factor is 1.04. Determine the overall diameter of the impeller and the power required to drive the compressor when the mass flow is 2.5 kg/s. Solution: Since the vanes are radial, using the Stanitz formula to find the slip factor: s ¼ 1 2 0:63p n ¼ 1 2 0:63p 19 ¼ 0:8958 The overall pressure ratio P 03 P 01 ¼ 1 þ h c scU 2 2 C p T 01 _ _ g= g21 ð Þ ; or 4:5 ¼ 1 þ ð0:84Þð0:8958Þð1:04ÞðU 2 2 Þ ð1005Þð293Þ _ _ 3:5 ; so U 2 ¼ 449:9 m/s The impeller diameter, D ¼ 60U 2 pN ¼ ð60Þð449:9Þ pð17; 000Þ ¼ 0:5053 m ¼ 50:53 cm. Centrifugal Compressors and Fans 163 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The work done on the air W ¼ csU 2 2 1000 ¼ ð0:8958Þð1:04Þð449:9 2 Þ 1000 ¼ 188:57 kJ/kg Power required to drive the compressor: P ¼ _ mW ¼ ð2:5Þð188:57Þ ¼ 471:43 kW Design Example 4.9: Repeat problem 4.8, assuming the air density at the impeller tip is 1.8 kg/m 3 and the axial width at the entrance to the diffuser is 12 mm. Determine the radial velocity at the impeller exit and the absolute Mach number at the impeller tip. Solution: Slip factor: s ¼ C w2 U 2 ; or C w2 ¼ ð0:8958Þð449:9Þ ¼ 403 m/s Using the continuity equation, _ m ¼ r 2 A 2 C r2 ¼ r 2 2pr 2 b 2 C r2 where: b 2 ¼ axial width r 2 ¼ radius Therefore: C r2 ¼ 2:5 ð1:8Þð2pÞð0:25Þð0:012Þ ¼ 73:65 m/s Absolute velocity at the impeller exit C 2 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 2 r2 þC 2 w2 _ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 73:65 2 þ 403 2 _ ¼ 409:67 m/s The temperature equivalent of work done: T 02 2T 01 ¼ 188:57/C p ¼ 188:57/1:005 ¼ 187:63 K Therefore, T 02 ¼ 293 þ 187.63 ¼ 480.63 K Hence the static temperature at the impeller exit is: T 2 ¼ T 02 2 C 2 2 2C p ¼ 480:63 2 409:67 2 ð2Þð1005Þ ¼ 397 K Now, the Mach number at the impeller exit is: M 2 ¼ C 2 ffiffiffiffiffiffiffiffiffiffiffi gRT 2 p ¼ 409:67 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1:4Þð287Þð397 p Þ ¼ 1:03 Design Example 4.10: A centrifugal compressor is required to deliver 8 kg/s of air with a stagnation pressure ratio of 4 rotating at 15,000 rpm. The air enters the compressor at 258C and 1 bar. Assume that the air enters axially with velocity of 145 m/s and the slip factor is 0.89. If the compressor isentropic Chapter 4 164 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved efficiency is 0.89, find the rise in stagnation temperature, impeller tip speed, diameter, work input, and area at the impeller eye. Solution: Inlet stagnation temperature: T 01 ¼ T a þ C 2 1 2C p ¼ 298 þ 145 2 ð2Þð1005Þ ¼ 308:46 K Using the isentropic P–T relation for the compression process, T 03 0 ¼ T 01 P 03 P 01 _ _ g21 ð Þ=g ¼ ð308:46Þ 4 ð Þ 0:286 ¼ 458:55K Using the compressor efficiency, T 02 2T 01 ¼ T 02 0 2T 01 _ _ h c ¼ 458:55 2 308:46 ð Þ 0:89 ¼ 168:64 K Hence, work done on the air is given by: W ¼ C p T 02 2T 01 ð Þ ¼ ð1:005Þð168:64Þ ¼ 169:48 kJ/kg But, W ¼ sU 2 2 ¼ ð0:89ÞðU 2 Þ 1000 ; or :169:48 ¼ 0:89U 2 2 /1000 or: U 2 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1000Þð169:48Þ 0:89 _ ¼ 436:38 m/s Hence, the impeller tip diameter D ¼ 60U 2 pN ¼ ð60Þð436:38Þ pð15; 000Þ ¼ 0:555 m The air density at the impeller eye is given by: r 1 ¼ P 1 RT 1 ¼ ð1Þð100Þ ð0:287Þð298Þ ¼ 1:17 kg/m 3 Using the continuity equation in order to find the area at the impeller eye, A 1 ¼ _ m r 1 C 1 ¼ 8 ð1:17Þð145Þ ¼ 0:047 m 2 The power input is: P ¼ _ m W ¼ ð8Þð169:48Þ ¼ 1355:24 kW Centrifugal Compressors and Fans 165 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Design Example 4.11: The following data apply to a double-sided centrifugal compressor (Fig. 4.15): Impeller eye tip diameter: 0:28 m Impeller eye root diameter: 0:14 m Impeller tip diameter: 0:48 m Mass flow of air: 10 kg/s Inlet stagnation temperature: 290 K Inlet stagnation pressure: 1 bar Air enters axially with velocity: 145 m/s Slip factor: 0:89 Power input factor: 1:03 Rotational speed: 15; 000 rpm Calculate (1) the impeller vane angles at the eye tip and eye root, (2) power input, and (3) the maximum Mach number at the eye. Solution: (1) Let U er be the impeller speed at the eye root. Then the vane angle at the eye root is: a er ¼ tan 21 C a U er _ _ and U er ¼ pD er N 60 ¼ pð0:14Þð15; 000Þ 60 ¼ 110 m/s Hence, the vane angle at the impeller eye root: a er ¼ tan 21 C a U er _ _ ¼ tan 21 145 110 _ _ ¼ 52 8 48 0 Figure 4.15 The velocity triangle at the impeller eye. Chapter 4 166 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Impeller velocity at the eye tip: U et ¼ pD et N 60 ¼ pð0:28Þð15; 000Þ 60 ¼ 220 m/s Therefore vane angle at the eye tip: a et ¼ tan 21 C a U et _ _ ¼ tan 21 145 220 _ _ ¼ 33 8 23 0 (2) Work input: W ¼ _ mcsU 2 2 ¼ ð10Þð0:819Þð1:03U 2 2 Þ but: U 2 ¼ pD 2 N 60 ¼ pð0:48Þð15; 000Þ 60 ¼ 377:14 m/s Hence, W ¼ ð10Þð0:89Þð1:03Þð377:14 2 Þ 1000 ¼ 1303:86 kW (3) The relative velocity at the eye tip: V 1 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi U 2 et þC 2 a _ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 220 2 þ 145 2 _ ¼ 263:5 m/s Hence, the maximum relative Mach number at the eye tip: M 1 ¼ V 1 ffiffiffiffiffiffiffiffiffiffiffi gRT 1 p ; where T 1 is the static temperature at the inlet T 1 ¼ T 01 2 C 2 1 2C p ¼ 290 2 145 2 ð2Þð1005Þ ¼ 279:54 K The Mach number at the inlet then is: M 1 ¼ V 1 ffiffiffiffiffiffiffiffiffiffiffi gRT 1 p ¼ 263/5 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1:4Þð287Þð279:54 p Þ ¼ 0:786 Design Example 4.12: Recalculate the maximum Mach number at the impeller eye for the same data as in the previous question, assuming prewhirl angle of 208. Centrifugal Compressors and Fans 167 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: Figure 4.16 shows the velocity triangle with the prewhirl angle. From the velocity triangle: C 1 ¼ 145 cosð20 8 Þ ¼ 154:305 m/s Equivalent dynamic temperature: C 2 1 2C p ¼ 154:305 2 ð2Þð1005Þ ¼ 11:846 K C w1 ¼ tanð208Þ C a1 ¼ ð0:36Þð145Þ ¼ 52:78 m/s Relative velocity at the inlet: V 2 1 ¼ C 2 a þ U e 2C w1 ð Þ 2 ¼ 145 2 þ ð220 2 52:78Þ 2 ; or V 1 ¼ 221:3 m/s Therefore the static temperature at the inlet: T 1 ¼ T 01 2 C 2 1 2C p ¼ 290 2 11:846 ¼ 278:2 K Hence, M 1 ¼ V 1 ffiffiffiffiffiffiffiffiffiffiffi gRT 1 p ¼ 221:3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1:4Þð287Þð278:2 p Þ ¼ 0:662 Note the reduction in Mach number due to prewhirl. Figure 4.16 The velocity triangle at the impeller eye. Chapter 4 168 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Design Example 4.13: The following data refers to a single-sided centrifugal compressor: Ambient Temperature: 288 K Ambient Pressure: 1 bar Hub diameter: 0:125 m Eye tip diameter: 0:25 m Mass flow: 5:5 kg/s Speed: 16; 500 rpm Assume zero whirl at the inlet and no losses in the intake duct. Calculate the blade inlet angle at the root and tip and the Mach number at the eye tip. Solution: Let: r h ¼ hub radius r t ¼ tip radius The flow area of the impeller inlet annulus is: A 1 ¼ p r 2 t 2r 2 h _ _ ¼ p 0:125 2 2 0:0625 2 _ _ ¼ 0:038 m 2 Axial velocity can be determined from the continuity equation but since the inlet density (r 1 ) is unknown a trial and error method must be followed. Assuming a density based on the inlet stagnation condition, r 1 ¼ P 01 RT 01 ¼ ð1Þð10 5 Þ ð287Þð288Þ ¼ 1:21 kg/m 3 Using the continuity equation, C a ¼ _ m r 1 A 1 ¼ 5:5 ð1:21Þð0:038Þ ¼ 119:6 m/s Since the whirl component at the inlet is zero, the absolute velocity at the inlet is C 1 ¼ C a . The temperature equivalent of the velocity is: C 2 1 2C p ¼ 119:6 2 ð2Þð1005Þ ¼ 7:12 K Therefore: T 1 ¼ T 01 2 C 2 1 2C p ¼ 288 2 7:12 ¼ 280:9 K Centrifugal Compressors and Fans 169 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Using isentropic P–T relationship, P 1 P 01 ¼ T 1 T 01 _ _ g= g21 ð Þ ; or P 1 ¼ 10 5 280:9 288 _ _ 3:5 ¼ 92 kPa and: r 1 ¼ P 1 RT 1 ¼ ð92Þð10 3 Þ ð287Þð280:9Þ ¼ 1:14 kg/m 3 ; and C a ¼ 5:5 ð1:14Þð0:038Þ ¼ 126:96 m/s Therefore: C 2 1 2C p ¼ ð126:96Þ 2 2ð1005Þ ¼ 8:02 K T 1 ¼ 288 2 8:02 ¼ 279:988 K P 1 ¼ 10 5 279:98 288 _ _ 3:5 ¼ 90:58 kPa r 1 ¼ ð90:58Þð10 3 Þ ð287Þð279:98Þ ¼ 1:13 kg/m 3 Further iterations are not required and the value of r 1 ¼ 1.13 kg/m 3 may be taken as the inlet density and C a ¼ C 1 as the inlet velocity. At the eye tip: U et ¼ 2pr et N 60 ¼ 2pð0:125Þð16; 500Þ 60 ¼ 216 m/s The blade angle at the eye tip: b et ¼ tan 21 U et C a _ _ ¼ tan 21 216 126:96 _ _ ¼ 59:568 At the hub, U eh ¼ 2pð0:0625Þð16; 500Þ 60 ¼ 108 m/s The blade angle at the hub: b eh ¼ tan 21 108 126:96 _ _ ¼ 40:398 Chapter 4 170 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The Mach number based on the relative velocity at the eye tip using the inlet velocity triangle is: V 1 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 2 a þU 2 1 _ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 126:96 2 þ 216 2 _ ; or V 1 ¼ 250:6 m/s The relative Mach number M ¼ V 1 ffiffiffiffiffiffiffiffiffiffiffi gRT 1 p ¼ 250:6 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1:4Þð287Þð279:98 p Þ ¼ 0:747 Design Example 4.14: A centrifugal compressor compresses air at ambient temperature and pressure of 288 K and 1 bar respectively. The impeller tip speed is 364 m/s, the radial velocity at the exit from the impeller is 28 m/s, and the slip factor is 0.89. Calculate the Mach number of the flow at the impeller tip. If the impeller total-to-total efficiency is 0.88 and the flow area from the impeller is 0.085 m 2 , calculate the mass flow rate of air. Assume an axial entrance at the impeller eye and radial blades. Solution: The absolute Mach number of the air at the impeller tip is: M 2 ¼ C 2 ffiffiffiffiffiffiffiffiffiffiffi gRT 2 p where T 2 is the static temperature at the impeller tip. Let us first calculate C 2 and T 2 . Slip factor: s ¼ C w2 U 2 Or: C w2 ¼ sU 2 ¼ ð0:89Þð364Þ ¼ 323:96 m/s From the velocity triangle, C 2 2 ¼ C 2 r2 þC 2 w2 ¼ 28 2 þ 323:96 2 ¼ ð1:06Þð10 5 Þ m 2 /s 2 With zero whirl at the inlet W m ¼ sU 2 2 ¼ C p T 02 2T 01 ð Þ Hence, T 02 ¼ T 01 þ sU 2 2 C p ¼ 288 þ ð0:89Þð364 2 Þ 1005 ¼ 405:33 K Centrifugal Compressors and Fans 171 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Static Temperature T 2 ¼ T 02 2 C 2 2 2C p ¼ 405:33 2 106000 ð2Þð1005Þ ¼ 352:6 K Therefore, M 2 ¼ ð1:06Þð10 5 Þ ð1:4Þð287Þð352:6Þ _ _1 2 ¼ 0:865 Using the isentropic P–T relation: P 02 P 01 _ _ ¼ 1 þh c T 02 T 01 2 1 _ _ _ _ g= g21 ð Þ ¼ 1 þ 0:88 405:33 288 2 1 _ _ _ _ 3:5 ¼ 2:922 P 2 P 02 _ _ ¼ T 2 T 02 _ _ 3:5 ¼ 352:6 405:33 _ _ 3:5 ¼ 0:614 Therefore, P 2 ¼ P 2 P 02 _ _ P 02 P 01 _ _ P 01 ¼ ð0:614Þð2:922Þð1Þð100Þ ¼ 179:4 kPa r 2 ¼ 179:4ð1000Þ 287ð352:6Þ ¼ 1:773 kg/m 3 Mass flow: _ m ¼ ð1:773Þð0:085Þð28Þ ¼ 4:22 kg/s Design Example 4.15: The impeller of a centrifugal compressor rotates at 15,500 rpm, inlet stagnation temperature of air is 290 K, and stagnation pressure at inlet is 101 kPa. The isentropic efficiency of impeller is 0.88, diameter of the impellar is 0.56 m, axial depth of the vaneless space is 38 mm, and width of the vaneless space is 43 mm. Assume ship factor as 0.9, power input factor 1.04, mass flow rate as 16 kg/s. Calculate 1. Stagnation conditions at the impeller outlet, assume no fore whirl at the inlet, 2. Assume axial velocity approximately equal to 105 m/s at the impeller outlet, calculate the Mach number and air angle at the impeller outlet, Chapter 4 172 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 3. The angle of the diffuser vane leading edges and the Mach number at this radius if the diffusion in the vaneless space is isentropic. Solution: 1. Impeller tip speed U 2 ¼ pD 2 N 60 ¼ p £ 0:56 £ 15500 60 U 2 ¼ 454:67 m/s Overall stagnation temperature rise T 03 2T 01 ¼ csU 2 2 1005 ¼ 1:04 £ 0:9 £ 454:67 2 1005 ¼ 192:53K Since T 03 ¼ T 02 Therefore, T 02 2 T 01 ¼ 192.53K and T 02 ¼ 192.53 þ 290 ¼ 482.53K Now pressure ratio for impeller p 02 p 01 ¼ T 02 T 01 _ _ 3:5 ¼ 482:53 290 _ _ 3:5 ¼ 5:94 then, p 02 ¼ 5.94 £ 101 ¼ 600 KPa 2. s ¼ C w2 U 2 C w2 ¼ sU 2 or C w2 ¼ 0:9 £ 454:67 ¼ 409 m/s Let C r2 ¼ 105 m/s Outlet area normal to periphery A 2 ¼ pD 2 £ impeller depth ¼ p £ 0:56 £ 0:038 A 2 ¼ 0:0669 m 2 From outlet velocity triangle C 2 2 ¼ C 2 r2 þC 2 w2 ¼ 105 2 þ 409 2 C 2 2 ¼ 178306 Centrifugal Compressors and Fans 173 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved i:e: C 2 ¼ 422:26 m/s T 2 ¼ T 02 2 C 2 2 2C p ¼ 482:53 2 422:26 2 2 £ 1005 T 2 ¼ 393:82 K Using isentropic P–T relations P 2 ¼ P 02 T 2 T 02 _ _ g g21 ¼ 600 393:82 482:53 _ _ 3:5 ¼ 294:69 kPa From equation of state r 2 ¼ P 2 RT 2 ¼ 293:69 £ 10 3 287 £ 393:82 ¼ 2:61 kg/m 3 The equation of continuity gives C r2 ¼ _ m A 2 P 2 ¼ 16 0:0669 £ 2:61 ¼ 91:63 m/s Thus, impeller outlet radial velocity ¼ 91.63 m/s Impeller outlet Mach number M 2 ¼ C 2 ffiffiffiffiffiffiffiffiffiffiffi gRT 2 p ¼ 422:26 1:4 £ 287 £ 393:82 ð Þ 0:5 M 2 ¼ 1:06 From outlet velocity triangle Cosa 2 ¼ C r2 C 2 ¼ 91:63 422:26 ¼ 0:217 i.e., a 2 ¼ 77.478 3. Assuming free vortex flow in the vaneless space and for convenience denoting conditions at the diffuser vane without a subscript (r ¼ 0.28 þ 0.043 ¼ 0.323) C w ¼ C w2 r 2 r ¼ 409 £ 0:28 0:323 C w ¼ 354:55 m/s Chapter 4 174 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The radial component of velocity can be found by trial and error. Choose as a first try, C r ¼ 105 m/s C 2 2C p ¼ 105 2 þ 354:55 2 2 £ 1005 ¼ 68 K T ¼ 482.53 2 68 (since T ¼ T 02 in vaneless space) T ¼ 414.53K p ¼ p 02 T 2 T 02 _ _ 3:5 ¼ 600 419:53 482:53 _ _ 3:5 ¼ 352:58 kPa r ¼ p 2 RT 2 ¼ 294:69 287 £ 393:82 r ¼ 2:61 kg/m 3 The equation of continuity gives A ¼ 2pr £ depth of vanes ¼ 2p £ 0:323 £ 0:038 ¼ 0:0772 m 2 C r ¼ 16 2:61 £ 0:0772 ¼ 79:41 m/s Next try C r ¼ 79.41 m/s C 2 2C p ¼ 79:41 2 þ 354:55 2 2 £ 1005 ¼ 65:68 T ¼ 482:53 2 65:68 ¼ 416:85K p ¼ p 02 T T 02 _ _ 3:5 ¼ 600 416:85 482:53 _ _ 3:5 p ¼ 359:54 Pa r ¼ 359:54 416:85 £ 287 ¼ 3 kg/m 3 Centrifugal Compressors and Fans 175 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved C r ¼ 16 3:0 £ 0:772 ¼ 69:08 m/s Try C r ¼ 69:08 m/s C 2 2C p ¼ 69:08 2 þ 354:55 2 2 £ 1005 ¼ 64:9 T ¼ 482:53 2 64:9 ¼ 417:63 K p ¼ p 02 T T 02 _ _ 3:5 ¼ 600 417:63 482:53 _ _ 3:5 p ¼ 361:9 Pa r ¼ 361:9 417:63 £ 287 ¼ 3:02 kg/m 3 C r ¼ 16 3:02 £ 0:772 ¼ 68:63 m/s Taking C r as 62.63 m/s, the vane angle tan a ¼ C w C r ¼ 354:5 68:63 ¼ 5:17 i.e. a ¼ 798 Mach number at vane M ¼ 65:68 £ 2 £ 1005 1:4 £ 287 £ 417:63 _ _ 1=2 ¼ 0:787 Design Example 4.16: The following design data apply to a double-sided centrifugal compressor: Impeller eye root diameter: 18 cm Impeller eye tip diameter: 31:75 cm Mass flow: 18:5 kg/s Impeller speed: 15500 rpm Inlet stagnation pressure: 1:0 bar Inlet stagnation temperature: 288 K Axial velocity at inlet ðconstantÞ: 150 m/s Chapter 4 176 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Find suitable values for the impeller vane angles at root and tip of eye if the air is given 208 of prewhirl at all radii, and also find the maximum Mach number at the eye. Solution: At eye root, C a ¼ 150 m/s [C 1 ¼ C a cos208 ¼ 150 cos208 ¼ 159:63 m/s and C w1 ¼ 150 tan 208 ¼ 54.6 m/s Impeller speed at eye root U er ¼ pD er N 60 ¼ p £ 0:18 £ 15500 60 U er ¼ 146 m/s From velocity triangle tan b er ¼ C a U er 2C w1 ¼ 150 146 2 54:6 ¼ 150 91:4 ¼ 1:641 i:e:; b er ¼ 58:648 At eye tip from Fig. 4.17(b) U et pD et N 60 ¼ p £ 0:3175 £ 15500 60 U et ¼ 258 m/s Figure 4.17 Velocity triangles at (a) eye root and (b) eye tip. Centrifugal Compressors and Fans 177 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved tan a et ¼ 150 258 2 54:6 ¼ 150 203:4 ¼ 0:7375 i:e: a et ¼ 36:418 Mach number will be maximum at the point where relative velocity is maximum. Relative velocity at eye root is: V er ¼ C a sin b er ¼ 150 sin 58:648 ¼ 150 0:8539 V er ¼ 175:66 m/s Relative velocity at eye tip is: V et ¼ C a sin a et ¼ 150 sin 36:418 ¼ 150 0:5936 V et ¼ 252:7 m/s Relative velocity at the tip is maximum. Static temperature at inlet: T 1 ¼ T 01 ¼ V 2 et 2C p ¼ 288 2 252:7 2 2 £ 1005 ¼ 288 2 31:77 T 1 ¼ 256:23 K M max ¼ V et gRT 1 _ _ 1=2 ¼ 252:7 ð1:4 £ 287 £ 256:23Þ 1=2 ¼ 252:7 320:86 M max ¼ 0:788 Design Example 4.17: In a centrifugal compressor air enters at a stagnation temperature of 288 K and stagnation pressure of 1.01 bar. The impeller has 17 radial vanes and no inlet guide vanes. The following data apply: Mass flow rate: 2:5 kg/s Impeller tip speed: 475 m/s Mechanical efficiency: 96% Absolute air velocity at diffuser exit: 90 m/s Compressor isentropic efficiency: 84% Absolute velocity at impeller inlet: 150 m/s Chapter 4 178 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Diffuser efficiency: 82% Axial depth of impeller: 6:5 mm Power input factor: 1:04 g for air: 1:4 Determine: 1. shaft power 2. stagnation and static pressure at diffuser outlet 3. radial velocity, absolute Mach number and stagnation and static pressures at the impeller exit, assume reaction ratio as 0.5, and 4. impeller efficiency and rotational speed Solution: 1. Mechanical efficiency is h m ¼ Work transferred to air Work supplied to shaft or shaft power ¼ W h m for vaned impeller, slip factor, by Stanitz formula is s ¼ 1 2 0:63p n ¼ 1 2 0:63 £ p 17 s ¼ 0:884 Work input per unit mass flow W ¼ csU 2 C w2 Since C w1 ¼ 0 ¼ csU 2 2 ¼ 1:04 £ 0:884 £ 475 2 Work input for 2.5 kg/s W ¼ 1:04 £ 0:884 £ 2:5 £ 475 2 W ¼ 518:58K Centrifugal Compressors and Fans 179 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Hence; Shaft Power ¼ 518:58 0:96 ¼ 540:19kW 2. The overall pressure ratio is p 03 p 01 ¼ 1 þ h c csU 2 2 C p T 01 _ _ g= g21 ð Þ ¼ 1 þ 0:84 £ 1:04 £ 0:884 £ 475 2 1005 £ 288 _ _ 3:5 ¼ 5:2 Stagnation pressure at diffuser exit P 03 ¼ p 01 £ 5:20 ¼ 1:01 £ 5:20 P 03 ¼ 5:25 bar p 3 p 03 ¼ T 3 T 03 _ _ g=g21 W ¼ m £ C p T 03 2T 01 ð Þ [ T 03 ¼ W mC p þT 01 ¼ 518:58 £ 10 3 2:5 £ 1005 þ 288 ¼ 494:4 K Static temperature at diffuser exit T 3 ¼ T 03 2 C 2 3 2C p ¼ 494:4 2 90 2 2 £ 1005 T 3 ¼ 490:37 K Static pressure at diffuser exit p 3 ¼ p 03 T 3 T 03 _ _ g=g21 ¼ 5:25 490:37 494:4 _ _ 3:5 p 3 ¼ 5:10 bar 3. The reaction is 0:5 ¼ T 2 2T 1 T 3 2T 1 Chapter 4 180 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved and T 3 2T 1 ¼ ðT 03 2T 01 Þ þ C 2 1 2C 2 3 2C p _ _ ¼ W mC p þ 150 2 2 90 2 2 £ 1005 ¼ 518:58 £ 10 3 2:5 £ 1005 þ 7:164 ¼ 213:56 K Substituting T 2 2T 1 ¼ 0:5 £ 213:56 ¼ 106:78 K Now T 2 ¼ T 01 2 C 2 1 2C p þ T 2 2T 1 ð Þ ¼ 288 2 11:19 þ 106:78 T 2 ¼ 383:59 K At the impeller exit T 02 ¼ T 2 þ C 2 2 2C p or T 03 ¼ T 2 þ C 2 2 2C p ðSince T 02 ¼ T 03 Þ Therefore, C 2 2 ¼ 2C p ½ðT 03 2T 01 Þ þ ðT 01 2T 2 Þ ¼ 2 £ 1005ð206:4 þ 288 þ 383:59Þ C 2 ¼ 471:94 m/s Mach number at impeller outlet M 2 ¼ C 2 1:4 £ 287 £ 383:59 ð Þ 1=2 M 2 ¼ 1:20 Radial velocity at impeller outlet C 2 r2 ¼ C 2 2 2C 2 w2 ¼ ð471:94Þ 2 2 ð0:884 £ 475Þ 2 Centrifugal Compressors and Fans 181 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved C 2 r2 ¼ 215:43 m/s Diffuser efficiency is given by h D ¼ h 3 0 2h 2 h 3 2h 2 ¼ isentropic enthalpy increase actual enthalpy increase ¼ T 3 0 2T 2 T 3 2T 2 ¼ T 2 T 3 0 T 2 2 1 _ _ T 3 2T 2 ¼ T 2 p 3 p 2 _ _ g21=g 21 _ _ T 3 2T 2 ð Þ Therefore p 3 p 2 ¼ 1 þh D T 3 2T 2 T 2 _ _ _ _ 3:5 ¼ 1 þ 0:821 £ 106:72 383:59 _ _ 3:5 ¼ 2:05 or p 2 ¼ 5:10 2:05 ¼ 2:49 bar From isentropic P–T relations p 02 ¼ p 2 T 02 T 2 _ _ 3:5 ¼ 2:49 494:4 383:59 _ _ 3:5 p 02 ¼ 6:05 bar 4. Impeller efficiency is h i ¼ T 01 p 02 p 01 _ _g21 g 21 _ _ T 03 2T 01 ¼ 288 6:05 1:01 _ _ 0:286 21 _ _ 494:4 2 288 ¼ 0:938 r 2 ¼ p 2 RT 2 ¼ 2:49 £ 10 5 287 £ 383:59 r 2 ¼ 2:27 kg/m 3 Chapter 4 182 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved _ m ¼ r 2 A 2 C r2 ¼ 2pr 2 r 2 b 2 But U 2 ¼ pND 2 60 ¼ pN _ m r 2 pC r2 b 2 £ 60 N ¼ 475 £ 2:27 £ 246:58 £ 0:0065 £ 60 2:5 N ¼ 41476 rpm PROBLEMS 4.1 The impeller tip speed of a centrifugal compressor is 450 m/s with no prewhirl. If the slip factor is 0.90 and the isentropic efficiency of the compressor is 0.86, calculate the pressure ratio, the work input per kg of air, and the power required for 25 kg/s of airflow. Assume that the compressor is operating at standard sea level and a power input factor of 1. (4.5, 182.25 kJ/kg, 4556.3 kW) 4.2 Air with negligible velocity enters the impeller eye of a centrifugal compressor at 158C and 1 bar. The impeller tip diameter is 0.45 m and rotates at 18,000 rpm. Find the pressure and temperature of the air at the compressor outlet. Neglect losses and assume g ¼ 1.4. (5.434 bar, 467 K) 4.3 A centrifugal compressor running at 15,000 rpm, overall diameter of the impeller is 60 cm, isentropic efficiency is 0.84 and the inlet stagnation temperature at the impeller eye is 158C. Calculate the overall pressure ratio, and neglect losses. (6) 4.4 Acentrifugal compressor that runs at 20,000 rpm has 20 radial vanes, power input factor of 1.04, and inlet temperature of air is 108C. If the pressure ratio is 2 and the impeller tip diameter is 28 cm, calculate the isentropic efficiency of the compressor. Take g ¼ 1.4 (77.4%) 4.5 Derive the expression for the pressure ratio of a centrifugal compressor: P 03 P 01 ¼ 1 þ h c scU 2 2 C p T 01 _ _ g= g21 ð Þ 4.6 Explain the terms “slip factor” and “power input factor.” Centrifugal Compressors and Fans 183 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 4.7 What are the three main types of centrifugal compressor impellers? Draw the exit velocity diagrams for these three types. 4.8 Explain the phenomenon of stalling, surging and choking in centrifugal compressors. 4.9 A centrifugal compressor operates with no prewhirl and is run with a tip speed of 475 the slip factor is 0.89, the work input factor is 1.03, compressor efficiency is 0.88, calculate the pressure ratio, work input per kg of air and power for 29 airflow. Assume T 01 ¼ 290 K and C p ¼ 1.005 kJ/kg K. (5.5, 232.4 kJ/kg, 6739 kW) 4.10 A centrifugal compressor impeller rotates at 17,000 rpm and compresses 32 kg of air per second. Assume an axial entrance, impeller trip radius is 0.3 m, relative velocity of air at the impeller tip is 105 m/s at an exit angle of 808. Find the torque and power required to drive this machine. (4954 Nm, 8821 kW) 4.11 A single-sided centrifugal compressor designed with no prewhirl has the following dimensions and data: Total head/ pressure ratio: 3:8:1 Speed: 12; 000 rpm Inlet stagnation temperature: 293 K Inlet stagnation pressure: 1:03 bar Slip factor: 0:9 Power input factor: 1:03 Isentropic efficiency: 0:76 Mass flow rate: 20 kg/s Assume an axial entrance. Calculate the overall diameter of the impeller and the power required to drive the compressor. (0.693 m, 3610 kW) 4.12 A double-entry centrifugal compressor designed with no prewhirl has the following dimensions and data: Impeller root diameter: 0:15 m Impeller tip diameter: 0:30 m Rotational speed: 15; 000 rpm Mass flow rate: 18 kg/s Chapter 4 184 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Ambient temperature: 258C Ambient pressure: 1:03 bar Density of air at eye inlet: 1:19 kg/m 3 Assume the axial entrance and unit is stationary. Find the inlet angles of the vane at the root and tip radii of the impeller eye and the maximum Mach number at the eye. (a 1 at root ¼ 50.78, a 1 ¼ 31.48 at tip, 0.79) 4.13 In Example 4.12, air does not enter the impeller eye in an axial direction but it is given a prewhirl of 208 (from the axial direction). The remaining values are the same. Calculate the inlet angles of the impeller vane at the root and tip of the eye. (a 1 at root ¼ 65.58, a 1 at tip ¼ 38.1 8 , 0.697) NOTATION C absolute velocity r radius U impeller speed V relative velocity a vane angle s slip factor v angular velocity c power input factor SUFFIXES 1 inlet to rotor 2 outlet from the rotor 3 outlet from the diffuser a axial, ambient r radial w whirl Centrifugal Compressors and Fans 185 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 5 Axial FlowCompressors and Fans 5.1 INTRODUCTION As mentioned in Chapter 4, the maximum pressure ratio achieved in centrifugal compressors is about 4:1 for simple machines (unless multi-staging is used) at an efficiency of about 70–80%. The axial flow compressor, however, can achieve higher pressures at a higher level of efficiency. There are two important characteristics of the axial flow compressor—high-pressure ratios at good efficiency and thrust per unit frontal area. Although in overall appearance, axial turbines are very similar, examination of the blade cross-section will indicate a big difference. In the turbine, inlet passage area is greater than the outlet. The opposite occurs in the compressor, as shown in Fig. 5.1. Thus the process in turbine blades can be described as an accelerating flow, the increase in velocity being achieved by the nozzle. However, in the axial flow compressor, the flow is decelerating or diffusing and the pressure rise occurs when the fluid passes through the blades. As mentioned in the chapter on diffuser design (Chapter 4, Sec. 4.7), it is much more difficult to carry out efficient diffusion due to the breakaway of air molecules from the walls of the diverging passage. The air molecules that break away tend to reverse direction and flow back in the direction of the pressure gradient. If the divergence is too rapid, this may result in the formation of eddies and reduction in useful pressure rise. During acceleration in a nozzle, there is a natural tendency for the air to fill the passage Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved walls closely (only the normal friction loss will be considered in this case). Typical blade sections are shown in Fig. 5.2. Modern axial flow compressors may give efficiencies of 86–90%—compressor design technology is a well-developed field. Axial flow compressors consist of a number of stages, each stage being formed by a stationary row and a rotating row of blades. Figure 5.3 shows how a few compressor stages are built into the axial compressor. The rotating blades impart kinetic energy to the air while increasing air pressure and the stationary row of blades redirect the air in the proper direction and convert a part of the kinetic energy into pressure. The flow of air through the compressor is in the direction of the axis of the compressor and, therefore, it is called an axial flow compressor. The height of the blades is seen to decrease as the fluid moves through the compressor. As the pressure increases in the direction of flow, the volume of air decreases. To keep the air velocity the same for each stage, the blade height is decreased along the axis of the compressor. An extra row of fixed blades, called the inlet guide vanes, is fitted to the compressor inlet. These are provided to guide the air at the correct angle onto the first row of moving blades. In the analysis of the highly efficient axial flow compressor, the 2-D flow through the stage is very important due to cylindrical symmetry. Figure 5.1 Cutaway sketch of a typical axial compressor assembly: the General Electric J85 compressor. (Courtesy of General Electric Co.) Chapter 5 188 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 5.3 Schematic of an axial compressor section. Figure 5.2 Compressor and turbine blade passages: turbine and compressor housing. Axial Flow Compressors and Fans 189 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The flow is assumed to take place at a mean blade height, where the blade peripheral velocities at the inlet and outlet are the same. No flow is assumed in the radial direction. 5.2 VELOCITY DIAGRAM The basic principle of axial compressor operation is that kinetic energy is imparted to the air in the rotating blade row, and then diffused through passages of both rotating and stationary blades. The process is carried out over multiple numbers of stages. As mentioned earlier, diffusion is a deceleration process. It is efficient only when the pressure rise per stage is very small. The blading diagram and the velocity triangle for an axial flow compressor stage are shown in Fig. 5.4. Air enters the rotor blade with absolute velocity C 1 at an angle a 1 measured from the axial direction. Air leaves the rotor blade with absolute velocity C 2 at an angle a 2 . Air passes through the diverging passages formed between the rotor blades. As work is done on the air in the rotor blades, C 2 is larger than C 1 . The rotor row has tangential velocity U. Combining the two velocity vectors gives the relative velocity at inlet V 1 at an angle b 1 . V 2 is the relative velocity at the rotor outlet. It is less than V 1 , showing diffusion of the relative velocity has taken place with some static pressure rise across the rotor blades. Turning of the air towards the axial direction is brought about by the camber of the blades. Euler’s equation Figure 5.4 Velocity diagrams for a compressor stage. Chapter 5 190 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved provides the work done on the air: W c ¼ UðC w2 2C w1 Þ ð5:1Þ Using the velocity triangles, the following basic equations can be written: U C a ¼ tan a 1 þ tan b 1 ð5:2Þ U C a ¼ tan a 2 þ tan b 2 ð5:3Þ in which C a ¼ C a1 ¼ C 2 is the axial velocity, assumed constant through the stage. The work done equation [Eq. (5.1)] may be written in terms of air angles: W c ¼ UC a ðtan a 2 2 tan a 1 Þ ð5:4Þ also, W c ¼ UC a ðtan b 1 2 tan b 2 Þ ð5:5Þ The whole of this input energy will be absorbed usefully in raising the pressure and velocity of the air and for overcoming various frictional losses. Regardless of the losses, all the energy is used to increase the stagnation temperature of the air, KT 0s . If the velocity of air leaving the first stage C 3 is made equal to C 1 , then the stagnation temperature rise will be equal to the static temperature rise, KT s . Hence: T 0s ¼ DT s ¼ UC a C p ðtan b 1 2 tan b 2 Þ ð5:6Þ Equation (5.6) is the theoretical temperature rise of the air in one stage. In reality, the stage temperature rise will be less than this value due to 3-D effects in the compressor annulus. To find the actual temperature rise of the air, a factor l, which is between 0 and 100%, will be used. Thus the actual temperature rise of the air is given by: T 0s ¼ lUC a C p ðtan b 1 2 tan b 2 Þ ð5:7Þ If R s is the stage pressure ratio and h s is the stage isentropic efficiency, then: R s ¼ 1 þ h s DT 0s T 01 _ _ g= g21 ð Þ ð5:8Þ where T 01 is the inlet stagnation temperature. Axial Flow Compressors and Fans 191 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 5.3 DEGREE OF REACTION The degree of reaction, L, is defined as: L ¼ Static enthalpy rise in the rotor Static enthalpy rise in the whole stage ð5:9Þ The degree of reaction indicates the distribution of the total pressure rise into the two types of blades. The choice of a particular degree of reaction is important in that it affects the velocity triangles, the fluid friction and other losses. Let: DT A ¼ the static temperature rise in the rotor DT B ¼ the static temperature rise in the stator Using the work input equation [Eq. (5.4)], we get: W c ¼ C p ðDT A þDT B Þ ¼ DT S ¼ UC a ðtan b 1 2 tan b 2 Þ ¼ UC a ðtan a 2 2 tan a 1 Þ _ ð5:10Þ But since all the energy is transferred to the air in the rotor, using the steady flow energy equation, we have: W c ¼ C p DT A þ 1 2 ðC 2 2 2 C 2 1 Þ ð5:11Þ Combining Eqs. (5.10) and (5.11), we get: C p DT A ¼ UC a ðtan a 2 2 tan a 1 Þ 2 1 2 ðC 2 2 2C 2 1 Þ from the velocity triangles, C 2 ¼ C a cos a 2 and C 1 ¼ C a cos a 1 Therefore, C p DT A ¼ UC a ðtan a 2 2 tan a 1 Þ 2 1 2 C 2 a ðsec 2 a 2 2 sec 2 a 1 Þ ¼ UC a ðtan a 2 2 tan a 1 Þ 2 1 2 C 2 a ðtan 2 a 2 2 tan 2 a 1 Þ Using the definition of degree of reaction, L ¼ DT A DT A þDT B ¼ UC a ðtan a 2 2 tan a 1 Þ 2 1 2 C 2 a ðtan 2 a 2 2 tan 2 a 1 Þ UC a ðtan a 2 2 tan a 1 Þ ¼ 1 2 C a U ðtan a 2 þ tan a 1 Þ Chapter 5 192 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved But from the velocity triangles, adding Eqs. (5.2) and (5.3), 2U C a ¼ ðtan a 1 þ tan b 1 þ tan a 2 þ tan b 2 Þ Therefore, L ¼ C a 2U 2U C a 2 2U C a þ tan b 1 þ tan b 2 _ _ ¼ C a 2U ðtan b 1 þ tan b 2 Þ ð5:12Þ Usually the degree of reaction is set equal to 50%, which leads to this interesting result: ðtan b 1 þ tan b 2 Þ ¼ U C a : Again using Eqs. (5.1) and (5.2), tan a 1 ¼ tan b 2 ; i:e:; a 1 ¼ b 2 tan b 1 ¼ tan a 2 ; i:e:; a 2 ¼ b 1 As we have assumed that C a is constant through the stage, C a ¼ C 1 cos a 1 ¼ C 3 cos a 3 : Since we know C 1 ¼ C 3 , it follows that a 1 ¼ a 3 . Because the angles are equal, a 1 ¼ b 2 ¼ a 3 , and b 1 ¼ a 2 . Under these conditions, the velocity triangles become symmetric. In Eq. (5.12), the ratio of axial velocity to blade velocity is called the flow coefficient and denoted by F. For a reaction ratio of 50%, (h 2 2 h 1 ) ¼ (h 3 2 h 1 ), which implies the static enthalpy and the temperature increase in the rotor and stator are equal. If for a given value of C a =U, b 2 is chosen to be greater than a 2 (Fig. 5.5), then the static pressure rise in the rotor is greater than the static pressure rise in the stator and the reaction is greater than 50%. Figure 5.5 Stage reaction. Axial Flow Compressors and Fans 193 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Conversely, if the designer chooses b 2 less than b 1 , the stator pressure rise will be greater and the reaction is less than 50%. 5.4 STAGE LOADING The stage-loading factor C is defined as: C ¼ W c mU 2 ¼ h 03 2 h 01 U 2 ¼ lðC w2 2C w1 Þ U ¼ lC a U ðtan a 2 2 tan a 1 Þ C ¼ lF ðtan a 2 2 tan a 1 Þ ð5:13Þ 5.5 LIFT-AND-DRAG COEFFICIENTS The stage-loading factor C may be expressed in terms of the lift-and-drag coefficients. Consider a rotor blade as shown in Fig. 5.6, with relative velocity vectors V 1 and V 2 at angles b 1 and b 2 . Let tan ðb m Þ ¼ ðtan ðb 1 Þ þ tan ðb 2 ÞÞ/2. The flowon the rotor blade is similar to flowover an airfoil, so lift-and-drag forces will be set up on the blade while the forces on the air will act on the opposite direction. The tangential force on each moving blade is: F x ¼ Lcos b m þDsin b m F x ¼ Lcos b m 1 þ C D C L _ _ tan b m _ _ ð5:14Þ where: L ¼ lift and D ¼ drag. Figure 5.6 Lift-and-drag forces on a compressor rotor blade. Chapter 5 194 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The lift coefficient is defined as: C L ¼ L 0:5rV 2 m A ð5:15Þ where the blade area is the product of the chord c and the span l. Substituting V m ¼ C a cosb m into the above equation, F x ¼ rC 2 a clC L 2 sec b m 1 þ C D C L _ _ tan b m _ _ ð5:16Þ The power delivered to the air is given by: UF x ¼ m h 03 2h 01 ð Þ ¼ rC a ls h 03 2 h 01 ð Þ ð5:17Þ considering the flow through one blade passage of width s. Therefore, ¼ h 03 2h 01 U 2 ¼ F x rC a lsU ¼ 1 2 C a U _ _ c s _ _ sec b m ðC L þ C D tan b m Þ ¼ 1 2 c s _ _ sec b m ðC L þ C D tan b m Þ ð5:18Þ For a stage in which b m ¼ 458, efficiency will be maximum. Substituting this back into Eq. (5.18), the optimal blade-loading factor is given by: C opt ¼ w ffiffiffi 2 p c s _ _ C L þ C D ð Þ ð5:19Þ For a well-designed blade, C D is much smaller than C L , and therefore the optimal blade-loading factor is approximated by: C opt ¼ w ffiffiffi 2 p c s _ _ C L ð5:20Þ 5.6 CASCADE NOMENCLATURE AND TERMINOLOGY Studying the 2-D flow through cascades of airfoils facilitates designing highly efficient axial flowcompressors. Acascade is a rowof geometrically similar blades arranged at equal distance from each other and aligned to the flow direction. Figure 5.7, which is reproduced from Howell’s early paper on cascade theory and performance, shows the standard nomenclature relating to airfoils in cascade. Axial Flow Compressors and Fans 195 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved a 1 0 and a 2 0 are the camber angles of the entry and exit tangents the camber line makes with the axial direction. The blade camber angle u ¼ a 0 1 2a 0 2 . The chord c is the length of the perpendicular of the blade profile onto the chord line. It is approximately equal to the linear distance between the leading edge and the trailing edge. The stagger angle j is the angle between the chord line and the axial direction and represents the angle at which the blade is set in the cascade. The pitch s is the distance in the direction of rotation between corresponding points on adjacent blades. The incidence angle i is the difference between the air inlet angle (a 1 ) and the blade inlet angle a 0 1 _ _ . That is, i ¼ a 1 2a 0 1 . The deviation angle (d) is the difference between the air outlet angle (a 2 ) and the blade outlet angle a 0 2 _ _ . The air deflection angle, 1 ¼ a 1 2 a 2 , is the difference between the entry and exit air angles. A cross-section of three blades forming part of a typical cascade is shown in Fig. 5.7. For any particular test, the blade camber angle u, its chord c, and the pitch (or space) s will be fixed and the blade inlet and outlet angles a 0 1 and a 0 2 are determined by the chosen setting or stagger angle j. The angle of incidence, i, is then fixed by the choice of a suitable air inlet angle a 1 , since i ¼ a 1 2a 0 1 . An appropriate setting of the turntable on which the cascade is mounted can accomplish this. With the cascade in this position the pressure and direction measuring instruments are then traversed along the blade row in the upstream and downstream position. The results of the traverses are usually presented as shown Figure 5.7 Cascade nomenclature. Chapter 5 196 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved in Fig. 5.8. The stagnation pressure loss is plotted as a dimensionless number given by: Stagnation pressure loss coefficient ¼ P 01 2 P 02 0:5rC 2 1 ð5:21Þ This shows the variation of loss of stagnation pressure and the air deflection, 1 ¼ a 1 2 a 2 , covering two blades at the center of the cascade. The curves of Fig. 5.8 can nowbe repeated for different values of incidence angle, and the whole set of results condensed to the form shown in Fig. 5.9, in which the mean loss and mean deflection are plotted against incidence for a cascade of fixed geometrical form. The total pressure loss owing to the increase in deflection angle of air is marked when i is increased beyond a particular value. The stalling incidence of the cascade is the angle at which the total pressure loss is twice the minimum cascade pressure loss. Reducing the incidence i generates a negative angle of incidence at which stalling will occur. Knowing the limits for air deflection without very high (more than twice the minimum) total pressure loss is very useful for designers in the design of efficient compressors. Howell has defined nominal conditions of deflection for Figure 5.8 Variation of stagnation pressure loss and deflection for cascade at fixed incidence. Axial Flow Compressors and Fans 197 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved a cascade as 80% of its stalling deflection, that is: 1* ¼ 0:81 s ð5:22Þ where 1 s is the stalling deflection and 1* is the nominal deflection for the cascade. Howell and Constant also introduced a relation correlating nominal deviation d * with pitch chord ratio and the camber of the blade. The relation is given by: d* ¼ mu s l _ _ n ð5:23Þ For compressor cascade, n ¼ 1 2 , and for the inlet guide vane in front of the compressor, n ¼ 1. Hence, for a compressor cascade, nominal deviation is given by: d* ¼ mu s l _ _1 2 ð5:24Þ The approximate value suggested by Constant is 0.26, and Howell suggested a modified value for m: m ¼ 0:23 2a l _ _ 2 þ0:1 a * 2 50 _ _ ð5:25Þ where the maximum camber of the cascade airfoil is at a distance a from the leading edge and a * 2 is the nominal air outlet angle. Figure 5.9 Cascade mean deflection and pressure loss curves. Chapter 5 198 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Then, a * 2 ¼ b 2 þd * ¼ b 2 þmu s l _ _ 1 2 and, a * 1 2a * 2 ¼ 1* or: a * 1 ¼ a * 2 þ1* Also, i* ¼ a * 1 2b 1 ¼ a * 2 þ1* 2b 1 5.7 3-D CONSIDERATION So far, all the above discussions were based on the velocity triangle at one particular radius of the blading. Actually, there is a considerable difference in the velocity diagram between the blade hub and tip sections, as shown in Fig. 5.10. The shape of the velocity triangle will influence the blade geometry, and, therefore, it is important to consider this in the design. In the case of a compressor with high hub/tip ratio, there is little variation in blade speed from root to tip. The shape of the velocity diagram does not change much and, therefore, little variation in pressure occurs along the length of the blade. The blading is of the same section at all radii and the performance of the compressor stage is calculated from the performance of the blading at the mean radial section. The flow along the compressor is considered to be 2-D. That is, in 2-D flow only whirl and axial flow velocities exist with no radial velocity component. In an axial flow compressor in which high hub/tip radius ratio exists on the order of 0.8, 2-D flow in the compressor annulus is a fairly reasonable assumption. For hub/tip ratios lower than 0.8, the assumption of two-dimensional flow is no longer valid. Such compressors, having long blades relative to the mean diameter, have been used in aircraft applications in which a high mass flow requires a large annulus area but a small blade tip must be used to keep down the frontal area. Whenever the fluid has an angular velocity as well as velocity in the direction parallel to the axis of rotation, it is said to have “vorticity.” The flow through an axial compressor is vortex flow in nature. The rotating fluid is subjected to a centrifugal force and to balance this force, a radial pressure gradient is necessary. Let us consider the pressure forces on a fluid element as shown in Fig. 5.10. Now, resolve Axial Flow Compressors and Fans 199 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved the forces in the radial direction Fig. 5.11: du ðP þ dPÞðr þ drÞ 2Pr du 2 2 P þ dP 2 _ _ dr du 2 ¼ r dr r du C 2 w r ð5:26Þ or ðP þ dPÞðr þ drÞ 2 Pr 2 P þ dP 2 _ _ dr ¼ r dr C 2 w where: P is the pressure, r, the density, C w , the whirl velocity, r, the radius. After simplification, we get the following expression: Pr þPdr þr dP þ dPdr 2Pr þr dr 2 1 2 dPdr ¼ r dr C 2 w or: r dP ¼ r dr C 2 w Figure 5.10 Variation of velocity diagram along blade. Chapter 5 200 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved That is, 1 r dP dr ¼ C 2 w r ð5:27Þ The approximation represented by Eq. (5.27) has become known as radial equilibrium. The stagnation enthalpy h 0 at any radius r where the absolute velocity is C may be rewritten as: h 0 ¼ h þ 1 2 C 2 a þ 1 2 C 2 w ; h ¼ c p T; and C 2 ¼ C 2 a þ C 2 w _ _ Differentiating the above equation w.r.t. r and equating it to zero yields: dh 0 dr ¼ g g 2 1 £ 1 r dP dr þ 1 2 0 þ 2C w dC w dr _ _ Figure 5.11 Pressure forces on a fluid element. Axial Flow Compressors and Fans 201 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved or: g g 2 1 £ 1 r dP dr þC w dC w dr ¼ 0 Combining this with Eq. (5.27): g g 2 1 C 2 w r þ C w dC w dr ¼ 0 or: dC w dr ¼ 2 g g 2 1 C w r Separating the variables, dC w C w ¼ 2 g g 2 1 dr r Integrating the above equation _ dC w C w ¼ 2 g g 2 1 _ dr r 2 g g 2 1 ln C w r ¼ c where c is a constant: Taking antilog on both sides, g g 2 1 £ C w £ r ¼ e c Therefore, we have C w r ¼ constant ð5:28Þ Equation (5.28) indicates that the whirl velocity component of the flow varies inversely with the radius. This is commonly known as free vortex. The outlet blade angles would therefore be calculated using the free vortex distribution. 5.8 MULTI-STAGE PERFORMANCE An axial flow compressor consists of a number of stages. If R is the overall pressure ratio, R s is the stage pressure ratio, and N is the number of stages, then the total pressure ratio is given by: R ¼ ðR s Þ N ð5:29Þ Equation (5.29) gives only a rough value of R because as the air passes through the compressor the temperature rises continuously. The equation used to Chapter 5 202 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved find stage pressure is given by: R s ¼ 1 þ h s DT 0s T 01 _ _ g g21 ð5:30Þ The above equation indicates that the stage pressure ratio depends only on inlet stagnation temperature T 01 , which goes on increasing in the successive stages. To find the value of R, the concept of polytropic or small stage efficiency is very useful. The polytropic or small stage efficiency of a compressor is given by: h 1;c ¼ g 2 1 g _ _ n n 2 1 _ _ or: n n 2 1 _ _ ¼ h s g g 2 1 _ _ where h s ¼ h 1,c ¼ small stage efficiency. The overall pressure ratio is given by: R ¼ 1 þ NDT 0s T 01 _ _ n n21 ð5:31Þ Although Eq. (5.31) is used to find the overall pressure ratio of a compressor, in actual practice the step-by-step method is used. 5.9 AXIAL FLOW COMPRESSOR CHARACTERISTICS The forms of characteristic curves of axial flow compressors are shown in Fig. 5.12. These curves are quite similar to the centrifugal compressor. However, axial flow compressors cover a narrower range of mass flow than the centrifugal compressors, and the surge line is also steeper than that of a centrifugal compressor. Surging and choking limit the curves at the two ends. However, the surge points in the axial flow compressors are reached before the curves reach a maximum value. In practice, the design points is very close to the surge line. Therefore, the operating range of axial flow compressors is quite narrow. Illustrative Example 5.1: In an axial flow compressor air enters the compressor at stagnation pressure and temperature of 1 bar and 292K, respectively. The pressure ratio of the compressor is 9.5. If isentropic efficiency of the compressor is 0.85, find the work of compression and the final temperature at the outlet. Assume g ¼ 1.4, and C p ¼ 1.005 kJ/kg K. Axial Flow Compressors and Fans 203 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: T 01 ¼ 292K; P 01 ¼ 1 bar; h c ¼ 0:85: Using the isentropic P–T relation for compression processes, P 02 P 01 ¼ T 0 02 T 01 _ _ g g21 where T 02 0 is the isentropic temperature at the outlet. Therefore, T 0 02 ¼ T 01 P 02 P 01 _ _g21 g ¼ 292ð9:5Þ 0:286 ¼ 555:92 K Now, using isentropic efficiency of the compressor in order to find the actual temperature at the outlet, T 02 ¼ T 01 þ T 0 02 2 T 01 _ _ h c ¼ 292 þ 555:92 2 292 ð Þ 0:85 ¼ 602:49 K Figure 5.12 Axial flow compressor characteristics. Chapter 5 204 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Work of compression: W c ¼ C p ðT 02 2 T 01 Þ ¼ 1:005ð602:49 2 292Þ ¼ 312 kJ/kg Illustrative Example 5.2: In one stage of an axial flow compressor, the pressure ratio is to be 1.22 and the air inlet stagnation temperature is 288K. If the stagnation temperature rise of the stages is 21K, the rotor tip speed is 200 m/s, and the rotor rotates at 4500 rpm, calculate the stage efficiency and diameter of the rotor. Solution: The stage pressure ratio is given by: R s ¼ 1 þ h s DT 0s T 01 _ _ g g21 or 1:22 ¼ 1 þ h s ð21Þ 288 _ _ 3:5 that is, h s ¼ 0:8026 or 80:26% The rotor speed is given by: U ¼ pDN 60 ; or D ¼ ð60Þð200Þ pð4500Þ ¼ 0:85 m Illustrative Example 5.3: An axial flow compressor has a tip diameter of 0.95 m and a hub diameter of 0.85 m. The absolute velocity of air makes an angle of 288 measured from the axial direction and relative velocity angle is 568. The absolute velocity outlet angle is 568 and the relative velocity outlet angle is 288. The rotor rotates at 5000 rpm and the density of air is 1.2 kg/m 3 . Determine: 1. The axial velocity. 2. The mass flow rate. 3. The power required. 4. The flow angles at the hub. 5. The degree of reaction at the hub. Solution: 1. Rotor speed is given by: U ¼ pDN 60 ¼ pð0:95Þð5000Þ 60 ¼ 249 m/s Axial Flow Compressors and Fans 205 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Blade speed at the hub: U h ¼ pD h N 60 ¼ pð0:85Þð5000Þ 60 ¼ 223 m/s From the inlet velocity triangle (Fig. 5.13), tan a 1 ¼ C w1 C a and tan b 1 ¼ U 2 C w1 ð Þ C a Adding the above two equations: U C a ¼ tan a 1 þ tan b 1 or: U ¼ C a ðtan 288 þ tan 568Þ ¼ C a ð2:0146Þ Therefore, C a ¼ 123.6 m/s (constant at all radii) 2. The mass flow rate: _ m ¼ pðr 2 t 2 r 2 h Þr C a ¼ pð0:475 2 2 0:425 2 Þð1:2Þð123:6Þ ¼ 20:98 kg/s 3. The power required per unit kg for compression is: W c ¼ lUC a ðtan b 1 2 tan b 2 Þ ¼ ð1Þð249Þð123:6Þðtan 568 2 tan 28 8 Þ10 23 ¼ ð249Þð123:6Þð1:483 2 0:53Þ ¼ 29:268 kJ/kg Figure 5.13 Inlet velocity triangle. Chapter 5 206 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The total power required to drive the compressor is: W c ¼ _ mð29:268Þ ¼ ð20:98Þð29:268Þ ¼ 614 kW 4. At the inlet to the rotor tip: C w1 ¼ C a tan a 1 ¼ 123:6 tan 28 8 ¼ 65:72 m/s Using free vortex condition, i.e., C w r ¼ constant, and using h as the subscript for the hub, C w1h ¼ C w1t r t r h ¼ ð65:72Þ 0:475 0:425 ¼ 73:452 m/s At the outlet to the rotor tip, C w2t ¼ C a tan a 2 ¼ 123:6 tan 56 8 ¼ 183:24 m/s Therefore, C w2h ¼ C w2t r t r h ¼ ð183:24Þ 0:475 0:425 ¼ 204:8 m/s Hence the flow angles at the hub: tan a 1 ¼ C w1h C a ¼ 73:452 123:6 ¼ 0:594 or; a 1 ¼ 30:728 tan b 1 ¼ U h ð Þ C a 2 tan a 1 ¼ 223 123:6 2 0:5942 ¼ 1:21 i.e., b 1 ¼ 50.438 tan a 2 ¼ C w2h C a ¼ 204:8 123:6 ¼ 1:657 i.e., a 2 ¼ 58.898 tan b 2 ¼ U h ð Þ C a 2 tan a 2 ¼ 223 123:6 2 tan 58:59 8 ¼ 0:1472 i.e., b 2 ¼ 8.378 5. The degree of reaction at the hub is given by: L h ¼ C a 2U h ðtan b 1 þ tan b 2 Þ ¼ 123:6 ð2Þð223Þ ðtan 50:438 þ tan 8:378Þ ¼ 123:6 ð2Þð223Þ ð1:21 þ 0:147Þ ¼ 37:61% Axial Flow Compressors and Fans 207 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Illustrative Example 5.4: An axial flow compressor has the following data: Blade velocity at root: 140 m/s Blade velocity at mean radius: 185 m/s Blade velocity at tip: 240 m/s Stagnation temperature rise in this stage: 15K Axial velocity ðconstant from root to tipÞ: 140 m/s Work done factor: 0:85 Degree of reaction at mean radius: 50% Calculate the stage air angles at the root, mean, and tip for a free vortex design. Solution: Calculation at mean radius: From Eq. (5.1), W c ¼ U(C w2 2C w1 ) ¼ UKC w or: C p ðT 02 2 T 01 Þ ¼ C p DT 0s ¼ lUDC w So: DC w ¼ C p DT 0s lU ¼ ð1005Þð15Þ ð0:85Þð185Þ ¼ 95:87 m/s Since the degree of reaction (Fig. 5.14) at the mean radius is 50%, a 1 ¼ b 2 and a 2 ¼ b 1 . From the velocity triangle at the mean, U ¼ DC w þ 2C w1 or: C w1 ¼ U 2DC w 2 ¼ 185 2 95:87 2 ¼ 44:57 m/s Hence, tan a 1 ¼ C w1 C a ¼ 44:57 140 ¼ 0:3184 that is, a 1 ¼ 17:668 ¼ b 2 Chapter 5 208 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved and tan b 1 ¼ DC w þ C w1 ð Þ C a ¼ 95:87 þ 44:57 ð Þ 140 ¼ 1:003 i.e., b 1 ¼ 45:098 ¼ a 2 Calculation at the blade tip: Using the free vortex diagram (Fig. 5.15), ðDC w £ UÞ t ¼ ðDC w £ UÞ m Therefore, DC w ¼ ð95:87Þð185Þ 240 ¼ 73:9 m/s Whirl velocity component at the tip: C w1 £ 240 ¼ ð44:57Þð185Þ Therefore: C w1 ¼ ð44:57Þð185Þ 240 ¼ 34:36 m/s tan a 1 ¼ C w1 C a ¼ 34:36 140 ¼ 0:245 Figure 5.14 Velocity triangle at the mean radius. Axial Flow Compressors and Fans 209 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Therefore, a 1 ¼ 13:798 From the velocity triangle at the tip, x 2 þDC w þ C w1 ¼ U or: x 2 ¼ U 2DC w 2 C w1 ¼ 240 2 73:9 2 34:36 ¼ 131:74 tan b 1 ¼ DC w þx 2 C a ¼ 73:9 þ 131:74 140 ¼ 1:469 i.e., b 1 ¼ 55.758 tan a 2 ¼ C w1 þDC w ð Þ C a ¼ 34:36 þ 73:9 ð Þ 140 ¼ 0:7733 i.e., a 2 ¼ 37.718 tan b 2 ¼ x 2 C a ¼ 131:74 140 ¼ 0:941 i.e., b 2 ¼ 43.268 Calculation at the blade root: ðDC w £ UÞ r ¼ ðDC w £ UÞ m Figure 5.15 Velocity triangles at tip. Chapter 5 210 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved or: DC w £ 140 ¼ ð95:87Þð185Þ and DC w ¼ 126:69 m/s Also: ðC w1 £ UÞ r ¼ ðC w1 £ UÞ m or: C w1 £ 140 ¼ ð44:57Þð185Þ and C w1 ¼ 58:9 m/s and ðC w2 £ UÞ t ¼ ðC w2 £ UÞ r so: C w2;tip ¼ C a tana 2 ¼ 140 tan 37:71 8 ¼ 108:24 m/s Therefore: C w2;root ¼ ð108:24Þð240Þ 140 ¼ 185:55 m/s tan a 1 ¼ 58:9 140 ¼ 0:421 i.e., a 1 ¼ 22.828 From the velocity triangle at the blade root, (Fig. 5.16) or: x 2 ¼ C w2 2 U ¼ 185:55 2 140 ¼ 45:55 Figure 5.16 Velocity triangles at root. Axial Flow Compressors and Fans 211 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Therefore: tan b 1 ¼ U 2 C w1 C a ¼ 140 2 58:9 140 ¼ 0:579 i.e., b 1 ¼ 30.088 tan a 2 ¼ C w2 C a ¼ 185:55 140 ¼ 1:325 i.e., a 2 ¼ 52.968 tan b 2 ¼ 2 x 2 C a ¼ 2 45:55 140 ¼ 20:325 i.e., b 2 ¼ 2188 Design Example 5.5: From the data given in the previous problem, calculate the degree of reaction at the blade root and tip. Solution: Reaction at the blade root: L root ¼ C a 2U r ðtanb 1r þtanb 2r Þ ¼ 140 ð2Þð140Þ ðtan30:08 8 þtanð218 8 ÞÞ ¼ 140 ð2Þð140Þ ð0:57920:325Þ ¼0:127; or 12:7% Reaction at the blade tip: L tip ¼ C a 2U t ðtanb 1t þtanb 2t Þ ¼ 140 ð2Þð240Þ ðtan55:75 8 þtan43:26 8 Þ ¼ 140 ð2Þð240Þ ð1:469 þ0:941Þ ¼ 0:7029; or 70:29% Illustrative Example 5.6: An axial flow compressor stage has the following data: Air inlet stagnation temperature: 295K Blade angle at outlet measured from the axial direction: 328 Flow coefficient: 0:56 Relative inlet Mach number: 0:78 Degree of reaction: 0:5 Find the stagnation temperature rise in the first stage of the compressor. Chapter 5 212 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: Since the degree of reaction is 50%, the velocity triangle is symmetric as shown in Fig. 5.17. Using the degree of reaction equation [Eq. (5.12)]: L ¼ C a 2U ðtan b 1 þ tan b 2 Þ and w ¼ C a U ¼ 0:56 Therefore: tanb 1 ¼ 2L 0:56 2 tan 328 ¼ 1:16 i.e., b 1 ¼ 49.248 Now, for the relative Mach number at the inlet: M r1 ¼ V 1 gRT 1 _ _1 2 or: V 2 1 ¼ gRM 2 r1 T 01 2 C 2 1 2C p _ _ From the velocity triangle, V 1 ¼ C a cosb 1 ; and C 1 ¼ C a cosa 1 and: a 1 ¼ b 2 ðsince L ¼ 0:5Þ Therefore: C 1 ¼ C a cos328 ¼ C a 0:848 Figure 5.17 Combined velocity triangles for Example 5.6. Axial Flow Compressors and Fans 213 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved and: V 1 ¼ C a cos 49:248 ¼ C a 0:653 Hence: C 2 1 ¼ C 2 a 0:719 ; and V 2 1 ¼ C 2 a 0:426 Substituting for V 1 and C 1 , C 2 a ¼ 104:41 295 2 C 2 a 1445 _ _ ; so : C a ¼ 169:51 m/s The stagnation temperature rise may be calculated as: T 02 2 T 01 ¼ C 2 a C p w ðtan b 1 2 tan b 2 Þ ¼ 169:51 2 ð1005Þð0:56Þ ðtan 49:248 2 tan 328Þ ¼ 27:31K Design Example 5.7: An axial flow compressor has the following design data: Inlet stagnation temperature: 290K Inlet stagnation pressure: 1 bar Stage stagnation temperature rise: 24K Mass flow of air: 22kg/s Axialvelocity through the stage: 155:5m/s Rotational speed: 152rev/s Work done factor: 0:93 Mean blade speed: 205m/s Reaction at the mean radius: 50% Determine: (1) the blade and air angles at the mean radius, (2) the mean radius, and (3) the blade height. Chapter 5 214 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: (1) The following equation provides the relationship between the temperature rise and the desired angles: T 02 2T 01 ¼ lUC a C p ðtan b 1 2 tan b 2 Þ or: 24 ¼ ð0:93Þð205Þð155:5Þ 1005 ðtan b 1 2 tan b 2 Þ so: tan b 1 2 tan b 2 ¼ 0:814 Using the degree of reaction equation: L ¼ C a 2U ðtan b 1 þ tan b 2 Þ Hence: tan b 1 þ tan b 2 ¼ ð0:5Þð2Þð205Þ 155:5 ¼ 1:318 Solving the above two equations simultaneously for b 1 and b 2 , 2 tan b 1 ¼ 2:132; so : b 1 ¼ 46:838 ¼ a 2 ðsince the degree of reaction is 50%Þ and: tan b 2 ¼ 1:318 2 tan 46:838 ¼ 1:318 2 1:066; so : b 2 ¼ 14:148 ¼ a 1 (2) The mean radius, r m , is given by: r m ¼ U 2pN ¼ 205 ð2pÞð152Þ ¼ 0:215m (3) The blade height, h, is given by: m ¼ rAC a, where A is the annular area of the flow. C 1 ¼ C a cosa 1 ¼ 155:5 cos14:148 ¼ 160:31 m/s T 1 ¼ T 01 2 C 2 1 2C p ¼ 290 2 160:31 2 ð2Þð1005Þ ¼ 277:21 K Axial Flow Compressors and Fans 215 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Using the isentropic P–T relation: P 1 P 01 ¼ T 1 T 01 _ _ g g21 Static pressure: P 1 ¼ ð1Þ 277:21 290 _ _ 3:5 ¼ 0:854 bar Then: r 1 ¼ P 1 RT 1 ¼ ð0:854Þð100Þ ð0:287Þð277:21Þ ¼ 1:073 kg/m 3 From the continuity equation: A ¼ 22 ð1:073Þð155:5Þ ¼ 0:132m 2 and the blade height: h ¼ A 2pr m ¼ 0:132 ð2pÞð0:215Þ ¼ 0:098m Illustrative Example 5.8: An axial flow compressor has an overall pressure ratio of 4.5:1, and a mean blade speed of 245 m/s. Each stage is of 50% reaction and the relative air angles are the same (308) for each stage. The axial velocity is 158 m/s and is constant through the stage. If the polytropic efficiency is 87%, calculate the number of stages required. Assume T 01 ¼ 290K. Solution: Since the degree of reaction at the mean radius is 50%, a 1 ¼ b 2 and a 2 ¼ b 1 . From the velocity triangles, the relative outlet velocity component in the x-direction is given by: V x2 ¼ C a tan b 2 ¼ 158tan 308 ¼ 91:22 m/s V 1 ¼ C 2 ¼ ðU 2 V x2 Þ 2 þ C 2 a _ ¸1 2 ¼ ð245 2 91:22Þ 2 þ 158 2 _ ¸1 2 ¼ 220:48 m/s cos b 1 ¼ C a V 1 ¼ 158 220:48 ¼ 0:7166 so: b 1 ¼ 44.238 Chapter 5 216 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Stagnation temperature rise in the stage, DT 0s ¼ UC a C p ðtan b 1 2 tan b 2 Þ ¼ ð245Þð158Þ 1005 ðtan 44:238 2 tan 308Þ ¼ 15:21K Number of stages R ¼ 1 þ NDT 0s T 01 _ _ n n21 n n 2 1 ¼ h 1 g g 2 1 ¼ 0:87 1:4 0:4 ¼ 3:05 Substituting: 4:5 ¼ 1 þ N15:21 290 _ _ 3:05 Therefore, N ¼ 12 stages: Design Example 5.9: In an axial flow compressor, air enters at a stagnation temperature of 290K and 1 bar. The axial velocity of air is 180 m/s (constant throughout the stage), the absolute velocity at the inlet is 185 m/s, the work done factor is 0.86, and the degree of reaction is 50%. If the stage efficiency is 0.86, calculate the air angles at the rotor inlet and outlet and the static temperature at the inlet of the first stage and stage pressure ratio. Assume a rotor speed of 200 m/s. Solution: For 50% degree of reaction at the mean radius (Fig. 5.18), a 1 ¼ b 2 and a 2 ¼ b 1 . From the inlet velocity triangle, cos a 1 ¼ C a C 1 ¼ 180 185 ¼ 0:973 i.e., a 1 ¼ 13.358 ¼ b 2 From the same velocity triangle, C w1 ¼ C 2 1 2 C 2 a _ _1 2 ¼ 185 2 2 180 2 _ _1 2 ¼ 42:72 m/s Axial Flow Compressors and Fans 217 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Therefore, tan b 1 ¼ U 2 C w1 ð Þ C a ¼ 200 2 42:72 ð Þ 180 ¼ 0:874 i.e., b 1 ¼ 41.158 ¼ a 2 Static temperature at stage inlet may be determined by using stagnation and static temperature relationship as given below: T 1 ¼ T 01 2 C 1 2C p ¼ 290 2 185 2 2ð1005Þ ¼ 273 K Stagnation temperature rise of the stage is given by DT 0s ¼ lUC a C p tanb 1 2 tanb 2 _ _ ¼ 0:86ð200Þð180Þ 1005 0:874 2 0:237 ð Þ ¼ 19:62 K Stage pressure ratio is given by R s ¼ 1 þ h s DT 0s T 01 _ _ g=g21 ¼ 1 þ 0:86 £ 19:62 290 _ _ 3:5 ¼ 1:22 Illustrative Example 5.10: Find the isentropic efficiency of an axial flow compressor from the following data: Pressure ratio: 6 Polytropic efficiency: 0:85 Inlet temperature: 285 K Figure 5.18 Velocity triangles (a) inlet, (b) outlet. Chapter 5 218 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: Using the isentropic P–T relation for the compression process, T 02 0 ¼ T 01 P 02 P 01 _ _g21 g ¼ 285 6 ð Þ 0:286 ¼ 475:77 K Using the polytropic P–T relation for the compression process: n 2 1 n ¼ g 2 1 gh 1;c ¼ 0:4 1:4ð0:85Þ ¼ 0:336 Actual temperature rise: T 02 ¼ T 01 p 02 p 01 _ _ ðn21Þ=n ¼ 285 6 ð Þ 0:336 ¼ 520:36 K The compressor isentropic efficiency is given by: h c ¼ T 02 0 2 T 01 T 02 2 T 01 ¼ 475:77 2 285 520 2 285 ¼ 0:8105; or 81:05% Design Example 5.11: In an axial flow compressor air enters the compressor at 1 bar and 290K. The first stage of the compressor is designed on free vortex principles, with no inlet guide vanes. The rotational speed is 5500 rpm and stagnation temperature rise is 22K. The hub tip ratio is 0.5, the work done factor is 0.92, and the isentropic efficiency of the stage is 0.90. Assuming an inlet velocity of 145 m/s, calculate 1. The tip radius and corresponding rotor air angles, if the Mach number relative to the tip is limited to 0.96. 2. The mass flow at compressor inlet. 3. The stagnation pressure ratio and power required to drive the compressor. 4. The rotor air angles at the root section. Solution: (1) As no inlet guide vanes a 1 ¼ 0; C w1 ¼ 0 Stagnation temperature, T 01 , is given by T 01 ¼ T 1 þ C 1 2C p2 or T 1 ¼ T 01 2 C 1 2C p ¼ 290 2 145 2 2 £ 1005 ¼ 288:9K Axial Flow Compressors and Fans 219 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The Mach number relative to tip is M ¼ V 1 ffiffiffiffiffiffiffiffiffiffiffi gRT 1 p or V 1 ¼ 0:96 1:4 £ 287 £ 288:9 ð Þ 0:5 ¼ 340:7 m/s i.e., relative velocity at tip ¼ 340.7 m/s From velocity triangle at inlet (Fig. 5.3) V 2 1 ¼ U 2 t þ C 2 1 or U t ¼ 340:7 2 2 145 2 _ _ 0:5 ¼ 308:3 m/s or tip speed, U t ¼ 2pr t N 60 or r t ¼ 308:3 £ 60 2p £ 5500 ¼ 0:535m: tanb 1 ¼ U t C a ¼ 308:3 145 ¼ 2:126 i:e:; b 1 ¼ 64:818 Stagnation temperature rise DT 0s ¼ tUC a C p tan b 1 2 tan b 2 _ _ Substituting the values, we get 22 ¼ 0:92 £ 308:3 £ 145 1005 tan b 1 2 tan b 2 _ _ or tanb 1 2 tanb 2 _ _ ¼ 0:538 (2) Therefore, tan b 2 ¼ 1.588 and b 2 ¼ 57.88 root radius/tip radius ¼ r m 2 h/2 r m þ h/2 ¼ 0:5 (where subscript m for mean and h for height) or r m 2 h/2 ¼ 0.5 r m þ 0.25 h [ r m ¼ 1.5 h but r t ¼ r m þ h/2 ¼ 1.5 h þ h/2 Chapter 5 220 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved or 0.535 ¼ 2 h or h ¼ 0.268 m and r m ¼ 1.5 h ¼ 0.402 m Area, A ¼ 2pr m h ¼ 2p £ 0.402 £ 0.268 ¼ 0.677 m 2 Now, using isentropic relationship for p–T p 1 p 01 ¼ T 1 T 01 _ _ g/ g21 ð Þ or p 1 ¼ 1 £ 288:9 290 _ _ 3:5 ¼ 0:987 bar and r 1 ¼ p 1 RT 1 ¼ 0:987 £ 10 5 287 £ 288:9 ¼ 1:19 kg/m 3 Therefore, the mass flow entering the stage _ m ¼ rAC a ¼ 1:19 £ 0:677 £ 145 ¼ 116:8 kg/s (3) Stage pressure ratio is R s ¼ 1 þ h s DT 0s T 01 _ _ g/ g21 ð Þ ¼ 1 þ 0:90 £ 22 290 _ _ 3:5 ¼ 1:26 Now, W ¼ C p DT 0s ¼ 1005 £ 22 ¼ 22110J/kg Power required by the compressor P ¼ _ mW ¼ 116:8 £ 22110 ¼ 2582:4 kW (4) In order to find out rotor air angles at the root section, radius at the root can be found as given below. r r ¼ r m 2h/2 ¼ 0:402 2 0:268/2 ¼ 0:267m: Impeller speed at root is U r ¼ 2pr r N 60 ¼ 2 £ p £ 0:267 £ 5500 60 ¼ 153:843 m/s Axial Flow Compressors and Fans 221 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Therefore, from velocity triangle at root section tanb 1 ¼ U r C a ¼ 153:843 145 ¼ 1:061 i:e:; b 1 ¼ 46:9 8 For b 2 at the root section DT 0s ¼ tU r C a C p tanb 1 2 tanb 2 _ _ or 22 ¼ 0:92 £ 153:843 £ 145 1005 tanb 1 2 tanb 2 _ _ or tanb 1 2 tanb 2 _ _ ¼ 1:078 [ b 2 ¼ 20:9748 Design Example 5.12: The following design data apply to an axial flow compressor: Overall pressure ratio: 4:5 Mass flow: 3:5kg/s Polytropic efficiency: 0:87 Stagnation temperature rise per stage: 22k Absolute velocity approaching the last rotor: 160m/s Absolute velocity angle; measured fromthe axial direction: 208 Work done factor: 0:85 Mean diameter of the last stage rotor is: 18:5cm Ambient pressure: 1:0bar Ambient temperature: 290K Calculate the number of stages required, pressure ratio of the first and last stages, rotational speed, and the length of the last stage rotor blade at inlet to the stage. Assume equal temperature rise in all stages, and symmetrical velocity diagram. Chapter 5 222 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: If N is the number of stages, then overall pressure rise is: R ¼ 1 þ NDT 0s T 01 _ _n21 n where n 2 1 n ¼ h ac g g 2 1 (where h ac is the polytropic efficiency) substituting values n 2 1 n ¼ 0:87 £ 1:4 0:4 ¼ 3:05 overall pressure ratio, R is R ¼ 1 þ N £ 22 290 _ _ 3:05 or 4:5 ð Þ 1 3:05 ¼ 1 þ N £ 22 290 _ _ [ N ¼ 8:4 Hence number of stages ¼ 8 Stagnation temperature rise, DT 0s , per stage ¼ 22K, as we took 8 stages, therefore DT 0s ¼ 22 £ 8:4 8 ¼ 23:1 From velocity triangle cos a 8 ¼ C a8 C 8 or C a8 ¼ 160 £ cos20 ¼ 150:35 m/s Using degree of reaction, L ¼ 0.5 Then, 0:5 ¼ C a8 2U tanb 8 þ tanb 9 _ _ Axial Flow Compressors and Fans 223 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved or 0:5 ¼ 150:35 2U tanb 8 þ tanb 9 _ _ ðAÞ Also, DT 0s ¼ tUC a8 C p tanb 8 2 tanb 9 _ _ Now, DT 0s ¼ 22K for one stage. As we took 8 stages, therefore; DT 0s ¼ 22 £ 8:4 8 ¼ 23:1 K [ 23:1 ¼ 0:85 £ U £ 150:35 1005 tanb 8 2 tan20 _ _ ðBÞ Because of symmetry, a 8 ¼ b 9 ¼ 208 From Eq. (A) U ¼ 150:35 tanb 8 þ 0:364 _ _ ðCÞ From Eq. (B) U ¼ 181:66 tanb 8 2 0:364 ðDÞ Comparing Eqs. (C) and (D), we have 150:35 tanb 8 þ 0:364 _ _ ¼ 181:66 tanb 8 2 0:364 _ _ or tan 2 b 8 2 0:364 2 _ _ ¼ 181:66 150:35 ¼ 1:21 or tan 2 b 8 ¼ 1:21 þ 0:1325 ¼ 1:342 [ tanb 8 ¼ ffiffiffiffiffiffiffiffiffiffiffi 1:342 p ¼ 1:159 i:e:; b 8 ¼ 49:20 8 Substituting in Eq. (C) U ¼ 150:35 tan 49:208 þ 0:364 ð Þ ¼ 228:9 m/s Chapter 5 224 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The rotational speed is given by N ¼ 228:9 2p £ 0:0925 ¼ 393:69rps In order to find the length of the last stage rotor blade at inlet to the stage, it is necessary to calculate stagnation temperature and pressure ratio of the last stage. Stagnation temperature of last stage: Fig. 5.19 T o8 ¼ T 01 þ 7 £ T 0s ¼ 290 þ 7 £ 23:1 ¼ 451:7 K Pressure ratio of the first stage is: R ¼ 1 þ 1 £ 23:1 451:7 _ _ 3:05 Now, p 08 /p 09 ¼ 1:1643 p 09 p 01 ¼ 4; and p 09 ¼ 4bar p 08 ¼ 4 1:1643 ¼ 3:44bar and T 08 ¼ T 8 þ C 2 8 2C p or T 8 ¼ T 08 2 C 2 8 2C p ¼ 451:7 2 160 2 2 £ 1005 ¼ 438:96 K Figure 5.19 Velocity diagram of last stage. Axial Flow Compressors and Fans 225 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Using stagnation and static isentropic temperature relationship for the last stage, we have p 8 p 08 ¼ T 8 T 08 _ _ 1:4/0:4 Therefore, p 8 ¼ 3:44 438:96 451:7 _ _ 3:5 ¼ 3:112bar and r 8 ¼ p 8 RT 8 ¼ 3:112 £ 10 5 287 £ 438:9 ¼ 2:471 kg/m 3 Using mass flow rate _ m ¼ r 8 A 8 C a8 or 3:5 ¼ 2:471 £ A 8 £ 150:35 [ A 8 ¼ 0:0094m 2 ¼ 2prh or h ¼ 0:0094 2p £ 0:0925 ¼ 0:0162m i.e., length of the last stage rotor blade at inlet to the stage, h ¼ 16.17 mm. Design Example 5.13: A 10-stage axial flow compressor is designed for stagnation pressure ratio of 4.5:1. The overall isentropic efficiency of the compressor is 88% and stagnation temperature at inlet is 290K. Assume equal temperature rise in all stages, and work done factor is 0.87. Determine the air angles of a stage at the design radius where the blade speed is 218 m/s. Assume a constant axial velocity of 165 m/s, and the degree of reaction is 76%. Solution: No. of stages ¼ 10 The overall stagnation temperature rise is: T 0 ¼ T 01 R g21 g 2 1 _ _ h c ¼ 290 4:5 0:286 2 1 _ _ 0:88 ¼ 155:879 K Chapter 5 226 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The stagnation temperature rise of a stage T 0s ¼ 155:879 10 ¼ 15:588 K The stagnation temperature rise in terms of air angles is: T 0s ¼ tUC a C p tan a 2 2 tan a 1 ð Þ or tan a 2 2 tan a 1 ð Þ ¼ T 0s £ C p tUC a ¼ 15:588 £ 1005 0:87 £ 218 £ 165 ¼ 0:501 ðAÞ From degree of reaction ^ ¼ 1 2 C a 2U tan a 2 þ tan a 1 ð Þ _ _ or 0:76 ¼ 1 2 165 2 £ 218 tan a 2 þ tan a 1 ð Þ _ _ [ tan a 2 þ tan a 1 ð Þ ¼ 0:24 £ 2 £ 218 165 ¼ 0:634 ðBÞ Adding (A) and (B), we get 2 tan a 2 ¼ 1.135 or tan a 2 ¼ 0.5675 i.e., a 2 ¼ 29.578 and tan a 1 ¼ 0.634 2 0.5675 ¼ 0.0665 i.e., a 1 ¼ 3.808 Similarly, for b 1 and b 2 , degree of reaction tan b 1 þ tan b 2 ¼ 2.01 and tan b 1 2 tan b 2 ¼ 0.501 [ 2 tan b 1 ¼ 2.511 i.e., b 1 ¼ 51.468 and tan b 2 ¼ 1.1256 2 0.501 ¼ 0.755 i.e., b 2 ¼ 37.038 Axial Flow Compressors and Fans 227 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Design Example 5.14: An axial flow compressor has a tip diameter of 0.9 m, hub diameter of 0.42 m, work done factor is 0.93, and runs at 5400 rpm. Angles of absolute velocities at inlet and exit are 28 and 588, respectively and velocity diagram is symmetrical. Assume air density of 1.5 kg/m 3 , calculate mass flow rate, work absorbed by the compressor, flow angles and degree of reaction at the hub for a free vortex design. Solution: Impeller speed is U ¼ 2prN 60 ¼ 2p £ 0:45 £ 5400 60 ¼ 254:57 m/s From velocity triangle U ¼ C a tan a 1 þ tan b 1 _ _ C a ¼ U tan a 1 þ tan b 1 ¼ 254:57 tan 288 þ tan 588 ð Þ ¼ 119:47 m/s Flow area is A ¼ p r tip 2 r root _ ¸ ¼ p 0:45 2 2 0:42 2 _ ¸ ¼ 0:0833 m 2 Mass flow rate is _ m ¼ rAC a ¼ 1:5 £ 0:0833 £ 119:47 ¼ 14:928 kg/s Power absorbed by the compressor ¼ tU C w2 2 C w1 ð Þ ¼ tUC a tan a 2 2 tan a 1 ð Þ ¼ 0:93 £ 254:57 £ 119:47 tan 588 2 tan 288 ð Þ ¼ 30213:7 Nm Total Power; P ¼ _ m £ 30213:7 1000 kW ¼ 451 kW and whirl velocity at impeller tip C wt ¼ C a tan a 1 ¼ 119.47 £ tan 288 ¼ 63.52 m/s Now using free vortex condition r C w ¼ constant [ r h C w1h ¼ r t C w1t (where subscripts h for hub and t for tip) Chapter 5 228 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved or C w1h ¼ r t C w1t r h ¼ 0:45 £ 63:52 0:4 ¼ 71:46 m/s Similarly C w2t ¼ C a tana 2 ¼ 119:47 tan588 ¼ 191:2 m/s and r h C w2h ¼ r t C w2t or C w2h ¼ r t C w2t r h ¼ 0:45 £ 191:2 0:4 ¼ 215:09 m/s Therefore, the flow angles at the hub are tan a 1 ¼ C w1h C a ðwhere C a is constantÞ ¼ 71:46 119:47 ¼ 0:598 i.e., a 1 ¼ 30.888 tanb 1 ¼ U h 2 C a tana 1 C a where U h at the hub is given by U h ¼ 2pr h N ¼ 2 £ p £ 0:4 £ 5400 60 ¼ 226:29 m/s [ tanb 1 ¼ 226:29 2 119:47 tan30:888 119:47 i.e., b 1 ¼ 52.348 tana 2 ¼ C w2h C a ¼ 215:09 119:47 ¼ 1:80 i.e., a 2 ¼ 60.958 Similarly, tanb 2 ¼ U h 2 C a tana 2 C a ¼ 226:29 2 119:47 tan 60:958 119:47 i.e., b 2 ¼ 5.368 Axial Flow Compressors and Fans 229 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Finally, the degree of reaction at the hub is ^ ¼ C a 2U h tanb 1 þ tanb 2 _ _ ¼ 119:47 2 £ 226:29 tan52:348 þ tan5:368 ð Þ ¼ 0:367 or 36:7%: Design Example 5.15: An axial flow compressor is to deliver 22 kg of air per second at a speed of 8000 rpm. The stagnation temperature rise of the first stage is 20 K. The axial velocity is constant at 155 m/s, and work done factor is 0.94. The mean blade speed is 200 m/s, and reaction at the mean radius is 50%. The rotor blade aspect ratio is 3, inlet stagnation temperature and pressure are 290 K and 1.0 bar, respectively. Assume C p for air as 1005 J/kg K and g ¼ 1.4. Determine: 1. The blade and air angles at the mean radius. 2. The mean radius. 3. The blade height. 4. The pitch and chord. Solution: 1. Using Eq. (5.10) at the mean radius T 02 2T 01 ¼ tUC a C p tan b 1 2 tanb 2 _ _ 20 ¼ 0:94 £ 200 £ 155 1005 tanb 1 2 tanb 2 _ _ tanb 1 2 tanb 2 _ _ ¼ 0:6898 Using Eq. (5.12), the degree of reaction is ^ ¼ C a 2U tanb 1 þ tanb 2 _ _ or tanb 1 þ tanb 2 _ _ ¼ 0:5 £ 2 £ 200 155 ¼ 1:29 Solving above two equations simultaneously 2 tanb 1 ¼ 1:98 [ b 1 ¼ 44:718 ¼ a 2 ðas the diagram is symmetricalÞ tanb 2 ¼ 1:29 2 tan44:718 i.e., b 2 ¼ 16:708 ¼ a 1 Chapter 5 230 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 2. Let r m be the mean radius r m ¼ U 2pN ¼ 200 £ 60 2p £ 8000 ¼ 0:239m 3. Using continuity equation in order to find the annulus area of flow C 1 ¼ C a cosa 1 ¼ 155 cos16:708 ¼ 162 m/s T 1 ¼ T 01 2 C 2 1 2C p ¼ 290 2 162 2 2 £ 1005 ¼ 276:94 K Using isentropic relationship at inlet p 1 p 01 ¼ T 1 T 01 _ _ g g21 Static pressure is p 1 ¼ 1:0 276:94 290 _ _ 3:5 ¼ 0:851bars Density is r 1 ¼ p 1 RT 1 ¼ 0:851 £ 100 0:287 £ 276:94 ¼ 1:07 kg/m 3 From the continuity equation, A ¼ 22 1:07 £ 155 ¼ 0:133m 2 Blade height is h ¼ A 2pr m ¼ 0:133 2 £ p £ 0:239 ¼ 0:089m: 4. At mean radius, and noting that blades b, an equivalent to cascade, a, nominal air deflection is 1 ¼ b 1 2b 2 ¼ 44:718 2 16:708 ¼ 28:018 Using Fig. 5.20 for cascade nominal deflection vs. air outlet angle, the solidity, s c ¼ 0:5 Blade aspect ratio ¼ span chord Axial Flow Compressors and Fans 231 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Blade chord, C ¼ 0:089 3 ¼ 0:03m Blade pitch, s ¼ 0:5 £ 0:03 ¼ 0:015 m: PROBLEMS 5.1 An axial flow compressor has constant axial velocity throughout the compressor of 152 m/s, a mean blade speed of 162 m/s, and delivers 10.5 kg of air per second at a speed of 10,500 rpm. Each stage is of 50% reaction and the work done factor is 0.92. If the static temperature and pressure at the inlet to the first stage are 288K and 1 bar, respectively, and the stagnation stage temperature rise is 15K, calculate: 1 the mean diameter of the blade row, (2) the blade height, (3) the air exit angle from the rotating blades, and (4) the stagnation pressure ratio of the stage with stage efficiency 0.84. (0.295 m, 0.062 m, 11.378, 1.15) 5.2 The following design data apply to an axial flow compressor: Stagnation temperature rise of the stage: 20 K Work done factor: 0:90 Blade velocity at root: 155 m/s Figure 5.20 Cascade nominal deflection versus air outlet angle. Chapter 5 232 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Blade velocity at mean radius: 208 m/s Blade velocity at tip: 255 m/s Axial velocity ðconstant through the stageÞ: 155 m/s Degree of reaction at mean radius: 0:5 Calculate the inlet and outlet air and blade angles at the root, mean radius and tip for a free vortex design. (188, 45.58, 14.848, 54.078, 39.718, 39.188, 23.568, 29.428, 53.758, 2208) 5.3 Calculatethedegreeofreactionat thetipandroot for thesamedataasProb. 5.2. (66.7%, 10%) 5.4 Calculate the air and blade angles at the root, mean and tip for 50% degree of reaction at all radii for the same data as in Prob. [5.2]. (47.868, 28.378, 43.988, 1.728) 5.5 Show that for vortex flow, C w £ r ¼ constant that is, the whirl velocity component of the flow varies inversely with the radius. 5.6 The inlet and outlet angles of an axial flow compressor rotor are 50 and 158, respectively. The blades are symmetrical; mean blade speed and axial velocity remain constant throughout the compressor. If the mean blade speed is 200 m/s, work done factor is 0.86, pressure ratio is 4, inlet stagnation temperature is equal to 290 K, and polytropic efficiency of the compressor is 0.85, find the number of stages required. (8 stages) 5.7 In an axial flow compressor air enters at 1 bar and 158C. It is compressed through a pressure ratio of four. Find the actual work of compression and temperature at the outlet from the compressor. Take the isentropic efficiency of the compressor to be equal to 0.84 . (167.66 kJ/kg, 454.83 K) 5.8 Determine the number of stages required to drive the compressor for an axial flow compressor having the following data: difference between the tangents of the angles at outlet and inlet, i.e., tan b 1 - tan b 2 ¼ 0.55. The isentropic efficiency of the stage is 0.84, the stagnation temperature at the compressor inlet is 288K, stagnation pressure at compressor inlet is 1 bar, the overall stagnation pressure rise is 3.5 bar, and the mass flow rate is 15 kg/s. Assume C p ¼ 1.005 kJ/kg K, g ¼ 1.4, l ¼ 0.86, h m ¼ 0.99 (10 stages, 287.5 kW) Axial Flow Compressors and Fans 233 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 5.9 From the data given below, calculate the power required to drive the compressor and stage air angles for an axial flow compressor. Stagnation temperature at the inlet: 288 K Overall pressure ratio: 4 Isentropic efficiency of the compressor: 0:88 Mean blade speed: 170 m/s Axial velocity: 120 m/s Degree of reaction: 0:5 (639.4 kW, b 1 ¼ 77.88, b 2 ¼ 272.698) 5.10 Calculate the number of stages from the data given below for an axial flow compressor: Air stagnation temperature at the inlet: 288 K Stage isentropic efficiency: 0:85 Degree of reaction: 0:5 Air angles at rotor inlet: 408 Air angle at the rotor outlet: 108 Meanblade speed: 180 m/s Work done factor: 0:85 Overall pressure ratio: 6 (14 stages) 5.11 Derive the expression for polytropic efficiency of an axial flow compressor in terms of: (a) n and g (b) inlet and exit stagnation temperatures and pressures. 5.12 Sketch the velocity diagrams for an axial flow compressor and derive the expression: P 02 P 01 ¼ 1 þ h s DT 0s T 01 _ _ g g21 5.13 Explain the term “degree of reaction”. Why is the degree of reaction generally kept at 50%? 5.14 Derive an expression for the degree of reaction and show that for 50% reaction, the blades are symmetrical; i.e., a 1 ¼ b 2 and a 2 ¼ b 1 . 5.16 What is vortex theory? Derive an expression for vortex flow. 5.17 What is an airfoil? Define, with a sketch, the various terms used in airfoil geometry. Chapter 5 234 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved NOTATION C absolute velocity C L lift coefficient C p specific heat at constant pressure D drag F x tangential force on moving blade h blade height, specific enthalpy L lift N number of stage, rpm n number of blades R overall pressure ratio, gas constant R s stage pressure ratio U tangential velocity V relative velocity a angle with absolute velocity, measured from the axial direction a * 2 nominal air outlet angle b angle with relative velocity, measure from the axial direction DT A static temperature rise in the rotor DT B static temperature rise in the stator DT 0s stagnation temperature rise DT s static temperature rise D * nominal deviation e * nominal deflection e s stalling deflection w flow coefficient L degree of reaction l work done factor c stage loading factor SUFFIXES 1 inlet to rotor 2 outlet from the rotor 3 inlet to second stage a axial, ambient m mean r radial, root t tip w whirl Axial Flow Compressors and Fans 235 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 6 Steam Turbines 6.1 INTRODUCTION In a steam turbine, high-pressure steam from the boiler expands in a set of stationary blades or vanes (or nozzles). The high-velocity steam from the nozzles strikes the set of moving blades (or buckets). Here the kinetic energy of the steam is utilized to produce work on the turbine rotor. Low-pressure steam then exhausts to the condenser. There are two classical types of turbine stage designs: the impulse stage and the reaction stage. Steam turbines can be noncondensing or condensing. In noncondensing turbines (or backpressure turbines), steam exhausts at a pressure greater than atmospheric. Steam then leaves the turbine and is utilized in other parts of the plant that use the heat of the steam for other processes. The backpressure turbines have very high efficiencies (range from 67% to 75%). A multi-stage condensing turbine is a turbine in which steam exhausts to a condenser and is condensed by air-cooled condensers. The exhaust pressure from the turbine is less than the atmospheric. In this turbine, cycle efficiency is low because a large part of the steam energy is lost in the condenser. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 6.2 STEAM NOZZLES The pressure and volume are related by the simple expression, PV g = constant, for a perfect gas. Steam deviates from the laws of perfect gases. The P-V relationship is given by: PV n = constant where: n = 1:135 for saturated steam n = 1:3 for superheated steam For wet steam, the Zeuner relation, n = 1:035 - x 10 _ _ (where x is the initial dryness fraction of the steam) may be used. All nozzles consist of an inlet section, a throat, and an exit. The velocity through a nozzle is a function of the pressure-differential across the nozzle. Consider a nozzle as shown in Fig. 6.1. Assume that the flow occurs adiabatically under steady conditions. Since no work is transferred, the velocity of the fluid at the nozzle entry is usually very small and its kinetic energy is negligible compared with that at the outlet. Hence, the equation reduces to: C 2 = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 h 1 2h 2 ( ) ¦ ¦ _ (6:1) where h 1 and h 2 are the enthalpies at the inlet and outlet of the nozzle, respectively. As the outlet pressure decreases, the velocity increases. Eventually, a point is reached called the critical pressure ratio, where the velocity is equal to the velocity of sound in steam. Any further reduction in pressure will not produce any further increases in the velocity. The temperature, pressure, and density are called critical temperature, critical pressure, and critical Figure 6.1 Nozzle. Chapter 6 238 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved density, respectively. The ratio between nozzle inlet temperature and critical temperature is given by: T 1 T c = 2 n - 1 (6:2) where T c is the critical temperature at which section M = 1. Assuming isentropic flow in the nozzle, the critical pressure ratio is: P 1 P c = T 1 T / c _ _ n n21 (6:3) where T c / is the temperature, which would have been reached after an isentropic expansion in the nozzle. The critical pressure ratio is approximately 0.55 for superheated steam. When the outlet pressure is designed to be higher than the critical pressure, a simple convergent nozzle may be used. In a convergent nozzle, shown in Fig. 6.2, the outlet cross-sectional area and the throat cross-sectional areas are equal. The operation of a convergent nozzle is not practical in high- pressure applications. In this case, steam tends to expand in all directions and is very turbulent. This will cause increased friction losses as the steamflows through the moving blades. To allow the steam to expand without turbulence, the convergent –divergent nozzle is used. In this type of nozzle, the area of the section from the throat to the exit gradually increases, as shown in Fig. 6.1. The increase in area causes the steam to emerge in a uniform steady flow. The size of the throat and the length of the divergent section of every nozzle must be specifically designed for the pressure ratio for which the nozzle will be used. If a nozzle is designed to operate so that it is just choked, any other operating condition is an off-design condition. In this respect, the behavior of convergent and convergent –divergent nozzles is different. The temperature at the throat, i.e., the critical temperature, can be found from steam tables at the value of P c and s c = s 1 . The critical velocity is given by the equation: C c = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 h 1 2h c ( ) ¦ ¦ _ (6:4) where h c is read from tables or the h–s chart at P c and s c . Figure 6.2 Convergent nozzle. Steam Turbines 239 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 6.3 NOZZLE EFFICIENCY The expansion process is irreversible due to friction between the fluid and walls of the nozzle, and friction within the fluid itself. However, it is still approximately adiabatic as shown in Fig. 6.3. 1–2 / is the isentropic enthalpy drop and 1–2 is the actual enthalpy drop in the nozzle. Then the nozzle efficiency is defined as h n = h 1 2h 2 h 1 2h 2 / 6.4 THE REHEAT FACTOR Consider a multi-stage turbine as shown by the Mollier diagram, Fig. 6.4. The reheat factor is defined by: R:F: = Cumulative stage isentropic enthalpy drop Turbine isentropic enthalpy drop = Dh / _ ¸ stage Dh / [ ] turbine = h 1 2h / 2 _ _ - h 2 2h / 3 _ _ - h 3 2h / 4 _ _ h 1 2h " 4 _ _ (6:5) Figure 6.3 Nozzle expansion process for a vapor. Chapter 6 240 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Since the isobars diverge, R.F. . 1. The reheat factor may be used to relate the stage efficiency and the turbine efficiency. Turbine isentropic efficiency is given by: h t = Dh Dh / (6:6) where Dh is the actual enthalpy drop and Dh / is the isentropic enthalpy drop. From diagram 6.4 it is clear that: Dh = Dh [ ] stage Dh 1–4 = h 1 2h 2 ( ) - h 2 2h 3 ( ) - h 3 2h 4 ( ) if h s (stage efficiency) is constant, then: h t = h s Dh / _ ¸ stage Dh / [ ] turbine = h s Dh / _ ¸ stage Dh / [ ] turbine or h t = h s £ (R:F): (6:7) Equation 6.7 indicates that the turbine efficiency is greater than the stage efficiency. The reheat factor is usually of the order of 1.03–1.04. 6.5 METASTABLE EQUILIBRIUM As shown in Fig. 6.5, slightly superheated steam at point 1 is expanded in a convergent –divergent nozzle. Assume reversible and adiabatic processes. 1–2 is Figure 6.4 Mollier chart for a multi-stage turbine. Steam Turbines 241 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved the path followed by the steam and point 2 is on the saturated vapor line. Here, we can expect condensation to occur. But, if point 2 is reached in the divergent section of the nozzle, then condensation could not occur until point 3 is reached. At this point, condensation occurs very rapidly. Although the steam between points 2–3 is in the vapor state, the temperature is below the saturation temperature for the given pressure. This is known as the metastable state. In fact, the change of temperature and pressure in the nozzle is faster than the condensation process under such conditions. The condensation lags behind the expansion process. Steam does not condense at the saturation temperature corresponding to the pressure. Degree of undercooling is the difference between the saturation temperature corresponding to pressure at point 3 and the actual temperature of the superheated vapor at point 3. Degree of supersaturation is the actual pressure at point 3 divided by the saturation pressure corresponding to the actual temperature of the superheated vapor at point 3. Illustrative Example 6.1: Dry saturated steam at 2 MPa enters a steam nozzle and leaves at 0.2 MPa. Find the exit velocity of the steam and dryness fraction. Assume isentropic expansion and neglect inlet velocity. Solution: From saturated steam tables, enthalpy of saturated vapor at 2 MPa: h 1 = h g = 2799:5 kJ/kg and entropy s 1 = s g = 6:3409 kJ/kg K Figure 6.5 Phenomenon of supersaturation on T–S diagram. Chapter 6 242 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Since the expansion is isentropic, s 1 = s 2 : i.e., s 1 = s 2 = 6.3409 = s f2 - x 2 s fg2 , where x 2 is the dryness fraction after isentropic expansion, s f2 is the entropy of saturated liquid at 0.2 MPa, s fg2 is the entropy of vaporization at 0.2 MPa. Using tables: x 2 = 6:3409 21:5301 5:5970 = 0:8595 Therefore, h 2 = h f2 - x 2 h fg2 = 504.7 - 0.8595 £ 2201.9 = 2397.233 kJ/kg Using the energy equation: C 2 = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2(h 1 2h 2 ) ¦ ¦ _ = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2) £ (1000) £ (2799:5 22397:233) ¦ ¦ _ or: C 2 = 897 m/s Illustrative Example 6.2: Dry saturated steam is expanded in a nozzle from 1.3 MPa to 0.1 MPa. Assume friction loss in the nozzle is equal to 10% of the total enthalpy drop; calculate the mass of steam discharged when the nozzle exit diameter is 10 mm. Solution: Enthalpy of dry saturated steam at 1.3 MPa, using steam tables, h 1 = 2787:6 kJ/kg; and entropy s 1 = 6:4953 kJ/kg K: Since the expansion process is isentropic, s 1 = s 2 = s f2 - x 2 s fg2 , hence dryness fraction after expansion: x 2 = 6:4953 21:3026 6:0568 = 0:857 Now, the enthalpy at the exit: h 2 = h f2 - x 2 h fg2 = 417:46 - (0:857) £ (2258) = 2352:566 kJ/kg Therefore enthalpy drop from 1.3 MPa to 0.1 MPa = h 1 –h 2 = 2787:6–2352:566 = 435:034 kJ/kg Actual enthalpy drop due to friction loss in the nozzle = 0:90 £ 435:034 = 391:531 kJ/kg Hence, the velocity of steam at the nozzle exit: C 2 = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2) £ (1000) £ (391:531) ¦ ¦ _ = 884:908 m/s Steam Turbines 243 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Specific volume of steam at 0.1 MPa: = x 2 v g 2 = (0:857) £ (1:694) = 1:4517 m 3 =kg (since the volume of the liquid is usually negligible compared to the volume of dry saturated vapor, hence for most practical problems, v = xv g ) Mass flow rate of steam at the nozzle exit: = AC 2 x 2 v g 2 = (p) £ (0:01) 2 £ (884:908) £ (3600) (4) £ (1:4517) = 172:42 kg=h: Illustrative Example 6.3: Steam at 7.5 MPa and 5008C expands through an ideal nozzle to a pressure of 5 MPa. What exit area is required to accommodate a flow of 2.8 kg/s? Neglect initial velocity of steam and assume isentropic expansion in the nozzle. Solution: Initial conditions: P 1 = 7.5 MPa, 5008C h 1 = 3404.3 kJ/kg s 1 = 6.7598 kJ/kg K (h 1 and s 1 from superheated steam tables) At the exit state, P 2 . P c = (0:545) £ (7:5) = 4:0875 MPa; and therefore the nozzle is convergent. State 2 is fixed by P 2 = 5 MPa, s 1 = s 2 = 6.7598 kJ/kg K T 2 = 4358K, v 2 = 0.06152 m 3 /kg, h 2 = 3277.9 kJ/kg (from the super- heated steam tables or the Mollier Chart). The exit velocity: C 2 = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2) £ (1000) £ (h 1 2h 2 ) ¦ ¦ _ = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2) £ (1000) £ (3404:3 23277:9) ¦ ¦ _ = 502:8 m/s Using the continuity equation, the exit area is A 2 = mv 2 C 2 = (2:8) £ (0:06152) 502:8 = (3:42) £ (10 24 ) m 2 Illustrative Example 6.4: Consider a convergent –divergent nozzle in which steam enters at 0.8 MPa and leaves the nozzle at 0.15 MPa. Assuming isentropic expansion and index n = 1.135, find the ratio of cross-sectional area, the area at the exit, and the area at the throat for choked conditions (i. e. , for maximum mass flow). Chapter 6 244 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: Critical pressure for maximum mass flow is given by Fig. 6.6: P c = P 2 = P 1 2 n - 1 _ _ n n21 = 0:8 2 2:135 _ _ 8:41 = 0:462 MPa From the Mollier chart: h 1 = 2769 kJ/kg h 2 = 2659 kJ/kg h 3 = 2452 kJ/kg Enthalpy drop from 0.8 MPa to 0.15 MPa: Dh 123 = h 1 2h 3 = 2769 22452 = 317 kJ/kg Enthalpy drop from 0.8 MPa to 0.462 MPa: Dh 1–2 = h 1 2h 2 = 2769 22659 = 110 kJ/kg Dryness fraction: x 2 = 0.954 Dryness fraction: x 3 = 0.902 The velocity at the exit, C 3 = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2) £ (1000) £ (Dh 1–3 ) ¦ ¦ _ = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2) £ (1000) £ (317) ¦ ¦ _ = 796m/s The velocity at the throat C 2 = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2) £ (1000) £ (Dh 1–2 ) ¦ ¦ _ = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2) £ (1000) £ (110) ¦ ¦ _ = 469m/s Figure 6.6 Convergent –divergent nozzle. Steam Turbines 245 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Mass discharged at the throat: _ m 2 = A 2 C 2 x 2 v g 2 Mass discharged at the exit _ m 3 = A 3 C 3 x 3 v g 3 Therefore A 3 C 3 x 3 v g 3 = A 2 C 2 x 2 v g 2 Hence, A 3 A 2 = C 2 C 3 _ _ x 3 v g3 x 2 v g2 _ _ = 469 796 _ _ (0:902)(1:1593) (0:954)(0:4038) _ _ = 1:599 Illustrative Example 6.5: Dry saturated steam enters the convergent – divergent nozzle and leaves the nozzle at 0.1 MPa; the dryness fraction at the exit is 0.85. Find the supply pressure of steam. Assume isentropic expansion (see Fig. 6.7). Figure 6.7 h–s diagram for Example 6.5. Chapter 6 246 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: At the state point 2, the dryness fraction is 0.85 and the pressure is 0.1 MPa. This problem can be solved easily by the Mollier chart or by calculations. Enthalpy and entropy may be determined using the following equations: h 2 = h f2 - x 2 h fg2 and s 2 = s f2 - x 2 s fg2 ; i.e.: h 2 = 417.46 - (0.85) £ (2258) = 2336.76 kJ/kg and s 2 = 1.3026 - (0.85) £ (6.0568) = 6.451 kJ/kg K Since s 1 = s 2 , the state 1 is fixed by s 1 = 6.451 kJ/kg K, and point 1 is at the dry saturated line. Therefore pressure P 1 may be determined by the Mollier chart or by calculations: i.e.: P 1 = 1.474 MPa. 6.6 STAGE DESIGN A turbine stage is defined as a set of stationary blades (or nozzles) followed by a set of moving blades (or buckets or rotor). Together, the two sets of blades allow the steam to perform work on the turbine rotor. This work is then transmitted to the driven load by the shaft on which the rotor assembly is carried. Two turbine stage designs in use are: the impulse stage and reaction stage. The first turbine, designated by DeLaval in 1889, was a single-stage impulse turbine, which ran at 30,000 rpm. Because of its high speed, this type of turbine has very limited applications in practice. High speeds are extremely undesirable due to high blade tip stresses and large losses due to disc friction, which cannot be avoided. In large power plants, the single-stage impulse turbine is ruled out, since alternators usually run speeds around 3000 rpm. Photographs of actual steam turbines are reproduced in Figs. 6.8–6.10. Figure 6.8 Steam turbine. Steam Turbines 247 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 6.7 IMPULSE STAGE In the impulse stage, the total pressure drop occurs across the stationary blades (or nozzles). This pressure drop increases the velocity of the steam. However, in the reaction stage, the total pressure drop is divided equally across the stationary blades and the moving blades. The pressure drop again results in a corresponding increase in the velocity of the steam flow. As shown in Figs. 6.10 and 6.11, the shape of the stationary blades or nozzles in both stage designs is very similar. However, a big difference exists in the shapes of the moving blades. In an impulse stage, the shape of the moving blades or buckets is like a cup. The shape of the moving blades in a reaction stage is more like that of an airfoil. These blades look very similar to the stationary blades or nozzles. 6.8 THE IMPULSE STEAM TURBINE Most of the steam turbine plants use impulse steam turbines, whereas gas turbine plants seldom do. The general principles are the same whether steam or gas is the working substance. As shown in Fig. 6.12, the steam supplied to a single-wheel impulse turbine expands completely in the nozzles and leaves with absolute velocity C 1 at an angle a 1 , and by subtracting the blade velocity vector U, the relative velocity vector at entry to the rotor V 1 can be determined. The relative velocity V 1 makes an angle of b 1 with respect to U. The increase in value of a 1 decreases the value of the useful component, C 1 cos a 1 and increases the value of the axial or flow component C a sin a 1 . The two points of particular interest are the inlet and exit of the blades. As shown in Fig. 6.12, these velocities are V 1 and V 2 , respectively. Figure 6.9 Pressure velocity-compounded impulse turbine. Chapter 6 248 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 6.10 Steam turbine cross-sectional view. S t e a m T u r b i n e s 2 4 9 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Vectorially subtracting the blade speed results in absolute velocity C 2 . The steam leaves tangentially at an angle b 2 with relative velocity V 2 . Since the two velocity triangles have the same common side U, these triangles can be combined to give a single diagram as shown in Fig. 6.13. Figure 6.11 Impulse and reaction stage design. Figure 6.12 Velocity triangles for turbine stage. Chapter 6 250 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved If the blade is symmetrical then b 1 = b 2 and neglecting the friction effects of blades on the steam, V 1 = V 2 . In the actual case, the relative velocity is reduced by friction and expressed by a blade velocity coefficient k. That is: k = V 2 V 1 From Euler’s equation, work done by the steam is given by: W t = U(C w1 - C w2 ) Since C w2 is in the negative r direction, the work done per unit mass flow is given by: W t = U(C w1 - C w2 ) (6:9) If C a1 – C a2 , there will be an axial thrust in the flow direction. Assume that C a is constant. Then: W t = UC a (tan a 1 - tan a 2 ) (6:10) W t = UC a (tan b 1 - tan b 2 ) (6:11) Equation (6.11) is often referred to as the diagram work per unit mass flow and hence the diagram efficiency is defined as: h d = Diagram work done per unit mass flow Work available per unit mass flow (6:12) Referring to the combined diagram of Fig. 6.13: DC w is the change in the velocity of whirl. Therefore: The driving force on the wheel = _ mC w (6:13) Figure 6.13 Combined velocity diagram. Steam Turbines 251 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The product of the driving force and the blade velocity gives the rate at which work is done on the wheel. From Eq. (6.13): Power output = _ mUDC w (6:14) If C a1 2 C a2 = DC a , the axial thrust is given by: Axial thrust : F a = _ mDC a (6:15) The maximum velocity of the steam striking the blades C 1 = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2(h 0 2h 1 ) ¦ ¦ _ (6:16) where h 0 is the enthalpy at the entry to the nozzle and h 1 is the enthalpy at the nozzle exit, neglecting the velocity at the inlet to the nozzle. The energy supplied to the blades is the kinetic energy of the jet, C 2 1 =2 and the blading efficiency or diagram efficiency: h d = Rate of work performed per unit mass flow Energy supplied per unit mass of steam h d = (UDC w ) £ 2 C 2 1 = 2UDC w C 2 1 (6:17) Using the blade velocity coefficient k = V 2 V 1 _ _ and symmetrical blades (i.e., b 1 = b 2 ), then: DC w = 2V 1 cos a 1 2U Hence DC w = 2 C 1 cos a 1 2U ( ) (6:18) And the rate of work performed per unit mass = 2(C 1 cos a 1 2 U)U Therefore: h d = 2 C 1 cos a 1 2U ( )U £ 2 C 2 1 h d = 4 C 1 cos a 1 2U ( )U C 2 1 h d = 4U C 1 cos a 1 2 U C1 _ _ (6:19) where U C 1 is called the blade speed ratio. Differentiating Eq. (6.19) and equating it to zero provides the maximum diagram efficiency: d h d _ _ d U C 1 _ _ = 4 cos a 1 2 8U C 1 = 0 Chapter 6 252 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved or U C 1 = cos a 1 2 (6:20) i.e., maximum diagram efficiency = 4 cos a 1 2 cos a 1 2 cos a 1 2 _ _ or: h d = cos 2 a 1 (6:21) Substituting this value into Eq. (6.14), the power output per unit mass flow rate at the maximum diagram efficiency: P = 2U 2 (6:22) 6.9 PRESSURE COMPOUNDING (THE RATEAU TURBINE) A Rateau-stage impulse turbine uses one row of nozzles and one row of moving blades mounted on a wheel or rotor, as shown in Fig. 6.14. The total pressure drop is divided in a series of small increments over the stages. In each stage, which consists of a nozzle and a moving blade, the steam is expanded and the kinetic energy is used in moving the rotor and useful work is obtained. The separating walls, which carry the nozzles, are known as diaphragms. Each diaphragm and the disc onto which the diaphragm discharges its steam is known as a stage of the turbine, and the combination of stages forms a pressure compounded turbine. Rateau-stage turbines are unable to extract a large amount of energy from the steam and, therefore, have a low efficiency. Although the Rateau turbine is inefficient, its simplicity of design and construction makes it well suited for small auxiliary turbines. 6.10 VELOCITY COMPOUNDING (THE CURTIS TURBINE) In this type of turbine, the whole of the pressure drop occurs in a single nozzle, and the steam passes through a series of blades attached to a single wheel or rotor. The Curtis stage impulse turbine is shown in Fig. 6.15. Fixed blades between the rows of moving blades redirect the steam flow into the next row of moving blades. Because the reduction of velocity occurs over two stages for the same pressure decreases, a Curtis-stage turbine can extract Steam Turbines 253 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved more energy from the steam than a Rateau-stage turbine. As a result, a Curtis- stage turbine has a higher efficiency than a Rateau-stage turbine. 6.11 AXIAL FLOW STEAM TURBINES Sir Charles Parsons invented the reaction steam turbine. The reaction turbine stage consists of a fixed row of blades and an equal number of moving blades fixed on a wheel. In this turbine pressure drop or expansion takes place both in the fixed blades (or nozzles) as well as in the moving blades. Because the pressure drop from inlet to exhaust is divided into many steps through use of alternate rows of fixed and moving blades, reaction turbines that have more than one stage are classified as pressure-compounded turbines. In a reaction turbine, a reactive force is produced on the moving blades when the steam increases in velocity and when the steam changes direction. Reaction turbines are normally used as Figure 6.14 Rateau-stage impulse turbine. Chapter 6 254 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 6.15 The Curtis-stage impulse turbine. Steam Turbines 255 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved low-pressure turbines. High-pressure reaction turbines are very costly because they must be constructed from heavy and expensive materials. For a 50% reaction, the fixed and moving blades have the same shape and, therefore, the velocity diagram is symmetrical as shown in Fig. 6.16. 6.12 DEGREE OF REACTION The degree of reaction or reaction ratio (L) is a parameter that describes the relation between the energy transfer due to static pressure change and the energy transfer due to dynamic pressure change. The degree of reaction is defined as the ratio of the static pressure drop in the rotor to the static pressure drop in the stage. It is also defined as the ratio of the static enthalpy drop in the rotor to the static enthalpy drop in the stage. If h 0 , h 1 , and h 2 are the enthalpies at the inlet due to the fixed blades, at the entry to the moving blades and at the exit from the moving blades, respectively, then: L = h 1 2h 2 h 0 2h 2 (6:23) The static enthalpy at the inlet to the fixed blades in terms of stagnation enthalpy and velocity at the inlet to the fixed blades is given by h 0 = h 00 2 C 2 0 2C p Similarly, h 2 = h 02 2 C 2 2 2C p Figure 6.16 Velocity triangles for 50% reaction design. Chapter 6 256 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Substituting, L = h 1 2h 2 ( ) h 00 2 C 2 0 2C p _ _ 2 h 02 2 C 2 2 2C p _ _ But for a normal stage, C 0 = C 2 and since h 00 = h 01 in the nozzle, then: L = h 1 2h 2 h 01 2h 02 (6:24) We know that h 01Re1 = h 02Re2 . Then: h 01Re1 2h 02Re2 = h 1 2h 2 ( ) - V 2 1 2V 2 2 _ _ 2 = 0 Substituting for (h 1 2 h 2 ) in Eq. (6.24): L = V 2 2 2V 2 1 _ _ 2 h 01 2h 02 ( ) [ ] L = V 2 2 2V 2 1 _ _ 2U C w1 - C w2 ( ) [ ] (6:25) Assuming the axial velocity is constant through the stage, then: L = V 2 w2 2V 2 w1 _ _ 2U U - V w1 - V w2 2U ( ) [ ] L = V w2 2V w1 ( ) V w2 - V w1 ( ) 2U V w1 - V w2 ( ) [ ] L = C a tanb 2 2tanb 1 _ _ 2U (6:26) From the velocity triangles, it is seen that C w1 = U - V w1 ; and C w2 = V w2 2U Therefore, Eq. (6.26) can be arranged into a second form: L = 1 2 - C a 2U tan b 2 2tan a 2 _ _ (6:27) Putting L = 0 in Eq. (6.26), we get b 2 = b 1 and V 1 = V 2 ; and for L = 0:5; b 2 = a 1 : Zero Reaction Stage: Let us first discuss the special case of zero reaction. According to the definition of reaction, when L = 0, Eq. (6.23) reveals that h 1 = h 2 and Eq. (6.26) that b 1 = b 2 . The Mollier diagram and velocity triangles for L = 0 are shown in Figs. 6.17 and 6.18: Steam Turbines 257 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Now, h 01r01 = h 02r01 and h 1 = h 2 for L = 0. Then, V 1 = V 2 . In the ideal case, there is no pressure drop in the rotor, and points 1, 2 and 2s on the Mollier chart shouldcoincide. But due to irreversibility, there is a pressure drop through the rotor. The zero reaction in the impulse stage, by definition, means there is no pressure drop through the rotor. The Mollier diagramfor an impulse stage is shown in Fig. 6.18, where it can be observed that the enthalpy increases through the rotor. From Eq. (6.23), it is clear that the reaction is negative for the impulse turbine stage when irreversibility is taken into account. Fifty-Percent Reaction Stage From Eq. (6.23), Fig. (6.19) for L = 0.5, a 1 = b 2 , and the velocity diagram is symmetrical. Because of symmetry, it is also clear that a 2 = b 1 . For L = 1/2, the enthalpy drop in the nozzle row equals the enthalpy drop in the rotor. That is: h 0 2h 1 = h 1 2h 2 Figure 6.17 Zero reaction (a) Mollier diagram and (b) velocity diagram. Figure 6.18 Mollier diagram for an impulse stage. Chapter 6 258 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Substituting b 2 = tan a 2 - U C a into Eq. (6.27) gives L = 1 - C a 2U tan a 2 2tan a 1 ( ) (6:28) Thus, when a 2 = a 1 , the reaction is unity (also C 1 = C 2 ). The velocity diagram for L = 1 is shown in Fig. 6.20 with the same value of C a , U, and W used for L = 0 and L = 1 2 . It is obvious that if L exceeds unity, then C 1 , C 0 (i.e., nozzle flow diffusion). Choice of Reaction and Effect on Efficiency: Eq. (6.24) can be rewritten as: L = 1 - C w2 2C w1 2U : C w2 can be eliminated by using this equation: C w2 = W U 2C w1 ; Figure 6.19 A 50% reaction stage (a) Mollier diagram and (b) velocity diagram. Figure 6.20 Velocity diagram for 100% reaction turbine stage. Steam Turbines 259 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 6.21 Influence of reaction on total-to-static efficiency with fixed values of stage-loading factor. Figure 6.22 Blade loading coefficient vs. flow coefficient. Chapter 6 260 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved yielding: L = 1 - W 2U 2 2 C w1 U (6:29) In Fig. 6.21 the total-to-static efficiencies are shown plotted against the degree of reaction. When W U 2 = 2, h ts is maximum at L = 0. With higher loading, the optimum h ts is obtained with higher reaction ratios. As shown in Fig. 6.22 for a high total-to-total efficiency, the blade-loading factor should be as small as possible, which implies the highest possible value of blade speed is consistent with blade stress limitations. It means that the total-to-static efficiency is heavily dependent upon the reaction ratio and h ts can be optimized by choosing a suitable value of reaction. 6.13 BLADE HEIGHT IN AXIAL FLOW MACHINES The continuity equation, _ m = rAC, may be used to find the blade height h. The annular area of flow = pDh. Thus, the mass flow rate through an axial flow compressor or turbine is: _ m = rpDhC a (6:30) Blade height will increase in the direction of flow in a turbine and decrease in the direction of flow in a compressor. Illustrative Example 6.6: The velocity of steam leaving a nozzle is 925 m/s and the nozzle angle is 208. The blade speed is 250 m/s. The mass flow through the turbine nozzles and blading is 0.182 kg/s and the blade velocity coefficient is 0.7. Calculate the following: 1. Velocity of whirl. 2. Tangential force on blades. 3. Axial force on blades. 4. Work done on blades. 5. Efficiency of blading. 6. Inlet angle of blades for shockless inflow of steam. Assume that the inlet and outlet blade angles are equal. Solution: From the data given, the velocity diagram can be constructed as shown in Fig. 6.23. The problem can be solved either graphically or by calculation. Steam Turbines 261 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Applying the cosine rule to the KABC, V 2 1 = U 2 - C 2 1 22UC 1 cosa 1 = 250 2 - 925 2 2(2) £ (250) £ (925) £ cos208 so: V 1 = 695:35 m/s But, k = V 2 V 1 ; orV 2 = (0:70) £ (695:35) = 487m/s: Velocity of whirl at inlet: C w1 = C 1 cosa 1 = 925cos208 = 869:22m/s Axial component at inlet: C a1 = BD = C 1 sina 1 = 925sin208 = 316:37m/s Blade angle at inlet: tanb 1 = C a1 C w1 2U = 316:37 619:22 = 0:511 Therefore, b 1 = 27.068 = b 2 = outlet blade angle. cos b 2 = C w2 - U V 2 ; or: C w2 = V 2 cos b 2 2U = 487 £ cos 27:068 2250 = 433:69 2250 = 183:69 m/s and: C a2 = FE = (U - C w2 ) tan b 2 = 433.69 tan 27.068 = 221.548 m/s Figure 6.23 Velocity triangles for Example 6.6. Chapter 6 262 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 1. Velocity of whirl at inlet, C w1 = 869.22 m/s; Velocity of whirl at outlet, C w2 = 183.69 m/s 2. Tangential force on blades = m˙ (C w1 - C w2 ) = (0.182) (1052.9) = 191.63 N. 3. Axial force on blades = _ m(C a1 2C a2 ) = (0:182) (316:372221:548) =17:26N 4. Work done on blades = tangential force on blades £ blade velocity = (191.63) £ (250)/1000 = 47.91 kW. 5. Efficiency of blading = Work done on blades Kinetic energy supplied = 47:91 1 2 mC 2 1 = (47:91)(2)(10 3 ) (0:182)(925 2 ) = 0:6153 or 61:53% 6. Inlet angle of blades b 1 = 27.068 = b 2 . Design Example 6.7: The steam velocity leaving the nozzle is 590 m/s and the nozzle angle is 208. The blade is running at 2800 rpm and blade diameter is 1050 mm. The axial velocity at rotor outlet = 155 m/s, and the blades are symmetrical. Calculate the work done, the diagram efficiency and the blade velocity coefficient. Solution: Blade speed U is given by: U = pDN 60 = (p £ 1050) £ (2800) (1000) £ (60) = 154 m/s The velocity diagram is shown in Fig. 6.24. Applying the cosine rule to the triangle ABC, V 2 1 = U 2 - C 2 1 22UC 1 cosa 1 = 154 2 - 590 2 2(2) £ (154) £ (590) cos 208 i.e. V 1 = 448:4m/s: Steam Turbines 263 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Applying the sine rule to the triangle ABC, C 1 sin (ACB) = V 1 sin(a 1 ) But sin (ACB) = sin (1808 2 b 1 ) = sin (b 1 ) Therefore, sin(b 1 ) = C 1 sin(a 1 ) V 1 = 590 sin (208) 448:4 = 0:450 and: b 1 = 26.758 From triangle ABD, C w1 = C 1 cos (a 1 ) = 590 cos (208) = 554:42 m/s From triangle CEF, C a2 U - C w2 = tan(b 2 ) = tan(b 1 ) = tan(26:758) = 0:504 or: U - C w2 = C a2 0:504 = 155 0:504 = 307:54 so : C w2 = 307:54 2154 = 153:54 m/s Therefore, DC w = C w1 - C w2 = 554:42 - 153:54 = 707:96 m/s Relative velocity at the rotor outlet is: V 2 = C a2 sin(b 2 ) = 155 sin(26:758) = 344:4 m/s Figure 6.24 Velocity diagram for Example 6.7. Chapter 6 264 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Blade velocity coefficient is: k = V 2 V 1 = 344:4 448:4 = 0:768 Work done on the blades per kg/s: DC w2 U = (707:96) £ (154) £ (10 23 ) = 109 kW The diagram efficiency is: h d = 2UDC w C 2 1 = (2) £ (707:96) £ (154) 590 2 = 0:6264 or, h d = 62:64% Illustrative Example 6.8: In one stage of an impulse turbine the velocity of steam at the exit from the nozzle is 460 m/s, the nozzle angle is 228 and the blade angle is 338. Find the blade speed so that the steam shall pass on without shock. Also find the stage efficiency and end thrust on the shaft, assuming velocity coefficient = 0.75, and blades are symmetrical. Solution: From triangle ABC (Fig. 6.25): C w1 = C 1 cos 228 = 460 cos 228 = 426:5 m/s and: C a1 = C 1 sin 228 = 460 sin 228 = 172:32 m/s Figure 6.25 Velocity triangles for Example 6.8. Steam Turbines 265 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Now, from triangle BCD: BD = C a1 tan 338 ( ) = 172:32 0:649 = 265:5 Hence, blade speed is given by: U = C w1 2BD = 426:5 2265:5 = 161 m/s From Triangle BCD, relative velocity at blade inlet is given by: V 1 = C a1 sin(338) = 172:32 0:545 = 316:2 m/s Velocity coefficient: k = V 2 V 1 ; or V 2 = kV 1 = (0:75) £ (316:2) = 237:2 m/s From Triangle BEF, BF = V 2 cos (338) = 237:2 £ cos (338) = 198:9 and C w2 = AF = BF 2U = 198:9 2161 = 37:9 m/s C a2 = V 2 sin (338) = 237:2 sin (338) = 129:2 m/s The change in velocity of whirl: DC w = C w1 - C w2 = 426:5 - 37:9 = 464:4 m/s Diagram efficiency: h d = 2UDC w C 2 1 = (2) £ (464:4) £ (161) 460 2 = 0:7067; or 70:67%: End thrust on the shaft per unit mass flow: C a1 2C a2 = 172:32 2129:2 = 43:12 N Design Example 6.9: In a Parson’s turbine, the axial velocity of flow of steam is 0.5 times the mean blade speed. The outlet angle of the blade is 208, diameter of the ring is 1.30 m and the rotational speed is 3000 rpm. Determine the inlet angles of the blades and power developed if dry saturated steam at 0.5 MPa passes through the blades where blade height is 6 cm. Neglect the effect of the blade thickness. Chapter 6 266 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: The blade speed, U = pDN 60 = p £ (1:30) £ (3000) 60 = 204 m/s Velocity of flow, C a = (0.5) £ (204) = 102 m/s Draw lines AB and CD parallel to each other Fig. 6.26 at the distance of 102 m/s, i.e., velocity of flow, C a1 = 102 m/s. At any point B, construct an angle a 2 = 208 to intersect line CD at point C. Thus, the velocity triangle at the outlet is completed. For Parson’s turbine, a 1 = b 2 ; b 1 = a 2 ; C 1 = V 2 ; and V 1 = C 2 : By measurement, DC w = C w1 - C w2 = 280:26 - 76:23 = 356:5 m/s The inlet angles are 53.228.Specific volume of vapor at 0.5 MPa, from the steam tables, is v g = 0:3749 m 3 /kg Therefore the mass flow is given by: _ m = AC 2 x 2 v g 2 = p £ (1:30) £ (6) £ (102) (100) £ (0:3749) = 66:7 kg/s Power developed: P = _ mUDC w 1000 = (66:7) £ (356:5) £ (102) 1000 = 2425:4 kW Figure 6.26 Velocity triangles for Example 6.9. Steam Turbines 267 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Design Example 6.10: In an impulse turbine, steam is leaving the nozzle with velocity of 950 m/s and the nozzle angle is 208. The nozzle delivers steam at the rate of 12 kg/min. The mean blade speed is 380 m/s and the blades are symmetrical. Neglect friction losses. Calculate (1) the blade angle, (2) the tangential force on the blades, and (3) the horsepower developed. Solution: With the help of a 1 , U and C 1 , the velocity triangle at the blade inlet can be constructed easily as shown in Fig. 6.27. Applying the cosine rule to the triangle ABC, V 2 1 = U 2 - C 2 1 22UC 1 cosa 1 = 950 2 - 380 2 2(2) £ (950) £ (380) £ cos208 = 607m/s Now, applying the sine rule to the triangle ABC, V 1 sin(a 1 ) = C 1 sin(1808 2b 1 ) = C 1 sin(b 1 ) or: sin(b 1 ) = C 1 sin(a 1 ) V 1 = (950) £ (0:342) 607 = 0:535 so: b 1 = 32:368 From Triangle ACD, C w1 = C 1 cos(a 1 ) = 950 £ cos(208) = (950) £ (0:9397) = 892:71m/s Figure 6.27 Velocity triangles for Example 6.10. Chapter 6 268 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved As b 1 = b 2 , using triangle BEF and neglecting friction loss, i.e,: V 1 = V 2 BF = V 2 cos b 2 = 607 £ cos 32:368 = 512:73 Therefore, C w2 = BF 2U = 512:73 2380 = 132:73 m/s Change in velocity of whirl: DC w = C w1 - C w2 = 892:71 - 132:73 = 1025:44 m/s Tangential force on blades: F = _ mDC w = (12) £ (1025:44) 60 = 205N Horsepower, P = _ mUDC w = (12) £ (1025:44) £ (380) (60) £ (1000) £ (0:746) = 104:47hp Design Example 6.11: In an impulse turbine, the velocity of steam at the exit from the nozzle is 700 m/s and the nozzles are inclined at 228 to the blades, whose tips are both 348. If the relative velocity of steam to the blade is reduced by 10% while passing through the blade ring, calculate the blade speed, end thrust on the shaft, and efficiency when the turbine develops 1600 kW. Solution: Velocity triangles for this problem are shown in Fig. 6.28. From the triangle ACD, C a1 = C 1 sin a 1 = 700 £ sin 228 = 262:224 m/s Figure 6.28 Velocity triangles for Example 6.11. Steam Turbines 269 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved and V 1 = C a1 sin(b 1 ) = 262:224 sin348 = 469:32 m/s Whirl component of C 1 is given by C w1 = C 1 cos(a 1 ) = 700 cos(228) = 700 £ 0:927 = 649 m/s Now, BD = C w1 2 U = V 1 cos b 1 = (469.32) £ (0.829) = 389 Hence, blade speed U = 649 2389 = 260 m/s Using the velocity coefficient to find V 2 : i:e:; V 2 = (0:90) £ (469:32) = 422:39 m/s From velocity triangle BEF, C a2 = V 2 sin(b 2 ) = 422:39 sin 348 = 236:2 m/s And U - C w2 = V 2 cos 348 = (422:39) £ (0:829) = 350:2 m/s Therefore, C w2 = 350:2 2260 = 90:2 m/s Then, DC w = C w1 - C w2 = 649 - 90:2 = 739:2 m/s Mass flow rate is given by: P = _ mUDC w or _ m = (1600) £ (1000) (739:2) £ (260) = 8:325 kg/s Thrust on the shaft, F = _ m C a1 2C a2 ( ) = 8:325(262:224 2236:2) = 216:65 N Diagram efficiency: h d = 2UDC w C 2 1 = (2) £ (739:2) £ (260) 700 2 = 0:7844; or 78:44%: Illustrative Example 6.12: The moving and fixed blades are identical in shape in a reaction turbine. The absolute velocity of steam leaving the fixed blade is 105 m/s, and the blade velocity is 40 m/s. The nozzle angle is 208. Assume axial Chapter 6 270 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved velocity is constant through the stage. Determine the horsepower developed if the steam flow rate is 2 kg/s. Solution: For 50% reaction turbine Fig. 6.29, a 1 = b 2 , and a 2 = b 1 . From the velocity triangle ACD, C w1 = C 1 cos a 1 = 105 cos 208 = 98:67 m/s Applying cosine rule to the Triangle ABC: V 2 1 = C 2 1 - U 2 22C 1 Ucosa 1 so: V 1 = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 105 2 - 40 2 2(2) £ (105) £ (40) £ cos208 _ = 68:79 m/s Now, BD = C w1 2U = V 1 cos b 1 = 98:67 240 = 58:67 Hence, cosb 1 = 58:67 68:79 = 0:853; and b 1 = 31:478 Change in the velocity of whirl is: DC w = C w1 - C w2 = 98:67 - 58:67 = 157:34 m/s Figure 6.29 Velocity triangles for Example 6.12. Steam Turbines 271 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Horsepower developed is: P = _ mUDC w = (2) £ (157:34) £ (40) (0:746) £ (1000) = 16:87 hp Illustrative Example 6.13: The inlet and outlet angles of blades of a reaction turbine are 25 and 188, respectively. The pressure and temperature of steam at the inlet to the turbine are 5 bar and 2508C. If the steam flow rate is 10 kg/s and the rotor diameter is 0.72 m, find the blade height and power developed. The velocity of steam at the exit from the fixed blades is 90 m/s. Solution: Figure 6.30 shows the velocity triangles. a 1 = b 2 = 188; and a 2 = b 1 = 258 C 1 = 90 m/s From the velocity triangle, C w1 = C 1 cos(a 1 ) = 90 cos 188 = 85:6 m/s C a1 = CD = C 1 sin a 1 = 90 sin 188 = 27:8 m/s From triangle BDC BD = C a1 sin(b 1 ) = 27:8 sin(258) = 27:8 0:423 = 65:72m/s Hence blade velocity is given by: U = C w1 2BD = 85:6 265:62 = 19:98m/s: Applying the cosine rule, V 2 1 = C 2 1 - U 2 22C 1 Ucosa 1 = 90 2 - 19:98 2 2(2) £ (90) £ (19:98) cos188 V 1 = 71:27m/s Figure 6.30 Velocity triangles for Example 6.13. Chapter 6 272 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved From triangle AEF, C w2 = C 2 cos(a 2 ) = 71:27 cos 258 = 64:59 m/s Change in the velocity of whirl: DC w = C w1 - C w2 = 85:6 - 64:59 = 150:19 m/s Power developed by the rotor: P = _ mUDC w = (10) £ (19:98) £ (150:19) 1000 = 30kW From superheated steam tables at 5 bar, 2508C, the specific volume of steam is: v = 0:4744 m 3 /kg Blade height is given by the volume of flow equation: v = pDhC a where C a is the velocity of flow and h is the blade height.Therefore, 0:4744 = p £ (0:72) £ (h) £ (27:8); and h = 0:0075 m or 0:75 cm Design Example 6.14: From the following data, for 50% reaction steam turbine, determine the blade height: RPM: 440 Power developed: 5:5 MW Steam mass flow rate: 6:8 kg/kW - h Stage absolute pressure: 0:90 bar Steam dryness fraction: 0:95 Exit angles of the blades: 708 (angle measured from the axial flow direction). The outlet relative velocity of steam is 1.2 times the mean blade speed. The ratio of the rotor hub diameter to blade height is 14.5. Steam Turbines 273 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution: Figure 6.31 shows the velocity triangles. From the velocity diagram, V 2 = 1:2U C a2 = V 2 cos(b 2 ) = 1:2Ucos708 = 0:41U m/s At mean diameter, U = pDN 60 = 2pN D h - h ( ) (60) £ (2) where D h is the rotor diameter at the hub and h is the blade height. Substituting the value of U in the above equation, C a2 = (0:41) £ (2p) £ (440) 14:5h - h ( ) (2) £ (60) = 146:45 h m/s Annular area of flow is given by: A = ph(D h - h) = ph(14:5h - h) or A = 15:5ph 2 Specific volume of saturated steam at 0.90 bar, v g = 1.869 m 3 /kg. Figure 6.31 Velocity triangles for Example 6.14. Chapter 6 274 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Then the specific volume of steam = (1.869) £ (0.95) = 1.776 m 3 /kg. The mass flow rate is given by: _ m = (5:5) £ (10 3 ) £ (6:8) 3600 = 10:39 kg/s But, _ m = C a2 A v = C a2 15:5ph 2 v Therefore: 10:39 = (146:45) £ (h) £ (15:5) £ (ph 2 ) 1:776 or: h 3 = 0:00259; and h = 0:137 m Design Example 6.15: From the following data for a two-row velocity compounded impulse turbine, determine the power developed and the diagram efficiency: Blade speed: 115 m/s Velocity of steam exiting the nozzle: 590 m/s Nozzle efflux angle: 188 Outlet angle from first moving blades: 378 Blade velocity coefficient (all blades): 0:9 Solution: Figure 6.32 shows the velocity triangles. Graphical solution: U = 115 m/s C 1 = 590 m/s a 1 = 188 b 2 = 208 The velocity diagrams are drawn to scale, as shown in Fig. 6.33, and the relative velocity: V 1 = 482 m/s using the velocity coefficient V 2 = (0.9) £ (482) = 434 m/s Steam Turbines 275 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The absolute velocity at the inlet to the second row of moving blades, C 3 , is equal to the velocity of steam leaving the fixed row of blades. i:e:; : C 3 = kC 2 = (0:9) £ (316:4) = 284:8 Driving force = m˙ DC w For the first row of moving blades, m˙ DC w1 = (1) £ (854) = 854 N. For the second row of moving blades, m˙ DC w2 = (1) £ (281.46) N = 281.46 N where DC w1 and DC w2 are scaled from the velocity diagram. Figure 6.33 Velocity diagram for Example 6.16. Figure 6.32 Velocity triangle for Example 6.15. Chapter 6 276 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Total driving force = 854 - 281.46 = 1135.46 N per kg/s Power = driving force £ blade velocity = (1135:46) £ (115) 1000 = 130:58 kW per kg/s Energy supplied to the wheel = mC 2 1 2 = (1) £ (590 2 ) (2) £ (10 3 ) = 174:05kW per kg/s Therefore, the diagram efficiency is: h d = (130:58) £ (10 3 ) £ (2) 590 2 = 0:7502; or 75:02% Maximum diagram efficiency: = cos 2 a 1 = cos 2 88 = 0:9045; or 90:45% Axial thrust on the first row of moving blades (per kg/s): = _ m(C a1 2C a2 ) = (1) £ (182:32 2148:4) = 33:9 N Axial thrust on the second row of moving blades (per kg/s): = _ m(C a3 2C a4 ) = (1) £ (111:3 297:57) = 13:73 N Total axial thrust: = 33:9 - 13:73 = 47:63 N per kg/s Design Example 6.16: In a reaction stage of a steam turbine, the blade angles for the stators and rotors of each stage are: a 1 = 258, b 1 = 608, a 2 = 71.18, b 2 = 328. If the blade velocity is 300 m/s, and the steam flow rate is 5 kg/s, find the power developed, degree of reaction, and the axial thrust. Solution: Figure 6.34 shows the velocity triangles. The velocity triangles can easily be constructed as the blade velocity and blade angles are given.From velocity triangles, work output per kg is given by: W t = U(C w1 - C w2 ) = (300) £ (450 cos 258 - 247 cos 71:18) = 14; 6; 354 J Steam Turbines 277 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Power output: _ mW t = (5) £ (1; 46; 354) 1000 = 732 kW Degree of reaction is given by: L = V 2 2 2V 2 1 2 £ W t = 443 2 2220 2 (2) £ (14; 6; 354) = 0:5051; or 50:51% Axial thrust: F = _ m(C a1 2C a2 ) = (5) £ (190:5 2234) = 2217:5 N The thrust is negative because its direction is the opposite to the fluid flow. Design Example 6.17: Steam enters the first row of a series of stages at a static pressure of 10 bars and a static temperature of 3008C. The blade angles for the rotor and stator of each stage are: a 1 = 258, b 1 = 608, a 2 = 70.28, b 2 = 328. If the blade speed is 250 m/s, and the rotor efficiency is 0.94, find the degree of reaction and power developed for a 5.2 kg/s of steam flow. Also find the static pressures at the rotor inlet and exit if the stator efficiency is 0.93 and the carry- over efficiency is 0.89. Solution: Using the given data, the velocity triangles for the inlet and outlet are shown in Fig. 6.34. By measurement, C 2 = 225 m/s, V 2 = 375 m/s, C 1 = 400 m/s, V 1 = 200 m/s. Figure 6.34 Velocity diagram for Example 6.17. Chapter 6 278 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Work done per unit mass flow: W t = (250) £ (400 cos 258 - 225 cos 70:28) = 1; 09; 685 J/kg Degree of reaction [Eq. (6.25)] L = V 2 2 2V 2 1 2 £ W t = 375 2 2200 2 (2) £ (1; 09; 685) = 0:4587; or 45:87% Power output: P = _ mW = (5:2) £ (1; 09; 685) 1000 = 570:37 kW Isentropic static enthalpy drop in the stator: Dh s / = C 2 1 2C 2 2 _ _ h s = 400 2 2(0:89) £ (225 2 ) _ _ 0:93 = 1; 23; 595 J/kg; or 123:6 kJ/kg Isentropic static enthalpy drops in the rotor: Dh r / = W h r h s = 1; 09; 685 (0:94) £ (0:93) = 1; 25; 469 J/kg; or 125:47 kJ/kg Since the state of the steam at the stage entry is given as 10 bar, 3008C, Enthalpy at nozzle exit: h 1 2 Dh / _ ¸ stator = 3051:05 2123:6 = 2927:5kJ/kg Enthalpy at rotor exit: h 1 2 Dh / _ ¸ rotor = 3051:05 2125:47 = 2925:58kJ/kg The rotor inlet and outlet conditions can be found by using the Mollier Chart. Rotor inlet conditions: P 1 = 7 bar, T 1 = 2358C Rotor outlet conditions: P 2 = 5 bar, T 2 = 2208C PROBLEMS 6.1 Dry saturated steam is expanded in a steam nozzle from 1 MPa to 0.01 MPa. Calculate dryness fraction of steam at the exit and the heat drop. (0.79, 686 kJ/kg) Steam Turbines 279 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 6.2 Steam initially dry and at 1.5 MPa is expanded adiabatically in a nozzle to 7.5 KPa. Find the dryness fraction and velocity of steamat the exit. If the exit diameter of the nozzles is 12.5 mm, find the mass of steam discharged per hour. (0.756, 1251.26 m/s, 0.376 kg/h) 6.3 Dry saturated steam expands isentropically in a nozzle from 2.5 MPa to 0.30 MPa. Find the dryness fraction and velocity of steam at the exit from the nozzle. Neglect the initial velocity of the steam. (0.862, 867.68 m/s) 6.4 The nozzles receive steam at 1.75 MPa, 3008C, and exit pressure of steam is 1.05 MPa. If there are 16 nozzles, find the cross-sectional area of the exit of each nozzle for a total discharge to be 280 kg/min. Assume nozzle efficiency of 90%. If the steam has velocity of 120 m/s at the entry to the nozzles, by how much would the discharge be increased? (1.36 cm 2 , 33.42%) 6.5 The steam jet velocity of a turbine is 615 m/s and nozzle angle is 228, The blade velocity coefficient = 0.70 and the blade is rotating at 3000 rpm. Assume mean blade radius = 600 mm and the axial velocity at the outlet = 160 m/s. Determine the work output per unit mass flowof steamand diagram efficiency. (93.43 kW, 49.4%) 6.6 Steam is supplied from the nozzle with velocity 400 m/s at an angle of 208 with the direction of motion of moving blades. If the speed of the blade is 200 m/s and there is no thrust on the blades, determine the inlet and outlet blade angles, and the power developed by the turbine. Assume velocity coefficient = 0.86, and mass flow rate of steam is 14 kg/s. (378 50 / , 458, 31 / , 1234.8 kW) 6.7 Steam expands isentropically in the reaction turbine from 4 MPa, 4008C to 0.225 MPa. The turbine efficiency is 0.84 and the nozzle angles and blade angles are 20 and 368 respectively. Assume constant axial velocity throughout the stage and the blade speed is 160 m/s. How many stages are there in the turbine? (8 stages) 6.8 Consider one stage of an impulse turbine consisting of a converging nozzle and one ring of moving blades. The nozzles are inclined at 208 to the blades, whose tip angles are both 338. If the velocity of the steam at the exit from the nozzle is 650 m/s, find the blade speed so that steam passes through without shock and find the diagram efficiency, neglecting losses. (273 m/s, 88.2%) Chapter 6 280 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 6.9 One stage of an impulse turbine consists of a converging nozzle and one ring of moving blades. The nozzle angles are 228 and the blade angles are 358. The velocity of steam at the exit from the nozzle is 650 m/s. If the relative velocity of steam to the blades is reduced by 14% in passing through the blade ring, find the diagram efficiency and the end thrust on the shaft when the blade ring develops 1650 kW. (79.2%, 449 N) 6.10 The following refer to a stage of a Parson’s reaction turbine: Mean diameter of the blade ring: 92 cm Blade speed: 3000 rpm Inlet absolute velocity of steam: 310 m/s Blade outlet angle: 208 Steam flow rate: 6:9 kg/s Determine the following: (1) blade inlet angle, (2) tangential force, and (3) power developed. (388, 2.66 kW, 384.7 kW) NOTATION C absolute velocity, velocity of steam at nozzle exit D diameter h enthalpy, blade height h 0 stagnation enthalpy, static enthalpy at the inlet to the fixed blades h 1 enthalpy at the entry to the moving blades h 2 enthalpy at the exit from the moving blades h 00 stagnation enthalpy at the entry to the fixed blades h 0l stagnation enthalpy at the entry to the fixed blades h 02 stagnation enthalpy at the exit from the moving blade k blade velocity coefficient N rotational speed R. F. reheat factor U blade speed V relative velocity a angle with absolute velocity b angle with relative velocity DC w change in the velocity of whirl Dh actual enthalpy drop Dh / isentropic enthalpy drop Steam Turbines 281 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved h d diffuser efficiency h n nozzle efficiency h s stage efficiency h t turbine efficiency h ts total - to - static efficiency h tt total - to - total efficiency L degree of reaction SUFFIXES 0 inlet to fixed blades 1 inlet to moving blades 2 outlet from the moving blades a axial, ambient r radial w whirl Chapter 6 282 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 7 Axial Flow and Radial Flow Gas Turbines 7.1 INTRODUCTION TO AXIAL FLOW TURBINES The axial flow gas turbine is used in almost all applications of gas turbine power plant. Development of the axial flow gas turbine was hindered by the need to obtain both a high-enough flow rate and compression ratio from a compressor to maintain the air requirement for the combustion process and subsequent expansion of the exhaust gases. There are two basic types of turbines: the axial flow type and the radial or centrifugal flow type. The axial flow type has been used exclusively in aircraft gas turbine engines to date and will be discussed in detail in this chapter. Axial flow turbines are also normally employed in industrial and shipboard applications. Figure 7.1 shows a rotating assembly of the Rolls- Royce Nene engine, showing a typical single-stage turbine installation. On this particular engine, the single-stage turbine is directly connected to the main and cooling compressors. The axial flow turbine consists of one or more stages located immediately to the rear of the engine combustion chamber. The turbine extracts kinetic energy from the expanding gases as the gases come from the burner, converting this kinetic energy into shaft power to drive the compressor and the engine accessories. The turbines can be classified as (1) impulse and (2) reaction. In the impulse turbine, the gases will be expanded in the nozzle and passed over to the moving blades. The moving blades convert this kinetic energy into mechanical energy and also direct the gas flow to the next stage Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved (multi-stage turbine) or to exit (single-stage turbine). Fig. 7.1 shows the axial flow turbine rotors. In the case of reaction turbine, pressure drop of expansion takes place in the stator as well as in the rotor-blades. The blade passage area varies continuously to allow for the continued expansion of the gas stream over the rotor-blades. The efficiency of a well-designed turbine is higher than the efficiency of a compressor, and the design process is often much simpler. The main reason for this fact, as discussed in compressor design, is that the fluid undergoes a pressure rise in the compressor. It is much more difficult to arrange for an efficient deceleration of flow than it is to obtain an efficient acceleration. The pressure drop in the turbine is sufficient to keep the boundary layer fluid well behaved, and separation problems, or breakaway of the molecules from the surface, which often can be serious in compressors, can be easily avoided. However, the turbine designer will face much more critical stress problem because the turbine rotors must operate in very high-temperature gases. Since the design principle and concepts of gas turbines are essentially the same as steam turbines, additional Figure 7.1 Axial flow turbine rotors. (Courtesy Rolls-Royce.) Chapter 7 284 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved information on turbines in general already discussed in Chapter 6 on steam turbines. 7.2 VELOCITY TRIANGLES AND WORK OUTPUT The velocity diagram at inlet and outlet from the rotor is shown in Fig. 7.2. Gas with an absolute velocity C 1 making an angle a 1 , (angle measured from the axial direction) enters the nozzle (in impulse turbine) or stator blades (in reaction turbine). Gas leaves the nozzles or stator blades with an absolute velocity C 2 , which makes and an a 2 with axial direction. The rotor-blade inlet angle will be chosen to suit the direction b 2 of the gas velocity V 2 relative to the blade at inlet. b 2 and V 2 are found by subtracting the blade velocity vector U from the absolute velocity C 2 . It is seen that the nozzles accelerate the flow, imparting an increased tangential velocity component. After expansion in the rotor-blade passages, the gas leaves with relative velocity V 3 at angle b 3 . The magnitude and direction of the absolute velocity at exit from the rotor C 3 at an angle a 3 are found by vectorial addition of U to the relative velocity V 3 . a 3 is known as the swirl angle. Figure 7.2 Velocity triangles for an axial flow gas turbine. Axial Flow and Radial Flow Gas Turbines 285 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The gas enters the nozzle with a static pressure p 1 and temperature T 1 . After expansion, the gas pressure is p 2 and temperature T 2 . The gas leaves the rotor- blade passages at pressure p 3 and temperature T 3 . Note that the velocity diagram of the turbine differs from that of the compressor, in that the change in tangential velocity in the rotor, DC w , is in the direction opposite to the blade speed U. The reaction to this change in the tangential momentum of the fluid is a torque on the rotor in the direction of motion. V 3 is either slightly less than V 2 (due to friction) or equal to V 2 . But in reaction stage, V 3 will always be greater than V 2 because part of pressure drop will be converted into kinetic energy in the moving blade. The blade speed U increases from root to tip and hence velocity diagrams will be different for root, tip, and other radii points. For short blades, 2-D approach in design is valid but for long blades, 3-D approach in the designing must be considered. We shall assume in this section that we are talking about conditions at the mean diameter of the annulus. Just as with the compressor blading diagram, it is more convenient to construct the velocity diagrams in combined form, as shown in Fig. 7.3. Assuming unit mass flow, work done by the gas is given by W ¼ U C w2 þ C w3 ð Þ ð7:1Þ From velocity triangle U Ca ¼ tan a 2 2 tan b 2 ¼ tan b 3 2 tan a 3 ð7:2Þ In single-stage turbine, a 1 ¼ 0 and C 1 ¼ Ca 1 . In multi-stage turbine, a 1 ¼ a 3 and C 1 ¼ C 3 so that the same blade shape can be used. In terms of air angles, the stage Figure 7.3 Combined velocity diagram. Chapter 7 286 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved work output per unit mass flow is given by W ¼ U C w2 þC w3 ð Þ ¼ UCa tan a 2 þ tan a 3 ð Þ ð7:3Þ or W ¼ UCa tan b 2 þ tan b 3 _ _ ð7:4Þ Work done factor used in the designing of axial flow compressor is not required because in the turbine, flow is accelerating and molecules will not break away from the surface and growth of the boundary layer along the annulus walls is negligible. The stagnation pressure ratio of the stage p 01 /p 03 can be found from DT 0s ¼ h s T 01 1 2 1 p 01 /p 03 _ _ g21 ð Þ/g _ _ ð7:5Þ where h s is the isentropic efficiency given by h s ¼ T 01 2T 03 T 01 2T 0 03 ð7:6Þ The efficiency given by Eq. (7.6) is based on stagnation (or total) temperature, and it is known as total-to-total stage efficiency. Total-to-total stage efficiency term is used when the leaving kinetics energy is utilized either in the next stage of the turbine or in propelling nozzle. If the leaving kinetic energy from the exhaust is wasted, then total-to-static efficiency term is used. Thus total- to-static efficiency, h ts ¼ T 01 2T 03 T 01 2T 0 3 ð7:7Þ where T 0 3 in Eq. (7.7) is the static temperature after an isentropic expansion from p 01 to p 3 . 7.3 DEGREE OF REACTION (L) Degree of reaction is defined as L ¼ Enthalpy drop in the moving blades Enthalpy drop in the stage ¼ h 2 2h 3 h 1 2h 3 ¼ Ca 2U tan b 1 2 tan b 2 _ _ ð7:8Þ This shows the fraction of the stage expansion, which occurs in the rotor, and it is usual to define in terms of the static temperature drops, namely L ¼ T 2 2T 3 T 1 2T 3 ð7:9Þ Axial Flow and Radial Flow Gas Turbines 287 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Assuming that the axial velocity is constant throughout the stage, then Ca 2 ¼ Ca 3 ¼ Ca 1 ; and C 3 ¼ C 1 From Eq. (7.4) C p T 1 2T 3 ð Þ ¼ C p T 01 2T 03 ð Þ ¼ UCa tan b 2 þ tan b 3 _ _ ð7:10Þ Temperature drop across the rotor-blades is equal to the change in relative velocity, that is C p T 2 2T 3 ð Þ ¼ 1 2 V 2 3 2V 2 2 _ _ ¼ 1 2 Ca 2 sec 2 b 3 2sec 2 b 2 _ _ ¼ 1 2 Ca 2 tan 2 b 3 2tan 2 b 2 _ _ Thus L ¼ Ca 2U tan b 3 2 tan b 2 _ _ ð7:11Þ 7.4 BLADE-LOADING COEFFICIENT The blade-loading coefficient is used to express work capacity of the stage. It is defined as the ratio of the specific work of the stage to the square of the blade velocity—that is, the blade-loading coefficient or temperature-drop coefficient c is given by c ¼ W 1 2 U 2 ¼ 2C p DT os U 2 ¼ 2Ca U tan b 2 þ tan b 3 _ _ ð7:12Þ Flow Coefficient (f) The flow coefficient, f, is defined as the ratio of the inlet velocity Ca to the blade velocity U, i.e., f ¼ Ca U ð7:13Þ This parameter plays the same part as the blade-speed ratio U/C 1 used in the design of steam turbine. The two parameters, c and f, are dimensionless and Chapter 7 288 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved useful to plot the design charts. The gas angles in terms of c, L, and f can be obtained easily as given below: Eqs. (7.11) and (7.12) can be written as c ¼ 2f tan b 2 þ tan b 3 _ _ ð7:14Þ L ¼ f 2 tan b 3 2 tan b 2 _ _ ð7:15Þ Now, we may express gas angles b 2 and b 3 in terms of c, L, and f as follows: Adding and subtracting Eqs. (7.14) and (7.15), we get tan b 3 ¼ 1 2f 1 2 c þ 2L _ _ ð7:16Þ tan b 2 ¼ 1 2f 1 2 c 22L _ _ ð7:17Þ Using Eq. (7.2) tan a 3 ¼ tan b 3 2 1 f ð7:18Þ tan a 2 ¼ tan b 2 þ 1 f ð7:19Þ It has been discussed in Chapter 6 that steam turbines are usually impulse or a mixture of impulse and reaction stages but the turbine for a gas-turbine power plant is a reaction type. In the case of steam turbine, pressure ratio can be of the order of 1000:1 but for a gas turbine it is in the region of 10:1. Now it is clear that a very long steam turbine with many reaction stages would be required to reduce the pressure by a ratio of 1000:1. Therefore the reaction stages are used where pressure drop per stage is low and also where the overall pressure ratio of the turbine is low, especially in the case of aircraft engine, which may have only three or four reaction stages. Let us consider 50% reaction at mean radius. Substituting L ¼ 0.5 in Eq. (7.11), we have 1 f ¼ tan b 3 2 tan b 2 ð7:20Þ Comparing this with Eq. (7.2), b 3 ¼ a 2 and b 2 ¼ a 3 , and hence the velocity diagram becomes symmetrical. Now considering C 1 ¼ C 3 , we have a 1 ¼ a 3 ¼ b 2 , and the stator and rotor-blades then have the same inlet and outlet angles. Finally, for L ¼ 0.5, we can prove that c ¼ 4ftan b 3 22 ¼ 4ftan a 2 22 ð7:21Þ and c ¼ 4ftan b 2 þ 2 ¼ 4ftan a 3 þ 2 ð7:22Þ and hence all the gas angles can be obtained in terms of c and f. Axial Flow and Radial Flow Gas Turbines 289 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The low values of f and c imply low gas velocities and hence reduced friction losses. But a low value of c means more stages for a given overall turbine output, and low f means larger turbine annulus area for a given mass flow. In industrial gas turbine plants, where lowsfc is required, a large diameter, relatively long turbine, of low flow coefficient and low blade loading, would be accepted. However, for the gas turbine used in an aircraft engine, the primary consideration is to have minimum weight, and a small frontal area. Therefore it is necessary to use higher values of c and f but at the expense of efficiency (see Fig. 7.4). 7.5 STATOR (NOZZLE) AND ROTOR LOSSES A T–s diagram showing the change of state through a complete turbine stage, including the effects of irreversibility, is given in Fig. 7.5. In Fig. 7.5, T 02 ¼ T 01 because no work is done in the nozzle, ðp 01 2p 02 Þ represents the pressure drop due to friction in the nozzle. ðT 01 2 T 0 2 Þ represents the ideal expansion in the nozzle, T 2 is the temperature at the nozzle exit due to friction. Temperature, T 2 at the nozzle exit is higher than T 0 2 . The nozzle loss coefficient, l N , in terms of temperature may be defined as l N ¼ T 2 2T 0 2 C 2 2 /2C p ð7:23Þ Figure 7.4 Total-to-static efficiency of a 50% reaction axial flow turbine stage. Chapter 7 290 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Nozzle loss coefficient in term of pressure y N ¼ p 01 2p 02 p 01 2p 2 ð7:24Þ l N and y N are not very different numerically. From Fig. 7.5, further expansion in the rotor-blade passages reduces the pressure to p 3 . T 0 3 is the final temperature after isentropic expansion in the whole stage, and T 00 3 is the temperature after expansion in the rotor-blade passages alone. Temperature T 3 represents the temperature due to friction in the rotor-blade passages. The rotor-blade loss can be expressed by l R ¼ T 3 2T 00 3 V 2 3 /2C p ð7:25Þ As we know that no work is done by the gas relative to the blades, that is, T 03rel ¼ T 02rel . The loss coefficient in terms of pressure drop for the rotor-blades is defined by l R ¼ p 02 rel 2p 03 rel p 03 rel 2p 3 ð7:26Þ Figure 7.5 T–s diagram for a reaction stage. Axial Flow and Radial Flow Gas Turbines 291 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The loss coefficient in the stator and rotor represents the percentage drop of energy due to friction in the blades, which results in a total pressure and static enthalpy drop across the blades. These losses are of the order of 10–15% but can be lower for very low values of flow coefficient. Nozzle loss coefficients obtained from a large number of turbine tests are typically 0.09 and 0.05 for the rotor and stator rows, respectively. Figure 7.4 shows the effect of blade losses, determined with Soderberg’s correlation, on the total-to- total efficiency of turbine stage for the constant reaction of 50%. It is the evident that exit losses become increasingly dominant as the flow coefficient is increased. 7.6 FREE VORTEX DESIGN As pointed out earlier, velocity triangles vary from root to tip of the blade because the blade speed U is not constant and varies from root to tip. Twisted blading designed to take account of the changing gas angles is called vortex blading. As discussed in axial flow compressor (Chapter 5) the momentum equation is 1 r dP dr ¼ C 2 w r ð7:27Þ For constant enthalpy and entropy, the equation takes the form dh 0 dr ¼ Ca dCa dr þ C w dC w dr þ C 2 w r ð7:28Þ For constant stagnation enthalpy across the annulus (dh 0 /dr ¼ 0) and constant axial velocity (dCa/dr ¼ 0) then the whirl component of velocity C w is inversely proportional to the radius and radial equilibrium is satisfied. That is, C w £ r ¼ constant ð7:29Þ The flow, which follows Eq. (7.29), is called a “free vortex.” Now using subscript m to denote condition at mean diameter, the free vortex variation of nozzle angle a 2 may be found as given below: C w2 r ¼ rCa 2 tan a 2 ¼ constant Ca 2 ¼ constant Therefore a 2 at any radius r is related to a 2m at the mean radius r m by tan a 2 ¼ r m r _ _ 2 tan a 2m ð7:30Þ Similarly, a 3 at outlet is given by tan a 3 ¼ r m r _ _ 3 tan a 3m ð7:31Þ Chapter 7 292 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The gas angles at inlet to the rotor-blade, from velocity triangle, tan b 3 ¼ tan a 2 2 U Ca ¼ r m r _ _ 2 tan a 2m 2 r r m _ _ 2 U m Ca 2 ð7:32Þ and b 3 is given by tan b 2 ¼ r m r _ _ 3 tan a 3m þ r r m _ _ 3 U m Ca 3 ð7:33Þ 7.7 CONSTANT NOZZLE ANGLE DESIGN As before, we assume that the stagnation enthalpy at outlet is constant, that is, dh 0 /dr ¼ 0. If a 2 is constant, this leads to the axial velocity distribution given by C w2 r sin 2 a 2 ¼ constant ð7:34Þ and since a 2 is constant, then Ca 2 is proportional to C w1 . Therefore C a2 r sin 2 a 2 ¼ constant ð7:35Þ Normally the change in vortex design has only a small effect on the performance of the blade while secondary losses may actually increase. Illustrative Example 7.1 Consider an impulse gas turbine in which gas enters at pressure ¼ 5.2 bar and leaves at 1.03 bar. The turbine inlet temperature is 1000 K and isentropic efficiency of the turbine is 0.88. If mass flow rate of air is 28 kg/s, nozzle angle at outlet is 578, and absolute velocity of gas at inlet is 140 m/s, determine the gas velocity at nozzle outlet, whirl component at rotor inlet and turbine work output. Take, g ¼ 1.33, and C pg ¼ 1.147 kJ/kgK (see Fig. 7.6). Figure 7.6 T-s diagram for Example 7.1. Axial Flow and Radial Flow Gas Turbines 293 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution From isentropic p–T relation for expansion process T 0 02 T 01 ¼ p 02 p 01 _ _ ðg21Þ/g or T 0 02 ¼ T 01 p 02 p 01 _ _ ðg21Þ/g ¼ 1000 1:03 5:2 _ _ ð0:248Þ ¼ 669 K Using isentropic efficiency of turbine T 02 ¼ T 01 2h t T 01 2T 0 02 _ _ ¼ 1000 20:88 1000 2669 ð Þ ¼ 708:72 K Using steady-flow energy equation 1 2 C 2 2 2C 2 1 _ _ ¼ C p T 01 2T 02 ð Þ Therefore, C 2 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2Þð1147Þ 1000 2708:72 ð Þ þ 19600 ½ _ ¼ 829:33 m/s From velocity triangle, velocity of whirl at rotor inlet C w2 ¼ 829:33 sin 578 ¼ 695:5 m/s Turbine work output is given by W t ¼ mC pg T 01 2T 02 ð Þ ¼ ð28Þð1:147Þ 1000 2708:72 ð Þ ¼ 9354:8 kW Design Example 7.2 In a single-stage gas turbine, gas enters and leaves in axial direction. The nozzle efflux angle is 688, the stagnation temperature and stagnation pressure at stage inlet are 8008C and 4 bar, respectively. The exhaust static pressure is 1 bar, total-to-static efficiency is 0.85, and mean blade speed is 480 m/s, determine (1) the work done, (2) the axial velocity which is constant through the stage, (3) the total-to-total efficiency, and (4) the degree of reaction. Assume g ¼ 1.33, and C pg ¼ 1.147 kJ/kgK. Solution (1) The specific work output W ¼ C pg T 01 2T 03 ð Þ ¼ h ts C pg T 01 1 2 1/4 ð Þ 0:33/1:33 _ ¸ ¼ ð0:85Þð1:147Þð1073Þ 1 2ð0:25Þ 0:248 _ ¸ ¼ 304:42 kJ/kg Chapter 7 294 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved (2) Since a 1 ¼ 0, a 3 ¼ 0, C w1 ¼ 0 and specific work output is given by W ¼ UC w2 or C w2 ¼ W U ¼ 304:42 £ 1000 480 ¼ 634:21 m/s From velocity triangle sin a 2 ¼ C w2 C 2 or C 2 ¼ C w2 sin a 2 ¼ 634:21 sin 688 ¼ 684 m/s Axial velocity is given by Ca 2 ¼ 684 cos 688 ¼ 256:23 m/s (3) Total-to-total efficiency, h tt , is h tt ¼ T 01 2T 03 T 01 2T 0 03 ¼ w s T 01 2 T 3 þ C 2 3 2C pg _ _ ¼ w s w s h ts 2 C 2 3 2C pg ¼ 304:42 304:42 0:85 2 256:23 ð Þ 2 2 £ 1147 ¼ 92:4% (4) The degree of reaction L ¼ Ca 2U tan b 3 2 tan b 2 _ _ ¼ Ca 2U £ U Ca _ _ 2 Ca 2U tan a 2 _ _ þ U Ca £ Ca 2U _ _ (from velocity triangle) L ¼ 1 2 Ca 2U tan a 2 ¼ 1 2 256:23 ð2Þð480Þ tan 688 ¼ 33:94% Design Example 7.3 In a single-stage axial flow gas turbine gas enters at stagnation temperature of 1100 K and stagnation pressure of 5 bar. Axial velocity is constant through the stage and equal to 250 m/s. Mean blade speed is 350 m/s. Mass flow rate of gas is 15 kg/s and assume equal inlet and outlet velocities. Nozzle efflux angle is 638, stage exit swirl angle equal to 98. Determine the rotor- blade gas angles, degree of reaction, and power output. Axial Flow and Radial Flow Gas Turbines 295 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution Refer to Fig. 7.7. Ca 1 ¼ Ca 2 ¼ Ca 3 ¼ Ca ¼ 250 m/s From velocity triangle (b) C 2 ¼ Ca 2 cos a 2 ¼ 250 cos 638 ¼ 550:67 m/s From figure (c) C 3 ¼ Ca 3 cos a 3 ¼ 250 cos 98 ¼ 253 m/s C w3 ¼ Ca 3 tan a 3 ¼ 250 tan 98 ¼ 39:596 m/s tan b 3 ¼ U þC w3 Ca 3 ¼ 350 þ 39:596 250 ¼ 1:5584 i:e:; b 3 ¼ 57:318 From figure (b) C w2 ¼ Ca 2 tan a 2 ¼ 250 tan 638 ¼ 490:65 m/s Figure 7.7 Velocity triangles for Example 7.3. Chapter 7 296 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved and tan b 2 ¼ C w2 2U Ca 2 ¼ 490:65 2350 250 ¼ 0:5626 [ b 2 ¼ 29821 0 Power output W ¼ mUCað tan b 2 þ tan b 3 Þ ¼ ð15Þð350Þð250Þð0:5626 þ 1:5584Þ/1000 ¼ 2784 kW The degree of reaction is given by L ¼ Ca 2U tan b 3 2 tan b 2 _ _ ¼ 250 2 £ 350 1:5584 20:5626 ð Þ ¼ 35:56% Design Example 7.4 Calculate the nozzle throat area for the same data as in the precious question, assuming nozzle loss coefficient, T N ¼ 0.05. Take g ¼ 1.333, and C pg ¼ 1.147 kJ/kgK. Solution Nozzle throat area, A ¼ m/r 2 Ca 2 and r 2 ¼ p 2 RT 2 T 2 ¼ T 02 2 C 2 2 2C p ¼ 1100 2 550:67 ð Þ 2 ð2Þð1:147Þð1000Þ T 01 ¼ T 02 ð Þ i.e., T 2 ¼ 967:81 K From nozzle loss coefficient T 0 2 ¼ T 2 2l N C 2 2 2C p ¼ 967:81 2 0:05 £ 550:67 ð Þ 2 ð2Þð1:147Þð1000Þ ¼ 961:2 K Using isentropic p–T relation for nozzle expansion p 2 ¼ p 01 = T 01 /T 0 2 _ _ g/ g21 ð Þ ¼ 5/ 1100/961:2 ð Þ 4 ¼ 2:915 bar Axial Flow and Radial Flow Gas Turbines 297 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Critical pressure ratio p 01 /p c ¼ g þ 1 2 _ _ g/ g21 ð Þ ¼ 2:333 2 _ _ 4 ¼ 1:852 or p 01 /p 2 ¼ 5/2:915 ¼ 1:715 Since p 01 p 2 , p 01 p c , and therefore nozzle is unchoked. Hence nozzle gas velocity at nozzle exit C 2 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2C pg T 01 2T 2 ð Þ _ ¸ _ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2Þð1:147Þð1000Þ 1100 2967:81 ð Þ ½ _ ¼ 550:68 m/s Therefore, nozzle throat area A ¼ m r 2 C 2 ; and r 2 ¼ p 2 RT 2 ¼ ð2:915Þð10 2 Þ ð0:287Þð967:81Þ ¼ 1:05 kg/m 3 Thus A ¼ 15 ð1:05Þð550:68Þ ¼ 0:026 m 2 Design Example 7.5 In a single-stage turbine, gas enters and leaves the turbine axially. Inlet stagnation temperature is 1000 K, and pressure ratio is 1.8 bar. Gas leaving the stage with velocity 270 m/s and blade speed at root is 290 m/s. Stage isentropic efficiency is 0.85 and degree of reaction is zero. Find the nozzle efflux angle and blade inlet angle at the root radius. Solution Since L ¼ 0, therefore L ¼ T 2 2T 3 T 1 2T 3 ; hence T 2 ¼ T 3 From isentropic p–T relation for expansion T 0 03 ¼ T 01 p 01 /p 03 _ _ g21 ð Þ/g ¼ 1000 1:8 ð Þ 0:249 ¼ 863:558 K Chapter 7 298 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Using turbine efficiency T 03 ¼ T 01 2h t T 01 2T 0 03 _ _ ¼ 1000 20:85ð1000 2863:558Þ ¼ 884 K In order to find static temperature at turbine outlet, using static and stagnation temperature relation T 3 ¼ T 03 2 C 2 3 2C pg ¼ 884 2 270 2 ð2Þð1:147Þð1000Þ ¼ 852 K ¼ T 2 Dynamic temperature C 2 2 2C pg ¼ 1000 2T 2 ¼ 1000 2852 ¼ 148 K C 2 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2Þð1:147Þð148Þð1000Þ ½ _ ¼ 582:677 m/s Since, C pg DT os ¼ U C w3 þC w2 ð Þ ¼ UC w2 ðC w3 ¼ 0Þ Therefore, C w2 ¼ ð1:147Þð1000Þ 1000 2884 ð Þ 290 ¼ 458:8 m/s From velocity triangle sin a 2 ¼ C w2 C 2 ¼ 458:8 582:677 ¼ 0:787 That is, a 2 ¼ 51854 0 tan b 2 ¼ C w2 2U Ca 2 ¼ 458:8 2290 C 2 cos a 2 ¼ 458:8 2290 582:677 cos 51:908 ¼ 0:47 i.e., b 2 ¼ 2589 0 Design Example 7.6 In a single-stage axial flowgas turbine, gas enters the turbine at a stagnation temperature and pressure of 1150 K and 8 bar, respectively. Isentropic efficiency of stage is equal to 0.88, mean blade speed is 300 m/s, and rotational speed is 240 rps. The gas leaves the stage with velocity 390 m/s. Assuming inlet and outlet velocities are same and axial, find the blade height at the outlet conditions when the mass flowof gas is 34 kg/s, and temperature drop in the stage is 145 K. Axial Flow and Radial Flow Gas Turbines 299 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Solution Annulus area A is given by A ¼ 2 pr m h where h ¼ blade height r m ¼ mean radius As we have to find the blade height from the outlet conditions, in this case annulus area is A 3 . [ h ¼ A 3 2 pr m U m ¼ pD m N or D m ¼ ðU m Þ pN ¼ 300 ð pÞð240Þ ¼ 0:398 i.e., r m ¼ 0:199 m Temperature drop in the stage is given by T 01 2T 03 ¼ 145 K Hence T 03 ¼ 1150 2145 ¼ 1005 K T 3 ¼ T 03 2 C 2 3 2C pg ¼ 1005 2 390 2 ð2Þð1:147Þð1000Þ ¼ 938:697 K Using turbine efficiency to find isentropic temperature drop T 0 03 ¼ 1150 2 145 0:88 ¼ 985:23 K Using isentropic p–T relation for expansion process p 03 ¼ p 01 T 01 /T 0 03 _ _ g/ g21 ð Þ ¼ 8 1150/985:23 ð Þ 4 ¼ 8 1:856 i.e., p 03 ¼ 4:31 bar Also from isentropic relation p 3 ¼ p 03 T 0 03 /T 3 _ _ g/ g21 ð Þ ¼ 4:31 985:23/938:697 ð Þ 4 ¼ 4:31 1:214 ¼ 3:55 bar r 3 ¼ p 3 RT 3 ¼ ð3:55Þð100Þ ð0:287Þð938:697Þ ¼ 1:32 kg/m 3 A 3 ¼ m r 3 Ca 3 ¼ 34 ð1:32Þð390Þ ¼ 0:066 m 2 Chapter 7 300 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Finally, h ¼ A 3 2 pr m ¼ 0:066 ð2 pÞð0:199Þ ¼ 0:053 m Design Example 7.7 The following data refer to a single-stage axial flow gas turbine with convergent nozzle: Inlet stagnation temperature; T 01 1100 K Inlet stagnation pressure; p 01 4 bar Pressure ratio; p 01 /p 03 1:9 Stagnation temperature drop 145 K Mean blade speed 345 m/s Mass flow; m 24 kg/s Rotational speed 14; 500 rpm Flow coefficient; F 0:75 Angle of gas leaving the stage 128 C pg ¼ 1147 J/kg K; g ¼ 1:333; l N ¼ 0:05 Assuming the axial velocity remains constant and the gas velocity at inlet and outlet are the same, determine the following quantities at the mean radius: (1) The blade loading coefficient and degree of reaction (2) The gas angles (3) The nozzle throat area Solution ð1Þ C ¼ C pg T 01 2T 03 ð Þ U 2 ¼ ð1147Þð145Þ 345 2 ¼ 1:4 Using velocity diagram U/Ca ¼ tan b 3 2 tan a 3 or tan b 3 ¼ 1 F þ tan a 3 ¼ 1 0:75 þ tan 128 b 3 ¼ 57:18 From Equations (7.14) and (7.15), we have C ¼ F tan b 2 þ tan b 3 _ _ Axial Flow and Radial Flow Gas Turbines 301 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved and L ¼ F 2 tan b 3 2 tan b 2 _ _ From which tan b 3 ¼ 1 2F Cþ 2L ð Þ Therefore tan 57:18 ¼ 1 2 £ 0:75 1:4 þ 2L ð Þ Hence L ¼ 0:4595 ð2Þ tan b 2 ¼ 1 2F C22L ð Þ ¼ 1 2 £ 0:75 1:4 2½2½0:459 ð Þ b 2 ¼ 17:88 tan a 2 ¼ tan b 2 þ 1 F ¼ tan 17:88 þ 1 0:75 ¼ 0:321 þ 1:33 ¼ 1:654 a 2 ¼ 58:88 ð3Þ Ca 1 ¼ UF ¼ ð345Þð0:75Þ ¼ 258:75 m/s C 2 ¼ Ca 1 cos a 2 ¼ 258:75 cos 58:88 ¼ 499:49 m/s T 02 2T 2 ¼ C 2 2 2C p ¼ 499:49 2 ð2Þð1147Þ ¼ 108:76 K T 2 2T 2s ¼ ðT N Þð499:49 2 Þ ð2Þð1147Þ ¼ ð0:05Þð499:49 2 Þ ð2Þð1147Þ ¼ 5:438 K T 2s ¼ T 2 25:438 T 2 ¼ 1100 2108:76 ¼ 991:24 K T 2s ¼ 991:24 25:438 ¼ 985:8 K p 01 p 2 ¼ T 01 T 2s _ _ g=ðg21Þ Chapter 7 302 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved p 2 ¼ 4 £ 985:8 1100 _ _ 4 ¼ 2:58 r 2 ¼ p 2 RT 2 ¼ ð2:58Þð100Þ ð0:287Þð991:24Þ ¼ 0:911 kg /m 3 ð4Þ Nozzle throat area ¼ m r 1 C 1 ¼ 24 ð0:907Þð499:49Þ ¼ 0:053 m 2 A 1 ¼ m r 1 Ca 1 ¼ 24 ð0:907Þð258:75Þ ¼ 0:102 m 2 Design Example 7.8 A single-stage axial flow gas turbine with equal stage inlet and outlet velocities has the following design data based on the mean diameter: Mass flow 20 kg/s Inlet temperature; T 01 1150K Inlet pressure 4 bar Axial flow velocity constant through the stage 255 m/s Blade speed; U 345 m/s Nozzle efflux angle; a 2 608 Gas-stage exit angle 128 Calculate (1) the rotor-blade gas angles, (2) the degree of reaction, blade- loading coefficient, and power output and (3) the total nozzle throat area if the throat is situated at the nozzle outlet and the nozzle loss coefficient is 0.05. Solution (1) From the velocity triangles C w2 ¼ Ca tan a 2 ¼ 255 tan 608 ¼ 441:67 m/s C w3 ¼ Ca tan a 3 ¼ 255 tan 128 ¼ 55:2 m/s V w2 ¼ C w2 2U ¼ 441:67 2345 ¼ 96:67 m/s b 2 ¼ tan 21 V w2 Ca ¼ tan 21 96:67 255 ¼ 20:88 Also V w3 ¼ C w3 þU ¼ 345 þ 55:2 ¼ 400:2 m/s [ b 3 ¼ tan 21 V w3 Ca ¼ tan 21 400:2 255 ¼ 57:58 Axial Flow and Radial Flow Gas Turbines 303 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved ð2Þ L ¼ F 2 tan b 3 2 tan b 2 _ _ ¼ 255 2 £ 345 tan 57:58 2 tan 20:88 ð Þ ¼ 0:44 C ¼ Ca U tan b 2 þ tan b 3 _ _ ¼ 255 345 tan 20:88 þ tan 57:58 ð Þ ¼ 1:44 Power W ¼ mU C w2 þ C w3 ð Þ ¼ ð20Þð345Þ 441:67 þ 54:2 ð Þ ¼ 3421:5 kW ð3Þ l N ¼ C p T 2 2T 0 2 _ _ 1 2 C 2 2 ; C 2 ¼ Ca seca 2 ¼ 255sec608 ¼ 510 m/s or T 2 2T 0 2 ¼ ð0:05Þð0:5Þð510 2 Þ 1147 ¼ 5:67 T 2 ¼ T 02 2 C 2 2 2C p ¼ 1150 2 510 2 ð2Þð1147Þ ¼ 1036:6 K T 0 2 ¼ 1036:6 25:67 ¼ 1030:93 K p 01 p 2 ¼ T 01 T 2 _ _ g= g21 ð Þ ¼ 1150 1030:93 _ _ 4 ¼ 1:548 p 2 ¼ 4 1:548 ¼ 2:584 bar r 2 ¼ p 2 RT 2 ¼ 2:584 £ 100 0:287 £ 1036:6 ¼ 0:869 kg/m 3 m ¼ r 2 A 2 C 2 A 2 ¼ 20 0:869 £ 510 ¼ 0:045 m 2 Chapter 7 304 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Illustrative Example 7.9 A single-stage axial flow gas turbine has the following data Mean blade speed 340 m/s Nozzle exit angle 158 Axial velocity ðconstantÞ 105 m/s Turbine inlet temperature 9008C Turbine outlet temperature 6708C Degree of reaction 50% Calculate the enthalpy drop per stage and number of stages required. Solution At 50%, a 2 ¼ b 3 a 3 ¼ b 2 C 2 ¼ U cos 158 ¼ 340 cos 158 ¼ 351:99 m/s Heat drop in blade moving row ¼ C 2 2 2C 2 3 2C p ¼ ð351:99Þ 2 2ð105Þ 2 ð2Þð1147Þ ¼ 123896:96 211025 ð2Þð1147Þ ¼ 49:2 K Therefore heat drop in a stage ¼ ð2Þð49:2Þ ¼ 98:41 K Number of stages ¼ 1173 2943 98:41 ¼ 230 98:4 ¼ 2 Design Example 7.10 The following particulars relate to a single-stage turbine of free vortex design: Inlet temperature; T 01 1100K Inlet pressure; p 01 4 bar Mass flow 20 kg/s Axial velocity at nozzle exit 250 m/s Blade speed at mean diameter 300 m/s Nozzle angle at mean diameter 258 Ratio of tip to root radius 1:4 Axial Flow and Radial Flow Gas Turbines 305 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The gas leaves the stage in an axial direction, find: (1) The total throat area of the nozzle. (2) The nozzle efflux angle at root and tip. (3) The work done on the turbine blades. Take C pg ¼ 1:147 kJ/kg K; g ¼ 1:33 Solution For no loss up to throat p * p 01 ¼ 2 g þ 1 _ _ g=ðg21Þ ¼ 2 2:33 _ _ 4 ¼ 0:543 p * ¼ 4 £ 0:543 ¼ 2:172 bar Also T * ¼ 1100 2 2:33 _ _ 4 ¼ 944 K T 01 ¼ T * þ C 2 2C pg C * ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2C pg T 01 2T * _ _ _ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2Þð1147Þ 1100 2944 ð Þ _ ¼ 598 m/s r * ¼ p * RT * ¼ ð2:172Þð100Þ ð0:287Þð944Þ ¼ 0:802 kg/m 3 (1) Throat area A ¼ m rC * ¼ 20 ð0:802Þð598Þ ¼ 0:042 m 2 (2) Angle a 1 , at any radius r and a 1m at the design radius r m are related by the equation tan a 1 ¼ r m r 1 tan a 1m Chapter 7 306 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Given Tip radius Root radius ¼ r t r r ¼ 1:4 [ Mean radius Root radius ¼ 1:2 a 1m ¼ 258 tan a 1r ¼ r mean r root £ tan a 1m ¼ 1:2 £ tan 258 ¼ 0:5596 [ a 1r ¼ 29:238 tan a 1t ¼ r r r t £ tan a 1r ¼ 1 1:4 _ _ ð0:5596Þ ¼ 0:3997 [ a 1t ¼ 21:798 ð3Þ C w2 ¼ r m r r xC w2m ¼ r m r r Ca 2 tan a 2m ¼ 1:2x 250 tan 258 ¼ 643 m/s W ¼ mUC w2 ¼ ð20Þð300Þð643Þ 1000 ¼ 3858 kW 7.8 RADIAL FLOW TURBINE In Sec. 7.1 “Introduction to Axial Flow Turbines”, it was pointed out that in axial flow turbines the fluid moves essentially in the axial direction through the rotor. In the radial type the fluid motion is mostly radial. The mixed flow machine is characterized by a combination of axial and radial motion of the fluid relative to the rotor. The choice of turbine depends on the application, though it is not always clear that any one type is superior. For small mass flows, the radial machine can be made more efficient than the axial one. The radial turbine is capable of a high-pressure ratio per stage than the axial one. However, multi- staging is very much easier to arrange with the axial turbine, so that large overall pressure ratios are not difficult to obtain with axial turbines. The radial flow turbines are used in turbochargers for commercial (diesel) engines and fire pumps. They are very compact, the maximum diameter being about 0.2 m, and run at very high speeds. In inward flow radial turbine, gas enters in the radial direction and leaves axially at outlet. The rotor, which is usually manufactured of Axial Flow and Radial Flow Gas Turbines 307 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 7.9 Elements of a 908 inward flow radial gas turbine with inlet nozzle ring. Figure 7.8 Radial turbine photograph of the rotor on the right. Chapter 7 308 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved cast nickel alloy, has blades that are curved to change the flow from the radial to the axial direction. Note that this turbine is like a single-faced centrifugal compressor with reverse flow. Figures 7.8–7.10 show photographs of the radial turbine and its essential parts. 7.9 VELOCITY DIAGRAMS AND THERMODYNAMIC ANALYSIS Figure 7.11 shows the velocity triangles for this turbine. The same nomenclature that we used for axial flow turbines, will be used here. Figure 7.12 shows the Mollier diagram for a 908 flow radial turbine and diffuser. As no work is done in the nozzle, we have h 01 ¼ h 02 . The stagnation pressure drops fromp 01 to p 1 due to irreversibilities. The work done per unit mass flow is given by Euler’s turbine equation W t ¼ U 2 C w2 2U 3 C w3 ð Þ ð7:36Þ If the whirl velocity is zero at exit then W t ¼ U 2 C w2 ð7:37Þ Figure 7.10 A 908 inward flow radial gas turbine without nozzle ring. Axial Flow and Radial Flow Gas Turbines 309 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure 7.12 Mollier chart for expansion in a 908 inward flow radial gas turbine. Figure 7.11 Velocity triangles for the 908 inward flow radial gas turbine. Chapter 7 310 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved For radial relative velocity at inlet W t ¼ U 2 2 ð7:38Þ In terms of enthalpy drop h 02 2h 03 ¼ U 2 C w2 2U 3 C w3 Using total-to-total efficiency h tt ¼ T 01 2T 03 T 01 2T 03 ss ; efficiency being in the region of 80–90% 7.10 SPOUTING VELOCITY It is that velocity, which has an associated kinetic energy equal to the isentropic enthalpy drop from turbine inlet stagnation pressure p 01 to the final exhaust pressure. Spouting velocities may be defined depending upon whether total or static conditions are used in the related efficiency definition and upon whether or not a diffuser is included with the turbine. Thus, when no diffuser is used, using subscript 0 for spouting velocity. 1 2 C 2 0 ¼ h 01 2h 03 ss ð7:39Þ or 1 2 C 2 0 ¼ h 01 2h 3 ss ð7:40Þ for the total and static cases, respectively. Now for isentropic flow throughout work done per unit mass flow W ¼ U 2 2 ¼ C 2 0 /2 ð7:41Þ or U 2 /C 0 ¼ 0:707 ð7:42Þ In practice, U 2 /C 0 lies in the range 0:68 , U 2 C 0 , 0:71. 7.11 TURBINE EFFICIENCY Referring to Fig. 7.12, the total-to-static efficiency, without diffuser, is defined as h ts ¼ h 01 2h 03 h 01 2h 3 ss ¼ W W þ 1 2 C 2 3 þ h 3 2h 3ss ð Þ þ h 3s 2h 3 ss ð Þ ð7:43Þ Axial Flow and Radial Flow Gas Turbines 311 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Nozzle loss coefficient, j n , is defined as j n ¼ Enthalpy loss in nozzle Kinetic energy at nozzle exit ¼ h 3s 2h 3 ss 0:5C 2 2 T 3 /T 2 ð Þ ð7:44Þ Rotor loss coefficient, j r , is defined as j r ¼ h 3 2h 3s 0:5V 2 3 ð7:45Þ But for constant pressure process, T ds ¼ dh; and, therefore h 3s 2h 3ss ¼ h 2h 2s ð Þ T 3 /T 2 ð Þ Substituting in Eq. (7.43) h ts ¼ 1 þ 1 2 C 2 3 þ V 2 3 j r þ C 2 j n T 3 /T 2 _ _ W _ _ 21 ð7:46Þ Using velocity triangles C 2 ¼ U 2 cosec a 2 ; V 3 ¼ U 3 cosec b 3 ; C 3 ¼ U 3 cot b 3 ; W ¼ U 2 2 Substituting all those values in Eq. (7.44) and noting that U 3 ¼ U 2 r 3 /r 2 , then h ts ¼ 1 þ 1 2 j n T 3 T 2 cosec 2 a 2 þ r 3 r 2 _ _ 2 j r cosec 2 b 3 þ cot 2 b 3 _ _ _ _ _ _ 21 ð7:47Þ Taking mean radius, that is, r 3 ¼ 1 2 r 3t þ r 3h ð Þ Using thermodynamic relation for T 3 /T 2 , we get T 3 T 2 ¼ 1 2 1 2 g 21 _ _ U 2 a 2 _ _ 2 1 2cot 2 a 2 þ r 3 r 2 _ _ 2 cot 2 b 3 _ _ But the above value of T 3 /T 2 is very small, and therefore usually neglected. Thus h ts ¼ 1 þ 1 2 j n cosec 2 a 2 þ r 3 av r 2 _ _ 2 j r cosec 2 b 3 av þ cot 2 b 3 av _ _ _ _ _ _ 21 ð7:48Þ Chapter 7 312 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Equation (7.46) is normally used to determine total-to-static efficiency. The h ts can also be found by rewriting Eq. (7.43) as h ts ¼ h 01 2h 03 h 01 2h 3ss ¼ h 01 2h 3ss ð Þ 2 h 03 2h 3 ð Þ 2 h 3 2h 3s ð Þ 2 h 3s 2h 3ss ð Þ h 01 2h 3ss ð Þ ¼ 1 2 C 2 3 þj n C 2 2 þj r V 2 3 _ _ /C 2 0 ð7:49Þ where spouting velocity C 0 is given by h 01 2h 3ss ¼ 1 2 C 2 0 ¼C p T 01 1 2 p 3 /p 01 _ _ g21=g _ _ ð7:50Þ The relationship between h ts and h tt can be obtained as follows: W ¼U 2 2 ¼h ts W ts ¼h ts h 01 2h 3ss ð Þ; then h tt ¼ W W ts 2 1 2 C 2 3 ¼ 1 1 h ts 2 C 2 3 2W [ 1 h tt ¼ 1 h ts 2 C 2 3 2W ¼ 1 h ts 2 1 2 r 3av r 2 2cot b 3av _ _ 2 ð7:51Þ Loss coefficients usually lie in the following range for 908 inward flow turbines j n ¼ 0:063–0:235 and j r ¼ 0:384–0:777 7.12 APPLICATION OF SPECIFIC SPEED We have already discussed the concept of specific speed N s in Chapter 1 and some applications of it have been made already. The concept of specific speed was applied almost exclusively to incompressible flow machines as an important parameter in the selection of the optimum type and size of unit. The volume flow rate through hydraulic machines remains constant. But in radial flow gas turbine, volume flow rate changes significantly, and this change must be taken into account. According to Balje, one suggested value of volume flow rate is that at the outlet Q 3 . Using nondimensional form of specific speed N s ¼ NQ 1/2 3 ðDh 0 0 Þ 3/4 ð7:52Þ where N is in rev/s, Q 3 is in m 3 /s and isentropic total-to-total enthalpy drop (from turbine inlet to outlet) is in J/kg. For the 908 inward flow radial turbine, Axial Flow and Radial Flow Gas Turbines 313 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved U 2 ¼ pND 2 and Dh 0s ¼ 1 2 C 2 0 ; factorizing the Eq. (7.52) N s ¼ Q 1/2 3 1 2 C 2 0 _ _ 3/4 U 2 pD 2 _ _ U 2 pND 2 _ _ 1/2 ¼ ffiffiffi 2 p p _ _3/2 U 2 C 0 _ _ 3/2 Q 3 ND 3 2 _ _ 1/2 ð7:53Þ For 908 inward flow radial turbine, U 2 /C 0 ¼ 1 ffiffi 2 p ¼ 0:707; substituting this value in Eq. (7.53), N s ¼ 0:18 Q 3 ND 3 2 _ _ 1/2 ; rev ð7:54Þ Equation (7.54) shows that specific speed is directly proportional to the square root of the volumetric flow coefficient. Assuming a uniform axial velocity at rotor exit C 3 , so that Q 3 ¼ A 3 C 3 , rotor disc area A d ¼ pD 2 2 /4, then N ¼ U 2 / pD 2 ð Þ ¼ C 0 ffiffiffi 2 p 2 pD 2 Q 3 ND 3 2 ¼ A 3 C 3 2 pD 2 ffiffiffi 2 p C 0 D 2 2 ¼ A 3 A d C 3 C 0 p 2 2 ffiffiffi 2 p Therefore, N s ¼ 0:336 C 3 C 0 _ _1 2 A 3 A d _ _1 2 ; rev ð7:55Þ ¼ 2:11 C 3 C 0 _ _1 2 A 3 A d _ _1 2 ; rad ð7:56Þ Suggested values for C 3 /C o and A 3 /A d are as follows: 0:04 , C 3 /C 0 , 0:3 0:1 , A 3 /A d , 0:5 Then 0:3 , N s , 1:1; rad Thus the N s range is very small and Fig. 7.13 shows the variation of efficiency with N s , where it is seen to match the axial flowgas turbine over the limited range of N s . Design Example 7.11 A small inward radial flow gas turbine operates at its design point with a total-to-total efficiency of 0.90. The stagnation pressure and temperature of the gas at nozzle inlet are 310 kPa and 1145K respectively. The flow leaving the turbine is diffused to a pressure of 100 kPa and the velocity of Chapter 7 314 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved flow is negligible at that point. Given that the Mach number at exit from the nozzles is 0.9, find the impeller tip speed and the flow angle at the nozzle exit. Assume that the gas enters the impeller radially and there is no whirl at the impeller exit. Take C p ¼ 1:147 kJ/kg K; g ¼ 1:333: Solution The overall efficiency of turbine from nozzle inlet to diffuser outlet is given by h tt ¼ T 01 2T 03 T 01 2T 03 ss Turbine work per unit mass flow W ¼ U 2 2 ¼ C p T 01 2T 03 ð Þ; C w3 ¼ 0 ð Þ Now using isentropic p–T relation T 01 1 2 T 03ss T 01 _ _ ¼ T 01 1 2 p 03 p 01 _ _ g21=g _ _ Therefore U 2 2 ¼ h tt C p T 01 1 2 p 03 p 01 _ _ g21=g _ _ ¼ 0:9 £ 1147 £ 1145 1 2 100 310 _ _ 0:2498 _ _ [ Impeller tip speed, U 2 ¼ 539.45 m/s Figure 7.13 Variation of efficiency with dimensionless specific speed. Axial Flow and Radial Flow Gas Turbines 315 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The Mach number of the absolute flow velocity at nozzle exit is given by M ¼ C 1 a 1 ¼ U 1 a 1 sin a 1 Since the flow is adiabatic across the nozzle, we have T 01 ¼ T 02 ¼ T 2 þ C 2 2 2C p ¼ T 2 þ U 2 2 2C p sin 2 a 2 or T 2 T 01 ¼ 1 2 U 2 2 2C p T 01 sin 2 a 2 ; but C p ¼ gR g 21 [ T 2 T 01 ¼ 1 2 U 2 2 g 21 _ _ 2gRT 01 sin 2 a 2 ¼ 1 2 U 2 2 g 21 _ _ 2a 2 01 sin 2 a 2 But T 2 T 01 _ _ 2 ¼ a 2 a 01 ¼ a 2 a 02 since T 01 ¼ T 02 and a 2 a 02 ¼ U 2 M 2 a 02 sin a 2 [ U 2 M 2 a 02 sin a 2 _ _ 2 ¼ 1 2 U 2 2 g 21 _ _ 2a 2 02 sin 2 a 2 and 1 ¼ U 2 a 02 sin a 2 _ _ 2 g 21 _ _ 2 þ 1 M 2 2 _ _ or sin 2 a 2 ¼ U 2 a 02 _ _ 2 g 21 _ _ 2 þ 1 M 2 2 _ _ But a 2 02 ¼ gRT 02 ¼ ð1:333Þð287Þð1145Þ ¼ 438043 m 2 / s 2 [ sin 2 a 2 ¼ 539:45 2 438043 0:333 2 þ 1 0:9 2 _ _ ¼ 0:9311 Therefore nozzle angle a 2 ¼ 758 Illustrative Example 7.12 The following particulars relate to a small inward flow radial gas turbine. Rotor inlet tip diameter 92 mm Rotor outlet tip diameter 64 mm Rotor outlet hub diameter 26 mm Ratio C 3 /C 0 0:447 Chapter 7 316 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Ratio U 2 /C 0 ðidealÞ 0:707 Blade rotational speed 30; 500 rpm Density at impeller exit 1:75 kg/m 3 Determine (1) The dimensionless specific speed of the turbine. (2) The volume flow rate at impeller outlet. (3) The power developed by the turbine. Solution (1) Dimensionless specific speed is N s ¼ 0:336 C 3 C 0 _ _1 2 A 3 A d _ _1 2 ; rev Now A 3 ¼ p D 2 3t 2D 2 3h _ _ 4 ¼ p 0:064 2 20:026 2 _ _ 4 ¼ ð2:73Þð10 23 Þ m 2 A d ¼ pD 2 2 4 ¼ p 4 _ _ ð0:092 2 Þ ¼ ð6:65Þð10 23 Þ m 2 Dimensionless specific speed N s ¼ 0:336 ½0:447½2:73 6:65 _ _1 2 ¼ 0:144 rev ¼ 0:904 rad (2) The flow rate at outlet for the ideal turbine is given by Eq. (7.54). N s ¼ 0:18 Q 3 ND 3 2 _ _ 1=2 0:144 ¼ 0:18 ½Q 3 ½60 ½30; 500½0:092 3 _ _ 1=2 Hence Q 3 ¼ 0:253 m 3 /s Axial Flow and Radial Flow Gas Turbines 317 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved (3) The power developed by the turbine is given by W t ¼ _ mU 2 3 ¼ r 3 Q 3 U 2 3 ¼ 1:75 £ 0:253 £ pND 2 60 _ _ 2 ¼ 1:75 £ 0:253 £ ½ p½30; 500½0:092 60 _ _ 2 ¼ 9:565 kW PROBLEMS 7.1 A single-stage axial flow gas turbine has the following data: Inlet stagnation temperature 1100K The ratio of static pressure at the nozzle exit to the stagnation pressure at the nozzle inlet 0:53 Nozzle efficiency 0:93 Nozzle angle 208 Mean blade velocity 454 m/s Rotor efficiency 0:90 Degree of reaction 50% C pg ¼ 1:147 kJ/kgK; g ¼ 1:33 Find (1) the work output per kg/s of air flow, (2) the ratio of the static pressure at the rotor exit to the stagnation pressure at the nozzle inlet, and (3) the total-to-total stage efficiency. (282 kW, 0.214, 83.78%) 7.2 Derive an equation for the degree of reaction for a single-stage axial flow turbine and show that for 50% reaction blading a 2 ¼ b 3 and a 3 ¼ b 2 . 7.3 For a free-vortex turbine blade with an impulse hub show that degree of reaction L ¼ 1 2 r h r _ _ 2 where r h is the hub radius and r is any radius. Chapter 7 318 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 7.4 A 50% reaction axial flow gas turbine has a total enthalpy drop of 288 kJ/kg. The nozzle exit angle is 708. The inlet angle to the rotating blade row is inclined at 208 with the axial direction. The axial velocity is constant through the stage. Calculate the enthalpy drop per row of moving blades and the number of stages required when mean blade speed is 310 m/s. Take C pg ¼ 1:147 kJ/kgK; g ¼ 1:33: (5 stages) 7.5 Show that for zero degree of reaction, blade-loading coefficient, C ¼ 2. 7.6 The inlet stagnation temperature and pressure for an axial flow gas turbine are 1000K and 8 bar, respectively. The exhaust gas pressure is 1.2 bar and isentropic efficiency of turbine is 85%. Assume gas is air, find the exhaust stagnation temperature and entropy change of the gas. (644K, 20.044 kJ/kgK) 7.7 The performance date from inward radial flow exhaust gas turbine are as follows: Stagnation pressure at inlet to nozzles; p 01 705 kPa Stagnation temperature at inlet to nozzles; T 01 1080K Static pressure at exit from nozzles; p 2 515 kPa Static temperature at exit from nozzles; T 2 1000K Static pressure at exit from rotor; p 3 360 kPa Static temperature at exit from rotor; T 3 923K Stagnation temperature at exit from rotor; T 03 925K Ratio r 2 av r 2 0:5 Rotational speed; N 25; 500 rpm The flow into the rotor is radial and at exit the flow is axial at all radii. Calculate (1) the total-to-static efficiency of the turbine, (2) the impeller tip diameter, (3) the enthalpy loss coefficient for the nozzle and rotor rows, (4) the blade outlet angle at the mean diameter, and (5) the total-to-total efficiency of the turbine. [(1) 93%, (2) 0.32 m, (3) 0.019, 0.399, (4) 72.28, (5) 94%] NOTATION A area C absolute velocity Axial Flow and Radial Flow Gas Turbines 319 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved C 0 spouting velocity h enthalpy, blade height N rotation speed N s specific speed P pressure r m mean radius T temperature U rotor speed V relative velocity Y N nozzle loss coefficient in terms of pressure a angle with absolute velocity b angle with relative velocity DT 0s stagnation temperature drop in the stage DT s static temperature drop in the stage 1 n nozzle loss coefficient in radial flow turbine 1 r rotor loss coefficient in radial flow turbine f flow coefficient h s isentropic efficiency of stage L degree of reaction Chapter 7 320 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 8 Cavitation in Hydraulic Machinery 8.1 INTRODUCTION Cavitation is caused by local vaporization of the fluid, when the local static pressure of a liquid falls below the vapor pressure of the liquid. Small bubbles or cavities filled with vapor are formed, which suddenly collapse on moving forward with the flow into regions of high pressure. These bubbles collapse with tremendous force, giving rise to as high a pressure as 3500 atm. In a centrifugal pump, these low-pressure zones are generally at the impeller inlet, where the fluid is locally accelerated over the vane surfaces. In turbines, cavitation is most likely to occur at the downstream outlet end of a blade on the low-pressure leading face. When cavitation occurs, it causes the following undesirable effects: 1. Local pitting of the impeller and erosion of the metal surface. 2. Serious damage can occur from prolonged cavitation erosion. 3. Vibration of machine; noise is also generated in the form of sharp cracking sounds when cavitation takes place. 4. A drop in efficiency due to vapor formation, which reduces the effective flow areas. The avoidance of cavitation in conventionally designed machines can be regarded as one of the essential tasks of both pump and turbine designers. This cavit- ation imposes limitations on the rate of discharge and speed of rotation of the pump. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 8.2 STAGES AND TYPES OF CAVITATION The term incipient stage describes cavitation that is just barely detectable. The discernible bubbles of incipient cavitation are small, and the zone over which cavitation occurs is limited. With changes in conditions (pressure, velocity, temperature) toward promoting increased vaporization rates, cavitation grows; the succeeding stages are distinguished fromthe incipient stage by the termdeveloped. Traveling cavitation is a type composed of individual transient cavities or bubbles, which form in the liquid, as they expand, shrink, and then collapse. Such traveling transient bubbles may appear at the low-pressure points along a solid boundary or in the liquid interior either at the cores of moving vortices or in the high-turbulence region in a turbulent shear field. The term fixed cavitation refers to the situation that sometimes develops after inception, in which the liquid flow detaches from the rigid boundary of an immersed body or a flow passage to form a pocket or cavity attached to the boundary. The attached or fixed cavity is stable in a quasi-steady sense. Fixed cavities sometimes have the appearance of a highly turbulent boiling surface. In vortex cavitation, the cavities are found in the cores of vortices that form in zones of high shear. The cavitation may appear as traveling cavities or as a fixed cavity. Vortex cavitation is one of the earliest observed types, as it often occurs on the blade tips of ships’ propellers. In fact, this type of cavitation is often referred to as “tip” cavitation. Tip cavitation occurs not only in open propellers but also in ducted propellers such as those found in propeller pumps at hydrofoil tips. 8.2.1 Cavitation on Moving Bodies There is no essential difference between cavitation in a flowing stream and that on a body moving through a stationary liquid. In both cases, the important factors are the relative velocities and the absolute pressures. When these are similar, the same types of cavitation are found. One noticeable difference is that the turbulence level in the stationary liquid is lower. Many cases of cavitation in a flowing stream occur in relatively long flow passages in which the turbulence is fully developed before the liquid reaches the cavitation zone. Hydraulic machinery furnishes a typical example of a combination of the two conditions. In the casing, the moving liquid flows past stationary guide surfaces; in the runner, the liquid and the guide surfaces are both in motion. 8.2.2 Cavitation Without Major Flow—Vibratory Cavitation The types of cavitation previously described have one major feature in common. It is that a particular liquid element passes through the cavitation zone only once. Vibratory cavitation is another important type of cavitation, which does not have Chapter 8 322 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved this characteristic. Although it is accompanied sometimes by continuous flow, the velocity is so low that a given element of liquid is exposed to many cycles of cavitation (in a time period of the order of milliseconds) rather than only one. In vibratory cavitation, the forces causing the cavities to form and collapse are due to a continuous series of high-amplitude, high-frequency pressure pulsations in the liquid. These pressure pulsations are generated by a submerged surface, which vibrates normal to its face and sets up pressure waves in the liquid. No cavities will be formed unless the amplitude of the pressure variation is great enough to cause the pressure to drop to or below the vapor pressure of the liquid. As the vibratory pressure field is characteristic of this type of cavitation, the name “vibratory cavitation” follows. 8.3 EFFECTS AND IMPORTANCE OF CAVITATION Cavitation is important as a consequence of its effects. These may be classified into three general categories: 1. Effects that modify the hydrodynamics of the flow of the liquid 2. Effects that produce damage on the solid-boundary surfaces of the flow 3. Extraneous effects that may or may not be accompanied by significant hydrodynamic flow modifications or damage to solid boundaries Unfortunately for the field of applied hydrodynamics, the effects of cavitation, with very few exceptions, are undesirable. Uncontrolled cavitation can produce serious and even catastrophic results. The necessity of avoiding or controlling cavitation imposes serious limitations on the design of many types of hydraulic equipment. The simple enumeration of some types of equipment, structures, or flow systems, whose performance may be seriously affected by the presence of cavitation, will serve to emphasize the wide occurrence and the relative importance of this phenomenon. In the field of hydraulic machinery, it has been found that all types of turbines, which form a low-specific-speed Francis to the high-specific-speed Kaplan, are susceptible to cavitation. Centrifugal and axial-flow pumps suffer from its effects, and even the various types of positive-displacement pumps may be troubled by it. Although cavitation may be aggravated by poor design, it may occur in even the best-designed equipment when the latter is operated under unfavorable condition. 8.4 CAVITATION PARAMETER FOR DYNAMIC SIMILARITY The main variables that affect the inception and subsequent character of cavitation in flowing liquids are the boundary geometry, the flow variables Cavitation in Hydraulic Machinery 323 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved of absolute pressure and velocity, and the critical pressure p crit at which a bubble can be formed or a cavity maintained. Other variables may cause significant variations in the relation between geometry, pressure, and velocity and in the value of the critical pressure. These include the properties of the liquid (such as viscosity, surface tension, and vaporization characteristics), any solid, or gaseous contaminants that may be entrained or dissolved in the liquid, and the condition of the boundary surfaces, including cleanliness and existence of crevices, which might host undissolved gases. In addition to dynamic effects, the pressure gradients due to gravity are important for large cavities whether they be traveling or attached types. Finally, the physical size of the boundary geometry may be important, not only in affecting cavity dimensions but also in modifying the effects of some of the fluid and boundary flow properties. Let us consider a simple liquid having constant properties and develop the basic cavitation parameter. A relative flow between an immersed object and the surrounding liquid results in a variation in pressure at a point on the object, and the pressure in the undisturbed liquid at some distance from the object is proportional to the square of the relative velocity. This can be written as the negative of the usual pressure coefficient C p , namely, 2C p = ( p 0 2p) d rV 2 0 =2 (8:1) where r is the density of liquid, V 0 the velocity of undisturbed liquid relative to body, p 0 the pressure of undisturbed liquid, p the pressure at a point on object, and ( p 0 2p) d the pressure differential due to dynamic effects of fluid motion. This is equivalent to omitting gravity. However, when necessary, gravity effects can be included. At some location on the object, p will be a minimum, p min , so that (2C p ) min = p 0 2p min rV 2 0 =2 (8:2) In the absence of cavitation (and if Reynolds-number effects are neglected), this value will depend only on the shape of the object. It is easy to create a set of conditions such that p min drops to some value at which cavitation exists. This can be accomplished by increasing the relative velocity V 0 for a fixed value of the pressure p 0 or by continuously lowering p 0 with V 0 held constant. Either procedure will result in lowering of the absolute values of all the local pressures on the surface of the object. If surface tension is ignored, the pressure p min will be the pressure of the contents of the cavitation cavity. Denoting this as a bubble pressure p b , we can define a cavitation parameter by replacing p min ; thus K b = p 0 2p b rV 2 0 =2 (8:3) Chapter 8 324 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved or, in terms of pressure head (in feet of the liquid), K b = ( p 0 2p b )=g V 2 0 =2g (8:4) where p 0 is the absolute-static pressure at some reference locality, V 0 the reference velocity, p b the absolute pressure in cavity or bubble, and g the specific weight of liquid. If we now assume that cavitation will occur when the normal stresses at a point in the liquid are reduced to zero, p b will equal the vapor pressure p v . Then, we write K b = p 0 2p v rV 2 0 =2 (8:5) The value of K at which cavitation inception occurs is designated as K i . A theoretical value of K i is the magnitude }(2C p ) min } for any particular body. The initiation of cavitation by vaporization of the liquid may require that a negative stress exist because of surface tension and other effects. However, the presence of such things as undissolved gas particles, boundary layers, and turbulence will modify and often mask a departure of the critical pressure p crit from p v . As a consequence, Eq. (8.5) has been universally adopted as the parameter for comparison of vaporous cavitation events. The beginning of cavitation means the appearance of tiny cavities at or near the place on the object where the minimum pressure is obtained. Continual increase in V 0 (or decrease in p 0 ) means that the pressure at other points along the surface of the object will drop to the critical pressure. Thus, the zone of cavitation will spread from the location of original inception. In considering the behavior of the cavitation parameter during this process, we again note that if Reynolds- number effects are neglected the pressure coefficient (2C p ) min depends only on the object’s shape and is constant prior to inception. After inception, the value decreases as p min becomes the cavity pressure, which tends to remain constant, whereas either V 0 increases or p 0 decreases. Thus, the cavitation parameter assumes a definite value at each stage of development or “degree” of cavitation on a particular body. For inception, K = K i ; for advanced stages of cavitation, K , K i : K i and values of K at subsequent stages of cavitation depend primarily on the shape of the immersed object past which the liquid flows. We should note here that for flow past immersed objects and curved boundaries, K i will always be finite. For the limiting case of parallel flow of an ideal fluid, K i will be zero since the pressure p 0 in the main stream will be the same as the wall pressure (again with gravity omitted and the assumption that cavitation occurs when the normal stresses are zero). Cavitation in Hydraulic Machinery 325 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 8.4.1 The Cavitation Parameter as a Flow Index The parameter K b or K can be used to relate the conditions of flow to the possibility of cavitation occurring as well as to the degree of postinception stages of cavitation. For any system where the existing or potential bubble pressure ( p b or p v ) is fixed, the parameter (K b or K) can be computed for the full range of values of the reference velocity V 0 and reference pressure p 0 . On the other hand, as previously noted, for any degree of cavitation from inception to advanced stages, the parameter has a characteristic value. By adjusting the flow conditions so that K is greater than, equal to, or less than K i , the full range of possibilities, from no cavitation to advanced stages of cavitation, can be established. 8.4.2 The Cavitation Parameter in Gravity Fields As the pressure differences in the preceding relations are due to dynamic effects, the cavitation parameter is defined independently of the gravity field. For large bodies going through changes in elevation as they move, the relation between dynamic pressure difference ( p 0 2 p min ) d and the actual pressure difference ( p 0 2 p min ) actual is ( p 0 2p min ) d = ( p 0 2p min ) actual -g(h 0 2h min ) where g is the liquid’s specific weight and h is elevation. Then, in terms of actual pressures, we have, instead of Eq. (8.5), K = ( p 0 -gh 0 ) 2( p v -gh min ) rV 0 =2 (8:6) For h 0 = h min ; Eq. (8.6) reduces to Eq. (8.5). 8.5 PHYSICAL SIGNIFICANCE AND USES OF THE CAVITATION PARAMETER A simple physical interpretation follows directly when we consider a cavitation cavity that is being formed and then swept from a low-pressure to a high- pressure region. Then the numerator is related to the net pressure or head, which tends to collapse the cavity. The denominator is the velocity pressure or head of the flow. The variations in pressure, which take place on the surface of the body or on any type of guide passage, are basically due to changes in the velocity of the flow. Thus, the velocity head may be considered to be a measure of the pressure reductions that may occur to cause a cavity to form or expand. The basic importance of cavitation parameter stems fromthe fact that it is an index of dynamic similarity of flow conditions under which cavitation occurs. Its use, therefore, is subject to a number of limitations. Full dynamic similarity Chapter 8 326 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved between flows in two systems requires that the effects of all physical conditions be reproduced according to unique relations. Thus, even if identical thermodynamics and chemical properties and identical boundary geometry are assumed, the variable effects of contaminants in the liquid-omitted dynamic similarity require that the effects of viscosity, gravity, and surface tension be in unique relationship at each cavitation condition. In other words, a particular cavitation condition is accurately reproduced only if Reynolds number, Froude number, Weber number, etc. as well as the cavitation parameter K have particular values according to a unique relation among themselves. 8.6 THE RAYLEIGH ANALYSIS OF A SPHERICAL CAVITY IN AN INVISCID INCOMPRESSIBLE LIQUID AT REST AT INFINITY The mathematical analysis of the formation and collapse of spherical cavities, which are the idealized form of the traveling transient cavities, has proved interesting to many workers in the field. Furthermore, it appears that as more experimental evidence is obtained on the detailed mechanics of the cavitation process, the role played by traveling cavities grows in importance. This is especially true with respect to the process by which cavitation produces physical damage on the guiding surfaces. Rayleigh first set up an expression for the velocity u, at any radial distance r, where r is greater than R, the radius of the cavity wall. U is the cavity-wall velocity at time t. For spherical symmetry, the radial flow is irrotational with velocity potential, and velocity is given by f = UR 2 r and u U = R 2 r 2 (8:7) Next, the expression for the kinetic energy of the entire body of liquid at time t is developed by integrating kinetic energy of a concentric fluid shell of thickness dr and density r. The result is (KE) liq = r 2 _ 1 R u 2 4pr 2 dr = 2prU 2 R 3 (8:8) The work done on the entire body of fluid as the cavity is collapsing from the initial radius R 0 to R is a product of the pressure p 1 at infinity and the change in volume of the cavity as no work is done at the cavity wall where the pressure is assumed to be zero, i.e., 4pp 1 3 R 3 0 2R 3 _ _ (8:9) If the fluid is inviscid as well as incompressible, the work done appears as kinetic Cavitation in Hydraulic Machinery 327 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved energy. Therefore, Eq. (8.8) can be equated to Eq. (8.9), which gives U 2 = 2p 1 3r R 3 0 R 3 21 _ _ (8:10) An expression for the time t required for a cavity to collapse from R 0 to R can be obtained from Eq. (8.10) by substituting for the velocity U of the boundary, its equivalent dR/dt and performing the necessary integration. This gives t = ffiffiffiffiffiffiffiffi 3r 2p 1 ¸ _ R 0 R R 3=2 dR R 3 0 2R 3 _ _ 1=2 = R 0 ffiffiffiffiffiffiffiffi 3r 2p 1 ¸ _ 1 b b 3=2 db (1 2b 3 ) 1=2 (8:11) The new symbol b is R/R 0 . The time t of complete collapse is obtained if Eq. (8.11) is evaluated for b = 0: For this special case, the integration may be performed by means of functions with the result that t becomes t = R 0 ffiffiffiffiffiffiffiffi r 6p 1 _ £ G 3 6 _ _ G 1 2 _ _ G 4 3 _ _ = 0:91468R 0 ffiffiffiffiffiffi r p 1 _ (8:12) Rayleigh did not integrate Eq. (8.11) for any other value of b. In the detailed study of the time history of the collapse of a cavitation bubble, it is convenient to have a solution for all values of b between 0 and 1.0. Table 8.1 gives values of the dimensionless time t = t=R 0 ffiffiffiffiffiffiffiffiffiffiffi r=p 1 _ over this range as obtained from a numerical solution of a power series expansion of the integral in Eq. (8.11). Equation (8.10) shows that as R decreases to 0, the velocity U increases to infinity. In order to avoid this, Rayleigh calculated what would happen if, instead of having zero or constant pressure within the cavity, the cavity is filled with a gas, which is compressed isothermally. In such a case, the external work done on the system as given by Eq. (8.9) is equated to the sum of the kinetic energy of the liquid given by Eq. (8.8) and the work of compression of the gas, which is 4pQR 3 0 ln(R 0 =R); where Q is the initial pressure of the gas. Thus Eq. (8.10) is replaced by U 2 = 2p 1 3r R 3 0 R 3 21 _ _ 2 2Q r £ R 3 0 R 3 ln 0 R 0 R (8:13) For any real (i.e., positive) value of Q, the cavity will not collapse completely, but U will come to 0 for a finite value of R. If Q is greater than p 1 , the first movement of the boundary is outward. The limiting size of the cavity can be obtained by setting U = 0 in Eq. (8.13), which gives p 1 z 21 z 2Qln z = 0 (8:14) Chapter 8 328 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved in which z denotes the ratio of the volume R 3 0 =R 3 : Equation (8.14) indicates that the radius oscillates between the initial value R 0 and another, which is determined by the ratio p 1 =Q from this equation. If p 1 =Q . 1; the limiting size is a minimum. Although Rayleigh presented this example only for isothermal Table 8.1 Values of the Dimensionless Time t / = t=R 0 ffiffiffiffiffiffiffiffiffiffiffi r=p 1 _ from Eq. (8.11) (Error Less Than 10 26 for 0 # b # 0:96) b t ffiffiffiffiffiffiffiffi p1=r _ R0 b t ffiffiffiffiffiffiffiffi p1=r _ R0 b t ffiffiffiffiffiffiffiffi p1=r _ R0 0.99 0.016145 0.64 0.733436 0.29 0.892245 0.98 0.079522 0.63 0.741436 0.28 0.894153 0.97 0.130400 0.62 0.749154 0.27 0.895956 0.96 0.174063 0.61 0.756599 0.26 0.897658 0.95 0.212764 0.60 0.763782 0.25 0.899262 0.94 0.247733 0.59 0.770712 0.24 0.900769 0.93 0.279736 0.58 0.777398 0.23 0.902182 0.92 0.309297 0.57 0.783847 0.22 0.903505 0.91 0.336793 0.56 0.790068 0.21 0.904738 0.90 0.362507 0.55 0.796068 0.20 0.905885 0.89 0.386662 0.54 0.801854 0.19 0.906947 0.88 0.409433 0.53 0.807433 0.18 0.907928 0.87 0.430965 0.52 0.812810 0.17 0.908829 0.86 0.451377 0.51 0.817993 0.16 0.909654 0.85 0.470770 0.50 0.822988 0.15 0.910404 0.84 0.489229 0.49 0.827798 0.14 0.911083 0.83 0.506830 0.48 0.832431 0.13 0.911692 0.82 0.523635 0.47 0.836890 0.12 0.912234 0.81 0.539701 0.46 0.841181 0.11 0.912713 0.80 0.555078 0.45 0.845308 0.10 0.913130 0.79 0.569810 0.44 0.849277 0.09 0.913489 0.78 0.583937 0.43 0.853090 0.08 0.913793 0.77 0.597495 0.42 0.856752 0.07 0.914045 0.76 0.610515 0.41 0.860268 0.06 0.914248 0.75 0.623027 0.40 0.863640 0.05 0.914406 0.74 0.635059 0.39 0.866872 0.04 0.914523 0.73 0.646633 0.38 0.869969 0.03 0.914604 0.72 0.657773 0.37 0.872933 0.02 0.914652 0.71 0.668498 0.36 0.875768 0.01 0.914675 0.70 0.678830 0.35 0.878477 0.00 0.914680 0.69 0.688784 0.34 0.887062 0.68 0.698377 0.33 0.883528 0.67 0.707625 0.32 0.885876 0.66 0.716542 0.31 0.222110 0.65 0.725142 0.30 0.890232 Cavitation in Hydraulic Machinery 329 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved compression, it is obvious that any other thermodynamic process may be assumed for the gas in the cavity, and equations analogous to Eq. (8.13) may be formulated. As another interesting aspect of the bubble collapse, Rayleigh calculated the pressure field in the liquid surrounding the bubble reverting to the empty cavity of zero pressure. He set up the radial acceleration as the total differential of the liquid velocity u, at radius r, with respect to time, equated this to the radial pressure gradient, and integrated to get the pressure at any point in the liquid. Hence, a r = 2 du dt = 2 ›u ›t 2u ›u ›t = 1 r ›p ›r (8:15) Expressions for ›u=›t and u(›u=›r) as functions of R and r are obtained from Eqs. (8.7) and (8.10), the partial differential of Eq. (8.7) being taken with respect to r and t, and the partial differential of Eq. (8.7) with respect to t. Substituting these expressions in Eq. (8.15) yields: 1 p 1 ›p ›r = R 3r 2 (4z 24)R 3 r 3 2(z 24) _ _ (8:16) in which z = *(R 0 =R) 3 and r # R always. By integration, this becomes 1 p 1 _ p p 1 dp = R 3 (4z 24)R 3 _ r 1 dr r 5 2(z 24) _ r 1 dr r 2 _ _ (8:17) which gives p p 1 21 = R 3r (z 24) 2 R 4 3r 4 (z 21) (8:18) The pressure distribution in the liquid at the instant of release is obtained by substituting R = R 0 in Eq. (8.18), which gives p = p 1 1 2 R 0 r _ _ (8:19) In Eq. (8.18), z = 1 at the initiation of the collapse and increases as collapse proceeds. Figure 8.1 shows the distribution of the pressure in the liquid according to Eq. (8.18). It is seen that for 1 , z , 4; p max = p 1 and occurs at R=r = 0; where r !1: For 4 , z , 1; p max . p 1 and occurs at finite r/R. This location moves toward the bubble with increasing z and approaches r=R = 1:59 as z approaches infinity. The location r m of the maximum pressure in the liquid may be found by setting dp/dr equal to zero in Eq. (8.16). This gives a maximum value for p when r 3 m R 3 = 4z 24 z 24 (8:20) Chapter 8 330 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved When r m is substituted for r in Eq. (8.18), the maximum value of p is obtained as p max p 1 = 1 - (z 24)R 4r m = 1 - (z 24) 4=3 4 4=3 (z 21) 1=3 (8:21) As cavity approaches complete collapse, z becomes great, and Eqs. (8.20) and (8.21) may be approximated by r m = 4 1=3 R = 1:587R (8:22) and p max p 1 = z 4 4=3 = R 3 0 4 4=3 R 3 (8:23) Equations (8.22) and (8.23) taken together show that as the cavity becomes very small, the pressure in the liquid near the boundary becomes very great in spite of the fact that the pressure at the boundary is always zero. This would suggest the possibility that in compressing the liquid some energy can be stored, which would add an additional term to Eq. (8.10). This would invalidate the assumption of incompressibility. Rayleigh himself abandoned this assumption in considering what happens if the cavity collapses on an absolute rigid sphere of radius R. In this treatment, the assumption of incompressibility is abandoned only at Figure 8.1 Rayleigh analysis: pressure profile near a collapsing bubble. Cavitation in Hydraulic Machinery 331 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved the instant that the cavity wall comes in contact with the rigid sphere. From that instant, it is assumed that the kinetic energy of deformation of the same particle is determined by the bulk modulus of elasticity of the fluid, as is common in water- hammer calculations. On this basis, it is found that (P / ) 2 2E 1 2 = rU 2 = p 1 3 R 3 0 R 3 21 _ _ = p 1 3 (z 21) (8:24) where P / is the instantaneous pressure on the surface of the rigid sphere and E is the bulk modulus of elasticity. Both must be expressed in the same units. It is instructive to compare the collapse of the cavity with the predicted collapse based on this simple theory. Figure 8.2 shows this comparison. This similarity is very striking, especially when it is remembered that there was some variation of pressure p 1 during collapse of the actual cavity. It will be noted that the actual collapse time is greater than that predicted by Eq. (8.12). Figure 8.2 Comparison of measured bubble size with the Rayleigh solution for an empty cavity in an incompressible liquid with a constant pressure field. Chapter 8 332 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 8.7 CAVITATION EFFECTS ON PERFORMANCE OF HYDRAULIC MACHINES 8.7.1 Basic Nature of Cavitation Effects on Performance The effects of cavitation on hydraulic performance are many and varied. They depend upon the type of equipment or structure under consideration and the purpose it is designed to fulfill. However, the basic elements, which together make up these effects on performance, are stated as follows: 1. The presence of a cavitation zone can change the friction losses in a fluid flow system, both by altering the skin friction and by varying the form resistance. In general, the effect is to increase the resistance, although this is not always true. 2. The presence of a cavitation zone may result in a change in the local direction of the flow due to a change in the lateral force, which a given element of guiding surface can exert on the flow as it becomes covered by cavitation. 3. With well-developed cavitation the decrease in the effective cross- section of the liquid-flow passages may become great enough to cause partial or complete breakdown of the normal flow. The development of cavitation may seriously affect the operation of all types of hydraulic structures and machines. For example, it may change the rate of discharge of gates or spillways, or it may lead to undesirable or destructive pulsating flows. It may distort the action of control valves and other similar flow devices. However, the most trouble from cavitation effects has been experienced in rotating machinery; hence, more is known about the details of these manifestations. Study of these details not only leads to a better understanding of the phenomenon in this class of equipment but also sheds considerable light on the reason behind the observed effects of cavitation in many types of equipment for which no such studies have been made. Figures 8.3 and 8.4 display the occurrence of cavitation and its effect on the performance of a centrifugal pump. 8.8 THOMA’S SIGMA AND CAVITATION TESTS 8.8.1 Thoma’s Sigma Early in the study of the effects of cavitation on performance of hydraulic machines, a need developed for a satisfactory way of defining the operating conditions with respect to cavitation. For example, for the same machine operating under different heads and at different speeds, it was found desirable Cavitation in Hydraulic Machinery 333 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved to specify the conditions under which the degree of cavitation would be similar. It is sometimes necessary to specify similarity of cavitation conditions between two machines of the same design but of different sizes, e.g., between model and prototype. The cavitation parameter commonly accepted for this purpose was Figure 8.3 Cavitation occurs when vapor bubbles form and then subsequently collapse as they move along the flow path on an impeller. Figure 8.4 Effect of cavitation on the performance of a centrifugal pump. Chapter 8 334 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved proposed by Thoma and is now commonly known as the Thoma sigma, s T . For general use with pumps or turbines, we define sigma as s sv = H sv H (8:25) where H sv , the net positive suction head at some location = total absolute head less vapor-pressure head = [( p atm =g) - ( p=g) - (V 2 =2g) 2( p v =g)]; H is the head produced (pump) or absorbed (turbine), and g is the specific weight of fluid. For turbines with negative static head on the runner, H sv = H a 2H s 2H v - V 2 e 2g - H f (8:26) where H a is the barometric-pressure head, H s the static draft head defined as elevation of runner discharge above surface of tail water, H v the vapor-pressure head, V e the draft-tube exit average velocity (tailrace velocity), and H f the draft- tube friction loss. If we neglect the draft-tube friction loss and exit-velocity head, we get sigma in Thoma’s original form: s T = H a 2H s 2H v H (8:27) Thus s T = s sv 2 V 2 e =2g - H f H (8:28) Sigma (s sv or s T ) has a definite value for each installation, known as the plant sigma. Every machine will cavitate at some critical sigma (s sv c or s T c ). Clearly, cavitation will be avoided only if the plant sigma is greater than the critical sigma. The cavitation parameter for the flow passage at the turbine runner discharge is, say, K d = H d 2H v V 2 d =2g (8:29) where H d is the absolute-pressure head at the runner discharge and V d the average velocity at the runner discharge. Equation (8.29) is similar in form to Eq. (8.25) but they do not have exactly the same significance. The numerator of K d is the actual cavitation-suppression pressure head of the liquid as it discharges from the runner. (This assumes the same pressure to exist at the critical location for cavitation inception.) Its relation to the numerator of s T is H d 2H v = H sv 2 V 2 d 2g (8:30) For a particular machine operating at a particular combination of the operating variables, flow rate, head speed, and wicket-gate setting, Cavitation in Hydraulic Machinery 335 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved V 2 d 2g = C 1 H (8:31) Using the previous relations, it can be shown that Eq. (8.29) may be written as K d = s T C 1 2 1 2 H f C 1 H _ _ - V 2 e =2g V 2 d =2g The term in parenthesis is the efficiency of the draft tube, h dt , as the converter of the entering velocity head to pressure head. Thus the final expression is K d = s T C 1 2h dt - V 2 e V 2 d (8:32) C 1 is a function of both design of the machine and the setting of the guide vane; h dt is a function of the design of the draft tube but is also affected by the guide- vane setting. If a given machine is tested at constant guide-vane setting and operating specific speed, both C 1 and h dt tend to be constant; hence s T and K d have a linear relationship. However, different designs usually have different values of C 1 even for the same specific speed and vane setting, and certainly for different specific speeds. K d , however, is a direct measure of the tendency of the flow to produce cavitation, so that if two different machines of different designs cavitated at the same value of K d it would mean that their guiding surfaces in this region had the dame value of K i . However, sigma values could be quite different. From this point of view, sigma is not a satisfactory parameter for the comparison of machines of different designs. On the other hand, although the determination of the value of K d for which cavitation is incipient is a good measure of the excellence of the shape of the passages in the discharge region, it sheds no light on whether or not the cross-section is an optimum as well. In this respect, sigma is more informative as it characterizes the discharge conditions by the total head rather than the velocity head alone. Both K d and sigma implicitly contain one assumption, which should be borne in mind because at times it may be rather misleading. The critical cavitation zone of the turbine runner is in the discharge passage just downstream from the turbine runner. Although this is usually the minimum-pressure point in the system, it is not necessarily the cross-section that may limit the cavitation performance of the machine. The critical cross-section may occur further upstream on the runner blades and may frequently be at the entering edges rather than trailing edges. However, these very real limitations and differences do not alter the fact that K d and s T are both cavitation parameters and in many respects, they can be used in the same manner. Thus K d (or K evaluated at any location in the machine) can be used to measure the tendency of the flow to cavitate, the conditions of the flow at which cavitation first begins (K i ), or the conditions of the flow corresponding to a certain degree of development of cavitation. Chapter 8 336 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Likewise, s T can be used to characterize the tendency of the flow through a machine to cause cavitation, the point of inception of cavitation, the point at which cavitation first affects the performance, or the conditions for complete breakdown of performance. K i is a very general figure of merit, as its numerical value gives directly the resistance of a given guiding surface to the development of cavitation. Thoma’s sigma can serve the same purpose for the entire machine, but in a much more limited sense. Thus, for example, s T can be used directly to compare the cavitation resistance of a series of different machines, all designed to operate under the same total head. However, the numerical value of s T , which characterizes a very good machine, for one given head may indicate completely unacceptable performance for another. Naturally, there have been empirical relations developed through experience, which show how the s T for acceptable performance varies with the design conditions. Figure 8.5 shows such a relationship. Here, the specific speed has been taken as the characteristic that describes the design type. It is defined for turbines as N s = N ffiffiffiffiffi hp _ H 5=4 (8:33) where N is the rotating speed, hp the power output, and H the turbine head. The ordinate is plant sigma (s T = s plant ): Both sigma and specific speed are based on rated capacity at the design head. In the use of such diagrams, it is always necessary to understand clearly the basis for their construction. Thus, in Fig. 8.5, the solid lines show the minimum- plant sigma for each specific speed at which a turbine can reasonably be expected to perform satisfactorily; i.e., cavitation will be absent or so limited as not to cause efficiency loss, output loss, undesirable vibration, unstable flow, or excessive pitting. Another criterion of satisfactory operation might be that cavitation damage should not exceed a specific amount, measured in pounds of metal removed per year. Different bases may be established to meet other needs. A sigma curve might be related to hydraulic performance by showing the limits of operation for a given drop in efficiency or for a specific loss in power output. Although the parameter sigma was developed to characterize the performance of hydraulic turbines, it is equally useful with pumps. For pumps, it is used in the form of Eq. (8.25). In current practice, the evaluation of H sv varies slightly depending on whether the pump is supplied directly from a forebay with a free surface or forms a part of a closed system. In the former case, H sv is calculated by neglecting forebay velocity and the friction loss between the forebay and the inlet, just as the tailrace velocity and friction loss between the turbine-runner discharge and tail water are neglected. In the latter case, H sv is calculated from the pressure measured at the inlet. Velocity is assumed to be the average velocity, Q/A. Because of this difference in meaning, if the same Cavitation in Hydraulic Machinery 337 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved machine was tested under both types of installation, the results would apparently show a slightly poor cavitation performance with the forebay. 8.8.2 Sigma Tests Most of the detailed knowledge of the effect of cavitation on the performance of hydraulic machines has been obtained in the laboratory, because of the difficulty encountered in nearly all field installations in varying the operating conditions over a wide enough range. In the laboratory, the normal procedure is to obtain data for the plotting of a group of s T curves. Turbine cavitation tests are best Figure 8.5 Experience limits of plant sigma vs. specific speed for hydraulic turbines. Chapter 8 338 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved run by operating the machine at fixed values of turbine head, speed, and guide- vane setting. The absolute-pressure level of the test system is the independent variable, and this is decreased until changes are observed in the machine performance. For a turbine, these changes will appear in the flow rate, the power output, and the efficiency. In some laboratories, however, turbine cavitation tests are made by operating at different heads and speeds, but at the same unit head and unit speed. The results are then shown as changes in unit power, unit flow rate, and efficiency. If the machine is a pump, cavitation tests can be made in two ways. One method is to keep the speed and suction head constant and to increase the discharge up to a cutoff value at which it will no longer pump. The preferable method is to maintain constant speed and flow rate and observe the effect of suction pressure on head, power (or torque), and efficiency as the suction pressure is lowered. In such cases, continual small adjustments in flow rate may be necessary to maintain it at constant value. Figure 8.6 shows curves for a turbine, obtained by operating at constant head, speed, and gate. Figure 8.7 shows curves for a pump, obtained from tests at constant speed and flow rate. These curves are typical in that each performance characteristic shows little or no deviation from its normal value Figure 8.6 Sigma curves for a hydraulic turbine under constant head, speed, and gate opening. (Normal torque, head, and discharge are the values at best efficiency and high sigma.) Cavitation in Hydraulic Machinery 339 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved (at high submergence) until low sigmas are reached. Then deviations appear, which may be gradual or abrupt. In nearly all cases, the pressure head across a pump or turbine is so small in comparison with the bulk modulus of the liquid such that change in system pressure during a sigma test produces no measurable change in the density of the liquid. Thus, in principle, until inception is reached, all quantities should remain constant and the s curves horizontal. Figure 8.8 shows some of the experimental sigma curves obtained from tests of different pumps. It will be noted that the first deviation of head H observed for machines A and C is downward but that for machine B is upward. In each case, the total deviation is considerably in excess of the limits of accuracy of measurements. Furthermore, only machine A shows no sign of change in head until a sharp break is reached. The only acceptable conclusion is, therefore, that the inception point occurs at much higher value of sigma than might be assumed and the effects of cavitation on the performance develop very slowly until a certain degree of cavitation has been reached. Figure 8.7 Sigma curves for a centrifugal pump at constant speed and discharge. (Normal head and discharge are the values at best efficiency and high sigma.) Chapter 8 340 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 8.8.3 Interpretation of Sigma Tests The sigma tests described are only one specialized use of the parameter. For example, as already noted, sigma may be used as a coordinate to plot the results of several different types of experience concerning the effect of cavitation of machines. Even though sigma tests are not reliable in indicating the actual inception of cavitation, attempts have often been made to use them for this purpose on the erroneous assumption that the first departure from the noncavitating value of any of the pertinent parameters marks the inception of cavitation. The result of this assumption has frequently been that serious cavitation damage has been observed in machines whose operation had always been limited to the horizontal portion of the sigma curve. Considering strictly from the effect of cavitation on the operating characteristics, the point where the sigma curve departs from the horizontal may Figure 8.8 Comparison of sigma curves for different centrifugal pumps at constant speed and discharge. (Normal head and discharge are the values at best efficiency and high sigma.) Cavitation in Hydraulic Machinery 341 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved be designed as the inception of the effect. For convenience in operation, points could be designated as s iP , s i , s iH , or s iQ , which would indicate the values of s i for the specified performance characteristics. In Fig. 8.8, such points are marked in each curve. For pumps A and C, the indicated s iH is at the point where the head has decreased by 0.5% from its high sigma value. For pump B, s iH is shown where the head begins to increase from its high sigma value. The curves of Fig. 8.8 show that at some lower limiting sigma, the curve of performance finally becomes nearly vertical. The knee of this curve, where the drop becomes very great, is called the breakdown point. There is remarkable similarity between these sigma curves and the lift curves of hydrofoil cascades. It is interesting to note that the knee of the curve for the cascade corresponds roughly to the development of a cavitation zone over about 10% of the length of the profile and the conditions for heavy vibrations do not generally develop until after the knee has been passed. 8.8.4 Suction Specific Speed It is unfortunate that sigma varies not only with the conditions that affect cavitation but also with the specific speed of the unit. The suction specific speed represents an attempt to find a parameter, which is sensitive only to the factors that affect cavitation. Specific speed as used for pumps is defined as N s = N ffiffiffiffi Q _ H 3=4 (8:34) where N is the rotating speed, Q the volume rate of flow, and H the head differential produced by pump. Suction specific speed is defined as S = N ffiffiffiffi Q _ H 3=4 sv (8:35) where H sv is the total head above vapor at pump inlet. Runners in which cavitation depends only on the geometry and flow in the suction region will develop cavitation at the same value of S. Presumably, for changes in the outlet diameter and head produced by a Francis-type pump runner, the cavitation behavior would be characterized by S. The name “suction specific speed” follows from this concept. The parameter is widely used for pumping machinery but has not usually been applied to turbines. It should be equally applicable to pumps and turbines where cavitation depends only on the suction region of the runner. This is more likely to be the case in low-specific-speed Francis turbines. The following relation between specific speed (as used for pumps), suction specific speed, and Chapter 8 342 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved sigma is obtained from Eqs. (8.34) and (8.35). N s–pump S = H sv H _ _ 3=4 = s 3=4 sv (8:36) A corresponding relation between specific speed as used for turbines, suction specific speed, and sigma can be obtained from Eqs. (8.33) and (8.35) together with the expression hp = h t gQH 550 where h t is the turbine efficiency. Then N s–turb S = s 3=4 sv h t g 550 _ _ 1=2 (8:37) It is possible to obtain empirical evidence to show whether or not S actually possesses the desirable characteristic for which it was developed, i.e., to offer a cavitation parameter that varies only with the factors that affect the cavitation performance of hydraulic machines and is independent of other design characteristics such as total head and specific speed. For example, Fig. 8.9 Figure 8.9 Sigma vs. specific speed for centrifugal, mixed-flow, and propeller pumps. Cavitation in Hydraulic Machinery 343 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved shows a logarithmic diagram of sigma vs. specific speed on which are plotted points showing cavitation limits of individual centrifugal, mixed-flow, and propeller pumps. In the same diagram, straight lines of constant S are shown, each with a slope of (log s sv )=(log N s ) = 3=4 [Eq. (8.36)]. It should be noted that s sv and S vary in the opposite direction as the severity of the cavitation condition changes, i.e., as the tendency to cavitate increases, s sv decreases, but S increases. If it is assumed that as the type of machine and therefore the specific speed change, all the best designs represent an equally close approach to the ideal design to resist cavitation, then a curve passing through the lowest point for each given specific speed should be a curve of constant cavitation performance. Currently, the limit for essentially cavitation-free operation is approximately S = 12,000 for standard pumps in general industrial use. With special designs, pumps having critical S values in the range of 18,000–20,000 are fairly common. For cavitating inducers and other special services, cavitation is expected and allowed. In cases where the velocities are relatively low (such as condensate pumps), several satisfactory designs have been reported for S in the 20,000– 35,000 range. As was explained, Fig. 8.5 shows limits that can be expected for satisfactory performance of turbines. It is based on experience with installed units and presumably represents good average practice rather than the optimum. The line for Francis turbines has been added to Fig. 8.9 for comparison with pump experience. Note that allowable S values for turbines operating with little or no cavitation tend to be higher than those for pumps when compared at their respective design conditions. Note also that the trend of limiting sigma for turbines is at a steeper slope than the constant S lines. This difference of slope can be taken to indicate that either the parameter S is affected by factors other than those involved in cavitation performance or the different specific-speed designs are not equally close to the optimum as regards cavitation. The latter leads to the conclusion that it is easier to obtain a good design from the cavitation point of view for the lower specific speeds. NOTATION a Acceleration A Area C p Pressure coefficient E Modulus of elasticity h Elevation H Head K Cavitation parameter KE Kinetic energy N Rotation speed Chapter 8 344 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved N s Specific speed p Pressure Q Flow rate r Radial distance r m Mean radius R Radius of cavity wall S Suction specific speed t Time u Velocity V Velocity Z Dimensionless volume of bubble r Density h t Turbine efficiency g Specific weight t / Dimensionless time s Cavitation parameter SUFFIXES 0 Undisturbed fluid properties atm Atmospheric values d Dynamic effects e Exit f Friction i Inception properties min Minimum r Radial s Static v vapor Cavitation in Hydraulic Machinery 345 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Appendix THE INTERNATIONAL SYSTEM OF UNITS (SI) Table 1 SI Base Units Quantity Name of unit Symbol Length meter m Mass kilogram kg Time second s Electric current ampere A Thermodynamic temperature kelvin K Luminous intensity candela cd Amount of a substance mole mol Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 2 SI Defined Units Quantity Name of unit Defining equation Capacitance farad, f 1 F ¼ 1As/V Electrical resistance ohm, V 1 V ¼ 1V/A Force newton, N 1 N ¼ 1 kg m/s 2 Potential difference volt, V 1 V ¼ 1 W/A Power watt, W 1 W¼ 1 J/s Pressure pascal, Pa 1 Pa ¼ 1 N/m 2 Temperature kelvin, K K ¼ 8C þ 273.15 Work, heat, energy joule, J 1 J ¼ 1 Nm Table 3 SI Derived Units Quantity Name of unit Symbol Acceleration meter per second square m/s 2 Area square meter m 2 Density kilogram per cubic meter kg/m 3 Dynamic viscosity newton-second per square meter N s/m 2 Force newton N Frequency hertz Hz Kinematic viscosity square meter per second m 2 /s Plane angle radian rad Power watt W Radiant intensity watt per steradian W/sr Solid angle steradian sr Specific heat joule per kilogram-kelvin J/kg K Thermal conductivity watt per meter-kelvin W/m K Velocity meter per second m/s Volume cubic meter m 3 Appendix 348 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 4 Physical Constants in SI Units Quantity Symbol Value — e 2.718281828 — P 3.141592653 — g c 1.00000 kg m N 21 s 22 Avogadro constant N A 6.022169 £ 10 26 kmol 21 Boltzmann constant k 1.380622 £ 10 223 J K 21 First radiation constant C 1 ¼ 2 p hc 2 3.741844 £ 10 216 W m 2 Gas constant R u 8.31434 £ 10 3 J kmol 21 K 21 Gravitational constant G 6.6732 £ 10 211 N m 2 kg 22 Planck constant h 6.626196 £ 10 234 Js Second radiation constant C 2 ¼ hc/k 1.438833 £ 10 22 m K Speed of light in a vacuum c 2.997925 £ 10 8 ms 21 Stefan-Boltzmann constant s 5.66961 £ 10 28 Wm 22 K 24 Table 5 SI Prefixes Multiplier Symbol Prefix Multiplier Symbol Prefix 10 12 T tera 10 22 c centi 10 9 G giga 10 23 m milli 10 6 M mega 10 26 m micro 10 3 k kilo 10 29 n nano 10 2 h hecto 10 212 p pico 10 1 da deka 10 215 f femto 10 21 d deci 10 218 a atto Appendix 349 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 6A Conversion Factors Physical quantity Symbol Conversion factor Area A 1 ft 2 ¼ 0.0929 m 2 1 in. 2 ¼ 6.452 £ 10 24 m 2 Density r 1 lb m /ft 3 ¼ 16.018 kg/m 3 1 slug/ft 3 ¼ 515.379 kg/m 3 Energy, heat Q 1 Btu ¼ 1055.1 J 1 cal ¼ 4.186 J 1 (ft)(lb f ) ¼ 1.3558 J 1 (hp)(h) ¼ 2.685 £ 10 6 J Force F 1 lb f ¼ 4.448 N Heat flow rate q 1 Btu/h ¼ 0.2931 W 1 Btu/s ¼ 1055.1 W Heat flux q 00 1 Btu/(h)(ft 2 ) ¼ 3.1525 W/m 2 Heat generation per unit volume q G 1 Btu/(h)(ft 3 ) ¼ 10.343 W/m 3 Heat transfer coefficient h 1 Btu/(h)(ft 2 )(8F) ¼ 5.678 W/m 2 K Length L 1 ft ¼ 0.3048 m 1 in. ¼ 2.54 cm ¼ 0.0254 m 1 mile ¼ 1.6093 km ¼ 1609.3 m Mass m 1 lb m ¼ 0.4536 kg 1 slug ¼ 14.594 kg Mass flow rate _ m 1 lb m /h ¼ 0.000126 kg/s 1 lb m /s ¼ 0.4536 kg/s Power W 1 hp ¼ 745.7 W 1 (ft)(lb f )/s ¼ 1.3558 W 1 Btu/s ¼ 1055.1 W 1 Btu/h ¼ 0.293 W Pressure p 1 lb f /in. 2 ¼ 6894.8 N/m 2 (Pa) 1 lb f /ft 2 ¼ 47.88 N/m 2 (Pa) 1 atm ¼ 101,325 N/m 2 (Pa) Specific energy Q/m 1 Btu/lb f ¼ 2326.1 J/kg Specific heat capacity c 1 Btu/(lb f )(8F) ¼ 4188 J/kg K Temperature T T(8R) ¼ (9/5) T (K) T(8F) ¼ [T(8C)](9/5) þ 32 T(8F) ¼ [T(K) 2 273.15](9/5) þ 32 Thermal conductivity k 1 Btu/(h)(ft)(8F) ¼ 1.731 W/m K Thermal diffusivity a 1 ft 2 /s ¼ 0.0929 m 2 /s 1 ft 2 /h ¼ 2.581 £ 10 25 m 2 /s Thermal resistance R t 1 (h)(8F)/Btu ¼ 1.8958 K/W (continued) Appendix 350 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 6A Continued Physical quantity Symbol Conversion factor Velocity U 1 ft/s ¼ 0.3048 m/s 1 mph ¼ 0.44703 m/s Viscosity, dynamic m 1 lb m /(ft)(s) ¼ 1.488 N s/m 2 1 centipoise ¼ 0.00100 N s/m 2 Viscosity, kinematic n 1 ft 2 /s ¼ 0.0929 m 2 /s 1 ft 2 /h ¼ 2,581 £ 10 25 m 2 /s Volume V 1 ft 3 ¼ 0.02832 m 3 1 in. 3 ¼ 1.6387 £ 10 25 m 3 1 gal(U.S. liq.) ¼ 0.003785 m 3 Table 6B Temperature Conversion Table K 8C 8F K 8C 8F K 8C 8F 220 253 263 335 62 144 450 177 351 225 248 254 340 67 153 455 182 360 230 243 245 345 72 162 460 187 369 235 238 236 350 77 171 465 192 378 240 233 227 355 82 180 470 197 387 245 228 218 360 87 189 475 202 396 250 223 29 365 92 198 480 207 405 255 218 0 370 97 207 485 212 414 260 213 9 375 102 216 490 217 423 265 28 18 380 107 225 495 222 432 270 23 27 385 112 234 500 227 441 275 2 36 390 117 243 505 232 450 280 7 45 395 122 252 510 237 459 285 12 54 400 127 261 515 242 468 290 17 63 405 132 270 520 247 477 295 22 72 410 137 279 525 252 486 300 27 81 415 142 288 530 257 495 305 32 90 420 147 297 535 262 504 310 37 99 425 152 306 540 267 513 315 42 108 430 157 315 545 272 522 320 47 117 435 162 324 550 277 531 325 52 126 440 167 333 555 282 540 330 57 135 445 172 342 560 287 549 Appendix 351 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 7 SI Saturated Water Specific volume (m 3 /kg) Internal energy (KJ/kg) Enthalpy (KJ/kg) Entropy (KJ/kg K) Temp C T Pressure Kpa P Saturated liquid (v f ) Evap. (v fg ) Saturated vapor (v g ) Saturated liquid (u f ) Evap. (u fg ) Saturated vapor (u g ) Saturated liquid (h f ) Evap. (h fg ) Saturated vapor (h g ) Saturated liquid (s f ) Evap. (s fg ) Saturated vapor (s g ) 0.01 0.6113 0.001000 206.131 206.131 0 2375.33 2375.33 0.00 2501.35 2501.35 0 9.1562 9.1562 5 0.8721 0.001000 147.117 147.118 20.97 2361.27 2382.24 20.98 2489.57 2510.54 0.0761 8.9496 9.0257 10 1.2276 0.001000 106.376 106.377 41.99 2347.16 2389.15 41.99 2477.75 2519.74 0.1510 8.7498 8.9007 15 1.705 0.001001 77.924 77.925 62.98 2333.06 2396.04 62.98 2465.93 2528.91 0.2245 8.5569 8.7813 20 2.339 0.001002 57.7887 57.7897 83.94 2318.98 2402.91 83.94 2454.12 2538.06 0.2966 8.3706 8.6671 25 3.169 0.001003 43.3583 43.3593 104.86 2304.90 2409.76 104.87 2442.30 2547.17 0.3673 8.1905 8.5579 30 4.246 0.001004 32.8922 32.8932 125.77 2290.81 2416.58 125.77 2430.48 2556.25 0.4369 8.0164 8.4533 35 5.628 0.001006 25.2148 25.2158 146.65 2276.71 2423.36 146.66 2418.62 2565.28 0.5052 7.8478 8.3530 40 7.384 0.001008 19.5219 19.5229 167.53 2262.57 2430.11 167.54 2406.72 2574.26 0.5724 7.6845 8.2569 45 9.593 0.001010 15.2571 15.2581 188.41 2248.40 2436.81 188.42 2394.77 2583.19 0.6386 7.5261 8.1649 50 12.350 0.001012 12.0308 12.0318 209.30 2234.17 2443.47 209.31 2382.75 2592.06 0.7037 7.3725 8.0762 55 15.758 0.001015 9.56734 9.56835 230.19 2219.12 2450.08 230.20 2370.66 2600.86 0.7679 7.2234 7.9912 60 19.941 0.001017 7.66969 7.67071 251.09 2205.54 2456.63 251.11 2358.48 2609.59 0.8311 7.0784 7.9095 65 25.03 0.001020 6.19554 6.19656 272.00 2191.12 2463.12 272.03 2346.21 2618.24 0.8934 6.9375 7.8309 70 31.19 0.001023 5.04114 5.04217 292.93 2176.62 2469.55 292.96 2333.85 2626.80 0.9548 6.8004 7.7552 75 38.58 0.001026 4.13021 4.13123 331.87 2162.03 2475.91 313.91 2321.37 2635.28 1.0154 6.6670 7.6824 80 47.39 0.001029 3.40612 3.40715 334.84 2147.36 2482.19 334.88 2308.77 2643.66 1.0752 6.5369 7.6121 85 57.83 0.001032 2.82654 2.82757 355.82 2132.58 2488.40 355.88 2296.05 2651.93 1.1342 6.4102 7.5444 90 70.14 0.001036 2.35953 2.36056 376.82 2117.70 2494.52 376.90 2283.19 2660.09 1.1924 6.2866 7.4790 95 84.55 0.001040 1.98082 1.98186 397.86 2102.70 2500.56 397.94 2270.19 2668.13 1.2500 6.1659 7.4158 100 101.3 0.001044 1.67185 1.67290 418.91 2087.58 2506.50 419.02 2257.03 2676.05 1.3068 6.0480 7.3548 105 120.8 0.001047 1.41831 1.41936 440.00 2072.34 2512.34 440.13 2243.70 2683.83 1.3629 5.9328 7.2958 THERMODYNAMIC PROPERTIES OF WATER A p p e n d i x 3 5 2 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 110 143.3 0.001052 1.20909 1.21014 461.12 2056.96 2518.09 461.27 2230.20 2691.47 1.4184 5.8202 7.2386 115 169.1 0.001056 1.03552 1.03658 482.28 2041.44 2523.72 482.46 2216.50 2698.96 1.4733 5.7100 7.1832 120 198.5 0.001060 0.89080 0.89186 503.48 2055.76 2529.24 503.69 2202.61 2706.30 1.5275 5.6020 7.1295 125 232.1 0.001065 0.76953 0.77059 524.72 2009.91 2534.63 524.96 2188.50 2713.46 1.5812 5.4962 7.0774 130 270.1 0.001070 0.66744 0.66850 546.00 1993.90 2539.90 546.29 2174.16 2720.46 1.6343 5.3925 7.0269 135 313.0 0.001075 0.58110 0.58217 567.34 1977.69 2545.03 567.67 2159.59 2727.26 1.6869 5.2907 6.9777 140 361.3 0.001080 0.50777 0.50885 588.72 1961.30 2550.02 589.11 2144.75 2733.87 1.7390 5.1908 6.9298 145 415.4 0.001085 0.44524 0.44632 610.16 1944.69 2554.86 610.61 2129.65 2740.26 1.7906 5.0926 6.8832 150 475.9 0.001090 0.39169 0.39278 631.66 1927.87 2559.54 632.18 2114.26 2746.44 1.8419 4.9960 6.8378 155 543.1 0.001096 0.34566 0.34676 653.23 1910.82 2564.04 653.82 2098.56 2752.39 1.8924 4.9010 6.7934 160 617.8 0.001102 0.30596 0.30706 674.85 1893.52 2568.37 675.53 2082.55 2758.09 1.9426 4.8075 6.7501 165 700.5 0.001108 0.27158 0.27269 969.55 1875.97 2572.51 697.32 2066.20 2763.53 1.9924 4.7153 6.7078 170 791.7 0.001114 0.24171 0.24283 718.31 1858.14 2576.46 719.20 2049.50 2768.70 2.0418 4.6244 6.6663 175 892.0 0.001121 0.21568 0.21680 740.16 1840.03 2580.19 741.16 2032.42 2773.58 2.0909 4.5347 6.6256 180 1002.2 0.001127 0.19292 0.19405 762.08 1821.62 2583.70 763.21 2014.96 2778.16 2.1395 4.4461 6.5857 185 1122.7 0.001134 0.17295 0.17409 784.08 1802.90 2586.98 785.36 1997.07 2782.43 2.1878 4.3586 6.5464 190 1254.4 0.001141 0.15539 0.15654 806.17 1783.84 2590.01 807.61 1978.76 2786.37 2.2358 4.2720 6.5078 195 1397.8 0.001149 0.13990 0.14105 828.36 1764.43 2592.79 829.96 1959.99 2789.96 2.2835 4.1863 6.4697 200 1553.8 0.001156 0.12620 0.12736 850.64 1744.66 2595.29 852.43 1940.75 2793.18 2.3308 4.1014 6.4322 205 1723.0 0.001164 0.11405 0.11521 873.02 1724.49 2597.52 875.03 1921.00 2796.03 2.3779 4.0172 6.3951 210 1906.3 0.001173 0.10324 0.10441 895.51 1703.93 2599.44 897.75 1900.73 2798.48 2.4247 3.9337 6.3584 215 2104.2 0.001181 0.09361 0.09497 918.12 1682.94 2601.06 920.61 1879.91 2800.51 2,4713 3.8507 6.3221 220 2317.8 0.001190 0.08500 0.08619 940.85 1661.49 2602.35 943.61 1858.51 2802.12 2.5177 3.7683 6.2860 225 2547.7 0.001199 0.07729 0.07849 963.72 1639.58 2603.30 966.77 1836.50 2803.27 2,5639 3.6863 6.2502 230 2794.9 0.001209 0.07037 0.07158 986.72 1617.17 2603.89 990.10 1813.85 2803.95 2.6099 3.6047 6.2146 235 3060.1 0.001219 0.06415 0.06536 1009.88 1594.24 2604.11 1013.61 1790.53 2804.13 2.6557 3.5233 6.1791 240 3344.2 0.001229 0.05853 0.05976 1033.19 1570.75 2603.95 1037.31 1766.50 2803.81 2.7015 3.4422 6.1436 245 3648.2 0.001240 0.05346 0.05470 1056.69 1546.68 2603.37 1061.21 1741.73 2802.95 2.7471 3.3612 6.1083 250 3973.0 0.001251 0.04887 0.05013 1080.37 1522.00 2602.37 1085.34 1716.18 2801.52 2.7927 3.2802 6.0729 255 4319.5 0.001263 0.04471 0.04598 1104.26 1496.66 2600.93 1109.42 1689.80 2799.51 2.8382 3.1992 6.0374 (continued) A p p e n d i x 3 5 3 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 7 Continued Specific volume (m 3 /kg) Internal energy (KJ/kg) Enthalpy (KJ/kg) Entropy (KJ/kg K) Temp C T Pressure Kpa P Saturated liquid (v f ) Evap. (v fg ) Saturated vapor (v g ) Saturated liquid (u f ) Evap. (u fg ) Saturated vapor (u g ) Saturated liquid (h f ) Evap. (h fg ) Saturated vapor (h g ) Saturated liquid (s f ) Evap. (s fg ) Saturated vapor (s g ) 260 4688.6 0.001276 0.04093 0.04220 1128.37 1470.64 2599.01 1134.35 1662.54 2796.89 2.8837 3.1181 6.0018 265 5081.3 0.001289 0.03748 0.03877 1152.72 1443.87 2596.60 1159.27 1634.34 2793.61 2.9293 3.0368 5.9661 270 5498.7 0.001302 0.03434 0.03564 1177.33 1416.33 2593.66 1184.49 1605.16 2789.65 2.9750 2.9551 5,9301 275 5941.8 0.001317 0.03147 0.03279 1202.23 1387.94 2590.17 1210.05 1574.92 2784.97 3,0208 2.8730 5.8937 280 6411.7 0.001332 0.02884 0.03017 1227.41 1358.66 2586.09 1235.97 1543.55 2779.53 3.0667 2.7903 5.8570 285 6909.4 0.001348 0.02642 0.02777 1252.98 1328.41 2581.38 1262.29 1510.97 2773.27 3.1129 2.7069 5.8198 290 7436.0 0.001366 0.02420 0.02557 1278.89 1297.11 2575.99 1289.04 1477.08 2766.13 3.1593 2.6227 5.7821 295 7992.8 0.001384 0.02216 0.02354 1305.21 1264.67 2569.87 1316.27 1441.08 2758.05 3.2061 2.5375 5.7436 300 8581.0 0.001404 0.02027 0.02167 1331.97 1230.9 2562.96 1344.01 1404.93 2748.94 3.2533 2.4511 5.7044 305 9201.8 0.001425 0.01852 0.01995 1359.22 1195.94 2555.16 1372.33 1366.38 2738.72 3.3009 2.3633 5.6642 310 9856.6 0.001447 0.01690 0.01835 1387.03 1159.94 2546.40 1401.29 1325.97 2727.27 3.3492 2.2737 5.6229 315 10547 0.001472 0.01539 0.01687 1415.44 1159.37 2536.55 1430.97 1283.48 2714.44 3.3981 2.1812 5.5803 320 11274 0.001499 0.01399 0.01549 1444.55 1121.11 2525.48 1461.45 1238.64 2700.08 3.4479 2.0882 5.5361 325 12040 0.001528 0.01267 0.01420 1474.44 1080.93 2513.01 1492.84 1191.13 2683.97 3.4987 1.9913 5.4900 330 12845 0.001561 0.01144 0.01300 1505.24 1038.57 2498.91 1525.29 1140.56 2665.85 3.5506 1.8909 5.4416 335 13694 0.001597 0.01027 0.01186 1537.11 993.66 2482.88 1558.98 1086.37 2645.35 3.6040 1.7863 5.3903 340 14586 0.001638 0.00916 0.01080 1570.26 945.77 2464.53 1594.15 1027.86 2622.01 3.6593 1.6763 5.3356 345 15525 0.001685 0.00810 0.00978 1605.01 894.26 2443.30 1631.17 964.02 2595.19 3.7169 1.5594 5,2763 350 16514 0.001740 0.00707 0.00881 1641.81 838.29 2418.39 1670.54 893.38 2563.92 3.7776 1.4336 5.2111 355 17554 0.001807 0.00607 0.00787 1681.41 776.58 2388.52 1713.13 813.59 2526.72 3.8427 1.2951 5.1378 360 18561 0.001892 0.00505 0.00694 1725.19 707.11 2351.47 1760.54 720.52 2481.00 3.9146 1.1379 5.0525 365 19807 0.002011 0.00398 0.00599 1776.13 626.29 2302.67 1815.96 605.44 2421.40 3.9983 0.9487 4.9470 370 21028 0.002213 0.00271 0.00493 1843.84 526.54 2228.53 1890.37 441.75 2332.12 4.1104 0.6868 4.7972 374.1 22089 0.003155 0 0.00315 2029.58 384.69 2029.58 2099.26 0 2099.26 4.4297 0 4.4297 A p p e n d i x 3 5 4 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 8 SI Saturated Water Pressure Entry Pressure Kpa P Temp T C Specific volume (m 3 /kg) Internal energy (KJ/kg) Enthalpy (KJ/kg) Entropy (KJ/kg K) Saturated liquid (v f ) Evap. (v fg ) Saturated vapor (v g ) Saturated liquid (u f ) Evap. (u fg ) Saturated vapor (u g ) Saturated liquid (h f ) Evap. (h fg ) Saturated vapor (h g ) Saturated liquid (s f ) Evap. (s fg ) Saturated vapor (s g ) 0.6613 0.01 0.0001000 206.131 206.132 0 2375.3 2375.3 0.00 2501.30 2501.30 0 9.1562 9.1562 1 6.98 0.0001000 129.20702 129.2080 29.29 2355.69 2384.98 29.29 2484.89 2514.18 0.1059 8.8697 8.9756 1.5 13.03 0.0001001 87.97913 87.98013 54.70 2338.63 2393.32 54.70 2470.59 2525.30 0.1956 8.6322 8.8278 2 17.50 0.0001001 67.00285 67.00385 73.47 2326.02 2399.48 73.47 2460.02 2533.49 0.2607 8.4629 8.7236 2.5 21.08 0.0001002 54.25285 54.25385 88.47 2315.93 2404.40 88.47 2451.56 2540.03 0.3120 8.3311 8.6431 3 24.08 0.0001003 45.66402 45.66502 101.43 2307.48 2408.51 101.03 2444.47 2545.50 0.3545 8.2231 8.5775 4 28.96 0.0001004 34.799915 34.80015 121.44 2293.73 2415.17 121.44 2432.93 2554.37 0.4226 8.0520 8.4746 5 32.88 0.0001005 28.19150 28.19251 137.79 2282.70 2420.49 137.79 2423.66 2561.45 0.4763 7.9187 8.3950 7.5 40.29 0.0001008 19.23674 19.23775 168.86 2261.74 2430.50 168.77 2406.02 2574.79 0.5763 7.6751 8.2514 10 45.81 0.0001010 14.67254 14.67355 191.76 2246.10 2437.89 191.81 2392.82 2584.63 0.6492 7.5010 8.1501 15 53.97 0.0001014 10.02117 10.02218 225.90 2222.83 2448.73 225.91 2373.14 2599.06 0.7548 7.2536 8.0084 20 60.06 0.0001017 7.64835 7.64937 251.35 2205.36 2456.71 251.38 2358.33 2609.70 0.8319 7.0766 7.9085 25 64.97 0.0001020 6.20322 6.20424 271.88 2191.21 2463.08 271.90 2346.29 2618.19 0.8930 6.9383 7.8313 30 69.10 0.0001022 5.22816 5.22918 289.18 2179.21 2468.40 289.21 2336.07 2625.28 0.9439 6.8247 7.7686 40 75.87 0.0001026 3.99243 3.99345 317.51 2159.49 2477.00 317.55 2319.19 2636.74 1.0258 6.6441 7.6700 50 81.33 0.0001030 3.23931 3.24034 340.52 2143.43 2483.85 340.47 2305.40 2645.87 1.0910 6.5029 7.5939 75 91.77 0.0001037 2.21607 2.21711 384.29 2112.39 2496.67 384.36 2278.59 2662.96 1.2129 6.2434 7.4563 100 99.62 0.0001043 1.69296 1.69400 417.33 2088.72 2506.06 417.44 2258.02 2675.46 1.3025 6.0568 7.3593 125 105.99 0.0001048 1.37385 1.37490 444.16 2069.32 2513.48 444.30 2241.05 2685.35 1.3739 5.9104 7.2843 150 111.37 0.0001053 1.15828 1.15933 466.92 2052.72 2519.64 467.08 2226.46 2693.54 1.4335 5.7897 7.2232 175 116.06 0.0001057 1.00257 1.00363 186.78 2038.12 2524.60 486.97 2213.57 2700.53 1.4848 5.6868 7.1717 200 120.23 0.0001061 0.88467 0.88573 504.47 2025.02 2529.49 504.68 2201.96 2706.63 1.5300 5.5960 7.1271 225 124.00 0.0001064 0.79219 0.79325 520.45 2013.10 2533.56 520.69 2191.35 2712.04 1.5705 5.5173 7.0878 250 127.43 0.0001067 0.71765 0.71871 535.08 2002.14 2537.21 535.34 2181.55 2716.89 1.6072 5.4455 7.0526 275 130.60 0.0001070 0.65624 0.65731 548.57 1991.95 2540.53 548.87 2172.42 2721.29 1.6407 5.3801 7.0208 300 133.55 0.0001073 0.60475 0.60582 561.13 1982.43 2543.55 561.45 2163.85 2725.30 1.6717 5.3201 6.9918 (continued) A p p e n d i x 3 5 5 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 8 Continued Pressure Kpa P Temp T C Specific volume (m 3 /kg) Internal energy (KJ/kg) Enthalpy (KJ/kg) Entropy (KJ/kg K) Saturated liquid (v f ) Evap. (v fg ) Saturated vapor (v g ) Saturated liquid (u f ) Evap. (u fg ) Saturated vapor (u g ) Saturated liquid (h f ) Evap. (h fg ) Saturated vapor (h g ) Saturated liquid (s f ) Evap. (s fg ) Saturated vapor (s g ) 325 136.30 0.0001076 0.56093 0.56201 572.88 1973.46 2546.34 573.23 2155.76 2728.99 1.7005 5.2646 6.9651 350 138.88 0.0001079 0.52317 0.52425 583.93 1964.98 2548.92 584.31 2148.10 2732.40 1.7274 5.2130 6.9404 375 141.32 0.0001081 0.49029 0.49137 594.38 1956.93 2551.31 594.79 2140.79 2735.58 1.7527 5.1647 6.9174 400 143.63 0.0001084 0.46138 0.42646 604.29 1949.26 2553.55 604.73 2133.81 2738.53 1.7766 5.1193 6.8958 450 147.93 0.0001088 0.41289 0.41398 622.75 1934.87 2557.62 623.24 2120.67 2743.91 1.8206 5.0359 6.8565 500 151.86 0.0001093 0.37380 0.37489 639.66 1921.57 2561.23 640.21 2108.47 2748.67 1.8606 4.9606 6.8212 550 155.48 0.0001097 0.34159 0.34268 655.30 1909.17 2564.47 655.91 2097.04 2752.94 1.8972 4.8920 6.7892 600 158.48 0.0001101 0.31457 0.31567 669.88 1897.52 2567.70 670.54 2086.26 2756.80 1.9311 4.8289 6.7600 650 162.01 0.0001104 0.29158 0.29268 683.55 1886.51 2570.06 684.26 2076.04 2760.30 1.9627 4.7704 6.7330 700 164.97 0.0001108 0.27126 0.27286 696.43 1876.07 2572.49 697.20 2066.30 2763.50 1.9922 4.7158 6.7080 750 167.77 0.0001111 0.25449 0.25560 708.62 1866.11 2574.73 709.45 2056.98 2766.43 2.0199 4.6647 6.6846 800 170.43 0.0001115 0.23931 0.24043 720.20 1856.58 2576.79 721.10 2048.04 2769.13 2.0461 4.6166 6.6627 850 172.96 0.001118 0.22586 0.22698 731.25 1847.45 2578.69 732.20 2039.43 2771.63 2.0709 4.5711 6.6421 900 175.38 0.001121 0.21385 0.21497 741.81 1838.65 2580.46 742.82 2031.12 2773.94 2.0946 4.5280 6.6225 950 177.69 0.001124 0.20306 0.20419 751.94 1830.17 2582.11 753.00 2023.08 2776.08 2.1171 4.4869 6.6040 1000 179.91 0.001127 0.19332 0.19444 761.67 1821.97 2583.64 762.79 2015.29 2778.08 2.1386 4.4478 6.5864 1100 184.09 0.001133 0.17639 0.17753 780.08 1806.32 2586.40 781.32 2000.36 2781.68 2.1791 4.3744 6.5535 1200 187.99 0.001139 0.16220 0.16333 797.27 1791.55 2588.82 798.64 1986.19 2784.82 2.2165 4.3067 6.5233 1300 191.64 0.001144 0.15011 0.15125 813.42 1777.53 2590.95 814.91 1972.67 2787.58 2.2514 4.2438 6.4953 1400 195.07 0.001149 0.13969 0.14084 828.68 1764.15 2592.83 830.29 1959.72 2790.00 2.2842 4.1850 6.4692 1500 198.32 0.001154 0.13062 0.13177 843.14 1751.30 2594.50 844.87 1947.28 2792.15 2.3150 4.1298 6.4448 1750 205.76 0.001166 0.11232 0.11349 876.44 1721.39 2597.83 878.48 1917.95 2796.43 2.3851 4.0044 6.3895 2000 212.42 0.001177 0.09845 0.09963 906.42 1693.84 2600.26 908.77 1890.74 2799.51 2.4473 3.8935 6.3408 2250 218.45 0.001187 0.08756 0.08875 933.81 1668.18 2601.98 936.48 1865.19 2801.67 2.5034 3.7938 6.2971 2500 223.99 0.001197 0.07878 0.07998 959.09 1644.04 2603.13 962.09 1840.98 2803.07 2.5546 3.7028 6.2574 2750 229.12 0.001207 0.07154 0.07275 982.65 1621.16 2603.81 985.97 1817.89 2803.86 2.6018 3.6190 6.2208 3000 233.90 0.001216 0.06546 0.06668 1004.76 1599.34 2604.10 1008.41 1795.73 2804.14 2.6456 3.5412 6.1869 A p p e n d i x 3 5 6 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 3250 238.38 0.001226 0.06029 0.06152 1025.62 1578.43 2604.04 1029.60 1774.37 2803.97 2.6866 3.4685 6.1551 3500 242.60 0.001235 0.04483 0.05707 1045.41 1558.29 2603.70 1049.73 1753.70 2803.43 2.7252 3.4000 6.1252 4000 250.40 0.001252 0.04853 0.04978 1082.28 1519.99 2602.27 1087.29 1714.09 2801.38 2.7963 3.2737 6.0700 5000 263.99 0.001286 0.03815 0.03944 1147.78 1449.34 2597.12 1154.21 1640.12 2794.33 2.9201 3.0532 5.9733 6000 275.44 0.001319 0.03112 0.03244 1205.41 1384.27 2589.69 1213.32 1571.00 2784.33 3.0266 2.8625 5.8891 7000 285.88 0.001351 0.02602 0.02737 1257.51 1322.97 2580.48 1266.97 1505.10 2772.07 3.1210 2.6922 5.8132 8000 295.06 0.001384 0.02213 0.02352 1305.54 1264.25 2569.79 1316.61 1441.33 2757.94 3.2067 2.5365 5.7431 9000 303.44 0.001418 0.01907 0.02048 1350.47 1207.28 2557.75 1363.23 1378.88 2742.11 3.2857 2.3915 5.6771 10000 311.06 0.001452 0.01657 0.01803 1393.00 1151.40 2544.41 1407.53 1317.14 2724.67 3.3595 2.2545 5.6140 11000 318.15 0.001489 0.01450 0.01599 1433.68 1096.06 2529.74 1450.05 1255.55 2705.60 3.4294 2.1233 5.5527 12000 324.75 0.001527 0.01274 0.01426 1472.92 1040.76 2513.67 1491.24 1193.59 2684.83 3.4961 1.9962 5.4923 13000 330.95 0.001567 0.01121 0.01278 1511.09 984.99 2496.08 1531.46 1130.76 2662.22 3.5604 1.8718 5.4323 14000 336.75 0.001611 0.00987 0.01149 1548.53 928.23 2476.76 1571.08 1066.47 2637.55 3.6231 1.7485 5.3716 15000 342.24 0.001658 0.00868 0.01034 1585.58 869.85 2455.43 1610.45 1000.04 2610.49 3.6847 1.6250 5.3097 16000 347.43 0.001711 0.00760 0.00931 1622.63 809.07 2431.70 1650.00 930.59 2580.49 3.7460 1.4995 5.2454 17000 352.37 0.001770 0.00659 0.00836 1660.16 744.80 2404.96 1690.25 856.90 2547.15 3.8078 1.3698 5.1776 18000 357.06 0.001840 0.00565 0.00749 1698.86 675.42 2374.28 1731.97 777.13 2509.09 3.8713 1.2330 5.1044 19000 361.54 0.001924 0.00473 0.00666 1739.87 598.18 2338.05 1776.43 688.11 2464.54 3.9387 1.0841 5.0227 20000 365.81 0.002035 0.00380 0.00583 1758.47 507.58 2293.05 1826.18 583.56 2409.74 4.0137 0.9132 4.9269 21000 369.89 0.002206 0.00275 0.00495 1841.97 388.74 2230.71 1888.30 446.42 2334.72 4.1073 0.6942 4.8015 22000 373.80 0.002808 0.00072 0.00353 1973.16 108.24 2081.39 2034.92 124.04 2158.97 4.3307 0.1917 4.5224 22089 374.14 0.003155 0 0.00315 2029.58 0 2029.58 2099.26 0 2099.26 4.4297 0 4.4297 A p p e n d i x 3 5 7 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 9 SI Superheated Vapor Water Temp (C) v (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) V (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) P ¼ 10 KPa (45.81) P ¼ 50KPa (81.33) Saturated 14.67355 2437.89 2584.63 8.1501 3.24034 2483.85 2645.87 7.5939 50 14.86920 2443.87 2592.56 8.1749 — — — — 100 17.19561 2515.50 2687.56 8.4479 3.41833 2511.61 2682.52 7.6947 150 19.51251 2587.86 2782.99 8.6881 3.88937 2585.61 2780.08 7.9400 200 21.82507 2661.27 2879.52 8.9037 4.35595 2659.85 2877.64 8.1579 250 24.13559 2735.95 2977.31 9.1002 4.82045 2734.97 2975.99 8.3555 300 26.44508 2812.06 3076.51 9.2812 5.28391 2811.33 3075.52 8.5372 400 31.06252 2968.89 3279.51 9.6076 6.20929 2968.43 3278.89 8.8641 500 35.67896 3132.26 3489.05 9.8977 7.13364 3131.94 3488.62 9.1545 600 40.29498 3302.45 3705.40 10.1608 8.05748 3302.22 3705.10 9.4177 700 44.91052 3479.63 3928.73 10.4028 8.98104 3479.45 3928.51 9.6599 800 49.52599 3663.84 4159.10 10.6281 9.90444 3663.70 4158.92 9.8852 900 54.14137 3855.03 4396.44 10.8395 10.82773 3854.91 4396.30 10.0967 1000 58.75669 4053.01 4640.58 11.0392 11.75097 4052.91 4640.46 10.2964 1100 63.37198 4257.47 4891.19 11.2287 12.67418 4257.37 4891.08 10.4858 1200 67.98724 4467.91 5147.78 11.4090 13.59737 4467.82 5147.69 10.6662 1300 72.60250 4683.68 5409.70 11.5810 14.52054 4683.58 5409.61 10.8382 P ¼ 100 KPa (99.62) P ¼ 200 KPa (120.23) Saturated 1.69400 2506.06 2675.46 7.3593 0.88573 2529.49 2706.63 7.1271 150 1.93636 2582.75 2776.38 7.6133 0.95964 2576.87 2768.80 7.2795 200 2.17226 2658.05 2875.27 7.8342 1.08034 2654.39 2870.46 7.5066 A p p e n d i x 3 5 8 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 250 2.40604 2733.73 2974.33 8.0332 1.19880 2731.22 2970.98 7.7085 300 2.63876 2810.41 3074.28 8.2157 1.31616 2808.55 3071.79 7.8926 400 3.01263 2967.85 3278.11 8.5434 1.54930 2966.69 3276.55 8.2217 500 3.56547 3131.54 3488.09 8.8314 1.78139 3130.75 3487.03 8.5132 600 4.02781 3301.94 3704.72 9.0975 2.01297 3301.36 3703.96 8.7769 700 4.48986 3479.24 3928.23 9.3398 2.24426 3478.81 3927.66 9.0194 800 4.95174 3663.53 4158.71 9.5652 2.47539 3663.19 4158.27 9.2450 900 5.41353 3854.77 4396.12 9.7767 2.70643 3854.49 4395.77 9.4565 1000 5.87526 4052.78 4640.31 9.9764 2.93740 4052.53 4640.01 9.6563 1100 6.33696 4257.25 4890.95 10.1658 3.16834 4257.53 4890.68 9.8458 1200 6.79863 4467.70 5147.56 10.3462 3.39927 4467.46 5147.32 10.0262 1300 7.26030 4683.47 5409.49 10.5182 3.63018 4683.23 5409.26 10.1982 P ¼ 300 KPa (133.5) P ¼ 400 KPa (143.63) Saturated 0.60582 2543.55 2775.30 6.9918 0.46246 2553.55 2738.53 6.8958 150 0.63388 2570.79 2760.95 7.0778 0.47084 2564.48 2752.82 6.9299 200 0.71629 2650.65 2865.54 7.3115 0.53422 2646.83 2860.51 7.1706 250 0.79636 2728.69 2967.59 7.5165 0.59512 2726.11 2964.16 7.3788 300 0.87529 2806.69 3069.28 7.7022 0.65484 2804.81 3066.75 7.5661 400 1.03151 2965.53 3274.98 8.0329 0.77232 2964.36 3273.41 7.8984 500 1.18669 3129.95 3485.96 8.3250 0.88934 3129.15 3484.89 8.1912 600 1.34136 3300.79 3703.20 8.5892 1.00555 3300.22 3702.44 8.4557 700 1.49573 3478.38 3927.10 8.8319 1.12147 3477.95 3926.53 8.6987 800 1.64994 3662.85 4157.83 9.0575 1.23422 3662.51 4157.40 8.9244 (continued) A p p e n d i x 3 5 9 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 9 Continued Temp (C) v (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) V (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) P ¼ 300 KPa (133.5) P ¼ 400 KPa (143.63) 900 1.80406 3854.20 4395.42 9.2691 1.35288 3853.91 4395.06 9.1361 1000 1.95812 4052.27 4639.71 9.4689 1.46847 4052.02 4639.41 9.3360 1100 2.11214 4256.77 4890.41 9.6585 1.58404 4256.53 4890.15 9.5255 1200 2.26614 4467.23 5147.07 9.8389 1.69958 4466.99 5146.83 9.7059 1300 2.42013 4682.99 5409.03 10.0109 1.81511 4682.75 5408.80 9.8780 P ¼ 500 KPa (151.86) P ¼ 600 KPa (158.85) Saturated 0.37489 2561.23 2748.67 6.8212 0.31567 2567.40 2756.80 6.7600 200 0.42492 2642.91 2855.37 7.0592 0.35202 2638.91 2850.12 6.9665 250 0.47436 2723.50 2960.68 7.2708 0.39383 2720.86 2957.16 7.1816 300 0.52256 2802.91 3064.20 7.4598 0.43437 2801.00 3061.63 7.3723 350 0.57012 2882.59 3167.65 7.6528 0.47424 2881.12 3156.66 7.5463 400 0.61728 2963.19 3271.83 7.7937 0.51372 2962.02 3270.25 7.7058 500 0.71093 3128.35 3483.82 8.0872 0.59199 3127.55 3482.75 8.0020 600 0.80406 3299.64 3701.67 8.3521 0.66974 3299.07 3700.91 8.2673 700 0.89691 3477.52 3925.97 8.5952 0.74720 3477.08 3925.41 8.5107 800 0.98959 3662.17 4156.96 8.8211 0.82450 3661.83 4156.52 8.7367 900 1.08217 3853.63 4394.71 9.0329 0.90169 3853.34 4394.36 8.9485 1000 1.17469 4051.76 4639.11 9.2328 0.97883 4051.51 4638.81 9.1484 1100 1.26718 4256.29 4889.88 9.4224 1.05594 4256.05 4889.61 9.3381 A p p e n d i x 3 6 0 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 1200 1.35964 4464.76 5164.58 9.6028 1.13302 446.52 5146.34 9.5185 1300 1.45210 4682.52 5408.57 9.7749 1.21009 4682.28 5408.34 9.6906 P ¼ 800 KPa (170.43) P ¼ 1000 KPa (179.91) Saturated 0.24043 2576.79 2769.13 6.6627 0.19444 2583.64 2778.08 6.5864 200 0.26080 2630.61 2839.25 6.8158 0.20596 2621.90 2827.86 6.6939 250 0.29314 2715.46 2949.97 7.0384 0.23268 2709.91 2942.59 6.9246 300 0.32411 2797.14 3056.43 7.2327 0.25794 2793.21 3051.15 7.1228 350 0.35439 2878.16 3161.68 7.4088 0.28247 2875.18 3157.65 7.3010 400 0.38426 2959.66 3267.07 7.5715 0.30659 2957.29 3263.88 7.4650 500 0.44331 3125.95 3480.60 7.8672 0.35411 3124.34 3478.44 7.7621 600 0.50184 3297.91 3699.38 8.1332 0.40109 3296.76 3697.85 8.0289 700 0.56007 3476.22 3924.27 8.3770 0.44779 3475.35 3923.14 8.2731 800 0.61813 3661.14 4155.65 8.6033 0.49432 3660.46 4154.78 8.4669 900 0.67610 3852.77 4393.65 8.8153 0.54075 3852.19 4392.94 8.7118 1000 0.73401 4051.00 4638.20 9.0153 0.58712 4050.49 4637.60 8.9119 1100 0.79188 4255.57 4889.08 9.2049 0.63345 4255.09 4888.55 9.1016 1200 0.84974 4466.05 5145.85 9.3854 0.67977 4465.58 5145.36 9.2821 1300 0.90758 4681.81 5407.87 9.5575 0.72608 4681.33 5407.41 9.4542 P ¼ 1200 KPa (187.99) P ¼ 1400 KPa (195.07) Saturated 0.16333 2588.82 2784.82 6.5233 0.14084 2592.83 2790.00 6.4692 200 0.16930 2612.74 2815.90 6.5898 0.14302 2603.09 2803.32 6.4975 250 0.19235 2704.20 2935.01 6.8293 0.16350 2698.32 2927.22 6.7467 300 0.21382 2789.22 3045.80 7.0316 0.18228 2785.16 3040.35 6.9533 350 0.23452 2872.16 3153.59 7.2120 0.20026 2869.12 3149.49 7.1359 (continued) A p p e n d i x 3 6 1 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 9 Continued Temp (C) v (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) V (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) P ¼ 1200 KPa (187.99) P ¼ 1400 KPa (195.07) 400 0.25480 2954.90 3260.66 7.3773 0.21780 2952.50 3257.42 7.3025 500 0.29463 3122.72 3476.28 7.6758 0.25215 3121.10 3474.11 7.6026 600 0.33393 3295.60 3696.32 7.9434 0.28596 3294.44 3694.78 7.8710 700 0.37294 3474.48 3922.01 8.1881 0.31947 3473.61 3920.87 8.1660 800 0.41177 3659.77 4153.90 8.4149 0.35281 3659.09 4153.03 8.3431 900 0.45051 3851.62 4392.23 8.6272 0.38606 3851.05 4391.53 8.5555 1000 0.48919 4049.98 4637.00 8.8274 0.41926 4049.47 4636.41 8.7558 1100 0.52783 4254.61 4888.02 9.0171 0.45239 4254.14 4887.49 8.9456 1200 0.56646 4465.12 5144.87 9.1977 0.48552 4464.65 5144.38 9.1262 1300 0.60507 4680.86 5406.95 9.3698 0.51864 4680.39 5406.49 9.2983 P ¼ 1600 KPa (201.40) P ¼ 1800 KPa (207.15) Saturated 0.12380 2595.95 2794.02 6.4217 0.11042 2598.38 2797.13 6.3793 250 0.14148 2692.26 2919.20 6.6732 0.12497 2686.02 2910.96 6.6066 300 0.15862 2781.03 3034.83 6.8844 0.14021 2776.83 3029.21 6.8226 350 0.17456 2866.05 3145.35 7.0693 0.15457 2862.95 3141.18 7.0099 400 0.19005 2950.09 3254.17 7.2373 0.16847 2947.66 3250.90 7.1793 500 0.22029 3119.47 3471.93 7.5389 0.19550 3117.84 3469.75 7.4824 600 0.24998 3293.27 3693.23 7.8080 0.22199 3292.10 3691.69 7.7523 700 0.27937 3472.74 3919.73 8.0535 0.24818 3471.87 3918.59 7.9983 800 0.30859 3658.40 4152.15 8.2808 0.27420 3657.71 4151.27 8.2258 900 0.33772 3850.47 4390.82 8.4934 0.30012 3849.90 4390.11 8.4386 A p p e n d i x 3 6 2 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 1000 0.36678 4048.96 4635.81 8.6938 0.32598 4048.45 4635.21 8.6390 1100 0.39581 4253.66 4886.95 8.8837 0.35180 4253.18 4886.42 8.8290 1200 0.42482 4464.18 5143.89 9.0642 0.37761 4463.71 5143.40 9.0096 1300 0.45382 4679.92 5406.02 9.2364 0.40340 4679.44 5405.56 9.1817 P ¼ 2000 KPa (212.42) P ¼ 2500 KPa (223.99) Saturated 0.09963 2600.26 2799.51 6.3408 0.07998 2603.13 2803.07 6.2574 250 0.11144 2679.58 2902.46 6.5452 0.08700 2662.55 2880.06 6.4084 300 0.12547 2772.56 3023.50 6.7663 0.09890 2761.56 3008.81 6.6437 350 0.13857 2859.81 3136.96 6.9562 0.10976 2851.84 3126.24 6.8402 400 0.15120 2945.21 3247.60 7.1270 0.12010 2939.03 3239.28 7.0147 450 0.16353 3030.41 3357.48 7.2844 0.13014 3025.43 3350.77 7.1745 500 0.17568 3116.20 3467.58 7.4316 0.13998 3112.08 3462.04 7.3233 600 0.19960 3290.93 3690.14 7.7023 0.15930 3287.99 3686.25 7.5960 700 0.22323 3470.99 3917.45 7.9487 0.17832 3468.80 3914.59 7.8435 800 0.24668 3657.03 4150.40 8.1766 0.19716 3655.30 4148.20 8.0720 900 0.27004 3849.33 4389.40 8.3895 0.21590 3847.89 4387.64 8.2853 1000 0.29333 4047.94 4634.61 8.5900 0.23458 4046.67 4633.12 8.4860 1100 0.31659 4252.71 4885.89 8.7800 0.25322 4251.52 4884.57 8.6761 1200 0.33984 4463.25 5142.92 8.9606 0.27185 4462.08 5141.70 8.8569 1300 0.36306 4678.97 5405.10 9.1328 0.29046 4677.80 5403.95 9.0921 P ¼ 3000 KPa (233.90) P ¼ 3500 KPa (242.60) Saturated 0.06668 2604.10 2804.14 6.1869 0.05707 2603.70 2803.43 6.1252 250 0.07058 2644.00 2855.75 6.2871 0.05873 2623.65 2829.19 6.1748 300 0.08114 2750.05 2993.43 6.5389 0.06842 2737.99 2977.46 6.4460 (continued) A p p e n d i x 3 6 3 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 9 Continued Temp (C) v (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) V (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) P ¼ 3000 KPa (233.90) P ¼ 3500 KPa (242.60) 350 0.09053 2843.66 3115.25 6.7427 0.07678 2835.27 3103.99 6.6578 400 0.09936 2932.75 3230.82 6.9211 0.08453 2926.37 3222.24 6.8404 450 0.10787 3020.38 3344.00 7.0833 0.09196 3015.28 3337.15 7.0051 500 0.11619 3107.92 3456.48 7.2337 0.09918 3103.73 3450.87 7.1571 600 0.13243 3285.03 3682.34 7.5084 0.11324 3282.06 3678.40 7.4338 700 0.14838 3466.59 3911.72 7.7571 0.12699 3464.37 3908.84 7.6837 800 0.16414 3653.58 4146.00 7.9862 0.14056 3651.84 4143.80 7.9135 900 0.17980 3846.46 4385.87 8.1999 0.15402 3845.02 4384.11 8.1275 1000 0.19541 4045.40 4631.63 8.4009 0.16743 4044.14 4630.14 8.3288 1100 0.21098 4250.33 4883.26 8.5911 0.18080 4249.14 4881.94 8.5191 1200 0.22652 4460.92 5140.49 8.7719 0.19415 4459.76 5139.28 8.7000 1300 0.24206 4676.63 5402.81 8.9942 0.20749 4675.45 5401.66 8.8723 P ¼ 4000 KPa (250.40) P ¼ 4500 KPa (257.48) Saturated 0.04978 2602.27 2801.38 6.0700 0.04406 2600.03 2798.29 6.0198 300 0.05884 2725.33 2960.68 6.3614 0.05135 2712.00 2943.07 6.2827 350 0.06645 2826.65 3092.43 6.5820 0.05840 2817.78 3080.57 6.5130 400 0.07341 2919.88 3213.51 6.7689 0.06475 2913.29 3204.65 6.7046 450 0.08003 3010.13 3330.23 6.9362 0.07074 3004.91 3323.23 6.8745 500 0.08643 3099.49 3445.21 7.0900 0.07661 3095.23 3439.51 7.0300 600 0.09885 3279.06 3674.44 7.3688 0.08765 3276.04 3670.47 7.3109 700 0.11095 3462.15 3905.94 7.6198 0.09847 3459.91 3903.04 7.5631 A p p e n d i x 3 6 4 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 800 0.12287 3650.11 4141.59 7.8502 0.10911 3648.37 4139.38 7.7942 900 0.13469 3843.59 4382.34 8.0647 0.11965 3842.15 4380.58 8.0091 1000 0.14645 4042.87 4328.65 8.2661 0.13013 4041.61 4627.17 8.2108 1100 0.15817 4247.96 4880.63 8.4566 0.14056 4246.78 4879.32 8.4014 1200 0.16987 4458.60 5138.07 8.6376 0.15098 4457.45 5136.87 8.5824 1300 0.18156 4674.29 5400.52 8.8099 0.16139 4673.12 5399.38 8.7548 P ¼ 5000 KPa (263.99) P ¼ 6000 KPa (275.64) Saturated 0.03944 2597.12 2794.33 5.9733 0.03244 2589.69 2784.33 5.8891 300 0.04532 2697.94 2924.53 6.2083 0.03616 2667.22 2884.19 6.0673 350 0.05194 2808.67 3068.39 6.4492 0.04223 2789.61 3042.97 6.3334 400 0.05781 2906.58 3195.64 6.6458 0.04739 2892.81 3177.17 6.5407 450 0.06330 2999.64 3316.15 6.8185 0.05214 2988.90 3301.76 6.7192 500 0.06857 3090.92 3433.76 6.9758 0.05665 3082.20 3422.12 6.8802 550 0.07368 3181.82 3550.23 7.1217 0.06101 3174.57 3540.62 7.0287 600 0.07879 3273.01 3666.47 7.2588 0.06525 3266.89 3658.40 7.1676 700 0.08849 3457.67 3900.13 7.5122 0.07352 3453.15 3894.28 7.4234 800 0.09811 3646.62 4137.17 7.7440 0.08160 3643.12 4132.74 7.6566 900 0.10762 3840.71 4378.82 7.9593 0.08958 3837.84 4375.29 7.8727 1000 0.11707 4040.35 4625.69 8.1612 0.09749 4037.83 4622.74 8.0751 1100 0.12648 4245.61 4878.02 8.3519 0.10536 4243.26 4875.42 8.2661 1200 0.13587 4456.30 5135.67 8.5330 0.11321 4454.00 5133.28 8.4473 1300 0.14526 4671.96 5398.24 8.7055 0.12106 4669.64 5397.97 8.6199 (continued) A p p e n d i x 3 6 5 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 9 Continued Temp (C) v (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) V (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) P ¼ 7000 KPa (285.88) P ¼ 8000 KPa (295.06) Saturated 0.02737 2580.48 2772.07 5.8132 0.02352 2569.79 2757.94 5.7431 300 0.02947 2632.13 2838.40 5.9304 0.02426 2590.93 2784.98 5.7905 350 0.03524 2769.34 3016.02 6.2282 0.02995 2747.67 2987.30 6.1300 400 0.03993 2878.55 3158.07 6.4477 0.03432 2863.75 3138.28 6.3633 450 0.04416 2977.91 3287.04 6.6326 0.03817 2966.66 3271.99 6.5550 500 0.04814 3073.33 3410.29 6.7974 0.04175 3064.30 3398.27 6.7239 550 0.05195 3167.21 3530.87 6.9486 0.04516 3159.76 3521.01 6.8778 600 0.05565 3260.69 3650.26 7.0894 0.04845 3254.43 3642.03 7.0205 700 0.06283 3448.60 3888.39 7.3476 0.05481 3444.00 3882.47 7.2812 800 0.06981 3639.61 4128.30 7.5822 0.06097 3636.08 4123.84 7.5173 900 0.07669 3834.96 4371.77 7.7991 0.06702 3832.08 4368.26 7.7350 1000 0.08350 4035.31 4619.80 8.0020 0.07301 4032.81 4616.87 7.9384 1100 0.09027 4240.92 4872.83 8.1933 0.07896 4238.60 4870.25 8.1299 1200 0.09703 4451.72 5130.90 8.3747 0.08489 4449.45 5128.54 8.3115 1300 0.10377 4667.33 5393.71 8.5472 0.09080 4665.02 5391.46 8.4842 P ¼ 9000 KPa (303.40) P ¼ 10000 KPa (311.06) Saturated 0.02048 2557.75 2742.11 5.6771 0.01803 2544.41 2724.67 5.6140 300 0.02580 2724.38 2956.55 6.0361 0.02242 2699.16 2923.39 5.9442 350 0.02993 2848.38 3117.76 6.2853 0.02641 2832.38 3096.46 6.2119 400 0.03350 2955.13 3256.59 6.4846 0.02975 2943.32 3240.83 6.4189 450 0.03677 3055.12 3386.05 6.6575 0.03279 3045.77 3373.63 6.5965 A p p e n d i x 3 6 6 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 500 0.03987 3152.20 3511.02 6.8141 0.03564 3144.54 3500.92 6.7561 550 0.04285 3248.09 3633.73 6.9588 0.03837 3241.68 3625.34 6.9028 600 0.04574 3343.65 3755.32 7.0943 0.04101 3338.22 3748.27 7.0397 700 0.04857 3439.38 3876.51 7.2221 0.04358 3434.72 3870.52 7.1687 800 0.05409 3632.53 4119.38 7.4597 0.04859 3628.97 4114.91 7.4077 900 0.05950 3829.20 4364.74 7.6782 0.05349 3826.32 4361.24 7.6272 1000 0.06482 4030.30 4613.95 7.8821 0.05832 4027.81 4611.04 7.8315 1100 0.07016 4236.28 4867.69 8.0739 0.06312 4233.97 4865.14 8.0236 1200 0.07544 4447.18 5126.18 8.2556 0.06789 4444.93 5123.84 8.2054 1300 0.08072 4662.73 5389.22 8.4283 0.07265 4660.44 5386.99 8.3783 P ¼ 12500 KPa (327.89) P ¼ 15000 KPa (342.24) Saturated 0.01350 2505.08 2673.77 5.4623 0.01034 2455.43 2610.49 5.3097 350 0.01613 2624.57 2826.15 5.7117 0.01147 2520.36 2692.41 5.4420 400 0.02000 2789.25 3039.30 6.0416 0.01565 2740.70 2975.44 5.8810 450 0.02299 2912.44 3199.78 6.2718 0.01845 2879.47 3156.15 6.1403 500 0.02560 3021.68 3341.72 6.4617 0.02080 2996.52 3308.53 6.3442 550 0.02801 3124.94 3475.13 6.6289 0.02293 3104.71 3448.61 6.5198 600 0.03029 3225.37 3604.75 6.7810 0.02491 3208.64 3582.30 6.6775 650 0.03248 3324.43 3730.44 6.9218 0.02680 3310.37 3712.32 6.8223 700 0.03460 3422.93 3855.41 7.0536 0.02861 3410.94 3840.12 6.9572 800 0.03869 3620.02 4103.69 7.2965 0.03210 3610.99 4092.43 7.2040 900 0.04267 3819.11 4352.48 7.5181 0.03546 3811.89 4343.75 7.4279 1000 0.04658 4021.59 4603.81 7.7237 0.03875 4015.41 4596.63 7.6347 1100 0.05045 4228.23 4858.82 7.9165 0.04200 4222.55 4852.56 7.8282 1200 0.05430 4439.33 5118.02 8.0987 0.04523 4433.78 5112.27 8.0108 (continued) A p p e n d i x 3 6 7 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 9 Continued Temp (C) v (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) V (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) P ¼ 12500 KPa (327.89) P ¼ 15000 KPa (342.24) 1300 0.05813 4654.76 5381.44 8.2717 0.04845 4649.12 5375.94 8.1839 P ¼ 17500 KPa (354.75) P ¼ 20000 KPa (365.81) Saturated 0.00792 2390.19 2528.79 5.1418 0.00583 2293.05 2409.74 4.9269 400 0.01245 2684.98 2902.82 5.7212 0.00994 2619.22 2818.07 5.5539 450 0.01517 2844.15 3109.69 6.0182 0.01270 2806.16 3060.06 5.9016 500 0.01736 2970.25 3274.02 6.2382 0.01477 2942.82 3238.18 6.1400 550 0.01929 3083.84 3421.37 6.4229 0.01656 3062.34 3393.45 6.3347 600 0.02106 3191.51 3560.13 6.5866 0.01818 3174.00 3537.57 6.5048 650 0.02274 3296.04 3693.94 6.7356 0.01969 3281.46 3675.32 6.6582 700 0.02434 3398.78 3824.67 6.8736 0.02113 3386.46 3809.09 6.7993 750 0.02588 3500.56 3953.48 7.0026 0.02251 3490.01 3940.27 6.9308 800 0.02738 3601.89 4081.13 7.1245 0.02385 3592.73 4069.80 7.0544 900 0.03031 3804.67 4335.05 7.3507 0.02645 3797.44 4326.37 7.2830 1000 0.03316 4009.25 4589.37 7.5588 0.02897 4003.12 4582.45 7.4925 1100 0.03597 4216.90 4846.37 7.7530 0.03145 4211.30 4840.24 7.6874 1200 0.03876 4428.28 5106.59 7.9359 0.03391 4421.81 5100.96 7.8706 1300 0.04154 4643.52 5370.50 8.1093 0.03636 4637.95 5365.10 8.0441 P ¼ 25000 KPa P ¼ 30000 KPa 375 0.001973 1798.60 1847.93 4.0319 0.001789 1737.75 1791.43 3.9303 A p p e n d i x 3 6 8 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 400 0.006004 2430.05 2580.16 5.1418 0.002790 2067.34 2151.04 4.4728 425 0.007882 2609.21 2806.16 5.4722 0.005304 2455.06 2614.17 5.1503 450 0.009162 2720.65 2949.70 5.6743 0.006735 2619.30 2821.35 5.4423 500 0.011124 2884.29 3162.39 5.9592 0.008679 2820.67 3081.03 5.7904 550 0.012724 3017.51 3335.62 6.1764 0.010168 2970.31 3275.36 6.0342 600 0.014138 3137.51 3491.36 6.3602 0.011446 3100.53 3443.91 6.2330 650 0.015433 3251.64 3637.46 6.5229 0.012596 3221.04 3598.93 6.4057 700 0.016647 3361.39 3777.56 6.6707 0.013661 3335.84 3745.67 6.5606 800 0.018913 3574.26 4047.08 6.9345 0.015623 3555.60 4024.31 6.8332 900 0.021045 3782.97 4309.09 7.1679 0.017448 3768.48 4291.93 7.0717 1000 0.023102 3990.92 4568.47 7.3801 0.019196 3978.79 4554.68 7.2867 1100 0.025119 4200.18 4828.15 7.5765 0.020903 4189.18 4816.28 7.4845 1200 0.027115 4412.00 5089.86 7.7604 0.022589 4401.29 5078.97 7.6691 1300 0.029101 4626.91 5354.44 7.9342 0.024266 4615.96 5343.95 7.8432 P ¼ 35000 KPa P ¼ 40000 KPa 375 0.001700 1702.86 1762.37 3.8721 0.001641 1677.09 1742.71 3.8283 400 0.002100 1914.02 1987.52 4.2124 0.001908 1854.52 1930.83 4.1134 425 0.003428 2253.42 2373.41 4.7747 0.002532 2096.83 2198.11 4.5028 450 0.004962 2498.71 2672.36 5.1962 0.003693 2365.07 2512.79 4.9459 500 0.006927 2751.88 2994.34 5.6281 0.005623 2678.36 2903.26 5.4699 550 0.008345 2920.94 3213.01 5.9025 0.006984 2869.69 3149.05 5.7784 600 0.009527 3062.03 3395.49 6.1178 0.008094 3022.61 3346.38 6.0113 650 0.010575 3189.79 3559.49 6.3010 0.009064 3158.04 3520.58 6.2054 700 0.011533 3309.89 3713.54 6.4631 0.009942 3283.63 3681.29 6.3750 800 0.013278 3536.81 4001.54 6.7450 0.011523 3517.89 3978.80 6.6662 900 0.014883 3753.96 4274.87 6.9886 0.012963 3739.42 4257.93 6.9150 (continued) A p p e n d i x 3 6 9 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 9 Continued Temp (C) v (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) V (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) P ¼ 35000 KPa P ¼ 40000 KPa 1000 0.016410 3966.70 4541.05 7.2063 0.014324 3954.64 4527.59 7.1356 1100 0.017895 4178.25 4804.59 7.4056 0.015643 4167.38 4793.08 7.3364 1200 0.019360 4390.67 5068.26 7.5910 0.016940 4380.11 5057.72 7.5224 1300 0.020815 4605.09 5333.62 7.7652 0.018229 4594.28 5323.45 7.6969 P ¼ 50000 KPa P ¼ 60000 KPa 375 0.001559 1638.55 1716.52 3.7638 0.001503 1609.34 1699.51 3.7140 400 0.001731 1788.04 1874.58 4.0030 0.001633 1745.34 1843.35 3.9317 425 0.002007 1959.63 2059.98 4.2733 0.001817 1892.66 2001.65 4.1625 450 0.002486 2159.60 2283.91 4.5883 0.002085 2053.86 2178.96 4.4119 500 0.003892 2525.45 2720.07 5.1725 0.002956 2390.53 2567.88 4.9320 550 0.005118 2763.61 3019.51 5.5485 0.003957 2658.76 2896.16 5.3440 600 0.006112 2941.98 3247.59 5.8177 0.004835 2861.14 3151.21 5.6451 650 0.006966 3093.56 3441.84 6.0342 0.005595 3028.83 3364.55 5.8829 700 0.007727 3230.54 3616.91 6.2189 0.006272 3177.25 3553.56 6.0824 800 0.009076 3479.82 3933.62 6.5290 0.007459 3441.60 3889.12 6.4110 900 0.010283 3710.26 4224.41 6.7882 0.008508 3680.97 4191.47 6.6805 1000 0.011411 3930.53 4501.09 7.0146 0.009480 3906.36 4475.16 6.9126 1100 0.012497 4145.72 4770.55 7.2183 0.010409 4124.07 4748.61 7.1194 1200 0.013561 4359.12 5037.15 7.4058 0.011317 4338.18 5017.19 7.3082 1300 0.014616 4572.77 5303.56 7.5807 0.012215 4551.35 5284.28 7.4837 A p p e n d i x 3 7 0 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 10 SI Compressed Liquid Water Temp (C) v (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) v (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) P ¼ 5000 KPa (263.99) P ¼ 10000 KPa (311.06) Saturated 0.00129 1147.78 1154.21 2.9201 0.001452 1393.00 1407.53 3.3595 0 0.00099 0.03 5.02 0.0001 0.000995 0.10 10.05 0.0003 20 0.00100 83.64 88.64 0.2955 0.00997 83.35 93.62 0.2945 40 0.00101 166.93 171.95 0.5705 0.00100 166.33 176.36 0.5685 60 0.00101 250.21 255.28 0.5284 0.00101 249.34 259.47 0.8258 80 0.00103 333.69 338.83 1.0719 0.00102 332.56 324.81 1.0687 100 0.00104 417.50 422.71 1.3030 0.00104 416.09 426.48 1.2992 120 0.00106 501.79 507.07 1.5232 0.00105 500.07 510.61 1.5188 140 0.00108 586.74 592.13 1.7342 0.00107 584.67 595.40 1.7291 160 0.00110 672.61 678.10 1.9374 0.00110 670.11 681.07 1.9316 180 0.00112 759.62 765.24 2.1341 0.00112 756.63 767.83 2.1274 200 0.00115 848.08 853.85 2.3254 0.00115 844.49 855.97 2.3178 220 0.00119 938.43 944.36 2.5128 0.00118 934.07 945.88 2.5038 240 0.00123 1031.34 1037.47 2.6978 0.00122 1025.94 1038.13 2.6872 260 0.00127 1127.92 1134.30 2.8829 0.00126 1121.03 1133.68 2.8698 280 0.00132 1220.90 1234.11 3.0547 300 0.00140 1328.34 1342.31 3.2468 P ¼ 15000 KPa (342.24) P ¼ 20000 KPa (365.81) Saturated 0.001658 1585.58 1610.45 3.6847 0.002035 1785.47 1826.18 4.0137 0 0.000993 0.15 15.04 0.0004 0.00099 0.20 20.00 0.0004 (continued) A p p e n d i x 3 7 1 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 10 Continued Temp (C) v (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) v (m 3 /kg) u (KJ/kg) h (KJ/kg) s (KJ/kg K) P ¼ 15000 KPa (342.24) P ¼ 20000 KPa (365.81) 20 0.000995 83.05 97.97 0.2934 0.000993 82.75 102.61 0.2922 40 0.00100 165.73 180.75 0.5665 0.00100 165.15 185.14 0.5646 60 0.00101 248.49 263.65 0.8231 0.00101 247.66 267.82 0.8205 80 0.00102 331.46 346.79 1.0655 0.00102 330.38 350.78 1.0623 100 0.00104 414.72 430.26 1.2954 0.00103 413.37 434.04 1.2917 120 0.00105 498.39 514.17 1.5144 0.00105 496.75 517.74 1.5101 140 0.00107 582.64 598.70 1.7241 0.00107 580.67 602.03 1.7192 160 0.00109 667.69 684.07 1.9259 0.00109 665.34 687.11 1.9203 180 0.00112 753.74 770.48 2.1209 0.00111 750.94 773.18 2.1146 200 0.00114 841.04 858.18 2.3103 0.00114 837.90 860.47 2.3031 220 0.00117 929.89 947.52 2.4952 0.00117 925.89 949.27 2.4869 240 0.00121 1020.82 1038.99 2.6770 0.00120 1015.94 1040.04 2.6673 260 0.00126 1114.59 1133.41 2.8575 0.00125 1108.53 1133.45 2.8459 280 0.00131 1212.47 1232.09 3.0392 0.00130 1204.69 1230.62 3.0248 300 0.00138 1316.58 1337.23 3.2259 0.00136 1306.10 1333.29 3.2071 320 0.00147 1431.05 1453.13 3.4246 0.00144 1415.66 1444.53 3.3978 340 0.00163 1567.42 1591.88 3.6545 0.00157 1539.64 1571.01 3.6074 360 0.00182 1702.78 1739.23 3.8770 A p p e n d i x 3 7 2 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved P ¼ 30000 KPa P ¼ 50000 KPa 0 0.000986 0.25 29.82 0.0001 0.000977 0.20 49.03 20.0014 20 0.000989 82.16 111.82 0.2898 0.000980 80.98 130.00 0.2847 40 0.000995 164.01 193.87 0.5606 0.000987 161.84 211.20 0.5526 60 0.001004 246.03 276.16 0.8153 0.000996 242.96 292.77 0.8051 80 0.001016 328.28 358.75 1.0561 0.001007 324.32 374.68 1.0439 100 0.001029 410.76 441.63 1.2844 0.001020 405.86 456.87 1.2703 120 0.001044 493.58 524.91 1.5017 0.001035 487.63 539.37 1.4857 140 0.001062 576.86 608.73 1.7097 0.001052 569.76 622.33 1.6915 160 0.001082 660.81 693.27 1.9095 0.001070 652.39 705.91 1.8890 180 0.001105 745.57 778.71 2.1024 0.001091 735.68 790.24 2.0793 200 0.001130 831.34 865.24 2.2892 0.001115 819.73 875.46 2.2634 220 0.001159 918.32 953.09 2.4710 0.001141 904.67 961.71 2.4419 240 0.001192 1006.84 1042.60 2.6489 0.001170 990.69 1049.20 2.6158 260 0.001230 1097.38 1134.29 2.8242 0.001203 1078.06 1138.23 2.7860 280 0.001275 1190.69 1228.96 2.9985 0.001242 1167.19 1229.26 2.9536 300 0.001330 1287.89 1327.80 3.1470 0.001286 1258.66 1322.95 3.1200 320 0.001400 1390.64 1462.63 3.3538 0.001339 1353.23 1420.17 3.2867 340 0.001492 1501.71 1546.47 3.5425 0.001403 1451.91 1522.07 3.4556 360 0.001627 1626.57 1675.36 3.7492 0.001484 1555.97 1630.16 3.6290 380 0.001869 1781.35 1837.43 4.0010 0.001588 1667.13 1746.54 3.8100 A p p e n d i x 3 7 3 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 11 SI Saturated Solid—Saturated Vapor Water Temp C T Pressure KPa P Specific volume (m 3 /kg) Internal energy (KJ/kg) Enthalpy (KJ/kg) Entropy (KJ/kg K) Saturated solid (v i ) Evap. (v fg ) Saturated vapor (v g ) Saturated solid (u i ) Evap. (u ig ) Saturated vapor (u g ) Saturated solid (h i ) Evap. (v fg ) Saturated vapor (v g ) Saturated solid (s i ) Evap. (s ig ) Saturated vapor (s g ) 0.01 0.6113 0.0010908 206.152 2036.153 2333.40 2708.7 2375.3 2333.40 2834.7 2501.3 21.2210 10.3772 9.1562 0 0.6108 0.0010908 206.314 206.315 2333.42 2708.7 2375.3 2333.42 2834.8 2501.3 21.2211 10.3776 9.1565 22 0.5177 0.0010905 241.662 241.663 2337.61 2710.2 2372.5 2337.61 2835.3 2497.6 21.2369 10.4562 9.2193 24 0.4376 0.0010901 283.798 283.799 2341.78 2711.5 2369.8 2341.78 2835.7 2494.0 21.2526 10.5368 9.2832 26 0.3689 0.0010898 334.138 334.139 2345.91 2712.9 2367.0 2345.91 2836.2 2490.3 21.2683 10.6165 9.3482 28 0.3102 0.0010894 394.413 394.414 2350.02 2714.2 2364.2 2350.02 2836.6 2486.6 21.2939 10.6982 9.4143 210 0.2601 0.0010891 466.756 466.757 2354.09 2715.5 2361.4 2354.09 2837.0 2482.9 21.2995 10.7809 9.4815 212 0.2176 0.0010888 553.802 553.803 2358.14 2716.8 2358.7 2358.14 2837.3 2479.2 21.3150 10.8648 9.5498 214 0.1815 0.0010884 658.824 658.824 2362.16 2718.0 2355.9 2362.16 2837.6 2475.5 21.3306 10.9498 9.6192 216 0.1510 0.0010881 785.906 785.907 2366.14 2719.2 2353.1 2366.14 2837.9 2471.8 21.3461 11.0359 9.6898 218 0.1252 0.0010878 940.182 940.183 2370.10 2720.4 2350.3 2370.10 2838.2 2468.1 21.3617 11.1233 9.7616 220 0.10355 0.0010874 1128.112 1128.113 2374.03 2721.6 2347.5 2374.03 2838.4 2464.3 21.3772 11.2120 9.8348 222 0.08535 0.0010871 1357.863 1357.864 2377.93 2722.7 2344.7 2377.93 2838.6 2460.6 21.3928 11.3020 9.9093 224 0.07012 0.0010868 1639.752 1639.753 2381.80 2723.7 2342.0 2381.80 2838.7 2456.9 21.4083 11.3935 9.9852 226 0.05741 0.0010864 1986.775 1986.776 2385.64 2724.8 2339.2 2385.64 2838.9 2453.2 21.4239 11.4864 10.0625 228 0.04684 0.0010861 2145.200 2145.201 2389.45 2725.8 2336.4 2389.45 2839.0 2449.5 21.4394 11.5808 10.1413 230 0.03810 0.0010858 2945.227 2945.228 2393.23 2726.8 2333.6 2393.23 2839.0 2445.8 21.4550 11.6765 10.2215 232 0.03090 0.0010854 3601.822 3601.823 2396.98 2727.8 2330.8 2396.98 2839.1 2442.1 21.4705 11.7733 10.3028 234 0.02499 0.0010851 4416.252 4416.253 2400.71 2728.7 2328.0 2400.71 2839.1 2438.4 21.4860 11.8713 10.3853 236 0.02016 0.0010848 5430.115 5430.116 2404.40 2729.6 2325.2 2404.40 2839.1 2434.7 21.5014 11.9704 10.4690 238 0.01618 0.0010844 6707.021 6707.022 2408.06 2730.5 2322.4 2408.06 2839.0 2431.0 21.5168 12.0714 10.5546 240 0.01286 0.0010841 8366.395 8366.396 2411.70 2731.3 2319.6 2411.70 2839.9 2427.2 21.5321 12.1768 10.6447 A p p e n d i x 3 7 4 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved THERMODYNAMIC PROPERTIES OF LIQUIDS Table 12 Water at Saturation Pressure Temp T Density r (kg/m 3 ) Coefficient of thermal expansion, b £ 10 4 (l/K) Specific heat, c p (J/kg K) Thermal conductivity, k (W/m k) Thermal Diffusivity, a £ 10 6 (m 2 /s) Absolute viscosity, m £ 10 6 (N s/m 2 ) Kinematic viscosity, n £ 10 6 (m 2 /s) Prandtl number, Pr gb/v 2 £ 10 29 (l/K m 3 ) 8F K 8C £ 6.243 £ 10 22 ¼ (lb m /ft 3 ) £ 0.5556 ¼ (l/R) £ 2.388 £ 10 24 ¼ (Btu/lb m 8F) £ 0.5777 ¼ (Btu/h ft 8F) £ 3.874 £ 10 4 ¼ (ft 2 /h) £ 0.6720 ¼ (lb m /ft s) £ 3.874 £ 10 4 ¼ (ft 2 /h) £ 1.573 £ 10 22 ¼ (l/R ft 3 ) 32 273 0 999.9 20.7 4266 0.558 0.131 1794 1.789 13.7 — 41 278 5 1000.0 — 4206 0.568 0.135 1535 1.535 11.4 — 50 283 10 999.7 0.95 4195 0.577 0.137 1296 1.300 9.5 0.551 59 288 15 999.1 — 4187 0.585 0.141 1136 1.146 8.1 — 68 293 20 998.2 2.1 4182 0.597 0.143 993 1.006 7.0 2.035 77 298 25 997.1 — 4178 0.606 0.146 880.6 0.884 6.1 — 86 303 30 995.7 3.0 4176 0.615 0.149 792.4 0.805 5.4 4.540 95 308 35 994.1 — 4175 0.624 0.150 719.8 0.725 4.8 — (continued) A p p e n d i x 3 7 5 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 12 Continued Temp T Density r (kg/m 3 ) Coefficient of thermal expansion, b £ 10 4 (l/K) Specific heat, c p (J/kg K) Thermal conductivity, k (W/m k) Thermal Diffusivity, a £ 10 6 (m 2 /s) Absolute viscosity, m £ 10 6 (N s/m 2 ) Kinematic viscosity, n £ 10 6 (m 2 /s) Prandtl number, Pr gb/v 2 £ 10 29 (l/K m 3 ) 8F K 8C £ 6.243 £ 10 22 ¼ (lb m /ft 3 ) £ 0.5556 ¼ (l/R) £ 2.388 £ 10 24 ¼ (Btu/lb m 8F) £ 0.5777 ¼ (Btu/h ft 8F) £ 3.874 £ 10 4 ¼ (ft 2 /h) £ 0.6720 ¼ (lb m /ft s) £ 3.874 £ 10 4 ¼ (ft 2 /h) £ 1.573 £ 10 22 ¼ (l/R ft 3 ) 104 313 40 992.2 3.9 4175 0.633 0.151 658.0 0.658 4.3 8.333 113 318 45 990.2 — 4176 0.640 0.155 605.1 0.611 3.9 — 122 323 50 988.1 4.6 4178 0.647 0.157 555.1 0.556 3.55 14.59 167 348 75 974.9 — 4190 0.671 0.164 376.6 0.366 2.23 — 212 373 100 958.4 7.5 4211 0.682 0.169 277.5 0.294 1.75 85.09 248 393 120 943.5 8.5 4232 0.685 0.171 235.4 0.244 1.43 140.0 284 412 140 926.3 9.7 4257 0.684 0.172 201.0 0.212 1.23 211.7 320 433 160 907.6 10.8 4285 0.680 0.173 171.6 0.191 1.10 290.3 356 453 180 886.6 12.1 4396 0.673 0.172 152.0 0.173 1.01 396.5 396 473 200 862.8 13.5 4501 0.665 0.170 139.3 0.160 0.95 517.2 428 493 220 837.0 15.2 4605 0.652 0.167 124.5 0.149 0.90 671.4 464 513 240 809.0 17.2 4731 0.634 0.162 113.8 0.141 0.86 848.5 500 533 260 779.0 20.0 4982 0.613 0.156 104.9 0.135 0.86 1076.0 536 553 280 750.0 23.8 5234 0.588 0.147 98.07 0.131 0.89 1360.0 572 573 300 712.5 29.5 5694 0.564 0.132 92.18 0.128 0.98 1766.0 A p p e n d i x 3 7 6 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 13 Water at Saturation Temperature Saturation Temperature T Saturation Pressure P £ 10 25 (N/m 2 ) Specific volume of Vapor v g (m 3 /kg) Enthalpy h f (KJ/kg) h g (KJ/kg) £ 0.430 ¼ (Btu/lb m ) h fg (KJ/kg) 8F K 8C £ 1.450 £ 10 24 ¼ (psi) £ 16.02 ¼ (ft 3 /lb m ) £ 0.430 ¼ (Btu/lb m ) £ 0.430 ¼ (Btu/lb m ) £ 0.430 ¼ (Btu/lb m ) 32 273 0 0.0061 206.3 20.04 2501 2501 60 283 10 0.0122 106.4 41.99 2519 2477 68 293 20 0.0233 57.833 83.86 2537 2453 86 303 30 0.0424 32.929 125.66 2555 2430 104 313 40 0.0737 19.548 167.45 2574 2406 122 323 50 0.1233 12.048 209.26 2591 2382 140 333 60 0.1991 7.680 251.09 2609 2358 158 343 70 0.3116 5.047 292.97 2626 2333 176 353 80 0.4735 3.410 334.92 2643 2308 194 363 90 0.7010 2.362 376.94 2660 2283 212 373 100 1.0132 1.673 419.06 2676 2257 248 393 120 1.9854 0.892 503.7 2706 2202 284 413 140 3.6136 0.508 589.1 2734 2144 320 433 160 6.1804 0.306 675.5 2757 2082 356 453 180 10.027 0.193 763.1 2777 2014 392 473 200 15.551 0.127 852.4 2791 1939 428 493 220 23.201 0.0860 943.7 2799 1856 464 513 240 33.480 0.0596 1037.6 2801 1764 500 533 260 46.940 0.0421 1135.0 2795 1660 536 553 280 64.191 0.0301 1237.0 2778 1541 572 573 300 85.917 0.0216 1345.4 2748 1403 A p p e n d i x 3 7 7 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 14 Unused Engine Oil (Saturated Liquid) Temp T Density r (kg/m 3 ) Coefficient of thermal expansion, b £ 10 4 (l/K) Specific heat, c p (J/kg K) Thermal conductivity k (W/m k) Thermal diffusivity, a £ 10 10 (m 2 /s) Absolute viscosity, m £ 10 3 (N s/m 2 ) Kinematic viscosity, v £ 10 6 (m 2 /s) Prandtl number, Pr gb/v 2 (l/K m 3 ) 8F K 8C £ 6.243 £ 10 22 ¼ (lb m /ft 3 ) £ 0.5556 ¼ (l/R) £ 2.388 £ 10 24 ¼ (Btu/lb m 8F) £ 0.5777 ¼ (Btu/h ft 8F) £ 3.874 £ 10 4 ¼ (ft 2 /h) £ 0.6720 ¼ (lb m /ft s) £ 3.874 £ 10 4 ¼ (ft 2 /h) £ 1.573 £ 10 22 ¼ (l/R ft 3 ) 32 273 0 899.1 1796 0.147 911 3848.0 4280.0 471.0 68 293 20 888.2 0.70 1880 0.145 872 799.0 900.0 104.0 8475 104 313 40 876.1 1964 0.144 834 210.0 240.0 28.7 140 333 60 864.0 2047 0.140 800 72.5 83.9 10.5 176 353 80 852.0 2121 0.138 769 32.0 37.5 4.90 212 373 100 840.0 2219 0.137 738 17.1 20.3 2.76 248 393 120 829.0 2307 0.135 710 10.3 12.4 1.75 284 413 140 816.9 2395 0.133 686 6.54 8.0 1.16 320 433 160 805.9 2483 0.132 663 4.51 5.6 0.84 A p p e n d i x 3 7 8 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 15 Mercury (Saturated Liquid) Temp T Density r (kg/m 3 ) Coefficient of thermal expansion, b £ 10 4 (l/K) Specific heat, c p (J/kg K) Thermal conductivity k (W/m k) Thermal diffusivity, a £ 10 10 (m 2 /s) Absolute viscosity, m £ 10 4 (N s/m 2 ) Kinematic viscosity, v £ 10 6 (m 2 /s) Prandtl number, Pr gb/v 2 £ 10 210 (l/K m 3 ) 8F K 8C £ 6.243 £ 10 22 ¼ (lb m /ft 3 ) £ 0.556 ¼ (l/R) £ 2.388 £ 10 24 ¼ (Btu/lb m 8F) £ 0.5777 ¼ (Btu/h ft 8F) £ 3.874 £ 10 4 ¼ (ft 2 /h) £ 0.6720 ¼ (lb m /ft s) £ 3.874 £ 10 4 ¼ (ft 2 /h) £ 1.573 £ 10 22 32 273 0 13,628 140.3 8.20 42.99 16.90 0.124 0.0288 68 293 20 13,579 1082 139.4 8.69 46.06 15.48 0.114 0.0249 13.73 122 323 50 13,506 138.6 9.40 50.22 14.05 0.104 0.0207 212 373 100 13,385 137.3 10.51 57.16 12.42 0.0928 0,0162 302 423 150 13,264 136.5 11.49 63.54 11.31 0.0853 0.0134 392 473 200 13,145 157.0 12.34 69.08 10.54 0.0802 0.0116 482 523 250 13,026 135.7 13.07 74.06 9.96 0.0765 0.0103 600 588.7 315.5 12,847 134.0 14.02 81.50 8.65 0.0673 0.0083 A p p e n d i x 3 7 9 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 16 Sodium Temp T Density r (kg/m 3 ) Coefficient of thermal expansion, b £ 10 4 (l/K) Specific heat, c p (J/kg K) Thermal conductivity k (W/m k) Thermal diffusivity, a £ 10 5 (m 2 /s) Absolute viscosity, m £ 10 4 (N s/m 2 ) Kinematic viscosity, v £ 10 7 (m 2 /s) Prandtl number, Pr gb/v 2 £ 10 28 (l/K m 3 ) 8F K 8C £ 6.243 £ 10 22 ¼ (lb m /ft 3 ) £ 0.5556 ¼ (l/R) £ 2.388 £ 10 24 ¼ (Btu/lb m 8F) £ 0.5777 ¼ (Btu/h ft 8F) £ 3.874 £ 10 4 ¼ (ft 2 /h) £ 0.6720 ¼ (lb m /ft s) £ 3.874 £ 10 4 ¼ (ft 2 /h) £ 1.573 £ 10 22 ¼ (l/R ft 3 ) 220 367 94 929 0.27 1382 86.2 6.71 6.99 7.31 0.0110 4.96 400 487 205 902 0.36 1340 80.3 6.71 4.32 4.60 0.0072 16.7 700 644 371 860 1298 72.4 6.45 2.83 3.16 0.0051 1000 811 538 820 1256 65.4 6.19 2.08 2.44 0.0040 1200 978 705 778 1256 59.7 6.19 1.79 2.26 0.0038 A p p e n d i x 3 8 0 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 17 Dry Air at Atmospheric Pressure Temp T Density r (kg/m 3 ) Coefficient of thermal expansion, b £ 10 4 (l/K) Specific heat, c p (J/kg K) Thermal conductivity k (W/m k) Thermal diffusivity, a £ 10 6 (m 2 /s) Absolute viscosity, m £ 10 6 (N s/m 2 ) Kinematic viscosity, v £ 10 6 (m 2 /s) Prandtl number, Pr gb/v 2 £ 10 28 (l/K m 3 ) 8F K 8C £ 6.243 £ 10 22 ¼ (lb m /ft 3 ) £ 0.5556 ¼ (l/R) £ 2.388 £ 10 24 ¼ (Btu/lb m 8F) £ 0.5777 ¼ (Btu/h ft 8F) £ 3.874 £ 10 4 ¼ (ft 2 /h) £ 0.6720 ¼ (lb m /ft s) £ 3.874 £ 10 4 ¼ (ft 2 /h) £ 1.573 £ 10 22 ¼ (l/R ft 3 ) 32 273 0 1,252 3.66 1011 0.0237 19.2 17,456 13.9 0.71 1.85 68 293 20 1,164 3.41 1012 0.0251 22.0 18,240 15.7 0.71 1.36 104 313 40 1,092 3.19 1014 0.0265 24.8 19,123 17.6 0.71 1.01 140 333 60 1,025 3.00 1017 0.0279 27.6 19,907 19.4 0.71 0.782 176 353 80 0,968 2.83 1019 0.0293 30.6 20,790 21.5 0.71 0.600 212 373 100 0,916 2.68 1022 0.0307 33.6 21,673 23.6 0.71 0.472 392 473 200 0,723 2.11 1035 0.0370 49.7 35,693 35.5 0.71 0.164 572 573 300 0,596 1.75 1047 0.0429 68.9 29,322 49.2 0.71 0.0709 752 673 400 0,508 1.49 1059 0.0485 89.4 32,754 64.6 0.72 0.0350 932 773 500 0,442 1.29 1076 0.0540 113.2 35,794 81.0 0.72 0.0193 1832 1273 1000 0,268 0.79 1139 0.0762 240 48,445 181 0.74 0.00236 A p p e n d i x 3 8 1 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved THERMODYNAMIC PROPERTIES OF AIR Table 18 Ideal Gas Properties of Air, Standard Entropy at 0.1 MPa (1 bar) Pressure T K u KJ/kg h KJ/kg s8 KJ/kg P r v r 200 142.768 200.174 6.46260 0.27027 493.466 220 157.071 220.218 6.55812 0.37700 389.150 240 171.379 240.267 6.64535 0.51088 313.274 260 185.695 260.323 6.72562 0.67573 256.584 280 200.022 280.390 6.79998 0.87556 213.257 290 207.191 290.430 6.83521 0.98990 195.361 298.15 213.036 298.615 6.86305 1.09071 182.288 300 214.364 300.473 6.86926 1.11458 179.491 320 228.726 320.576 6.93412 1.39722 152.728 340 243.113 340.704 6.99515 1.72814 131.200 360 257.532 360.863 7.05276 2.11226 113.654 380 271.988 381.060 7.10735 2.55479 99.1882 400 286.487 401.299 7.15926 3.06119 87.1367 420 301.035 421.589 7.20875 3.63727 77.0025 440 315.640 441.934 7.25607 4.28916 68.4088 460 330.306 462.340 7.30142 5.02333 61.0658 480 345.039 482.814 7.34499 5.84663 54.7479 500 359.844 503.360 7.38692 6.76629 49.2777 520 374.726 523.982 7.42736 7.78997 44.5143 540 389.689 544.686 7.46642 8.92569 40.3444 560 404.736 565.474 7.50422 10.18197 36.6765 580 419.871 586.350 7.54084 11.56771 33.4358 600 435.097 607.316 7.57638 13.09232 30.5609 620 450.415 628.375 7.61090 14.76564 28.0008 640 465.828 649.528 7.64448 16.59801 25.7132 660 481.335 670.776 7.67717 18.60025 23.6623 680 496.939 692.120 7.70903 20.78637 21.8182 700 512.639 713.561 7.74010 23.16010 20.1553 720 528.435 735.098 7.77044 25.74188 18.6519 740 544.328 756.731 7.80008 28.54188 17.2894 760 560.316 778.460 7.82905 31.57347 16.0518 780 574.600 800.284 7.85740 34.85061 14.9250 (continued) Appendix 382 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 18 Continued T K u KJ/kg h KJ/kg s8 KJ/kg P r v r 800 592.577 822.202 7.88514 38.38777 13.8972 850 633.422 877.397 7.95207 48.46828 11.6948 900 674.824 933.152 8.01581 60.51977 9.91692 950 716.756 989.436 8.07667 74.81519 8.46770 1000 759.189 1046.221 8.13493 91.65077 7.27604 1050 802.095 1103.478 8.19081 111.3467 6.28845 1100 845.445 1161.180 8.24449 134.2478 5.46408 1150 889.211 1219.298 8.29616 160.7245 4.77141 1200 933.367 1277.805 8.34596 191.1736 4.18568 1250 977.888 1336.677 8.39402 226.0192 3.68804 1300 1022.751 1395.892 8.44046 265.7145 3.26257 1350 1067.936 1455.429 8.48539 310.7426 2.89711 1400 1113.426 1515.270 8.52891 361.6192 2.58171 1450 1159.202 1575.398 8.57111 418.8942 2.30831 1500 1205.253 1635.800 8.61208 483.1554 2.07031 1550 1251.547 1696.446 8.65185 554.9577 1.86253 1600 1298.079 1757.329 8.69051 634.9670 1.68035 1650 1344.834 1818.436 8.72811 723.8560 1.52007 1700 1391.801 1879.755 8.76472 822.3320 1.37858 1750 1438.970 1941.275 8.80039 931.1376 1.25330 1800 1486.331 2002.987 8.83516 1051.051 1.14204 1850 1533.873 2064.882 8.86908 1182.888 1.04294 1900 1581.591 2126.951 8.90219 1327.498 0.95445 1950 1629.474 2189.186 8.93452 1485.772 0.87521 2000 1677.518 2251.581 8.96611 1658.635 0.80410 2050 1725.714 2314.128 8.99699 1847.077 0.74012 2100 1774.057 2376.823 9.02721 2052.109 0.68242 2150 1822.541 2439.659 9.05678 2274.789 0.63027 2200 1871.161 2502.630 9.08573 2516.217 0.58305 2250 1919.912 2565.733 9.11409 2777.537 0.54020 2300 1968.790 2628.962 9.14189 3059.939 0.50124 2350 2017.789 2692.313 9.16913 3364.658 0.46576 2400 2066.907 2755.782 9.19586 3692.974 0.43338 2450 2116.138 2819.366 9.22208 4046.215 0.40378 2500 2165.480 2883.059 9.24781 4425.759 0.37669 2550 2214.133 2946.859 9.27308 4833.031 0.35185 (continued) Appendix 383 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Table 18 Continued T K u KJ/kg h KJ/kg s8 KJ/kg P r v r 2600 2264.481 3010.763 9.29790 5269.505 0.32903 2650 2314.133 3074.767 9.32228 5736.707 0.30805 2700 2363.883 3138.868 9.34625 6236.215 0.28872 2750 2413.727 3203.064 9.36980 6769.657 0.27089 2800 2463.663 3267.351 9.39297 7338.715 0.25443 2850 2513.687 3331.726 9.41576 7945.124 0.23291 2900 2563.797 3396.188 9.43818 8590.676 0.22511 2950 2613.990 3460.733 9.46025 9277.216 0.21205 3000 2664.265 3525.359 9.48198 10006.645 0.19992 Appendix 384 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure A.1 Temperature-Entropy Diagram for Water A p p e n d i x 3 8 5 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Figure A.2 Enthalpy-Entropy Diagram for Water A p p e n d i x 3 8 6 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Bibliography 1. Ackers, P., White, W. R., Perkins, J. A., Harrison, A. J. (1978). Weirs and Flumes for Flow Measurement. New York: John Wiley and Sons, Inc. 2. Aerodynamic Design of Axial-Flow Compressors—Volumes, I, II and III, NACA R.M. E56B03, 1956. 3. ASME Research Committee on Fluid Meters (1961). Flowmeter Computation Handbook. New York: The American Society of Mechanical Engineers. 4. Albertson, M. L., Barton, J. R., Simons, D. B. (1960). Fluid Mechanics for Engineers. Englewood Cliffs, NJ: Prentice-Hall. 5. Allen, J. (1952). Scale Models in Hydraulic Engineering. London: Longmans, Green and Co. Ltd. 6. Abbot, I. H. (1932). The Drag of Two Streamline Bodies as Affected by Protuberances and Appendages, NACA Report. 451. 7. Allen, R. P., Butter, E. A. (1967). An axial flow reversing gas turbine for marine propulsion. Trans. ASME, Journal of Basic Power, Jan. 8. Anderson, L. R., et al. (1970). Axisymmetric one-dimensional compressible flow- theory and application. 37(4), Trans. ASME series E, Dec. 9. Air Compressor Handbook. Caterpillar Tractor Co., 1969. 10. Alleman, M., Walther, R. Centrifugal Compressors for Special Applications. Sulzer Technical Review, Reprint. 11. Alley, D. G., Mathieson, G. C. R. (1951). A Method of Performance Estimation for Axial-Flow Turbines, R and M 2974. 12. Atkins, P. W. (1984). The Second Law. New York: Scientific American Library. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 13. Bridgman, P. (1931). Dimensional Analysis. New Haven, Conn: Yale University Press. 14. Bolz, R. E., True, G. L. (1973). Handbook of Tables for Applied Engineering Science. Chemical Rubber Company. 15. Bradshaw, P. (1964). Experimental Fluid Mechanics. New York: Pergamon Press. 16. Bernoulli, D. (1968). Hydrodynamics. In: Bernoulli, J., Camody, T., Kobus, H., ed. Hydraulics. New York: Dover Publications, Inc. 17. Beckwith, T. G., Buck, N. L. (1961). Mechanical Measurements. Reading, Mass: Addison-Wesley. 18. Baumeister, T., Marks, L. S. (1967). Standard Handbook for Mechanical Engineers. 7th ed. New York: McGraw Hill. 19. Binder, R. C. (1958). Advanced Fluid Mechanics. Vol. 1. Prentice-Hall. 20. Bright, R. H. (1945). The development of gas turbine power plants for traction purposes. Proc. Inst. Mech. Eng., E. 21. Bowen, J. T., Sabersky, R. H., Rannic, W. D. (1951). Investigation of axial flow compressors. Trans. ASME 73. 22. Brgg, S. L., Hawthorne, W. R. (1950). Some exact solutions of the flow through annular cascade discs. J. Aeronaut. Sci. 17. 23. Batchelor, G. K. (1967). An Introduction to Fluid Dynamics. Cambridge: Cambridge University Press. 24. Barter, E. F., King, H. W. (1976). Handbook of Applied Hydraulics. 6th ed. New York: McGraw Hill Book Co. 25. Bogdonoff, S. M. (1948). NACA cascade data for the blade design of high performance axial flow compressors. J. Aeronaut. Sci, Feb. 26. Bullock, R. O., Wilcox, W. W., Moses, J. J. (1946). Experimental and Theoretical Studies of Surging in Continuous Flow Compressors, NACA Report No. 861. 27. Csanady, G. T. (1946). Theory of Turbomachines. McGraw-Hill. 28. Church, F. E. (1950). Steam Turbines. McGraw-Hill. 29. Cohen, H, White, E. M. (1943). The Theoretical Determination of the Three- dimensional Flow in an Axial Compressor with Special Reference to Constant Reaction Blading. A.R.C. Report, 6842. 30. Carmichael, A. D. (1958). Stall Propagation in Compressors. Ph.D. Thesis, Cambridge University. 31. Carter, A. D. S. (1957). The Effect of Reynolds Number on the Performance of a Single-Stage Compressor. Aeronautical Research Council, R and M. 3184. 32. Cocrell, D. J., Markland, E. (1963). Review of incompressible diffuser flow. Aircraft Eng. 35(10):286–292. 33. Church, A. H. (1944). Centrifugal Pumps and Blowers. New York: John Wiley and Sons, Inc. 34. Clauser, F. H. (1956). The turbulent boundary layer. Advances in Applied Mechanics. Vol. 4. New York: Pergamon Press. 35. Cheshire, L. J. (1945). The design and development of centrifugal compressors for aircraft gas turbines. Proc. Inst. Mech. Eng. 153:426–440. 36. Came, P. M. (1978). The development, application and experimental evaluation of a design procedure for centrifugal compressors. Proc. Inst. Mech. Eng. 192(5):49–67. Bibliography 388 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 37. Constant, H. (1945). The early history of the axial type of gas turbine engine. Proc. Inst. Mech. Eng. 153, W.E.P. No. 12. 38. Daily, J. W. (1950). Hydraulic machinery. In: Rouse, H., ed. Engineering Hydraulics. New York: John Wiley and Sons, Inc. 39. Daugherty, R. L. (1915). Centrifugal Pumps. New York: McGraw-Hill Book Company. 40. Daugherty, R. L., Franzini, J. B. (1957). Fluid Mechanics with Engineering Applications. New York: McGraw Hill Book Company. 41. Daily, J. W., Harleman, D. R. F. (1966). Fluid Dynamics. Reading, Mass: Addison- Wesley Publishing Company. 42. Ennis, W. D. (1912). Applied Thermodynamics for Engineers. 3rd ed. New York: D. Van Nostrand. 43. Eckert, E. R. G., Drake, R. M. (1972). Analysis of Heat and Mass Transfer. New York: McGraw-Hill Book Co. 44. Fox, R. W., Kline, S. J. (1961). Flow regimes in curved subsonic diffusers. Trans. ASME 84D:303–316. 45. Fujie, K. (1962). A study of the flow through the rotor of an axial flow compressor. Bull. J.S.M.E. 5:18. 46. Fielding, D., Topps, J.E.C. (1959) Thermodynamic Data for the Calculation of Gas Turbine Performance. H.M.S.O., A.R.C., R and M., No 3099. 47. Ferguson, T. B. (1963). The Centrifugal Compressor Stage. Butterworths Scientific Publishing Company. 48. Freeman, C., Stow, P. The application of computational fluid mechanics to aero gas turbine compressor design and development. Inst. Mech. Engineers. Conference Publications, (1984-3), C70/84. 49. Goldstein, S. (1965). Modern Development in Fluid Dynamics. New York: Dover Publications Inc. 50. Gastelow, J. P. (1964). Potential Flow Through Cascades—A Comparison Between Exact and Appropriate Solutions, A.R.C.C.P. 808. 51. Gastelow, J. P., Horlock, J. H., Marsh, H. (1968). Recent Developments in the Aerodynamic Design of Axial Flow Compressors, Proc. Inst. Mech. Eng. 1833N 52. Horlock, J. H. (1963). Annulus wall boundary layers in axial flow compressor stages. Trans. A.S.M.E. 85D(1):55. 53. Horlock, J. H., Dixon, S. L. (1965). The Off Design Performance of Free Vortex Turbine and Compressor Stages. A.R.C. 27:612. 54. Horlock, J. H. (1966). Axial Flow Turbines. Butterworths Scientific Publication. 55. Howell, W. T. (1963). Approximate three-dimensional flow theory for axial turbomachines. Aeronaut. Q. 14:125. 56. Hydraulic Institute (1975). Standards of the Hydraulic Institute. 13th ed. New York: Hydraulic Institute. 57. Herbert, M. V. (1980). A Method of Performance Prediction for Centrifugal Compressors, H.M.S.O., A.R.C., R and M. No. 3840. 58. Hicks, T. G. (1958). Pump Operation and Maintenance. New York: McGraw Hill Book Company. Bibliography 389 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 59. Hydraulic Institute, 1969. Hydraulic Institute Standards. 12th ed. New York: Hydraulic Institute. 60. Holman, J. P. (1971). Experimental Methods for Engineers. New York: McGraw- Hill Book Company. 61. Hickox, G. H. (1952). Hydraulic Models. In: Davis, C. V., ed. Handbook of Applied Hydraulics. 2nd ed. New York: McGraw-Hill Book Company. 62. Horlock, J. H. (1960). Losses and efficiencies in axial-flow turbines. Int. J. Mech. Eng. Sci. 2:48–75. 63. Hunsaker, J. C., Rightmire, B. G. Engineering Applications of Fluid Mechanics. Pergamon Press. 64. Howell, A. R. (1945). Fluid dynamics of axial compressors and design of axial compressors. Proc. Inst. Mech. Eng. 153:441–452. 65. Hawthorne, W. R., Ringros, J. Actuator Disc Theory of the Incompressible Flow in Free Vortex. 66. Kerrebrack, J. L. (1981). Flow in transonic compressors. A.I.A.A. J. 19:4–19. 67. Kacker, S. C., Okapuu, U. (1981). A Mean Line Prediction Method for Axial Flow Turbine Efficiency, A.S.M.E. Paper 21-GT-58. 68. Kearton, W. J. (1926). Turbo Blowers and Compressors. London: Sir Issac Pitman and Sons Ltd. 69. Keller, C. (1937). Performance of Axial Flow Fans. New York: McGraw- Hill Book Co. 70. Kramer, J. J., Stanitz, J. D. (1953). A Note on Secondary Flow in Rotating Radial Channels, NACA T.N. 301. 71. Keratin, W. J. (1962). Steam Turbines Theory and Practice. London: The English Language Book Society. 72. King, H. W., Brater, E. F. (1963). Handbook of Hydraulics. 5th ed. New York: McGraw-Hill. 73. King, R. C., Crocker, S. (1967). Piping Handbook. 5th ed. NewYork: McGraw-Hill. 74. Lamb, H. (1945). Hydrodynamics. 6th ed. New York: Dover Publications. 75. Liang hear, H. L. (1951). Dimensional Analysis and Theory of Models. New York: John Wiley and Sons, reprint by kreiger. 76. Louis, J. F. (1957). Stalling Phenomena in Axial-Flow Compressors. Ph.D. Thesis, Cambridge University. 77. Loeb, W. A. (1949). A study of the supersonic axial flow compressors. J. Appl. Mech. 16. 78. Lighthill, M. (1945). A Mathematical Method of Cascade Design, ARC R and M No. 2104. 79. Lazar Kiewics, S., Troskolanski, A. T. (1965). Impeller Pumps. Pergamon Press, pp. 410–411. 80. Li, W. H., Lam, S. H. (1964). Principles of Fluid Mechanics. Reading, Mass: Addison-Wesley Publishing Company Inc. 81. Moody, L. F., Zowski, T. (1969). Hydraulic machinery. In: Davis, C. V., Sorensen, K. E., eds. Handbook of Applied Hydraulics. Sec. 26. McGraw-Hill, 26–76. 82. McNally, W. D., Sockol, P. M. (1985). Reviewof computational methods for internal flows with emphasis on turbomachinery, Trans. A.S.M.E. J. Fluids Eng. 107:6–22. Bibliography 390 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 83. Mallinson, D. H., Lewis, W. G. E. (1948). The part load performance of various gas turbine engine schemes. Inst. Mech. Eng. 159:198–219. 84. Mattingley, J. D., Heiser, W. H., Daley, D. H. (1987). Aircraft Engine Design. AIAA Education Series. 85. Marble, F. E. (1948). The flow of a perfect fluid through an axial turbomachine with predescribed blade loading. J. Aeronaut. Sci., Aug. 86. Marble, F. E., Michelson, I. (1951). Analytical Investigation of Some Three Dimensional Flow Problems in Turbomachines, NACA T.N. 2614. 87. Moult, E. S., Pearson, H. (1951). The relative merits of centrifugal and axial compressors for aircraft gas turbines. J. Aeronaut. Soc. 55. 88. Norrie, D. H. (1963). Incompressible Flow Machines. American Elseveir, 1006. 89. Naehlizba, M. In: Mayer, Tr. C., Evans, A.G., ed. Hydraulic Turbines, Their Design and Equipment. Prague, R.A.: Artia. 90. Dates, G. C. (1984). Aerothermodynamics of Gas Turbine and Rocket Propulsion. AIAA Education Series. 91. Ostuka, S. (1956). A Theory of the Secondary Flow in Cascades. Transportation Tech. Research Inst. Rep. No. 15. 92. Pearson, H. (1953). The Aerodynamics of Compressor Blade Vibration. London: Anglo American Aeronautical Conference Royal Aeronautical Society. 93. Pearson, H., Bowmer, T. (1949). Surging in axial compressors. Aeronaut. Q., Nov. 94. Prandtl, L., Tietyens, O. G. (1934). Fundamentals of Hydro and Aeromechanics. New York: McGraw-Hill. 95. Prandtl, L. (1950). Essentials of Fluid Dynamics. London: Blacjie and Son. 96. Railly, J. W. (1951). The flow of an incompressible fluid through an axial turbomachine with any number of rows. Aeronaut. Q. 3. 97. Rannie, W. D., Bowon, J. J., Sabersky, R. H. (1951). Investigations of axial flow compressors. Trans. A.S.M.E. 73. 98. Rannie, W. D., Iura, T. (1954). Experimental investigations of propagating stall in axial flow compressors. Trans. A.S.M.E 76. 99. Rhoden, P. (1944). Investigation of Axial Flow Fans, NACA T.M. 1062. 100. Robinson, E. L. (1928). Report on reheat factors. Mech. Eng. 50:154. 101. Rouse, H., ed. (1950). Engineering Hydraulics. New York: John Wiley and Sons Inc. 102. Rouse, H., et al. (1959). Advanced Mechanics of Fluids. New York: John Wiley and Sons Inc. 103. Streeter, V. L., ed. (1961). Handbook of Fluid Dynamics. New York: McGraw-Hill Book Company. 104. Shames, I. (1962). Mechanics of Fluids. New York: McGraw-Hill Book Company. 105. Shapiro, A. H. (1953). The Dynamics and Thermodynamics of Compressible Fluid Flow. Vols. 1 & 2. New York: The Ronald Press Company. 106. Sorensen, H. A. (1951). Gas Turbines. New York: The Ronald Press Company. 107. Spamhake, W. (1934). Centrifugal Pumps, Turbines and Propellers. Cambridge, Mass: The Technology Press of the Massachusetts Institute of Technology. 108. Stepanoff, A. J. (1957). Centrifugal and Axial Flow Pumps. 2nd ed. New York: John Wiley and Sons Inc. Bibliography 391 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 109. Shepherd, D. C. (1956). Principles of Turbomachinery. London: The Macmillan Company. 110. Thomson, P. A. (1971). Compressible Fluid Dynamics. New York: McGraw-Hill Book Co. 111. Vallentine, H. R. (1980). Applied Hydrodynamics. 2nd ed. New York: John Wiley and Sons Inc. 112. Van Wylen, G. J., Sonntag, R. E. (1978). Fundamentals of Classical Thermodynamics. New York: John Wiley and Sons, Inc. 113. Wood, M. D., Horlock, J. H., Armstrong, E. K. (1953). Experimental Investigations of the Stalled Flow in a Single Stage Axial Compressor, ARC Report No. 17280. 114. Wu, C. H. (1953). Subsonic Flow of Air Through a Single Stage and Seven Stage Compressor, NACA T.N. 2961. 115. Yalin, M. S. (1971). Theory of Hydraulic Models. London: The Macmillan Company. 116. Wilson, D. G. (1984). Design of High Efficiency Turbomachinery and Gas Turbines. MIT Press. 117. Zamansky, M. W. (1937). Heat and Thermodynamics. New York: McGraw-Hill. 118. Zucrow, M. J. (1958). Aircraft and Missile Propulsion. Vols. 1 & 2. New York: John Wiley and Sons. Inc. Bibliography 392 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Although great care has been taken to provide accurate and current information, neither the author(s) nor the publisher, nor anyone else associated with this publication, shall be liable for any loss, damage, or liability directly or indirectly caused or alleged to be caused by this book. The material contained herein is not intended to provide specific advice or recommendations for any specific situation. Trademark notice: Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress. ISBN: 0-8247-0980-2 This book is printed on acid-free paper. Headquarters Marcel Dekker, Inc., 270 Madison Avenue, New York, NY 10016, U.S.A. tel: 212-696-9000; fax: 212-685-4540 Distribution and Customer Service Marcel Dekker, Inc., Cimarron Road, Monticello, New York 12701, U.S.A. tel: 800-228-1160; fax: 845-796-1772 Eastern Hemisphere Distribution Marcel Dekker AG, Hutgasse 4, Postfach 812, CH-4001 Basel, Switzerland tel: 41-61-260-6300; fax: 41-61-260-6333 World Wide Web http://www.dekker.com The publisher offers discounts on this book when ordered in bulk quantities. For more information, write to Special Sales/Professional Marketing at the headquarters address above. Copyright q 2003 by Marcel Dekker, Inc. All Rights Reserved. Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage and retrieval system, without permission in writing from the publisher. Current printing (last digit): 10 9 8 7 6 5 4 3 2 1 PRINTED IN THE UNITED STATES OF AMERICA Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved MECHANICAL ENGINEERING A Series of Textbooks and Reference Books Founding Editor L. L. Faulkner Columbus Division, Battelle Memorial Institute and Department of Mechanical Engineering The Ohio State University Columbus, Ohio 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 2 1. 22. 23. 24. 25. Spring Designer's Handbook, Harold Carlson Computer-Aided Graphics and Design, Daniel L. Ryan Lubrication Fundamentals,J. George Wills Solar Engineering for Domestic Buildings,William A. Himmelmant Applied Engineering Mechanics: Statics and Dynamics, G. Booithroyd and C. Poli Centrifugal Pump Clinic, lgor J. Karassik Computer-Aided Kinetics for Machine Design, Daniel L. Ryan Plastics Products Design Handbook, Part A: Materials and Components; Part B: Processes and Design for Processes,edited by Edward Miller Turbomachinery:Basic Theory and Applications, Earl Logan, Jr. Vibrations of Shells and Plates, Werner Soedel Flat and Corrugated Diaphragm Design Handbook, Mario Di Giovanni Practical Stress Analysis in Engineering Design, Alexander Blake An lntroduction to the Design and Behavior of Bolted Joints, John H. Bickford Optimal Engineering Design: Principles and Applications,James N, Siddall Spring Manufacturing Handbook, Harold Carlson lndustrial Noise Control: Fundamentals and Applications, edited by Lewis H. Bell Gears and Their Vibration: A Basic Approach to Understanding Gear Noise, J. Derek Smith Chains for Power Transmission and Material Handling: Design and Applications Handbook,American Chain Association Corrosion and Corrosion Protection Handbook, edited by Philip A. Schweitzer Gear Drive Systems: Design and Application, Peter Lynwander Controlling In-Plant Airborne Contaminants: Systems Design and Calculations, John D. Constance CAD/CAM Systems Planning and Implementation,Charles S. Knox Probabilistic Engineering Design: Principles and Applications, James N . Siddall Traction Drives: Selection and Application, Frederick W. Heilich Ill and Eugene E. Shube Finite Element Methods: An lntroduction, Ronald L. Huston anld Chris E. Passerello Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 26. Mechanical Fastening of Plastics: An Engineering Handbook, Brayton Lincoln, Kenneth J. Gomes, and James F. Braden 27. Lubrication in Practice: Second Edition, edited by W. S. Robertson 28. Principles of Automated Drafting, Daniel L. Ryan 29. Practical Seal Design, edited by Leonard J. Martini 30. Engineering Documentation for CAD/CAM Applications, Charles S. Knox 31. Design Dimensioning with Computer Graphics Applications, Jerome C. Lange 32. Mechanism Analysis: Simplified Graphical and Analytical Techniques,Lyndon 0. Barton 33. CAD/CAM Systems: Justification, Implementation, Productivity Measurement, Edward J. Preston, George W. Crawford, and Mark E. Coticchia 34. Steam Plant Calculations Manual, V. Ganapathy 35. Design Assurance for Engineers and Managers, John A. Burgess 36. Heat Transfer Fluids and Systems for Process and Energy Applications, Jasbir Singh 37. Potential Flows: Computer Graphic Solutions, Robert H. Kirchhoff 38. Computer-Aided Graphics and Design: Second Edition, Daniel L. Ryan 39. Electronically Controlled Proportional Valves: Selection and Application, Michael J. Tonyan, edited by Tobi Goldoftas 40. Pressure Gauge Handbook,AMETEK, U.S. Gauge Division, edited by Philip W. Harland 41. Fabric Filtration for Combustion Sources: Fundamentals and Basic Technology, R. P. Donovan 42. Design of Mechanical Joints, Alexander Blake 43. CAD/CAM Dictionary, Edward J. Preston, George W. Crawford, and Mark E. Coticchia 44. Machinery Adhesives for Locking, Retaining, and Sealing,Girard S. Haviland 45. Couplings and Joints: Design, Selection, and Application,Jon R. Mancuso 46. Shaft Alignment Handbook, John Piotrowski 47. BASIC Programs for Steam Plant Engineers: Boilers, Combustion, Fluid Flow, and Heat Transfer,V. Ganapathy 48. Solving Mechanical Design Problems with Computer Graphics, Jerome C. Lange 49. Plastics Gearing: Selection and Application, Clifford E. Adams 50. Clutches and Brakes: Design and Selection,William C. Orthwein 51. Transducersin Mechanical and Electronic Design, Harry L. Trietley 52. Metallurgical Applications of Shock-Wave and High-Strain-Rate Phenomena, edited by Lawrence E. Murr, Karl P. Staudhammer, and Marc A. Meyers 53. Magnesium Products Design, Robert S. Busk 54. How to Integrate CAD/CAM Systems: Management and Technology, William D. Engelke 55. Cam Design and Manufacture: Second Edition; with cam design software for the IBM PC and compatibles, disk included, Preben W. Jensen 56. Solid-stateAC Motor Controls: Selection and Application,Sylvester Campbell 57. Fundamentals of Robotics, David D. Ardayfio 58. Belt Selection and Application for Engineers, edited by Wallace D. Erickson 59. Developing Three-DimensionalCAD Software with the ISM PC, C. Stan Wei 60. Organizing Data for CIM Applications, Charles S. Knox, with contributions by Thomas C. Boos, Ross S. Culverhouse, and Paul F. Muchnicki Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 61. Computer-Aided Simulation in Railway Dynamics, by Rao V. Dukkipati and Joseph R. Amyot 62. Fiber-Reinforced Composites: Materials, Manufacturing, and Design, P. K. Mallick 63. Photoelectric Sensors and Controls: Selection and Application, Scott M. Juds 64. Finite Element Analysis with Personal Computers, Edward R. Champion, Jr., and J. Michael Ensminger 65. Ultrasonics: Fundamentals, Technology, Applications: Second Edition, Revised and Expanded, Dale Ensminger 66. Applied Finite Element Modeling: Practical Problem Solving for Engineers, Jeffrey M. Steele 67. Measurement and Instrumentation in Engineering: Principles and Basic Laboratory Experiments, Francis S. Tse and Ivan E. Morse 68. Centrifugal Pump Clinic: Second Edition, Revised and Expanded, lgor J. Karassik 69. Practical Stress Analysis in Engineering Design: Second Edition, Revised and Expanded, Alexander Blake 70. An Introduction to the Design and Behavior of Bolted Joints: Second Edition, Revised and Expanded, John H. Bickford 71. High Vacuum Technology:A Practical Guide, Marsbed H. Hablanian 72. Pressure Sensors: Selection and Application, Duane Tandeske 73. Zinc Handbook: Properties, Processing, and Use in Design, Frank Porter 74. Thermal Fatigue of Metals, Andrzej Weronski and Tadeusz Hejwowski 75. Classical and Modern Mechanisms for Engineers and Inventors, Preben W. Jensen 76. Handbook of Electronic Package Design, edited by Michael Pecht 77. Shock- Wave and High-Strain-Rate Phenomena in Materials, edited by Marc A. Meyers, Lawrence E. Murr, and Karl P. Staudhammer 78. Industrial Refrigeration: Principles, Design and Applications, P. C. Koelet 79. Applied Combustion, Eugene L. Keating 80. Engine Oils and Automotive Lubrication, edited by Wilfried J. Barb 81. Mechanism Analysis: Simplified and Graphical Techniques, Second Edition, Revised and Expanded, Lyndon 0. Barton 82. Fundamental Fluid Mechanics for the Practicing Engineer, James W. Murdock 83. Fiber-Reinforced Composites: Materials, Manufacturing, and Design, Second Edition, Revised and Expanded, P. K. Mallick 84. Numerical Methods for Engineering Applications, Edward R. Champion, Jr. 85. Turbomachinery: Basic Theory and Applications, Second Edition, Revised and Expanded, Earl Logan, Jr. 86. Vibrations of Shells and Plates: Second Edition, Revised and Expanded, Werner Soedel 87. Steam Plant Calculations Manual: Second Edition, Revised and Expanded, V. Ganapathy 88. Industrial Noise Control: Fundamentals and Applications, Second Edition, Revised and Expanded, Lewis H. Bell and Douglas H. Bell 89. Finite Elements: Their Design and Performance, Richard H. MacNeal 90. Mechanical Properties of Polymers and Composites: Second Edition, Revised and Expanded, Lawrence E. Nielsen and Robert F. Landel 91. Mechanical Wear Prediction and Prevention, Raymond G. Bayer Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 92. Mechanical Power Transmission Components, edited by David W. South and Jon R. Mancuso 93. Handbook of Turbomachinery,edited by Earl Logan, Jr. 94. Engineering Documentation Control Practices and Procedures, Ray E. Monahan 95. Refractory Linings ThermomechanicalDesign and Applications, Charles A. Schacht 96. Geometric Dimensioning and Tolerancing:Applications and Techniques for Use in Design, Manufacturing, and Inspection, James D. Meadows 97. An lntroduction to the Design and Behavior of Bolted Joints: Third Edition, Revised and Expanded, John H. Bickford 98. Shaft Alignment Handbook: Second Edition, Revised and Expanded, John Piotrowski 99. Computer-Aided Design of Polymer-Matrix Composite Structures,edited by Suong Van Hoa 100. Friction Science and Technology,Peter J. Blau 101, lntroduction to Plastics and Composites: Mechanical Properties and Engineering Applications, Edward MiIler 102. Practical Fracture Mechanics in Design, Alexander Blake 103. Pump Characteristics and Applications, Michael W. Volk 104. Optical Principles and Technology for Engineers, James E. Stewart 105. Optimizing the Shape of Mechanical Elements and Structures,A. A. Seireg and Jorge Rodriguez 106. Kinematics and Dynamics of Machinery, Vladimir Stejskal and Michael ValaSek 107. Shaft Seals for Dynamic Applications, Les Home 108. Reliability-Based Mechanical Design, edited by Thomas A. Cruse 109. Mechanical Fastening, Joining, and Assembly, James A. Speck 110. TurbomachineryFluid Dynamics and Heat Transfer,edited by Chunill Hah 111. High-Vacuum Technology: A Practical Guide, Second Edition, Revised and Expanded, Marsbed H. Hablanian 112. Geometric Dimensioning and Tolerancing: Workbook and Answerbook, James D. Meadows 113. Handbook of Materials Selection for Engineering Applications, edited by G. T. Murray 114. Handbook of Thermoplastic Piping System Design, Thomas Sixsmith and Reinhard Hanselka 115. Practical Guide to Finite Elements: A Solid Mechanics Approach, Steven M. Lepi 116. Applied Computational Fluid Dynamics, edited by Vijay K. Garg 117. Fluid Sealing Technology,Heinz K. Muller and Bernard S. Nau 118. Friction and Lubrication in Mechanical Design, A. A. Seireg 119. lnfluence Functions and Matrices, Yuri A. Melnikov 120. Mechanical Analysis of Electronic Packaging Systems, Stephen A. McKeown 121. Couplings and Joints: Design, Selection, and Application, Second Edition, Revised and Expanded, Jon R. Mancuso 122. Thermodynamics:Processes and Applications, Earl Logan, Jr. 123. Gear Noise and Vibration,J. Derek Smith 124. Practical Fluid Mechanics for Engineering Applications, John J. Bloomer 125. Handbook of Hydraulic Fluid Technology,edited by George E. Totten 126. Heat Exchanger Design Handbook,T. Kuppan Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 127. Designing for Product Sound Quality, Richard H. Lyon 128. Probability Applications in Mechanical Design, Franklin E. Fisher and Joy R. Fisher 129. Nickel Alloys, edited by Ulrich Heubner 130. Rotating Machinery Vibration: Problem Analysis and Troubleshooting, Maurice L. Adams, Jr. 131. Formulas for Dynamic Analysis, Ronald L. Huston and C. Q. Liu 132. Handbook of Machinery Dynamics, Lynn L. Faulkner and Earl Logan, Jr. 133. Rapid Prototyping Technology: Selection and Application, Kenneth G. Cooper 134. Reciprocating Machinery Dynamics: Design and Analysis, Abdulla S. Rangwala 135. Maintenance Excellence: Optimizing Equipment Life-Cycle Decisions, edited by John D. Campbell and Andrew K. S. Jardine 136. Practical Guide to Industrial Boiler Systems, Ralph L. Vandagriff 137. Lubrication Fundamentals: Second Edition, Revised and Expanded, D. M. Pirro and A. A. Wessol 138. Mechanical Life Cycle Handbook: Good Environmental Design and Manufacturing, edited by Mahendra S. Hundal 139. Micromachining of Engineering Materials, edited by Joseph McGeough 140. Control Strategies for Dynamic Systems: Design and Implementation, John H. Lumkes, Jr. 141. Practical Guide to Pressure Vessel Manufacturing, Sunil Pullarcot 142. Nondestructive Evaluation: Theory, Techniques, and Applications, edited by Peter J. Shull 143. Diesel Engine Engineering: Thermodynamics, Dynamics, Design, and Control, Andrei Makartchouk 144. Handbook of Machine Tool Analysis, loan D. Marinescu, Constantin Ispas, and Dan Boboc 145. Implementing Concurrent Engineering in Small Companies, Susan Carlson Skalak 146. Practical Guide to the Packaging of Electronics: Thermal and Mechanical Design and Analysis, Ali Jamnia 147. Bearing Design in Machinery: Engineering Tribology and Lubrication, Avraham Harnoy 148. Mechanical Reliability Improvement: Probability and Statistics for Experimental Testing, R. E. Little 149. Industrial Boilers and Heat Recovery Steam Generators: Design, Applications, and Calculations, V. Ganapathy 150. The CAD Guidebook: A Basic Manual for Understanding and Improving Computer-Aided Design, Stephen J. Schoonmaker 151. Industrial Noise Control and Acoustics, Randall F. Barron 152. Mechanical Properties of Engineered Materials, Wole Soboyejo 153. Reliability Verification, Testing, and Analysis in Engineering Design, Gary S. Wasserman 154. Fundamental Mechanics of Fluids: Third Edition, I. G. Currie 155. Intermediate Heat Transfer, Kau-Fui Vincent W ong 156. HVAC Water Chillers and Cooling Towers: Fundamentals, Application, and Operation, Herbert W. Stanford Ill 157. Gear Noise and Vibration: Second Edition, Revised and Expanded, J. Derek Smith Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 158. Handbook of Turbomachinery: Second Edition, Revised and Expanded, Earl Logan, Jr., and Ramendra Roy 159. Piping and Pipeline Engineering: Design, Construction, Maintenance, Integrity, and Repair, George A. Antaki 160. Turbomachinery: Design and Theory, Rama S. R. Gorla and Aijaz Ahmed Khan Additional Volumes in Preparation Target Costing: Market-Driven Product Design, M. Bradford Clifton, VVesley P. Townsend, Henry M. B. Bird, and Robert E. Albano Theory of Dimensioning: An Introduction to Parameterizing Geometric Models, Vijay Srinivasan Fluidized Bed Combustion, Simeon N. Oka Structural Analysis of Polymeric Composite Materials, Mark E. Tuttle Handbook of Pneumatic Conveying Engineering, David Mills, Mark G. Jones, and Vijay K. Agarwal Handbook of Mechanical Design Based on Material Composition, Geolrge E. Totten, Lin Xie, and Kiyoshi Funatani Mechanical Wear Fundamentals and Testing: Second Edition, Revised and Expanded, Raymond G. Bayer Engineering Design for Wear: Second Edition, Revised and Expanded, Raymond G. Bayer Clutches and Brakes: Design and Selection, Second Edition, William C. Orthwein Progressing Cavity Pumps, Downhole Pumps, and Mudmotors, Lev Nelik Mechanical Engineering Sofmare Spring Design with an ISM PC, A Dietrich 1 Mechanical Design Failure Analysis: With Failure Analysis System SolyWare for the IBM PC, David G. Ullman Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved A. Sheema. Inc. R. who encouraged me to strive for excellence in education —R. Copyright 2003 by Marcel Dekker.To my parents. and Afifa —A. All Rights Reserved . and to my daughters. S. Shumaila. G. Tirupelamma and Subba Reddy Gorla. K. Tahseen Ara. To my wife. chemical engineering. centrifugal compressors and fans. Each chapter includes a large number of solved illustrative and design example problems. A brief discussion of cavitation in hydraulic machinery is presented. fluid dynamics. We applied the basic principles to the study of hydraulic pumps. and axial flow and radial flow gas turbines. In addition. design engineering. This book is also a valuable reference to practicing engineers in the fields of propulsion and turbomachinery. steam turbines. aerospace engineering. An introduction to dimensional analysis is included. which will allow students to gain more experience. A basic knowledge of thermodynamics. hydraulic turbines. we have provided several exercise problems at the end of each chapter. axial flow compressors and fans.Preface Turbomachinery: Design and Theory offers an introduction to the subject of turbomachinery and is intended to be a text for a single-semester course for senior undergraduate and beginning graduate students in mechanical engineering. with particular attention to the proper use of units. An intuitive and systematic approach is used in the solution of these example problems. All Rights Reserved . and heat transfer is assumed. We urge students to take these exercise problems seriously: they are designed to help students fully grasp each topic Copyright 2003 by Marcel Dekker. Inc. and manufacturing engineering. which will help students understand the subject matter easily. We have introduced the relevant concepts from these topics and reviewed them as applied to turbomachines in more detail. K. for her patience and perfect skills in the preparation of figures.D. Gorla Aijaz A.S. a civil engineer who provided numerous suggestions for enhancement of the material on hydraulic turbomachines. who typed some portions of the manuscript. Rama S.K.A. a practicing computer engineer. and we are grateful to John Corrigan. Vijaya Lakshmi. is also indebted to Aftab Ahmed. without including irrelevant details. University of Engineering and Technology.G. A. All Rights Reserved . Khan Copyright 2003 by Marcel Dekker.E. would like to extend special recognition to his daughter. Our goal is to offer an engineering textbook on turbomachinery that will be read by students with enthusiasm and interest— we have made special efforts to touch students’ minds and assist them in exploring the exciting subject matter. Inc.R.vi Preface and to lead them toward a more concrete understanding and mastery of the techniques presented. R.. R. A. for her support and understanding during the preparation of this book. Associate Professor of Mechanical Engineering at N. for encouragement and assistance. This book has been written in a straightforward and systematic manner. editor at Marcel Dekker. Shumaila. Inc. to Sheema Aijaz. We would like to thank Shirley Love for her assistance in typing portions of the manuscript. for his many helpful discussions during the writing of this book. We also thank the reviewers for their helpful comments.A. would like to express thanks to his wife. and to M. Sadiq. 15 Basic Thermodynamics. Inc.6 Hydraulic Machines 1. and Definitions of Efficiency Copyright 2003 by Marcel Dekker.13 Properties Involving the Mass or Weight of the Fluid 1. All Rights Reserved .10 Kinematic Similarity 1.Contents Preface 1.5 The Buckingham P Theorem 1.12 Prototype and Model Efficiency 1.1 Introduction to Turbomachinery 1.4 Dimensions and Equations 1. Fluid Mechanics.8 Model Testing 1.7 The Reynolds Number 1.11 Dynamic Similarity 1.2 Types of Turbomachines 1.14 Compressible Flow Machines 1.3 Dimensional Analysis 1.9 Geometric Similarity 1. Introduction: Dimensional Analysis—Basic Thermodynamics and Fluid Mechanics 1. 3 Slip Factor 2.24 1.18 1.22 1.16 1.25 1. Inc.viii Contents 1.19 1.4 Pelton Wheel (Losses and Efficiencies) Examples 3.9 Cavitation in Pumps 2.10 Suction Specific Speed 2.4 Pump Losses 2.14 Changing Pump Speed 2.7 Vaneless Diffuser 2.21 1. Hydraulic Turbines 3.5 Reaction Turbine Copyright 2003 by Marcel Dekker.26 1.20 1. Hydraulic Pumps 2.12 Pumping System Design 2.13 Life Cycle Analysis 2.15 Multiple Pump Operation Examples Problems Notation 3.8 Vaned Diffuser 2.1 Introduction 2.28 Continuity Equation The First Law of Thermodynamics Newton’s Second Law of Motion The Second Law of Thermodynamics: Entropy Efficiency and Losses Steam and Gas Turbines Efficiency of Compressors Polytropic or Small-Stage Efficiency Nozzle Efficiency Diffuser Efficiency Energy Transfer in Turbomachinery The Euler Turbine Equation Components of Energy Transfer Examples Problems Notation 2. All Rights Reserved .1 Introduction 3.6 Volute or Scroll Collector 2.17 1.23 1.2 Centrifugal Pumps 2.2 Pelton Wheel 3.11 Axial Flow Pump 2.5 The Effect of Impeller Blade Shape on Performance 2.3 Velocity Triangles 3.27 1. 5 5.2 Centrifugal Compressor 4.10 Centrifugal Compressor Characteristics 4. Inc.7 5.1 Introduction 6. Steam Turbines 6.2 Steam Nozzles 6.9 Flow Compressors and Fans Introduction Velocity Diagram Degree of Reaction Stage Loading Lift-and-Drag Coefficients Cascade Nomenclature and Terminology 3-D Consideration Multi-Stage Performance Axial Flow Compressor Characteristics Examples Problems Notation 5.11 Stall 4.2 5.7 Diffuser 4.12 Surging 4.Contents ix 3.3 Nozzle Efficiency Copyright 2003 by Marcel Dekker. Centrifugal Compressors and Fans 4.8 3.9 Mach Number in the Diffuser 4. 6.4 5.4 Velocity Diagrams 4.5 Slip Factor 4.1 5.13 Choking Examples Problems Notation Axial 5.9 Turbine Losses Turbine Characteristics Axial Flow Turbine Cavitation Examples Problems Notation 4.3 The Effect of Blade Shape on Performance 4.6 5. All Rights Reserved .8 Compressibility Effects 4.8 5.6 Work Done 4.6 3.3 5.7 3.1 Introduction 4. 5 6.6 6.2 Velocity Triangles and Work Output 7. Copyright 2003 by Marcel Dekker.x Contents 6. Inc.13 The Reheat Factor Metastable Equilibrium Examples Stage Design Impulse Stage The Impulse Steam Turbine Pressure Compounding (The Rateau Turbine) Velocity Compounding (The Curtis Turbine) Axial Flow Steam Turbines Degree of Reaction Blade Height in Axial Flow Machines Examples Problems Notation 7.10 6.1 Introduction to Axial Flow Turbines 7.2 Stages and Types of Cavitation 8.1 Introduction 8.10 Spouting Velocity 7.12 Application of Specific Speed Examples Problems Notation Cavitation in Hydraulic Machinery 8.8 6. Axial Flow and Radial Flow Gas Turbines 7.11 6. All Rights Reserved .5 Physical Significance and Uses of the Cavitation Parameter 8.8 Radial Flow Turbine 7.7 6.4 Blade-Loading Coefficient 7.9 Velocity Diagrams and Thermodynamic Analysis 7.11 Turbine Efficiency 7.5 Stator (Nozzle) and Rotor Losses 7.4 6.4 Cavitation Parameter for Dynamic Similarity 8.3 Degree of Reaction (L 7.12 6.6 Free Vortex Design 7.3 Effects and Importance of Cavitation 8.9 6.7 Constant Nozzle Angle Design Examples 7. Contents xi 8.7 8.6 8. Inc. All Rights Reserved .8 The Rayleigh Analysis of a Spherical Cavity in an Inviscid Incompressible Liquid at Rest at Infinity Cavitation Effects on Performance of Hydraulic Machines Thoma’s Sigma and Cavitation Tests Notation Appendix The International System of Units (SI) Thermodynamic Properties of Water Thermodynamic Properties of Liquids Thermodynamic Properties of Air Bibliography Copyright 2003 by Marcel Dekker. as well as those types that produce a head or pressure. 1. The turbomachine extracts energy from or imparts energy to a continuously moving stream of fluid. such as centrifugal pumps and compressors. we shall deal with incompressible and compressible fluid flow machines. Copyright 2003 by Marcel Dekker. such as turbines. steam turbines.1 INTRODUCTION TO TURBOMACHINERY A turbomachine is a device in which energy transfer occurs between a flowing fluid and a rotating element due to dynamic action. windmills. They can be classified as: 1. Turbomachines include all those machines that produce power. In this text. it is intermittent. centrifugal and axial flow compressors. water wheels. and hydraulic turbines. The turbomachine as described above covers a wide range of machines. and results in a change in pressure and momentum of the fluid. However in a positive displacement machine. usually in a steady-flow process. Turbomachines in which (i) work is done by the fluid and (ii) work is done on the fluid. Inc. Mechanical energy transfer occurs inside or outside of the turbomachine.2 TYPES OF TURBOMACHINES There are different types of turbomachines.1 Introduction: Dimensional Analysis—Basic Thermodynamics and Fluid Mechanics 1. All Rights Reserved . such as gas turbines. centrifugal pumps. 1 Types and shapes of turbomachines.2 Chapter 1 Figure 1. Inc. Copyright 2003 by Marcel Dekker. All Rights Reserved . 1. Some of these machines are shown in Fig. that the path of the fluid in the rotor may be substantially axial. (Courtesy of the Buffalo Forge Corp. usually the gas may be considered to be incompressible. All Rights Reserved . 1. and secondly. or in some cases a combination of both. Pumps: Machines that increase the pressure or head of flowing fluid.2 –1. Turbomachines in which fluid moves through the rotating member in axial direction with no radial movement of the streamlines.Basic Thermodynamics and Fluid Mechanics 3 2. it is called a radial flow or centrifugal flow machine. Figure 1. and photographs of actual machines are shown in Figs. substantially radial. that the main element is a rotor or runner carrying blades or vanes.) Copyright 2003 by Marcel Dekker.2 Radial flow fan rotor. Inc.1. Fans: Machines that impart only a small pressure-rise to a continuously flowing gas.6. Such machines are called axial flow machines whereas if the flow is essentially radial. Two primary points will be observed: first. Turbomachines can further be classified as follows: Turbines: Machines that produce power by expansion of a continuously flowing fluid to a lower pressure or head. (Courtesy of Rolls-Royce.) Copyright 2003 by Marcel Dekker. Inc.4 Chapter 1 Figure 1.) Figure 1.4 Centrifugal pump rotor (open type impeller). (Courtesy of the IngersollRand Co.3 Centrifugal compressor rotor (the large double-sided impellar on the right is the main compressor and the small single-sided impellar is an auxiliary for cooling purposes). Ltd. All Rights Reserved . 5 Multi-stage axial flow compressor rotor.6 Axial flow pump rotor. All Rights Reserved . Inc.) Figure 1. (Courtesy of the Worthington Corp.Basic Thermodynamics and Fluid Mechanics 5 Figure 1. (Courtesy of the Westinghouse Electric Corp.) Copyright 2003 by Marcel Dekker. 1 for reference. this is achieved by imparting kinetic energy to the air in the impeller and then this kinetic energy is converted into pressure energy in the diffuser. the properties of interest in regard to turbomachine are the power output. the efficiency.6 Chapter 1 Compressors: Machines that impart kinetic energy to a gas by compressing it and then allowing it to rapidly expand.1 Physical Properties and Dimensions Property Surface Volume Density Velocity Acceleration Momentum Force Energy and work Power Moment of inertia Angular velocity Angular acceleration Angular momentum Torque Modules of elasticity Surface tension Viscosity (absolute) Viscosity (kinematic) Dimension L2 L3 M/L3 L/T L/T2 ML/T ML/T2 ML2/T2 ML2/T3 ML2 I/T I/T2 ML2/T ML2/T2 M/LT2 M/T2 M/LT L2/T Copyright 2003 by Marcel Dekker. The performance of turbomachines depends on one or more of several variables. 1. or a combination of both types. A summary of the physical properties and dimensions is given in Table 1. centrifugal. and the head. All Rights Reserved . The use of dimensional analysis reduces the variables to a number of manageable dimensional groups. in order to produce the highly compressed air. a large number of variables are involved. Inc. Dimensional analysis applied to turbomachines has two more important uses: (1) prediction of a prototype’s performance from tests conducted on a scale Table 1. Usually.3 DIMENSIONAL ANALYSIS To study the performance characteristics of turbomachines. In a dynamic compressor. Compressors can be axial flow. L 2 represents the dimensional characteristics of area. on the basis of maximum efficiency.” In this case. F. temperature and time. D. English system: mass. H. where n is the number of variables involved and m is the number of dimensionless parameters included in the variables.4 DIMENSIONS AND EQUATIONS The variables involved in engineering are expressed in terms of a limited number of basic dimensions. Suppose.5 THE BUCKINGHAM P THEOREM In 1915. time and force. the basic dimensions are: 1. m. SI system: mass. 2.6 HYDRAULIC MACHINES Consider a control volume around the pump through which an incompressible fluid of density r flows at a volume flow rate of Q. 1. (1. and T ) involved. temperature. The dimensions of pressure can be designated as follows P¼ F L2 ð1:1Þ Equation (1. Inc. m. It is assumed here that the student has acquired the basic techniques of forming nondimensional groups. Since the flow enters at one point and leaves at another point the volume flow rate Q can be independently adjusted by means of a throttle valve. v. and flow rate. 1.Basic Thermodynamics and Fluid Mechanics 7 model (similitude). 1. speed. The discharge Q of a pump is given by Q ¼ f ðN. and (2) determination of the most suitable type of machine. the drag force F of a flowing fluid past a sphere is known to be a function of the velocity (v) mass density (r) viscosity (m) and diameter (D). length. The left hand side of Eq. Then. for a specified range of head. and D) and three basic dimensions (L. Buckingham showed that the number of independent dimensionless group of variables (dimensionless parameters) needed to correlate the unknown variables in a given process is equal to n 2 m.1) must have the same dimensions as the right hand side. Then we have five variables (F.1) reads as follows: “The dimension of P equals force per length squared. All Rights Reserved . length. g. there are 5 2 3 ¼ 2 basic grouping of variables that can be used to correlate experimental results. r. for example. For most engineering problems. rÞ ð1:2Þ Copyright 2003 by Marcel Dekker. À Áf P2 ¼ ðNÞd ðDÞe r ðgÞ Q ND 3 ð1:3Þ M0 L0 T0 ¼ ðT21 Þd ðLÞe ðML23 Þf ðLT22 Þ Now. All Rights Reserved . P1 ¼ N 21 D 23 r 0 Q ¼ Similarly. for L. g ð1:4Þ P2 ¼ N 22 D 21 r 0 g ¼ 2 N D Similarly. and T on both sides of the equation: for M. Taking N. N is the revolution. 0 ¼ c or c ¼ 0. 0 ¼ 2 a 2 1 or a ¼ 2 1. and r as repeating variables. for T. Copyright 2003 by Marcel Dekker. Inc. equating the exponents: for M. H D ð1:5Þ À Ál P4 ¼ ðNÞj ðDÞk r ðmÞ M0 L0 T0 ¼ ðT21 Þ j ðLÞk ðML23 Þl ðML21 T21 Þ Equating the exponents: for M. for L. (1. equating the powers of M. D. 0 ¼ 2 j 2 1 or j ¼ 2 1. r is the density of fluid. 0 ¼ b 2 3c þ 3 or b ¼ 2 3. P3 ¼ N 0 D 21 r 0 H ¼ and. 0 ¼ h 2 3i þ 1 or h ¼ 2 1. 0 ¼ 2 g or g ¼ 0. Therefore. 0 ¼ l þ 1 or l ¼ 2 1. for T. primary dimensions are only four. g is the acceleration due to gravity. we get À Ác P1 ¼ ðNÞa ðDÞb r ðQÞ M0 L0 T0 ¼ ðT21 Þa ðLÞb ðML23 Þc ðL3 T21 Þ For dimensional homogeneity.2). À Ái P3 ¼ ðNÞg ðDÞh r ðHÞ M0 L0 T0 ¼ ðT21 Þg ðLÞh ðML23 Þi ðLÞ Equating the exponents: for M. for T.8 Chapter 1 where H is the head. for L. 0 ¼ e 2 3f þ 1 or e ¼ 2 1. 0 ¼ f or f ¼ 0. 0 ¼ i or i ¼ 0. L. In Eq. Thus. for L. D is the diameter of impeller. and m is the viscosity of fluid. for T. 0 ¼ 2 d 2 2 or d ¼ 2 2. Thus. 0 ¼ k-3l 2 1 or k ¼ 2 2. It is the non-dimensional term. It is constant for similar rotors. It shows the relation between power.Basic Thermodynamics and Fluid Mechanics 9 Thus. ¼0 ND 3 N 2 D D ND 2 r Since the product of two P terms is dimensionless. 2 2. N 2 D 2 ND 2 r ¼0 ð1:7Þ A dimensionless term of extremely great importance that may be obtained by manipulating the discharge and head coefficients is the specific speed. P4 ¼ N 21 D 22 r 21 m ¼ m ND 2 r ð1:6Þ The functional relationship may be written as Q g H m f . operating at unit speed. The head coefficient: The term gH/N 2D 2 is called the specific head. . All Rights Reserved . ¼0 ND 3 N D ND 2 r or Q ¼ ND 3 f gH m . À Á À Á c ¼ H/ U 2 /g ¼ gH/ p 2 N 2 D 2 ð1:9Þ 3. It is constant for similar impellers. defined by the equation rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiÀ Á3/4 Flow coefficient ð1:8Þ ¼ N Q gH Ns ¼ Head coefficient The following few dimensionless terms are useful in the analysis of incompressible fluid flow machines: 1. 4. fluid density. The flow coefficient and speed ratio: The term Q/(ND 3) is called the flow coefficient or specific capacity and indicates the volume flow rate of fluid through a turbomachine of unit diameter runner. therefore replace the terms P2 and P3 by gh/N 2D 2 Q gH m f . It is the kinetic energy of the fluid spouting under the head H divided by the kinetic energy of the fluid running at the rotor tangential speed. 2. Power coefficient or specific power: The dimensionless quantity P/(rN 2D 2) is called the power coefficient or the specific power. Specific speed: The most important parameter of incompressible fluid flow machinery is specific speed. speed and wheel diameter. Inc. All turbomachineries operating under the same conditions of flow and head Copyright 2003 by Marcel Dekker. . m.12) is used for the specific speeds of turbines. The functional equation may be written as P ¼ f ðr. Q. which develops 1 hp under a head of 1 meter of water.10) is a nondimensional quantity. (1. 1. N. In flow through turbomachines. The head difference across the control volume is H. Consider a control volume around a hydraulic turbine through which an incompressible fluid of density r flows at a volume flow rate of Q. Inc.7 THE REYNOLDS NUMBER Reynolds number is represented by Re ¼ D 2 N/n where y is the kinematic viscosity of the fluid. Various other losses such as those due to shock at entry. impact. which is controlled by a valve.P and H. it varies between 4 (for very high head Pelton wheel) and 1000 (for the low-head propeller on Kaplan turbines).11) is used for specifying the specific speeds of pumps and Eq. and leakage affect the machine characteristics along with various friction losses.10 Chapter 1 having the same specific speed. gHÞ ð1:13Þ Copyright 2003 by Marcel Dekker. (1. All Rights Reserved . Q and H or N. which cause similar flows in turbomachines that are geometrically similar. irrespective of the actual physical size of the machines. the turbine develops a shaft power P at a speed of rotation N. the dimensionless parameter D 2N/n is not as important since the viscous resistance alone does not determine the machine losses. and if the control volume represents a turbine of diameter D. These are pffiffiffiffi ð1:11Þ N s ¼ N Q/H 3/4 and pffiffiffi N s ¼ N P/H 5/4 ð1:12Þ Equation (1. It is clear that Ns is a dimensional quantity. The turbine specific speed may be defined as the speed of a geometrically similar turbine. however. turbulence. The specific speed represented by Eq. Since the quantity D 2N is proportional to DV for similar machines that have the same speed ratio. In metric units. Specific speed can be expressed in this form pffiffiffiffi pffiffiffi À Á5/4 ð1:10Þ N s ¼ N Q/ðgHÞ3/4 ¼ N P/ b r 1/2 gH c The specific speed parameter expressing the variation of all the variables N. D. It can also be expressed in alternate forms. (1. c Equation (1. Reynolds number is usually very high. P ¼ const: rN D . In flow through hydraulic turbomachinery. and T on both sides of the equation: for M. and d in terms of c. b. Copyright 2003 by Marcel Dekker. such that À Áf ð1:14Þ P ¼ const: r a N b m c D d Q e gH Substituting the respective dimensions in the above Eq. Therefore the viscous action of the fluid has very little effect on the power output of the machine and the power coefficient remains only a function of f and c.16). L.Basic Thermodynamics and Fluid Mechanics 11 Equation (1. Eqution (1. e. There are six variables and only three equations. (1. f. and f we have: a¼12c b ¼ 3 2 c 2 e 2 2f d ¼ 5 2 2c 2 3e 2 2f Substituting the values of a. À Á ML2 /T3 ¼ const:ðM/L3 Þa ð1/TÞb ðM/LTÞc ðLÞd ðL3 /TÞe ðL2 /T2 Þ f ð1:15Þ Equating the powers of M. the flow coefficient Q/ND 3 ¼ f .13) may be written as the product of all the variables raised to a power and a constant. b. . ð1:16Þ rND 2 ND 3 N 2D 2 In Eq. for T. 2 3 ¼ 2 b 2 c 2 e 2 2f. 2 ¼ 2 3a 2 c þ d þ 3e þ 2f. (1.14). this term can be inverted and Eq. the second term in the brackets is the inverse of the Reynolds number. " c # À 3 5Á m Q e gH f .18) indicates that the power coefficient of a hydraulic machine is a function of Reynolds number.16) may be written as " c # rND 2 Q e gH f 3 5 P/rN D ¼ const: . Solving for a.17) can be expressed in the following form: À Á ð1:18Þ P ¼ f Re. It is therefore necessary to solve for three of the indices in terms of the remaining three. . 1 ¼ a þ c. and collecting like indices into separate brackets. and the À Á head coefficient gH/N 2 D 2 ¼ c . and d in Eq. flow coefficient and head coefficient. ð1:17Þ ND 3 N 2D 2 m In Eq.17) each group of variables is dimensionless and all are used in hydraulic turbomachinery practice. and are known by the following names: the Á À Á À power coefficient P/rN 3 D 5 ¼ P . (1. All Rights Reserved . Since the value of c is unknown. for L.13). Inc. (1. 8 MODEL TESTING Some very large hydraulic machines are tested in a model form before making the full-sized machine. 1. After the result is obtained from the model. Q.7 Performance characteristics of hydraulic machines: (a) hydraulic turbine. Therefore if the curves shown in Fig 1. Dp.7 have been obtained for a completely similar model. the ratios of all the corresponding linear dimensions should be equal. 1. Or. Typical dimensionless characteristic curves for a hydraulic turbine and pump are shown in Fig. one may transpose the results from the model to the full-sized machine. these same curves would apply to the full-sized prototype machine. All Rights Reserved . and H for a given machine or for any other geometrically similar machine.9 GEOMETRIC SIMILARITY For geometric similarity to exist between the model and prototype. and Lm. the breadth of the prototype. for geometric similarity between the model and the prototype. respectively. both of them should be identical in shape but differ only in size. and Dm the corresponding dimensions of Copyright 2003 by Marcel Dekker. the depth of the prototype. Bp. Let Lp be the length of the prototype.12 Chapter 1 Figure 1. in other words. (b) hydraulic pump. N. These characteristic curves are also the curves of any other combination of P. 1. Inc.7 (a) and (b). Bm. both model and prototype have identical motions or velocities. Now from similarity laws. Vr ¼ 1. hp and hm are the prototype and model efficiencies. Then the force ratio to establish dynamic similarity between the prototype and the model is given by Fr ¼ F p1 F p2 ¼ F m1 F m2 ð1:23Þ 1. and v2 is the velocity of liquid in the model at point 2.11 DYNAMIC SIMILARITY If model and prototype have identical forces acting on them. v1. For geometric similarity. Inc.10 KINEMATIC SIMILARITY For kinematic similarity. Let F1 be the forces acting on the prototype at point 1. 2 2 Hp Hp Np Dp Hm or ¼ À Á2 ¼ 2 Hm Nm Dm ðN m Dm Þ N p Dp Copyright 2003 by Marcel Dekker.12 PROTOTYPE AND MODEL EFFICIENCY Let us suppose that the similarity laws are satisfied. V2. linear ratio (or scale ratio) is given by Lp Bp Dp ¼ ¼ Lm Bm Dm Similarly. All Rights Reserved . the area ratio between prototype and model is given by 2 2 2 Lp Bp Dp ¼ ¼ Ar ¼ Lm Bm Dm Lr ¼ and the volume ratio 3 3 3 Lp Bp Dp Vr ¼ ¼ ¼ Lm Bm Dm ð1:19Þ ð1:20Þ ð1:21Þ 1.Basic Thermodynamics and Fluid Mechanics 13 the model. then dynamic similarity will exist. and F2 be the forces acting on the prototype at point 2. representing the model and prototype by subscripts m and p respectively. the velocity of liquid in the model at point 1. then the velocity ratio of the prototype to the model is V1 V2 ¼ ð1:22Þ v1 v2 where V1 is the velocity of liquid in the prototype at point 1. respectively. the velocity of liquid in the prototype at point 2. If the ratio of the corresponding points is equal. I. All Rights Reserved .2 Mass Density (r) The mass per unit volume is mass density. is given the greek symbol r (rho). of course. the units are kilograms per cubic meter or NS2/m4.13. fluid machines.I.4 Viscosity (m) We define viscosity as the property of a fluid. it is dimensionless and. units. The energy loss due to friction in a flowing Copyright 2003 by Marcel Dekker. The mass density of water at 15. the specific weight of water is taken as 1000 l/m3. Inc. ¼ 1: hp Pp Qm Hm ht ¼ Thus.80 kN/m3. 1. the efficiencies of the model and prototype are the same providing the similarity laws are satisfied.14 Chapter 1 Qp Qm ¼ 3 N p Dp N m D3 m Pp Pm ¼ 3 3 N p D5 N m D5 p m or or Qp ¼ Qm 3 Np Dp Nm Dm 3 5 Pp Np Dp ¼ Pm Nm Dm Turbine efficiency is given by Power transferred from fluid P ¼ Fluid power available: rgQH Qp Hp hm Pm ¼ Hence. For the purpose of all calculations relating to hydraulics. the specific weight of water is taken as 9. independent of system of units used. 1.gr. The standard reference temperature for water is often taken as 48C Because specific gravity is a ratio of specific weights.13. 1. often simply called density. systems.) The ratio of the specific weight of a given liquid to the specific weight of water at a standard reference temperature is defined as specific gravity.13.13.58 is 1000 kg/m3. which offers resistance to the relative motion of fluid molecules. 1. In S. Mass density.13 PROPERTIES INVOLVING THE MASS OR WEIGHT OF THE FLUID 1.3 Specific Gravity (sp. In S.1 Specific Weight (g) The weight per unit volume is defined as specific weight and it is given the symbol g (gamma). some new variables must be added to those already discussed in the case of hydraulic machines and changes must be made in some of the definitions used. or other common liquids. denoted by the symbol t (tau) can be defined as the force required to slide on unit area layers of a substance over another.25) is called a Newtonian fluid. Inc. Copyright 2003 by Marcel Dekker. The important parameters in compressible flow machines are pressure and temperature.Basic Thermodynamics and Fluid Mechanics 15 fluid is due to the viscosity. This fact can be stated mathematically as Dv t¼m ð1:24Þ Dy where Dv is the velocity gradient and the constant of proportionality m is called Dy the dynamic viscosity of fluid. Shearing stress. alcohol. In a fluid such as water. oil. a shearing stress develops in it. centrifugal and axial flow compressors. When a fluid moves.14 COMPRESSIBLE FLOW MACHINES Compressible fluids are working substances in gas turbines. we find that the magnitude of the shearing stress is directly proportional to the change of velocity between different positions in the fluid. Units for Dynamic Viscosity Solving for m gives m¼ t Dy ¼t Dv/Dy Dv Substituting the units only into this equation gives N m N£s ¼ £ m2 m/s m2 Since Pa is a shorter symbol representing N/m2. All Rights Reserved . 1. (1. we can also express m as m¼ m ¼ Pa · s 1. The magnitude of the shearing stress depends on the viscosity of the fluid. To include the compressibility of these types of fluids (gases).13. Thus t is a force divided by an area and can be measured in units N/m2 or Pa. It is defined as n¼ m kg m3 m2 £ ¼ ¼ mð1/rÞ ¼ s r ms kg ð1:25Þ Any fluid that behaves in accordance with Eq.5 Kinematic Viscosity (n) The ratio of the dynamic viscosity to the density of the fluid is called the kinematic viscosity y (nu). deleting density and combining R with T. Inc.16 Chapter 1 Figure 1. respectively. D. However. (b) expansion. mÞ ð1:27Þ Substituting the basic dimensions and equating the indices. Using the perfect gas equation.8 Compression and expansion in compressible flow machines: (a) compression. N. m ð1:26Þ The pressure ratio P02/P01 replaces the head H. P01 . can be expressed as follows À Á P02 ¼ f D. r01 . RT 02 . (1. R is constant and may be eliminated. T 02 . The Reynolds number in most cases is very high and the flow is turbulent and therefore changes in this parameter over the usual operating range may be neglected. the following fundamental relationship may be obtained 0 1/2 1 0 1 m P02 ND C B RT 02 B RT 01 C ¼ f@ . 1 and 2 refer to the inlet and outlet states of the gas. ReA ð1:28Þ A. P01 RT 01 P01 D 2 ðRT 01 Þ1/2 In Eq. density may be written as r ¼ P/RT. r02 . The pressure at the outlet. m.8 T-s charts for compression and expansion processes are shown. N. 1. T 01 . due to Copyright 2003 by Marcel Dekker. All Rights Reserved . Isentropic compression and expansion processes are represented by s and the subscript 0 refers to stagnation or total conditions. In Fig. m.@ . P02. the functional relationship can be written as P02 ¼ f ðP01 .28). while the mass flow rate m (kg/s) replaces Q. RT 01 . Now. and hence Eq. (b) efficiency. the diameter D may be ignored. Inc. Copyright 2003 by Marcel Dekker.10 Axial flow gas turbine characteristics: (a) pressure ratio. For a constant diameter machine. large changes of density.9 Axial flow compressor characteristics: (a) pressure ratio. All Rights Reserved . For a particular machine.Basic Thermodynamics and Fluid Mechanics 17 Figure 1. a significant reduction in Re can occur which must be taken into consideration. it is typical to plot P02 /P01 and T 02 /T 01 against the mass flow Figure 1. .29) some of the terms are new and no longer dimensionless. (1. (1. (b) efficiency.28) becomes 1/2 P02 T 02 mT 01 N ¼f . ð1:29Þ 1/2 P01 T 01 P01 T 01 In Eq. These laws are: 1. Inc.17 THE FIRST LAW OF THERMODYNAMICS According to the First Law of Thermodynamics. All Rights Reserved .28) must be used if it is required to change the size of the machine.18 Chapter 1 1/2 rate parameter mT 1/2 /P01 for different values of the speed parameter N/T 01 . C1. 01 Equation (1. 1 and 2 along a passage respectively. so that much of the elementary discussion and analysis of these laws need not be repeated here. if a system is taken through a complete cycle during which heat is supplied and work is done. 1. the velocity at section 1. The continuity equation.9 and 1. This occurs because the impeller velocity v / ND and the acoustic velocity a01 / RT01. FLUID MECHANICS. The term ND/ðRT 01 Þ1/2 indicates the Mach number effect. AND DEFINITIONS OF EFFICIENCY In this section. then _ m ¼ r1 A1 C1 ¼ r2 A2 C 2 ¼ constant ð1:31Þ where r1. If A1 and A2 are the flow areas at Secs. while the Mach number M ¼ V/a01 ð1:30Þ The performance curves for an axial flow compressor and turbine are shown in Figs. 3. the basic physical laws of fluid mechanics and thermodynamics will be discussed. 1. is the velocity at section 2. Newton’s Second Law of Motion. 1. is the density at section 1.10. the density at section 2. The Second Law of Thermodynamics. 2. 1.15 BASIC THERMODYNAMICS. 4. The First Law of Thermodynamics. and C2.16 CONTINUITY EQUATION For steady flow through a turbomachine. r2. then I ðdQ 2 dW Þ ¼ 0 ð1:32Þ H H where dQ represents the heat supplied to the system during this cycle and dW Copyright 2003 by Marcel Dekker. m remains constant. The above items are comprehensively dealt with in books on thermodynamics with engineering applications. E may be written as: E ¼ Internal Energy þ Kinetic Energy þ Potential Energy E ¼ U þ K:E: þ P:E: ð1:35Þ where U is the internal energy.Basic Thermodynamics and Fluid Mechanics 19 the work done by the system during the cycle. The units of heat and work are taken to be the same. Since the terms comprising E are point functions. Thus. The total energy of a system. (1. In many applications.) ¼ 0).) ¼ K(P. or changes in. K(K.38) indicates that there are differences between. Inc. there is a change in the internal energy of the system Z 2 U2 2 U1 ¼ ðdQ 2 dW Þ ð1:33Þ 1 For an infinitesimal change of state dU ¼ dQ 2 dW ð1:34Þ 1. and the first law relation Q122 ¼ U 2 2 U 1 þ Copyright 2003 by Marcel Dekker. the changes in kinetic and potential energies are negligible (i. All Rights Reserved . similar forms of energy entering or leaving the unit.17. Most closed systems encountered in practice are stationary. i.35) in the following form dE ¼ dU þ dðK:E:Þ þ dðP:E:Þ ð1:36Þ The First Law of Thermodynamics for a change of state of a system may therefore be written as follows dQ ¼ dU þ dðKEÞ þ dðPEÞ þ dW ð1:37Þ Let subscript 1 represents the system in its initial state and 2 represents the system in its final state. these differences are insignificant and can be ignored.E. During a change of state from 1 to 2.E. for stationary closed systems. they do not involve any changes in their velocity or the elevation of their centers of gravity during a process.e.1 The Steady Flow Energy Equation The First Law of Thermodynamics can be applied to a system to find the change in the energy of the system when it undergoes a change of state.e. the energy equation at the inlet and outlet of any device may be written mðC 2 2 C 2 Þ 2 1 þ mgðZ 2 2 Z 1 Þ þ W 1 – 2 ð1:38Þ 2 Equation (1. we can write Eq. (1. Equations. 1. For work Most turbomachinery flow processes are adiabatic. 0. Inc. the initial and final states are identical. and so Q ˙ producing machines. For a control volume with fluid entering with Copyright 2003 by Marcel Dekker.17. (1. so that _ _ W ¼ mðh01 2 h02 Þ ð1:46Þ For work absorbing machines (compressors) W . Then the first law relation for a cycle simplifies to Q 2 W ¼ 0ðkJÞ ð1:44Þ That is. and dE is the rate of change dt of total energy. W the power. the first law relation on a unit-mass basis is q 2 w ¼ DeðkJ/kgÞ ð1:40Þ Dividing Eq. 0. the net heat transfer and the net work done during a cycle must be equal. DE ¼ E2 2 E1 .40) and (1.41) can be expressed in differential form dQ 2 dW ¼ dEðkJÞ ð1:42Þ dq 2 dw ¼ deðkJ/kgÞ ð1:43Þ For a cyclic process. For example.2 Other Forms of the First Law Relation The first law can be written in various forms. so that _ _ _ W ! 2W ¼ mðh02 2 h01 Þ _ _ _ Q 2 W ¼ mðh02 2 h01 Þ ð1:47Þ 1.39) by the time interval Dt and taking the limit as Dt ! 0 yields the rate form of the first law dE _ _ ð1:41Þ Q2W ¼ dt ˙ ˙ where Q is the rate of net heat transfer.18 NEWTON’S SECOND LAW OF MOTION Newton’s Second Law states that the sum of all the forces acting on a control volume in a particular direction is equal to the rate of change of momentum of the fluid across the control volume. the steady flow energy equation becomes ð1:45Þ ˙ ¼ 0. Defining the stagnation enthalpy by: h0 ¼ h þ 1 c 2 and assuming g (Z2 2 Z1) is 2 negligible. W .20 Chapter 1 reduces to Q 2 W ¼ DE ð1:39Þ If the initial and final states are specified the internal energies 1 and 2 can easily be determined from property tables or some thermodynamic relations. therefore. All Rights Reserved . Basic Thermodynamics and Fluid Mechanics 21 uniform velocity C1 and leaving with uniform velocity C2. Z 2 _ dQ _ # mðs2 2 s1 Þ ð1:53Þ T 1 For adiabatic process. For steady flow through a control volume in which the fluid experiences a change of state from inlet 1 to outlet 2. 1. then I dQR ¼0 ð1:50Þ T The property called entropy.48) is the one-dimensional form of the steady flow momentum equation. and applies for linear momentum. is then given by Z 2 dQR ð1:51Þ S2 2 S1 ¼ T 1 For an incremental change of state dS ¼ mds ¼ dQR T ð1:52Þ where m is the mass of the fluid. Inc. If all the processes in the cycle are reversible. Therefore. then X _ F ¼ mðC2 2 C1 Þ ð1:48Þ Equation (1. All Rights Reserved . so that dQ ¼ dQR.19 THE SECOND LAW OF THERMODYNAMICS: ENTROPY This law states that for a fluid passing through a cycle involving heat exchanges I dQ #0 ð1:49Þ T where dQ is an element of heat transferred to the system at an absolute temperature T. turbomachines have impellers that rotate. However. dQ ¼ 0 so that s2 $ s1 For reversible process s2 ¼ s1 ð1:55Þ ð1:54Þ Copyright 2003 by Marcel Dekker. angular momentum is the most descriptive parameter for this system. and the power output is expressed as the product of torque and angular velocity. for a finite change of state. All Rights Reserved . Then: Power wasted ¼ DQðgHÞðkWÞ For volumetric efficiency. let DQ be the amount of water leaking from the tail race. which is not providing useful work. (1. we have ð1:58Þ hn ¼ Q 2 DQ Q ð1:59Þ Net power supplied to turbine ¼ ðQ 2 DQÞgHðkWÞ ð1:60Þ If Hr is the runner head. This is the amount of water. the first law of thermodynamics. (1. hh is given by ð1:61Þ hh ¼ Hydraulic output power ðQ 2 DQÞgH r H r ¼ ¼ Hydraulic input power ðQ 2 DQÞgH H ð1:62Þ Copyright 2003 by Marcel Dekker.34) becomes Tds ¼ du þ pdv Putting h ¼ u þ pv and dh ¼ du þ pdv þ vdp in Eq.20 EFFICIENCY AND LOSSES Let H be the head parameter (m).22 Chapter 1 In the absence of motion. ¼ QgHðin kWÞ Now. Eq. then the hydraulic power generated by the runner is given by Ph ¼ ðQ 2 DQÞgH r ðkWÞ The hydraulic efficiency.56) gives Tds ¼ dh 2 vdp ð1:57Þ ð1:56Þ 1. Q discharge (m3/s) The waterpower supplied to the machine is given by P ¼ rQgHðin wattsÞ and letting r ¼ 1000 kg/m3 . Inc. gravity and other effects. The line 1 –2s represents isentropic expansion.11 shows an enthalpy– entropy or Mollier diagram. All Rights Reserved . Thus Ps Ph ¼ hm h0 ¼ WP WP ¼ hm WPðQ 2 DQÞ ¼ hm h v hh WPQDH ð1:65Þ 1. The actual Figure 1.21 STEAM AND GAS TURBINES Figure 1. Inc. (b) compression process.Basic Thermodynamics and Fluid Mechanics 23 If Pm represents the power loss due to mechanical friction at the bearing. The process is represented by line 1– 2 and shows the expansion from pressure P1 to a lower pressure P2. then the available shaft power is given by Ps ¼ Ph 2 Pm Mechanical efficiency is given by ð1:63Þ hm ¼ Ps Ps ¼ Ph Pm 2 P s ð1:64Þ The combined effect of all these losses may be expressed in the form of overall efficiency. Copyright 2003 by Marcel Dekker.11 Enthalpy – entropy diagrams for turbines and compressors: (a) turbine expansion process. the ideal or isentropic turbine work is that obtained between static points 01 and 2s. in turbojet engines. 1. h ts. In this case. the kinetic energy leaving one stage is utilized in the next stage. the isentropic turbine rotor specific work between the same two pressures is Á 1À ð1:67Þ W 0t ¼ h01 2 h02s ¼ ðh1 2 h2s Þ þ C2 2 C2 2s 2 1 Efficiency can be expressed in several ways. Eq. the total-to-static efficiency. Inc.22 EFFICIENCY OF COMPRESSORS Isentropic work h02s 2 h01 ¼ h02 2 h01 Actual work The isentropic efficiency of the compressor is defined as hc ¼ ð1:70Þ If the difference between inlet and outlet kinetic energies is small. For the above two cases. All Rights Reserved . Thus h ts ¼ h01 2 h02 h01 2 h02 ¼ h01 2 h2s h01 2 h02s þ 1 C2 2 2s ð1:69Þ If the difference between inlet and outlet kinetic energies is small. The choice of definitions depends largely upon whether the kinetic energy at the exit is usefully utilized or wasted. 1 C 2 ¼ 1 C 2 2 1 2 2 and h2s 2 h1 hc ¼ ð1:71Þ h2 2 h1 Copyright 2003 by Marcel Dekker. (1. the energy in the gas exhausting through the nozzle is used for propulsion.69) becomes h ts ¼ h1 2 h2 h1 2 h2s þ 1 C 2 2 1s An example where the outlet kinetic energy is wasted is a turbine exhausting directly to the atmosphere rather than exiting through a diffuser. In multistage gas turbines. is used. the turbine isentropic efficiency htt is defined as h tt ¼ Wt h01 2 h02 ¼ Wt0 h01 2 h02s ð1:68Þ When the exhaust kinetic energy is not totally used but not totally wasted either. Similarly.24 Chapter 1 turbine-specific work is given by 1 W t ¼ h01 2 h02 ¼ ðh1 2 h2 Þ þ ðC2 2 C 2 Þ ð1:66Þ 2 2 1 Similarly. This is made more apparent in the following argument. Isentropic efficiency of compressors tends to decrease and isentropic efficiency of turbines tends to increase as the pressure ratios for which the machines are designed are increased. Then the overall temperature rise can be expressed by X DT 0 s DT ¼ hs ¼ 1 X 0 DTs hs Figure 1. Consider an axial flow compressor. Inc.12. 1. as shown in Fig.23 POLYTROPIC OR SMALL-STAGE EFFICIENCY Isentropic efficiency as described above can be misleading if used for compression and expansion processes in several stages. then the isentropic efficiency of the whole machine will be different from the small stage efficiency. which is made up of several stages.Basic Thermodynamics and Fluid Mechanics 25 1. Copyright 2003 by Marcel Dekker. and this difference is dependent upon the pressure ratio of the machine.12 Compression process in stages. Turbomachines may be used in large numbers of very small stages irrespective of the actual number of stages in the machine. All Rights Reserved . each stage having equal values of hc. If each small stage has the same efficiency. 73) is the functional relation n between P and T for a polytropic process. The relationship between a polytropic efficiency. The above discussions have led to the concept of polytropic efficiency. It is by clear from Fig. DT 0 . substituting dT 0 from the previous equation. P 0 Also. if we write gh1c as n21. hs and the difference will increase with increasing pressure ratio. which is constant through the compressor. DT ¼ D T /hc P definition of hc . h1. small stage efficiency) is less than the overall efficiency of the turbine. and subscript s is for stage temperature). Copyright 2003 by Marcel Dekker.72) can also be written in the form g gh21 T2 P2 1c ¼ T1 P1 The relation between h1c and hc is given by ðT 0 /T 1 Þ 2 1 ðP2 /P1 Þ g 2 1 ¼ hc ¼ 2 g ðT 2 /T 1 Þ 2 1 ðP2 /P1 Þgh21 2 1 1c g21 ð1:73Þ ð1:74Þ g21 From Eq. and thus: hs /hc ¼ DT s 0 /DT 0 .74). which is defined as the isentropic efficiency of an elemental stage in the process such that it is constant throughout the entire process. (1. Eq.e. For compression. h1c ¼ But. (1.26 Chapter 1 (Prime symbol is used for isentropic temperature rise. and the overall efficiency hc may be obtained for a gas of constant specific heat. and thus it is clear that the non isentropic process is polytropic. All Rights Reserved . we get h1c lnðP2 /P1 Þ g ¼ lnðT 2 /T 1 Þ g21 ð1:72Þ Equation (1. we have dT 0 g 2 1 dP ¼ dT g P h1c Integrating the above equation between the inlet 1 and outlet 2. hc . Hence. 1. T p ðg21Þ/g dT 0 ¼ constant dT ¼ constant for an isentropic process.12 that DT 0 s .. which in differential form is dT 0 g 2 1 dP ¼ dT g P Now. Inc. The opposite effect is obtained in a turbine where hs (i. Overall isentropic efficiencies have been Figure 1. the following relations can be developed between the inlet 1 and outlet 2: T1 ¼ T2 and 12 1 P1 /P2 h1t ðgg21Þ P1 P2 h1t ðgg21Þ ðg21Þ g ð1:75Þ ht ¼ 12 1 P1 /P2 where h1t is the small-stage or polytropic efficiency for the turbine.Basic Thermodynamics and Fluid Mechanics 27 Similarly. the isentropic efficiency for an expansion process exceeds the small-stage efficiency. for an isentropic expansion and polytropic expansion.13 pressure ratio. All Rights Reserved . As mentioned earlier. and Copyright 2003 by Marcel Dekker. Inc. polytropic efficiency. for varying polytropic efficiencies and for varying pressure ratios. Figure 1.4. Relationships among overall efficiency.13 shows the overall efficiency related to the polytropic efficiency for a constant value of g ¼ 1. All Rights Reserved . The difference in the enthalpy change between the actual process and the ideal process is due to friction. 1. From Fig.15. the entropy will increase. calculated for a range of pressure ratios and different polytropic efficiencies. the path is curved as illustrated by line 1 –2. As a result.14 Turbine isentropic efficiency against pressure ratio for various polytropic efficiencies (g ¼ 1.28 Chapter 1 Figure 1. when nozzle flow is accompanied by friction. This ratio is known as the nozzle adiabatic efficiency and is called nozzle efficiency (hn) or jet Copyright 2003 by Marcel Dekker.14. Inc. Now. 1.24 NOZZLE EFFICIENCY The function of the nozzle is to transform the high-pressure temperature energy (enthalpy) of the gasses at the inlet position into kinetic energy. These relationships are shown in Fig. Such a process is illustrated as the path 1– 2s. This is achieved by decreasing the pressure and temperature of the gasses in the nozzle. it is clear that the maximum amount of transformation will result when we have an isentropic process between the pressures at the entrance and exit of the nozzle.4). 1. Inc. On the other hand. If the rate of diffusion is too rapid. 1.25 DIFFUSER EFFICIENCY The diffuser efficiency hd is defined in a similar manner to compressor efficiency (see Fig. All Rights Reserved . pipe efficiency (hj). The diffusion is difficult to achieve and is rightly regarded as one of the main problems of turbomachinery design.Basic Thermodynamics and Fluid Mechanics 29 Figure 1.15 Comparison of ideal and actual nozzle expansion on a T-s or h – s plane.16): hd ¼ Isentropic enthalpy rise Actual enthalpy rise h2s 2 h1 ¼ h2 2 h1 ð1:77Þ The purpose of diffusion or deceleration is to convert the maximum possible kinetic energy into pressure energy. This efficiency is given by: hj ¼ cp ðT 01 2 T 02 Þ Dh h01 2 h02 ¼ ¼ 0 h0 cp ðT 01 2 T 02 0 Þ h01 2 h02 D ð1:76Þ 1. large losses in stagnation pressure are inevitable. if Copyright 2003 by Marcel Dekker. This problem is due to the growth of boundary layers and the separation of the fluid molecules from the diverging part of the diffuser. the fluid is exposed to an excessive length of wall and friction losses become predominant. 1. let v (omega) be the angular velocity about the axis A – A. and the mass flow rate m is constant.30 Chapter 1 Figure 1.27 THE EULER TURBINE EQUATION The fluid flows through the turbomachine rotor are assumed to be steady over a long period of time. the rate of diffusion is very low. Fluid enters the rotor at point 1 and leaves at point 2.26 ENERGY TRANSFER IN TURBOMACHINERY This section deals with the kinematics and dynamics of turbomachines by means of definitions. To minimize these two effects. The flow in and across Copyright 2003 by Marcel Dekker.17. 1.16 Mollier diagram for the diffusion process. 1. In the designing of blade shapes. In turbomachine flow analysis. and dimensionless parameters. the most important variable is the fluid velocity and its variation in the different coordinate directions. Turbulence and other losses may then be neglected. The kinematics and dynamic factors depend on the velocities of fluid flow in the machine as well as the rotor velocity itself and the forces of interaction due to velocity changes. there must be an optimum rate of diffusion. diagrams. All Rights Reserved . As shown in Fig. Inc. velocity vector diagrams are very useful. Similarly. Ca1 — Axial velocity in a direction parallel to the axis of the rotating shaft. which must be taken by a thrust bearing to the stationary rotor casing. The whirl or tangential components Cw produce the rotational effect.. Inc. C). Cw1 — whirl or tangential velocity in the direction normal to a radius.. The change in magnitude of the axial velocity components through the rotor gives rise to an axial force. and Cw2. Neither has any effect on the angular motion of the rotor. Ca2. the stators. It is necessary to restrict the flow to a steady flow.Basic Thermodynamics and Fluid Mechanics 31 Figure 1.e. the mass flow rate is constant (no accumulation of fluid in the rotor).17 Velocity components for a generalized rotor. The point 2 is at a radial distance r2 from the axis A – A. the absolute velocities are of interest (i. Cr2. The flow velocities across the rotor relative to the rotating blade must be considered.e. which is at a radial distance r1 from the axis A –A. i. Cr1 — Radial velocity in the direction normal to the axis of the rotating shaft. that is. At point 2 the fluid leaves with absolute velocity (that velocity relative to an outside observer). The fluid enters with velocity C1. This may be expressed in general as follows: The unit mass of fluid entering at section 1 and leaving in any unit of time produces: The angular momentum at the inlet: Cw1r1 The angular momentum at the outlet: Cw2r2 And therefore the rate of change of angular momentum ¼ Cw1r1 – Cw2r2 Copyright 2003 by Marcel Dekker. The rotating disc may be either a turbine or a compressor.. The change in magnitude of the radial velocity components produces radial force. All Rights Reserved . viz. exit velocity C2 can be resolved into three components. The velocity C1 at the inlet to the rotor can be resolved into three components. W ¼ ðC w1 U 1 2 C w2 U 2 Þ. and U1 and U2 are the rotor speeds at the inlet and the exit respectively.18. Inc. as shown in Fig. and combustion products. Thus the vector sum of U and V is equal to the vector C. and the rotor speeds U1 and U2 or the velocities V1. To calculate torque from the Euler turbine equation. Cr2 as well as U1 and U2. This means the work produced by the turbine is positive and the work absorbed by the compressors and pumps is negative.. Equation (1. W is the energy transferred per unit mass. the net torque of the rotor t (tau). V2. v r1 ¼ U1 and v r2 ¼ U2. and are usually combined into one diagram. energy will be given by: W ¼ vðC w1 r 1 2 C w2 r 2 Þ ¼ ðC w1 r 1 v 2 Cw2 r 2 vÞ But.78) is referred to as Euler’s turbine equation. so: W ¼ tv ¼ mvðC w1 r 1 2 C w2 r 2 Þ For unit mass flow. this is equal to the summation of all the applied forces on the rotor. is the product of the torque and the angular velocity of the rotor v (omega). Cw2. the two velocities U and V are relative to one another. Therefore. The flow through a turbomachine rotor. 1. it is necessary to know the velocity components Cw1. i. Hence.e. W.32 Chapter 1 By Newton’s laws of motion. so that the tail of V is at the head of U. using mass flow rate m. ð1:78Þ where. steam. All Rights Reserved . The standard thermodynamic sign convention is that work done by a fluid is positive. the energy transfer equations can be written separately as W ¼ ðC w1 U 1 2 C w2 U 2 Þ for turbine and W ¼ ðC w2 U 2 2 C w1 U 1 Þ for compressor and pump: The Euler turbine equation is very useful for evaluating the flow of fluids that have very small viscosities. Cr1. air. like water. These quantities can be determined easily by drawing the velocity triangles at the rotor inlet and outlet. These triangles are usually drawn as a vector triangle: Since these are vector triangles. The velocity triangles are key to the analysis of turbomachinery problems. Under steady flow conditions. the torque exerted by or acting on the rotor will be: t ¼ mðC w1 r1 2 C w2 r2 Þ Therefore the rate of energy transfer. and work done on a fluid is negative. the absolute velocities C1 and C2 as well as the relative velocities V1 and V2 can have three Copyright 2003 by Marcel Dekker. but also useful in Copyright 2003 by Marcel Dekker.Basic Thermodynamics and Fluid Mechanics 33 Figure 1. the two velocity components.28 COMPONENTS OF ENERGY TRANSFER The Euler equation is useful because it can be transformed into other forms. The velocity components Cr1 and Cr2 are the flow velocity components. However. which passes through the point under consideration and the turbomachine axis.18 Velocity triangles for a rotor. one tangential to the rotor (Cw) and another perpendicular to it are sufficient. which may be axial or radial depending on the type of machine. Inc. All Rights Reserved . 1. components as mentioned earlier. which are not only convenient to certain aspects of design. The component Cr is called the meridional component. 1 C 2 . represents the energy transfer due to change of 1 2 2 absolute kinetic energy of the fluid during its passage between the entrance and exit sections. is due to the centrifugal forces that are developed 1 2 2 as the fluid particles move outwards towards the rim of the machine. By simple geometry. 1 ðC 2 2 C 2 Þ. à 1 E ¼ ðC 2 2 C 2 Þ þ ðU 2 2 U 2 Þ þ ðV 2 2 V 2 Þ ð1:79Þ 1 2 1 2 1 2 2 The first term.” 1 2 2 The other two terms of Eq. This effect is produced if the fluid changes radius as it flows from the entrance to the exit section. All Rights Reserved . i. 1 ðU 2 2 U 2 Þ. Normally. In a turbine. C2 ¼ C2 2 C2 r2 2 w2 and C 2 ¼ V 2 2 ðU 2 2 C w2 Þ2 r2 2 Equating the values of C 2 and expanding. it acts like a diffuser.e. 1 U 1 Cw1 ¼ ðC2 þ U 2 2 V 2 Þ 1 1 2 1 Inserting these values in the Euler equation. is sometimes called an “external effect. the passage acts like a nozzle and if V2 . 1 ðV 2 2 V 2 Þ. which is external to the rotor. Consider the fluid velocities at the inlet and outlet of the turbomachine. it can be called a “virtual pressure rise” or “a pressure rise” which is possible to attain. The amount of pressure rise in the diffuser depends. and hence they are called “internal diffusion. As this absolute kinetic energy change can be used to accomplish rise in pressure. respectively.. this term. of course. the discharge kinetic energy from the rotor. may be considerable.” The centrifugal effect. V1. If V2 . again designated by the subscripts 1 and 2. it is static head or pressure that is 2 2 required as useful energy. The third term. Since this pressure rise comes from the diffuser. it is Á 1À 2 C2 þ U 2 2 V 2 2 2 2 Copyright 2003 by Marcel Dekker. the change in absolute kinetic energy represents the power transmitted from the fluid to the rotor due to an impulse effect. 1 ðC2 2 C 2 Þ. Usually the kinetic energy at the rotor outlet is converted into a static pressure head by passing the fluid through a diffuser. From the above discussions.79) are factors that produce pressure rise within the rotor itself. Inc. r2 C 2 2 C2 ¼ V 2 2 U 2 þ 2U 2 Cw2 2 C 2 2 w2 2 2 w2 and U 2 Cw2 ¼ Similarly.34 Chapter 1 understanding the basic physical principles of energy transfer. represents the energy transfer due to the 1 2 2 change of the relative kinetic energy of the fluid. V1. (1. In a pump or compressor. on the efficiency of the diffuser. equating head.1: A radial flow hydraulic turbine produces 32 kW under a head of 16 m and running at 100 rpm. and power coefficients. where r1 ¼ r2 : 3 5 r1 N 1 D 1 r2 N 3 D 5 2 2 1 3 1 3 3 D2 P2 5 N 1 5 D2 0:032 5 N 1 5 N1 5 Then. Illustrative Example 1. (1.19) that gH 1 gH 2 ¼ ðgravity remains constantÞ 2 ð N 1 D1 Þ ð N 2 D 2 Þ2 Then D2 ¼ D1 1 1 H2 2 N1 6 2 N1 ¼ H1 N2 N2 16 Equating the diameter ratios. It should be noted that the turbine derives power from the same effects. find the diameter ratio between the model and prototype. using subscripts 1 for the prototype and 2 for the model. If model efficiency is assumed to be 92%. the volume flow rate through the model. ¼ or ¼ ¼ 0:238 D1 P1 N2 D1 N2 N2 42 À Also. P1 P2 Á¼À Á . the centrifugal effects are not utilized at all. (1.Basic Thermodynamics and Fluid Mechanics 35 apparent that in a turbocompresser. and speed of the model. we have from Eq. However. we know from Eq. All Rights Reserved . in axial flow compressors. A geometrically similar model producing 42 kW and a head of 6 m is to be tested under geometrically similar conditions.19). pressure rise occurs due to both external effects and internal diffusion effect. Inc. This is why the pressure rise per stage is less than in a machine that utilizes all the kinetic energy effects available. we get 3 1 N1 5 6 2 N1 ¼ 0:238 N2 N2 16 or N2 N1 2 5 ¼ 0:612 ¼ 2:57 0:238 Therefore the model speed is N 2 ¼ 100 £ ð2:57Þ2 ¼ 1059 rpm 5 Copyright 2003 by Marcel Dekker. flow. Solution: Assuming constant fluid density. equating head coefficients for both cases gives [Eq. 0:92 ¼ or. we get [Eq.3)] Q1 Q2 ¼ 3 N 1 D1 N 2 D 3 2 or 2:5 Q2 ¼ 3 2010 £ ð0:125Þ 2210 £ ð0:104Þ3 Solving the above equation. 9:81 £ 14 9:81 £ H 2 ¼ ð2010 £ 125Þ2 ð2210 £ 104Þ2 Therefore.72 m of water. H2 ¼ 11. let us assume that dynamic similarity exists between the two pumps. Q¼ 42 £ 103 ¼ 0:776 m3 /s 0:92 £ 103 £ 9:81 £ 6 42 £ 103 .9)] gH 1 /N 2 D 2 ¼ gH 2 /N 2 D 2 1 1 2 2 Substituting the given values.2: A centrifugal pump delivers 2. (1. Copyright 2003 by Marcel Dekker. If a 104 mm diameter impeller is fitted and the pump runs at a speed of 2210 rpm. (1. rgQH Power output Water power input Illustrative Example 1. what is the volume rate? Determine also the new pump head. Inc. the volume flow rate of the second pump is Q2 ¼ 2:5 £ 2210 £ ð0:104Þ3 ¼ 1:58 m3 /s 2010 £ ð0:125Þ3 Now.5 m3/s under a head of 14 m and running at a speed of 2010 rpm.36 Chapter 1 Model scale ratio is given by 3 D2 100 5 ¼ ð0:238Þ ¼ 0:238ð0:094Þ0:6 ¼ 0:058: D1 1059 Model efficiency is hm ¼ or. Equating the flow coefficients. Solution: First of all. The impeller diameter of the pump is 125 mm. All Rights Reserved . Q. and c are constants. If specific weight of the liquid is w. for M. What is the correct speed at which the compressor must run? If an entry pressure of 65 kPa is obtained at the point where the mass flow rate would be 64 kg/s. Now.013 bar. 1 ¼ a or a ¼ 1. for L. and for T. 65 m2 ¼ 64 £ 101:3 291 298 0:5 ¼ 40:58 kg/s Illustrative Example 1. The performance characteristic of the compressor is obtained at the atmosphere temperature of 258C. b. so c ¼ 1. pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi m1 T 01 m2 T 02 ¼ p01 p02 Therefore. the gas constant R and diameter can be cancelled from the operating equations. 2 3 ¼ 2 2a 2 b or b ¼ 1. 2 ¼ 2 2a þ 3b þ c. the correct speed is 5060 rpm. respectively. considering the mass flow parameter.Basic Thermodynamics and Fluid Mechanics 37 Illustrative Example 1. HÞ ¼ kw a Q b H c where k. Inc.3: An axial flow compressor handling air and designed to run at 5000 rpm at ambient temperature and pressure of 188C and 1. N1 N2 pffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffi T 01 T 02 Therefore. Solution: Since the machine is the same in both cases. Solution: Let Power P be given by: P ¼ f ðw.4: A pump discharges liquid at the rate of Q against a head of H. Substituting the respective dimensions in the above equation. ML2 T23 ¼ kðML22 T22 Þa ðL3 T21 Þb ðLÞc Equating corresponding indices. All Rights Reserved . 273 þ 25 N 2 ¼ 5000 273 þ 18 1 2 298 0:5 ¼ 5000 ¼ 5060 rpm 291 Hence. Copyright 2003 by Marcel Dekker. find the expression for the pumping power. calculate the expected mass flow rate obtained in the test. Using first the speed parameter. a. and r as repeating variables. k. there are only two primary dimensions. Thus. L. l is the linear dimension. l. Thus. V. gÞ ¼ 0 In the above equation. 0 ¼ 2 d or d ¼ 0. r. Taking V. l. and for L. P ¼ kwQH Illustrative Example 1. 0 ¼ d þ e 2 3f þ 1 or e ¼ 2 1. we get: À Ác P1 ¼ ðVÞa ðlÞb r F Mo Lo To ¼ ðLT21 Þa ðLÞb ðML23 Þc ðMLT22 Þ Equating the powers of M. the fluid density. m ¼ 2. l V2 2 2 Prove that the drag force F on a partially where V is the velocity of the body. l. r. À Áf P2 ¼ ðVÞd ðlÞe r ðkÞ M0 L0 T0 ¼ ðLT21 Þd ðLÞe ðML23 Þ f ðLÞ for M. k is the rms height of surface roughness. for T. F P1 ¼ ðVÞ22 ðlÞ22 ðrÞ21 F ¼ 2 2 V l r Similarly. 0 ¼ 2 a 2 2 or a ¼ 2 2. and for L. for T. Inc. Therefore. r. 0 ¼ c þ 1 or c ¼ 2 1. k. and T on both sides of the equation.38 Chapter 1 Therefore.5: submerged body is given by: k lg F ¼ V l rf . All Rights Reserved . Solution: Let the functional relation be: F ¼ f ðV. 0 ¼ f or f ¼ 0. and g is the gravitational acceleration. for M. Copyright 2003 by Marcel Dekker. 0 ¼ a þ b2 3c þ 1 or b ¼ 2 2. k P2 ¼ ðVÞ0 ðlÞ21 ðrÞ0 k ¼ l Therefore. gÞ Or in the general form: F ¼ f ðF. 6: Consider an axial flow pump. for T. 2 l V Illustrative Example 1. and (3) input power? Solution: Using the geometric and dynamic similarity equations. F ¼ V 2l 2r f k lg . what are its (1) flow rate.5 m3/min while running at 1450 rpm. 0 ¼ g þ h 2 3i þ 1 or h ¼ 1. (2) change in total pressure. The corresponding energy input is 120 J/kg. 0 ¼ 2 g –2 or g ¼ 2 2. lg Therefore. for M. All Rights Reserved . Inc. for L.Basic Thermodynamics and Fluid Mechanics 39 and À Ái P3 ¼ ðVÞg ðlÞh r ðgÞ M0 L0 T0 ¼ ðLT21 Þg ðLÞh ðML23 Þi ðLT22 Þ Equating the exponents gives. . P3 ¼ V 22 l 1 r 0 g ¼ 2 V Now the functional relationship may be written as: F k lg f . 0 ¼ i or i ¼ 0. W2 ¼ The change in total pressure is: DP ¼ W 2 htt r ¼ ð226:88Þð0:78Þð1000Þ N/m2 ¼ ð226:88Þð0:78Þð1000Þ1025 ¼ 1:77 bar Copyright 2003 by Marcel Dekker. Q1 Q2 ¼ N 1 D2 N 2 D2 1 2 Therefore. If a second geometrically similar pump with diameter of 22 cm operates at 2900 rpm. Q2 ¼ Q1 N 2 D2 ð2:5Þð2900Þð0:22Þ2 2 ¼ ¼ 2:363 m3 /min N 1 D2 ð1450Þð0:32Þ2 1 W 1 N 2 D2 ð120Þð2900Þ2 ð0:22Þ2 2 2 ¼ ¼ 226:88 J/kg N 2 D2 ð1450Þ2 ð0:32Þ2 1 1 As the head coefficient is constant. which has rotor diameter of 32 cm that discharges liquid water at the rate of 2. 2 ¼0 V 2l 2r l V Therefore. and the total efficiency is 78%. 86 kJ/kg Illustrative Example 1. Inc. Take g ¼ 1. The turbine operates with a total pressure ratio of 4:1. Solution: Using the isentropic P– T relation: 0:286 ðg21Þ P02 2 1 0 T02 ¼ T 01 ¼ ð1050Þ ¼ 706:32K 4 P01 Using total-to-total efficiency.5 m/s. If the total-to-total efficiency is 0.40 Chapter 1 Input power is given by _ P ¼ mW 2 ¼ ð1000Þð2:363Þð0:22688Þ ¼ 8:94 kW 60 Illustrative Example 1. 500Þ2 ð0:30Þ2 1 1 ¼ 84.500 rpm. Solution: Let Velocity of the model: V m Length of the model: Lm ¼ 160 mm Length of the prototype: Lp ¼ 1000 mm Velocity of the prototype: V p ¼ 40:5 m/s Copyright 2003 by Marcel Dekker. find the power output per kg per second of airflow if the rotor diameter is reduced to 20 cm and the rotational speed is 12. Air is under atmospheric pressure in the wind tunnel.8: At what velocity should tests be run in a wind tunnel on a model of an airplane wing of 160 mm chord in order that the Reynolds number should be the same as that of the prototype of 1000 mm chord moving at 40.4.85.7: Consider an axial flow gas turbine in which air enters at the stagnation temperature of 1050 K. All Rights Reserved . 862 J/kg [ Power output ¼ 84. The rotor turns at 15500 rpm and the overall diameter of the rotor is 30 cm. 500Þ2 ð0:20Þ2 2 2 ¼ N 2 D2 ð15. À Á 0 T01 2 T 02 ¼ T 01 2 T02 h tt ¼ ð343:68Þð0:85Þ ¼ 292:13 K and W 1 ¼ cp DT 0 ¼ ð1:005Þð292:13Þ ¼ 293:59 kJ/kg W2 ¼ W 1 N 2 D2 ð293:59 £ 103 Þð12. Illustrative Example 1. or K:E: ¼ kV a m b : Dimensionally. we get: F: 1 ¼ b. mÞ. or V m ¼ Lp V p /Lm ¼ 40:5 £ 1000/160 ¼ 253:13 m/s Illustrative Example 1. FLT0 ¼ ðLTÿ1 Þa ðFT2 Lÿ1 Þb Equating the exponents of F. All Rights Reserved .Basic Thermodynamics and Fluid Mechanics 41 According to the given conditions: ðReÞm ¼ ðReÞp V m Lm V p Lp ¼ . Find the torque per unit mass flow rate. So. Solution: Since the kinetic energy of a body depends on its mass and velocity. T: 0 ¼ 2 a þ 2b This gives b ¼ 1 and a ¼ 2. respectively. Solution: Here. respectively.9: Show that the kinetic energy of a body equals kmV 2 using the method of dimensional analysis.E. and T. r 2 ¼ 0:07 m Cw2 ¼ 50 m/s Copyright 2003 by Marcel Dekker. K. L. where k is a constant. L: 1 ¼ a 2 b. ¼ kV2m. Inc. K:E: ¼ f ðV. Therefore.10: Consider a radial inward flow machine. and the inlet and the outlet radii are 14 cm and 7 cm. r 1 ¼ 0:14 m Cw1 ¼ 340 m/s. the radial and tangential velocity components are 340 m/s and 50 m/s. vm ¼ vp ¼ vair nm np Hence V m Lm ¼ V p Lp . r. (7 turbines). All Rights Reserved .2 Develop an expression for the drag force on a smooth sphere of diameter D immersed in a liquid (of density r and dynamic viscosity m) moving with velocity V. Q ¼ rate of discharge. kÞ where L ¼ the length of the aircraft. show that dimensionless expression KP is given by: DP ¼ 4 f V 2r l 2D 1. diameter D of the rotor and discharge Q. estimate how many turbines should be used. Determine the torque equation.000 kW of energy is generated. m. w ¼ specific weight of liquid. and efficiency. The specific speed of a Kaplan turbine is 450 when working under a head of 12 m at 150 rpm. The resisting force F of a supersonic plane in flight is given by: F ¼ f ðL.6 1. 1. Inc. r ¼ air density. and H ¼ head dimension. dynamic viscosity m of the fluid. By using Buckingham’s P theorem. If under this head.1 Show that the power developed by a pump is given by P ¼ kwQH where k ¼ constant. The efficiency of a fan depends on density r.5 Show that the resisting force is a function of Reynolds number and Mach number. specific weight w of water. head H.7 1. 1.8 Copyright 2003 by Marcel Dekker. angular velocity v. and k ¼ the bulk modulus of air.42 Chapter 1 Torque is given by: T ¼ r 1 Cw1 2 r 2 Cw2 ¼ ð0:14 £ 340 2 0:07 £ 50Þ ¼ ð47:6 2 3:5Þ ¼ 44:1 N-m per kg/s PROBLEMS 1.3 1. The torque of a turbine is a function of the rate of flow Q. 30. V.4 1. V ¼ velocity. Express efficiency in terms of dimensionless parameters. angular velocity v. m ¼ air viscosity. Inc. and r ¼ density of the fluid.Basic Thermodynamics and Fluid Mechanics 43 where KP ¼ pressure drop in a pipe.L.L. turbine runner. V ¼ mean velocity of the flow.9 If Hf is the head loss due to friction (KP/w) and w is the specific weight of the fluid. ML21 T21 . 1. All Rights Reserved .E. or pump depth of the prototype energy transfer by a rotor or absorbed by the rotor force force ratio local acceleration due to gravity head specific enthalpy stagnation enthalpy kinetic energy length length of prototype scale ratio Mach number mass rate of flow speed specific speed power hydraulic power power loss due to mechanical friction at the bearing shaft power potential energy Copyright 2003 by Marcel Dekker. show that Hf ¼ 4 f V 2l 2gD (other symbols have their usual meaning). L Lp Lr M m N Ns P Ph Pm Ps P.T and F.E. area ratio sonic velocity breadth of prototype velocity of gas. m ¼ viscosity of the fluid. FT2 L21 . (2) dynamic viscosity.10 Determine the dimensions of the following in M. D ¼ diameter of the pipe. FTL22 .T systems: (1) mass. and (3) shear stress. k ¼ average roughness of the pipe. ML21 T22 . l ¼ length of the pipe. absolute velocity of turbo machinery diameter of pipe. FL23 NOTATION Ar a Br C D Dp E F Fr g H h h0 K. À Á M. 1. 44 Chapter 1 p p0 Q R Re r s sp . torque exerted by or acting on the rotor angular velocity Copyright 2003 by Marcel Dekker. Inc. heat transfer gas constant Reynolds number radius of rotor specific entropy specific gravity of fluid temperature. velocity ratio actual turbine work output isentropic turbine work output absolute air angle relative air angle specific weight. specific heat ratio efficiency polytropic efficiency of compressor polytropic efficiency of turbine compressor efficiency diffuser efficiency hydraulic efficiency jet pipe or nozzle efficiency mechanical efficiency overall efficiency prototype efficiency isentropic efficiency turbine efficiency total-to-static efficiency total-to-total efficiency volumetric efficiency absolute or dynamic viscosity kinematic viscosity dimensionless parameter mass density shear stress. All Rights Reserved . time stagnation temperature time rotor speed relative velocity. mean velocity work volume ratio. gr T T0 t U V W Vr Wt Wt0 a b g h h/c h/t hc hd hh hj hm ho hp hs ht hts htt hv m n P r t v fluid pressure stagnation pressure volume rate of flow. Inc.Basic Thermodynamics and Fluid Mechanics 45 SUFFIXES 0 1 2 3 a h r t w stagnation conditions inlet to rotor outlet from the rotor outlet from the diffuser axial hub radial tip whirl or tangential Copyright 2003 by Marcel Dekker. All Rights Reserved . 2. Centrifugal and axial flow pumps are very common hydraulic pumps. but other liquids can also be used.2 Hydraulic Pumps 2. this assumption will not cause a significant error in calculations. Most of the theory applicable to hydraulic pumps has been derived using water as the working fluid. Both work on the principle that the energy of the liquid is increased by imparting kinetic energy to it as it flows through the pump.2 CENTRIFUGAL PUMPS The three important parts of centrifugal pumps are (1) the impeller. Unless the change in pressure in a particular situation is very great. we will assume that liquids are totally incompressible unless otherwise specified. (2) the volute casing. The centrifugal and axial flow pumps will be discussed separately in the following sections. This means that the density of liquids will be considered constant no matter how much pressure is applied. In this chapter. This energy is supplied by the impeller. All Rights Reserved . which is driven by an electric motor or some other drive.1 INTRODUCTION Hydraulics is defined as the science of the conveyance of liquids through pipes. Inc. and (3) the diffuser. The pump is often used to raise water from a low level to a high level where it can be stored in a tank. Copyright 2003 by Marcel Dekker. Copyright 2003 by Marcel Dekker. 2. The flow rate is E¼ Q ¼ 2pr 1 Cr1 b1 ¼ 2pr 2 Cr2 b2 ð2:2Þ Where Cr is the radial component of absolute velocity and is perpendicular to the tangent at the inlet and outlet and b is the width of the blade.1 shows a centrifugal pump impeller with the velocity triangles at inlet and outlet. The functions of a volute casing can be summarized as follows: It collects water and conveys it to the pump outlet. the work done on the water by the pump consists of the following three parts: 1. and. the fluid that is drawn into the blade passages at the impeller inlet or eye is accelerated as it is forced radially outwards. Inc.48 Chapter 2 2.2. The water coming out of the impeller is then lead through the pump casing under high pressure. For the best efficiency of the pump. water enters and leaves the vane tips in a direction parallel to their relative velocities at the two tips. i. The stationary blade passages have an increasing cross-sectional area. the static pressure at the outer radius is much higher than at the eye inlet radius. more and more water is added from the outlet periphery of the impeller.e. Vaneless diffuser passages may also be used. a1 ¼ 908 and Cw1 ¼ 0.1 Impeller The centrifugal pump is used to raise liquids from a lower to a higher level by creating the required pressure with the help of centrifugal action. the work done per second on the water per unit mass of fluid flowing W ¼ ðU 2 Cw2 2 U 1 Cw1 Þ ð2:1Þ m Where Cw is the component of absolute velocity in the tangential direction. The fluid moves from the diffuser blades into the volute casing. to recover this kinetic energy by changing it into pressure energy. As discussed in Chapter 1.. For shockless entry and exit to the vanes. Using Euler’s pump equation. Figure 2. As the impeller rotates. The shape of the casing is such that its area of cross-section gradually increases towards the outlet of the pump. The part (C2 – C2)/2 represents the change in kinetic energy of the 2 1 liquid. diffusion action takes place and hence the kinetic energy is converted into pressure energy. In this way. The fluid has a very high velocity at the outer radius of the impeller. As the fluid moves through them. All Rights Reserved . it is assumed that water enters the impeller radially. E is referred to as the Euler head and represents the ideal or theoretical head developed by the impeller only. diffuser blades mounted on a diffuser ring may be used. As the flowing water progresses towards the delivery pipe. Whirling motion is imparted to the liquid by means of backward curved blades mounted on a wheel known as the impeller. The part (U2 – U2)/2 represents the effect of the centrifugal head or 2 1 energy produced by the impeller. b2 0 is the angle at which the fluid leaves the impeller.3 SLIP FACTOR From the preceding section. Inc. s ¼ Copyright 2003 by Marcel Dekker. The slip factor is defined as C w2 0 Cw2 According to Stodola’s theory.1 Velocity triangles for centrifugal pump impeller. and Cw2 and C w2 0 are the tangential components of absolute velocity corresponding to the angles b2 and b2 0 .2 shows the velocity triangles at impeller tip. In Fig. slip in centrifugal pumps and compressors is due to relative rotation of fluid in a direction opposite to that of impeller with the same Slip factor. All Rights Reserved . The part (V2 2 V2)/2 represents the change in static pressure of the 2 1 liquid. it may be seen that there is no assurance that the actual fluid will follow the blade shape and leave the impeller in a radial direction. Figure 2.2. Thus. There is usually a slight slippage of the fluid with respect to the blade rotation. and b2 is the actual blade angle. respectively. 2. 2. 3. if the losses in the impeller are neglected.Hydraulic Pumps 49 Figure 2. Cw2 is reduced to C w2 0 and the difference DCw is defined as the slip. 2. is shown in Fig.4. Due to this fact. Pressure distribution on impeller vane. where there is a high-pressure region while on the trailing side of the blade there is a low-pressure region. The net result of the previous discussion is that the fluid is discharged from the impeller at an angle relative to the impeller.3 pressure.2 Velocity triangle at impeller outlet with slip. Figure 2.3 shows the leading side of a blade. Thus. the impeller itself has an angular velocity v so that. so that low velocity on the highpressure side and high velocity on the low-pressure side and velocity distribution is not uniform at any radius. there will be a higher velocity and a velocity gradient across the passage. Cw2 is less than Cw20 and the difference is defined as the slip.50 Chapter 2 Figure 2. This pressure distribution is associated with the existence of circulation around the blade. Due to the lower pressure on the trailing face. HP ¼ high Copyright 2003 by Marcel Dekker. Figure 2. Inc. relative to the impeller. All Rights Reserved . LP ¼ low pressure. angular velocity as that of an impeller. which is less than the vane angle as mentioned earlier. the flow may separate from the suction surface of the blade. the result of this being a circulatory motion relative to the channel or relative eddy. Another way of looking at this effect. as given by Stodola. the fluid must have an angular velocity of 2 v. Eddy losses at entrance and exit of impeller. 2. while the slip occurs even if the fluid is ideal. Thus. The above losses are known as hydraulic losses. the slip factor lies in the region of 0.Hydraulic Pumps 51 Figure 2. which are often used in centrifugal compressors. frictional and eddy losses in the diffuser.4 PUMP LOSSES The following are the various losses occurring during the operation of a centrifugal pump. b2 will be 908 and the Stodola slip factor becomes p ð2:4Þ s¼12 n where n is the number of vanes. Hence. the slip factor s is defined as C w2 ð2:3Þ C w2 For purely radial blades. 1. 2. the Euler pump equation becomes s¼12 ð2:5Þ W ¼ sU 2 Cw2 2 U 1 C w1 ð2:6Þ m Typically. the energy supplied by the prime mover to Copyright 2003 by Marcel Dekker. Inc. Mechanical losses are losses due to friction of the main bearings.9. All Rights Reserved . friction losses in the impeller. The Stanitz slip factor is given by 0 s¼ 0:63p n When applying a slip factor. if provided. 3. and stuffing boxes. Losses in the suction and delivery pipe.4 Relative eddy in impeller channel. All Rights Reserved . Inc. H.52 Chapter 2 impeller is equal to the energy produced by impeller plus mechanical losses. total head developed by the pump. the overall efficiency ho is given by: ho ¼ Fluid power developed by pump rgQH ¼ Shaft power input Ps ð2:7Þ Casing efficiency hc is given by: hc ¼ Fluid power at casing outlet/fluid power at casing inlet ¼ Fluid power at casing outlet/ðfluid power developed by impeller 2 leakage lossÞ ¼ rgQH/rgQH i ¼ H/H i Impeller efficiency hi is given by: ð2:8Þ hi ¼ Fluid power at impeller exit/fluid power supplied to impeller ¼ Fluid power at impeller exit/ðfluid power developed by impeller þ impeller lossÞ Â Ã ¼ rgQi H i / rgQi ðH i þ hi Þ ¼ H i /ðH i þ hi Þ Volumetric efficiency hv is given by: ð2:9Þ hv ¼ Flow rate through pump/flow rate through impeller ¼ Q/ðQ þ qÞ Mechanical efficiency hm is given by: ð2:10Þ hm ¼ Fluid power supplied to the impeller/power input to the shaft ¼ rgQi ðhi þ H i Þ/Ps Therefore. Let r be the density of liquid. and hi. Then. head loss in the impeller. ð2:11Þ ho ¼ hc hi hv h m ð2:12Þ Copyright 2003 by Marcel Dekker. A number of efficiencies are associated with these losses. Q. flow rate. shaft power input. total head across the impeller. Ps. Hi. 2. Figure 2. while forward-curved vanes have a high value of energy transfer. radial vanes (b2 ¼ 908) have some particular advantages for very highspeed compressors where the highest possible pressure is required. Radial vanes are relatively easy to manufacture and introduce no complex bending stresses (Fig. However. The blade shapes can be classified as: 1. 908) 2.5 THE EFFECT OF IMPELLER BLADE SHAPE ON PERFORMANCE The various blade shapes utilized in impellers of centrifugal pumps/compressors are shown in Fig.5 Centrifugal pump outlet velocity triangles for varying blade outlet angle. 2.6). 2. Forward-curved blades (b2 . and its reduction to static pressure by diffusion in a fixed casing is difficult to perform in a reasonable sized casing. 908) As shown in Fig. Copyright 2003 by Marcel Dekker.5. Therefore. All Rights Reserved . such rotors have a low energy transfer for a given impeller tip speed. the value of Cw2 (whirl component at outlet) is much reduced. but the velocity diagrams show that this also leads to a very high value of C2. and thus. it is desirable to design for high values of b2 (over 908). Radial blades (b2 ¼ 908) 3. Inc.Hydraulic Pumps 53 A hydraulic efficiency may be defined as hh ¼ Actual head developed by pump Theoretical head developed by impeller H ¼ ðH i þ hi Þ ð2:13Þ The head H is also known as manometric head. 2. for backward-curved vanes.5. High kinetic energy is seldom required. Backward-curved blades (b2 . 7 Volute or scroll collector. and.7).54 Chapter 2 Figure 2. The cross-sectional area increases as the increment of discharge increases around the periphery of the impeller. The advantage of volute is its simplicity and low cost. Figure 2. 2. Inc. if the velocity is constant in the volute. Copyright 2003 by Marcel Dekker.6 Characteristics for varying outlet blade angle.6 VOLUTE OR SCROLL COLLECTOR A volute or scroll collector consists of a circular passage of increasing crosssectional area (Fig. All Rights Reserved . 2. It consists simply of an annular passage without vanes surrounding the impeller. A vaneless diffuser passage is shown in Fig. All Rights Reserved . 2. The cross-sectional shape of the volute is generally similar to that shown in Fig. Inc.Hydraulic Pumps 55 Figure 2. with the sidewalls diverging from the impeller tip and joined by a semicircular outer wall. The circular section is used to reduce the losses due to friction and impact when the fluid hits the casing walls on exiting from the impeller. the vaneless diffuser is reasonably efficient and is best suited for a wide range of operations. Assuming the flow is frictionless in the diffuser.8 Cross-section of volute casing. angular momentum Copyright 2003 by Marcel Dekker. The mass flow rate in any radius is given by m ¼ rAC r ¼ 2prbrCr where b is the width of the diffuser passage. Any deviation in capacity (i.. 2. then the static pressure is likewise constant and the radial thrust will be zero. 2. flow rate) from the design condition will result in a radial thrust which if allowed to persist could result in shaft bending. Cr ¼ r 2 b2 r2 Cr2 rbr ð2:15Þ ð2:14Þ where subscripted variables represent conditions at the impeller outlet and the unsubscripted variables represent conditions at any radius r in the vaneless diffuser.7 VANELESS DIFFUSER For the diffusion process.e.9.8. The size of the diffuser can be determined by using the continuity equation. 56 Chapter 2 Figure 2.9 Vaneless diffuser.10 Logarithmic spiral flow in vaneless space. All Rights Reserved . Figure 2. Copyright 2003 by Marcel Dekker. Inc. 8 VANED DIFFUSER The vaned diffuser is advantageous where small size is important. For an incompressible flow. 2. Because of the long flow path with this type of diffuser. for an incremental radius dr. when the local static pressure of a liquid falls below the vapor pressure of the liquid. Thus. where the fluid is locally accelerated over the vane surfaces. and. In a centrifugal pump.10. these low-pressure zones are generally at the impeller inlet. As shown in Fig. All Rights Reserved . 2. the fluid moves through angle du. Integrating we have u 2 u2 ¼ tan a logðr/r2 Þ ð2:17Þ Substituting a ¼ 788 and (r/r2) ¼ 2. and hence it is possible to reduce the length of flow path and diameter.Hydraulic Pumps 57 is constant and Cw ¼ ðC w2 r 2 Þ/r ð2:16Þ But the tangential velocity component (Cw) is usually very much larger than the radial velocity component Cr. The collector and diffuser operate at their maximum efficiency at the design point only. the number of diffuser vanes should have no common factor with the number of impeller vanes. and. cavitation is most likely Copyright 2003 by Marcel Dekker. therefore. which suddenly collapse on moving forward with the flow into regions of high pressure. In turbines. friction effects are high and the efficiency is low. Small bubbles or cavities filled with vapor are formed. rCr is constant. then rdu ¼ dr tan a. therefore. Any deviation from the design discharge will change the outlet velocity triangle and the subsequent flow in the casing. the flow path traces a logarithmic spiral. the flow maintains a constant inclination to radial lines. Inc. In this type of diffuser. vanes are used to diffuse the outlet kinetic energy of the fluid at a much higher rate than is possible by a simple increase in radius. and the diffuser becomes more efficient. tan a ¼ Cw/Cr ¼ constant. but greater is the friction. The cross section of the diffuser passage should be square to give a maximum hydraulic radius. C3 r2 It means that for a large reduction in the outlet kinetic energy. the angle of divergence is smaller.9 CAVITATION IN PUMPS Cavitation is caused by local vaporization of the fluid. The vane number. a diffuser with a large radius is required. 2. However. giving rise to pressure as high as 3500 atm. the change in angle of the diffuser is equal to 1808. These bubbles collapse with tremendous force. the ratio of the inlet to outlet diffuser velocities C2 ¼ r3 . 18) is a suction head and is called the net positive suction = Figure 2. Serious damage can occur from this prolonged cavitation erosion.11 Cavitation limits for radial flow pumps. When cavitation occurs. Local pitting of the impeller and erosion of the metal surface. 2. A cavitation parameter is defined as sc ¼ pump total inlet head above vapor pressure/head developed by the pump or at the inlet flange p1 V 2 pv 1 þ 2 sc ¼ H ð2:18Þ rg 2g rg The numerator of Eq. Inc. The avoidance of cavitation in conventionally designed machines can be regarded as one of the essential tasks of both pump and turbine designers.58 Chapter 2 to occur at the downstream outlet end of a blade on the low-pressure leading face. Vibration of machine and noise is also generated in the form of sharp cracking sounds when cavitation takes place. A drop in efficiency due to vapor formation. 3. it causes the following undesirable effects: 1. 4. (2. All Rights Reserved . which reduces the effective flow areas. This cavitation imposes limitations on the rate of discharge and speed of rotation of the pump. Copyright 2003 by Marcel Dekker. 12 PUMPING SYSTEM DESIGN Proper pumping system design is the most important single element in minimizing the life cycle cost. and H is the manometric head. pipe installation. 2. All Rights Reserved . À Á h ¼ f Q.12).11 for any condition of operation. 2. a driver. All pumping systems are comprised of a pump. The characteristics of the piping system must be calculated in order to determine required pump performance. 2. and number of vanes. The advantage of an axial flow pump is its compact construction as well as its ability to run at extremely high speeds. pressure is developed by flow of liquid over blades of airfoil section.10 SUCTION SPECIFIC SPEED The efficiency of the pump is a function of flow coefficient and suction specific speed. The cavitation parameter is a function of specific speed. which is defined as  Ã23/4 N suc ¼ NQ 1/2 gðNPSHÞ Thus. Figure 2. A number of installation and operational costs are directly dependent on the piping diameter and the components in the piping system. efficiency of the pump.11 AXIAL FLOW PUMP In an axial flow pump. N suc N s /N suc ¼ ðNPSHÞ3/4 /H 3/4 3/4 ¼ sc The cavitation parameter may also be determined by the following equation ð2:19Þ 2. and operating controls. 2.Hydraulic Pumps 59 head (NPSH) of the pump.11 shows the relationship between sc and Ns. It is a measure of the energy available on the suction side of the pump. The flow area is the same at inlet and outlet and the minimum head for this type of pump is the order of 20 m. Both procurement costs and the operational costs make up the total cost of an installation during its lifetime. Inc. Proper design considers the interaction between the pump and the rest of the system and the calculation of the operating duty point(s) (Fig. This applies to both simple systems as well as to more complex (branched) systems. It may be necessary in the selection of pumps that the value of sc does not fall below the given value by the plots in Fig. Copyright 2003 by Marcel Dekker. It consists of a propeller-type of impeller running in a casing. in particular. Plant standard pipe diameters. and in conjunction with the piping system. which together.. Costs for electrical supply systems will therefore increase. an optimum pipeline size may be found.g. . Required minimum internal diameter for the application (e. the pump workload is divided between the pumps. solid handling) . Inc. Because of this. Pump installation procurement costs will increase as a result of increased flow losses with consequent requirements for higher head pumps and larger motors. A pump application might need to cover several Copyright 2003 by Marcel Dekker. All Rights Reserved . Economy of the whole installation (pumps and system) .12 The duty point of the pump is determined by the intersection of the system curve and the pump curve as shown above. In systems with several pumps. control valves in throttle-regulated installations. Some costs increase with increasing pipeline size and some decrease. avoid sedimentation) . Required lowest flow velocity for the application (e.60 Chapter 2 Figure 2. A considerable amount of the pressure losses in the system are caused by valves. Operating costs will increase as a result of higher energy usage due to increased friction losses. deliver the required flow.. . Maximum flow velocity to minimize erosion in piping and fittings . The piping diameter is selected based on the following factors: .g. Piping and component procurement and installation costs will decrease. Decreasing the pipeline diameter has the following effects: . based on minimizing costs over the life of the system. The pump user must carefully consider the duration of operation at the individual duty points to properly select the number of pumps in the installation and to select output control. Procure pumps and systems using life cycle cost considerations. Do not oversize the pump.12. Minimize system losses needed to balance the flow rates. All Rights Reserved . Assemble a complete document inventory of the items in the pumping system. . One consists of observing the operation of the actual piping system. . .Hydraulic Pumps 61 duty points. Match the equipment to the system needs for maximum benefit. . Determine the flow rates required for each load in the system. Match the driver type to the intended duty. and the second consists of performing detailed calculations using fluid analysis techniques. Balance the system to meet the required flow rates of each load. . . . . Consider all relevant costs to determine the life cycle cost. One of two methods can be used to analyze existing pumping systems. of which the largest flow and/or head will determine the rated duty for the pump.1 Methods for Analyzing Existing Pumping Systems The following steps provide an overall guideline to improve an existing pumping system. The following is a checklist of some useful means to reduce the life cycle cost of a pumping system. Specify motors to have high efficiency. Copyright 2003 by Marcel Dekker. differential pressures. The first method relies on observation of the operating piping system (pressures. Consider the duration of the individual pump duty points. . Inc. . Optimize total cost by considering operational costs and procurement costs. Match the pump type to the intended duty. . Identify pumps with high maintenance cost. the second deals with creating an accurate mathematical model of the piping system and then calculating the pressure and flow rates with the model. . and flow rates). . Affect changes to the pump to minimize excessive pump head in the balanced system. 2. Monitor and sustain the pump and the system to maximize benefit. . Consider the energy wasted using control valves. . . . Inc. defining the Head supplied by the pump and the Discharge of the system. being the difference in elevation between the upstream and downstream controls (generally represented by reservoir levels). The total static head.62 Chapter 2 Figure 2. . The efficiency of the pump under these conditions will also be defined.13 Typical pump characteristics. 2. The interaction point of these curves represents the actual operating point (as shown later). 2. 2. The energy losses in the system (generally pipe friction).16).13 LIFE CYCLE ANALYSIS Over the economic life of a pumped supply system.13). 2. System behavior for all possible operating environments is needed (Fig.14 and 2. a number of design parameter will change. 2. . All Rights Reserved . Note that the efficiency of the pump at this operating point is the critical parameter in pump selection and suitability for a particular system (Figs. Parameters that will change include: Copyright 2003 by Marcel Dekker. which are normally expressed as a function of velocity head.12.15). The system characteristics will consist of: .2 Pump System Interaction The actual operating point on the pump system characteristic curve is defined by its interaction with the operating characteristics of the installed system (Fig. 14 Pump– system interaction point and pump efficiency. Figure 2.15 Selection of pump type. Copyright 2003 by Marcel Dekker. Inc.Hydraulic Pumps 63 Figure 2. All Rights Reserved . it is necessary to maintain pump operation close to peak efficiency. Copyright 2003 by Marcel Dekker.64 Chapter 2 Figure 2.16 Variations in demand and operating characteristics. All Rights Reserved . For all operating conditions. Inc. . Seasonal variations in demand. This can be achieved using multiple pumps and timed pumping periods. . Water demand increases as the system population expands. Increasing pipe friction as the system ages. . Composite characteristics (head – discharge curves) are obtained by combining the individual curves. Inc. Occasionally situations will be encountered where pumps are operated in series to boost outputs. 2. Where pumps operate in parallel.e. Analysis of system operation will require the head –dischargeefficiency characteristic for the particular operating speed.15 MULTIPLE PUMP OPERATION The most common type of pumping station has two or more pumps operating in parallel. with the number of pairs of poles in the motor (N) reducing the pump speed to 3000/N revolutions per minute.14 CHANGING PUMP SPEED The most common pump – motor combination employed in water supply operations is a close coupled system powered by an electric motor. the corresponding characteristics for any speed N0 can be established as follows: 0 N Q0 ¼ Q flow points ð2:20Þ N H0 ¼ 0 2 N H N head points ð2:21Þ ð2:22Þ h0 ¼ h efficiency points The data set for the new pump speed can then be matched to the system characteristics. This is generally a temporary measure as any failure of one unit will severely affect system operation. This provides flexibility of operation in coping with a range of flow conditions and allows maintenance of individual units while other units are in operation. These units can only operate at a speed related to the frequency of the A.C.Hydraulic Pumps 65 2.17). the composite curve is obtained by adding the flow rates for a given head. Pumps driven through belt drives or powered by petrol or diesel motors are more flexible allowing the pump speed to be adjusted to suit the operational requirements. Copyright 2003 by Marcel Dekker. All Rights Reserved . the composite is obtained by adding heads for a given flow rate. Given the head – discharge-efficiency characteristics for speed N (in tabular form). intersection with system curve) s an individual curve (Fig. The composite curve is employed in the same manner (i. Where pumps operate in series.. 2. supply (50 cycles/s or 3000 cycles/min). 1 Net Positive Suction Head Simply. 2. the datum line for pump NPSH is the centerline of the inlet. The NPSH required is reported in head of fluid (being pumped) required at the pump inlet. Sufficient NPSH allows for pumping without liquid vaporizing in the pump first-stage impeller eye as the fluid pressure drops due to pump suction losses (Fig. Usually. causes a pressure rise.66 Chapter 2 Figure 2. 2.15. NPSH is the minimum suction condition (pressure) required to prevent pump cavitation.17 Composite pump characteristics.18). This is satisfactory for small pumps. Inc. Conceptually. as it leaves the impeller. NPSH can be imagined as the pressure drop between the pump inlet flange and the point inside the pump where the fluid dynamic action. The NPSH Copyright 2003 by Marcel Dekker. All Rights Reserved . As such. NPSH required has units of length (height). the NPSH requirements reported by the manufacturer should be verified for the datum line being discussed. For larger pumps. The NPSH available must be large enough to eliminate head loss. available differs from NPSH required. (2. Our concern here is the process system side of determining what NPSH is available. Incorporating friction losses and restating the formula in a form familiar to process engineers.18 The elements of Eq. Determining the NPSH required is the job of the pump vendor. the NPSH available in a system can be expressed as: 2:31ðP þ Pa 2 Pv Þ V2 NPSHa ¼ ð2:23Þ þ S2B2Lþ g 2g where NPSHa is the net positive suction head available (ft). vapor pressure of liquid at pumping conditions (psia). atmospheric pressure (psi). Pv. Determining the NPSH available is the process engineer’s job. P. g. Pressure balance and NPSH available derive from Bernoulli’s equation for a stationary conduit where the total energy (head) in a system is the same for any point along a streamline (assuming no friction losses). Inc. All Rights Reserved . Except in rare circumstances.Hydraulic Pumps 67 Figure 2.24) are illustrated with a pump taking suction from a tower. centrifugal pumps require the NPSH available to be greater than NPSH required to avoid cavitation during operation. pressure above liquid (psi gage). The NPSH available is the excess of pressure over the liquid’s vapor pressure at the pump suction flange. Copyright 2003 by Marcel Dekker. specific gravity of liquid at pumping conditions. The NPSH required determined during the manufacturers test and shown on the vendor’s pump curve is based upon a 3% head pump differential loss. Pa. (2. ft. and H is the static height of liquid between liquid level and pump suction centerline (datum line).1: A centrifugal pump runs at a tip speed of 12 m/s and a flow velocity of 1. B.2: A fluid passes through an impeller of 0. and the outlet vane angle is set back at an angle of 228 to the tangent. b2 ¼ 228. Solution: From Fig. Assuming that the fluid enters in the axial direction and zero slip. Converting to absolute pressures. Ca1 ¼ 3. Since Cw1 ¼ 0. Assuming that the fluid enters radially with velocity of flow as 3. L. fluid density and resetting the datum line to the pump centerline results in: NPSHa ¼ 144ðPabs 2 Pv Þ V2 þH2Lþ r 2g ð2:24Þ where Pabs is the pressure above liquids (psia). Illustrative Example 2. fluid density (lb/ft3). calculate the head imparted to a fluid.1 m inlet diameter.68 Chapter 2 S.5 m/s ¼ Ca2 Copyright 2003 by Marcel Dekker.1). All Rights Reserved . The outlet blade angle is 288 to the tangent at the impeller periphery.6)/12 ¼ 349 Nm. for zero slip b2 ¼ b1.2 m and delivers 3. Illustrative Example 2. and V is the average liquid velocity at pump suction nozzle (ft/s). static height of liquid from grade (ft). a1 ¼ 908. suction system friction losses (ft of pumping liquid). Inc.8 m3/min of water. 2.5 m/s. the Euler head 2 H ¼ E ¼ (U2Cw2 2 U1Cw1)/g. The impeller is rotating at 1250 rpm. The impeller diameter is 1. head H is given by U 2 C w2 U 2 1:5 12 1:5 12 2 ¼ U2 2 H¼ ¼ tan 288 9:81 tan 288 g g ¼ 11:23 m Power delivered ¼ rgQH J 1000ð9:81Þð3:8Þð11:23Þ ¼ s 60ð1000Þ ¼ 6:98 kW Torque delivered ¼ Power/angular velocity ¼ 6980(0. Using Eq. calculate the torque delivered by the impeller.5 m/s. r. Cw1 ¼ 0.22 m outlet diameter and 0. as there is no inlet whirl component.2. Solution: Since fluid enters in the radial direction. distance of pump centerline (suction nozzle centerline for vertical pumps). calculate (1) the angle made by the absolute velocity of water at exit with the tangent. Solution: 1. from velocity diagram.Hydraulic Pumps 69 Head developed H ¼ Cw2U2/g Impeller tip speed.2 m. U 2 ¼ pDN ¼ pð0:22Þð1250Þ ¼ 14:40 m/s 60 60 Whirl velocity at impeller outlet. All Rights Reserved . C w2 ¼ U 2 2 ðC a2 /tan b2 Þ ¼ 14:40 2 ð3:5/tan 228 Þ ¼ 5:74 m/s Therefore. 2.19). Assuming a constant radial flow through the impeller at 2.19 Velocity triangle at outlet. Inc. Impeller tip speed is given by U2 ¼ pD2 N pð0:4Þð1400Þ ¼ ¼ 29:33 m/s 60 60 Whirl velocity at impeller tip Cw2 ¼ U 2 2 C r2 2:6 ¼ 29:33 2 ¼ 23:75 m/s tan b2 tan 258 Figure 2. the head imparted is given by H ¼ 5:74ð14:40Þ/9:81 ¼ 8:43 m Design Example 2. (2) the inlet vane angle. and (3) the work done per kg of water (Fig. and vanes angle at exit is 258. Copyright 2003 by Marcel Dekker.4 m and an internal diameter of 0.3: A centrifugal pump impeller runs at 1400 rpm. The impeller has an external diameter of 0.6 m/s. 2.20). from the velocity triangle at impeller tip. and (3) minimum speed to lift water against a head of 6. Work done per kg of water is given by C w2 U 2 ¼ 23:75ð29:33Þ ¼ 696:59 Nm ¼ 696:59 J: Design Example 2. Impeller velocity at inlet U1 ¼ pD1 N pð0:2Þ1400 ¼ ¼ 14:67 m/s 60 60 Cr1 2:6 tan b1 ¼ ¼ 0:177 ¼ U 1 14:67 Therefore. and ratio of external to internal diameter 2:1. Here.65 m2. b2 ¼ 258.2 m. 2.2 m. A ¼ 0:65 m2 . Inc. area at the outer periphery 0.258. Velocity of flow at impeller tip C r2 ¼ Q 1550 ¼ ¼ 2:385 m/s A 1000ð0:65Þ Figure 2. Assume that the pump discharges 1550 l/s (Fig. D2 /D1 ¼ 2.4: A centrifugal pump impeller has a diameter of 1. 3. Solution: 1. rpm 210. b1 ¼ 10. angle of vane at outlet 258. D2 ¼ 1:2 m. (2) power. All Rights Reserved . H ¼ 6:2 m.70 Chapter 2 Now. Copyright 2003 by Marcel Dekker. Calculate (1) the hydraulic efficiency. Q ¼ 1550 l/s. N ¼ 210 rpm.058. tan a2 ¼ C r2 2:6 ¼ 0:1095 ¼ Cw2 23:75 Therefore.20 Velocity triangle at impeller outlet. a2 ¼ 6. theoretical head is given by Cw2 U 2 8:09ð13:2Þ ¼ 10:89 m ¼ 9:81 9:81 Assuming slip factor. minimum speed is ¼ 12. All Rights Reserved . H ¼ 35m Discharge.6: A centrifugal pump impeller has 0.81)/1000 ¼ 165. Centrifugal head is the minimum head. Solution: Total head. U2 ¼ pð1:2Þð210Þ ¼ 13:2 m/s 60 C r2 2:385 ¼ 8:09 m/s ¼ 13:2 2 tan b2 tan 258 C w2 ¼ U 2 2 Assuming radial entry. P ¼ ho s 1000ho 9:81ð0:045Þð35Þ ¼ 0:746ð0:60Þ ¼ 34:5 hp Illustrative Example 2.Hydraulic Pumps 71 Impeller tip speed.. and the entry Copyright 2003 by Marcel Dekker. Inc. Power P ¼ (1550)(10. Find the horsepower of the pump.74(60)/p(1. Q ¼ 45l/s ¼ 0:045 m3 /s Overall efficiency. U2 ¼ 12. U2 2 U2 2 1 ¼ 6:2 2g It is given that U1 ¼ U2/2. Therefore. Illustrative Example 2.ho ¼ 60% ¼ 0:60 rgQH J rgQH ¼ kW Power.89)(9. Therefore.6 m external diameters. The pump runs at 950 rpm.59 kW. U 2 2 0:25U 2 ¼ 2ð9:81Þð6:2Þ 2 2 i. if the overall efficiency is 60%. s ¼ 1.5: A centrifugal pump is required to pump water against a total head of 35 m at the rate of 45 l/s. 3.3 m inlet diameter and 0. hydraulic efficiency is given by H¼ hh ¼ 6:2ð100Þ ¼ 56:9%: 10:89 2.74 m/s Hence.2) ¼ 203 rpm.e. 198. The impeller vanes are set back at angle of 468 to the outer rim. 2. The velocity of flow through the impeller is constant at 3. Cr1 ¼ Cr2 ¼ 3. using velocity triangle at outlet C w2 ¼ U 2 2 C r2 3:5 ¼ 29:86 2 ¼ 26:48 m/s tan 468 tan 468 C 2 ¼ 26:71 m/s and C 2 ¼ C 2 þ C 2 ¼ 3:52 þ 26:482 . All Rights Reserved . Tangential velocity of impeller at inlet U1 ¼ pD1 N pð0:3Þð950Þ ¼ ¼ 14:93 m/s 60 60 Cr1 3:5 ¼ 0:234 ¼ U 1 14:93 From inlet velocity triangle tan a1 ¼ Therefore.5 m/s.21). Calculate (1) the vane angle at inlet of a pump. Tangential velocity of impeller at outlet U2 ¼ pD2 N pð0:6Þð950Þ ¼ ¼ 29:86 m/s 60 60 For velocity of whirl at impeller outlet.72 Chapter 2 Figure 2. Let a1 be the vane angle at inlet. Let a2 be the direction of Copyright 2003 by Marcel Dekker. (2) the velocity direction of water at outlet. 2 r2 w2 where C2 is the velocity of water at outlet. Solution: 1. of the pump is radial.5 m/s. Velocity of flow. 2.21 Velocity triangle at impeller outlet and inlet. Inc. and (3) the work done by the water per kg of water (Fig. a1 ¼ 13. Solution: 1. and (3) minimum starting speed of the pump (Fig. therefore. Vanes are set back at outlet at an angle of 458.22 Velocity triangle (a) outlet.7: A centrifugal pump delivers water at the rate of 8.22). Determine (1) the manometric efficiency. The manometric efficiency is given by hman ¼ H ðCw2 U 2 /gÞ From outlet velocity triangle U2 ¼ pD2 N pð0:5Þð500Þ ¼ ¼ 13:0 m/s 60 60 Figure 2..e. and. (2) vane angle at inlet. Work done by the wheel per kg of water W ¼ Cw2 U 2 ¼ 26:48ð29:86Þ ¼ 790:69 Nm Design Example 2. All Rights Reserved . a2 is given by Cr2 3:5 ¼ 0:132 ¼ tan a2 ¼ C w2 26:48 i. 2. Inc. It has an impeller of 50 cm outer diameter and 25 cm inner diameter. The constant velocity of flow is 2 m/s. Copyright 2003 by Marcel Dekker.5 m3/min against a head of 10 m.538. (b) inlet.Hydraulic Pumps 73 water at outlet. a2 ¼ 7. and impeller is running at 500 rpm. 3. or tan 458 ¼ 2/(U2 2 Cw2). Cw2 ¼ U2.2 cm wide.8: A centrifugal pump impeller has 0.e. Vanes are radial at exit and 8.8.5 m/s.23 Velocity triangle for Example 2. 3. hman ¼ H 10ð9:81Þ ¼ 68:6% ¼ ðC w2 U 2 /gÞ 11ð13Þ 2. or 1 ¼ 2/(13 2 Cw2). Copyright 2003 by Marcel Dekker. b1 ¼ 178.5 m/s.. D2 ¼ 0.5 m/s. and velocity in the delivery pipe is 2. Inc.6 m. N ¼ 618 rpm. Hence. Neglecting other losses. Solution: 1. All Rights Reserved .6 m outside diameter and rotated at 550 rpm. calculate head and power of the pump (Fig. N ¼ 550 rpm. Cr2 ¼ 3.23). 2. Illustrative Example 2. Impeller speed at outlet U2 ¼ pD2 N pð0:6Þð550Þ ¼ ¼ 17:29 m/s: 60 60 Figure 2. tan b2 ¼ Cr2/(U2 2 Cw2).74 Chapter 2 Now. Velocity of flow through the impeller is 3. Cw2 ¼ 11 m/s. Vane angle at inlet b1 Cr1 and U 1 ¼ 0:5 £ U 2 ¼ 6:5 m/s U1 2 ¼ 0:308 [ tan b1 ¼ 6:5 tan b1 ¼ i. The minimum starting speed is ðU 2 2 2 U 2 Þ/2g 1 À pD N Á 2 À pD N Á 2 2 1 2 60 60 ¼ 10 ¼ H or 2ð9:81Þ Therefore. Power ¼ 1000 where Q ¼ pD2 b2 Cr2 ðwhere b2 is widthÞ ¼ pð0:6Þð0:082Þð3:5Þ ¼ 0:54 m3 /s Therefore. Water enters radially and discharges with a velocity whose radial component is 2. of the pump and (2) turning moment of the shaft (Fig. Inc. H ¼ C w2 U 2 2 g ¼ V2 ðneglecting all lossesÞ 2g ð17:29Þð17:29Þ 2:52 ¼ 30:47 2 0:319 2 2ð9:81Þ 9:81 ¼ 30:2 m of water: rgQH kW 2. If the discharge through the pump is 5.5 m3/min. Solution: 1. power is given by P¼ rgQH 1000ð9:81Þð0:54Þð30:2Þ kW ¼ ¼ 160 kW: 1000 1000 Illustrative Example 2.9.5 m/s. determine (1) h. Copyright 2003 by Marcel Dekker. Data D2 ¼ 1 m.9: A centrifugal pump impeller has a diameter of 1 m and speed of 11 m/s.24). 2. Backward vanes make an angle of 328 at exit. All Rights Reserved .Hydraulic Pumps 75 Figure 2. Head.p.24 Velocity triangles for Example 2. through which the water can be lifted. U2 ¼ 11 m/s. H2 ¼ ? Copyright 2003 by Marcel Dekker. T¼ ð7Þð1000Þð60Þ ¼ 318 Nm/s: 2pð210Þ Illustrative Example 2.5 m/s.76 Chapter 2 a1 ¼ 908. Q1 ¼ 110 l/min ¼ 1. First. Therefore. N2 ¼ 390 rpm.10: A centrifugal pump running at 590 rpm and discharges 110 l/min against a head of 16 m. Inc. Cr2 ¼ 2. What head will be developed and quantity of water delivered when the pump runs at 390 rpm? Solution: N1 ¼ 590. All Rights Reserved .83 l/s. T¼ h:p: £ 60 2pN But U 2 ¼ pD2 N or N ¼ 60£U 2 ¼ 60ð11Þ ¼ 210 rpm pD 2 pð1Þ 60 Hence.5 m3/min. consider outlet velocity triangle Cw2 ¼ U 2 2 C r2 tan b2 2:5 ¼ 7 m/s ¼ 11 2 tan 328 Power of the pump is given by P¼ P¼ rQC w2 U 2 kW 1000 ð1000Þð5:5Þð7Þð11Þ ¼ 7 kW: ð60Þð1000Þ 2. Q ¼ 5. H1 ¼ 16 m. h:p: ¼ 2pNT 60 where T is the torque of the shaft. b2 ¼ 328. Now. 97.370 m. head developed by the pump at 390 rpm ¼ 6.11) pffiffiffiffiffiffi pffiffiffiffiffiffi N 1 Q1 N 2 Q2 ¼ 3/4 H 3/4 H2 1 pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi 590 1:83 390 Q2 798:14 390 Q2 ¼ ¼ or 8 4:29 ð16Þ3/4 ð6:98Þ3/4 pffiffiffiffiffiffi Q2 ¼ 1:097 i.. (1. Calculate (1) ideal head developed with no slip and no hydraulic losses and (2) the hydraulic efficiency.203 l/s Illustrative Example 2. The impeller vanes are backward curved with an exit angle of 458. respectively. H2 ¼ 6. runs at 800 rpm.98 m Therefore. The radial velocity at the impeller exit is 2. its mechanical and volumetric effectiveness being 0. C 2:5 Therefore.98 m.e. using the Eq. In order to find discharge through the pump at 390 rpm. Inc. Solution: 1.5 m/s.5 m/s. All Rights Reserved . The power required to drive the pump is 8 hp.Hydraulic Pumps 77 As pffiffiffiffiffiffi H1 N1 ¼ pffiffiffiffiffiffi H2 N2 Then. Impeller tip speed U2 ¼ or U2 ¼ pD2 N 60 pð0:37Þð800Þ ¼ 15:5 m/s: 60 As the radial velocity at the impeller exit ¼ 2. The difference between the water levels at the overhead tank and the pump is 14 m. the head developed will be H¼ Cw2 U 2 ð13Þð15:5Þ ¼ 20:54 m ¼ 9:81 g Copyright 2003 by Marcel Dekker. and delivers 30 l/s of water. C w2 ¼ U 2 2 tanr2 2 ¼ 15:5 2 tan 458 ¼ 13 m/s.96 and 0. pffiffiffiffiffi pffiffiffiffiffiffi 16 ¼ H 2 590 390 Therefore. Q ¼ 1. b When there is no slip.11: The impeller of a centrifugal pump has outlet diameter of 0. Cw2 ¼ s £ 35. is given by Hi ¼ ð7:68Þð0:746Þ ¼ 18:84 m: ð9:81Þð0:031Þ 2.88. The blades are backward curved and they make an angle of 208 with the wheel tangent at the blade tip. Work input per kg of water flow W¼ Cw2 U 2 ð56Þð31:2Þ ¼ ¼ 1:75 kJ/kg: 1000 1000 2.12: The impeller of a centrifugal pump has outer diameter of 1. and delivers 0.06 m and speed is 56 m/s.88 £ 35. The impeller has Copyright 2003 by Marcel Dekker. Absolute velocity at impeller tip ffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qÀ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi À 2 Á Á C2 ¼ C r2 þ C 2 ¼ 7:52 þ 31:22 W2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 2 ¼ 56:25 þ 973:44 ¼ 32:09 m/s Design Example 2. Hi.4 ¼ 0. All Rights Reserved . Solution: 1.88.03 m3/s of water. If the radial velocity of the flow at the tip is 7.2 m/s. the power utilized by the pump will be: P ¼ ð0:96Þð8Þ ¼ 7:68 hp Theoretical flow rate ¼ Q 0:03 ¼ 0:031 m3 /s ¼ hv 0:97 Ideal head.78 Chapter 2 If there are no hydraulic internal losses. s ¼ 0. the velocity whirl at exit is. b2 ¼ 208 [ C w2 ¼ U 2 2 Cr2 7:5 ¼ 35:4 m/s ¼ 56 2 tan 208 tan b2 Using slip factor. Exit blade angle.4 ¼ 31.5 m/s and the slip factor is 0. Inc. Determine (1) the actual work input per kg of water flow and (2) the absolute velocity of fluid at the impeller. The hydraulic efficiency is hh ¼ H 14 ¼ 0:746 or ¼ H i 18:84 74:3%: Illustrative Example 2.13: A centrifugal pump impeller of 260 mm diameter runs at 1400 rpm. U2. and a slip factor of 0.20 m and 0. assuming constant radial velocity at 2.78 £ 30. and the number of impeller blades. (2) inlet vane angle. Copyright 2003 by Marcel Dekker.e. All Rights Reserved . no whirl at inlet) H¼ U 2 Cw2 ð19:07Þð15:88Þ ¼ 30:87 m ¼ 9:81 g Theoretical head with slip is H ¼ 0. Inc.08 m.14: A centrifugal pump impeller runs at 1500 rpm. Calculate (1) the angle absolute velocity of water at exit makes with the tangent. and has internal and external diameter of 0. Cw2 is given by C w2 ¼ U 2 2 C r2 1:84 ¼ 15:88 m/s ¼ 19:07 2 tan 308 tan 308 Using Euler’s equation. s ¼ 1 2 or 0:78 ¼ 1 2 or ¼12 n n n 0:78 ¼ 0:22 [ n¼ 0:63p ¼9 0:22 Number of blades required ¼ 9 Design Example 2.78 may be assumed.87 ¼ 24. the flow area is given by A ¼ impeller periphery £ blade depth ¼ p £ 0:26 £ 0:02 ¼ 0:0163 m2 Flow velocity is given by C r2 ¼ Q 0:03 ¼ ¼ 1:84 m/s A 0:0163 Impeller tip speed. and assuming Cw1 ¼ 0 (i. is U2 ¼ pD2 N pð0:26Þð1400Þ ¼ ¼ 19:07 m/s 60 60 Absolute whirl component.8 m/s and the vanes at the exit are set back at an angle of 308. Calculate the theoretical head developed by the impeller. and (3) the work done per kg of water. The blades are 20 mm in depth at the outlet. using Stanitz formula 0:63p 0:63p 0:63p Slip factor.Hydraulic Pumps 79 a backward curved facing blades inclined at 308 to the tangent at outlet. respectively. To find numbers of impeller blades.. Solution: Assuming the blades are of infinitesimal thickness.4 m. N ¼ 1500 rpm. Impeller speed at inlet pD1 N pð0:2Þð1500Þ ¼ ¼ 15:7 m/s U1 ¼ 60 60 tan b1 ¼ i.18.83.87. (2) the impeller hub diameter and tip diameter. D1 ¼ 0. and the overall pump efficiency ¼ 0. the flow velocity ¼ 4. D2 ¼ 0.e. Work done per kg of water C w2 U 2 ¼ 26:58 £ 31:43 ¼ 835:4 Nm: Design Example 2.15: An axial flow pump discharges water at the rate of 1.. is pD2 N pð0:4Þð1500Þ ¼ ¼ 31:43 m/s 60 60 Whirl component of absolute velocity at impeller exit is U2 ¼ C w2 ¼ U 2 2 tan a2 ¼ C r2 2:8 ¼ 26:58 m/s ¼ 31:43 2 tan 308 tan 308 2:8 ¼ 0:1053 26:58 i.8 m/s. Power delivered to the water P ¼ rgHQ/1000 kW ¼ ð9:81Þð1:30Þð10Þ ¼ 127:53 kW Power input to the pump P¼ 127:53 ¼ 153:65 kW: 0:83 2:8 ¼ 0:178 15:7 Copyright 2003 by Marcel Dekker. a2 ¼ 68.30 m3/s and runs at 550 rpm.e..5 m/s. hydraulic efficiency ¼ 0. and (3) the inlet and outlet blade angles for the rotor. 2. Cr2 ¼ 2. Impeller tip speed.2 m. Solution: 1. b2 ¼ 308. find (1) the power delivered to the water. Inc. and power input. U2.80 Chapter 2 Solution: 1. 3. Assume blade velocity ¼ 22 m/s.4 m. All Rights Reserved . The total head is 10 m. b1 ¼ 10. Rotor velocity at hub is given by D1 ð0:465Þð22Þ ¼ 13:39 m/s U2 ¼ D2 0:764 Since. and double suction centrifugal pump having the following data: Discharge Inner diameter Outer diameter Revolution/minute Head Width at inlet Width at outlet Absolute velocity angle at inlet Leakage losses Mechanical losses Contraction factor due to vane thickness Relative velocity angle measured from tangential direction Overall efficiency of the pump 72 l/s 90 mm 280 mm 1650 25 m 20 mm/side 18 mm/side 908 2 l/s 1:41 kW 0:85 358 0:56 Copyright 2003 by Marcel Dekker. Inc. 3.465 m. All Rights Reserved .16: A single stage. we have: rotor inlet angle at tip U1 ¼ Q 1:3 ¼ 0:216 m ¼ 0:7642 2 ðp/4Þ £ C a ðp/4Þð4:5Þ a1t ¼ tan 21 ðCa /U 1 Þ ¼ tan 21 ð4:5/13:39Þ ¼ 18:588 Rotor outlet angle a2t ¼ tan 21 ðCa /U 2 Þ ¼ tan 21 ð4:5/22Þ ¼ 11:568: Design Example 2. D1 ¼ 0.e.. radial flow. the axial velocity is constant. Rotor tip diameter is given by 60U 2 ð60Þð22Þ ¼ ¼ 0:764 m pN pð550Þ Rotor hub diameter D2 ¼ D2 ¼ D2 2 1 2 i.Hydraulic Pumps 81 2. 0067 m2. So. the velocity of flow at inlet C r1 ¼ Q 37 £ 1023 ¼ ¼ 7:708 m/s Area of flow 0:0048 From inlet velocity triangle tan b1 ¼ Cr1 7:708 ¼ 0:9907 ¼ 7:78 U1 b1 ¼ 44:738 2.82 Chapter 2 Determine (1) inlet vane angle.85) ¼ 0. (3) the absolute velocity of water leaving impeller. the velocity of flow at outlet C r2 ¼ Q 37 £ 1023 ¼ ¼ 5:522 m/s Area of flow 0:0067 The impeller speed at outlet U2 ¼ pD2 N pð0:28Þð1650Þ ¼ ¼ 24:2 m/s 60 60 Copyright 2003 by Marcel Dekker. All Rights Reserved . A2 ¼ (P)(0.009) (0. (4) manometric efficiency. Inc. Area of flow at outlet A2 ¼ P £ D2 £ b2 £ contraction factor Where b2 ¼ 18/2 ¼ 9 mm for one side. Impeller speed at inlet U1 ¼ pD1 N pð0:09Þð1650Þ ¼ ¼ 7:78 m/s 60 60 Flow area at inlet ¼ PD1 b1 £ contraction factor ¼ ðPÞð0:09Þð0:02Þð0:85Þ ¼ 0:0048 m2 Therefore.28)(0. Therefore. (2) the angle at which the water leaves the wheel. Solution: Total quantity of water to be handled by the pump Qt ¼ Qdel þ Qleak ¼ 72 þ 2 ¼ 74 Total quantity of water per side ¼ 74/2 ¼ 37 l/s 1. and (5) the volumetric and mechanical efficiencies. On test. tan a2 ¼ Cr2 5:522 ¼ 0:325 ¼ Cw2 16:99 a2 ¼ 188: 3. Copyright 2003 by Marcel Dekker. Q. compute the required diameter to be reduced.Hydraulic Pumps 83 Now using velocity triangle at outlet tan b2 ¼ C r2 5:522 ¼ U 2 2 Cw2 24:2 2 C w2 Cw2 ¼ 16:99 m/s Further.5 m for the designed discharge. If it is required to reduce the original diameter 32 cm without reducing the speed of the impeller. The volumetric efficiency Q2 72 ¼ 0:973 hv ¼ ¼ QTotal 74 Water power ¼ rgQH ¼ 1000 £ 9:81 £ 72 £ 25/1000 ¼ 17:66 kW Shaft power ¼ Water power 17:66 ¼ 31:54 kW ¼ 0:56 ho Mechanical efficiency is hm ¼ Ps 2 Ploss 31:54 2 1:41 ¼ 0:955 ¼ Ps 31:54 or 95:5%: Illustrative Example 2.17: A single stage centrifugal pump is designed to give a discharge of Q when working against a manometric head of 20 m. The manometric efficiency hmano ¼ ðgÞðH mano Þ 9:81 £ 25 ¼ 0:596: ¼ ðU 2 ÞðC W2 Þ 24:2 £ 16:99 5. it was found that head actually generated was 21. The absolute velocity of water leaving the impeller C2 ¼ Cw2 16:99 ¼ 17:8 m/s: ¼ Cos a2 Cos 188 4. Inc. All Rights Reserved . assuming no slip. find (1) the running speed in rpm and (2) the power required to run pump. If the overall efficiency is 76% and specific speed per stage about 38. D ¼ 32 cm So. 0 1/2 H 20 2 D ¼D ¼ 32 ¼ 30:86 cm H 21:5 0 Design Example 2.84 Chapter 2 Solution: Head generated by the pump H¼ or H / D2 H ¼ H0 2 D D0 U 2 ðpDN/60Þ2 ¼ 2g 2g H ¼ 21:5 m.18: A two stage centrifugal pump is designed to discharge 55 l/s at a head of 70 m. If the actual manometric head developed is 65% of the theoretical head.14 times the impeller tip speed at exit. Inc. and radial velocity at exit 0. Q ¼ 55 £ 1023 m3/s Power required to drive pump ¼ rgQH 1000 £ 9:81 £ 55 £ 1023 £ 70 ¼ 0:76 0:76 £ 1000 ¼ 49:7 kW Copyright 2003 by Marcel Dekker. find the diameter of impeller. H 0 ¼ 20 m. All Rights Reserved . Solution: 1. The specific speed is pffiffiffiffi N Q N s ¼ 3/4 H N¼ N s H 3/4 38ð70/2Þ3/4 546:81 pffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ¼ 2327 rpm: 23 0:235 Q 55 £ 10 2. the outlet angle of the blades 288. If the impeller in all the stages is identical and specific speed is 14. Each pump is required to produce a head of half the total and run at 1445 rpm. All Rights Reserved . and total head required is 845 m. 0:65 ¼ U 2 CW2 g U 2 CW2 ¼ CW2 ¼ 35 £ 9:81 0:65 ðBÞ 528:23 U2 Substituting for CW2 in Eq.0352 m3/s. (A) and solving U2 ¼ 3:798 U 2 2 528:23 U2 U 2 ¼ 26:78 m/s: Also. U2 ¼ pD2 N 60 21:97 cm: or 26:78 ¼ p£D2 £2327 60 D2 ¼ 0:2197 m or Design Example 2. From velocity triangle at outlet C r2 U 2 2 CW2 0:14U 2 or tan 288 ¼ U 2 2 C W2 tan b2 ¼ U2 0:5317 ¼ 3:798 ¼ 0:14 U 2 2 C W2 ðAÞ As the flow at entrance is radial and a1 ¼ 908.65 H. (2) The required impeller diameters assuming the speed ratio based on the outer Copyright 2003 by Marcel Dekker.Hydraulic Pumps 85 Hmano ¼ 0. Here b2 ¼ 288 and C r2 ¼ 0:14U 2 . determine (1) head developed per stage and the required number of stages in each pump. Inc.19: Two multistage centrifugal pumps are used in series to handle water flow rate of 0. the fundamental equation of pump would be H mano U 2 C W2 ¼ hmano g Where hmano manometric efficiency of pump which is 65%. 35 Therefore. Inc. Substituting. D2 ¼ p £ 60 £ 30:6 £ 1445. if the overall efficiency of each pump is 0.75. But U2 ¼ or D2 ¼ U 2 £ 60 30:6 £ 60 ¼ ¼ 0:4043 m or p £ 1445 p £ 1445 40:43 cm: pD2 N 60 Design Example 2. H 3/4 H ¼ 28 m Copyright 2003 by Marcel Dekker. H ¼ 105 N ¼ 900.96 and the shaft power input.86 Chapter 2 tip diameter to be 0. Determine the number of pumps required and how they should be connected? Solution: Specific speed for a single impeller is given by pffiffiffiffi N Q N s ¼ 3/4 H Given. N s ¼ 700. and Q ¼ 5500 ¼ 91:67 l/s 60 pffiffiffiffiffiffiffiffiffiffiffi 900 91:67 700 ¼ . All Rights Reserved .20: A centrifugal pump is required to be made to lift water through 105 m heights from a well. Number of identical pumps having their designed speed 900 rpm and specific speed 700 rpm with a rated discharge of 5500 l/min are available. Solution: Head developed in each stage is pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffi N Q 1445 0:0352 3/4 H ¼ ¼ Ns 14 H ¼ 51:93 m Total head required ¼ 845 m (of water) 845 Number of stages needed ¼ 51:93 ¼ 16 Number of stages in each pump ¼ 8 Impeller speed at tip is U 2 ¼ 0:96ð2gHÞ0:5 ¼ 0:96½2 £ 9:81 £ 51:930:5 ¼ 30:6 m/s Impeller diameter at tip. 45 m.5 m/s. H ¼ 5.5 m. Cr ¼ 2.45 m. pD1 N p £ 0:45 £ 120 ¼ ¼ 2:82 m/s 60 60 Cr1 2:5 tan a1 ¼ 0:8865 ¼ U 1 2:82 U1 ¼ i. Also. a ¼ 41:568: Copyright 2003 by Marcel Dekker. using velocity triangle at inlet. D2 ¼ 0. pffiffiffiffi N Q N s ¼ 3/4 H or pffiffiffiffiffiffiffiffiffiffi N 1193 1150 ¼ ð5:5Þ3/4 N¼ ð5:5Þ3/4 £ 1150 pffiffiffiffiffiffiffiffiffiffi ¼ 120 rpm 1193 In order to find vane angle at entry. The outer and inner diameters of the impeller are 0.Hydraulic Pumps 87 Hence number of stages required Total head to be developed ¼ Head per stage 105 ¼ 4 stages in series ¼ 28 Design Example 2.90 m and 0..5 m/s. D1 ¼ 0.21: The specific speed of an axial flow pump impeller is 1150 and velocity of flow is 2. All Rights Reserved . Ns ¼ 1150. Inc. respectively. Solution: Given. Q ¼ area of flow £ velocity of flow ¼ p ð0:92 2 0:452 Þ £ 2:5 ¼ 1:193 m3 /s 4 ¼ 1193 l/s Also.5 m. As discharge.9 m. calculate vane angle at the entry of the pump.e. Calculate the suitable speed of the pump to give a head of 5. The area of flow is constant from inlet to outlet and is 0.18 Nm) A centrifugal pump discharges 50 liters/second of water against a total head of 40 m. Calculate the manometric efficiency Copyright 2003 by Marcel Dekker. Show that the pressure rise in an impeller of a centrifugal pump is given by C2 þU 2 2C 2 cosec2 b2 r1 r2 2 2g 2. If the outlet flow area is 480 cm2. (11. A geometrically similar pump of 30 cm diameter is running at 2900 rpm. respectively.15 and 0.8 2. (148) A centrifugal pump having vane angles at inlet and outlet are 258 and 308.003C2 . find 1 the outlet blade angle. and loss of head in the pump can be taken as 0. A centrifugal pump discharges 0. and b2 ¼ blade angle at outlet).30 m. and discharging water 0. Cr2 ¼ velocity of flow at outlet.4 2. U2 ¼ blade velocity at outlet. Find the horsepower of the pump.12 m3/s. if the overall efficiency is 62%.9 Derive an expression for static head developed by a centrifugal pump having radial flow at inlet. (42 hp) A centrifugal pump delivers 26 l/s against a total head of 16 m at 1450 rpm. All Rights Reserved .5 m. If internal and external diameters of impeller are 0. 11. respectively. calculate the work done per kg of water.3 2. (197. The impeller has outer and inner diameter of 35 and 15 cm.15 m3/s of water against a head of 15 m. Assume velocity of flow constant.52 m. 2. Find the speed of the prototype.7 (where Cr1 ¼ velocity of flow at inlet. A model of this pump built to one-fourth its size is found to generate a head of 7 m when running at its best speed of 450 rpm and requires 13.5 2. (190 rpm) Derive the expression for power required for a pump when it discharges a liquid of specific weight w at the rate of Q against a head of H. Inc. The outlet vanes are set back at an angle 408.88 Chapter 2 PROBLEMS 2.23 l/s) A centrifugal pump is built to work against a head of 20 m. respectively.2 2. Calculate head and discharge required assuming equal efficiencies between the two pumps. Assuming that friction and other losses are neglected. The impeller diameter is 0.06 m2.1 A centrifugal pump of 25 cm impeller diameter running at 1450 rpm.5 hp to run it.6 2. develops a head of 15 m. Find (1) flow hub to tip and (2) efficiency is 84%. (76. If velocity flow is constant at 2. the inlet blade angle is 128 and the outlet blade angle is 158.35 times the peripheral velocity.4 m/s. 2. At the mean blade radius. (3) the shaft h. 0. Take slip factor ¼ 1. 705 l/s. 230 hp.4%) An axial flow pump running at 960 rpm. and estimate from them (1) the head the pump will generate.83) 2. The vanes are curved back at an angle of 308 to the tangent at outlet. (0. Assume a manometric or hydraulic efficiency of 88% and a gross or overall efficiency of 81%.11 Through flow velocity is 0.59 m. All Rights Reserved . input required to drive the pump. is required to deliver 1 m3/s at 7 m head while Its outer diameter is 50 and hub diameter is velocity. deduce an expression for the speed of a geometrically similar pump of such a size that when working against unit head. the rotor has an outer diameter of 75 cm and an inner diameter of 40 cm. it rotates at 500 rpm. 188) 2. (57.Hydraulic Pumps 89 and vane angle at inlet if the speed of the pump is 960 rpm. Sketch the corresponding velocity diagrams at inlet and outlet.10 A centrifugal pump of 35 cm diameter running at 1000 rpm develops a head of 18 m. (2) the discharge or rate of flow in l/s. 45) If an axial flow pump delivers a discharge Q against a head H when running at a speed N.13 In an axial flow pump. find the manometric efficiency of the pump.75 kW) 2. (19. 81.3%. 25 cm. and (4) the specific speed of the pump.8 m. which is assumed to be constant from power required to drive the pump if overall (6. Inc. Show that this value is proportional to the specific speed of the pump. it will transmit unit power to the water flowing through it.12 An axial flow pump has the following data: Rotational speed Discharge of water Head Hub to runner diameter ratio 750 rpm 1:75 m3 /s 7:5 m 0:45 2. Find the diameter and minimum speed ratio.791 m/s.p.14 Copyright 2003 by Marcel Dekker. 90 Chapter 2 NOTATION b Cw2 E H Hi Nsuc m n Ps Q r U V a b hc hR hi hm ho hv r s v width of the diffuser passage tangential components of absolute velocity corresponding to the angle b2 Euler head total head developed by the pump total head across the impeller Suction specific speed mass flow rate number of vanes shaft power input flow rate radius impeller speed relative velocity absolute velocity angle relative velocity angle casing efficiency hydraulic efficiency impeller efficiency mechanical efficiency overall efficiency volumetric efficiency density of liquid slip factor angular velocity SUFFIXES 1 2 3 a r w inlet to impeller outlet from the impeller outlet from the diffuser axial radial whirl Copyright 2003 by Marcel Dekker. Inc. All Rights Reserved . In the sections that follow. Water or hydraulic turbines convert kinetic and potential energies of the water into mechanical power.1) turbine in 1880. It is a pure impulse turbine in which a jet of fluid delivered is by the nozzle at a high velocity on the buckets. and the Kaplan or propeller type. which is characterized by a radial flow impeller. It operates under very high heads (up to 1800 m. The predominant type of impulse machine is the Pelton wheel. which is generally mounted on a horizontal shaft.2 PELTON WHEEL An American Engineer Lester A. All Rights Reserved .000 m. The reaction turbine is further subdivided into the Francis type. reaction. These buckets are fixed on the periphery of a circular wheel (also known as runner). and method of operation. water is used as the source of energy. 3. each type of hydraulic turbine will be studied separately in terms of the velocity triangles. The primary feature of the impulse Copyright 2003 by Marcel Dekker.1 INTRODUCTION In a hydraulic turbine. 3. efficiencies.) and requires comparatively less quantity of water. which is suitable for a range of heads of about 150 –2.3 Hydraulic Turbines 3. which is an axial-flow machine. Pelton discovered this (Fig. The main types of turbines are (1) impulse and (2) reaction turbines. Inc. The impact of water on the buckets causes the runner to rotate and thus develops mechanical energy. horizontal shaft Pelton turbine. The casing of a Pelton wheel does not perform any hydraulic function. The absolute Copyright 2003 by Marcel Dekker.2. After doing work on the buckets water is discharged in the tailrace. But it is necessary to safeguard the runner against accident and also to prevent the splashing water and lead the water to the tailrace. The buckets deflect the jet through an angle of about 160 and 1658 in the same plane as the jet. Inc. then the relative fluid velocity. is given by V 1 ¼ jet velocity 2 bucket speed ¼ C1 2 U 1 The angle turned through by the jet in the horizontal plane during its passage over the bucket surface is a and the relative velocity at exit is V2. as shown in Fig. 3.3 VELOCITY TRIANGLES The velocity diagrams for the Pelton wheel are shown in Fig. and the whole energy transfer from nozzle outlet to tailrace takes place at constant pressure. The buckets are so shaped that water enters tangentially in the middle and discharges backward and flows again tangentially in both the directions to avoid thrust on the wheel.92 Chapter 3 Figure 3. V1. Since the angle of entry of the jet is nearly zero. 3.1 Single-jet. the inlet velocity triangle is a straight line. turbine with respect to fluid mechanics is the power production as the jet is deflected by the moving vane(s). 3. All Rights Reserved . If the bucket is brought to rest.2. velocity. Therefore dE ¼ ð1 2 cos aÞðC 1 2 2UÞ/g ¼ 0 dU Then C1 ¼ 2U or U ¼ C1 /2 ð3:2Þ Copyright 2003 by Marcel Dekker. and equating it to zero. Eq.Hydraulic Turbines 93 Figure 3. V1 ¼ V2.2 Velocity triangles for a Pelton wheel. that is. (1. V2. (3. W ¼ U fðU þ V 1 Þ þ ½V 1 cosð180 2 aÞ 2 U g Neglecting loss due to friction across the bucket surface.78) W ¼ U 1 CW1 2 U 2 C W2 Since in this case CW2 is in the negative x direction.1) can be optimized by differentiating with respect to U. then W ¼ UðV 1 2 V 1 cos aÞ Therefore ð3:1Þ E ¼ UðC 1 2 UÞð1 2 cos aÞ/g the units of E being Watts per Newton per second weight of flow. Inc. Now using Euler’s turbine Eq. at exit can be obtained by adding bucket speed vector U2 and relative velocity. C2. at exit. All Rights Reserved . The frictional head loss. Inc. All Rights Reserved . Losses also occur in the nozzle and are expressed by the velocity coefficient.1) we get Emax ¼ C2 ð1 2 cos aÞ/4g 1 In practice.94 Chapter 3 Substituting Eq. The overall efficiency (ho) for large Pelton turbine is about 85– 90%. the maximum hydraulic efficiency is 100%.2) into Eq. Following efficiency is usually used for Pelton wheel. (3. Cv.4 PELTON WHEEL (LOSSES AND EFFICIENCIES) Head losses occur in the pipelines conveying the water to the nozzle due to friction and bend. 3. In practice. hf. The jet efficiency (hj) takes care of losses in the nozzle and the mechanical efficiency (hm) is meant for the bearing friction and windage losses. surface friction is always present and V1 – V2. Pipeline transmission efficiency ¼ Energy at end of the pipe Energy available at reservoir Figure 3. deflection angle is in the order of 160 – 1658.3 shows the total headline. (3. then Eq. where the water supply is from a reservoir at a head H1 above the nozzle.1) becomes E ¼ UðC 1 2 UÞð1 2 k cos aÞ/g where k ¼ V 2 V1 Introducing hydraulic efficiency as ð3:3Þ hh ¼ Energy Transferred Energy Available in jet E ðC 2 /2gÞ 1 ð3:4Þ i:e: hh ¼ if a ¼ 1808. Then the transmission efficiency is htrans ¼ ðH 1 2 hf Þ/H 1 ¼ H/H 1 The nozzle efficiency or jet efficiency is Energy at nozzle outlet ¼ C2 /2gH 1 Energy at nozzle inlet ð3:5Þ hj ¼ ð3:6Þ Copyright 2003 by Marcel Dekker. is the loss as the water flows through the pressure tunnel and penstock up to entry to the nozzle. (3. Inc. Nozzle velocity coefficient Cv ¼ pffiffiffiffiffiffiffiffiffi Actual jet velocity ¼ C 1 = 2gH Theoretical jet velocity Therefore the nozzle efficiency becomes hj ¼ C 2 =2gH ¼ C2 ð3:7Þ v 1 The characteristics of an impulse turbine are shown in Fig. speed at various nozzle settings.Hydraulic Turbines 95 Figure 3.4 shows the curves for constant head and indicates that the peak efficiency occurs at about the same speed ratio for any gate opening and that Figure 3. All Rights Reserved . Figure 3.3 Schematic layout of hydro plant. 3.4 Efficiency vs. Copyright 2003 by Marcel Dekker.4. 96 Chapter 3 Figure 3. Illustrative Example 3.5 shows the curves for power vs. Due to losses. H ¼ 220 m U1 ¼ U2 ¼ 14 m/s b2 ¼ 180 2 1608 ¼ 208 Refer to Fig. Inc. find out the power developed.1: A generator is to be driven by a Pelton wheel with a head of 220 m and discharge rate of 145 L/s. The mean peripheral velocity of wheel is 14 m/s. This happens as the nozzle velocity remaining constant in magnitude and direction as the flow rate changes. Solution: Dischargerate. Q ¼ 145 L/s Head.6 Using Euler’s equation. All Rights Reserved .5 Power vs. gives an optimum value of U/C1 at a fixed speed. speed. speed of various nozzle setting. such as windage. 3. and friction cause the small variation. Fig. If the outlet tip angle of the bucket is 1608. 3. work done per weight mass of water per sec. mechanical. Fixed speed condition is important because generators are usually run at constant speed. ¼ ðC w1 U 1 2 C w2 U 2 Þ But for Pelton wheel Cw2 is negative Copyright 2003 by Marcel Dekker. the peak values of efficiency do not vary much. Inc. work done per unit mass of water per sec. C1 ¼ 2gH ¼ 2 £ 9:81 £ 220 ¼ 65:7 m/s Relative velocity at inlet is V 1 ¼ C 1 2 U 1 ¼ 65:7 2 14 ¼ 51:7 m/s From outlet velocity triangle V 1 ¼ V 2 ¼ 51:7 m/s(neglecting friction) and cos b2 ¼ U 2 þCw2 or V2 cosð20Þ ¼ 14 þ C w2 51:7 Therefore C w2 ¼ 34:58 m/s Hence. ¼ ð65:7Þð14Þ þ ð34:58Þð14Þ ¼ 1403:92 Nm Power developed ¼ ð1403:92Þð145Þ ¼ 203:57 kW 1000 Copyright 2003 by Marcel Dekker.Hydraulic Turbines 97 Figure 3. All Rights Reserved .6 Inlet and outlet velocity triangles. Therefore Work done / s ¼ ðC w1 U 1 þ C w2 U 2 Þ Nm / s From inlet velocity triangle C2 C w1 ¼ C1 and 1 ¼ H 2g pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Hence. Dimensionless power specific speed of the wheel Solution: Power developed Overall efficiency ho ¼ Power available [ P ¼ rgQH ho J/s ¼ rgQH ho kW 1000 ¼ 9:81ð0:035Þð92Þð0:82Þ ¼ 25:9 kW Velocity coefficient C1 Cv ¼ pffiffiffiffiffiffiffiffiffi 2gH pffiffiffiffiffiffiffiffiffi or C1 ¼ Cv 2gH ¼ 0:95½ð2Þð9:81Þð92Þ1/2 ¼ 40:36 m/s 1.035 m3/s of water under a head of 92 m. d. Speed of the wheel 2.98 Chapter 3 Design Example 3. If D is the wheel diameter.45.2: A Pelton wheel is supplied with 0. Inc. then U¼ vD 2 or D¼ 2U ð2Þð18:16Þð60Þ ¼ 0:478 m ¼ v 725ð2pÞ Jet area A¼ Q 0:035 ¼ 0:867 £ 1023 m2 ¼ C 1 40:36 and Jet diameter.95. is given by 1/2 1/2 4A ð4Þð0:867 £ 1023 Þ ¼ ¼ 0:033 m d¼ p p Diameter ratio D 0:478 ¼ ¼ 14:48 d 0:033 Copyright 2003 by Marcel Dekker. Speed of the wheel is given by U ¼ 0:45ð40:36Þ ¼ 18:16 m/s 2. Wheel to jet diameter ratio 3. All Rights Reserved . The wheel rotates at 725 rpm and the velocity coefficient of the nozzle is 0. The efficiency of the wheel is 82% and the ratio of bucket speed to jet speed is 0. Determine the following: 1. Dimensionless specific speed is given by Eq. All Rights Reserved . assuming C v ¼ 0:98 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:98 ð2Þð9:81Þð45Þ ¼ 29:12 m/s Figure 3. If the jet is deflected by the buckets at an angle of 1608.3.7 U1 ¼ U2 ¼ 14 m/s Q ¼ 820 L/s ¼ 0. The water is supplied at the rate of 820 L/s against a head of 45 m. Copyright 2003 by Marcel Dekker.Hydraulic Turbines 99 3.7 Velocity triangle for Example 3.3: The speed of Pelton turbine is 14 m/s. Solution: Refer to Fig. Inc.10) N sp ¼ NP 1/2 r 1/2 ðgHÞ5/4 5/4 1/2 ¼ 725 £ ð25:9Þð1000Þ £ ð9:81Þ1£ ð92Þ 60 103 ¼ ð12:08Þð5:09Þð0:0002Þ ¼ 0:0123 rev ¼ ð0:0123Þð2pÞ rad ¼ 0:077 rad Illustrative Example 3. find the hP and the efficiency of the turbine. 3. (1.82 m3/s H ¼ 45 m b2 ¼ 180 2 1608 ¼ 208 Velocity of jet pffiffiffiffiffiffiffiffiffi C 1 ¼ Cv 2gH . 100 Chapter 3 Assuming b1 ¼ 1808 b2 ¼ 180 2 1608 ¼ 208 Cw1 ¼ C 1 ¼ 29:12 m/s V 1 ¼ C1 2 U 1 ¼ 29:12 2 14 ¼ 15:12 m/s From outlet velocity triangle. Find (1) discharge of the turbine. Inc.900 kW at 425 rpm under a head of 505 m. Power. Let Q be the discharge of the turbine Using the relation ho ¼ P 9:81QH Copyright 2003 by Marcel Dekker.46. P ¼ 12. H ¼ 505 m. N ¼ 425 rpm Efficiency. and ratio of bucket speed to jet speed ¼ 0. Assume Cv ¼ 0.98.4: A Pelton wheel develops 12. ho ¼ 84% 1. The efficiency of the machine is 84%. and (3) diameter of the nozzle. U 1 ¼ U 2 (neglecting losses on buckets) V 2 ¼ 15:12 m / s and U 2 ¼ 14 m/s Cw2 ¼ V 2 cosa2 2 U 2 ¼ 15:12 cos 208 2 14 ¼ 0:208 m/s Work done per weight mass of water per sec ¼ ðCw1 þ C w2 ÞU ¼ ð29:12 þ 0:208Þ £ ð14Þ ¼ 410:6 Nm / s 3 [ Power developed ¼ ð410:6Þð0:82 £ 10 Þ ¼ 336:7 kW 1000 ¼ 451 hP Power developed Efficiencyh1 ¼ Available Power ð1000Þð336:7Þ ¼ ¼ 0:930 or 93:0% ð1000Þð9:81Þð0:82Þð45Þ Illustrative Example 3. (2) diameter of the wheel. All Rights Reserved .900 kW Speed. Solution: Head. h ¼ 95% Therefore. power generated by the runner ¼ 12. Find the power developed by each runner if the generator is 95%.5: A double Overhung Pelton wheel unit is to operate at 12.000 kW Efficiency. Inc. 900 2:60 ¼ ð9:81Þð505ÞQ Q pDN . All Rights Reserved .000 kW generator.Hydraulic Turbines 101 or 0:84 ¼ or Q ¼ 3:1 m3 /s 2. Let d be the diameter of the nozzle The discharge through the nozzle must be equal to the discharge of the turbine. hence wheel diameter is 60 60U ð60Þð44:87Þ ¼ ¼ 2:016 m pN ðpÞð425Þ 3. Solution: Output power ¼ 12. 000 ¼ 12. Therefore Q ¼ p £ d2 £ C 4 3:1 ¼ ðpÞðd 2 Þð97:55Þ ¼ 76:65 d 2 4 qffiffiffiffiffiffiffiffiffiffiffi 3:1 [ d ¼ 76:65 ¼ 0:20 m Illustrative Example 3. Velocity of jet pffiffiffiffiffiffiffiffiffi C ¼ Cv 2gH ðassume Cv ¼ 0:98Þ or pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C ¼ 0:98 ð2Þð9:81Þð505Þ ¼ 97:55 m/s Tangential velocity of the wheel is given by U ¼ 0:46C ¼ ð0:46Þð97:55Þ ¼ 44:87 m/s and U¼ D¼ 12. 632 kW 0:95 Copyright 2003 by Marcel Dekker. The loss of head due to pipe friction between the reservoir and nozzle is 46 m. The bucket circle diameter of the wheel is 890 mm and there are two jets.102 Chapter 3 Since there are two runners.8. 3. a Pelton wheel produces 1260 kW under a head of 610 m. W/m ¼ 0.46. power developed by each runner ¼ 12. Cv ¼ 0. speed of rotation of the wheel. Inc.6: At the power station. Solution: Refer to Fig. while relative velocity of the water is reduced by 10% due to bucket friction. Assume nozzle velocity coefficient. 632 ¼ 6316 kW 2 Design Example 3. Find the theoretical hydraulic efficiency. and diameter of the nozzle if the actual hydraulic efficiency is 0. The buckets of the Pelton wheel deflect the jet through an angle of 1658.9 times that calculated above. The bucket/jet speed ratio is 0.46C1(C1 2 0.98. All Rights Reserved .9 cos 1658) Substitute the value of C1 W/m ¼ 5180:95 Theoretical hydraulic efficiency ¼ Power output Energy available in the jet 5180:95 ¼ ¼ 98% 0:5 £ 1032 Copyright 2003 by Marcel Dekker.46C1)(1 2 0. Power output P ¼ Hydraulic efficiency hh ¼ Energy available in the jet 0:5mC 2 1 At entry to nozzle H ¼ 610 2 46 ¼ 564 m Using nozzle velocity coefficient pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi C1 ¼ Cv 2gH ¼ 0:98 ð2Þð9:81Þð564Þ ¼ 103:1 m/s Now W ¼U C 2U C 1 w1 2 w2 m ¼ U fðU þ V 1 Þ 2 ½U 2 V 2 cosð1808 2 aÞg ¼ U ½ðC 1 2 U Þð1 2 k cos aÞ where V 2 ¼ kV 1 Therefore. 5 kg/s For nozzle diameter. using continuity equation. The main bucket speed is 12 m/s and the nozzle discharge is 1. Inc. Copyright 2003 by Marcel Dekker.98. d ¼ ðpÞð103 £ 103 Þ rC1 pd 2 4 Illustrative Example 3.6. If the bucket has an angle of 158 at the outlet and Cv ¼ 0.0 m3/s. All Rights Reserved .8 Velocity triangle for Example 3.Hydraulic Turbines 103 Figure 3. Actual hydraulic efficiency ¼ ð0:9Þð0:98Þ ¼ 0:882 Wheel bucket speed ¼ ð0:46Þð103Þ ¼ 47:38 m/s Wheel rotational speed ¼ N ¼ Actual hydraulic efficiency ¼ Therefore. m ¼ 134. m ¼ ð47:38Þð60Þ ¼ 1016 rpm ð0:445Þð2pÞ Actual power ð1260 £ 103 Þ ¼ energy in the jet 0:5 mC 2 1 ð1260 £ 103 Þ ¼ 269 kg/s ð0:882Þð0:5Þð1032 Þ For one nozzle. find the power of Pelton wheel and hydraulic efficiency.7: A Pelton wheel has a head of 90 m and head lost due to friction in the penstock is 30 m. m ¼ rC 1 A ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð134:5Þð4Þ ¼ 0:041 m ¼ 41 mm Hence. Inc. V1 ¼ C1 2 U1 ¼ 33. All Rights Reserved .7.104 Chapter 3 Figure 3.9 Velocity triangle for Example 3. 3.62 2 12 ¼ 21.9) Head ¼ 90 m Head lost due to friction ¼ 30 m Head available at the nozzle ¼ 90 2 30 ¼ 60 m Q ¼ 1 m3/s From inlet diagram pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi C1 ¼ Cv 2gH ¼ 0:98 £ ð2Þð9:81Þð60Þ ¼ 33:62 m/s Therefore. Solution: (Fig.62 m/s From outlet velocity triangle V 2 ¼ V 1 ¼ 21:16 m/s (neglecting losses) U 2 ¼ U 1 ¼ 12 m/s Cw2 ¼ V 2 cos a 2 U 2 ¼ 21:62 cos 158 2 12 ¼ 8:88 m/s Copyright 2003 by Marcel Dekker. its deflection inside the bucket is 1658 and its relative velocity is reduced by 12% due to friction. Water power is given by P ¼ rgQH ¼ ð9:81Þð3:096Þð515Þ ¼ 15641:5 kW 2. Q is given by p Q ¼ Area of jet £ Velocity ¼ £ ð0:2Þ2 ð98:5Þ ¼ 3:096 m3 /s 4 1.8: A single jet Pelton wheel turbine runs at 305 rpm against a head of 515 m. (3) shaft power if the mechanical losses are 4% of power supplied. Find (1) the waterpower. Bucket velocity. U1. C2 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 2 þ Cr 2 ¼ ð8:88Þ2 þ ð5:6Þ2 ¼ 10:5 m/s w2 2 C2 2 C2 ð33:62Þ2 2 ð10:5Þ2 1 2 ¼ ¼ 510 kJ/kg 2 2 [ Work done ¼ Note Work done can also be found by using Euler’s equation (Cw1U1 þ Cw2U2) Power ¼ 510 kW Hydraulic efficiency work done ð510Þð2Þ ¼ ¼ 90:24% kinetic energy ð33:62Þ2 hh ¼ Design Example 3.Hydraulic Turbines 105 and Cr 2 ¼ V 2 sin a ¼ 21:62 sin 158 ¼ 5:6 m/s Therefore. C 1 ¼ Cv 2gH ¼ 0:98 ð2Þð9:81Þð515Þ ¼ 98:5 m/s Discharge. Solution: (Fig. V1. Assume necessary data. (2) resultant force on the bucket. is given by pffiffiffiffiffiffiffiffiffi U 1 ¼ C v 2gH pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:46 ð2Þð9:81Þð515Þ ¼ 46 m/s ðassuming C v ¼ 0:46Þ Relative velocity. Inc. The jet diameter is 200 mm. 3. All Rights Reserved .10) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi Velocity of jet. at inlet is given by V 1 ¼ C1 2 U 1 ¼ 98:5 2 46 ¼ 52:5 m/s Copyright 2003 by Marcel Dekker. and (4) overall efficiency. Inc.8 ¼ 13279.106 Chapter 3 Figure 3.8. shaft power produced ¼ 0.96 £ 13832. and V 2 ¼ 0:88 £ 52:5 ¼ 46:2 m/s From the velocity diagram Cw2 ¼ U 2 2 V 2 cos 15 ¼ 46 2 46:2 £ 0:966 ¼ 1:37 m/s Therefore force on the bucket ¼ rQðCw1 2 C w2 Þ ¼ 1000 £ 3:096ð98:5 2 1:37Þ ¼ 300714 N 3.10 Velocity triangles for Example 3. Power produced by the Pelton wheel ¼ ð300714Þð46Þ ¼ 13832:8 kW 1000 Taking mechanical loss ¼ 4% Therefore. All Rights Reserved .5 kW 4. Overall efficiency ho ¼ 13279:5 ¼ 0:849 or 84:9% 15641:5 Copyright 2003 by Marcel Dekker. Inc. Therefore. Therefore. which are fixed around the periphery of runners. most of the initial energy of water is given to the runner. As the water flows over the curved blades. The water then passes immediately into the rotor where it moves radially through the rotor vanes and exits from the rotor blades at a smaller diameter. All Rights Reserved . Thus.13.5 REACTION TURBINE The radial flow or Francis turbine is a reaction machine. the total head is equal to pressure head plus velocity head. under a high station head. 3. the water is always at a pressure other than atmosphere. 3. Copyright 2003 by Marcel Dekker.Hydraulic Turbines 107 Figure 3. Fig. 3. water leaving the blade has a large relative velocity but small absolute velocity. has its pressure energy converted into kinetic energy in a nozzle. Thus. water from the reservoir enters the turbine casing through penstocks. part of the work done by the fluid on the rotor is due to reaction from the pressure drop. the water enters the runner or passes through the stationary vanes. In reaction turbine. The inlet and outlet velocity triangles for the runner are shown in Fig. the runner is enclosed in a casing and therefore. In a reaction turbine. and part is due to a change in kinetic energy. which carries water from the reservoir to the turbine casing. The penstock is a waterway. It helps in increasing the work done by the turbine by reducing pressure at the exit.12 shows an energy distribution through a hydraulic reaction turbine.11 Outlines of a Francis turbine. the water.11 shows a cross-section through a Francis turbine and Fig. The draft tube is a gradually increasing cross-sectional area passage. Hence. The pressure difference between entrance and exit points of the runner is known as reaction pressure. 3. In reaction turbines. which represents an impulse function. The essential difference between the reaction rotor and impulse rotor is that in the former. after which it turns through 908 into the draft tube. water leaves the runner at atmospheric pressure. the pressure head is transformed into velocity head. Copyright 2003 by Marcel Dekker.12 Reaction turbine installation.108 Chapter 3 Figure 3. Inc. All Rights Reserved . 13 turbine. (a) Francis turbine runner and (b) velocity triangles for inward flow reaction Copyright 2003 by Marcel Dekker.Hydraulic Turbines 109 Figure 3. Inc. All Rights Reserved . 3.6 TURBINE LOSSES Let Ps ¼ Shaft power output Pm ¼ Mechanical power loss Pr ¼ Runner power loss Pc ¼ Casing and draft tube loss Pl ¼ Leakage loss P ¼ Water power available Ph ¼ Pr þ Pc þ Pl ¼ Hydraulic power loss Runner power loss is due to friction. If hf is the head loss associated with a flow rate through the runner of Qr. U 2 . then Ps ¼ rgQr hf ðNm/sÞ ð3:8Þ Leakage power loss is due to leakage in flow rate. D2 . Inc.110 Chapter 3 Let C 1 ¼ Absolute velocity of water at inlet D1 ¼ Outer diameter of the runner N ¼ Revolution of the wheel per minute U 1 ¼ Tangential velocity of wheel at inlet V 1 ¼ Relative velocity at inlet C r1 ¼ radial velocity at inlet a1 ¼ Angle with absolute velocity to the direction of motion b1 ¼ Angle with relative velocity to the direction of motion H ¼ Total head of water under which turbine is working C 2 . Thus Q ¼ Qr þ q ð3:9Þ Copyright 2003 by Marcel Dekker. and flow separation. q. All Rights Reserved . V 2 . (1. C r2 ¼ Corresponding values at outlet Euler’s turbine equation Eq.78) and E is maximum when Cw2 (whirl velocity at outlet) is zero that is when the absolute and flow velocities are equal at the outlet. shock at impeller entry. past the runner and therefore not being handled by the runner. The part-load performance of the various types are compared in Fig. is given by hh ¼ or hh ¼ ð3:13Þ Eq. the leakage power loss becomes Pl ¼ rgH r q ðNm / sÞ ð3:10Þ Casing power loss. eddy. is due to friction.Hydraulic Turbines 111 If Hr is the head across the runner. hh.14 Figure 3.14 shows that machines of low specific speeds have a slightly higher efficiency. 3. is given by Shaft power output Fluid power available at inlet Ps rgQH Power available at runner Fluid power available at inlet ðPs þ Pm Þ rgQH ho ¼ or ho ¼ ð3:12Þ Hydraulic efficiency. Both these problems were solved by Paul Deriaz by designing a runner similar to Francis runner but with adjustable blades. Pc. ho. It has been experienced that the Francis turbine has unstable characteristics for gate openings between 30 to 60%. Therefore the maximum efficiency is hh ¼ U 1 Cw1 /gH ð3:14Þ 3. 3. If hc is the head loss in casing then Pc ¼ rgQhc ðNm / sÞ From total energy balance we have ð3:11Þ rgQH ¼ Pm þ rg ðhf Qr þ hc Q þ H r q þ Ps Þ Then overall efficiency.15 showing that the Kaplan and Pelton types are best adopted for a wide range of load but are followed fairly closely by Francis turbines of low specific speed. All Rights Reserved . causing pulsations in output and pressure surge in penstocks. (3. Inc. and flow separation losses in the casing and draft tube.7 TURBINE CHARACTERISTICS Part and overload characteristics of Francis turbines for specific speeds of 225 and 360 rpm are shown in Fig.13) is the theoretical energy transfer per unit weight of fluid. Copyright 2003 by Marcel Dekker. Figure 3. All Rights Reserved .15 Comparison of part-load efficiencies of various types of hydraulic turbine.14 Variation of efficiency with load for Francis turbines.112 Chapter 3 Figure 3. Copyright 2003 by Marcel Dekker. Inc. Hydraulic Turbines 113 3.8 AXIAL FLOW TURBINE In an axial flow reaction turbine, also known as Kaplan turbine, the flow of water is parallel to the shaft. A Kaplan turbine is used where a large quantity of water is available at low heads and hence the blades must be long and have large chords so that they are strong enough to transmit the very high torque that arises. Fig. 3.16 and 3.17 shows the outlines of the Kaplan turbine. The water from the scroll flows over the guide blades and then over the vanes. The inlet guide vanes are fixed and are situated at a plane higher than the runner blades such that fluid must turn through 908 to enter the runner in the axial direction. The function of the guide vanes is to impart whirl to the fluid so that the radial distribution of velocity is the same as in a free vortex. Fig. 3.18 shows the velocity triangles and are usually drawn at the mean radius, since conditions change from hub to tip. The flow velocity is axial at inlet and outlet, hence Cr1 ¼ Cr2 ¼ Ca C1 is the absolute velocity vector at anglea1 toU1, andV1 is the relative velocity at an angle b1. For maximum efficiency, the whirl component Cw2 ¼ 0, in which case the absolute velocity at exit is axial and then C2 ¼ Cr2 Using Euler’s equation E ¼ UðC w1 2 C w2 Þ/g and for zero whirl (Cw2 ¼ 0) at exit E ¼ UC w1 /g 3.9 CAVITATION In the design of hydraulic turbine, cavitation is an important factor. As the outlet velocity V2 increases, then p2 decreases and has its lowest value when the vapor pressure is reached. At this pressure, cavitation begins. The Thoma parameter s ¼ NPSH and H Fig. 3.19 give the permissible value of sc in terms of specific speed. The turbines of high specific speed have a high critical value of s, and must therefore be set lower than those of smaller specific speed (Ns). Illustrative Example 3.9: Consider an inward flow reaction turbine in which velocity of flow at inlet is 3.8 m/s. The 1 m diameter wheel rotates at 240 rpm and absolute velocity makes an angle of 168 with wheel tangent. Determine (1) velocity of whirl at inlet, (2) absolute velocity of water at inlet, (3) vane angle at inlet, and (4) relative velocity of water at entrance. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 114 Chapter 3 Figure 3.16 Kaplan turbine of water is available at low heads. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Hydraulic Turbines 115 Figure 3.17 Kaplan turbine runner. Solution: From Fig. 3.13b 1. From inlet velocity triangle (subscript 1) tana1 ¼ C r1 C r1 3:8 ¼ 13:3 m/s or C w1 ¼ ¼ C w1 tana1 tan168 2. Absolute velocity of water at inlet, C1, is sina1 ¼ 3. U1 ¼ C r1 C r1 3:8 ¼ 13:79 m/s or C 1 ¼ ¼ C1 sina1 sin168 ðpD1 ÞðNÞ ðpÞð1Þð240Þ ¼ ¼ 12:57 m/s 60 60 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 116 Chapter 3 Figure 3.18 Velocity triangles for an axial flow hydraulic turbine. and tan b1 ¼ C r1 3:8 3:8 ¼ ¼ ¼ 5:21 ðC w1 2 U 1 Þ ð13:3 2 12:57Þ 0:73 [ b1 ¼ 798 nearby 4. Relative velocity of water at entrance sin b1 ¼ C r1 C r1 3:8 ¼ 3:87m/s or V 1 ¼ ¼ V1 sin b1 sin 798 Illustrative Example 3.10: The runner of an axial flow turbine has mean diameter of 1.5 m, and works under the head of 35 m. The guide blades make an angle of 308 with direction of motion and outlet blade angle is 228. Assuming axial discharge, calculate the speed and hydraulic efficiency of the turbine. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Hydraulic Turbines 117 Figure 3.19 Cavitation limits for reaction turbines. Figure 3.20 Velocity triangles (a) inlet and (b) outlet. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 118 Chapter 3 Solution: Since this is an impulse turbine, assume coefficient of velocity ¼ 0.98 Therefore the absolute velocity at inlet is pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi C1 ¼ 0:98 2gH ¼ 0:98 ð2Þð9:81Þð35Þ ¼ 25:68 m/s The velocity of whirl at inlet Cw1 ¼ C1 cos a1 ¼ 25:68 cos 308 ¼ 22:24 m/s Since U1 ¼ U2 ¼ U Using outlet velocity triangle C2 ¼ U 2 tan b2 ¼ U tan b2 ¼ U tan 228 Hydraulic efficiency of turbine (neglecting losses) C w1 U 1 H 2 C 2 /2g 2 ¼ H gH hh ¼ 22:24U ðU tan 228Þ2 ¼H2 2g g or 22:24U ðU tan 22Þ2 þ ¼H g 2g or 22:24U þ 0:082U 2 2 9:81H ¼ 0 or 0:082U 2 þ 22:24U 2 9:81H ¼ 0 or U¼ 222:24 ^ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð22:24Þ2 þ ð4Þð0:082Þð9:81Þð35Þ ð2Þð0:082Þ As U is positive, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 494:62 U ¼ 222:24 þ 0:164 þ 112:62 ¼ 222:24 þ 24:64 ¼ 14:63 m/s 0:164 Now using relation U¼ pDN 60 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Hydraulic Turbines 119 or N¼ 60U ð60Þð14:63Þ ¼ ¼ 186 rpm ðpÞð1:5Þ pD C w1 U ð22:24Þð14:63Þ ¼ ¼ 0:948 or 94:8% gH ð9:81Þð35Þ Hydraulic efficiency hh ¼ Illustrative Example 3.11: A Kaplan runner develops 9000 kW under a head of 5.5 m. Assume a speed ratio of 2.08, flow ratio 0.68, and mechanical efficiency 85%. The hub diameter is 1/3 the diameter of runner. Find the diameter of the runner, and its speed and specific speed. Solution: U 1 ¼ 2:08 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi 2gH ¼ 2:08 ð2Þð9:81Þð5:5Þ ¼ 21:61 m/s pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi 2gH ¼ 0:68 ð2Þð9:81Þð5:5Þ ¼ 7:06 m/s Cr1 ¼ 0:68 Now power is given by 9000 ¼ ð9:81Þð5:5Þð0:85ÞQ Therefore, Q ¼ 196:24 m3 /s If D is the runner diameter and, d, the hub diameter p Q ¼ ðD 2 2 d 2 ÞC r1 4 or p 1 2 2 D 2 D 7:06 ¼ 196:24 4 9 Solving rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð196:24Þð4Þð9Þ ¼ 6:31 m ðpÞð7:06Þð8Þ pffiffiffi pffiffiffiffiffiffiffiffiffiffi P N s ¼ N 5/4 ¼ 65 9000 ¼ 732 rpm H 5:55/4 D¼ Design Example 3.12: A propeller turbine develops 12,000 hp, and rotates at 145 rpm under a head of 20 m. The outer and hub diameters are 4 m and 1.75 m, Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 120 Chapter 3 respectively. Calculate the inlet and outlet blade angles measured at mean radius if overall and hydraulic efficiencies are 85% and 93%, respectively. Solution: Mean diameter ¼ U1 ¼ 4 þ 1:75 ¼ 2:875 m 2 pDN ðpÞð2:875Þð145Þ ¼ ¼ 21:84 m/s 60 60 C w1 U 1 ðC w1 Þð21:84Þ ¼ ¼ 0:93Cw1 gH ð9:81Þð20Þ Using hydraulic efficiency hh ¼ or Cw1 ¼ 8:35 m/s Power ¼ ð12; 000Þð0:746Þ ¼ 8952 kW Power ¼ rgQH ho or 8952 ¼ 9:81 £ Q £ 20 £ 0:85 8952 Therefore, Q ¼ ð9:81Þð20Þð0:85Þ ¼ 53:68 m3 /s Discharge, Q ¼ 53:68 ¼ p ð42 2 1:752 ÞC r1 4 [ Cr1 ¼ 5:28 m/s tan b1 ¼ C r1 5:28 5:28 ¼ ¼ 0:3914 ¼ U 1 2 Cw1 21:84 2 8:35 13:49 b1 ¼ 21:388 and tan b2 ¼ C r2 5:28 ¼ 0:2418 ¼ U 2 21:84 b2 ¼ 13:598 Illustrative Example 3.13: An inward flow reaction turbine wheel has outer and inner diameter are 1.4 m and 0.7 m respectively. The wheel has radial vanes and discharge is radial at outlet and the water enters the vanes at an angle of 128. Assuming velocity of flow to be constant, and equal to 2.8 m/s, find Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved The vane angle at outlet. the velocity of whirl at inlet and outlet will be zero. D2 ¼ 1. Solution: Outer diameter. Tangential velocity of wheel at inlet.Hydraulic Turbines 121 Figure 3.13.21. The speed of the wheel. All Rights Reserved . D1 ¼ 0. as shown in Fig. and 2. C r1 ¼ C r2 ¼ 2:8 m/s As the vanes are radial at inlet and outlet end.21 Velocity triangles at inlet and outlet for Example 3. U 1 ¼ pD2 N or 60 N¼ 60U 1 ð60Þð13:15Þ ¼ 179 rpm ¼ ðpÞð1:4Þ pD2 Copyright 2003 by Marcel Dekker. a1 ¼ 128 Velocity of flow at inlet.7 m Angle at which the water enters the vanes. 1. U1 ¼ C r1 2:8 ¼ 13:15 m/s ¼ tan 128 0:213 Also.4 m Inner diameter. Inc. 3. Q ¼ 500 L/s ¼ 0.5 m3/s Velocity of flow at inlet.122 Chapter 3 Let b2 is the vane angle at outlet U2 ¼ pD1 N ðpÞð0:7Þð179Þ ¼ ¼ 6:56 m/s 60 60 C r2 2:8 ¼ 0:4268 i:e: b2 ¼ 23:118 ¼ U2 6:56 From Outlet triangle. All Rights Reserved . Cr1 ¼ Cr2 ¼ 1. The velocity periphery at inlet is 20 m/s and velocity of whirl at inlet is 15 m/s. U1 ¼ 20 m/s Velocity of whirl at inlet. Inc.14: Consider an inward flow reaction turbine in which water is supplied at the rate of 500 L/s with a velocity of flow of 1. Head of water on the wheel.5 m/s Velocity of periphery at inlet. Cw1 U 1 C2 C2 ¼ H 2 1 ¼ H 2 r1 g 2g 2g Copyright 2003 by Marcel Dekker.5 m/s Let b1 ¼ vane angle at inlet From inlet velocity triangle tan ð180 2 b1 Þ ¼ C r1 1:5 ¼ 0:3 ¼ U 1 2 C w1 20 2 15 [ ð180 2 b1 Þ ¼ 168410 or b1 ¼ 1808 2 168410 ¼ 1638190 Since the discharge is radial at outlet. Cr1 ¼ 1. Solution: Discharge. Vane angle at inlet. find 1. and 2. and velocity of flow to be constant. tan b2 ¼ Illustrative Example 3. ad so the velocity of whirl at outlet is zero Therefore. Cw1 ¼ 15 m/s As the velocity of flow is constant.5 m/s. Assuming radial discharge. e. U 1 ¼ pD1 N ¼ ðpÞð1Þð290Þ ¼ 15:19 m/s 60 60 From inlet triangle. tan b1 ¼ C C2 U ¼ 32:97 12 15:19 ¼ 17:78 ¼ 0:675 2 w1 1 i. Cr1 ¼ Cr2 ¼ 12 m/s Speed of wheel. All Rights Reserved . Therefore. Solution: Inner diameter of wheel. The water enters the vane at angle of 208 and leaves the vane radially. Assuming the velocity of flow remains constant at 12 m/s and wheel rotates at 290 rpm.Hydraulic Turbines 123 or ð15Þð20Þ 1:52 ¼H2 ð2Þð9:81Þ 9:81 [ H ¼ 30:58 2 0:1147 ¼ 30:47 m Design Example 3. N ¼ 290 rpm Vane angle at inlet ¼ b1 U1 is the velocity of periphery at inlet. Inc. velocity of whirl is given by 12 12 ¼ ¼ 32:97 m/s C w1 ¼ tan 20 0:364 12 r1 Hence. D1 ¼ 1 m Outer diameter of wheel. find the vane angles at inlet and outlet. b1 ¼ 348 Let b2 ¼ vane angle at outlet U2 ¼ velocity of periphery at outlet pD2 N ðpÞð2Þð290Þ ¼ ¼ 30:38 m/s 60 60 From the outlet triangle Therefore U 2 ¼ tan b2 ¼ C r2 12 ¼ 0:395 ¼ 30:38 U2 Copyright 2003 by Marcel Dekker.15: Inner and outer diameters of an outward flow reaction turbine wheel are 1 m and 2 m respectively. D2 ¼ 2 m a1 ¼ 208 Velocity of flow is constant That is. 16: An inward flow turbine is supplied with 245 L of water per second and works under a total head of 30 m. b2 ¼ 218330 Illustrative Example 3. then discharge is p Q ¼ D2 C 2 4 1 or C2 ¼ ð4Þð0:245Þ ¼ 3:98 m/s ðpÞð0:28Þ2 But C2 ¼ Cr1 ¼ Cr2 Neglecting losses.. we have Cw1 U 1 H 2 C2 /2g 2 ¼ gH H or Cw1 U 1 ¼ gH 2 C2 /2 2 ¼ ½ð9:81Þð30Þ 2 Power developed P ¼ ð286:38Þð0:245Þ kW ¼ 70:16 kW and Cw1 ¼ 286:38 ¼ 17:9 m/s 16 3:98 tan a1 ¼ ¼ 0:222 17:9 i. The vane angle at inlet 2. Inc.43 or b1 ¼ 648250 tan b1 ¼ ð3:98Þ2 ¼ 294:3 2 7:92 ¼ 286:38 2 Copyright 2003 by Marcel Dekker. a1 ¼ 128310 C r1 3:98 3:98 ¼ ¼ 2:095 ¼ C w1 2 U 1 17:9 2 16 1:9 i. calculate 1.124 Chapter 3 i. All Rights Reserved . The radial velocity is constant. b1 ¼ 64. Solution: If D1 is the diameter of pipe. Power. Neglecting friction. The outlet pipe of the turbine is 28 cm in diameter.e. The velocity of wheel periphery at inlet is 16 m/s. The guide blade angle 3.e.e. 45 m3/s. The height of the runner vanes at inlet is 0. 337 kW Hence overall efficiency. radius of the runner is 0. Solution: Velocity in casing at inlet to turbine Discharge Cc ¼ Cross 2 sectional area of casing ¼ 7:8 2 ¼ 9:93 m/s ðp/4Þð1Þ The net head on turbine C2 2 C2 ¼ Pressure head þ Head due to turbine position þ c 2g 1 2 2 2 ¼ 164 þ 5:4 þ ð9:93Þ 2g ð1:6Þ ¼ 164 þ 5:4 þ 98:6 2 2:56 ¼ 174:3 m of water 19:62 Waterpower supplied to turbine ¼ QgH kW ¼ ð7:8Þð9:81Þð174:3Þ ¼ 13. ho ¼ Shaft Power 12. 337 Design Example 3. 400 ¼ ¼ 0:93 or 93% Water Power 13. Inc. 400 kW Pressure head in scroll casing at the entrance to turbine ¼ 164 m of water Elevation of turbine casing above tail water level ¼ 5:4 m Diameter of turbine casing ¼ 1 m Velocity in tail race ¼ 1:6 m/s Calculate the effective head on the turbine and the overall efficiency of the unit.18: A Francis turbine wheel rotates at 1250 rpm and net head across the turbine is 125 m. and the angle of Copyright 2003 by Marcel Dekker.5 m.17: A reaction turbine is to be selected from the following data: Discharge ¼ 7:8 m3 /s Shaft power ¼ 12.035 m.Hydraulic Turbines 125 Design Example 3. The volume flow rate is 0. All Rights Reserved . Solution: For torque. All Rights Reserved . The torque exerted by the fluid is þ 2534 Nm Power exerted P ¼ Tv ¼ ð2534Þð2ÞðpÞð1250Þ ð60Þð1000Þ ¼ 331:83 kW Copyright 2003 by Marcel Dekker. then inlet area.126 Chapter 3 the inlet guide vanes is set at 708 from the radial direction. Inc. is A ¼ 2pr 1 h1 ¼ ð2ÞðpÞð0:5Þð0:035Þ ¼ 0:11m2 0:45 ¼ 4:1 m/s 0:11 Cr1 ¼ Q/A ¼ From velocity triangle. A. find the torque and power exerted by the water. Also calculate the hydraulic efficiency. using angular momentum equation T ¼ mðCw2 r 2 2 Cw1 r 1 Þ As the flow is radial at outlet. torque is given by T ¼ 2ð225Þð11:26Þ ¼ 22534 Nm Negative sign indicates that torque is exerted on the fluid. velocity of whirl Cw1 ¼ Cr1 tan708 ¼ ð4:1Þð2:75Þ ¼ 11:26m/s Substituting Cw1. Assume that the absolute flow velocity is radial at exit. Cw2 ¼ 0 and therefore T ¼ 2mC w1 r 1 ¼ 2rQC w1 r 1 ¼ 2ð103 Þð0:45Þð0:5Cw1 Þ ¼ 2225C w1 Nm If h1 is the inlet runner height. Solution: Hydraulic efficiency is Power deleloped hh ¼ Power available ¼ mðC w1 U 1 2 C w2 UÞ rgQH Since flow is radial at outlet.19: An inward radial flow turbine develops 130 kW under a head of 5 m. The runner rotates at 230 rpm while hydraulic losses accounting for 20% of the energy available. then Cw2 ¼ 0 and m ¼ rQ. Inc. The flow velocity is 4 m/s and the runner tangential velocity at inlet is 9. All Rights Reserved . Calculate the inlet guide vane exit angle. the inlet angle to the runner vane.Hydraulic Turbines 127 Hydraulic efficiency is given by Power exerted hh ¼ Power available 3 ¼ ð331:83Þð10 Þ rgQH 331:83 £ 103 ð103 Þð9:81Þð0:45Þð125Þ ¼ 0:6013 ¼ 60:13% ¼ Design Example 3. therefore hh ¼ C w1 U 1 gH ðCw1 Þð9:6Þ ð9:81Þð5Þ ð0:80Þð9:81Þð5Þ ¼ 4:09 m/s 9:6 0:80 ¼ Cw1 ¼ Radial velocity Cr1 ¼ 4 m/s tan a1 ¼ C r1 /C w1 ðfrom velocity triangleÞ 4 ¼ 4:09 ¼ 0:978 Copyright 2003 by Marcel Dekker.6 m/s. and the height of the runner at inlet. and overall efficiency equal to 72%. the runner diameter at the inlet. Assume radial discharge. 50 m and 2 m respectively.20: The blade tip and hub diameters of an axial hydraulic turbine are 4.028 Runner speed is U1 ¼ or D1 ¼ 60U 1 ð60Þð9:6Þ ¼ ðpÞð230Þ pN D1 ¼ 0:797 m Power output Power available ð130Þð103 Þ 0:72 pD1 N 60 Overall efficiency ho ¼ or rgQH ¼ or Q¼ But ð130Þð103 Þ ¼ 3:68 m3 /s ð0:72Þð103 Þð9:81Þð5Þ Q ¼ pD1 h1 C r1 ðwhere h1 is the height of runnerÞ Therefore.128 Chapter 3 i.. h1 ¼ 3:68 ¼ 0:367 m ðpÞð0:797Þð4Þ Illustrative Example 3.. Copyright 2003 by Marcel Dekker. All Rights Reserved . The turbine has a net head of 22 m across it and develops 22 MW at a speed of 150 rpm.988 or 1808 2 35. calculate the inlet and outlet blade angles at the mean radius assuming axial flow at outlet. inlet guide vane angle a1 ¼ 448210 tan b1 ¼ C r1 ðC w1 2 U 1 Þ 4 4 ¼ ¼ 20:726 ð4:09 2 9:6Þ 25:51 ¼ i.e.e. If the hydraulic efficiency is 92% and the overall efficiency 84%.98 ¼ 144. Inc. b1 ¼ 2 35. ho. All Rights Reserved . Q. is given by Dm ¼ Dh þ Dt 2 þ 4:50 ¼ ¼ 3:25 m 2 2 Overall efficiency. is given by ho ¼ Power develpoed Power available 22 ¼ 26:2 MW [ Power available ¼ 0:84 Also. Inc. available power ¼ rgQH ð26:2Þð106 Þ ¼ ð103 Þð9:81Þð22ÞQ Hence flow rate.Hydraulic Turbines 129 Solution: Mean diameter. is given by Q¼ ð26:2Þð106 Þ ¼ 121:4 m3 /s ð103 Þð9:81Þð22Þ Now rotor speed at mean diameter pDm N ðpÞð3:25Þð150Þ ¼ ¼ 25:54 m/s 60 60 Power given to runner ¼ Power available £ hh Um ¼ ¼ 26:2 £ 106 £ 0:92 ¼ 24:104 MW Theoretical power given to runner can be found by using P ¼ rQU m Cw1 ðC w2 ¼ 0Þ ð24:104Þð106 Þ ¼ ð103 Þð121:4Þð25:54ÞðC w1 Þ [ C w1 ¼ ð24:104Þð106 Þ ¼ 7:77 m/s ð103 Þð121:4Þð25:54Þ Axial velocity is given by Cr ¼ Q£4 ð121:4Þð4Þ ¼ 9:51 m/s ¼ pðD2 2 D2 Þ pð4:502 2 22 Þ t h Cr 9:51 ¼ U m 2 C w1 25:54 2 7:77 Using velocity triangle tan ð180 2 b1 Þ ¼ Copyright 2003 by Marcel Dekker. Dm. the inlet angle of the runner vane. is given by Power given to runner hh ¼ Water Power available w1 2 ¼ mðU 1 CrgQHU 2 C w2 Þ Since flow is radial at exit. and height of the runner at inlet.21: The following design data apply to an inward flow radial turbine: Overall efficiency Net head across the turbine Power output The runner tangential velocity Flow velocity Runner rotational speed Hydraulic losses 75% 6m 128 kW 10:6 m/s 4 m/s 235 rpm 18% of energy available Calculate the inlet guide vane angle. b2 ¼ 20:438 Design Example 3. hh.130 Chapter 3 Inlet angle. the runner diameter at inlet. Cw2 ¼ 0 and m ¼ rQ. Assume that the discharge is radial. Hence 9:51 ¼ 0:3724 25:54 tan b2 ¼ that is. Therefore U 1 Cw1 gH ð10:6ÞðCw1 Þ or Cw1 ¼ 4:6 m/s 0:82 ¼ ð9:81Þð6Þ hh ¼ Copyright 2003 by Marcel Dekker. All Rights Reserved . Inc. Solution: Hydraulic efficiency. b1 ¼ 151:858 At outlet tan b2 ¼ But Vcw2 Cr V cw2 equals to Um since Cw2 is zero. Copyright 2003 by Marcel Dekker. is given by 1808 2 33:698 ¼ 146:318 Runner speed at inlet U1 ¼ pD1 N 60 Figure 3.14.Hydraulic Turbines 131 Now tan a1 ¼ Cr1 /C w1 ¼ that is. Inc. 3.22 and 3. b1.22 Velocity triangles for Example 3. a1 ¼ 418 From Figs. All Rights Reserved .23 tan ð180 2 b1 Þ ¼ C r1 4 ¼ 0:667 ¼ U 1 2 Cw1 10:6 2 4:6 4 ¼ 0:8695 4:6 that is. b1 ¼ 33:698 Hence blade angle. Inc.15. Copyright 2003 by Marcel Dekker.24 Inlet velocity triangle.23 Velocity triangles at inlet and outlet for Example 3.132 Chapter 3 Figure 3. All Rights Reserved . Figure 3. Inc.000 kW Overall efficiency.Hydraulic Turbines 133 or D1 ¼ U 1 ð60Þ ð10:6Þð60Þ ¼ ¼ 0:86 m pN ðpÞð235Þ Overall efficiency ho ¼ Power output Power available ð128Þð103 Þ 0:75 rgQH ¼ From which flow rate Q¼ Also. P ¼ 10. All Rights Reserved .60. H ¼ 8 m. and flow ratio 0.86 Speed ratio pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi U1 2¼ . Q ¼ pD1 hC r1 where h1 is the height of runner Therefore.86. The hub diameter of the wheel is 0.22: A Kaplan turbine develops 10.35 times the outside diameter of the wheel. Power. Find the diameter and speed of the turbine. ho ¼ 0. h1 ¼ 2:9 ¼ 0:268 m ðpÞð0:86Þð4Þ ð128Þð103 Þ ¼ 2:9 m3 /s ð0:75Þð103 Þð9:81Þð6Þ Design Example 3. the speed ratio 2. Solution: Head. or U l ¼ 2 £ 9:81 £ 8 ¼ 25:06 m/s 1=2 ð2gHÞ Flow ratio pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C r1 ¼ 0:60 or C r1 ¼ 0:60 2 £ 9:81 £ 8 ¼ 7:52 m/s 1=2 ð2gHÞ Hub diameter. The overall efficiency is 0.0. D1 ¼ 0.35 D2 Copyright 2003 by Marcel Dekker.000 kW under an effective head 8 m. 134 Chapter 3 Overall efficiency.9 m a2 ¼ 908ðradial dischargeÞ a1 ¼ 128.23: An inward flow reaction turbine. Assuming the velocity of flow as constant and equal to 2.8 m/s. Inc.45 m Outer Diameter. Solution: Inner Diameter. Cr1 ¼ Cr2 ¼ 2:8 m/s Copyright 2003 by Marcel Dekker. find the speed of the wheel and the vane angle at outlet. The vanes are radial at inlet and the discharge is radial at outlet and the water enters the vanes at an angle of 128.45 m and 0. D2 ¼ 0. All Rights Reserved . having an inner and outer diameter of 0. ho ¼ Or P rgQH 0:86 ¼ 10000 1000 £ 9:81 £ Q £ 8 [ Q ¼ 148:16 m3 /s Now using the relation Q ¼ Cr1 £ Or 148:16 ¼ 7:52 £ [ D1 ¼ 5:35 m The peripheral velocity of the turbine at inlet 25:06 ¼ [ N¼ Áà p 2 À D 1 2 0:35D 2 1 4 à p 2 D1 2 D2 2 4 pD1 N p £ 5:35 £ N ¼ 60 60 60 £ 25:06 ¼ 89 rpm p £ 5:35 Design Example 3. D1 ¼ 0. respectively.90 m. 11).Hydraulic Turbines 135 From velocity triangle at inlet (see Fig. find the actual and theoretical hydraulic efficiencies of the turbine and the inlet angles of the guide and wheel vanes. Solution: Q ¼ 545 L/s ¼ 0. Cr2 ¼ 2.8 m/s. All Rights Reserved . D2 ¼ 40 cm H ¼ 14 m. The inner and outer diameters of the wheel are 40 and 80 cm. Cr2 ¼ C2 Work done/s by the turbine per kg of water ¼ Cw£U1 g Copyright 2003 by Marcel Dekker. U1 ¼ or N¼ 60U 1 60 £ 13:173 ¼ 279 rpm ¼ pD1 p £ 0:9 C r1 2:8 ¼ 13:173 m/s ¼ tan a1 tan 128 pD1 N 60 Considering velocity triangle at outlet peripheral velocity at outlet U2 ¼ pD2 N p £ 0:45 £ 279 ¼ ¼ 6:58 m/s 60 60 Cr2 2:8 ¼ 0:426 ¼ U 2 6:58 tan b2 ¼ [ b2 ¼ 23:058 Design Example 3. 3.24: An inward flow reaction turbine develops 70 kW at 370 rpm. Inc. respectively. The velocity of the water at exit is 2. Assuming that the discharge is radial and that the width of the wheel is constant. Turbine discharges 545 L/s under a head of 14 m.8 m/s As a2 ¼ 908. The peripheral velocity of the wheel at inlet U1 ¼ Now.545 m3/s D1 ¼ 80 cm. a2 ¼ 908 (radial discharge) b1 ¼ b2 Peripheral velocity of the wheel at inlet U1 ¼ pD1 N p £ 0:80 £ 370 ¼ ¼ 15:5 m/s 60 60 Velocity of flow at the exit. i.e. Inc.136 Chapter 3 But this is equal to the head utilized by the turbine. All Rights Reserved . Cw1 U 1 C2 ¼H2 g 2g (Assuming there is no loss of pressure at outlet) or Cw1 £ 15:5 ð2:8Þ2 ¼ 14 2 ¼ 13:6 m 9:81 2 £ 9:81 or Cw1 ¼ 13:6 £ 9:81 ¼ 8:6 m/s 15:5 Work done per second by turbine Q ¼ rg Cw1 U 1 ¼ 1000 £ 0:545 £ 8:6 £ 15:5 1000 ¼ 72:65kW rgQH Available power or water power ¼ 1000 ¼ 74:85 Actual available power ¼ 70 kW Overall turbine efficiency isht ¼ 70 £ 100 74:85 ht ¼ 93:52% This is the actual hydraulic efficiency as required in the problem. Hydraulic Efficiency is hh ¼ 72:65 £ 100 ¼ 97:06% 75:85 This is the theoretical efficiency Q ¼ pD1 b1 Cr1 ¼ pD2 b2 Cr2 (Neglecting blade thickness) Cr1 ¼ Cr2 D2 40 ¼ 1:4 m/s ¼ 2:8 £ D1 20 Drawing inlet velocity triangle tan b1 ¼ C r1 1:4 1:4 ¼ ¼ 0:203 ¼ U 1 2 C w1 15:5 2 8:6 6:9 Copyright 2003 by Marcel Dekker. e. rpm of the wheel. Ns ¼ 270. 2. a1 ¼ 5. P ¼ 5000 kW.58 £ 0. ho ¼ 0. b1 ¼ 11. specific speed is 270.25: An inward flow Francis turbine.76 m/s Copyright 2003 by Marcel Dekker. assume guide vane angle 308.90. Inc. All Rights Reserved .478 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi À Á0:5 C 1 ¼ Cw1 þ Cr1 ¼ 8:62 þ 1:42 ¼ 8:64 m/s and cosa1 ¼ i..Hydraulic Turbines 137 i. H ¼ 30 m.86 1. hh ¼ 0. The turbine is required to develop 5000 kW when operating under a net head of 30 m. hydraulic efficiency of 90%. the diameter and the width of the runner at inlet.866 ¼ 11. find 1. Velocity of Flow: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Cr1 ¼ 0:28 2 £ 9:81 £ 30 ¼ 6:79m/s From inlet velocity triangle C r1 ¼ C 1 sin a1 or 6:79 ¼ C 1 sin 308 or C1 ¼ 6:79 ¼ 13:58 m/s 0:5 Cw1 ¼ C1 cos308 ¼ 13. a1 ¼ 308. Specific speed of the turbine is pffiffiffi N P N s ¼ 5=4 H or N¼ Ns H5=4 270 £ ð30Þ1:25 18957 pffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffi ¼ 267 rpm ¼ 71 P 5000 C w1 8:6 ¼ 0:995 ¼ 8:64 C1 2..e. having an overall efficiency of 86%. the theoretical inlet angle of the runner vanes.58 Design Example 3. and radial velocity of flow at inlet pffiffiffiffiffiffiffiffiffi 0.28 2gH . Cr1 ¼ 0:28 2gH . and 3. Solution: pffiffiffiffiffiffiffiffiffi Power. mean diameter of the runner and specific speed of wheel. From inlet velocity triangle tan b1 ¼ i. All Rights Reserved . jet ratio 12. and speed ratio 0. C r1 6:79 6:79 ¼ ¼ 0:632 ¼ U 1 2 C w1 22:5 2 11:76 10:74 Copyright 2003 by Marcel Dekker. the generator is fed by two Pelton turbines. Assume Pelton wheel efficiency 84%. P ¼ rgQHho or 5000 ¼ 1000 £ 9. Inc.8 m3/s Also Q ¼ kpD1b1Cr1 (where k is the blade thickness coefficient and b1 is the breath of the wheel at inlet) or b1 ¼ Q 19:8 ¼ 0:61 m ¼ kpD1 C r1 0:95 £ p £ 1:61 £ 6:79 3.81 £ Q £ 30 £ 0.e. b1 ¼ 32.26: A 35 MW generator is to operate by a double overhung Pelton wheel. The effective head is 350 m at the base of the nozzle.96. Find the size of jet.138 Chapter 3 Work done per (sec) (kg) of water ¼ C w1 £ U 1 ¼ hh £ H g ¼ 0:9 £ 30 ¼ 27 mkg/s Peripheral Velocity. velocity coefficient of nozzle 0.45.308 Design Example 3. U1 ¼ 27 £ 9:81 ¼ 22:5 m/s 11:76 But U 1 ¼ pD1 N 60 or D1 ¼ 60U 1 60 £ 22:5 ¼ ¼ 1:61 m pN p £ 267 Power. Solution: In this case.86 or Q ¼ 19. 500 ¼ 20. Find the diameter of the nozzle if the coefficient of velocity for the nozzle is 0. All Rights Reserved . 500 N s ¼ 5=4 ¼ ¼ 16:6 H ð350Þ1:25 PROBLEMS 3. Inc. P ¼ rgQH P 20833 ¼ ¼ 6:07 m3 /s rgH 1000 £ 9:81 £ 350 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi Velocity of jet. 000 ¼ 17. (0.98. 833 kW 0:84 But.744 m Peripheral velocity of the wheel pffiffiffiffiffiffiffiffiffi U ¼ speed ratio 2gH pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:45 £ 2 £ 9:81 £ 350 ¼ 37:29 m/s DN But U ¼ p60 or 60U 60 £ 37:29 ¼ ¼ 190 rpm N¼ pD p £ 3:744 Specific speed.d ¼ 4A ¼ 4 £ 0:0763 ¼ 0:312 m p p d ¼ 31:2 cm Diameter of wheel D ¼ d £ jet ratio ¼ 0. A ¼ Cj ¼ 6:07 ¼ 0:0763 m2 79:6 0:5 0:5 [ Diameter of jet.Hydraulic Turbines 139 Power developed by each turbine.312 £ 12 ¼ 3.1 A Pelton wheel produces 4600 hP under a head of 95 m. PT ¼ 35. and with an overall efficiency of 84%.36 m) Copyright 2003 by Marcel Dekker. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi N PT 190 17.C j ¼ C v 2gH ¼ 0:96 2 £ 9:81 £ 350 Q¼ Cj ¼ 79:6 m/s Q Area of jet. 500 kW 2 Using Pelton wheel efficiency in order to find available power of each turbine P¼ 17. 3. 0.6 A Pelton wheel develops 740 kW under a head of 310 m.500 hP under a head of 1750 m while running at 760 rpm.9 A Kaplan turbine produces 16000 kW under a head of 20 m. and (3) the diameter ratio.5 m at inlet and 2. Its inlet vane angle is 1208.069 m) 3. 20.6) Show that in an inward flow reaction turbine.21 m. Assume 85% efficiency. Calculate (1) the turbine efficiency.5 3. Find the jet diameter if its efficiency is 86% and C v ¼ 0:98: (0.14 m.2 m while the hub diameter is 2 m.3 3. Copyright 2003 by Marcel Dekker. (0. the best peripheral velocity is pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2gH / 2 þ tan2 a where H is the head and a the angle of guide vane. Find the size of the jet and the specific speed. breadth of wheel constant and hydraulic efficiency of 88%. 3. The wheel rotates at 430 rpm.7 A reaction turbine runner diameter is 3.45 and specific speed is 17. calculate the power developed and speed of machine.5 m at outlet.104 m. 356 rpm) Show that in a Pelton wheel. The diameter of the runner is 4. Assume radial discharge at 14 m/s.4 3. while running at 166 rpm.2 Pelton wheel develops 13.500 kW under a head of 500 m.8 hh ¼ 3. when the velocity of flow is constant and wheel vane angle at entrance is 908.140 Chapter 3 3. (140 mm) A Pelton wheel of power station develops 30. 21) A Pelton wheel develops 2800 bhP under a head of 300 m at 84% efficiency. where the buckets deflect the water through an angle of (1808 2 a) degrees. (2) the jet diameter. Assume any necessary data and find the jet diameter. The ratio of peripheral velocity of wheel to jet velocity is 0. the hydraulic efficiency of the wheel is given by 2UðC 2 UÞð1 þ cos aÞ C2 where C is the velocity of jet and U is mean blade velocity. The turbine discharge 102 m3 per second of water under a head of 145 m. (2. (128 MW. Inc. the discharge being 120 m3/s. Calculate (1) the mean diameter of the runner. All Rights Reserved . h ¼ overall efficiency. (78%. All Rights Reserved . absolute nozzle velocity coefficient velocity of whirl wheel diameter diameter of nozzle energy transfer by bucket head across the runner frictional head loss specific speed water power available casing and draft tube losses hydraulic power loss leakage loss mechanical power loss runner power loss shaft power output bucket speed work done angle of the blade tip at outlet angle with relative velocity nozzle efficiency transmission efficiency relative velocity ratio Copyright 2003 by Marcel Dekker. 1. 0. NOTATION C Cv Cw D d E Hr hf Ns P Pc Ph Pl Pm Pr Ps U W a b hi htrans k jet velocity. (3) the speed ratio based on the tip diameter of the blade.10 Evolve a formula for the specific speed of a Pelton wheel in the following form pffiffiffiffiffiffi d ffi N s ¼ kÁ hÁ D where Ns ¼ specific speed. and k ¼ a constant. 497. Inc. D ¼ diameter of bucket circle. and (4) the flow ratio.Hydraulic Turbines 141 (2) specific speed.48) 3.84. d ¼ diameter of jet. All Rights Reserved . or a combination of the two. blowers. A fan causes only a small rise in stagnation pressure of the flowing fluid. and then this kinetic energy Copyright 2003 by Marcel Dekker.1 INTRODUCTION This chapter will be concerned with power absorbing turbomachines. A fan consists of a rotating wheel (called the impeller). The overall pressure rise may range from 1.4 Centrifugal Compressors and Fans 4. Inc. air is compressed in a series of successive stages and is often led through a diffuser located near the exit. which is surrounded by a stationary member known as the housing. which can be axial flow. providing airflow.5 atm with shaft speeds up to 30. In turbocompressors or dynamic compressors.2 CENTRIFUGAL COMPRESSOR The compressor. while the air exhausted from a fan is called forced draft.000 rpm or more. 4. There are three types of turbomachines: fans. The air feed into a fan is called induced draft. and compressors. centrifugal flow. produces the highly compressed air needed for efficient combustion. high pressure is achieved by imparting kinetic energy to the air in the impeller.5 to 2. Energy is transmitted to the air by the power-driven wheel and a pressure difference is created. used to handle compressible fluids. In blowers. There is renewed interest in the centrifugal stage. and the part of the centrifugal compressor containing the diverging passages is known as the diffuser. which rotates and imparts kinetic energy to the air and a number of diverging passages in which the air decelerates. Energy is imparted to the air by the rotating blades. It consists of a stationary casing containing an impeller. thereby increasing the static pressure as it moves from eye radius r1 to tip radius r2. Inc. Figure 4. Centrifugal compressors can be built with a double entry or a single entry impeller. The remainder of the static pressure rise is achieved in the diffuser. Air enters the impeller eye and is whirled around at high speed by the vanes on the impeller disc. for small turbofan and turboprop aircraft engines.144 Chapter 4 converts into pressure in the diffuser.1 Typical centrifugal compressor. and ratios of 6:1 are possible if materials such as titanium are used. used in conjunction with one or more axial stages. Velocities of airflow are quite high and the Mach number of the flow may approach unity at many points in the air stream. Figure 4.2 shows a double entry centrifugal compressor.1 shows part of a centrifugal compressor. The air leaving the diffuser is collected and delivered to the outlet. Copyright 2003 by Marcel Dekker. Nevertheless. Compressibility effects may have to be taken into account at every stage of the compressor. Pressure ratios of 4:1 are typical in a single stage. The normal practice is to design the compressor so that about half the pressure rise occurs in the impeller and half in the diffuser. the air passes through a diffuser in which kinetic energy is exchanged with pressure. All Rights Reserved . This process is known as diffusion. The deceleration converts kinetic energy into static pressure. The centrifugal compressor is not suitable when the pressure ratio requires the use of more than one stage in series because of aerodynamic problems. Figure 4. After leaving the impeller. two-stage centrifugal compressors have been used successfully in turbofan engines. The impellers tend to undergo high stress forces. 4. The straight radial Figure 4. (Courtesy of Rolls-Royce. (b) radial blades. Copyright 2003 by Marcel Dekker. there are three types of vanes used in impellers. such as those used in some fans and hydraulic pumps. All Rights Reserved .2 Double-entry main stage compressor with side-entry compressor for cooling air. They are: forward-curved. and radial vanes. Ltd.3.3 THE EFFECT OF BLADE SHAPE ON PERFORMANCE As discussed in Chapter 2. tend to straighten out due to centrifugal force and bending stresses are set up in the vanes.Centrifugal Compressors and Fans 145 Figure 4. Inc. as shown in Fig. backward-curved. Curved blades.3 Shapes of centrifugal impellar blades: (a) backward-curved blades. and (c) forward-curved blades.) 4. All Rights Reserved .4 represents the relative performance of these types of blades. It is clear that increased mass flow decreases the pressure on the backward blade. in the radial direction).3 shows the three types of impeller vanes schematically. Angle u is made by C1 and Ca1 and this angle is known as the angle of prewhirl. C1 ¼ Ca1. the radial blade less. The absolute velocity C1 has a whirl component Cw1. exerts the same pressure on the radial blade. and the least energy is transferred by the backward-curved blades. a given pressure ratio can be achieved from a smaller-sized machine than those with radial or backward-curved blades. Hence with forward-blade impellers. Figure 4.146 Chapter 4 Figure 4..5 shows the impeller and velocity diagrams at the inlet and outlet.5c shows the ideal velocity Copyright 2003 by Marcel Dekker. absolute velocity at the inlet. Figure 4. Figure 4. In the ideal case. the whirl component is exactly equal to the impeller tip velocity. In this case. for the three blade blades are not only free from bending stresses.4 shapes.e. along with the velocity triangles in the radial plane for the outlet of each type of vane. Figure 4. Figure 4. they may also be somewhat easier to manufacture than curved blades. 4. Pressure ratio or head versus mass flow or volume flow. That is.4 VELOCITY DIAGRAMS Figure 4. For a given tip speed. the forward-curved blade impeller transfers maximum energy.5b represents the velocity triangle at the inlet to the impeller eye and air enters through the inlet guide vanes. air comes out from the impeller tip after making an angle of 908 (i.5a represents the velocity triangle when the air enters the impeller in the axial direction. and increases the pressure on the forward blade. Inc. so Cw2 ¼ U2. and actual values of Cw1 are somewhat less than U2.5 Centrifugal impellar and velocity diagrams. this results in a higher static pressure on the leading face of a vane than on the trailing face. But there is some slip between the impeller and the fluid. Inc. As we have already noted in the centrifugal pump. All Rights Reserved . Hence.Centrifugal Compressors and Fans 147 Figure 4. the air is prevented from acquiring Copyright 2003 by Marcel Dekker. triangle. Cw2 is not equal to U2. Inc. by the above definition.5d represents the actual velocity triangle. 4.6 Centrifugal compressor impeller with radial vanes. The slip factor is nearly constant for any machine and is related to the number of vanes on the impeller. If radial exit velocities are to be achieved by the actual fluid. the slip factor is less than unity. All Rights Reserved .148 Chapter 4 a whirl velocity equal to the impeller tip speed. the exit blade angle must be curved forward about 10– 14 degrees.5 SLIP FACTOR From the above discussion. consequently. Various theoretical and empirical studies of the flow in an impeller channel have led to formulas for Figure 4.6 shows the phenomenon of fluid slip with respect to a radial blade. Copyright 2003 by Marcel Dekker. Thus. it is convenient to define a slip factor s as: s¼ C w2 U2 ð4:1Þ Figure 4. Figure 4. it may be seen that there is no assurance that the actual fluid will follow the blade shape and leave the compressor in a radial direction. In this case. t ¼ Cw2 r2 ð4:3Þ where. and Cp is mean specific heat over this temperature range. The velocity diagram indicates that Cw2 approaches U2 as the slip factor is increased. 2 Using the slip factor. although this is not always the case. A slip factor of about 0. disk friction. a power input factor can be introduced. T02 is stagnation temperature at the impeller exit.6 WORK DONE The theoretical torque will be equal to the rate of change of angular momentum experienced by the air. some of the power supplied by the impeller is used in overcoming losses that have a braking effect on the air carried round by the vanes. These include windage. As no work is done on the air in the diffuser. T03 ¼ T02.Centrifugal Compressors and Fans 149 slip factors: For radial vaned impellers. Considering a unit mass of air. Cw2 is whirl component of C2 and r2 is impeller tip radius. we have theoretical Wc ¼ sU2 (treating work done on the air as positive) In a real fluid.9 is typical for a compressor with 19 – 21 vanes. this torque is given by theoretical torque. The compressor isentropic efficiency (hc) may be defined as: hc ¼ T 030 2 T 01 T 03 2 T 01 Copyright 2003 by Marcel Dekker. Thus the actual work done on the air becomes: Wc ¼ csU 2 (assuming Cw1 ¼ 0. the formula for s is given by Stanitz as follows: 0:63p ð4:2Þ s¼12 n where n is the number of vanes. 4. Inc.04.) Temperature equivalent of work done on the air is given by: T 02 2 T 01 ¼ 2 ð4:4Þ csU 2 2 Cp where T01 is stagnation temperature at the impeller entrance. where T03 is the stagnation temperature at the diffuser outlet. Increasing the number of vanes may increase the slip factor but this will decrease the flow area at the inlet. This factor typically takes values between 1. Then the theoretical work done on the air may be written as: Theoretical work done Wc ¼ Cw2r2v ¼ Cw2U2. To take account of these losses. All Rights Reserved . and casing friction. Let v ¼ angular velocity.035 and 1. P03 is stagnation pressure at the diffuser exit.150 0 Chapter 4 (where T03 ¼ isentropic stagnation temperature at the diffuser outlet) or À Á T 01 T 030 =T 01 2 1 hc ¼ T 03 2 T 01 Let P01 be stagnation pressure at the compressor inlet and. The natural tendency of the air in a diffusion process is to break away from the walls of the diverging passage. Eddy formation during air deceleration causes loss by reducing the maximum pressure rise. All Rights Reserved . reverse its direction and flow back in the direction of the pressure gradient. but it is not under the control of the designer. Typical diffuser outlet velocities are in the region of 90 m/s. The centrifugal stresses in a rotating disc are proportional to the square of the rim. Then.7 Diffusing flow. Any increase in this angle leads to a loss of efficiency due to Figure 4. lower speeds must be used for double-sided impellers. Copyright 2003 by Marcel Dekker. Such speeds produce pressure ratios of about 4:1. To avoid disc loading. as shown in Fig.5) indicates that the pressure ratio also depends on the inlet temperature T01 and impeller tip speed U2.7 DIFFUSER The designing of an efficient combustion system is easier if the velocity of the air entering the combustion chamber is as low as possible. Inc. For single sided impellers of light alloy. the maximum permissible included angle of the vane diffuser passage is about 118. Any lowering of the inlet temperature T01 will clearly increase the pressure ratio of the compressor for a given work input.7. 4. Therefore. 4. using the isentropic P –T relationship. we get: ! T 030 g /ðg21Þ P03 hc ðT 03 2 T 01 Þ g /ðg21Þ ¼ ¼ 1þ P01 T 01 T 01 ¼ 1þ hc csU 2 2 C p T 01 !g /ðg21Þ ð4:5Þ Equation (4. U2 is limited to about 460 m/s by the maximum allowable centrifugal stresses in the impeller. covered in Chapter 2. leading to eddy formation and reduced pressure rise. One obvious disadvantage of prewhirl is that the work capacity of Figure 4. Copyright 2003 by Marcel Dekker. 4.5b). The flow theory of diffusion. and b) Compressibility effects.8a. The use of variable-angle diffuser vanes can control the efficiency loss. The relative Mach number at the inlet will be given by: V1 M1 ¼ pffiffiffiffiffiffiffiffiffiffiffi gRT 1 ð4:6Þ where T1 is the static temperature at the inlet. and hence the Mach number at this point. All Rights Reserved . diffusion is a very difficult process and there is always a tendency for the flow to break away from the surface.8 a) Breakaway commencing at the aft edge of the shock wave. 4. As shown in Fig. It is possible to reduce the Mach number by introducing the prewhirl. separation of flow causes excessive pressure losses. Now. Inc. As shown in Fig. 4. relative velocity is reduced as indicated by the dotted triangle.Centrifugal Compressors and Fans 151 boundary layer separation on the passage walls. consider the inlet velocity triangle again (Fig. 4. will be very important and a shock wave might occur.8b. the air breakaway from the convex face of the curved part of the impeller. is applicable here. The relative Mach number at the impeller inlet must be less than unity. The value of the Mach number cannot exceed the value at which shock waves occur. As mentioned earlier.8 COMPRESSIBILITY EFFECTS If the relative velocity of a compressible fluid reaches the speed of sound in the fluid. It should also be noted that any change from the design mass flow and pressure ratio would also result in a loss of efficiency. The prewhirl is given by a set of fixed intake guide vanes preceding the impeller. It is necessary to control the Mach number at certain points in the flow to mitigate this problem. All Rights Reserved . High Mach numbers at the inlet to the diffuser vanes will also cause high pressure at the stagnation points on the diffuser vane tips. Figure 4. Inc. Mach number greater than unity can be used at the impeller tip without loss of efficiency. Assuming a perfect gas.9 MACH NUMBER IN THE DIFFUSER The absolute velocity of the fluid becomes a maximum at the tip of the impeller and so the Mach number may well be in excess of unity. supersonic diffusion can occur without the formation of shock waves provided constant angular momentum is maintained with vortex motion in the vaneless space. The prewhirl is therefore gradually reduced to zero by twisting the inlet guide vanes. This may lead to early fatigue failure when the exciting frequency is of the same order as one of the natural frequencies of the impeller vanes. Copyright 2003 by Marcel Dekker. the Mach number at the impeller exit M2 can be written as: C2 M2 ¼ pffiffiffiffiffiffiffiffiffiffiffi gRT 2 ð4:7Þ However. To overcome this concern. it is a common a practice to use prime numbers for the impeller vanes and an even number for the diffuser vanes. which leads to a variation of static pressure around the circumference of the diffuser. It is not necessary to introduce prewhirl down to the hub because the fluid velocity is low in this region due to lower blade speed.152 Chapter 4 the compressor is reduced by an amount U1Cw1. 4. In addition. This pressure variation is transmitted upstream in a radial direction through the vaneless space and causes cyclic loading of the impeller.9 The theoretical centrifugal compressor characteristic. it has been found that as long as the radial velocity component (Cr2) is subsonic. All Rights Reserved . Therefore the stall may pass along the cascade in the direction of lift on the blades. Figure 4. in general. The blade A then stalls. Consider a valve placed in the delivery line of a compressor running at constant speed.Centrifugal Compressors and Fans 153 4. Point D indicates that the pressure rise is zero. but the flow on blade C is now at a lower incidence. which disrupts the velocity distribution and hence adversely affects the performance of the succeeding stages. A multistage compressor may operate stably in the unsurged region with one or more of the stages stalled. the pressure ratio increases. it is pffiffiffiffiffi 01 necessary to plot P03/P01 against the mass flow parameter m PT01 for fixed N speed intervals of pffiffiffiffiffi. or the breakaway of the flow from the suction side of the blade airfoil. Rotating stall may lead to vibrations resulting in fatigue failure in other parts of the gas turbine. Referring to the cascade of Fig. suppose that the valve is fully closed. The air now flows onto blade A at an increased angle of incidence due to blockage of channel AB. and at Point B. 4. Now. suppose that the valve is opened and airflow begins.9 shows an idealized fixed speed T 01 characteristic. However.10 CENTRIFUGAL COMPRESSOR CHARACTERISTICS The performance of compressible flow machines is usually described in terms of the groups of variables derived in dimensional analysis (Chapter 1). This is the design mass flow rate pressure ratio. Inc. and the rest of the stages unstalled. but the pressure has dropped slightly from the maximum possible value. Point C indicates the further increase in mass flow. 4. First.10. But the compressor efficiency at this pressure ratio will be below the maximum efficiency. it is supposed that some nonuniformity in the approaching flow or in a blade profile causes blade B to stall. the maximum pressure occurs. and blade C may unstall. is characterized by reverse flow near the blade tip. Then the pressure ratio will have some value as indicated by Point A. This pressure ratio is available from vanes moving the air about in the impeller. These characteristics are dependent on other variables such as the conditions of pressure and temperature at the compressor inlet and physical properties of the working fluid. The diffuser contributes to the pressure rise. To study the performance of a compressor completely. Further increases in mass flow will increase the slope of the curve until point D. the abovedescribed curve is not possible to obtain. Copyright 2003 by Marcel Dekker.11 STALL Stalling of a stage will be defined as the aerodynamic stall. Stall. In the mean time. All Rights Reserved . When the discharge is further reduced. 4. When the compressor is running at a particular speed and the discharge is gradually reduced. When the discharge pipe of the compressor is completely choked (mass flow is zero) the pressure ratio will have some value due to the centrifugal head produced by the impeller. Figure 4. The pressure ratio for a given speed. peaks at a maximum value. unlike the temperature ratio. there will be increase in mass flow. The phenomenon of surging should not be confused with the stalling of a compressor stage. the pressure ratio will first increase. and then decreased.10 Mechanism of stall propagation. if the downstream pressure drops below the compressor outlet pressure. This phenomenon of sudden drop in delivery pressure accompanied by pulsating flow is called surging.11 shows typical overall pressure ratios and efficiencies hc of a centrifugal compressor stage. The point on the curve where surging starts is called the surge point. If the downstream pressure does not drop quickly there will be backflow accompanied by further decrease in mass flow.12 SURGING Surging is marked by a complete breakdown of the continuous steady flow throughout the whole compressor. Copyright 2003 by Marcel Dekker.154 Chapter 4 Figure 4. is strongly dependent on mass flow rate. since the machine is usually at its peak value for a narrow range of mass flows. resulting in large fluctuations of flow with time and also in subsequent mechanical damage to the compressor. The pressure ratio is maximized when the isentropic efficiency has the maximum value. the pressure ratio drops due to fall in the isentropic efficiency. Inc. In the case of inlet flow passages.11 Centrifugal compressor characteristics. The line joining the surge points at different speeds gives the surge line. Between the zero mass flow and the surge point mass flow. 4. Inc. mass flow is constant. The choking behavior of rotating passages Copyright 2003 by Marcel Dekker. the flow becomes choked (air ceases to flow).13 CHOKING When the velocity of fluid in a passage reaches the speed of sound at any crosssection.Centrifugal Compressors and Fans 155 Figure 4. the operation of the compressor will be unstable. All Rights Reserved . The relative velocity is given by:  à gRT U2 2 V 2 ¼ a 2 ¼ gRT and T 01 ¼ T þ 2C p 2C p Therefore. which remain unchanged. we have: ! Á 2 ð12gÞ=ðg21Þ r P T0 1À ¼ ¼ 1þ g21 M T r0 P0 2 and when C ¼ a. ¼ ð4:12Þ T T 01 2 gþ1 U2 1þ 2C p T 01 Copyright 2003 by Marcel Dekker. and that the fluid is a perfect gas. 4. All Rights Reserved . The mass flow rate at choking is constant. M ¼ 1.13.2 Impeller When choking occurs in the impeller passages. then 2 C p T 0 ¼ Cp T þ 1 g RT. and 2 T gR 21 2 ð4:8Þ ¼ 1þ ¼ T0 2C p gþ1 Assuming isentropic flow.1 Inlet When the flow is choked. Inc. we have A ! _ m 2 ðgþ1Þ=2ðg21Þ ¼ r0 a0 gþ1 A ð4:10Þ ð4:9Þ ð4:11Þ where (r0 and a0 refer to inlet stagnation conditions. the relative velocity equals the speed of sound at any section. C 2 ¼ a 2 ¼ gRT. and therefore it is necessary to make separate analysis for impeller and diffuser. assuming one dimensional. adiabatic flow.156 Chapter 4 differs from that of the stationary passages. Since h0 ¼ h þ 1 C 2 . so that: #1=ðg21Þ " r 2 Á ¼ À r0 gþ1 À_Á  Ã1=2 Using the continuity equation. 4.13. m ¼ rC ¼ r gRT . 12). 4. Solution: From the velocity triangle (Fig. from the continuity equation: ¼ r01 T 01 ! _ m T ðgþ1Þ=2ðg21Þ ¼ r0 a01 A T 01 !ðgþ1Þ=2ðg21Þ 2 U2 ¼ r01 a01 1þ gþ1 2C p T 01 À Á 2 2 ðgþ1Þ=2ðg21Þ 2 þ g 2 1 U =a01 ¼ r01 a01 gþ1 ð4:13Þ Equation (4. All Rights Reserved .3 Diffuser For choking in the diffuser. Neglect power input factor and assume g ¼ 1.1: Air leaving the impeller with radial velocity 110 m/s makes an angle of 258300 with the axial direction. we use the stagnation conditions for the diffuser and not the inlet.80 and the mechanical efficiency is 0. and the mass flow rate is 3 kg/s. Copyright 2003 by Marcel Dekker.13) indicates that for rotating passages.13.96. Inc. 4. ! r T 1=ðg21Þ and.Centrifugal Compressors and Fans 157 Using isentropic conditions. The impeller tip speed is 475 m/s. overall pressure ratio. Find the slip factor. T01 ¼ 298 K. Illustrative Example 4. À Á U 2 2 C w2 tan b2 ¼ Cr2 tanð25:58Þ ¼ Therefore. The compressor efficiency is 0.4. mass flow is dependent on the blade speed. and power required to drive the compressor. C w2 475 2 C w2 110 ¼ 422:54 m/s. Thus: ðgþ1Þ=2ðg21Þ _ m 2 ¼ r02 a02 gþ1 A ð4:14Þ It is clear that stagnation conditions at the diffuser inlet are dependent on the impeller process. Now. Also.88. find the overall pressure ratio of the compressor.12 Velocity triangle at the impeller tip. s ¼ C w2 ¼ 422:54 ¼ 0:89 U2 475 The overall pressure ratio of the compressor: 2 P03 hc scU 2 ¼ 1þ P01 C p T 01 !g=ðg21Þ !3:5 ð0:80Þð0:89Þð4752 Þ ¼ 1þ ¼ 4:5 ð1005Þð298Þ The theoretical power required to drive the compressor: P¼ ! ! mscU 22 ð3Þð0:89Þð4752 Þ kW ¼ ¼ 602:42kW 1000 1000 Using mechanical efficiency. Solution: C w2 U2 Therefore: Cw2 ¼ U2s ¼ (0.90)(370) ¼ 333 m/s The absolute velocity at the impeller exit: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 2 ¼ C2 þ C 2 2 ¼ 3332 þ 352 ¼ 334:8 m/s v r2 Slip factor: s¼ _ The mass flow rate of air: m ¼ r2 A2 Cr2 ¼ 1:57* 0:18* 35 ¼ 9:89 kg/s Copyright 2003 by Marcel Dekker.57 kg/m3 and inlet stagnation temperature is 290 K. Illustrative Example 4. and the radial velocity component at the exit is 35 m/s. slip factor is 0. Neglect the work input factor.42/0. Assume air density ¼ 1.18 m2 and compressor efficiency is 0.2: The impeller tip speed of a centrifugal compressor is 370 m/s. Inc.96 ¼ 627. If the flow area at the exit is 0.90.52 kW. the actual power required to drive the compressor is: P ¼ 602.158 Chapter 4 Figure 4. All Rights Reserved . determine the mass flow rate of air and the absolute Mach number at the impeller tip. Centrifugal Compressors and Fans 159 The temperature equivalent of work done (neglecting c): sU 2 2 Cp 2 sU 2 Therefore. Inc. Air enters the compressor at stagnation temperature of 208C and 1 bar.13 Velocity triangle at exit.3: A centrifugal compressor is running at 16. Figure 4. The stagnation pressure ratio between the impeller inlet and outlet is 4. All Rights Reserved .13) of the impeller and calculate slip. 4. Assume axial entrance and rotor diameter at the outlet is 58 cm.82. T 02 2 T 01 ¼ T 2 ¼ T 02 2 C2 334:82 2 ¼ 356:83 K ¼ 412:6 2 2Cp ð2Þð1005Þ The Mach number at the impeller tip: C2 334:8 M2 ¼ pffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:884 gRT 2 ð1:4Þð287Þð356:83Þ The overall pressure ratio of the compressor (neglecting c): P03 hc scU 2 2 ¼ 1þ P01 C p T 01 !3:5 !3:5 ð0:88Þð0:9Þð3702 Þ ¼ 1þ ¼ 3:0 ð1005Þð290Þ Illustrative Example 4. If the impeller has radial blades at the exit such that the radial velocity at the exit is 136 m/s and the isentropic efficiency of the compressor is 0. Draw the velocity triangle at the exit (Fig. T 02 ¼ T 01 þ C 2 ¼ 290 þ ð0:90Þð370 Þ ¼ 412:6 K 1005 p The static temperature at the impeller exit. Copyright 2003 by Marcel Dekker.2.000 rpm. and the power input of a centrifugal compressor from the following given data: Impeller tip diameter ¼ 1 m Speed ¼ 5945 rpm Mass flow rate of air ¼ 28 kg/s Static pressure ratio p3 /p1 ¼ 2:2 Atmospheric pressure ¼ 1 bar Atmospheric temperature ¼ 258C Slip factor ¼ 0:90 Neglect the power input factor. All Rights Reserved . temperature of the air at the exit.160 Chapter 4 Solution: Impeller tip speed is given by: pDN ðpÞð0:58Þð16000Þ ¼ ¼ 486 m/s 60 60 Assuming isentropic flow between impeller inlet and outlet. then U2 ¼ T 020 ¼ T 01 ð4:2Þ0:286 ¼ 441:69 K Using compressor efficiency.4: Determine the adiabatic efficiency. the actual temperature rise À Á T 020 2 T 01 ð441:69 2 293Þ ¼ 181:33K T 02 2 T 01 ¼ ¼ 0:82 hc Since the flow at the inlet is axial. Solution: The impeller tip speed is given by: DN U 2 ¼ p60 ¼ ðpÞð1Þð5945Þ ¼ 311 m/s 60 2 2 The work input: W ¼ sU 2 ¼ ð0:9Þð311 Þ ¼ 87 kJ/kg 1000 Copyright 2003 by Marcel Dekker. Cw1 ¼ 0 W ¼ U 2 C w2 ¼ C p ðT 02 2 T 01 Þ ¼ 1005ð181:33Þ ð181:33Þ ¼ 375 m/s Therefore: C w2 ¼ 1005486 Slip ¼ 486 – 375 ¼ 111 m/s C w2 375 Slip factor: s ¼ ¼ 0:772 ¼ U2 486 Illustrative Example 4. Inc. From the velocity triangle. (2) Cw2. For radial exit. If Cw2 is 95% of U2 and hc ¼ 0. 5.14). (3) Cr2. Since the exit is radial and no slip. relative velocity is exactly perpendicular to rotational velocity U2. calculate the following for operation in standard sea level atmospheric conditions: (1) U2.914 m and a2 ¼ 208. All Rights Reserved . we get: 0:286 P3 T 30 ¼ T 1 ¼ ð298Þð2:2Þ0:286 ¼ 373:38 K P1 0 Hence the isentropic temperature rise: T30 2 T 1 ¼ 373:38 2 298 ¼ 75:38 K The temperature equivalent of work done: W T3 2 T1 ¼ ¼ 87/1:005 ¼ 86:57 K Cp The compressor adiabatic efficiency is given by: À Á T 30 2 T 1 75:38 ¼ ¼ 0:871 or 87:1% hc ¼ ðT 3 2 T 1 Þ 86:57 The air temperature at the impeller exit is: T 3 ¼ T 1 þ 86:57 ¼ 384:57 K Power input: _ P ¼ mW ¼ ð28Þð87Þ ¼ 2436 kW Illustrative Example 4. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C2 ¼ U 2 þ C 2 ¼ 4312 þ 156:872 ¼ 458:67 m/s 2 r2 Illustrative Example 4. Copyright 2003 by Marcel Dekker. DN 1.5: A centrifugal compressor impeller rotates at 9000 rpm. and (5) C2.88.87 m/s 4.6: A centrifugal compressor operates with no prewhirl is run with a rotor tip speed of 457 m/s.Centrifugal Compressors and Fans 161 Using the isentropic P – T relation and denoting isentropic temperature by T3 . 4. Cr2 ¼ U2 tan(a2) ¼ (431) (0. Impeller tip speed is given by U 2 ¼ p60 ¼ ðpÞð0:914Þð9000Þ ¼ 431 m/s 60 2. (4) b2. Inc.364) ¼ 156. Using the velocity triangle (Fig. Cw2 ¼ U2 ¼ 431 m/s 3. If the impeller tip diameter is 0. Thus the angle b2 is 908 for radial exit. 88. and (3) the power required for a flow of 29 k/s. The isentropic efficiency of the compressor is 0. The inlet stagnation temperature of air is 290 K and at the exit from the impeller tip the stagnation temperature is 440 K. calculate the following for operation in standard sea level air: (1) pressure ratio.000 rpm and air enters in the axial direction.162 Chapter 4 Figure 4. and the slip factor s ¼ 0. All Rights Reserved .85. The work per kg of air W ¼ U 2 C w2 ¼ ð457Þð0:95Þð457Þ ¼ 198:4 kJ/kg 3. overall pressure ratio. (2) work input per kg of air. The pressure ratio is given by (assuming s ¼ c ¼ 1): !g= g21 P03 hc scU 2 ð Þ 2 ¼ 1þ P01 C p T 01 ¼ 1þ !3:5 ð0:88Þð0:95Þð4572 Þ ¼ 5:22 ð1005Þð288Þ 2.14 Velocity triangle at impeller exit. Solution: 1. and power required to drive the compressor per unit mass flow rate of air.7: A centrifugal compressor is running at 10.04. Inc. work input factor c ¼ 1. Calculate the impeller tip diameter. The power for 29 kg/s of air _ P ¼ mW ¼ ð29Þð198:4Þ ¼ 5753:6 kW Illustrative Example 4. Copyright 2003 by Marcel Dekker. 000Þ Copyright 2003 by Marcel Dekker. using the Stanitz formula to find the slip factor: s¼12 0:63p 0:63p ¼12 ¼ 0:8958 n 19 !g=ðg21Þ .5 kg/s. so U 2 ¼ 449:9 m/s ð1005Þð293Þ The impeller diameter. or 4:5 The overall pressure ratio P03 hc scU 2 2 ¼ 1þ P01 C p T 01 !3:5 ð0:84Þð0:8958Þð1:04ÞðU 2 Þ 2 ¼ 1þ . D ¼ 60U 2 ð60Þð449:9Þ ¼ ¼ 0:5053 m ¼ 50:53 cm. All Rights Reserved .Centrifugal Compressors and Fans 163 Solution: Temperature equivalent of work done: scU 2 ð0:88Þð1:04ÞðU 2 Þ 2 2 : or 1005 Cp Therefore. U 2 ¼ 405:85 m/s T 02 2 T 01 ¼ and D¼ 60U 2 ð60Þð405:85Þ ¼ ¼ 0:775 m ðpÞð10. pN pð17. Determine the overall diameter of the impeller and the power required to drive the compressor when the mass flow is 2. Isentropic efficiency of the compressor is 0.000 rpm.84 and the work input factor is 1.04. 000Þ pN The overall pressure ratio is given by: !g= g21 P03 hc scU 2 ð Þ 2 ¼ 1þ P01 C p T 01 ¼ 1þ ð0:85Þð0:88Þð1:04Þð405:852 Þ ð1005Þð290Þ !3:5 ¼ 3:58 Power required to drive the compressor per unit mass flow: P ¼ mcsU 2 ¼ 2 ð1Þð0:88Þð1:04Þð405:852 Þ ¼ 150:75 kW 1000 Design Example 4. The impeller has 19 radial vanes and rotates at 17. Inc.8: Air enters axially in a centrifugal compressor at a stagnation temperature of 208C and is compressed from 1 to 4. Solution: Since the vanes are radial.5 bars. the Mach number at the impeller exit is: C2 409:67 M2 ¼ pffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 1:03 gRT 2 ð1:4Þð287Þð397Þ Design Example 4.9: Repeat problem 4. T02 ¼ 293 þ 187.8 kg/m3 and the axial width at the entrance to the diffuser is 12 mm. The air enters the compressor at 258C and 1 bar.63 K Hence the static temperature at the impeller exit is: T 2 ¼ T 02 2 C2 409:672 2 ¼ 397 K ¼ 480:63 2 2C p ð2Þð1005Þ Now.89. _ m ¼ r2 A2 C r2 ¼ r2 2pr 2 b2 C r2 where: b2 ¼ axial width r2 ¼ radius Therefore: 2:5 ¼ 73:65 m/s C r2 ¼ ð1:8Þð2pÞð0:25Þð0:012Þ Absolute velocity at the impeller exit qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 2 ¼ C2 þ C 2 ¼ 73:652 þ 4032 ¼ 409:67 m/s r2 w2 The temperature equivalent of work done: T 02 2 T 01 ¼ 188:57/C p ¼ 188:57/1:005 ¼ 187:63 K Therefore. or Cw2 ¼ ð0:8958Þð449:9Þ ¼ 403 m/s Slip factor: s ¼ U2 Using the continuity equation.164 Chapter 4 csU 2 ð0:8958Þð1:04Þð449:92 Þ 2 ¼ ¼ The work done on the air W ¼ 1000 1000 188:57 kJ/kg _ Power required to drive the compressor: P ¼ mW ¼ ð2:5Þð188:57Þ ¼ 471:43 kW Design Example 4. assuming the air density at the impeller tip is 1.000 rpm. If the compressor isentropic Copyright 2003 by Marcel Dekker. All Rights Reserved . Assume that the air enters axially with velocity of 145 m/s and the slip factor is 0.63 ¼ 480.10: A centrifugal compressor is required to deliver 8 kg/s of air with a stagnation pressure ratio of 4 rotating at 15. Solution: C w2 . Determine the radial velocity at the impeller exit and the absolute Mach number at the impeller tip.8. Inc. 000Þ P1 ð1Þð100Þ ¼ 1:17 kg/m3 ¼ RT 1 ð0:287Þð298Þ _ m 8 ¼ 0:047 m2 ¼ r1 C 1 ð1:17Þð145Þ Hence. All Rights Reserved . diameter. Inc. the impeller tip diameter D¼ The air density at the impeller eye is given by: r1 ¼ Using the continuity equation in order to find the area at the impeller eye. Solution: Inlet stagnation temperature: T 01 ¼ T a þ C2 1452 1 ¼ 308:46 K ¼ 298 þ 2C p ð2Þð1005Þ Using the isentropic P – T relation for the compression process. A1 ¼ The power input is: _ P ¼ m W ¼ ð8Þð169:48Þ ¼ 1355:24 kW Copyright 2003 by Marcel Dekker.89. W ¼ sU 2 ¼ 2 or: ð0:89ÞðU 2 Þ . work done on the air is given by: W ¼ C p ðT 02 2 T 01 Þ ¼ ð1:005Þð168:64Þ ¼ 169:48 kJ/kg But.Centrifugal Compressors and Fans 165 efficiency is 0. ðg21Þ=g P03 0 ¼ T 01 ¼ ð308:46Þð4Þ0:286 ¼ 458:55K T 03 P01 Using the compressor efficiency. and area at the impeller eye. À Á T020 2 T 01 ð458:55 2 308:46Þ ¼ 168:64 K ¼ T 02 2 T 01 ¼ hc 0:89 Hence. or :169:48 ¼ 0:89U 2 /1000 2 1000 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1000Þð169:48Þ U2 ¼ ¼ 436:38 m/s 0:89 60U 2 ð60Þð436:38Þ ¼ ¼ 0:555 m pN p ð15. work input. impeller tip speed. find the rise in stagnation temperature. 15 The velocity triangle at the impeller eye. Inc. and (3) the maximum Mach number at the eye.166 Chapter 4 Figure 4. (2) power input. Then the vane angle at the eye root is: Ca aer ¼ tan21 U er and pDer N pð0:14Þð15. Design Example 4. 000Þ ¼ ¼ 110 m/s 60 60 Hence. the vane angle at the impeller eye root: 0 21 C a 21 145 aer ¼ tan ¼ tan ¼ 528 48 110 U er U er ¼ Copyright 2003 by Marcel Dekker. 4. All Rights Reserved . Solution: (1) Let Uer be the impeller speed at the eye root. 000 rpm Calculate (1) the impeller vane angles at the eye tip and eye root.15): Impeller eye tip diameter: Impeller eye root diameter: Impeller tip diameter: Mass flow of air: Inlet stagnation temperature: Inlet stagnation pressure: Air enters axially with velocity: Slip factor: Power input factor: Rotational speed: 0:28 m 0:14 m 0:48 m 10 kg/s 290 K 1 bar 145 m/s 0:89 1:03 15.11: The following data apply to a double-sided centrifugal compressor (Fig. Inc. All Rights Reserved . 000Þ ¼ ¼ 220 m/s 60 60 Therefore vane angle at the eye tip: aet ¼ tan (2) 21 Ca U et 145 0 ¼ 338 23 ¼ tan 220 21 Work input: _ W ¼ mcsU 2 ¼ ð10Þð0:819Þð1:03U 2 Þ 2 2 but: U2 ¼ Hence.12: Recalculate the maximum Mach number at the impeller eye for the same data as in the previous question. 000Þ ¼ ¼ 377:14 m/s 60 60 The relative velocity at the eye tip: V1 ¼ Hence. Copyright 2003 by Marcel Dekker. assuming prewhirl angle of 208. the maximum relative Mach number at the eye tip: V1 M1 ¼ pffiffiffiffiffiffiffiffiffiffiffi .Centrifugal Compressors and Fans 167 Impeller velocity at the eye tip: U et ¼ pDet N pð0:28Þð15. gRT 1 where T1 is the static temperature at the inlet T 1 ¼ T 01 2 C2 1452 1 ¼ 279:54 K ¼ 290 2 2Cp ð2Þð1005Þ The Mach number at the inlet then is: V1 263/5 M1 ¼ pffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:786 gRT 1 ð1:4Þð287Þð279:54Þ Design Example 4. W¼ (3) ð10Þð0:89Þð1:03Þð377:142 Þ ¼ 1303:86 kW 1000 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi U 2 þ C2 ¼ 2202 þ 1452 ¼ 263:5 m/s et a pD2 N pð0:48Þð15. 168 Chapter 4 Figure 4. Solution: Figure 4.16 The velocity triangle at the impeller eye. All Rights Reserved .16 shows the velocity triangle with the prewhirl angle. or V 1 1 a ¼ 221:3 m/s Therefore the static temperature at the inlet: T 1 ¼ T 01 2 Hence. C2 1 ¼ 290 2 11:846 ¼ 278:2 K 2C p Copyright 2003 by Marcel Dekker. From the velocity triangle: C1 ¼ 145 ¼ 154:305 m/s cosð208 Þ Equivalent dynamic temperature: C2 154:3052 1 ¼ 11:846 K ¼ 2C p ð2Þð1005Þ C w1 ¼ tanð208Þ Ca1 ¼ ð0:36Þð145Þ ¼ 52:78 m/s Relative velocity at the inlet: V 2 ¼ C 2 þ ðU e 2 C w1 Þ2 ¼ 1452 þ ð220 2 52:78Þ2 . Inc. V1 221:3 M1 ¼ pffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:662 gRT 1 ð1:4Þð287Þð278:2Þ Note the reduction in Mach number due to prewhirl. Ca ¼ _ m 5:5 ¼ 119:6 m/s ¼ r1 A1 ð1:21Þð0:038Þ Since the whirl component at the inlet is zero. Solution: Let: rh ¼ hub radius rt ¼ tip radius The flow area of the impeller inlet annulus is: À Á À Á A1 ¼ p r 2 2 r 2 ¼ p 0:1252 2 0:06252 ¼ 0:038 m2 t h Axial velocity can be determined from the continuity equation but since the inlet density (r1) is unknown a trial and error method must be followed. The temperature equivalent of the velocity is: C2 119:62 1 ¼ 7:12 K ¼ 2Cp ð2Þð1005Þ Therefore: T 1 ¼ T 01 2 C2 1 ¼ 288 2 7:12 ¼ 280:9 K 2Cp Copyright 2003 by Marcel Dekker. r1 ¼ P01 ð1Þð105 Þ ¼ 1:21 kg/m3 ¼ RT 01 ð287Þð288Þ Using the continuity equation. Inc. Calculate the blade inlet angle at the root and tip and the Mach number at the eye tip. the absolute velocity at the inlet is C1 ¼ Ca. All Rights Reserved . 500 rpm Assume zero whirl at the inlet and no losses in the intake duct.13: The following data refers to a single-sided centrifugal compressor: Ambient Temperature: 288 K Ambient Pressure: Hub diameter: Eye tip diameter: Mass flow: Speed: 1 bar 0:125 m 0:25 m 5:5 kg/s 16. Assuming a density based on the inlet stagnation condition.Centrifugal Compressors and Fans 169 Design Example 4. 3:5 g=ðg21Þ P1 T1 5 280:9 ¼ ¼ 92 kPa . and ¼ RT 1 ð287Þð280:9Þ 5:5 ¼ 126:96 m/s ð1:14Þð0:038Þ r1 ¼ Ca ¼ Therefore: C2 ð126:96Þ2 1 ¼ 8:02 K ¼ 2C p 2ð1005Þ T 1 ¼ 288 2 8:02 ¼ 279:988 K 279:98 3:5 P1 ¼ 10 ¼ 90:58 kPa 288 5 r1 ¼ ð90:58Þð103 Þ ¼ 1:13 kg/m3 ð287Þð279:98Þ Further iterations are not required and the value of r1 ¼ 1.170 Chapter 4 Using isentropic P– T relationship. 500Þ ¼ ¼ 216 m/s 60 60 The blade angle at the eye tip: U et 216 ¼ 59:568 bet ¼ tan21 ¼ tan21 Ca 126:96 U et ¼ At the hub.13 kg/m3 may be taken as the inlet density and Ca ¼ C1 as the inlet velocity. U eh ¼ 2pð0:0625Þð16. 500Þ ¼ 108 m/s 60 The blade angle at the hub: 108 beh ¼ tan21 ¼ 40:398 126:96 Copyright 2003 by Marcel Dekker. Inc. At the eye tip: 2pr et N 2p ð0:125Þð16. or P1 ¼ 10 288 P01 T 01 and: P1 ð92Þð103 Þ ¼ 1:14 kg/m3 . All Rights Reserved . 14: A centrifugal compressor compresses air at ambient temperature and pressure of 288 K and 1 bar respectively. Slip factor: s¼ Or: C w2 U2 Cw2 ¼ sU 2 ¼ ð0:89Þð364Þ ¼ 323:96 m/s From the velocity triangle. T 02 ¼ T 01 þ sU 2 ð0:89Þð3642 Þ 2 ¼ 405:33 K ¼ 288 þ 1005 Cp Copyright 2003 by Marcel Dekker.Centrifugal Compressors and Fans 171 The Mach number based on the relative velocity at the eye tip using the inlet velocity triangle is: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi V 1 ¼ C2 þ U 2 ¼ 126:962 þ 2162 .085 m2. the radial velocity at the exit from the impeller is 28 m/s. and the slip factor is 0. Calculate the Mach number of the flow at the impeller tip. Inc. If the impeller total-to-total efficiency is 0. Let us first calculate C2 and T2. All Rights Reserved . Solution: The absolute Mach number of the air at the impeller tip is: C2 M2 ¼ pffiffiffiffiffiffiffiffiffiffiffi gRT 2 where T2 is the static temperature at the impeller tip. or V 1 ¼ 250:6 m/s a 1 The relative Mach number V1 250:6 M ¼ pffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:747 gRT 1 ð1:4Þð287Þð279:98Þ Design Example 4. The impeller tip speed is 364 m/s.89. C2 ¼ C 2 þ C2 ¼ 282 þ 323:962 ¼ ð1:06Þð105 Þ m2 /s2 2 r2 w2 With zero whirl at the inlet W ¼ sU 2 ¼ C p ðT 02 2 T 01 Þ 2 m Hence.88 and the flow area from the impeller is 0. calculate the mass flow rate of air. Assume an axial entrance at the impeller eye and radial blades. Calculate 1.9. Assume axial velocity approximately equal to 105 m/s at the impeller outlet.500 rpm. Copyright 2003 by Marcel Dekker. P2 ¼ P2 P02 P02 P01 P01 ¼ ð0:614Þð2:922Þð1Þð100Þ ¼ 179:4 kPa 179:4ð1000Þ ¼ 1:773 kg/m3 r2 ¼ 287ð352:6Þ Mass flow: _ m ¼ ð1:773Þð0:085Þð28Þ ¼ 4:22 kg/s Design Example 4. power input factor 1.88.172 Chapter 4 Static Temperature T 2 ¼ T 02 2 Therefore. diameter of the impellar is 0. mass flow rate as 16 kg/s. M2 ¼ C2 106000 2 ¼ 352:6 K ¼ 405:33 2 2C p ð2Þð1005Þ 1 2 ð1:06Þð105 Þ ð1:4Þð287Þð352:6Þ ¼ 0:865 Using the isentropic P– T relation: !g=ðg21Þ P02 T 02 21 ¼ 1 þ hc P01 T 01 !3:5 405:33 21 ¼ 1 þ 0:88 ¼ 2:922 288 P2 P02 ¼ T2 T 02 3:5 ¼ 352:6 405:33 3:5 ¼ 0:614 Therefore. calculate the Mach number and air angle at the impeller outlet. Inc. axial depth of the vaneless space is 38 mm. and stagnation pressure at inlet is 101 kPa.04. Assume ship factor as 0. inlet stagnation temperature of air is 290 K.56 m. Stagnation conditions at the impeller outlet. and width of the vaneless space is 43 mm.15: The impeller of a centrifugal compressor rotates at 15. All Rights Reserved . 2. The isentropic efficiency of impeller is 0. assume no fore whirl at the inlet. Centrifugal Compressors and Fans 173 3. The angle of the diffuser vane leading edges and the Mach number at this radius if the diffusion in the vaneless space is isentropic. Solution: 1. Impeller tip speed U2 ¼ pD2 N p £ 0:56 £ 15500 ¼ 60 60 U 2 ¼ 454:67 m/s Overall stagnation temperature rise T 03 2 T 01 ¼ csU 2 1:04 £ 0:9 £ 454:672 2 ¼ 1005 1005 ¼ 192:53K Since T03 ¼ T02 Therefore, T02 2 T01 ¼ 192.53K and T02 ¼ 192.53 þ 290 ¼ 482.53K Now pressure ratio for impeller 3:5 p02 T 02 482:53 3:5 ¼ ¼ ¼ 5:94 290 p01 T 01 then, p02 ¼ 5.94 £ 101 ¼ 600 KPa 2. s¼ C w2 U2 Cw2 ¼ sU 2 or Cw2 ¼ 0:9 £ 454:67 ¼ 409 m/s Let Cr2 ¼ 105 m/s Outlet area normal to periphery A2 ¼ pD2 £ impeller depth ¼ p £ 0:56 £ 0:038 A2 ¼ 0:0669 m2 From outlet velocity triangle 2 C 2 2 ¼ C r22 þ Cw2 ¼ 1052 þ 4092 C 2 2 ¼ 178306 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 174 Chapter 4 i:e: C 2 ¼ 422:26 m/s T 2 ¼ T 02 2 C 22 422:262 ¼ 482:53 2 2C p 2 £ 1005 T 2 ¼ 393:82 K Using isentropic P –T relations g g21 T2 393:82 3:5 ¼ 600 ¼ 294:69 kPa P2 ¼ P02 482:53 T 02 From equation of state r2 ¼ P2 293:69 £ 103 ¼ ¼ 2:61 kg/m3 RT 2 287 £ 393:82 _ m 16 ¼ 91:63 m/s ¼ A2 P2 0:0669 £ 2:61 The equation of continuity gives C r2 ¼ Thus, impeller outlet radial velocity ¼ 91.63 m/s Impeller outlet Mach number C2 422:26 M2 ¼ pffiffiffiffiffiffiffiffiffiffiffi ¼ gRT 2 ð1:4 £ 287 £ 393:82Þ0:5 M2 ¼ 1:06 From outlet velocity triangle Cosa2 ¼ C r2 91:63 ¼ 0:217 ¼ 422:26 C2 3. i.e., a2 ¼ 77.478 Assuming free vortex flow in the vaneless space and for convenience denoting conditions at the diffuser vane without a subscript (r ¼ 0.28 þ 0.043 ¼ 0.323) Cw ¼ Cw2 r 2 409 £ 0:28 ¼ 0:323 r C w ¼ 354:55 m/s Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Centrifugal Compressors and Fans 175 The radial component of velocity can be found by trial and error. Choose as a first try, Cr ¼ 105 m/s C2 1052 þ 354:552 ¼ 68 K ¼ 2C p 2 £ 1005 T ¼ 482.53 2 68 (since T ¼ T02 in vaneless space) T ¼ 414.53K T 2 3:5 419:53 3:5 p ¼ p02 ¼ 600 ¼ 352:58 kPa T 02 482:53 r¼ p2 294:69 ¼ RT 2 287 £ 393:82 r ¼ 2:61 kg/m3 The equation of continuity gives A ¼ 2pr £ depth of vanes ¼ 2p £ 0:323 £ 0:038 ¼ 0:0772 m2 Cr ¼ 16 ¼ 79:41 m/s 2:61 £ 0:0772 Next try Cr ¼ 79.41 m/s C2 79:412 þ 354:552 ¼ 65:68 ¼ 2Cp 2 £ 1005 T ¼ 482:53 2 65:68 ¼ 416:85K p ¼ p02 T T 02 3:5 416:85 3:5 ¼ 600 482:53 p ¼ 359:54 Pa r¼ 359:54 ¼ 3 kg/m3 416:85 £ 287 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 176 Chapter 4 16 ¼ 69:08 m/s 3:0 £ 0:772 Try Cr ¼ 69:08 m/s Cr ¼ C2 69:082 þ 354:552 ¼ 64:9 ¼ 2C p 2 £ 1005 T ¼ 482:53 2 64:9 ¼ 417:63 K T 3:5 417:63 3:5 p ¼ p02 ¼ 600 T 02 482:53 p ¼ 361:9 Pa r¼ 361:9 ¼ 3:02 kg/m3 417:63 £ 287 16 ¼ 68:63 m/s 3:02 £ 0:772 Cr ¼ Taking Cr as 62.63 m/s, the vane angle Cw Cr 354:5 ¼ 5:17 ¼ 68:63 i.e. a ¼ 798 Mach number at vane 65:68 £ 2 £ 1005 1=2 M¼ ¼ 0:787 1:4 £ 287 £ 417:63 tan a ¼ Design Example 4.16: The following design data apply to a double-sided centrifugal compressor: Impeller eye root diameter: Impeller eye tip diameter: Mass flow: Impeller speed: Inlet stagnation pressure: Inlet stagnation temperature: Axial velocity at inlet ðconstantÞ: 18 cm 31:75 cm 18:5 kg/s 15500 rpm 1:0 bar 288 K 150 m/s Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Centrifugal Compressors and Fans 177 Find suitable values for the impeller vane angles at root and tip of eye if the air is given 208 of prewhirl at all radii, and also find the maximum Mach number at the eye. Solution: At eye root, Ca ¼ 150 m/s Ca 150 ¼ 159:63 m/s ¼ cos208 cos208 and Cw1 ¼ 150 tan 208 ¼ 54.6 m/s Impeller speed at eye root [ C1 ¼ U er ¼ pDer N p £ 0:18 £ 15500 ¼ 60 60 U er ¼ 146 m/s Ca 150 150 ¼ ¼ 1:641 ¼ U er 2 C w1 146 2 54:6 91:4 From velocity triangle tan ber ¼ i:e:; ber ¼ 58:648 At eye tip from Fig. 4.17(b) U et pDet N p £ 0:3175 £ 15500 ¼ 60 60 U et ¼ 258 m/s Figure 4.17 Velocity triangles at (a) eye root and (b) eye tip. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 178 Chapter 4 tan aet ¼ 150 150 ¼ ¼ 0:7375 258 2 54:6 203:4 i:e: aet ¼ 36:418 Mach number will be maximum at the point where relative velocity is maximum. Relative velocity at eye root is: V er ¼ Ca 150 150 ¼ ¼ sin ber sin 58:648 0:8539 V er ¼ 175:66 m/s Relative velocity at eye tip is: V et ¼ Ca 150 150 ¼ ¼ sin aet sin 36:418 0:5936 V et ¼ 252:7 m/s Relative velocity at the tip is maximum. Static temperature at inlet: T 1 ¼ T 01 ¼ V2 252:72 et ¼ 288 2 31:77 ¼ 288 2 2C p 2 £ 1005 T 1 ¼ 256:23 K Mmax ¼ À V et gRT 1 Á1=2 ¼ 252:7 252:7 ¼ 1=2 320:86 ð1:4 £ 287 £ 256:23Þ Mmax ¼ 0:788 Design Example 4.17: In a centrifugal compressor air enters at a stagnation temperature of 288 K and stagnation pressure of 1.01 bar. The impeller has 17 radial vanes and no inlet guide vanes. The following data apply: Mass flow rate: Impeller tip speed: Mechanical efficiency: Compressor isentropic efficiency: Absolute velocity at impeller inlet: 2:5 kg/s 475 m/s 96% 84% 150 m/s Absolute air velocity at diffuser exit: 90 m/s Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Centrifugal Compressors and Fans 179 Diffuser efficiency: Power input factor: 82% 1:04 1:4 Axial depth of impeller: 6:5 mm g for air: Determine: 1. shaft power 2. stagnation and static pressure at diffuser outlet 3. radial velocity, absolute Mach number and stagnation and static pressures at the impeller exit, assume reaction ratio as 0.5, and 4. impeller efficiency and rotational speed Solution: 1. Mechanical efficiency is hm ¼ Work transferred to air Work supplied to shaft W hm for vaned impeller, slip factor, by Stanitz formula is or shaft power ¼ s¼12 0:63p 0:63 £ p ¼12 n 17 s ¼ 0:884 Work input per unit mass flow W ¼ csU 2 C w2 Since C w1 ¼ 0 ¼ csU 22 ¼ 1:04 £ 0:884 £ 4752 Work input for 2.5 kg/s W ¼ 1:04 £ 0:884 £ 2:5 £ 4752 W ¼ 518:58K Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 180 Chapter 4 Hence; Shaft Power ¼ 2. 518:58 ¼ 540:19kW 0:96 The overall pressure ratio is " #g=ðg21Þ hc csU 2 p03 2 ¼ 1þ p01 C p T 01 0:84 £ 1:04 £ 0:884 £ 4752 ¼ 1þ 1005 £ 288 Stagnation pressure at diffuser exit P03 ¼ p01 £ 5:20 ¼ 1:01 £ 5:20 P03 ¼ 5:25 bar p3 ¼ p03 T3 T 03 g=g21 !3:5 ¼ 5:2 W ¼ m £ C p ðT 03 2 T 01 Þ [ T 03 ¼ W 518:58 £ 103 þ 288 ¼ 494:4 K þ T 01 ¼ 2:5 £ 1005 mC p Static temperature at diffuser exit T 3 ¼ T 03 2 C 32 902 ¼ 494:4 2 2C p 2 £ 1005 T 3 ¼ 490:37 K Static pressure at diffuser exit g=g21 T3 490:37 3:5 p3 ¼ p03 ¼ 5:25 494:4 T 03 p3 ¼ 5:10 bar 3. The reaction is 0:5 ¼ T2 2 T1 T3 2 T1 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Inc.Centrifugal Compressors and Fans 181 and T 3 2 T 1 ¼ ðT 03 2 T 01 Þ þ ¼ Substituting T 2 2 T 1 ¼ 0:5 £ 213:56 ¼ 106:78 K Now T 2 ¼ T 01 2 2 C1 2 C3 2 W 1502 2 902 þ ¼ 2C p 2 £ 1005 mC p 518:58 £ 103 þ 7:164 ¼ 213:56 K 2:5 £ 1005 C 12 þ ðT 2 2 T 1 Þ 2Cp ¼ 288 2 11:19 þ 106:78 T 2 ¼ 383:59 K At the impeller exit T 02 ¼ T 2 þ or T 03 ¼ T 2 þ Therefore. All Rights Reserved . C 2 2 ¼ 2C p ½ðT 03 2 T 01 Þ þ ðT 01 2 T 2 Þ ¼ 2 £ 1005ð206:4 þ 288 þ 383:59Þ C 2 ¼ 471:94 m/s Mach number at impeller outlet M2 ¼ C2 ð1:4 £ 287 £ 383:59Þ1=2 C2 2 ðSince T 02 ¼ T 03 Þ 2C p C2 2 2C p M2 ¼ 1:20 Radial velocity at impeller outlet 2 C r22 ¼ C 2 2 2 Cw2 ¼ ð471:94Þ2 2 ð0:884 £ 475Þ2 Copyright 2003 by Marcel Dekker. Inc. All Rights Reserved .182 Chapter 4 C r22 ¼ 215:43 m/s Diffuser efficiency is given by hD ¼ h30 2 h2 isentropic enthalpy increase T 30 2 T 2 ¼ ¼ h3 2 h2 T3 2 T2 actual enthalpy increase ! g21=g T 0 3 T 2 p3 21 T 2 T2 2 1 p2 ¼ ¼ T3 2 T2 ðT 3 2 T 2 Þ T3 2 T2 T2 !3:5 Therefore p3 p2 ¼ 1 þ hD 0:821 £ 106:72 3:5 ¼ 1þ 383:59 ¼ 2:05 5:10 ¼ 2:49 bar or p2 ¼ 2:05 From isentropic P –T relations 3:5 T 02 494:4 3:5 p02 ¼ p2 ¼ 2:49 383:59 T2 p02 ¼ 6:05 bar 4. Impeller efficiency is g21 ! g T 01 p02 21 p01 hi ¼ T 03 2 T 01 " # 6:05 0:286 288 21 1:01 ¼ 494:4 2 288 ¼ 0:938 p2 2:49 £ 105 ¼ RT 2 287 £ 383:59 r2 ¼ r2 ¼ 2:27 kg/m3 Copyright 2003 by Marcel Dekker. 1 The impeller tip speed of a centrifugal compressor is 450 m/s with no prewhirl. (6) A centrifugal compressor that runs at 20. 467 K) A centrifugal compressor running at 15.4%) Derive the expression for the pressure ratio of a centrifugal compressor: P03 hc scU 2 2 ¼ 1þ P01 C p T 01 4. the work input per kg of air. 4556.90 and the isentropic efficiency of the compressor is 0. Take g ¼ 1. and neglect losses.4 4. isentropic efficiency is 0.45 m and rotates at 18.000 rpm has 20 radial vanes. If the pressure ratio is 2 and the impeller tip diameter is 28 cm.434 bar. and inlet temperature of air is 108C.4 (77. power input factor of 1.3 kW) Air with negligible velocity enters the impeller eye of a centrifugal compressor at 158C and 1 bar. The impeller tip diameter is 0.4. calculate the isentropic efficiency of the compressor. calculate the pressure ratio.86.” Copyright 2003 by Marcel Dekker.000 rpm. and the power required for 25 kg/s of airflow. If the slip factor is 0.5 Explain the terms “slip factor” and “power input factor. (5.Centrifugal Compressors and Fans _ m ¼ r2 A2 Cr2 183 ¼ 2p r 2 r2 b2 But U2 ¼ _ pND2 pN m ¼ 60 r2 pCr2 b2 £ 60 475 £ 2:27 £ 246:58 £ 0:0065 £ 60 N¼ 2:5 N ¼ 41476 rpm PROBLEMS 4.5. 182.84 and the inlet stagnation temperature at the impeller eye is 158C.04.6 !g=ðg21Þ 4. Neglect losses and assume g ¼ 1. Inc. Calculate the overall pressure ratio.25 kJ/kg. Assume that the compressor is operating at standard sea level and a power input factor of 1. overall diameter of the impeller is 60 cm.2 4. (4.3 4. Find the pressure and temperature of the air at the compressor outlet.000 rpm. All Rights Reserved . surging and choking in centrifugal compressors. (0. All Rights Reserved . the work input factor is 1.4 kJ/kg.9 What are the three main types of centrifugal compressor impellers? Draw the exit velocity diagrams for these three types.89.12 A double-entry centrifugal compressor designed with no prewhirl has the following dimensions and data: Impeller root diameter: Impeller tip diameter: Rotational speed: Mass flow rate: 0:15 m 0:30 m 15.5. Find the torque and power required to drive this machine. 8821 kW) A single-sided centrifugal compressor designed with no prewhirl has the following dimensions and data: Total head / pressure ratio: Speed: Inlet stagnation pressure: Slip factor: Power input factor: Isentropic efficiency: Mass flow rate: 3:8:1 12. calculate the pressure ratio.88. 232. Inc. (4954 Nm.03.3 m. 000 rpm 18 kg/s Copyright 2003 by Marcel Dekker.11 Inlet stagnation temperature: 293 K Assume an axial entrance. impeller trip radius is 0. Assume T01 ¼ 290 K and Cp ¼ 1. A centrifugal compressor operates with no prewhirl and is run with a tip speed of 475 the slip factor is 0. relative velocity of air at the impeller tip is 105 m/s at an exit angle of 808. 3610 kW) 4.000 rpm and compresses 32 kg of air per second.184 Chapter 4 4.8 4. work input per kg of air and power for 29 airflow. Explain the phenomenon of stalling.693 m.7 4.10 4. (5. compressor efficiency is 0. 000 rpm 1:03 bar 0:9 1:03 0:76 20 kg/s 4.005 kJ/kg K. Assume an axial entrance. Calculate the overall diameter of the impeller and the power required to drive the compressor. 6739 kW) A centrifugal compressor impeller rotates at 17. a1 ¼ 31. Calculate the inlet angles of the impeller vane at the root and tip of the eye. All Rights Reserved .12.Centrifugal Compressors and Fans 185 Ambient temperature: 258C Ambient pressure: Density of air at eye inlet: 1:03 bar 1:19 kg/m3 Assume the axial entrance and unit is stationary. (a1 at root ¼ 65.18.697) NOTATION C r U V a s v c absolute velocity radius impeller speed relative velocity vane angle slip factor angular velocity power input factor SUFFIXES 1 2 3 a r w inlet to rotor outlet from the rotor outlet from the diffuser axial.58. ambient radial whirl Copyright 2003 by Marcel Dekker. a1 at tip ¼ 38. Find the inlet angles of the vane at the root and tip radii of the impeller eye and the maximum Mach number at the eye.79) 4. air does not enter the impeller eye in an axial direction but it is given a prewhirl of 208 (from the axial direction).48 at tip. (a1 at root ¼ 50. 0. 0.78. Inc.13 In Example 4. The remaining values are the same. axial turbines are very similar. There are two important characteristics of the axial flow compressor—high-pressure ratios at good efficiency and thrust per unit frontal area. the maximum pressure ratio achieved in centrifugal compressors is about 4:1 for simple machines (unless multi-staging is used) at an efficiency of about 70– 80%. If the divergence is too rapid. examination of the blade cross-section will indicate a big difference. in the axial flow compressor.1. Inc. The axial flow compressor. it is much more difficult to carry out efficient diffusion due to the breakaway of air molecules from the walls of the diverging passage. the increase in velocity being achieved by the nozzle. The air molecules that break away tend to reverse direction and flow back in the direction of the pressure gradient. there is a natural tendency for the air to fill the passage Copyright 2003 by Marcel Dekker. However. can achieve higher pressures at a higher level of efficiency. As mentioned in the chapter on diffuser design (Chapter 4. the flow is decelerating or diffusing and the pressure rise occurs when the fluid passes through the blades. as shown in Fig. All Rights Reserved . inlet passage area is greater than the outlet. Sec.5 Axial Flow Compressors and Fans 5.1 INTRODUCTION As mentioned in Chapter 4.7). however. this may result in the formation of eddies and reduction in useful pressure rise. 5. The opposite occurs in the compressor. 4. Thus the process in turbine blades can be described as an accelerating flow. In the turbine. Although in overall appearance. During acceleration in a nozzle. Copyright 2003 by Marcel Dekker.3 shows how a few compressor stages are built into the axial compressor. These are provided to guide the air at the correct angle onto the first row of moving blades. each stage being formed by a stationary row and a rotating row of blades. called the inlet guide vanes.188 Chapter 5 Figure 5. Typical blade sections are shown in Fig. the 2-D flow through the stage is very important due to cylindrical symmetry. the volume of air decreases. All Rights Reserved . In the analysis of the highly efficient axial flow compressor. is fitted to the compressor inlet. An extra row of fixed blades.1 Cutaway sketch of a typical axial compressor assembly: the General Electric J85 compressor. it is called an axial flow compressor. Axial flow compressors consist of a number of stages. As the pressure increases in the direction of flow.2. therefore. To keep the air velocity the same for each stage.) walls closely (only the normal friction loss will be considered in this case). Figure 5. The height of the blades is seen to decrease as the fluid moves through the compressor. Modern axial flow compressors may give efficiencies of 86– 90%—compressor design technology is a well-developed field. The rotating blades impart kinetic energy to the air while increasing air pressure and the stationary row of blades redirect the air in the proper direction and convert a part of the kinetic energy into pressure. 5. the blade height is decreased along the axis of the compressor. The flow of air through the compressor is in the direction of the axis of the compressor and. (Courtesy of General Electric Co. Inc. Copyright 2003 by Marcel Dekker. Inc.2 Compressor and turbine blade passages: turbine and compressor housing.3 Schematic of an axial compressor section. All Rights Reserved .Axial Flow Compressors and Fans 189 Figure 5. Figure 5. C2 is larger than C1. The process is carried out over multiple numbers of stages. Air leaves the rotor blade with absolute velocity C2 at an angle a2.2 VELOCITY DIAGRAM The basic principle of axial compressor operation is that kinetic energy is imparted to the air in the rotating blade row.190 Chapter 5 The flow is assumed to take place at a mean blade height. Air passes through the diverging passages formed between the rotor blades. Euler’s equation Figure 5. It is efficient only when the pressure rise per stage is very small. Inc. It is less than V1. 5. V2 is the relative velocity at the rotor outlet. The blading diagram and the velocity triangle for an axial flow compressor stage are shown in Fig. As work is done on the air in the rotor blades. No flow is assumed in the radial direction. Combining the two velocity vectors gives the relative velocity at inlet V1 at an angle b1.4. Air enters the rotor blade with absolute velocity C1 at an angle a1 measured from the axial direction. diffusion is a deceleration process. The rotor row has tangential velocity U. where the blade peripheral velocities at the inlet and outlet are the same. showing diffusion of the relative velocity has taken place with some static pressure rise across the rotor blades.4 Velocity diagrams for a compressor stage. All Rights Reserved . and then diffused through passages of both rotating and stationary blades. Copyright 2003 by Marcel Dekker. Turning of the air towards the axial direction is brought about by the camber of the blades. As mentioned earlier. 5. Thus the actual temperature rise of the air is given by: T 0s ¼ lUC a ðtan b1 2 tan b2 Þ Cp ð5:7Þ If Rs is the stage pressure ratio and hs is the stage isentropic efficiency. the following basic equations can be written: U ¼ tan a1 þ tan b1 Ca U ¼ tan a2 þ tan b2 Ca ð5:2Þ ð5:1Þ ð5:3Þ in which Ca ¼ Ca1 ¼ C2 is the axial velocity. KT0s. then: Rs ¼ 1 þ hs DT 0s T 01 !g=ðg21Þ ð5:8Þ where T01 is the inlet stagnation temperature. If the velocity of air leaving the first stage C3 is made equal to C1. a factor l. W c ¼ UC a ðtan b1 2 tan b2 Þ ð5:5Þ ð5:4Þ The whole of this input energy will be absorbed usefully in raising the pressure and velocity of the air and for overcoming various frictional losses. All Rights Reserved . then the stagnation temperature rise will be equal to the static temperature rise. all the energy is used to increase the stagnation temperature of the air.6) is the theoretical temperature rise of the air in one stage.1)] may be written in terms of air angles: W c ¼ UC a ðtan a2 2 tan a1 Þ also. To find the actual temperature rise of the air. The work done equation [Eq.Axial Flow Compressors and Fans 191 provides the work done on the air: W c ¼ UðCw2 2 C w1 Þ Using the velocity triangles. (5. assumed constant through the stage. KTs. will be used. In reality. the stage temperature rise will be less than this value due to 3-D effects in the compressor annulus. Copyright 2003 by Marcel Dekker. Inc. Regardless of the losses. Hence: T 0s ¼ DT s ¼ UC a ðtan b1 2 tan b2 Þ Cp ð5:6Þ Equation (5. which is between 0 and 100%. Inc. L. is defined as: L¼ Static enthalpy rise in the rotor Static enthalpy rise in the whole stage ð5:9Þ The degree of reaction indicates the distribution of the total pressure rise into the two types of blades.11). C p DT A ¼ UC a ðtan a2 2 tan a1 Þ 2 1 C2 ðsec2 a2 2 sec2 a1 Þ 2 a ¼ UC a ðtan a2 2 tan a1 Þ 2 1 C2 ðtan 2 a2 2 tan 2 a1 Þ 2 a Using the definition of degree of reaction. (5. All Rights Reserved . C 2 ¼ C a cos a2 Therefore. we have: 1 W c ¼ C p DT A þ ðC 2 2 C 2 Þ 1 2 2 Combining Eqs. Let: DT A ¼ the static temperature rise in the rotor DT B ¼ the static temperature rise in the stator Using the work input equation [Eq. The choice of a particular degree of reaction is important in that it affects the velocity triangles. we get: 1 C p DT A ¼ UC a ðtan a2 2 tan a1 Þ 2 ðC 2 2 C 2 Þ 1 2 2 from the velocity triangles. we get: W c ¼ Cp ðDT A þ DT B Þ ¼ DT S ) ¼ UC a ðtan b1 2 tan b2 Þ ¼ UC a ðtan a2 2 tan a1 Þ ð5:10Þ But since all the energy is transferred to the air in the rotor. (5.10) and (5. using the steady flow energy equation.4)]. L¼ ¼ DT A DT A þ DT B UC a ðtan a2 2 tan a1 Þ 2 1 C 2 ðtan 2 a2 2 tan 2 a1 Þ 2 a UC a ðtan a2 2 tan a1 Þ and C 1 ¼ C a cos a1 ð5:11Þ ¼ 1 2 Ca ðtan a2 þ tan a1 Þ U Copyright 2003 by Marcel Dekker. the fluid friction and other losses.3 DEGREE OF REACTION The degree of reaction.192 Chapter 5 5. In Eq. (h2 2 h1) ¼ (h3 2 h1). (5. 5. the ratio of axial velocity to blade velocity is called the flow coefficient and denoted by F.12). Copyright 2003 by Marcel Dekker. 2U ¼ ðtan a1 þ tan b1 þ tan a2 þ tan b2 Þ Ca Therefore.5 Stage reaction.3). a1 ¼ b 2 a2 ¼ b 1 As we have assumed that Ca is constant through the stage. For a reaction ratio of 50%.1) and (5. which leads to this interesting result: ðtan b1 þ tan b2 Þ ¼ U : Ca Again using Eqs. tan a1 ¼ tan b2 . Ca ¼ C 1 cos a1 ¼ C 3 cos a3 : Since we know C1 ¼ C3. b2 is chosen to be greater than a2 (Fig. it follows that a1 ¼ a3. and b1 ¼ a2. C a 2U 2U L¼ 2 þ tan b1 þ tan b2 Ca 2U C a ¼ Ca ðtan b1 þ tan b2 Þ 2U ð5:12Þ Usually the degree of reaction is set equal to 50%. then the static pressure rise in the rotor is greater than the static pressure rise in the stator and the reaction is greater than 50%. which implies the static enthalpy and the temperature increase in the rotor and stator are equal. (5. tan b1 ¼ tan a2 . (5.Axial Flow Compressors and Fans 193 But from the velocity triangles. i:e:. Under these conditions.2). All Rights Reserved .2) and (5. Figure 5. Inc. If for a given value of C a =U. the velocity triangles become symmetric. adding Eqs.5). i:e:. Because the angles are equal. a1 ¼ b2 ¼ a3. so lift-and-drag forces will be set up on the blade while the forces on the air will act on the opposite direction. with relative velocity vectors V1 and V2 at angles b1 and b2. the stator pressure rise will be greater and the reaction is less than 50%. All Rights Reserved . Copyright 2003 by Marcel Dekker. if the designer chooses b2 less than b1.6 Lift-and-drag forces on a compressor rotor blade.5 LIFT-AND-DRAG COEFFICIENTS The stage-loading factor C may be expressed in terms of the lift-and-drag coefficients.194 Chapter 5 Conversely. The flow on the rotor blade is similar to flow over an airfoil. Consider a rotor blade as shown in Fig. ð5:14Þ Figure 5.4 STAGE LOADING The stage-loading factor C is defined as: Wc h03 2 h01 ¼ mU 2 U2 lðC w2 2 C w1 Þ ¼ U lC a ðtan a2 2 tan a1 Þ ¼ U C ¼ lF ðtan a2 2 tan a1 Þ C¼ ð5:13Þ 5. 5.6. The tangential force on each moving blade is: F x ¼ L cos bm þ D sin bm ! CD F x ¼ L cos bm 1 þ tan bm CL where: L ¼ lift and D ¼ drag. Let tan ðbm Þ ¼ ðtan ðb1 Þ þ tan ðb2 ÞÞ/2. 5. Inc. Substituting this back into Eq. CD is much smaller than CL. (5. and therefore the optimal blade-loading factor is approximated by: w c CL Copt ¼ pffiffiffi ð5:20Þ 2 s ð5:17Þ ð5:15Þ ð5:16Þ 5. Therefore. h03 2 h01 U2 Fx ¼ rCa lsU 1 C a c sec bm ðC L þ C D tan bm Þ ¼ 2 U s ð5:18Þ 1 c ¼ sec bm ðCL þ CD tan bm Þ 2 s For a stage in which bm ¼ 458.18).7. ! rC 2 clC L CD sec bm 1 þ Fx ¼ a tan bm 2 CL CL ¼ The power delivered to the air is given by: UF x ¼ mðh03 2 h01 Þ ¼ rCa lsðh03 2 h01 Þ considering the flow through one blade passage of width s.6 CASCADE NOMENCLATURE AND TERMINOLOGY Studying the 2-D flow through cascades of airfoils facilitates designing highly efficient axial flow compressors. shows the standard nomenclature relating to airfoils in cascade. the optimal blade-loading factor is given by: w c Copt ¼ pffiffiffi ðC L þ C D Þ ð5:19Þ 2 s ¼ For a well-designed blade. All Rights Reserved . Ca Substituting V m ¼ cosbm into the above equation. which is reproduced from Howell’s early paper on cascade theory and performance.Axial Flow Compressors and Fans 195 The lift coefficient is defined as: L 0:5rV 2 A m where the blade area is the product of the chord c and the span l. Inc. efficiency will be maximum. A cascade is a row of geometrically similar blades arranged at equal distance from each other and aligned to the flow direction. Figure 5. Copyright 2003 by Marcel Dekker. is 0 then fixed by the choice of a suitable air inlet angle a1. is the difference between the entry and exit air angles. An appropriate setting of the turntable on which the cascade is mounted can accomplish this. The deviation angle (d) À 0Á is the difference between the air outlet angle (a2) and the blade outlet angle a2 . That is. The incidence angle i is the difference between the air inlet angle À 0Á 0 (a1) and the blade inlet angle a1 . The air deflection angle. and the pitch 0 0 (or space) s will be fixed and the blade inlet and outlet angles a1 and a2 are determined by the chosen setting or stagger angle j. the blade camber angle u. 5. Inc. The chord c is the length of the perpendicular of the blade profile onto the chord line.7 Cascade nomenclature. For any particular test.7. i ¼ a1 2 a1 . The angle of incidence. a1 0 and a2 0 are the camber angles of the entry and exit tangents the camber 0 0 line makes with the axial direction. The results of the traverses are usually presented as shown Copyright 2003 by Marcel Dekker. i. A cross-section of three blades forming part of a typical cascade is shown in Fig. 1 ¼ a1 2 a2. All Rights Reserved . its chord c. With the cascade in this position the pressure and direction measuring instruments are then traversed along the blade row in the upstream and downstream position. The pitch s is the distance in the direction of rotation between corresponding points on adjacent blades. since i ¼ a1 2 a1 .196 Chapter 5 Figure 5. The blade camber angle u ¼ a1 2 a2 . The stagger angle j is the angle between the chord line and the axial direction and represents the angle at which the blade is set in the cascade. It is approximately equal to the linear distance between the leading edge and the trailing edge. The total pressure loss owing to the increase in deflection angle of air is marked when i is increased beyond a particular value.9. in which the mean loss and mean deflection are plotted against incidence for a cascade of fixed geometrical form. The stagnation pressure loss is plotted as a dimensionless number given by: Stagnation pressure loss coefficient ¼ P01 2 P02 0:5rC2 1 ð5:21Þ This shows the variation of loss of stagnation pressure and the air deflection. Reducing the incidence i generates a negative angle of incidence at which stalling will occur. covering two blades at the center of the cascade.Axial Flow Compressors and Fans 197 Figure 5. The stalling incidence of the cascade is the angle at which the total pressure loss is twice the minimum cascade pressure loss. and the whole set of results condensed to the form shown in Fig.8 can now be repeated for different values of incidence angle. 5. 5. 5. Inc. Howell has defined nominal conditions of deflection for Copyright 2003 by Marcel Dekker. 1 ¼ a1 2 a2. in Fig.8.8 Variation of stagnation pressure loss and deflection for cascade at fixed incidence. Knowing the limits for air deflection without very high (more than twice the minimum) total pressure loss is very useful for designers in the design of efficient compressors. The curves of Fig. All Rights Reserved . All Rights Reserved . 2 Copyright 2003 by Marcel Dekker.198 Chapter 5 Figure 5. and for the inlet guide vane in front of the 2 compressor. nominal deviation is given by: s1 2 d* ¼ mu ð5:24Þ l The approximate value suggested by Constant is 0. Inc.9 Cascade mean deflection and pressure loss curves. and Howell suggested a modified value for m: * 2 2a a m ¼ 0:23 þ0:1 2 ð5:25Þ l 50 where the maximum camber of the cascade airfoil is at a distance a from the leading edge and a* is the nominal air outlet angle. The relation is given by: sn ð5:23Þ d* ¼ mu l For compressor cascade. n ¼ 1. a cascade as 80% of its stalling deflection. n ¼ 1. Hence. for a compressor cascade. Howell and Constant also introduced a relation correlating nominal deviation d* with pitch chord ratio and the camber of the blade. that is: 1* ¼ 0:81s ð5:22Þ where 1s is the stalling deflection and 1* is the nominal deflection for the cascade.26. Now.” The flow through an axial compressor is vortex flow in nature. there is little variation in blade speed from root to tip. a* 2 a* ¼ 1* 1 2 or: a* ¼ a* þ 1* 1 2 Also. resolve Copyright 2003 by Marcel Dekker. 5. The shape of the velocity diagram does not change much and. it is important to consider this in the design. In an axial flow compressor in which high hub/tip radius ratio exists on the order of 0. For hub/tip ratios lower than 0. Such compressors.7 3-D CONSIDERATION So far.Axial Flow Compressors and Fans 199 Then.10. The blading is of the same section at all radii and the performance of the compressor stage is calculated from the performance of the blading at the mean radial section.8. i* ¼ a* 2 b1 ¼ a* þ 1* 2 b1 1 2 5. That is. having long blades relative to the mean diameter. In the case of a compressor with high hub/tip ratio. therefore.8. little variation in pressure occurs along the length of the blade. it is said to have “vorticity. The shape of the velocity triangle will influence the blade geometry. The rotating fluid is subjected to a centrifugal force and to balance this force. and. Actually. 2-D flow in the compressor annulus is a fairly reasonable assumption. the assumption of two-dimensional flow is no longer valid. a radial pressure gradient is necessary. All Rights Reserved . a* ¼ b2 þ d * 2 s1 2 ¼ b2 þ m u l and.10. as shown in Fig. therefore. have been used in aircraft applications in which a high mass flow requires a large annulus area but a small blade tip must be used to keep down the frontal area. Whenever the fluid has an angular velocity as well as velocity in the direction parallel to the axis of rotation. The flow along the compressor is considered to be 2-D. Let us consider the pressure forces on a fluid element as shown in Fig. Inc. all the above discussions were based on the velocity triangle at one particular radius of the blading. 5. there is a considerable difference in the velocity diagram between the blade hub and tip sections. in 2-D flow only whirl and axial flow velocities exist with no radial velocity component. 200 Chapter 5 Figure 5.10 Variation of velocity diagram along blade. the forces in the radial direction Fig. 5.11: dP du dr du ðP þ dPÞðr þ drÞ 2 Pr du 2 2 P þ 2 2 ¼ r dr r du or C2 w r ð5:26Þ dP ðP þ dPÞðr þ drÞ 2 Pr 2 P þ dr ¼ r dr C 2 w 2 where: P is the pressure, r, the density, Cw, the whirl velocity, r, the radius. After simplification, we get the following expression: 1 Pr þ P dr þ r dP þ dP dr 2 Pr þ r dr 2 dP dr ¼ r dr C 2 w 2 or: r dP ¼ r dr C2 w Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Axial Flow Compressors and Fans 201 Figure 5.11 Pressure forces on a fluid element. That is, 1 dP C 2 ¼ w r dr r ð5:27Þ The approximation represented by Eq. (5.27) has become known as radial equilibrium. The stagnation enthalpy h0 at any radius r where the absolute velocity is C may be rewritten as: À 1 1 h0 ¼ h þ C 2 þ C 2 ; h ¼ cp T; a w 2 2 and C 2 ¼ C2 þ C2 a w Á Differentiating the above equation w.r.t. r and equating it to zero yields: dh0 g 1 dP 1 dC w þ 0 þ 2C w ¼ £ g 2 1 r dr 2 dr dr Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 202 Chapter 5 or: g 1 dP dCw þ Cw £ ¼0 dr g 2 1 r dr Combining this with Eq. (5.27): g C2 dC w w þ Cw ¼0 g21 r dr or: dC w g Cw ¼2 g21 r dr Separating the variables, dC w g dr ¼2 Cw g21 r Integrating the above equation Z R dCw g dr ¼2 Cw g21 r g ln C w r ¼ c where c is a constant: 2 g21 Taking antilog on both sides, g £ Cw £ r ¼ e c g21 Therefore, we have C w r ¼ constant ð5:28Þ Equation (5.28) indicates that the whirl velocity component of the flow varies inversely with the radius. This is commonly known as free vortex. The outlet blade angles would therefore be calculated using the free vortex distribution. 5.8 MULTI-STAGE PERFORMANCE An axial flow compressor consists of a number of stages. If R is the overall pressure ratio, Rs is the stage pressure ratio, and N is the number of stages, then the total pressure ratio is given by: R ¼ ðRs ÞN ð5:29Þ Equation (5.29) gives only a rough value of R because as the air passes through the compressor the temperature rises continuously. The equation used to Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Axial Flow Compressors and Fans 203 find stage pressure is given by: !g hs DT 0s g21 Rs ¼ 1 þ T 01 ð5:30Þ The above equation indicates that the stage pressure ratio depends only on inlet stagnation temperature T01, which goes on increasing in the successive stages. To find the value of R, the concept of polytropic or small stage efficiency is very useful. The polytropic or small stage efficiency of a compressor is given by: g21 n h1;c ¼ n21 g or: n g ¼ hs n21 g21 where hs ¼ h1,c ¼ small stage efficiency. The overall pressure ratio is given by: !n NDT 0s n21 R¼ 1þ T 01 ð5:31Þ Although Eq. (5.31) is used to find the overall pressure ratio of a compressor, in actual practice the step-by-step method is used. 5.9 AXIAL FLOW COMPRESSOR CHARACTERISTICS The forms of characteristic curves of axial flow compressors are shown in Fig. 5.12. These curves are quite similar to the centrifugal compressor. However, axial flow compressors cover a narrower range of mass flow than the centrifugal compressors, and the surge line is also steeper than that of a centrifugal compressor. Surging and choking limit the curves at the two ends. However, the surge points in the axial flow compressors are reached before the curves reach a maximum value. In practice, the design points is very close to the surge line. Therefore, the operating range of axial flow compressors is quite narrow. Illustrative Example 5.1: In an axial flow compressor air enters the compressor at stagnation pressure and temperature of 1 bar and 292K, respectively. The pressure ratio of the compressor is 9.5. If isentropic efficiency of the compressor is 0.85, find the work of compression and the final temperature at the outlet. Assume g ¼ 1.4, and Cp ¼ 1.005 kJ/kg K. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 204 Chapter 5 Figure 5.12 Axial flow compressor characteristics. Solution: T 01 ¼ 292K; P01 ¼ 1 bar; hc ¼ 0:85: Using the isentropic P – T relation for compression processes, g 0 ! P02 T 02 g21 ¼ P01 T 01 where T 020 is the isentropic temperature at the outlet. Therefore, !g 21 P02 g 0 ¼ 292ð9:5Þ0:286 ¼ 555:92 K T 02 ¼ T 01 P01 Now, using isentropic efficiency of the compressor in order to find the actual temperature at the outlet, À 0 Á T 02 2 T 01 ð555:92 2 292Þ T 02 ¼ T 01 þ ¼ 602:49 K ¼ 292 þ 0:85 hc Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Axial Flow Compressors and Fans 205 Work of compression: W c ¼ Cp ðT 02 2 T 01 Þ ¼ 1:005ð602:49 2 292Þ ¼ 312 kJ/kg Illustrative Example 5.2: In one stage of an axial flow compressor, the pressure ratio is to be 1.22 and the air inlet stagnation temperature is 288K. If the stagnation temperature rise of the stages is 21K, the rotor tip speed is 200 m/s, and the rotor rotates at 4500 rpm, calculate the stage efficiency and diameter of the rotor. Solution: The stage pressure ratio is given by: !g hs DT 0s g21 Rs ¼ 1 þ T 01 or 1:22 ¼ 1 þ that is, hs ð21Þ 288 !3:5 hs ¼ 0:8026 or 80:26% The rotor speed is given by: U¼ pDN ; 60 or D ¼ ð60Þð200Þ ¼ 0:85 m pð4500Þ Illustrative Example 5.3: An axial flow compressor has a tip diameter of 0.95 m and a hub diameter of 0.85 m. The absolute velocity of air makes an angle of 288 measured from the axial direction and relative velocity angle is 568. The absolute velocity outlet angle is 568 and the relative velocity outlet angle is 288. The rotor rotates at 5000 rpm and the density of air is 1.2 kg/m3. Determine: 1. 2. 3. 4. 5. The The The The The axial velocity. mass flow rate. power required. flow angles at the hub. degree of reaction at the hub. Solution: 1. Rotor speed is given by: U¼ pDN pð0:95Þð5000Þ ¼ ¼ 249 m/s 60 60 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 206 Chapter 5 Figure 5.13 Inlet velocity triangle. Blade speed at the hub: pDh N pð0:85Þð5000Þ ¼ ¼ 223 m/s 60 60 From the inlet velocity triangle (Fig. 5.13), Uh ¼ tan a1 ¼ C w1 Ca and tan b1 ¼ ðU 2 C w1 Þ Ca Adding the above two equations: U ¼ tan a1 þ tan b1 Ca or: U ¼ C a ðtan 288 þ tan 568Þ ¼ C a ð2:0146Þ 2. Therefore, Ca ¼ 123.6 m/s (constant at all radii) The mass flow rate: _ m ¼ pðr 2 2 r 2 Þr C a t h ¼ pð0:4752 2 0:4252 Þð1:2Þð123:6Þ ¼ 20:98 kg/s 3. The power required per unit kg for compression is: W c ¼ lUC a ðtan b1 2 tan b2 Þ ¼ ð1Þð249Þð123:6Þðtan 568 2 tan 288 Þ1023 ¼ ð249Þð123:6Þð1:483 2 0:53Þ ¼ 29:268 kJ/kg Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Axial Flow Compressors and Fans 207 The total power required to drive the compressor is: _ W c ¼ mð29:268Þ ¼ ð20:98Þð29:268Þ ¼ 614 kW 4. At the inlet to the rotor tip: C w1 ¼ C a tan a1 ¼ 123:6 tan 288 ¼ 65:72 m/s Using free vortex condition, i.e., Cwr ¼ constant, and using h as the subscript for the hub, Cw1h ¼ C w1t rt 0:475 ¼ 73:452 m/s ¼ ð65:72Þ rh 0:425 At the outlet to the rotor tip, Cw2t ¼ C a tan a2 ¼ 123:6 tan 568 ¼ 183:24 m/s Therefore, Cw2h ¼ C w2t rt 0:475 ¼ 204:8 m/s ¼ ð183:24Þ 0:425 rh Hence the flow angles at the hub: tan a1 ¼ tan b1 ¼ i.e., b1 ¼ 50.438 tan a2 ¼ i.e., a2 ¼ 58.898 tan b2 ¼ ðU h Þ 223 2 tan 58:598 ¼ 0:1472 2 tan a2 ¼ Ca 123:6 C w2h 204:8 ¼ 1:657 ¼ Ca 123:6 C w1h 73:452 ¼ 0:594 or; a1 ¼ 30:728 ¼ Ca 123:6 ðU h Þ 223 2 0:5942 ¼ 1:21 2 tan a1 ¼ Ca 123:6 5. i.e., b2 ¼ 8.378 The degree of reaction at the hub is given by: Lh Ca 123:6 ðtan 50:438 þ tan 8:378Þ ðtan b1 þ tan b2 Þ ¼ 2U h ð2Þð223Þ 123:6 ¼ ð1:21 þ 0:147Þ ¼ 37:61% ð2Þð223Þ ¼ Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved tan a1 ¼ that is. (5.4: An axial flow compressor has the following data: Blade velocity at root: Blade velocity at mean radius: Blade velocity at tip: Stagnation temperature rise in this stage: 140 m/s 185 m/s 240 m/s 15K Axial velocity ðconstant from root to tipÞ: 140 m/s Work done factor: Degree of reaction at mean radius: 0:85 50% Calculate the stage air angles at the root. mean. From the velocity triangle at the mean. Solution: Calculation at mean radius: From Eq.208 Chapter 5 Illustrative Example 5. 5.1). All Rights Reserved . Inc. and tip for a free vortex design. U ¼ DCw þ 2C w1 or: Cw1 ¼ Hence.14) at the mean radius is 50%. C w1 44:57 ¼ 0:3184 ¼ Ca 140 U 2 DC w 185 2 95:87 ¼ ¼ 44:57 m/s 2 2 a1 ¼ 17:668 ¼ b2 Copyright 2003 by Marcel Dekker. Wc ¼ U(Cw2 2 Cw1) ¼ UKCw or: Cp ðT 02 2 T 01 Þ ¼ C p DT 0s ¼ lUDCw So: DC w ¼ C p DT 0s ð1005Þð15Þ ¼ ¼ 95:87 m/s lU ð0:85Þð185Þ Since the degree of reaction (Fig. a1 ¼ b2 and a2 ¼ b1. All Rights Reserved .14 Velocity triangle at the mean radius.e. ðDC w £ UÞt ¼ ðDCw £ UÞm Therefore. 5. Inc. DCw ¼ ð95:87Þð185Þ ¼ 73:9 m/s 240 Whirl velocity component at the tip: C w1 £ 240 ¼ ð44:57Þð185Þ Therefore: C w1 ¼ ð44:57Þð185Þ ¼ 34:36 m/s 240 Cw1 34:36 ¼ 0:245 ¼ 140 Ca tan a1 ¼ Copyright 2003 by Marcel Dekker. b1 ¼ 45:098 ¼ a2 Calculation at the blade tip: Using the free vortex diagram (Fig. and tan b1 ¼ ðDCw þ Cw1 Þ ð95:87 þ 44:57Þ ¼ 1:003 ¼ Ca 140 i..15).Axial Flow Compressors and Fans 209 Figure 5. Therefore.758 tan a2 ¼ i. a1 ¼ 13:798 From the velocity triangle at the tip..15 Velocity triangles at tip.268 Calculation at the blade root: ðDCw £ UÞr ¼ ðDC w £ UÞm Copyright 2003 by Marcel Dekker. Inc.210 Chapter 5 Figure 5.e...e. a2 ¼ 37.718 tan b2 ¼ x2 131:74 ¼ 0:941 ¼ Ca 140 ðC w1 þ DCw Þ ð34:36 þ 73:9Þ ¼ 0:7733 ¼ Ca 140 DCw þ x2 73:9 þ 131:74 ¼ 1:469 ¼ Ca 140 i. All Rights Reserved . b1 ¼ 55.e. x2 þ DC w þ Cw1 ¼ U or: x2 ¼ U 2 DC w 2 Cw1 ¼ 240 2 73:9 2 34:36 ¼ 131:74 tan b1 ¼ i. b2 ¼ 43. root ¼ tan a1 ¼ ð108:24Þð240Þ ¼ 185:55 m/s 140 and C w1 ¼ 58:9 m/s and DCw ¼ 126:69 m/s 58:9 ¼ 0:421 140 i.828 From the velocity triangle at the blade root. a1 ¼ 22.16 Velocity triangles at root. Inc. Copyright 2003 by Marcel Dekker..e.Axial Flow Compressors and Fans 211 or: DCw £ 140 ¼ ð95:87Þð185Þ Also: ðC w1 £ UÞr ¼ ðC w1 £ UÞm or: C w1 £ 140 ¼ ð44:57Þð185Þ and ðC w2 £ UÞt ¼ ðC w2 £ UÞr so: C w2. 5. (Fig.tip ¼ C a tana2 ¼ 140 tan 37:718 ¼ 108:24 m/s Therefore: C w2.16) or: x2 ¼ Cw2 2 U ¼ 185:55 2 140 ¼ 45:55 Figure 5. All Rights Reserved . 212 Chapter 5 Therefore: tan b1 ¼ i. b2 ¼ 2 188 Design Example 5.968 U 2 Cw1 140 2 58:9 ¼ 0:579 ¼ Ca 140 C w2 185:55 ¼ 1:325 ¼ 140 Ca x2 45:55 ¼ 20:325 ¼2 Ca 140 tan b2 ¼ 2 i.e. or 12:7% ¼ ð2Þð140Þ Reaction at the blade tip: Ltip ¼ Ca 140 ðtan 55:758 þ tan 43:268 Þ ðtan b1t þ tan b2t Þ ¼ 2U t ð2Þð240Þ 140 ð1:469 þ 0:941Þ ¼ 0:7029..e. Solution: Reaction at the blade root: Lroot ¼ Ca 140 ðtan b1r þ tan b2r Þ ¼ ðtan 30:088 þ tan ð2188 ÞÞ 2U r ð2Þð140Þ 140 ð0:579 2 0:325Þ ¼ 0:127.088 tan a2 ¼ i.. b1 ¼ 30. calculate the degree of reaction at the blade root and tip.5: From the data given in the previous problem. Inc. or 70:29% ¼ ð2Þð240Þ Illustrative Example 5..e. Copyright 2003 by Marcel Dekker.6: An axial flow compressor stage has the following data: Air inlet stagnation temperature: Flow coefficient: Relative inlet Mach number: Degree of reaction: 295K 0:56 0:78 0:5 Blade angle at outlet measured from the axial direction: 328 Find the stagnation temperature rise in the first stage of the compressor. a2 ¼ 52. All Rights Reserved . the velocity triangle is symmetric as shown in Fig.6. Inc. Copyright 2003 by Marcel Dekker.17 Combined velocity triangles for Example 5. for the relative Mach number at the inlet: M r1 ¼ À or: V1 gRT 1 Á1 2 C2 V 2 ¼ gRM 2 T 01 2 1 1 r1 2Cp Ca .Axial Flow Compressors and Fans 213 Solution: Since the degree of reaction is 50%. V1 ¼ and: and C1 ¼ a1 ¼ b2 ðsince L ¼ 0:5Þ Therefore: C1 ¼ Ca Ca ¼ cos328 0:848 Figure 5.e. cosb1 Ca cosa1 From the velocity triangle. b1 ¼ 49.12)]: L¼ Therefore: tanb1 ¼ 2L 2 tan 328 ¼ 1:16 0:56 Ca ðtan b1 þ tan b2 Þ and 2U w¼ Ca ¼ 0:56 U i. (5. All Rights Reserved . Using the degree of reaction equation [Eq. 5.17..248 Now. C2 a C2 a ¼ 104:41 295 2 . 1445 so : C a ¼ 169:51 m/s The stagnation temperature rise may be calculated as: T 02 2 T 01 ¼ ¼ C2 a ðtan b1 2 tan b2 Þ Cp w 169:512 ðtan 49:248 2 tan 328Þ ¼ 27:31K ð1005Þð0:56Þ Design Example 5.214 Chapter 5 and: V1 ¼ Hence: C2 ¼ 1 C2 a .7: An axial flow compressor has the following design data: Inlet stagnation temperature: Inlet stagnation pressure: Stage stagnation temperature rise: Mass flow of air: Axialvelocity through the stage: Rotational speed: Work done factor: Mean blade speed: Reaction at the mean radius: 290K 1 bar 24K 22kg/s 155:5m/s 152rev/s 0:93 205m/s 50% Determine: (1) the blade and air angles at the mean radius. Copyright 2003 by Marcel Dekker. 0:719 and V2 ¼ 1 C2 a 0:426 Ca Ca ¼ cos 49:248 0:653 Substituting for V1 and C1. (2) the mean radius. and (3) the blade height. All Rights Reserved . Inc. Axial Flow Compressors and Fans 215 Solution: (1) The following equation provides the relationship between the temperature rise and the desired angles: T 02 2 T 01 ¼ or: 24 ¼ so: tan b1 2 tan b2 ¼ 0:814 Using the degree of reaction equation: L¼ Hence: ð0:5Þð2Þð205Þ ¼ 1:318 155:5 Solving the above two equations simultaneously for b1 and b2. h. rm. so : b2 ¼ 14:148 ¼ a1 (2) The mean radius. Inc. C1 ¼ Ca 155:5 ¼ 160:31 m/s ¼ cosa1 cos14:148 C2 160:312 1 ¼ 277:21 K ¼ 290 2 2C p ð2Þð1005Þ T 1 ¼ T 01 2 Copyright 2003 by Marcel Dekker. tan b1 þ tan b2 ¼ 2 tan b1 ¼ 2:132. where A is the annular area of the flow. All Rights Reserved . is given by: m ¼ rACa. is given by: rm ¼ U 205 ¼ ¼ 0:215m 2pN ð2pÞð152Þ Ca ðtan b1 þ tan b2 Þ 2U lUC a ðtan b1 2 tan b2 Þ Cp ð0:93Þð205Þð155:5Þ ðtan b1 2 tan b2 Þ 1005 (3) The blade height. so : b1 ¼ 46:838 ¼ a2 ðsince the degree of reaction is 50%Þ and: tan b2 ¼ 1:318 2 tan 46:838 ¼ 1:318 2 1:066. The axial velocity is 158 m/s and is constant through the stage.8: An axial flow compressor has an overall pressure ratio of 4. Inc. calculate the number of stages required.216 Chapter 5 Using the isentropic P– T relation: P1 ¼ P01 Static pressure: 277:21 3:5 ¼ 0:854 bar P1 ¼ ð1Þ 290 Then: T1 T 01 g g21 r1 ¼ P1 ð0:854Þð100Þ ¼ 1:073 kg/m3 ¼ RT 1 ð0:287Þð277:21Þ From the continuity equation: A¼ 22 ¼ 0:132m2 ð1:073Þð155:5Þ and the blade height: h¼ A 0:132 ¼ 0:098m ¼ 2pr m ð2pÞð0:215Þ Illustrative Example 5. the relative outlet velocity component in the x-direction is given by: V x2 ¼ C a tan b2 ¼ 158tan 308 ¼ 91:22 m/s  Ã1 V 1 ¼ C2 ¼ ðU 2 V x2 Þ2 þ C2 2 a  Ã1 ¼ ð245 2 91:22Þ2 þ 1582 2 ¼ 220:48 m/s cos b1 ¼ so: b1 ¼ 44.238 Ca 158 ¼ 0:7166 ¼ V 1 220:48 Copyright 2003 by Marcel Dekker. From the velocity triangles. Solution: Since the degree of reaction at the mean radius is 50%. and a mean blade speed of 245 m/s. Assume T01 ¼ 290K. All Rights Reserved .5:1. If the polytropic efficiency is 87%. Each stage is of 50% reaction and the relative air angles are the same (308) for each stage. a1 ¼ b2 and a2 ¼ b1. DT 0s ¼ ¼ UC a ðtan b1 2 tan b2 Þ Cp ð245Þð158Þ ðtan 44:238 2 tan 308Þ ¼ 15:21K 1005 Number of stages !n NDT 0s n21 R¼ 1þ T 01 n g 1:4 ¼ 0:87 ¼ h1 ¼ 3:05 n21 g21 0:4 Substituting: N15:21 4:5 ¼ 1 þ 290 Therefore.Axial Flow Compressors and Fans 217 Stagnation temperature rise in the stage. À Á1 À Á1 C w1 ¼ C2 2 C2 2 ¼ 1852 2 1802 2 ¼ 42:72 m/s 1 a Copyright 2003 by Marcel Dekker.18)..e. From the inlet velocity triangle.9: In an axial flow compressor. calculate the air angles at the rotor inlet and outlet and the static temperature at the inlet of the first stage and stage pressure ratio.358 ¼ b2 From the same velocity triangle. a1 ¼ 13. the work done factor is 0. Assume a rotor speed of 200 m/s. a1 ¼ b2 and a2 ¼ b1. air enters at a stagnation temperature of 290K and 1 bar. and the degree of reaction is 50%. the absolute velocity at the inlet is 185 m/s.86. The axial velocity of air is 180 m/s (constant throughout the stage). If the stage efficiency is 0. N ¼ 12 stages: Design Example 5. All Rights Reserved . 5. Solution: For 50% degree of reaction at the mean radius (Fig. Inc.86. cos a1 ¼ C a 180 ¼ 0:973 ¼ C 1 185 !3:05 i. (b) outlet. Therefore.10: Find the isentropic efficiency of an axial flow compressor from the following data: Pressure ratio: Polytropic efficiency: Inlet temperature: 6 0:85 285 K Copyright 2003 by Marcel Dekker. All Rights Reserved .218 Chapter 5 Figure 5.. b1 ¼ 41.e. tan b1 ¼ ðU 2 C w1 Þ ð200 2 42:72Þ ¼ 0:874 ¼ Ca 180 i. Inc.158 ¼ a2 Static temperature at stage inlet may be determined by using stagnation and static temperature relationship as given below: T 1 ¼ T 01 2 C1 1852 ¼ 273 K ¼ 290 2 2Cp 2ð1005Þ Stagnation temperature rise of the stage is given by DT 0s ¼ ¼ Á lUC a À tanb1 2 tanb2 Cp 0:86ð200Þð180Þ ð0:874 2 0:237Þ ¼ 19:62 K 1005 Stage pressure ratio is given by ! ! hs DT 0s g=g21 0:86 £ 19:62 3:5 Rs ¼ 1 þ ¼ 1þ ¼ 1:22 290 T 01 Illustrative Example 5.18 Velocity triangles (a) inlet. with no inlet guide vanes. The tip radius and corresponding rotor air angles. calculate 1.c 1:4ð0:85Þ Actual temperature rise: ðn21Þ=n p02 T 02 ¼ T 01 ¼ 285ð6Þ0:336 ¼ 520:36 K p01 The compressor isentropic efficiency is given by: T 020 2 T 01 475:77 2 285 ¼ 0:8105. 4. g 21 P02 g ¼ 285ð6Þ0:286 ¼ 475:77 K T 020 ¼ T 01 P01 Using the polytropic P– T relation for the compression process: n21 g21 0:4 ¼ ¼ 0:336 ¼ n gh1.11: In an axial flow compressor air enters the compressor at 1 bar and 290K. Assuming an inlet velocity of 145 m/s. All Rights Reserved . The first stage of the compressor is designed on free vortex principles.96. if the Mach number relative to the tip is limited to 0. is given by T 01 ¼ T 1 þ or T 1 ¼ T 01 2 C1 1452 ¼ 288:9K ¼ 290 2 2Cp 2 £ 1005 C1 2C p2 Copyright 2003 by Marcel Dekker. The rotational speed is 5500 rpm and stagnation temperature rise is 22K. The mass flow at compressor inlet. ¼ T 02 2 T 01 520 2 285 hc ¼ or 81:05% Design Example 5.90.92. The stagnation pressure ratio and power required to drive the compressor. 2. T01. 3.Axial Flow Compressors and Fans 219 Solution: Using the isentropic P – T relation for the compression process. The rotor air angles at the root section. and the isentropic efficiency of the stage is 0. Inc. Solution: (1) As no inlet guide vanes a1 ¼ 0. the work done factor is 0. C w1 ¼ 0 Stagnation temperature.5. The hub tip ratio is 0. relative velocity at tip ¼ 340. tan b2 ¼ 1. b1 ¼ 64:818 Stagnation temperature rise DT 0s ¼ Á tUC a À tan b1 2 tan b2 Cp Substituting the values. Ut ¼ or rt ¼ 308:3 £ 60 ¼ 0:535m: 2p £ 5500 U t 308:3 ¼ 2:126 ¼ Ca 145 2pr t N 60 tanb1 ¼ i:e:. Inc.220 Chapter 5 The Mach number relative to tip is V1 M ¼ pffiffiffiffiffiffiffiffiffiffiffi gRT 1 or V 1 ¼ 0:96ð1:4 £ 287 £ 288:9Þ0:5 ¼ 340:7 m/s i. we get 22 ¼ or À Á 0:92 £ 308:3 £ 145 À tan b1 2 tan b2 1005 Á tanb1 2 tanb2 ¼ 0:538 r m 2 h/2 ¼ 0:5 r m þ h/2 (2) Therefore. 5. All Rights Reserved .25 h [ rm ¼ 1.5 h but rt ¼ rm þ h/2 ¼ 1.88 root radius/tip radius ¼ (where subscript m for mean and h for height) or rm 2 h/2 ¼ 0.5 rm þ 0..7 m/s From velocity triangle at inlet (Fig.3) À Á0:5 V 2 ¼ U 2 þ C 2 or U t ¼ 340:72 2 1452 ¼ 308:3 m/s 1 t 1 or tip speed.588 and b2 ¼ 57.5 h þ h/2 Copyright 2003 by Marcel Dekker.e. Inc.402 £ 0.5 h ¼ 0. rr ¼ r m 2 h/2 ¼ 0:402 2 0:268/2 ¼ 0:267m: Impeller speed at root is Ur 2pr r N 60 2 £ p £ 0:267 £ 5500 ¼ ¼ 153:843 m/s 60 ¼ Copyright 2003 by Marcel Dekker. All Rights Reserved . using isentropic relationship for p – T p1 ¼ p01 and T1 T 01 g/ðg21Þ or 288:9 3:5 p1 ¼ 1 £ ¼ 0:987 bar 290 r1 ¼ p1 0:987 £ 105 ¼ 1:19 kg/m3 ¼ RT 1 287 £ 288:9 Therefore.268 ¼ 0.535 ¼ 2 h or h ¼ 0.677 m2 Now. W ¼ C p DT 0s ¼ 1005 £ 22 ¼ 22110J/kg Power required by the compressor _ P ¼ mW ¼ 116:8 £ 22110 ¼ 2582:4 kW (4) In order to find out rotor air angles at the root section.Axial Flow Compressors and Fans 221 or 0.268 m and rm ¼ 1. A ¼ 2prmh ¼ 2p £ 0. radius at the root can be found as given below.402 m Area. the mass flow entering the stage _ m ¼ rAC a ¼ 1:19 £ 0:677 £ 145 ¼ 116:8 kg/s (3) Stage pressure ratio is Rs hs DT 0s ¼ 1þ T 01 !g/ðg21Þ !3:5 ¼ 1:26 0:90 £ 22 ¼ 1þ 290 Now. rotational speed. and the length of the last stage rotor blade at inlet to the stage. Inc. pressure ratio of the first and last stages. Assume equal temperature rise in all stages. and symmetrical velocity diagram. measured from the axial direction: 208 Work done factor: Mean diameter of the last stage rotor is: Ambient pressure: Ambient temperature: 0:85 18:5cm 1:0bar 290K Calculate the number of stages required. All Rights Reserved .12: The following design data apply to an axial flow compressor: Overall pressure ratio: Mass flow: Polytropic efficiency: Stagnation temperature rise per stage: Absolute velocity approaching the last rotor: 4:5 3:5kg/s 0:87 22k 160m/s Absolute velocity angle. b1 ¼ 46:98 For b2 at the root section DT 0s ¼ or 22 ¼ or À Á 0:92 £ 153:843 £ 145 À tanb1 2 tanb2 1005 Á tU r C a À tanb1 2 tanb2 Cp U r 153:843 ¼ 1:061 ¼ 145 Ca Á tanb1 2 tanb2 ¼ 1:078 [ b2 ¼ 20:9748 Design Example 5.222 Chapter 5 Therefore. Copyright 2003 by Marcel Dekker. from velocity triangle at root section tanb1 ¼ i:e:. R is ! N £ 22 3:05 R¼ 1þ 290 or ð4:5Þ [ 1 3:05 N £ 22 ¼ 1þ 290 ! N ¼ 8:4 Hence number of stages ¼ 8 Stagnation temperature rise.5 Then. 0:5 ¼ Á C a8 À tanb8 þ tanb9 2U Ca8 C8 Copyright 2003 by Marcel Dekker. per stage ¼ 22K. All Rights Reserved . then overall pressure rise is: !n21 NDT 0s n R¼ 1þ T 01 where n21 g ¼ ha c n g21 (where hac is the polytropic efficiency) substituting values n21 1:4 ¼ 0:87 £ ¼ 3:05 n 0:4 overall pressure ratio.Axial Flow Compressors and Fans 223 Solution: If N is the number of stages. therefore 22 £ 8:4 ¼ 23:1 8 From velocity triangle DT 0s ¼ cos a8 ¼ or C a8 ¼ 160 £ cos20 ¼ 150:35 m/s Using degree of reaction. DT0s. as we took 8 stages. L ¼ 0. Inc. DT 0s ¼ Á tUC a8 À tanb8 2 tanb9 Cp Á 150:35 À tanb8 þ tanb9 2U ðAÞ Now. All Rights Reserved . a8 ¼ b9 ¼ 208 From Eq.224 Chapter 5 or 0:5 ¼ Also. (C) and (D). (A) À Á U ¼ 150:35 tanb8 þ 0:364 [ 23:1 ¼ From Eq. (C) U ¼ 150:35ðtan 49:208 þ 0:364Þ ¼ 228:9 m/s Á 181:66 ¼ 1:21 tan 2 b8 2 0:3642 ¼ 150:35 Copyright 2003 by Marcel Dekker. (B) U¼ 181:66 tanb8 2 0:364 181:66 Á tanb8 2 0:364 ðCÞ ðDÞ Comparing Eqs. Inc. we have À Á 150:35 tanb8 þ 0:364 ¼ À or À or tan2 b8 ¼ 1:21 þ 0:1325 ¼ 1:342 pffiffiffiffiffiffiffiffiffiffiffi [ tanb8 ¼ 1:342 ¼ 1:159 i:e:. As we took 8 stages. b8 ¼ 49:208 Substituting in Eq. DT0s ¼ 22K for one stage. DT 0s ¼ 22 £ 8:4 ¼ 23:1 K 8 ðBÞ Á 0:85 £ U £ 150:35 À tanb8 2 tan20 1005 Because of symmetry. therefore. p01 p08 ¼ and T 08 ¼ T 8 þ or T 8 ¼ T 08 2 C2 8 2Cp C2 8 2C p and p09 ¼ 4bar 4 ¼ 3:44bar 1:1643 ¼ 451:7 2 1602 2 £ 1005 ¼ 438:96 K Figure 5. 5.19 T o8 ¼ T 01 þ 7 £ T 0s ¼ 290 þ 7 £ 23:1 ¼ 451:7 K Pressure ratio of the first stage is: ! 1 þ 1 £ 23:1 3:05 R¼ 451:7 Now. Inc. Copyright 2003 by Marcel Dekker. it is necessary to calculate stagnation temperature and pressure ratio of the last stage. p08 /p09 ¼ 1:1643 p09 ¼ 4. Stagnation temperature of last stage: Fig. All Rights Reserved .19 Velocity diagram of last stage.Axial Flow Compressors and Fans 225 The rotational speed is given by N¼ 228:9 ¼ 393:69rps 2p £ 0:0925 In order to find the length of the last stage rotor blade at inlet to the stage. Determine the air angles of a stage at the design radius where the blade speed is 218 m/s.17 mm. h¼ Design Example 5. of stages ¼ 10 The overall stagnation temperature rise is: g21 À Á T 01 R g 2 1 290 4:50:286 2 1 T0 ¼ ¼ hc 0:88 ¼ 155:879 K ¼ 0:0094m 2 ¼ 2prh Copyright 2003 by Marcel Dekker. length of the last stage rotor blade at inlet to the stage. Solution: No. All Rights Reserved . h ¼ 16. Assume a constant axial velocity of 165 m/s.. we have 1:4/0:4 p8 T8 ¼ p08 T 08 Therefore.226 Chapter 5 Using stagnation and static isentropic temperature relationship for the last stage. 438:96 3:5 ¼ 3:112bar p8 ¼ 3:44 451:7 and r8 ¼ ¼ p8 RT 8 3:112 £ 105 ¼ 2:471 kg/m3 287 £ 438:9 Using mass flow rate _ m ¼ r8 A8 C a8 or 3:5 ¼ 2:471 £ A8 £ 150:35 [ A8 or 0:0094 ¼ 0:0162m 2p £ 0:0925 i. and the degree of reaction is 76%.87.e.13: A 10-stage axial flow compressor is designed for stagnation pressure ratio of 4. The overall isentropic efficiency of the compressor is 88% and stagnation temperature at inlet is 290K. Assume equal temperature rise in all stages.5:1. and work done factor is 0. Inc. e.. a2 ¼ 29. we get 2 tan a2 ¼ 1.135 or tan a2 ¼ 0.5675 ¼ 0.e.e.511 i. b2 ¼ 37. b1 ¼ 51.578 and tan a1 ¼ 0.01 and tan b1 2 tan b2 ¼ 0.1256 2 0.501 ¼ 0. degree of reaction tan b1 þ tan b2 ¼ 2.038 Copyright 2003 by Marcel Dekker..5675 i. All Rights Reserved ..808 Similarly. a1 ¼ 3.755 i. for b1 and b2.e..501 [ 2 tan b1 ¼ 2. Inc.0665 i.468 and tan b2 ¼ 1.Axial Flow Compressors and Fans 227 The stagnation temperature rise of a stage T 0s ¼ 155:879 ¼ 15:588 K 10 The stagnation temperature rise in terms of air angles is: T 0s ¼ or tUC a ðtan a2 2 tan a1 Þ Cp T 0s £ C p 15:588 £ 1005 ¼ tUC a 0:87 £ 218 £ 165 ¼ 0:501 ðtan a2 2 tan a1 Þ ¼ ðAÞ From degree of reaction ^¼ 12 or ! Ca ðtan a2 þ tan a1 Þ 2U ! 165 0:76 ¼ 1 2 ðtan a2 þ tan a1 Þ 2 £ 218 [ ðtan a2 þ tan a1 Þ ¼ 0:24 £ 2 £ 218 ¼ 0:634 165 ðBÞ Adding (A) and (B).634 2 0. and runs at 5400 rpm.52 m/s Now using free vortex condition r Cw ¼ constant [ rhCw1h ¼ rtCw1t (where subscripts h for hub and t for tip) £ Copyright 2003 by Marcel Dekker.228 Chapter 5 Design Example 5. Angles of absolute velocities at inlet and exit are 28 and 588.14: An axial flow compressor has a tip diameter of 0. flow angles and degree of reaction at the hub for a free vortex design.9 m. calculate mass flow rate.93. Solution: Impeller speed is 2prN 2p £ 0:45 £ 5400 ¼ ¼ 254:57 m/s 60 60 From velocity triangle À Á U ¼ C a tan a1 þ tan b1 U¼ Ca ¼ Flow area is U 254:57 ¼ 119:47 m/s ¼ tan a1 þ tan b1 ðtan 288 þ tan 588Þ Â Ã A ¼ p r tip 2 r root  à ¼ p 0:452 2 0:422 ¼ 0:0833 m2 Mass flow rate is _ m ¼ rACa ¼ 1:5 £ 0:0833 £ 119:47 ¼ 14:928 kg/s Power absorbed by the compressor ¼ tU ðC w2 2 Cw1 Þ ¼ tUC a ðtan a2 2 tan a1 Þ ¼ 0:93 £ 254:57 £ 119:47ðtan 588 2 tan 288Þ ¼ 30213:7 Nm Total Power.5 kg/m3. respectively and velocity diagram is symmetrical. All Rights Reserved . Assume air density of 1.47 tan 288 ¼ 63. work absorbed by the compressor. hub diameter of 0. work done factor is 0.42 m. Inc. P ¼ _ m £ 30213:7 kW 1000 ¼ 451 kW and whirl velocity at impeller tip Cwt ¼ C a tan a 1 ¼ 119. .368 U h 2 C a tana2 226:29 2 119:47 tan 60:958 ¼ 119:47 Ca Cw2h 215:09 ¼ 1:80 ¼ Ca 119:47 226:29 2 119:47 tan30:888 119:47 Copyright 2003 by Marcel Dekker. a2 ¼ 60.Axial Flow Compressors and Fans 229 or C w1h ¼ r t C w1t 0:45 £ 63:52 ¼ 71:46 m/s ¼ rh 0:4 Similarly C w2t ¼ Ca tana2 ¼ 119:47 tan588 ¼ 191:2 m/s and r h C w2h ¼ r t C w2t or C w2h ¼ r t C w2t 0:45 £ 191:2 ¼ 215:09 m/s ¼ 0:4 rh Therefore.. Inc. the flow angles at the hub are tan a1 ¼ C w1h ðwhere C a is constantÞ Ca 71:46 ¼ 0:598 ¼ 119:47 i.958 Similarly.348 tana2 ¼ i...e. a1 ¼ 30.e.e. b1 ¼ 52.888 tanb1 ¼ U h 2 C a tana1 Ca 2 £ p £ 0:4 £ 5400 ¼ 226:29 m/s 60 where Uh at the hub is given by U h ¼ 2pr h N ¼ [ tanb1 ¼ i. b2 ¼ 5. tanb2 ¼ i.e. All Rights Reserved . and work done factor is 0.0 bar. (5. The stagnation temperature rise of the first stage is 20 K. Determine: 1. inlet stagnation temperature and pressure are 290 K and 1.10) at the mean radius Á tUC a À T 02 2 T 01 ¼ tan b1 2 tanb2 Cp Á 0:94 £ 200 £ 155 À tanb1 2 tanb2 1005 À Á tanb1 2 tanb2 ¼ 0:6898 20 ¼ Using Eq.e. the degree of reaction is Á Ca À ^¼ tanb1 þ tanb2 2U or À Á 0:5 £ 2 £ 200 ¼ 1:29 tanb1 þ tanb2 ¼ 155 Solving above two equations simultaneously 2 tanb1 ¼ 1:98 [ b1 ¼ 44:718 ¼ a2 ðas the diagram is symmetricalÞ tanb2 ¼ 1:29 2 tan44:718 i. The blade height. Inc. Using Eq. the degree of reaction at the hub is Á Ca À 119:47 ðtan52:348 þ tan5:368Þ tanb1 þ tanb2 ¼ ^¼ 2U h 2 £ 226:29 ¼ 0:367 or 36:7%: Design Example 5.230 Chapter 5 Finally. 3. respectively.94. 4. and reaction at the mean radius is 50%. The pitch and chord.12). The mean blade speed is 200 m/s. The mean radius. The blade and air angles at the mean radius.4. b2 ¼ 16:708 ¼ a1 Copyright 2003 by Marcel Dekker. The axial velocity is constant at 155 m/s. The rotor blade aspect ratio is 3. (5. Assume Cp for air as 1005 J/kg K and g ¼ 1.. 2. Solution: 1. All Rights Reserved .15: An axial flow compressor is to deliver 22 kg of air per second at a speed of 8000 rpm. the solidity. 5. an equivalent to cascade. and noting that blades b. Let rm be the mean radius rm ¼ U 200 £ 60 ¼ ¼ 0:239m 2pN 2p £ 8000 Ca 155 ¼ 162 m/s ¼ cosa1 cos16:708 C2 1622 1 ¼ 276:94 K ¼ 290 2 2Cp 2 £ 1005 3. A 0:133 ¼ 0:089m: ¼ 2pr m 2 £ p £ 0:239 At mean radius. s ¼ 0:5 c span Blade aspect ratio ¼ chord Copyright 2003 by Marcel Dekker.Axial Flow Compressors and Fans 231 2. 22 ¼ 0:133m2 1:07 £ 155 Blade height is A¼ h¼ 4. Inc. a. Using continuity equation in order to find the annulus area of flow C1 ¼ T 1 ¼ T 01 2 Using isentropic relationship at inlet g g21 p1 T1 ¼ p01 T 01 Static pressure is p1 ¼ 1:0 Density is 3:5 ¼ 0:851bars 276:94 290 r1 ¼ p1 0:851 £ 100 ¼ 1:07 kg/m3 ¼ RT 1 0:287 £ 276:94 From the continuity equation. air outlet angle.20 for cascade nominal deflection vs. All Rights Reserved . nominal air deflection is 1 ¼ b1 2 b2 ¼ 44:718 2 16:708 ¼ 28:018 Using Fig. 062 m. 0. respectively.295 m. and (4) the stagnation pressure ratio of the stage with stage efficiency 0.232 Chapter 5 Figure 5. All Rights Reserved .2 Copyright 2003 by Marcel Dekker. If the static temperature and pressure at the inlet to the first stage are 288K and 1 bar. and the stagnation stage temperature rise is 15K. and delivers 10.5 kg of air per second at a speed of 10.500 rpm.1 An axial flow compressor has constant axial velocity throughout the compressor of 152 m/s. Inc.92. calculate: 1 the mean diameter of the blade row. 1. (0. 11. Each stage is of 50% reaction and the work done factor is 0. (2) the blade height.20 Cascade nominal deflection versus air outlet angle. a mean blade speed of 162 m/s. C¼ Blade pitch.84. s ¼ 0:5 £ 0:03 ¼ 0:015 m: 0:089 ¼ 0:03m 3 PROBLEMS 5.378.15) The following design data apply to an axial flow compressor: Stagnation temperature rise of the stage: 20 K Work done factor: Blade velocity at root: 0:90 155 m/s 5. Blade chord. (3) the air exit angle from the rotating blades. 5 bar.86.3 5.4. i. (47.Axial Flow Compressors and Fans 233 Blade velocity at mean radius: Blade velocity at tip: Degree of reaction at mean radius: 208 m/s 255 m/s 0:5 Axial velocity ðconstant through the stageÞ: 155 m/s Calculate the inlet and outlet air and blade angles at the root.848. (167. 23. 5.e.378. Find the actual work of compression and temperature at the outlet from the compressor. find the number of stages required. the stagnation temperature at the compressor inlet is 288K. The isentropic efficiency of the stage is 0. 39. l ¼ 0. respectively.8 Copyright 2003 by Marcel Dekker. 287.55. C w £ r ¼ constant that is.5 kW) 5. g ¼ 1. 5.84 . 29. [5. mean radius and tip for a free vortex design. 10%) Calculate the air and blade angles at the root. work done factor is 0.83 K) Determine the number of stages required to drive the compressor for an axial flow compressor having the following data: difference between the tangents of the angles at outlet and inlet.568. Assume Cp ¼ 1. 45. and polytropic efficiency of the compressor is 0.005 kJ/kg K. tan b1 . 54.6 The inlet and outlet angles of an axial flow compressor rotor are 50 and 158. mean blade speed and axial velocity remain constant throughout the compressor.728) Show that for vortex flow.86. the whirl velocity component of the flow varies inversely with the radius. It is compressed through a pressure ratio of four. The blades are symmetrical.718.428. the overall stagnation pressure rise is 3.84. 14. If the mean blade speed is 200 m/s. 28.58.99 (10 stages. (8 stages) In an axial flow compressor air enters at 1 bar and 158C. (188.4 Calculate the degree ofreaction at the tip and root for the same data as Prob. hm ¼ 0.988.7 5. Take the isentropic efficiency of the compressor to be equal to 0. (66.85.66 kJ/kg.5 5.. pressure ratio is 4.tan b2 ¼ 0.188. 454.868. 1. mean and tip for 50% degree of reaction at all radii for the same data as in Prob.758. stagnation pressure at compressor inlet is 1 bar. 39. and the mass flow rate is 15 kg/s. Inc.2].078. All Rights Reserved . 53.7%. 2 208) 5. inlet stagnation temperature is equal to 290 K.2. 43. What is vortex theory? Derive an expression for vortex flow. a1 ¼ b2 and a2 ¼ b1. Stagnation temperature at the inlet: Overall pressure ratio: Isentropic efficiency of the compressor: Mean blade speed: Axial velocity: Degree of reaction: 288 K 4 0:88 170 m/s 120 m/s 0:5 (639. the various terms used in airfoil geometry.4 kW. with a sketch.88.12 Sketch the velocity diagrams for an axial flow compressor and derive the expression: !g P02 hs DT 0s g21 ¼ 1þ P01 T 01 Explain the term “degree of reaction”. i.698) 5.e. Why is the degree of reaction generally kept at 50%? Derive an expression for the degree of reaction and show that for 50% reaction. the blades are symmetrical. 5. b2 ¼ 2 72.14 5.10 Calculate the number of stages from the data given below for an axial flow compressor: Air stagnation temperature at the inlet: Stage isentropic efficiency: Degree of reaction: Air angles at rotor inlet: Air angle at the rotor outlet: Meanblade speed: Work done factor: Overall pressure ratio: 5. 5. b1 ¼ 77.9 From the data given below. calculate the power required to drive the compressor and stage air angles for an axial flow compressor.11 288 K 0:85 0:5 408 108 180 m/s 0:85 6 (14 stages) Derive the expression for polytropic efficiency of an axial flow compressor in terms of: (a) n and g (b) inlet and exit stagnation temperatures and pressures. Inc.17 Copyright 2003 by Marcel Dekker. All Rights Reserved .234 Chapter 5 5.. What is an airfoil? Define.13 5.16 5. measured from the axial direction nominal air outlet angle angle with relative velocity. Inc. rpm number of blades overall pressure ratio.Axial Flow Compressors and Fans 235 NOTATION C CL Cp D Fx h L N n R Rs U V a a* 2 b DTA DTB DT0s DTs D* e* es w L l c absolute velocity lift coefficient specific heat at constant pressure drag tangential force on moving blade blade height. specific enthalpy lift number of stage. measure from the axial direction static temperature rise in the rotor static temperature rise in the stator stagnation temperature rise static temperature rise nominal deviation nominal deflection stalling deflection flow coefficient degree of reaction work done factor stage loading factor SUFFIXES 1 2 3 a m r t w inlet to rotor outlet from the rotor inlet to second stage axial. ambient mean radial. root tip whirl Copyright 2003 by Marcel Dekker. All Rights Reserved . gas constant stage pressure ratio tangential velocity relative velocity angle with absolute velocity. Low-pressure steam then exhausts to the condenser. All Rights Reserved . A multi-stage condensing turbine is a turbine in which steam exhausts to a condenser and is condensed by air-cooled condensers. Steam turbines can be noncondensing or condensing. The high-velocity steam from the nozzles strikes the set of moving blades (or buckets).1 INTRODUCTION In a steam turbine.6 Steam Turbines 6. steam exhausts at a pressure greater than atmospheric. In noncondensing turbines (or backpressure turbines). Here the kinetic energy of the steam is utilized to produce work on the turbine rotor. There are two classical types of turbine stage designs: the impulse stage and the reaction stage. The exhaust pressure from the turbine is less than the atmospheric. Inc. In this turbine. Steam then leaves the turbine and is utilized in other parts of the plant that use the heat of the steam for other processes. Copyright 2003 by Marcel Dekker. cycle efficiency is low because a large part of the steam energy is lost in the condenser. high-pressure steam from the boiler expands in a set of stationary blades or vanes (or nozzles). The backpressure turbines have very high efficiencies (range from 67% to 75%). and an exit. for a perfect gas. Consider a nozzle as shown in Fig. Assume that the flow occurs adiabatically under steady conditions. Copyright 2003 by Marcel Dekker. the velocity increases. Steam deviates from the laws of perfect gases. a point is reached called the critical pressure ratio. Since no work is transferred. The temperature. respectively. All Rights Reserved . 6. PVg ¼ constant. Inc.2 STEAM NOZZLES The pressure and volume are related by the simple expression. The P-V relationship is given by: PV n ¼ constant where: n ¼ 1:135 for saturated steam n ¼ 1:3 for superheated steam For wet steam. pressure. Any further reduction in pressure will not produce any further increases in the velocity.1 Nozzle. The velocity through a nozzle is a function of the pressure-differential across the nozzle. critical pressure. a throat. Eventually. the velocity of the fluid at the nozzle entry is usually very small and its kinetic energy is negligible compared with that at the outlet. where the velocity is equal to the velocity of sound in steam. the Zeuner relation. the equation reduces to: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð6:1Þ C 2 ¼ f 2ð h1 2 h2 Þ g where h1 and h2 are the enthalpies at the inlet and outlet of the nozzle. As the outlet pressure decreases. and density are called critical temperature.1. and critical Figure 6. All nozzles consist of an inlet section. Hence. x n ¼ 1:035 þ 10 (where x is the initial dryness fraction of the steam) may be used.238 Chapter 6 6. steam tends to expand in all directions and is very turbulent. the critical pressure ratio is: P1 ¼ Pc T1 0 Tc n n21 ð6:3Þ where Tc0 is the temperature. In a convergent nozzle. the area of the section from the throat to the exit gradually increases.Steam Turbines 239 density. the outlet cross-sectional area and the throat cross-sectional areas are equal. All Rights Reserved . The critical pressure ratio is approximately 0. the critical temperature. 6. If a nozzle is designed to operate so that it is just choked. To allow the steam to expand without turbulence. In this type of nozzle.55 for superheated steam. The critical velocity is given by the equation: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð6:4Þ C c ¼ f 2ð h1 2 hc Þ g where hc is read from tables or the h – s chart at Pc and sc. i. Figure 6. The ratio between nozzle inlet temperature and critical temperature is given by: T1 2 ð6:2Þ ¼ Tc n þ 1 where Tc is the critical temperature at which section M ¼ 1. the convergent– divergent nozzle is used. respectively.1. Copyright 2003 by Marcel Dekker. the behavior of convergent and convergent –divergent nozzles is different. This will cause increased friction losses as the steam flows through the moving blades. The increase in area causes the steam to emerge in a uniform steady flow. shown in Fig. as shown in Fig. can be found from steam tables at the value of Pc and sc ¼ s1. Assuming isentropic flow in the nozzle. The temperature at the throat. In this case. which would have been reached after an isentropic expansion in the nozzle. When the outlet pressure is designed to be higher than the critical pressure. any other operating condition is an off-design condition.2.2 Convergent nozzle.e. Inc. a simple convergent nozzle may be used. The operation of a convergent nozzle is not practical in highpressure applications. In this respect. The size of the throat and the length of the divergent section of every nozzle must be specifically designed for the pressure ratio for which the nozzle will be used. 6.. All Rights Reserved . Fig. Copyright 2003 by Marcel Dekker.3 NOZZLE EFFICIENCY The expansion process is irreversible due to friction between the fluid and walls of the nozzle. Inc.4. it is still approximately adiabatic as shown in Fig.3.4 THE REHEAT FACTOR Consider a multi-stage turbine as shown by the Mollier diagram.240 Chapter 6 6. However. 6.3 Nozzle expansion process for a vapor. The reheat factor is defined by: R:F: ¼ ¼ Cumulative stage isentropic enthalpy drop Turbine isentropic enthalpy drop P 0 à Dh stage ½Dh0 turbine 0 0 0 h1 2h2 þ h2 2h3 þ h3 2h4 À Á ¼ h1 2h" 4 ð6:5Þ Figure 6. Then the nozzle efficiency is defined as hn ¼ h1 2 h2 h1 2 h2 0 6. and friction within the fluid itself. 6. 1 –20 is the isentropic enthalpy drop and 1 – 2 is the actual enthalpy drop in the nozzle. The reheat factor may be used to relate the stage efficiency and the turbine efficiency.5 METASTABLE EQUILIBRIUM As shown in Fig. All Rights Reserved . then: P  0à P à hs Dh stage hs Dh0 stage ht ¼ ¼ ½Dh0 turbine ½Dh0 turbine or ht ¼ hs £ ðR:FÞ: ð6:7Þ Equation 6. 1 – 2 is Copyright 2003 by Marcel Dekker. 6.7 indicates that the turbine efficiency is greater than the stage efficiency.03 –1. The reheat factor is usually of the order of 1. R. Assume reversible and adiabatic processes. 6.5. Turbine isentropic efficiency is given by: Dh ð6:6Þ Dh0 where Dh is the actual enthalpy drop and Dh0 is the isentropic enthalpy drop.4 Mollier chart for a multi-stage turbine. .4 it is clear that: P Dh ¼ ½Dhstage ht ¼ Dh1 – 4 ¼ ðh1 2 h2 Þ þ ðh2 2 h3 Þ þ ðh3 2 h4 Þ if hs (stage efficiency) is constant. From diagram 6. slightly superheated steam at point 1 is expanded in a convergent– divergent nozzle.04. Since the isobars diverge. Inc.Steam Turbines 241 Figure 6.F. 1. But.2 MPa.5 Phenomenon of supersaturation on T –S diagram. condensation occurs very rapidly. the change of temperature and pressure in the nozzle is faster than the condensation process under such conditions.242 Chapter 6 Figure 6. Solution: From saturated steam tables. Degree of undercooling is the difference between the saturation temperature corresponding to pressure at point 3 and the actual temperature of the superheated vapor at point 3. Find the exit velocity of the steam and dryness fraction. Steam does not condense at the saturation temperature corresponding to the pressure. enthalpy of saturated vapor at 2 MPa: h1 ¼ hg ¼ 2799:5 kJ/kg and entropy s1 ¼ sg ¼ 6:3409 kJ/kg K Copyright 2003 by Marcel Dekker. the path followed by the steam and point 2 is on the saturated vapor line. Illustrative Example 6. then condensation could not occur until point 3 is reached. Although the steam between points 2 –3 is in the vapor state.1: Dry saturated steam at 2 MPa enters a steam nozzle and leaves at 0. At this point. Here. This is known as the metastable state. Inc. Degree of supersaturation is the actual pressure at point 3 divided by the saturation pressure corresponding to the actual temperature of the superheated vapor at point 3. if point 2 is reached in the divergent section of the nozzle. The condensation lags behind the expansion process. In fact. the temperature is below the saturation temperature for the given pressure. Assume isentropic expansion and neglect inlet velocity. All Rights Reserved . we can expect condensation to occur. Solution: Enthalpy of dry saturated steam at 1. h1 ¼ 2787:6 kJ/kg. Assume friction loss in the nozzle is equal to 10% of the total enthalpy drop.3409 ¼ sf2 þ x2sfg2.3 MPa. sfg2 is the entropy of vaporization at 0.2 MPa.e.3 MPa to 0.2 MPa.2: Dry saturated steam is expanded in a nozzle from 1. Inc. h2 ¼ hf2 þ x2hfg2 ¼ 504. hence dryness fraction after expansion: 6:4953 2 1:3026 ¼ 0:857 x2 ¼ 6:0568 Now. the enthalpy at the exit: h2 ¼ hf2 þ x2 hfg2 ¼ 417:46 þ ð0:857Þ £ ð2258Þ ¼ 2352:566 kJ/kg Therefore enthalpy drop from 1. using steam tables. s1 ¼ s2: i. and entropy s1 ¼ 6:4953 kJ/kg K: Since the expansion process is isentropic. the velocity of steam at the nozzle exit: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 2 ¼ fð2Þ £ ð1000Þ £ ð391:531Þg ¼ 884:908 m/s Copyright 2003 by Marcel Dekker.3 MPa to 0.233 kJ/kg Using the energy equation: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 2 ¼ f2ðh1 2 h2 Þg pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ fð2Þ £ ð1000Þ £ ð2799:5 2 2397:233Þg or: C 2 ¼ 897 m/s Illustrative Example 6. All Rights Reserved . Using tables: x2 ¼ 6:3409 2 1:5301 ¼ 0:8595 5:5970 Therefore.1 MPa ¼ h1 – h2 ¼ 2787:6 – 2352:566 ¼ 435:034 kJ/kg Actual enthalpy drop due to friction loss in the nozzle ¼ 0:90 £ 435:034 ¼ 391:531 kJ/kg Hence.Steam Turbines 243 Since the expansion is isentropic.9 ¼ 2397.8595 £ 2201. sf2 is the entropy of saturated liquid at 0. s1 ¼ s2 ¼ sf2 þ x2 sfg2 . calculate the mass of steam discharged when the nozzle exit diameter is 10 mm.1 MPa.. s1 ¼ s2 ¼ 6.7 þ 0. where x2 is the dryness fraction after isentropic expansion. 5 MPa and 5008C expands through an ideal nozzle to a pressure of 5 MPa. . and therefore the nozzle is convergent.1 MPa: ¼ x2 vg2 ¼ ð0:857Þ £ ð1:694Þ ¼ 1:4517 m3 =kg (since the volume of the liquid is usually negligible compared to the volume of dry saturated vapor. h2 ¼ 3277.15 MPa. the exit area is A2 ¼ mv2 ð2:8Þ £ ð0:06152Þ ¼ ð3:42Þ £ ð1024 Þ m2 ¼ 502:8 C2 Illustrative Example 6. the area at the exit. P2 .7598 kJ/kg K T2 ¼ 4358K.4: Consider a convergent –divergent nozzle in which steam enters at 0.06152 m3/kg. Solution: Initial conditions: P1 ¼ 7. hence for most practical problems. e. Assuming isentropic expansion and index n ¼ 1. v2 ¼ 0. Pc ¼ ð0:545Þ £ ð7:5Þ ¼ 4:0875 MPa.135. for maximum mass flow). 5008C h1 ¼ 3404. The exit velocity: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 2 ¼ fð2Þ £ ð1000Þ £ ðh1 2 h2 Þg pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ fð2Þ £ ð1000Þ £ ð3404:3 2 3277:9Þg ¼ 502:8 m/s Using the continuity equation. What exit area is required to accommodate a flow of 2. v ¼ xvg) Mass flow rate of steam at the nozzle exit: ¼ AC 2 ðpÞ £ ð0:01Þ2 £ ð884:908Þ £ ð3600Þ ¼ 172:42 kg=h: ¼ ð4Þ £ ð1:4517Þ x2 vg2 Illustrative Example 6. find the ratio of cross-sectional area. Inc.244 Chapter 6 Specific volume of steam at 0. s1 ¼ s2 ¼ 6. Copyright 2003 by Marcel Dekker.5 MPa. and the area at the throat for choked conditions (i.8 kg/s? Neglect initial velocity of steam and assume isentropic expansion in the nozzle.8 MPa and leaves the nozzle at 0.3 kJ/kg s1 ¼ 6.9 kJ/kg (from the superheated steam tables or the Mollier Chart). State 2 is fixed by P2 ¼ 5 MPa.3: Steam at 7. All Rights Reserved .7598 kJ/kg K (h1 and s1 from superheated steam tables) At the exit state. 462 MPa: Dh1 – 2 ¼ h1 2 h2 ¼ 2769 2 2659 ¼ 110 kJ/kg Dryness fraction: x2 ¼ 0. Inc.15 MPa: Dh123 ¼ h1 2 h3 ¼ 2769 2 2452 ¼ 317 kJ/kg Enthalpy drop from 0.954 Dryness fraction: x3 ¼ 0.8 MPa to 0. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C3 ¼ fð2Þ £ ð1000Þ £ ðDh1 – 3 Þg pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ fð2Þ £ ð1000Þ £ ð317Þg ¼ 796m/s The velocity at the throat pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C2 ¼ fð2Þ £ ð1000Þ £ ðDh1 – 2 Þg ¼ fð2Þ £ ð1000Þ £ ð110Þg ¼ 469m/s Figure 6.6 Convergent – divergent nozzle.6: n n21 8:41 2 2 ¼ 0:8 ¼ 0:462 MPa P c ¼ P2 ¼ P 1 nþ1 2:135 From the Mollier chart: h1 ¼ 2769 kJ/kg h2 ¼ 2659 kJ/kg h3 ¼ 2452 kJ/kg Enthalpy drop from 0. 6.Steam Turbines 245 Solution: Critical pressure for maximum mass flow is given by Fig. Copyright 2003 by Marcel Dekker.8 MPa to 0.902 The velocity at the exit. All Rights Reserved . Figure 6. the dryness fraction at the exit is 0. All Rights Reserved . Find the supply pressure of steam. Assume isentropic expansion (see Fig. 6. Inc.7).5.5: Dry saturated steam enters the convergent – divergent nozzle and leaves the nozzle at 0.7 h – s diagram for Example 6.85.246 Chapter 6 Mass discharged at the throat: m2 ¼ _ A2 C 2 x2 vg2 A3 C 3 x3 vg3 Mass discharged at the exit m3 ¼ _ Therefore A3 C 3 A 2 C 2 ¼ x3 vg3 x2 vg2 Hence.1 MPa. A3 C2 ¼ A2 C3 ! ! ! ! x3 vg3 469 ð0:902Þð1:1593Þ ¼ ¼ 1:599 x2 vg2 796 ð0:954Þð0:4038Þ Illustrative Example 6. Copyright 2003 by Marcel Dekker. this type of turbine has very limited applications in practice. the two sets of blades allow the steam to perform work on the turbine rotor.10. In large power plants. 6. Figure 6. Two turbine stage designs in use are: the impulse stage and reaction stage. This work is then transmitted to the driven load by the shaft on which the rotor assembly is carried.3026 þ (0. All Rights Reserved .6 STAGE DESIGN A turbine stage is defined as a set of stationary blades (or nozzles) followed by a set of moving blades (or buckets or rotor). the dryness fraction is 0.451 kJ/kg K Since s1 ¼ s2. designated by DeLaval in 1889.8 Steam turbine.1 MPa.451 kJ/kg K. Because of its high speed. i. which cannot be avoided.0568) ¼ 6. This problem can be solved easily by the Mollier chart or by calculations. which ran at 30. and point 1 is at the dry saturated line.85) £ (2258) ¼ 2336. 6.76 kJ/kg and s2 ¼ 1.Steam Turbines 247 Solution: At the state point 2. Copyright 2003 by Marcel Dekker. Therefore pressure P1 may be determined by the Mollier chart or by calculations: i. The first turbine.: h2 ¼ 417. Together.: P1 ¼ 1. was a single-stage impulse turbine. the state 1 is fixed by s1 ¼ 6.474 MPa.e. the single-stage impulse turbine is ruled out. High speeds are extremely undesirable due to high blade tip stresses and large losses due to disc friction. since alternators usually run speeds around 3000 rpm. Inc.000 rpm. Enthalpy and entropy may be determined using the following equations: h2 ¼ hf2 þ x2 hfg2 and s2 ¼ sf2 þ x2 sfg2 .46 þ (0.85) £ (6.e.85 and the pressure is 0.8 – 6. Photographs of actual steam turbines are reproduced in Figs. C1 cos a1 and increases the value of the axial or flow component Ca sin a1. In an impulse stage. 6.248 Chapter 6 Figure 6.11. The pressure drop again results in a corresponding increase in the velocity of the steam flow. the steam supplied to a single-wheel impulse turbine expands completely in the nozzles and leaves with absolute velocity C1 at an angle a1. the total pressure drop occurs across the stationary blades (or nozzles). in the reaction stage.9 Pressure velocity-compounded impulse turbine. These blades look very similar to the stationary blades or nozzles. these velocities are V1 and V2. the total pressure drop is divided equally across the stationary blades and the moving blades. This pressure drop increases the velocity of the steam. and by subtracting the blade velocity vector U. the shape of the stationary blades or nozzles in both stage designs is very similar.12. 6. As shown in Fig. However. Inc. As shown in Fig. 6. a big difference exists in the shapes of the moving blades. 6. The shape of the moving blades in a reaction stage is more like that of an airfoil. As shown in Figs. respectively. 6. whereas gas turbine plants seldom do.10 and 6. The relative velocity V1 makes an angle of b1 with respect to U. The increase in value of a1 decreases the value of the useful component. Copyright 2003 by Marcel Dekker.12. the relative velocity vector at entry to the rotor V1 can be determined.7 IMPULSE STAGE In the impulse stage. The two points of particular interest are the inlet and exit of the blades. However. the shape of the moving blades or buckets is like a cup. The general principles are the same whether steam or gas is the working substance.8 THE IMPULSE STEAM TURBINE Most of the steam turbine plants use impulse steam turbines. All Rights Reserved . Inc.10 Steam turbine cross-sectional view. All Rights Reserved .Steam Turbines Figure 6. 249 Copyright 2003 by Marcel Dekker. Inc.12 Velocity triangles for turbine stage.13. these triangles can be combined to give a single diagram as shown in Fig. Vectorially subtracting the blade speed results in absolute velocity C2. The steam leaves tangentially at an angle b2 with relative velocity V2. Since the two velocity triangles have the same common side U.250 Chapter 6 Figure 6.11 Impulse and reaction stage design. Copyright 2003 by Marcel Dekker. 6. All Rights Reserved . Figure 6. In the actual case. Assume that Ca is constant. All Rights Reserved . work done by the steam is given by: W t ¼ UðC w1 þ Cw2 Þ Since Cw2 is in the negative r direction.Steam Turbines 251 Figure 6. That is: k¼ V2 V1 From Euler’s equation. Therefore: _ The driving force on the wheel ¼ mC w ð6:13Þ Copyright 2003 by Marcel Dekker. Inc. If the blade is symmetrical then b1 ¼ b2 and neglecting the friction effects of blades on the steam.13: DCw is the change in the velocity of whirl. there will be an axial thrust in the flow direction. the relative velocity is reduced by friction and expressed by a blade velocity coefficient k.13 Combined velocity diagram. 6. the work done per unit mass flow is given by: W t ¼ UðC w1 þ Cw2 Þ ð6:9Þ If Ca1 – Ca2. Then: W t ¼ UC a ðtan a1 þ tan a2 Þ W t ¼ UC a ðtan b1 þ tan b2 Þ ð6:10Þ ð6:11Þ Equation (6. V1 ¼ V2.11) is often referred to as the diagram work per unit mass flow and hence the diagram efficiency is defined as: hd ¼ Diagram work done per unit mass flow Work available per unit mass flow ð6:12Þ Referring to the combined diagram of Fig. . the axial thrust is given by: _ Axial thrust : F a ¼ mDC a The maximum velocity of the steam striking the blades pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 1 ¼ f2ðh0 2 h1 Þg where h0 is the enthalpy at the entry to the nozzle and h1 is the enthalpy at the nozzle exit. The energy supplied to the blades is the kinetic energy of the jet.e. Inc.252 Chapter 6 The product of the driving force and the blade velocity gives the rate at which work is done on the wheel. C1 Differentiating Eq.13): _ Power output ¼ mUDCw ð6:14Þ ð6:15Þ ð6:16Þ If Ca1 2 Ca2 ¼ DCa.19) and equating it to zero provides the maximum diagram efficiency: À Á d hd 8U ¼ 4 cos a1 2 ¼0 U C1 d C1 Copyright 2003 by Marcel Dekker. C2 =2 and the blading efficiency or 1 diagram efficiency: hd ¼ Rate of work performed per unit mass flow Energy supplied per unit mass of steam 2 2UDC w ¼ C2 C2 1 1 ð6:17Þ Using the blade velocity coefficient k ¼ V 2 and symmetrical blades V1 (i. then: DC w ¼ 2V 1 cos a1 2 U Hence hd ¼ ðUDC w Þ £ DC w ¼ 2ðC 1 cos a1 2 U Þ ð6:18Þ And the rate of work performed per unit mass ¼ 2(C1 cos a1 2 U )U Therefore: 2 hd ¼ 2ðC1 cos a1 2 U ÞU £ 2 C1 4ðC1 cos a1 2 U ÞU hd ¼ C2 1 4U U hd ¼ cos a1 2 ð6:19Þ C1 C1 U where is called the blade speed ratio. neglecting the velocity at the inlet to the nozzle. All Rights Reserved . (6. (6. b1 ¼ b2). From Eq. 15. Rateau-stage turbines are unable to extract a large amount of energy from the steam and. 6. Fixed blades between the rows of moving blades redirect the steam flow into the next row of moving blades. In each stage. its simplicity of design and construction makes it well suited for small auxiliary turbines.e. The separating walls. which consists of a nozzle and a moving blade. (6.10 VELOCITY COMPOUNDING (THE CURTIS TURBINE) In this type of turbine. the steam is expanded and the kinetic energy is used in moving the rotor and useful work is obtained. The Curtis stage impulse turbine is shown in Fig.14). Although the Rateau turbine is inefficient.14.Steam Turbines 253 or U cos a1 ¼ 2 C1 i. a Curtis-stage turbine can extract Copyright 2003 by Marcel Dekker. Inc. therefore. and the combination of stages forms a pressure compounded turbine. which carry the nozzles. the whole of the pressure drop occurs in a single nozzle. 6. maximum diagram efficiency 4 cos a1 cos a1 ¼ cos a1 2 2 2 or: ð6:20Þ hd ¼ cos2 a1 ð6:21Þ Substituting this value into Eq. have a low efficiency. and the steam passes through a series of blades attached to a single wheel or rotor. as shown in Fig. 6. the power output per unit mass flow rate at the maximum diagram efficiency: P ¼ 2U 2 ð6:22Þ 6. Because the reduction of velocity occurs over two stages for the same pressure decreases. The total pressure drop is divided in a series of small increments over the stages. All Rights Reserved .9 PRESSURE COMPOUNDING (THE RATEAU TURBINE) A Rateau-stage impulse turbine uses one row of nozzles and one row of moving blades mounted on a wheel or rotor. are known as diaphragms. Each diaphragm and the disc onto which the diaphragm discharges its steam is known as a stage of the turbine.. 11 AXIAL FLOW STEAM TURBINES Sir Charles Parsons invented the reaction steam turbine. All Rights Reserved . Inc. 6. more energy from the steam than a Rateau-stage turbine. Reaction turbines are normally used as Copyright 2003 by Marcel Dekker. In a reaction turbine. As a result. reaction turbines that have more than one stage are classified as pressure-compounded turbines. a reactive force is produced on the moving blades when the steam increases in velocity and when the steam changes direction. Because the pressure drop from inlet to exhaust is divided into many steps through use of alternate rows of fixed and moving blades. a Curtisstage turbine has a higher efficiency than a Rateau-stage turbine. In this turbine pressure drop or expansion takes place both in the fixed blades (or nozzles) as well as in the moving blades.254 Chapter 6 Figure 6. The reaction turbine stage consists of a fixed row of blades and an equal number of moving blades fixed on a wheel.14 Rateau-stage impulse turbine. Copyright 2003 by Marcel Dekker.Steam Turbines 255 Figure 6. All Rights Reserved .15 The Curtis-stage impulse turbine. Inc. the fixed and moving blades have the same shape and. respectively. then: h1 2 h2 ð6:23Þ h0 2 h2 The static enthalpy at the inlet to the fixed blades in terms of stagnation enthalpy and velocity at the inlet to the fixed blades is given by C2 h0 ¼ h00 2 0 2Cp Similarly. For a 50% reaction.16 Velocity triangles for 50% reaction design.12 DEGREE OF REACTION The degree of reaction or reaction ratio (L) is a parameter that describes the relation between the energy transfer due to static pressure change and the energy transfer due to dynamic pressure change. low-pressure turbines. at the entry to the moving blades and at the exit from the moving blades. h1. C2 h2 ¼ h02 2 2 2Cp L¼ Copyright 2003 by Marcel Dekker. the velocity diagram is symmetrical as shown in Fig. and h2 are the enthalpies at the inlet due to the fixed blades. It is also defined as the ratio of the static enthalpy drop in the rotor to the static enthalpy drop in the stage. High-pressure reaction turbines are very costly because they must be constructed from heavy and expensive materials. If h0. Inc.16. The degree of reaction is defined as the ratio of the static pressure drop in the rotor to the static pressure drop in the stage. therefore. 6.256 Chapter 6 Figure 6. All Rights Reserved . 6. L¼ ð h1 2 h2 Þ C2 2 h02 2 2C2p h00 2 C2 0 2C p But for a normal stage. Inc.18: Copyright 2003 by Marcel Dekker. (6. and C w2 ¼ V w2 2 U Therefore. it is seen that Cw1 ¼ U þ V w1 . (6.26). According to the definition of reaction. Then: À 2 Á V 2 V2 2 ¼0 h01Re1 2 h02Re2 ¼ ðh1 2 h2 Þ þ 1 2 Substituting for (h1 2 h2) in Eq.26) can be arranged into a second form: Á 1 Ca À tan b2 2 tan a2 L¼ þ 2 2U Putting L ¼ 0 in Eq. and for L ¼ 0:5. Eq. when L ¼ 0.24): À 2 Á V2 2 V2 1 L¼ ½2ðh01 2 h02 Þ À 2 Á V2 2 V2 1 L¼ ð6:25Þ ½2U ðC w1 þ Cw2 Þ Assuming the axial velocity is constant through the stage. C0 ¼ C2 and since h00 ¼ h01 in the nozzle. then: À 2 Á V w2 2 V 2 w1 L¼ ½2U ðU þ V w1 þ V w2 2 U Þ ðV w2 2 V w1 ÞðV w2 þ V w1 Þ ½2U ðV w1 þ V w2 Þ À Á Ca tanb2 2tanb1 L¼ 2U L¼ From the velocity triangles. (6. The Mollier diagram and velocity triangles for L ¼ 0 are shown in Figs.17 and 6. we get ð6:26Þ ð6:27Þ b2 ¼ b1 and V 1 ¼ V 2 . 6. (6. b2 ¼ a1 : Zero Reaction Stage: Let us first discuss the special case of zero reaction. All Rights Reserved . then: h1 2 h2 L¼ ð6:24Þ h01 2 h02 We know that h01Re1 ¼ h02Re2.Steam Turbines 257 Substituting. (6.23) reveals that h1 ¼ h2 and Eq.26) that b1 ¼ b2. Eq. 19) for L ¼ 0.18. (6. The zero reaction in the impulse stage. Fifty-Percent Reaction Stage From Eq.23). Fig. it is also clear that a2 ¼ b1. it is clear that the reaction is negative for the impulse turbine stage when irreversibility is taken into account. From Eq. (6. the enthalpy drop in the nozzle row equals the enthalpy drop in the rotor.17 Zero reaction (a) Mollier diagram and (b) velocity diagram. and points 1. a1 ¼ b2. (6. That is: h0 2 h1 ¼ h1 2 h2 Figure 6. Copyright 2003 by Marcel Dekker.5. All Rights Reserved . means there is no pressure drop through the rotor. and the velocity diagram is symmetrical. Then.23). there is a pressure drop through the rotor. where it can be observed that the enthalpy increases through the rotor. Because of symmetry. In the ideal case. Inc. by definition. Now. h01r01 ¼ h02r01 and h1 ¼ h2 for L ¼ 0. 6. 2 and 2s on the Mollier chart should coincide.18 Mollier diagram for an impulse stage. But due to irreversibility. For L ¼ 1/2. V1 ¼ V2.258 Chapter 6 Figure 6. there is no pressure drop in the rotor. The Mollier diagram for an impulse stage is shown in Fig. 6.. when a2 ¼ a1. Copyright 2003 by Marcel Dekker. U U into Eq.Steam Turbines 259 Figure 6. All Rights Reserved .24) can be rewritten as: C w2 2 C w1 : 2U can be eliminated by using this equation: L¼1þ Cw2 ¼ W 2 C w1 .19 A 50% reaction stage (a) Mollier diagram and (b) velocity diagram. The velocity diagram for L ¼ 1 is shown in Fig. nozzle 2 flow diffusion). (6.e.20 with the same value of Ca. then C1 . and W used for L ¼ 0 and L ¼ 1. Substituting b2 ¼ tan a2 þ L¼1þ Ca ðtan a2 2tan a1 Þ ð6:28Þ 2U Thus. Inc. the reaction is unity (also C1 ¼ C2). C0 (i. Choice of Reaction and Effect on Efficiency: Eq. It is obvious that if L exceeds unity.20 Velocity diagram for 100% reaction turbine stage. (6.27) gives Ca Cw2 Figure 6. U. Figure 6.21 Influence of reaction on total-to-static efficiency with fixed values of stage-loading factor.22 Blade loading coefficient vs. flow coefficient. Inc. All Rights Reserved .260 Chapter 6 Figure 6. Copyright 2003 by Marcel Dekker. Efficiency of blading. W When 2 ¼ 2. 6. Thus. Inc. All Rights Reserved . The mass flow through the turbine nozzles and blading is 0. Calculate the following: 1. which implies the highest possible value of blade speed is consistent with blade stress limitations. Copyright 2003 by Marcel Dekker.7. may be used to find the blade height h. Axial force on blades. Tangential force on blades. Work done on blades. 6. 5.22 for a high total-to-total efficiency. Velocity of whirl. m ¼ rAC.13 BLADE HEIGHT IN AXIAL FLOW MACHINES The continuity equation.23. 4.182 kg/s and the blade velocity coefficient is 0. The problem can be solved either graphically or by calculation. the velocity diagram can be constructed as shown in Fig. 2. 6. 6.21 the total-to-static efficiencies are shown plotted against the degree of reaction. It means that the total-to-static efficiency is heavily dependent upon the reaction ratio and hts can be optimized by choosing a suitable value of reaction.6: The velocity of steam leaving a nozzle is 925 m/s and the nozzle angle is 208. hts is maximum at L ¼ 0. the U optimum hts is obtained with higher reaction ratios. Illustrative Example 6. The _ annular area of flow ¼ pDh. 6. the blade-loading factor should be as small as possible. the mass flow rate through an axial flow compressor or turbine is: _ m ¼ rpDhC a ð6:30Þ Blade height will increase in the direction of flow in a turbine and decrease in the direction of flow in a compressor. 3. Solution: From the data given. The blade speed is 250 m/s. Assume that the inlet and outlet blade angles are equal. With higher loading. As shown in Fig. Inlet angle of blades for shockless inflow of steam.Steam Turbines 261 yielding: L¼1þ W C w1 2 2 U 2U ð6:29Þ In Fig. Inc.068 ¼ 221.6. V2 .69 tan 27. tanb1 ¼ cos b2 ¼ or: C w2 Cw2 þ U .262 Chapter 6 Figure 6. V 2 ¼ U 2 þ C 2 2 2UC 1 cosa1 1 1 ¼ 2502 þ 9252 2 ð2Þ £ ð250Þ £ ð925Þ £ cos208 so: V 1 ¼ 695:35 m/s But. orV 2 ¼ ð0:70Þ £ ð695:35Þ ¼ 487m/s: V1 Velocity of whirl at inlet: k¼ Cw1 ¼ C1 cosa1 ¼ 925cos208 ¼ 869:22m/s Axial component at inlet: Ca1 ¼ BD ¼ C 1 sina1 ¼ 925sin208 ¼ 316:37m/s Blade angle at inlet: C a1 316:37 ¼ 0:511 ¼ Cw1 2 U 619:22 Therefore. b1 ¼ 27.23 Velocity triangles for Example 6. Applying the cosine rule to the KABC. V2 ¼ V 2 cos b2 2U ¼ 487 £ cos 27:068 2 250 ¼ 433:69 2 250 ¼ 183:69 m/s and: Ca2 ¼ FE ¼ (U þ Cw2) tan b2 ¼ 433.548 m/s Copyright 2003 by Marcel Dekker.068 ¼ b2 ¼ outlet blade angle. All Rights Reserved . 63) £ (250)/1000 ¼ 47. All Rights Reserved .63 N. 2. 4. 6. the diagram efficiency and the blade velocity coefficient. Velocity of whirl at outlet. Cw2 ¼ 183. Velocity of whirl at inlet.068 ¼ b2. Axial force on blades ¼ m ðC a1 2 C a2 Þ ¼ ð0:182Þ ð316:37 2 221:548Þ ¼ 17:26N _ Work done on blades ¼ tangential force on blades £ blade velocity ¼ (191. Inc. V 2 ¼ U 2 þ C2 2 2UC 1 cos a1 1 1 ¼ 1542 þ 5902 2 ð2Þ £ ð154Þ £ ð590Þ cos 208 i. 5.7: The steam velocity leaving the nozzle is 590 m/s and the nozzle angle is 208. Cw1 ¼ 869.9) ¼ 191. The axial velocity at rotor outlet ¼ 155 m/s.24. Calculate the work done.91 kW. and the blades are symmetrical.69 m/s Tangential force on blades ˙ ¼ m (Cw1 þ Cw2) ¼ (0. Design Example 6. Work done on blades Efficiency of blading ¼ Kinetic energy supplied 47:91 ð47:91Þð2Þð103 Þ ¼ ¼1 2 ð0:182Þð9252 Þ 2 mC 1 ¼ 0:6153 or 61:53% 6.182) (1052.22 m/s. V 1 ¼ 448:4m/s: Copyright 2003 by Marcel Dekker. Inlet angle of blades b1 ¼ 27. Solution: Blade speed U is given by: U¼ pDN ðp £ 1050Þ £ ð2800Þ ¼ ¼ 154 m/s 60 ð1000Þ £ ð60Þ The velocity diagram is shown in Fig.Steam Turbines 263 1. 3. Applying the cosine rule to the triangle ABC. The blade is running at 2800 rpm and blade diameter is 1050 mm.e. DCw ¼ Cw1 þ C w2 ¼ 554:42 þ 153:54 ¼ 707:96 m/s Relative velocity at the rotor outlet is: V2 ¼ C a2 155 ¼ 344:4 m/s ¼ sin ðb2 Þ sin ð26:758Þ Copyright 2003 by Marcel Dekker. sin ðb1 Þ ¼ C1 sin ða1 Þ 590 sin ð208Þ ¼ 0:450 ¼ V1 448:4 and: b1 ¼ 26. Inc.24 Velocity diagram for Example 6. Ca2 ¼ tan ðb2 Þ ¼ tan ðb1 Þ ¼ tan ð26:758Þ ¼ 0:504 U þ C w2 Ca2 155 or: U þ C w2 ¼ 0:504 ¼ 0:504 ¼ 307:54 so : C w2 ¼ 307:54 2 154 ¼ 153:54 m/s Therefore.264 Chapter 6 Figure 6. C1 V1 ¼ sin ðACBÞ sin ða1 Þ But sin (ACB) ¼ sin (1808 2 b1) ¼ sin (b1) Therefore.7. All Rights Reserved . C w1 ¼ C 1 cos ða1 Þ ¼ 590 cos ð208Þ ¼ 554:42 m/s From triangle CEF. Applying the sine rule to the triangle ABC.758 From triangle ABD. Also find the stage efficiency and end thrust on the shaft. Solution: From triangle ABC (Fig. assuming velocity coefficient ¼ 0. Inc. Find the blade speed so that the steam shall pass on without shock.8: In one stage of an impulse turbine the velocity of steam at the exit from the nozzle is 460 m/s.Steam Turbines 265 Blade velocity coefficient is: V 2 344:4 ¼ 0:768 ¼ V 1 448:4 Work done on the blades per kg/s: k¼ DC w2 U ¼ ð707:96Þ £ ð154Þ £ ð1023 Þ ¼ 109 kW The diagram efficiency is: 2UDC w ð2Þ £ ð707:96Þ £ ð154Þ ¼ ¼ 0:6264 5902 C2 1 hd ¼ or. All Rights Reserved . 6.75.8. and blades are symmetrical.25 Velocity triangles for Example 6. hd ¼ 62:64% Illustrative Example 6. the nozzle angle is 228 and the blade angle is 338.25): C w1 ¼ C 1 cos 228 ¼ 460 cos 228 ¼ 426:5 m/s and: C a1 ¼ C1 sin 228 ¼ 460 sin 228 ¼ 172:32 m/s Figure 6. Copyright 2003 by Marcel Dekker. 9: In a Parson’s turbine.266 Chapter 6 Now. blade speed is given by: U ¼ C w1 2 BD ¼ 426:5 2 265:5 ¼ 161 m/s From Triangle BCD. diameter of the ring is 1. relative velocity at blade inlet is given by: V1 ¼ C a1 172:32 ¼ ¼ 316:2 m/s sin ð338Þ 0:545 Velocity coefficient: k¼ V2 . BF ¼ V 2 cos ð338Þ ¼ 237:2 £ cos ð338Þ ¼ 198:9 and Cw2 ¼ AF ¼ BF 2 U ¼ 198:9 2 161 ¼ 37:9 m/s Ca2 ¼ V 2 sin ð338Þ ¼ 237:2 sin ð338Þ ¼ 129:2 m/s The change in velocity of whirl: DC w ¼ C w1 þ Cw2 ¼ 426:5 þ 37:9 ¼ 464:4 m/s Diagram efficiency: 2UDC w ð2Þ £ ð464:4Þ £ ð161Þ ¼ ¼ 0:7067. the axial velocity of flow of steam is 0. All Rights Reserved .30 m and the rotational speed is 3000 rpm. or V 2 ¼ kV 1 ¼ ð0:75Þ £ ð316:2Þ ¼ 237:2 m/s V1 From Triangle BEF. from triangle BCD: BD ¼ Ca1 172:32 ¼ ¼ 265:5 tan ð338Þ 0:649 Hence. The outlet angle of the blade is 208. or 70:67%: 4602 C2 1 hd ¼ End thrust on the shaft per unit mass flow: Ca1 2 Ca2 ¼ 172:32 2 129:2 ¼ 43:12 N Design Example 6.5 times the mean blade speed.5 MPa passes through the blades where blade height is 6 cm. Inc. Neglect the effect of the blade thickness. Determine the inlet angles of the blades and power developed if dry saturated steam at 0. Copyright 2003 by Marcel Dekker. 228. 6. All Rights Reserved . U ¼ 60 60 Velocity of flow.Specific volume of vapor at 0.5 MPa. At any point B. velocity of flow. Ca ¼ (0. C1 ¼ V 2 . i. Thus. a 1 ¼ b2 .26 Velocity triangles for Example 6. the velocity triangle at the outlet is completed.. and V 1 ¼ C2 : DCw ¼ Cw1 þ C w2 ¼ 280:26 þ 76:23 ¼ 356:5 m/s The inlet angles are 53. For Parson’s turbine. is vg ¼ 0:3749 m3 /kg Therefore the mass flow is given by: _ m¼ AC 2 p £ ð1:30Þ £ ð6Þ £ ð102Þ ¼ 66:7 kg/s ¼ x2 vg2 ð100Þ £ ð0:3749Þ _ mUDC w ð66:7Þ £ ð356:5Þ £ ð102Þ ¼ 2425:4 kW ¼ 1000 1000 Power developed: P¼ Copyright 2003 by Marcel Dekker. Inc.5) £ (204) ¼ 102 m/s Draw lines AB and CD parallel to each other Fig. b 1 ¼ a2 . construct an angle a2 ¼ 208 to intersect line CD at point C.Steam Turbines 267 Figure 6. from the steam tables.9. Ca1 ¼ 102 m/s. By measurement.e. Solution: pDN p £ ð1:30Þ £ ð3000Þ ¼ ¼ 204 m/s The blade speed.26 at the distance of 102 m/s. steam is leaving the nozzle with velocity of 950 m/s and the nozzle angle is 208.27. V1 C1 C1 ¼ ¼ sinða1 Þ sinð1808 2 b1 Þ sinðb1 Þ or: sinðb1 Þ ¼ so: C 1 sinða1 Þ ð950Þ £ ð0:342Þ ¼ 0:535 ¼ V1 607 b1 ¼ 32:368 From Triangle ACD. and (3) the horsepower developed. The mean blade speed is 380 m/s and the blades are symmetrical. All Rights Reserved .10.10: In an impulse turbine. Calculate (1) the blade angle. The nozzle delivers steam at the rate of 12 kg/min. Cw1 ¼ C1 cos ða1 Þ ¼ 950 £ cos ð208Þ ¼ ð950Þ £ ð0:9397Þ ¼ 892:71m/s Copyright 2003 by Marcel Dekker.27 Velocity triangles for Example 6. the velocity triangle at the blade inlet can be constructed easily as shown in Fig. Neglect friction losses. Design Example 6. U and C1. V 2 ¼ U 2 þ C 2 2 2UC 1 cosa1 1 1 ¼ 9502 þ 3802 2 ð2Þ £ ð950Þ £ ð380Þ £ cos208 ¼ 607m/s Now. Solution: With the help of a1. Applying the cosine rule to the triangle ABC. Inc. applying the sine rule to the triangle ABC. (2) the tangential force on the blades.268 Chapter 6 Figure 6. 6. 11: In an impulse turbine.28 Velocity triangles for Example 6.11. whose tips are both 348. using triangle BEF and neglecting friction loss.e. Solution: Velocity triangles for this problem are shown in Fig. Inc. calculate the blade speed. and efficiency when the turbine develops 1600 kW.28. All Rights Reserved . i.: V1 ¼ V2 BF ¼ V 2 cos b2 ¼ 607 £ cos 32:368 ¼ 512:73 Therefore. From the triangle ACD. C w2 ¼ BF 2 U ¼ 512:73 2 380 ¼ 132:73 m/s Change in velocity of whirl: DCw ¼ Cw1 þ C w2 ¼ 892:71 þ 132:73 ¼ 1025:44 m/s Tangential force on blades: _ F ¼ mDCw ¼ ð12Þ £ ð1025:44Þ ¼ 205N 60 ð12Þ £ ð1025:44Þ £ ð380Þ ¼ 104:47 hp ð60Þ £ ð1000Þ £ ð0:746Þ Horsepower.Steam Turbines 269 As b1 ¼ b2. _ P ¼ mUDC w ¼ Design Example 6. the velocity of steam at the exit from the nozzle is 700 m/s and the nozzles are inclined at 228 to the blades. end thrust on the shaft. If the relative velocity of steam to the blade is reduced by 10% while passing through the blade ring. C a1 ¼ C1 sin a1 ¼ 700 £ sin 228 ¼ 262:224 m/s Figure 6. Copyright 2003 by Marcel Dekker. 6. or 78:44%: 7002 C2 1 Illustrative Example 6. DC w ¼ C w1 þ Cw2 ¼ 649 þ 90:2 ¼ 739:2 m/s Mass flow rate is given by: _ P ¼ mUDCw or _ m¼ ð1600Þ £ ð1000Þ ¼ 8:325 kg/s ð739:2Þ £ ð260Þ Thrust on the shaft.32) £ (0.829) ¼ 389 Hence. BD ¼ Cw1 2 U ¼ V1 cos b1 ¼ (469.12: The moving and fixed blades are identical in shape in a reaction turbine.270 Chapter 6 and V1 ¼ C a1 262:224 ¼ ¼ 469:32 m/s sin ðb1 Þ sin 348 Whirl component of C1 is given by Cw1 ¼ C1 cos ða1 Þ ¼ 700 cos ð228Þ ¼ 700 £ 0:927 ¼ 649 m/s Now. Cw2 ¼ 350:2 2 260 ¼ 90:2 m/s Then. Ca2 ¼ V 2 sin ðb2 Þ ¼ 422:39 sin 348 ¼ 236:2 m/s And U þ Cw2 ¼ V 2 cos 348 ¼ ð422:39Þ £ ð0:829Þ ¼ 350:2 m/s Therefore. V 2 ¼ ð0:90Þ £ ð469:32Þ ¼ 422:39 m/s From velocity triangle BEF. blade speed U ¼ 649 2 389 ¼ 260 m/s Using the velocity coefficient to find V2: i:e:. All Rights Reserved . _ F ¼ mðC a1 2 Ca2 Þ ¼ 8:325ð262:224 2 236:2Þ ¼ 216:65 N Diagram efficiency: hd ¼ 2UDC w ð2Þ £ ð739:2Þ £ ð260Þ ¼ ¼ 0:7844. The nozzle angle is 208. The absolute velocity of steam leaving the fixed blade is 105 m/s. Inc. Assume axial Copyright 2003 by Marcel Dekker. and the blade velocity is 40 m/s. Solution: For 50% reaction turbine Fig. Inc.29 Velocity triangles for Example 6. C w1 ¼ C 1 cos a1 ¼ 105 cos 208 ¼ 98:67 m/s Applying cosine rule to the Triangle ABC: V 2 ¼ C2 þ U 2 2 2C 1 U cos a1 1 1 so: V1 ¼ Now.29. From the velocity triangle ACD. All Rights Reserved . BD ¼ C w1 2 U ¼ V 1 cos b1 ¼ 98:67 2 40 ¼ 58:67 Hence. cosb1 ¼ 58:67 ¼ 0:853. velocity is constant through the stage.Steam Turbines 271 Figure 6. a1 ¼ b2. Determine the horsepower developed if the steam flow rate is 2 kg/s. and b1 ¼ 31:478 68:79 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1052 þ 402 2 ð2Þ £ ð105Þ £ ð40Þ £ cos208 ¼ 68:79 m/s Change in the velocity of whirl is: DCw ¼ Cw1 þ C w2 ¼ 98:67 þ 58:67 ¼ 157:34 m/s Copyright 2003 by Marcel Dekker.12. 6. and a2 ¼ b1. and a2 ¼ b1 ¼ 258 C1 ¼ 90 m/s From the velocity triangle. Inc. If the steam flow rate is 10 kg/s and the rotor diameter is 0.72 m.30 shows the velocity triangles. Cw1 ¼ C1 cos ða1 Þ ¼ 90 cos 188 ¼ 85:6 m/s Ca1 ¼ CD ¼ C 1 sin a1 ¼ 90 sin 188 ¼ 27:8 m/s From triangle BDC BD ¼ C a1 27:8 27:8 ¼ ¼ 65:72 m/s ¼ sin ðb1 Þ sin ð258Þ 0:423 Hence blade velocity is given by: U ¼ C w1 2 BD ¼ 85:6 2 65:62 ¼ 19:98 m/s: Applying the cosine rule. a1 ¼ b2 ¼ 188.13. The pressure and temperature of steam at the inlet to the turbine are 5 bar and 2508C. respectively. Solution: Figure 6. V 2 ¼ C 2 þ U 2 2 2C 1 U cos a1 1 1 ¼ 902 þ 19:982 2 ð2Þ £ ð90Þ £ ð19:98Þ cos 188 V 1 ¼ 71:27m/s Copyright 2003 by Marcel Dekker.13: The inlet and outlet angles of blades of a reaction turbine are 25 and 188.30 Velocity triangles for Example 6. Horsepower developed is: _ P ¼ mUDCw ¼ ð2Þ £ ð157:34Þ £ ð40Þ ¼ 16:87 hp ð0:746Þ £ ð1000Þ Illustrative Example 6. find the blade height and power developed.272 Chapter 6 Figure 6. The velocity of steam at the exit from the fixed blades is 90 m/s. All Rights Reserved . Therefore. determine the blade height: RPM: Power developed: Steam mass flow rate: Stage absolute pressure: Steam dryness fraction: 440 5:5 MW 6:8 kg/kW ÿ h 0:90 bar 0:95 Exit angles of the blades: 708 (angle measured from the axial flow direction). for 50% reaction steam turbine.2 times the mean blade speed. Inc. and h ¼ 0:0075 m or 0:75 cm Design Example 6. The outlet relative velocity of steam is 1.Steam Turbines 273 From triangle AEF.5. the specific volume of steam is: v ¼ 0:4744 m3 /kg Blade height is given by the volume of flow equation: v ¼ pDhC a where Ca is the velocity of flow and h is the blade height. 0:4744 ¼ p £ ð0:72Þ £ ðhÞ £ ð27:8Þ. All Rights Reserved . The ratio of the rotor hub diameter to blade height is 14. C w2 ¼ C 2 cosða2 Þ ¼ 71:27 cos 258 ¼ 64:59 m/s Change in the velocity of whirl: DCw ¼ Cw1 þ C w2 ¼ 85:6 þ 64:59 ¼ 150:19 m/s Power developed by the rotor: _ P ¼ mUDC w ¼ ð10Þ £ ð19:98Þ £ ð150:19Þ ¼ 30 kW 1000 From superheated steam tables at 5 bar.14: From the following data. 2508C. Copyright 2003 by Marcel Dekker. vg ¼ 1.90 bar.869 m3/kg. Substituting the value of U in the above equation.31 shows the velocity triangles. Copyright 2003 by Marcel Dekker. All Rights Reserved .31 Velocity triangles for Example 6. U¼ pDN 2pN ðDh þ hÞ ¼ 60 ð60Þ £ ð2Þ where Dh is the rotor diameter at the hub and h is the blade height.14. Ca2 ¼ ð0:41Þ £ ð2pÞ £ ð440Þð14:5h þ hÞ ¼ 146:45 h m/s ð2Þ £ ð60Þ Annular area of flow is given by: A ¼ phðDh þ hÞ ¼ phð14:5h þ hÞ or A ¼ 15:5ph 2 Specific volume of saturated steam at 0. From the velocity diagram. Figure 6. Inc.274 Chapter 6 Solution: Figure 6. V 2 ¼ 1:2U C a2 ¼ V 2 cos ðb2 Þ ¼ 1:2U cos 708 ¼ 0:41U m/s At mean diameter. as shown in Fig. Inc. and the relative velocity: V1 ¼ 482 m/s using the velocity coefficient V2 ¼ (0.95) ¼ 1.Steam Turbines 275 Then the specific volume of steam ¼ (1. 6.33. The mass flow rate is given by: _ m¼ ð5:5Þ £ ð103 Þ £ ð6:8Þ ¼ 10:39 kg/s 3600 But. Graphical solution: U ¼ 115 m/s C1 ¼ 590 m/s a1 ¼ 188 b2 ¼ 208 The velocity diagrams are drawn to scale.32 shows the velocity triangles. and h ¼ 0:137 m Design Example 6.9) £ (482) ¼ 434 m/s 115 m/s 590 m/s 188 378 0:9 Copyright 2003 by Marcel Dekker.869) £ (0.776 m3/kg. determine the power developed and the diagram efficiency: Blade speed: Velocity of steam exiting the nozzle: Nozzle efflux angle: Outlet angle from first moving blades: Blade velocity coefficient ðall bladesÞ: Solution: Figure 6. _ m¼ C a2 A Ca2 15:5ph 2 ¼ v v ð146:45Þ £ ðhÞ £ ð15:5Þ £ ðph 2 Þ 1:776 Therefore: 10:39 ¼ or: h 3 ¼ 0:00259.15: From the following data for a two-row velocity compounded impulse turbine. All Rights Reserved . 46) N ¼ 281. mDCw2 ¼ (1) £ (281. Copyright 2003 by Marcel Dekker. Inc.32 Velocity triangle for Example 6. ˙ For the second row of moving blades. Figure 6.33 Velocity diagram for Example 6. C3. mDCw1 ¼ (1) £ (854) ¼ 854 N.276 Chapter 6 Figure 6. All Rights Reserved .15. The absolute velocity at the inlet to the second row of moving blades.46 N where DCw1 and DCw2 are scaled from the velocity diagram. is equal to the velocity of steam leaving the fixed row of blades.16. i:e:. : C3 ¼ kC 2 ¼ ð0:9Þ £ ð316:4Þ ¼ 284:8 ˙ Driving force ¼ m DCw ˙ For the first row of moving blades. All Rights Reserved . and the steam flow rate is 5 kg/s. or 90:45% Axial thrust on the first row of moving blades (per kg/s): ¼ mðC a1 2 C a2 Þ ¼ ð1Þ £ ð182:32 2 148:4Þ ¼ 33:9 N _ Axial thrust on the second row of moving blades (per kg/s): ¼ mðC a3 2 C a4 Þ ¼ ð1Þ £ ð111:3 2 97:57Þ ¼ 13:73 N _ Total axial thrust: ¼ 33:9 þ 13:73 ¼ 47:63 N per kg/s Design Example 6. Solution: Figure 6.Steam Turbines 277 Total driving force ¼ 854 þ 281. or 75:02% 5902 Therefore.46 N per kg/s Power ¼ driving force £ blade velocity ð1135:46Þ £ ð115Þ ¼ 130:58 kW per kg/s ¼ 1000 Energy supplied to the wheel ¼ mC 2 ð1Þ £ ð5902 Þ 1 ¼ ¼ 174:05kW per kg/s 2 ð2Þ £ ð103 Þ ð130:58Þ £ ð103 Þ £ ð2Þ ¼ 0:7502. degree of reaction.34 shows the velocity triangles. 6. b2 ¼ 328. a2 ¼ 71. the diagram efficiency is: hd ¼ Maximum diagram efficiency: ¼ cos2 a1 ¼ cos2 88 ¼ 0:9045. work output per kg is given by: W t ¼ UðC w1 þ Cw2 Þ ¼ ð300Þ £ ð450 cos 258 þ 247 cos 71:18Þ ¼ 14.From velocity triangles.18. 354 J Copyright 2003 by Marcel Dekker. the blade angles for the stators and rotors of each stage are: a1 ¼ 258. Inc. If the blade velocity is 300 m/s.16: In a reaction stage of a steam turbine. b1 ¼ 608.46 ¼ 1135. and the axial thrust. find the power developed. The velocity triangles can easily be constructed as the blade velocity and blade angles are given. V2 ¼ 375 m/s.28. V1 ¼ 200 m/s. By measurement. 6. V2 2 V2 4432 2 2202 2 1 ¼ 0:5051.94. 354Þ ¼ 732 kW 1000 Degree of reaction is given by: L¼ Axial thrust: F ¼ mðC a1 2 Ca2 Þ ¼ ð5Þ £ ð190:5 2 234Þ ¼ 2217:5 N _ The thrust is negative because its direction is the opposite to the fluid flow. Also find the static pressures at the rotor inlet and exit if the stator efficiency is 0. and the rotor efficiency is 0.2 kg/s of steam flow.34 Velocity diagram for Example 6. Power output: _ mW t ¼ ð5Þ £ ð1. If the blade speed is 250 m/s.17. or 50:51% ¼ 2 £ Wt ð2Þ £ ð14. 354Þ Copyright 2003 by Marcel Dekker. The blade angles for the rotor and stator of each stage are: a1 ¼ 258. b2 ¼ 328. All Rights Reserved . 6. a2 ¼ 70.17: Steam enters the first row of a series of stages at a static pressure of 10 bars and a static temperature of 3008C. C1 ¼ 400 m/s. Design Example 6. the velocity triangles for the inlet and outlet are shown in Fig. Solution: Using the given data.93 and the carryover efficiency is 0.278 Chapter 6 Figure 6. 46.34. b1 ¼ 608. find the degree of reaction and power developed for a 5. C2 ¼ 225 m/s.89. Inc. 23. 469 J/kg. 09.01 MPa.1 Dry saturated steam is expanded in a steam nozzle from 1 MPa to 0. Inc. 595 J/kg. 09. All Rights Reserved . or 123:6 kJ/kg Isentropic static enthalpy drops in the rotor: Dhr 0 ¼ W 1. 685Þ ¼ 570:37 kW 1000 Isentropic static enthalpy drop in the stator: Á À 2 Á À C1 2 C2 4002 2 ð0:89Þ £ ð2252 Þ 2 0 Dhs ¼ ¼ 0:93 hs _ P ¼ mW ¼ ¼ 1. (0.25)] L¼ V2 2 V2 3752 2 2002 2 1 ¼ 0:4587. Calculate dryness fraction of steam at the exit and the heat drop. 685 ¼ hr hs ð0:94Þ £ ð0:93Þ ¼ 1. T1 ¼ 2358C Rotor outlet conditions: P2 ¼ 5 bar. or 125:47 kJ/kg Since the state of the steam at the stage entry is given as 10 bar. 09. 09. 685 J/kg Degree of reaction [Eq. 25.Steam Turbines 279 Work done per unit mass flow: W t ¼ ð250Þ £ ð400 cos 258 þ 225 cos 70:28Þ ¼ 1. Enthalpy at nozzle exit:  à h1 2 Dh0 stator ¼ 3051:05 2 123:6 ¼ 2927:5kJ/kg Enthalpy at rotor exit:  à h1 2 Dh0 rotor ¼ 3051:05 2 125:47 ¼ 2925:58kJ/kg The rotor inlet and outlet conditions can be found by using the Mollier Chart. (6. 3008C.79. T2 ¼ 2208C PROBLEMS 6. or 45:87% ¼ 2 £ Wt ð2Þ £ ð1. 685Þ Power output: ð5:2Þ £ ð1. 686 kJ/kg) Copyright 2003 by Marcel Dekker. Rotor inlet conditions: P1 ¼ 7 bar. 5 6.42%) The steam jet velocity of a turbine is 615 m/s and nozzle angle is 228. If there are 16 nozzles.8 kW) Steam expands isentropically in the reaction turbine from 4 MPa.225 MPa. determine the inlet and outlet blade angles.2%) 6. 4008C to 0. Assume nozzle efficiency of 90%.84 and the nozzle angles and blade angles are 20 and 368 respectively.5 mm.8 Copyright 2003 by Marcel Dekker. find the mass of steam discharged per hour. If the speed of the blade is 200 m/s and there is no thrust on the blades. whose tip angles are both 338. Assume mean blade radius ¼ 600 mm and the axial velocity at the outlet ¼ 160 m/s.70 and the blade is rotating at 3000 rpm.4 6. Find the dryness fraction and velocity of steam at the exit from the nozzle. Find the dryness fraction and velocity of steam at the exit. Neglect the initial velocity of the steam.86. All Rights Reserved . Assume constant axial velocity throughout the stage and the blade speed is 160 m/s. find the blade speed so that steam passes through without shock and find the diagram efficiency.3 6. Inc.2 Steam initially dry and at 1.43 kW. If the velocity of the steam at the exit from the nozzle is 650 m/s.7 6.6 6. 310 . by how much would the discharge be increased? (1. (273 m/s.280 Chapter 6 6. If the steam has velocity of 120 m/s at the entry to the nozzles.756.75 MPa.5 MPa to 0.376 kg/h) Dry saturated steam expands isentropically in a nozzle from 2.4%) Steam is supplied from the nozzle with velocity 400 m/s at an angle of 208 with the direction of motion of moving blades. 1251. (0. 1234.5 KPa. Assume velocity coefficient ¼ 0. and exit pressure of steam is 1. The blade velocity coefficient ¼ 0. (0.05 MPa. find the cross-sectional area of the exit of each nozzle for a total discharge to be 280 kg/min. The turbine efficiency is 0. 3008C.30 MPa. and mass flow rate of steam is 14 kg/s. The nozzles are inclined at 208 to the blades. 867. neglecting losses. 458. 0. 49. How many stages are there in the turbine? (8 stages) Consider one stage of an impulse turbine consisting of a converging nozzle and one ring of moving blades.5 MPa is expanded adiabatically in a nozzle to 7. Determine the work output per unit mass flow of steam and diagram efficiency. (378 500 . (93. and the power developed by the turbine. 88.68 m/s) The nozzles receive steam at 1. 33.26 m/s.36 cm2.862. If the exit diameter of the nozzles is 12. Steam Turbines 281 6.66 kW. (2) tangential force. The velocity of steam at the exit from the nozzle is 650 m/s. Inc. (388. and (3) power developed. (79. blade height stagnation enthalpy.10 Determine the following: (1) blade inlet angle. velocity of steam at nozzle exit diameter enthalpy. All Rights Reserved . find the diagram efficiency and the end thrust on the shaft when the blade ring develops 1650 kW. 449 N) The following refer to a stage of a Parson’s reaction turbine: Mean diameter of the blade ring: 92 cm Blade speed: Inlet absolute velocity of steam: Blade outlet angle: Steam flow rate: 3000 rpm 310 m/s 208 6:9 kg/s 6.7 kW) NOTATION C D h h0 h1 h2 h00 h0l h02 k N R. If the relative velocity of steam to the blades is reduced by 14% in passing through the blade ring. 384. U V a b DCw Dh Dh0 absolute velocity. 2. The nozzle angles are 228 and the blade angles are 358.2%.9 One stage of an impulse turbine consists of a converging nozzle and one ring of moving blades. static enthalpy at the inlet to the fixed blades enthalpy at the entry to the moving blades enthalpy at the exit from the moving blades stagnation enthalpy at the entry to the fixed blades stagnation enthalpy at the entry to the fixed blades stagnation enthalpy at the exit from the moving blade blade velocity coefficient rotational speed reheat factor blade speed relative velocity angle with absolute velocity angle with relative velocity change in the velocity of whirl actual enthalpy drop isentropic enthalpy drop Copyright 2003 by Marcel Dekker. F. ambient radial whirl Copyright 2003 by Marcel Dekker. Inc.total efficiency degree of reaction SUFFIXES 0 1 2 a r w inlet to fixed blades inlet to moving blades outlet from the moving blades axial. All Rights Reserved .282 Chapter 6 hd hn hs ht hts htt L diffuser efficiency nozzle efficiency stage efficiency turbine efficiency total .to .static efficiency total .to . All Rights Reserved . There are two basic types of turbines: the axial flow type and the radial or centrifugal flow type. In the impulse turbine. Axial flow turbines are also normally employed in industrial and shipboard applications.1 shows a rotating assembly of the RollsRoyce Nene engine. The moving blades convert this kinetic energy into mechanical energy and also direct the gas flow to the next stage Copyright 2003 by Marcel Dekker. The turbines can be classified as (1) impulse and (2) reaction.1 INTRODUCTION TO AXIAL FLOW TURBINES The axial flow gas turbine is used in almost all applications of gas turbine power plant. On this particular engine.7 Axial Flow and Radial Flow Gas Turbines 7. converting this kinetic energy into shaft power to drive the compressor and the engine accessories. The axial flow turbine consists of one or more stages located immediately to the rear of the engine combustion chamber. the gases will be expanded in the nozzle and passed over to the moving blades. The turbine extracts kinetic energy from the expanding gases as the gases come from the burner. Figure 7. the single-stage turbine is directly connected to the main and cooling compressors. Inc. showing a typical single-stage turbine installation. The axial flow type has been used exclusively in aircraft gas turbine engines to date and will be discussed in detail in this chapter. Development of the axial flow gas turbine was hindered by the need to obtain both a high-enough flow rate and compression ratio from a compressor to maintain the air requirement for the combustion process and subsequent expansion of the exhaust gases. The pressure drop in the turbine is sufficient to keep the boundary layer fluid well behaved. is that the fluid undergoes a pressure rise in the compressor.284 Chapter 7 Figure 7.1 shows the axial flow turbine rotors. The efficiency of a well-designed turbine is higher than the efficiency of a compressor. as discussed in compressor design. or breakaway of the molecules from the surface. (Courtesy Rolls-Royce. can be easily avoided. 7. and separation problems. However. the turbine designer will face much more critical stress problem because the turbine rotors must operate in very high-temperature gases. pressure drop of expansion takes place in the stator as well as in the rotor-blades. and the design process is often much simpler. In the case of reaction turbine. additional Copyright 2003 by Marcel Dekker. All Rights Reserved . It is much more difficult to arrange for an efficient deceleration of flow than it is to obtain an efficient acceleration.1 Axial flow turbine rotors. The main reason for this fact. Fig.) (multi-stage turbine) or to exit (single-stage turbine). which often can be serious in compressors. The blade passage area varies continuously to allow for the continued expansion of the gas stream over the rotor-blades. Since the design principle and concepts of gas turbines are essentially the same as steam turbines. Inc. Copyright 2003 by Marcel Dekker. which makes and an a2 with axial direction. 7.2 VELOCITY TRIANGLES AND WORK OUTPUT The velocity diagram at inlet and outlet from the rotor is shown in Fig.Axial Flow and Radial Flow Gas Turbines 285 information on turbines in general already discussed in Chapter 6 on steam turbines. The rotor-blade inlet angle will be chosen to suit the direction b2 of the gas velocity V2 relative to the blade at inlet. The magnitude and direction of the absolute velocity at exit from the rotor C3 at an angle a3 are found by vectorial addition of U to the relative velocity V3. All Rights Reserved . Figure 7.2. Gas leaves the nozzles or stator blades with an absolute velocity C2.2 Velocity triangles for an axial flow gas turbine. b2 and V2 are found by subtracting the blade velocity vector U from the absolute velocity C2. a3 is known as the swirl angle. 7. It is seen that the nozzles accelerate the flow. (angle measured from the axial direction) enters the nozzle (in impulse turbine) or stator blades (in reaction turbine). the gas leaves with relative velocity V3 at angle b3. After expansion in the rotor-blade passages. Inc. imparting an increased tangential velocity component. Gas with an absolute velocity C1 making an angle a1. Inc. is in the direction opposite to the blade speed U. V3 will always be greater than V2 because part of pressure drop will be converted into kinetic energy in the moving blade. it is more convenient to construct the velocity diagrams in combined form. In multi-stage turbine. the gas pressure is p2 and temperature T2. For short blades. V3 is either slightly less than V2 (due to friction) or equal to V2. DCw.3 Combined velocity diagram. as shown in Fig. After expansion. the stage Figure 7. and other radii points. 7. We shall assume in this section that we are talking about conditions at the mean diameter of the annulus. Assuming unit mass flow. The gas leaves the rotorblade passages at pressure p3 and temperature T3. work done by the gas is given by W ¼ U ðC w2 þ Cw3 Þ From velocity triangle U ¼ tan a2 2 tan b2 ¼ tan b3 2 tan a3 Ca ð7:2Þ ð7:1Þ In single-stage turbine. a1 ¼ 0 and C1 ¼ Ca1. Note that the velocity diagram of the turbine differs from that of the compressor. 2-D approach in design is valid but for long blades. 3-D approach in the designing must be considered. Copyright 2003 by Marcel Dekker. The reaction to this change in the tangential momentum of the fluid is a torque on the rotor in the direction of motion. But in reaction stage. Just as with the compressor blading diagram.3.286 Chapter 7 The gas enters the nozzle with a static pressure p1 and temperature T1. in that the change in tangential velocity in the rotor. a1 ¼ a3 and C1 ¼ C3 so that the same blade shape can be used. In terms of air angles. All Rights Reserved . The blade speed U increases from root to tip and hence velocity diagrams will be different for root. tip. and it is known as total-to-total stage efficiency. Thus totalto-static efficiency. Total-to-total stage efficiency term is used when the leaving kinetics energy is utilized either in the next stage of the turbine or in propelling nozzle. flow is accelerating and molecules will not break away from the surface and growth of the boundary layer along the annulus walls is negligible.3 DEGREE OF REACTION (L) Degree of reaction is defined as L¼ Enthalpy drop in the moving blades Enthalpy drop in the stage Á h2 2 h3 Ca À tan b1 2 tan b2 ¼ ¼ h1 2 h3 2U ð7:8Þ This shows the fraction of the stage expansion. The stagnation pressure ratio of the stage p01 /p03 can be found from " ðg21Þ/g # 1 DT 0s ¼ hs T 01 1 2 ð7:5Þ p01 /p03 where hs is the isentropic efficiency given by T 01 2 T 03 T 01 2 T 003 hs ¼ ð7:6Þ The efficiency given by Eq. namely L¼ T2 2 T3 T1 2 T3 ð7:9Þ Copyright 2003 by Marcel Dekker.7) is the static temperature after an isentropic expansion from p01 to p3.6) is based on stagnation (or total) temperature. and it is usual to define in terms of the static temperature drops. (7. All Rights Reserved . If the leaving kinetic energy from the exhaust is wasted. (7. 7. hts ¼ T 01 2 T 03 0 T 01 2 T 3 ð7:7Þ 0 where T 3 in Eq.Axial Flow and Radial Flow Gas Turbines 287 work output per unit mass flow is given by or W ¼ U ðC w2 þ C w3 Þ ¼ UCað tan a2 þ tan a3 Þ À Á W ¼ UCa tan b2 þ tan b3 ð7:3Þ ð7:4Þ Work done factor used in the designing of axial flow compressor is not required because in the turbine. which occurs in the rotor. Inc. then total-to-static efficiency term is used. c and f. (7.e. f. It is defined as the ratio of the specific work of the stage to the square of the blade velocity—that is.4) À Á C p ðT 1 2 T 3 Þ ¼ Cp ðT 01 2 T 03 Þ ¼ UCa tan b2 þ tan b3 ð7:10Þ Temperature drop across the rotor-blades is equal to the change in relative velocity. is defined as the ratio of the inlet velocity Ca to the blade velocity U. and C3 ¼ C 1 From Eq. that is C p ðT 2 2 T 3 Þ ¼ Á 1À 2 V 2 V2 2 2 3 À Á 1 ¼ Ca 2 sec2 b3 2 sec2 b2 2 À Á 1 ¼ Ca 2 tan2 b3 2 tan2 b2 2 Thus L¼ Á Ca À tan b3 2 tan b2 2U ð7:11Þ 7. then Ca2 ¼ Ca3 ¼ Ca1 .4 BLADE-LOADING COEFFICIENT The blade-loading coefficient is used to express work capacity of the stage.288 Chapter 7 Assuming that the axial velocity is constant throughout the stage. All Rights Reserved . the blade-loading coefficient or temperature-drop coefficient c is given by c¼1 Á 2C p DT os 2Ca À W tan b2 þ tan b3 ¼ ¼ 2 2 U U 2U ð7:12Þ Flow Coefficient (f) The flow coefficient. Inc. i.. are dimensionless and Copyright 2003 by Marcel Dekker. f¼ Ca U ð7:13Þ This parameter plays the same part as the blade-speed ratio U/C1 used in the design of steam turbine. The two parameters. (7. b3 ¼ a2 and b2 ¼ a3.15).11). and hence the velocity diagram becomes symmetrical.14) and (7. we may express gas angles b2 and b3 in terms of c. In the case of steam turbine. Inc. Therefore the reaction stages are used where pressure drop per stage is low and also where the overall pressure ratio of the turbine is low. we have 1 ¼ tan b3 2 tan b2 f ð7:20Þ Comparing this with Eq. Now it is clear that a very long steam turbine with many reaction stages would be required to reduce the pressure by a ratio of 1000:1. we get 1 1 c þ 2L ð7:16Þ tan b3 ¼ 2f 2 1 1 tan b2 ¼ c 2 2L ð7:17Þ 2f 2 Using Eq.2).5. especially in the case of aircraft engine. Substituting L ¼ 0.5 in Eq. c ¼ 4f tan b3 2 2 ¼ 4f tan a2 2 2 c ¼ 4f tan b2 þ 2 ¼ 4f tan a3 þ 2 ð7:21Þ ð7:22Þ Copyright 2003 by Marcel Dekker. we have a1 ¼ a3 ¼ b2. we can prove that and and hence all the gas angles can be obtained in terms of c and f. (7. for L ¼ 0. (7. All Rights Reserved . Finally. pressure ratio can be of the order of 1000:1 but for a gas turbine it is in the region of 10:1.11) and (7. and f can be obtained easily as given below: Eqs. Now considering C1 ¼ C3. L. (7. which may have only three or four reaction stages. The gas angles in terms of c. and the stator and rotor-blades then have the same inlet and outlet angles.Axial Flow and Radial Flow Gas Turbines 289 useful to plot the design charts.12) can be written as À Á ð7:14Þ c ¼ 2f tan b2 þ tan b3 À Á f ð7:15Þ tan b3 2 tan b2 L¼ 2 Now. Let us consider 50% reaction at mean radius. (7. and f as follows: Adding and subtracting Eqs.2) tan a3 ¼ tan b3 2 1 f 1 tan a2 ¼ tan b2 þ f ð7:18Þ ð7:19Þ It has been discussed in Chapter 6 that steam turbines are usually impulse or a mixture of impulse and reaction stages but the turbine for a gas-turbine power plant is a reaction type. L. Inc. a large diameter. and low f means larger turbine annulus area for a given mass flow.5. 7. The nozzle loss coefficient. The low values of f and c imply low gas velocities and hence reduced friction losses. T2 at the nozzle exit is higher than T 2 . and a small frontal area. In Fig. ðp01 2 p02 Þ represents the pressure drop due to friction in the nozzle.4 Total-to-static efficiency of a 50% reaction axial flow turbine stage. But a low value of c means more stages for a given overall turbine output. Temperature. All Rights Reserved . including the effects of irreversibility.290 Chapter 7 Figure 7. the primary consideration is to have minimum weight. in terms of temperature may be defined as T2 2 T2 lN ¼ 2 C 2 /2C p 0 ð7:23Þ Copyright 2003 by Marcel Dekker. where low sfc is required. ðT 01 2 0 T 2 Þ represents the ideal expansion in the nozzle. lN. T2 is the temperature at the 0 nozzle exit due to friction. is given in Fig.5. 7. 7. for the gas turbine used in an aircraft engine. Therefore it is necessary to use higher values of c and f but at the expense of efficiency (see Fig. In industrial gas turbine plants.5 STATOR (NOZZLE) AND ROTOR LOSSES A T –s diagram showing the change of state through a complete turbine stage. However.4). T02 ¼ T01 because no work is done in the nozzle. relatively long turbine. would be accepted. 7. of low flow coefficient and low blade loading. Temperature T3 represents the temperature due to friction in the rotor-blade passages. and T 3 is the temperature after expansion in the rotor-blade passages alone. Nozzle loss coefficient in term of pressure yN ¼ p01 2 p02 p01 2 p2 ð7:24Þ lN and yN are not very different numerically. Inc. T03rel ¼ T02rel. The loss coefficient in terms of pressure drop for the rotor-blades is defined by lR ¼ p02 rel 2 p03 rel p03 rel 2 p3 ð7:26Þ Copyright 2003 by Marcel Dekker. From Fig. 7.Axial Flow and Radial Flow Gas Turbines 291 Figure 7.5. further expansion in 0 the rotor-blade passages reduces the pressure to p3. that is. T 3 is the final temperature 00 after isentropic expansion in the whole stage.5 T– s diagram for a reaction stage. The rotor-blade loss can be expressed by lR ¼ T3 2 T3 V 2 /2C p 3 00 ð7:25Þ As we know that no work is done by the gas relative to the blades. All Rights Reserved . 4 shows the effect of blade losses. the equation takes the form dh0 dCa dCw C 2 þ Cw ¼ Ca þ w ð7:28Þ dr dr r dr For constant stagnation enthalpy across the annulus (dh0/dr ¼ 0) and constant axial velocity (dCa/dr ¼ 0) then the whirl component of velocity Cw is inversely proportional to the radius and radial equilibrium is satisfied. It is the evident that exit losses become increasingly dominant as the flow coefficient is increased. These losses are of the order of 10 – 15% but can be lower for very low values of flow coefficient. the free vortex variation of nozzle angle a2 may be found as given below: C w2 r ¼ rCa2 tan a2 ¼ constant Ca2 ¼ constant Therefore a2 at any radius r is related to a2m at the mean radius rm by r m tan a2 ¼ tan a2m r 2 Similarly. on the total-tototal efficiency of turbine stage for the constant reaction of 50%. respectively. 7.6 FREE VORTEX DESIGN As pointed out earlier. Nozzle loss coefficients obtained from a large number of turbine tests are typically 0. That is. Figure 7.09 and 0. C w £ r ¼ constant ð7:29Þ The flow.05 for the rotor and stator rows. Inc.” Now using subscript m to denote condition at mean diameter. velocity triangles vary from root to tip of the blade because the blade speed U is not constant and varies from root to tip.29). a3 at outlet is given by r m tan a3m tan a3 ¼ r 3 ð7:30Þ ð7:27Þ ð7:31Þ Copyright 2003 by Marcel Dekker. determined with Soderberg’s correlation. is called a “free vortex. which follows Eq.292 Chapter 7 The loss coefficient in the stator and rotor represents the percentage drop of energy due to friction in the blades. which results in a total pressure and static enthalpy drop across the blades. Twisted blading designed to take account of the changing gas angles is called vortex blading. (7. As discussed in axial flow compressor (Chapter 5) the momentum equation is 1 dP C 2 ¼ w r dr r For constant enthalpy and entropy. All Rights Reserved . 1. that is. dh0/dr ¼ 0. Therefore 2 a2 Normally the change in vortex design has only a small effect on the performance of the blade while secondary losses may actually increase.6 T-s diagram for Example 7. whirl component at rotor inlet and turbine work output. Illustrative Example 7. tan b3 ¼ tan a2 2 ¼ r r Um m tan a2m 2 r 2 r m 2 Ca2 U Ca ð7:32Þ and b3 is given by r r Um m tan a3m þ tan b2 ¼ r m 3 Ca3 r 3 ð7:33Þ 7. this leads to the axial velocity distribution given by Cw2 r sin Ca2 r sin 2 a2 ¼ constant ¼ constant ð7:34Þ ð7:35Þ and since a2 is constant. Copyright 2003 by Marcel Dekker.88.7 CONSTANT NOZZLE ANGLE DESIGN As before. If mass flow rate of air is 28 kg/s. g ¼ 1.6).1 Consider an impulse gas turbine in which gas enters at pressure ¼ 5. and absolute velocity of gas at inlet is 140 m/s. All Rights Reserved . nozzle angle at outlet is 578. from velocity triangle.Axial Flow and Radial Flow Gas Turbines 293 The gas angles at inlet to the rotor-blade. 7.2 bar and leaves at 1.33.147 kJ/kgK (see Fig. Take. If a2 is constant.03 bar. The turbine inlet temperature is 1000 K and isentropic efficiency of the turbine is 0. then Ca2 is proportional to Cw1. determine the gas velocity at nozzle outlet. and Cpg ¼ 1. Inc. Figure 7. we assume that the stagnation enthalpy at outlet is constant. velocity of whirl at rotor inlet C w2 ¼ 829:33 sin 578 ¼ 695:5 m/s Turbine work output is given by W t ¼ mC pg ðT 01 2 T 02 Þ ¼ ð28Þð1:147Þð1000 2 708:72Þ ¼ 9354:8 kW Design Example 7. Assume g ¼ 1. the stagnation temperature and stagnation pressure at stage inlet are 8008C and 4 bar.2 In a single-stage gas turbine. Inc. and (4) the degree of reaction. and mean blade speed is 480 m/s. (2) the axial velocity which is constant through the stage.33.294 Chapter 7 Solution From isentropic p –T relation for expansion process ðg21Þ/g 0 T 02 p02 ¼ T 01 p01 ðg21Þ/g p02 1:03 ð0:248Þ 0 or ¼ 1000 ¼ 669 K T 02 ¼ T 01 p01 5:2 Using isentropic efficiency of turbine 0 T 02 ¼ T 01 2 ht T 01 2 T 02 ¼ 1000 2 0:88ð1000 2 669Þ ¼ 708:72 K Using steady-flow energy equation Á 1À 2 C 2 2 C 2 ¼ C p ðT 01 2 T 02 Þ 1 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Therefore. All Rights Reserved . respectively. Solution (1) The specific work output W ¼ C pg ðT 01 2 T 03 Þ Â Ã ¼ hts C pg T 01 1 2 ð1/4Þ0:33/1:33  à ¼ ð0:85Þð1:147Þð1073Þ 1 2 ð0:25Þ0:248 ¼ 304:42 kJ/kg Copyright 2003 by Marcel Dekker. total-to-static efficiency is 0. The nozzle efflux angle is 688. gas enters and leaves in axial direction.147 kJ/kgK. (3) the total-to-total efficiency. determine (1) the work done.85. C2 ¼ ½ð2Þð1147Þð1000 2 708:72Þ þ 19600 ¼ 829:33 m/s From velocity triangle. and Cpg ¼ 1. The exhaust static pressure is 1 bar. Mean blade speed is 350 m/s. is T 01 2 T 03 T 01 2 T 003 w ws s ¼ ¼ C2 3 ws C2 T 01 2 T 3 þ 2Cpg 2 3 hts 2C pg ¼ 304:42 ¼ 92:4% 304:42 ð256:23Þ2 2 2 £ 1147 0:85 htt ¼ (4) The degree of reaction Á Ca À tan b3 2 tan b2 L¼ 2U Ca U Ca U Ca £ tan a2 þ £ ¼ 2 2U Ca 2U Ca 2U Ca 256:23 tan a2 ¼ 1 2 tan 688 ¼ 33:94% 2U ð2Þð480Þ (from velocity triangle) L¼12 Design Example 7. Mass flow rate of gas is 15 kg/s and assume equal inlet and outlet velocities. Axial velocity is constant through the stage and equal to 250 m/s. stage exit swirl angle equal to 98. Nozzle efflux angle is 638. Copyright 2003 by Marcel Dekker. htt. All Rights Reserved . and power output.3 In a single-stage axial flow gas turbine gas enters at stagnation temperature of 1100 K and stagnation pressure of 5 bar. Cw1 ¼ 0 and specific work output is given by W ¼ UC w2 or C w2 ¼ W 304:42 £ 1000 ¼ ¼ 634:21 m/s U 480 From velocity triangle Cw2 sin a2 ¼ C2 or C w2 634:21 ¼ 684 m/s ¼ C2 ¼ sin a2 sin 688 Axial velocity is given by Ca2 ¼ 684 cos 688 ¼ 256:23 m/s (3) Total-to-total efficiency.Axial Flow and Radial Flow Gas Turbines 295 (2) Since a1 ¼ 0. Determine the rotorblade gas angles. a3 ¼ 0. degree of reaction. Inc. 7 Velocity triangles for Example 7. All Rights Reserved . Solution Refer to Fig. b3 ¼ 57:318 C2 ¼ From figure (b) C w2 ¼ Ca2 tan a2 ¼ 250 tan 638 ¼ 490:65 m/s Copyright 2003 by Marcel Dekker. Ca1 ¼ Ca2 ¼ Ca3 ¼ Ca ¼ 250 m/s From velocity triangle (b) Ca2 250 ¼ 550:67 m/s ¼ cos a2 cos 638 From figure (c) Ca3 250 ¼ 253 m/s C3 ¼ ¼ cos a3 cos 98 C w3 ¼ Ca3 tan a3 ¼ 250 tan 98 ¼ 39:596 m/s U þ C w3 350 þ 39:596 tan b3 ¼ ¼ 1:5584 ¼ Ca3 250 i:e:. Inc.7. 7.3.296 Chapter 7 Figure 7. A ¼ m/r2Ca2 and r2 ¼ p2 RT 2 C2 ð550:67Þ2 2 ¼ 1100 2 2Cp ð2Þð1:147Þð1000Þ ðT 01 ¼ T 02 Þ T 2 ¼ T 02 2 i. Inc. Solution Nozzle throat area. Take g ¼ 1.333. assuming nozzle loss coefficient. All Rights Reserved .e.147 kJ/kgK.Axial Flow and Radial Flow Gas Turbines 297 and tan b2 ¼ Cw2 2 U 490:65 2 350 ¼ 0:5626 ¼ Ca2 250 0 [ b2 ¼ 29821 Power output W ¼ mUCað tan b2 þ tan b3 Þ ¼ ð15Þð350Þð250Þð0:5626 þ 1:5584Þ/1000 ¼ 2784 kW The degree of reaction is given by L¼ ¼ Á Ca À tan b3 2 tan b2 2U 250 ð1:5584 2 0:5626Þ 2 £ 350 ¼ 35:56% Design Example 7..4 Calculate the nozzle throat area for the same data as in the precious question. and Cpg ¼ 1. TN ¼ 0.05. T 2 ¼ 967:81 K From nozzle loss coefficient T 2 ¼ T 2 2 lN 0 C2 0:05 £ ð550:67Þ2 2 ¼ 961:2 K ¼ 967:81 2 2Cp ð2Þð1:147Þð1000Þ Using isentropic p –T relation for nozzle expansion g/ðg21Þ 0 p2 ¼ p01 = T 01 /T 2 ¼ 5/ð1100/961:2Þ4 ¼ 2:915 bar Copyright 2003 by Marcel Dekker. Stage isentropic efficiency is 0. Inc.8 bar. and r2 ¼ ¼ r2 C 2 RT 2 ð0:287Þð967:81Þ Design Example 7. Solution Since L ¼ 0. gas enters and leaves the turbine axially. All Rights Reserved . Copyright 2003 by Marcel Dekker. and therefore nozzle is unchoked. and pressure ratio is 1. therefore L¼ hence T2 ¼ T3 From isentropic p –T relation for expansion T 003 ¼ À T 01 1000 ¼ 863:558 K Áðg21Þ/g ¼ ð1:8Þ0:249 p01 /p03 T2 2 T3 T 1 2 T 3.5 In a single-stage turbine. p2 pc Hence nozzle gas velocity at nozzle exit qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  à 2C pg ðT 01 2 T 2 Þ C2 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ½ð2Þð1:147Þð1000Þð1100 2 967:81Þ ¼ 550:68 m/s Therefore.85 and degree of reaction is zero. p01 .298 Chapter 7 Critical pressure ratio p01 /pc ¼ or g þ 1 g/ðg21Þ 2:333 4 ¼ ¼ 1:852 2 2 p01 /p2 ¼ 5/2:915 ¼ 1:715 Since p01 . Gas leaving the stage with velocity 270 m/s and blade speed at root is 290 m/s. nozzle throat area A¼ Thus A¼ 15 ¼ 0:026 m 2 ð1:05Þð550:68Þ m p2 ð2:915Þð102 Þ ¼ 1:05 kg/m 3 . Find the nozzle efflux angle and blade inlet angle at the root radius. Inlet stagnation temperature is 1000 K. a2 ¼ 518540 tan b2 ¼ ¼ i.. and rotational speed is 240 rps. mean blade speed is 300 m/s. C pg DT os ¼ U ðC w3 þ C w2 Þ ¼ UC w2 ðC w3 ¼ 0Þ ð1:147Þð1000Þð1000 2 884Þ C w2 ¼ ¼ 458:8 m/s Therefore. Isentropic efficiency of stage is equal to 0. 290 From velocity triangle sin a2 ¼ That is. The gas leaves the stage with velocity 390 m/s. Assuming inlet and outlet velocities are same and axial. using static and stagnation temperature relation T 3 ¼ T 03 2 C2 2702 3 ¼ 852 K ¼ T 2 ¼ 884 2 2C pg ð2Þð1:147Þð1000Þ Dynamic temperature C2 2 ¼ 1000 2 T 2 ¼ 1000 2 852 ¼ 148 K 2Cpg C2 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½ð2Þð1:147Þð148Þð1000Þ ¼ 582:677 m/s Since. gas enters the turbine at a stagnation temperature and pressure of 1150 K and 8 bar.88. and temperature drop in the stage is 145 K.Axial Flow and Radial Flow Gas Turbines 299 Using turbine efficiency À Á T 03 ¼ T 01 2 ht T 01 2 T 003 ¼ 1000 2 0:85ð1000 2 863:558Þ ¼ 884 K In order to find static temperature at turbine outlet. All Rights Reserved .e. find the blade height at the outlet conditions when the mass flow of gas is 34 kg/s. Inc.6 In a single-stage axial flow gas turbine. respectively. Cw2 2 U 458:8 2 290 ¼ Ca2 C2 cos a2 458:8 2 290 ¼ 0:47 582:677 cos 51:908 C w2 458:8 ¼ 0:787 ¼ 582:677 C2 Copyright 2003 by Marcel Dekker. b2 ¼ 25890 Design Example 7. e. Inc. p03 ¼ 4:31 bar Also from isentropic relation p3 ¼ À T p03 Á 0 /T g/ðg21Þ 03 3 ¼ 4:31 4:31 ¼ 3:55 bar ¼ ð985:23/938:697Þ4 1:214 p3 ð3:55Þð100Þ ¼ 1:32 kg/m 3 ¼ RT 3 ð0:287Þð938:697Þ m 34 ¼ 0:066 m 2 A3 ¼ ¼ r3 Ca3 ð1:32Þð390Þ r3 ¼ Copyright 2003 by Marcel Dekker.e. A3 [ h¼ 2 pr m U m ¼ pDm N or Dm ¼ i. ðU m Þ 300 ¼ ¼ 0:398 pN ð pÞð240Þ r m ¼ 0:199 m Temperature drop in the stage is given by T 01 2 T 03 ¼ 145 K Hence T 03 ¼ 1150 2 145 ¼ 1005 K T 3 ¼ T 03 2 C2 3902 3 ¼ 938:697 K ¼ 1005 2 2C pg ð2Þð1:147Þð1000Þ Using turbine efficiency to find isentropic temperature drop 145 ¼ 985:23 K 0:88 Using isentropic p– T relation for expansion process T 003 ¼ 1150 2 p01 8 8 p03 ¼ À ¼ ¼ Á 0 g/ðg21Þ ð1150/985:23Þ4 1:856 T 01 /T 03 i..300 Chapter 7 Solution Annulus area A is given by A ¼ 2 pr m h where h ¼ blade height r m ¼ mean radius As we have to find the blade height from the outlet conditions. in this case annulus area is A3.. All Rights Reserved . h¼ A3 0:066 ¼ 0:053 m ¼ 2 pr m ð2 pÞð0:199Þ Design Example 7.Axial Flow and Radial Flow Gas Turbines 301 Finally. p01 Pressure ratio. p01 /p03 Stagnation temperature drop Mean blade speed Mass flow. All Rights Reserved . determine the following quantities at the mean radius: (1) The blade loading coefficient and degree of reaction (2) The gas angles (3) The nozzle throat area Solution C pg ðT 01 2 T 03 Þ ð1147Þð145Þ ¼ ¼ 1:4 U2 3452 Using velocity diagram ð1Þ C¼ U/Ca ¼ tan b3 2 tan a3 or tan b3 ¼ 1 þ tan a3 F 1 þ tan 128 ¼ 0:75 b3 ¼ 57:18 From Equations (7. we have À Á C ¼ F tan b2 þ tan b3 Copyright 2003 by Marcel Dekker. m Rotational speed Flow coefficient. lN ¼ 0:05 Assuming the axial velocity remains constant and the gas velocity at inlet and outlet are the same. g ¼ 1:333.7 The following data refer to a single-stage axial flow gas turbine with convergent nozzle: Inlet stagnation temperature.14) and (7. Inc. 500 rpm 0:75 128 Cpg ¼ 1147 J/kg K. T 01 Inlet stagnation pressure.15). F Angle of gas leaving the stage 1100 K 4 bar 1:9 145 K 345 m/s 24 kg/s 14. All Rights Reserved . Inc.302 Chapter 7 and L¼ From which Á FÀ tan b3 2 tan b2 2 1 ðC þ 2LÞ 2F 1 ð1:4 þ 2LÞ 2 £ 0:75 tan b3 ¼ Therefore tan 57:18 ¼ Hence L ¼ 0:4595 ð2Þ tan b2 ¼ 1 ðC 2 2LÞ 2F 1 ð1:4 2 ½2½0:459Þ ¼ 2 £ 0:75 1 F 1 ¼ 0:321 þ 1:33 ¼ 1:654 0:75 b2 ¼ 17:88 tan a2 ¼ tan b2 þ ¼ tan 17:88 þ a2 ¼ 58:88 ð3Þ Ca1 ¼ U F ¼ ð345Þð0:75Þ ¼ 258:75 m/s C2 ¼ T 02 2 T 2 ¼ T 2 2 T 2s ¼ Ca1 258:75 ¼ 499:49 m/s ¼ cos 58:88 cos a2 C2 499:492 2 ¼ 108:76 K ¼ 2Cp ð2Þð1147Þ ðT N Þð499:492 Þ ð0:05Þð499:492 Þ ¼ ¼ 5:438 K ð2Þð1147Þ ð2Þð1147Þ T 2s ¼ T 2 2 5:438 T 2 ¼ 1100 2 108:76 ¼ 991:24 K T 2s ¼ 991:24 2 5:438 ¼ 985:8 K g=ðg21Þ p01 T 01 ¼ p2 T 2s Copyright 2003 by Marcel Dekker. Inc. a2 Gas-stage exit angle 20 kg/s 1150K 4 bar 255 m/s 345 m/s 608 128 Calculate (1) the rotor-blade gas angles. and power output and (3) the total nozzle throat area if the throat is situated at the nozzle outlet and the nozzle loss coefficient is 0.05. All Rights Reserved . Solution (1) From the velocity triangles C w2 ¼ Ca tan a2 ¼ 255 tan 608 ¼ 441:67 m/s C w3 ¼ Ca tan a3 ¼ 255 tan 128 ¼ 55:2 m/s V w2 ¼ C w2 2 U ¼ 441:67 2 345 ¼ 96:67 m/s b2 ¼ tan 21 Also V w2 96:67 ¼ tan 21 ¼ 20:88 Ca 255 V w3 400:2 ¼ 57:58 ¼ tan 21 255 Ca V w3 ¼ C w3 þ U ¼ 345 þ 55:2 ¼ 400:2 m/s [ b3 ¼ tan 21 Copyright 2003 by Marcel Dekker. T 01 Inlet pressure Axial flow velocity constant through the stage Blade speed. U Nozzle efflux angle. (2) the degree of reaction. bladeloading coefficient.8 A single-stage axial flow gas turbine with equal stage inlet and outlet velocities has the following design data based on the mean diameter: Mass flow Inlet temperature.Axial Flow and Radial Flow Gas Turbines 303 p2 ¼ 4 £ 985:8 1100 4 ¼ 2:58 r2 ¼ ð4Þ A1 ¼ p2 ð2:58Þð100Þ ¼ 0:911 kg /m 3 ¼ RT 2 ð0:287Þð991:24Þ Nozzle throat area ¼ m 24 ¼ 0:053 m 2 ¼ r1 C 1 ð0:907Þð499:49Þ m 24 ¼ 0:102 m 2 ¼ r1 Ca1 ð0:907Þð258:75Þ Design Example 7. Inc. C 2 ¼ Ca seca2 ¼ 255sec608 ¼ 510 m/s 1 2 2 C2 T2 2 T2 ¼ 0 or ð0:05Þð0:5Þð5102 Þ ¼ 5:67 1147 C2 5102 2 ¼ 1036:6 K ¼ 1150 2 2Cp ð2Þð1147Þ T 2 ¼ T 02 2 0 T 2 ¼ 1036:6 2 5:67 ¼ 1030:93 K p01 ¼ p2 p2 ¼ g=ðg21Þ T 01 1150 4 ¼ ¼ 1:548 T2 1030:93 4 ¼ 2:584 bar 1:548 p2 2:584 £ 100 ¼ 0:869 kg/m3 ¼ RT 2 0:287 £ 1036:6 r2 ¼ m ¼ r2 A 2 C 2 A2 ¼ 20 ¼ 0:045 m 2 0:869 £ 510 Copyright 2003 by Marcel Dekker.304 Chapter 7 ð2Þ L¼ ¼ C¼ ¼ Á FÀ tan b3 2 tan b2 2 255 ð tan 57:58 2 tan 20:88Þ ¼ 0:44 2 £ 345 Á Ca À tan b2 þ tan b3 U 255 ð tan 20:88 þ tan 57:58Þ ¼ 1:44 345 Power W ¼ mU ðCw2 þ Cw3 Þ ¼ ð20Þð345Þð441:67 þ 54:2Þ ¼ 3421:5 kW ð3Þ À 0Á Cp T 2 2 T 2 lN ¼ . All Rights Reserved . All Rights Reserved .10 The following particulars relate to a single-stage turbine of free vortex design: Inlet temperature. a3 ¼ b2 C2 ¼ U 340 ¼ ¼ 351:99 m/s cos 158 cos 158 C2 2 C2 ð351:99Þ2 2 ð105Þ2 2 3 ¼ 2Cp ð2Þð1147Þ 123896:96 2 11025 ð2Þð1147Þ Heat drop in blade moving row ¼ ¼ ¼ 49:2 K Therefore heat drop in a stage ¼ ð2Þð49:2Þ ¼ 98:41 K Number of stages ¼ 1173 2 943 230 ¼ ¼2 98:41 98:4 Design Example 7. T 01 Inlet pressure.Axial Flow and Radial Flow Gas Turbines 305 Illustrative Example 7.9 A single-stage axial flow gas turbine has the following data Mean blade speed 340 m/s Nozzle exit angle Axial velocity ðconstantÞ Turbine inlet temperature Turbine outlet temperature Degree of reaction Solution At 50%. p01 Mass flow Axial velocity at nozzle exit Blade speed at mean diameter Nozzle angle at mean diameter Ratio of tip to root radius 1100K 4 bar 20 kg/s 250 m/s 300 m/s 258 1:4 Copyright 2003 by Marcel Dekker. Inc. a2 ¼ b3 158 105 m/s 9008C 6708C 50% Calculate the enthalpy drop per stage and number of stages required. Take C pg ¼ 1:147 kJ/kg K. (2) The nozzle efflux angle at root and tip. find: (1) The total throat area of the nozzle. (3) The work done on the turbine blades. g ¼ 1:33 Solution For no loss up to throat g=ðg21Þ p* 2 2 4 ¼ ¼ ¼ 0:543 p01 2:33 gþ1 p * ¼ 4 £ 0:543 ¼ 2:172 bar T * ¼ 1100 2 2:33 4 ¼ 944 K Also T 01 ¼ T * þ C2 2C pg C* ¼ ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2Cpg T 01 2 T * pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2Þð1147Þð1100 2 944Þ ¼ 598 m/s r* ¼ (1) p* ð2:172Þð100Þ ¼ 0:802 kg/m3 ¼ * ð0:287Þð944Þ RT Throat area A¼ m 20 ¼ 0:042 m 2 ¼ rC * ð0:802Þð598Þ (2) Angle a1. Inc. at any radius r and a1m at the design radius rm are related by the equation tan a1 ¼ rm tan a1m r1 Copyright 2003 by Marcel Dekker. All Rights Reserved .306 Chapter 7 The gas leaves the stage in an axial direction. All Rights Reserved . The rotor. In inward flow radial turbine. the maximum diameter being about 0. The radial flow turbines are used in turbochargers for commercial (diesel) engines and fire pumps.8 RADIAL FLOW TURBINE In Sec. it was pointed out that in axial flow turbines the fluid moves essentially in the axial direction through the rotor. and run at very high speeds. 7. so that large overall pressure ratios are not difficult to obtain with axial turbines. They are very compact. In the radial type the fluid motion is mostly radial.1 “Introduction to Axial Flow Turbines”. The radial turbine is capable of a high-pressure ratio per stage than the axial one. Inc.2 m. gas enters in the radial direction and leaves axially at outlet. multistaging is very much easier to arrange with the axial turbine. though it is not always clear that any one type is superior. The choice of turbine depends on the application. For small mass flows. The mixed flow machine is characterized by a combination of axial and radial motion of the fluid relative to the rotor.Axial Flow and Radial Flow Gas Turbines 307 Given Tip radius rt ¼ ¼ 1:4 Root radius r r [ Mean radius ¼ 1:2 Root radius a1m ¼ 258 tan a1r ¼ r mean £ tan a1m r root ¼ 1:2 £ tan 258 ¼ 0:5596 [ a1r ¼ 29:238 tan a1t ¼ rr £ tan a1r ¼ rt 1 ð0:5596Þ ¼ 0:3997 1:4 [ a1t ¼ 21:798 rm r m Ca2 250 ¼ 643 m/s xC w2m ¼ ¼ 1:2x rr r r tan a2m tan 258 ð20Þð300Þð643Þ ¼ 3858 kW 1000 ð3Þ C w2 ¼ W ¼ mUC w2 ¼ 7. the radial machine can be made more efficient than the axial one. which is usually manufactured of Copyright 2003 by Marcel Dekker. However. 308 Chapter 7 Figure 7. Inc. Figure 7. Copyright 2003 by Marcel Dekker.8 Radial turbine photograph of the rotor on the right.9 Elements of a 908 inward flow radial gas turbine with inlet nozzle ring. All Rights Reserved . 9 VELOCITY DIAGRAMS AND THERMODYNAMIC ANALYSIS Figure 7. The stagnation pressure drops from p01 to p1 due to irreversibilities.11 shows the velocity triangles for this turbine. Figure 7. we have h01 ¼ h02. Figures 7. Note that this turbine is like a single-faced centrifugal compressor with reverse flow. cast nickel alloy.10 show photographs of the radial turbine and its essential parts.Axial Flow and Radial Flow Gas Turbines 309 Figure 7. As no work is done in the nozzle. The same nomenclature that we used for axial flow turbines. All Rights Reserved . will be used here.10 A 908 inward flow radial gas turbine without nozzle ring.12 shows the Mollier diagram for a 908 flow radial turbine and diffuser. The work done per unit mass flow is given by Euler’s turbine equation W t ¼ ðU 2 Cw2 2 U 3 C w3 Þ If the whirl velocity is zero at exit then W t ¼ U 2 C w2 ð7:37Þ ð7:36Þ Copyright 2003 by Marcel Dekker. has blades that are curved to change the flow from the radial to the axial direction.8 –7. Inc. 7. 12 Mollier chart for expansion in a 908 inward flow radial gas turbine. Copyright 2003 by Marcel Dekker. Inc.11 Velocity triangles for the 908 inward flow radial gas turbine. Figure 7.310 Chapter 7 Figure 7. All Rights Reserved . 7. T 01 2 T 03 ss efficiency being in the region of 80– 90% ð7:38Þ htt ¼ 7. C0 U 2 /C 0 ¼ 0:707 ð7:41Þ ð7:42Þ 7. Spouting velocities may be defined depending upon whether total or static conditions are used in the related efficiency definition and upon whether or not a diffuser is included with the turbine. Thus.12. without diffuser. All Rights Reserved . using subscript 0 for spouting velocity. 1 2 C ¼ h01 2 h03 ss 2 0 or ð7:39Þ 1 2 C ¼ h01 2 h3 ss ð7:40Þ 2 0 for the total and static cases. 0:71. the total-to-static efficiency. Now for isentropic flow throughout work done per unit mass flow W ¼ U 2 ¼ C2 /2 2 0 or U2 In practice. which has an associated kinetic energy equal to the isentropic enthalpy drop from turbine inlet stagnation pressure p01 to the final exhaust pressure. . Inc. respectively.Axial Flow and Radial Flow Gas Turbines 311 For radial relative velocity at inlet W t ¼ U2 2 In terms of enthalpy drop h02 2 h03 ¼ U 2 Cw2 2 U 3 Cw3 Using total-to-total efficiency T 01 2 T 03 .10 SPOUTING VELOCITY It is that velocity.11 TURBINE EFFICIENCY Referring to Fig. is defined as h01 2 h03 h01 2 h3 ss W W þ 1 C2 þ ðh3 2 h3ss Þ þ ðh3s 2 h3 ss Þ 2 3 ð7:43Þ hts ¼ ¼ Copyright 2003 by Marcel Dekker. U2/C0 lies in the range 0:68 . when no diffuser is used. (7. is defined as jn ¼ ¼ Enthalpy loss in nozzle Kinetic energy at nozzle exit h3s 2 h3 ss 0:5C2 ðT 3 /T 2 Þ 2 ð7:44Þ Rotor loss coefficient.44) and noting that U3 ¼ U2 r3/r2. therefore h3s 2 h3ss ¼ ðh 2 h2s ÞðT 3 /T 2 Þ Substituting in Eq.312 Chapter 7 Nozzle loss coefficient. is defined as jr ¼ h3 2 h3s 0:5V 2 3 ð7:45Þ But for constant pressure process. (7. and therefore usually neglected. that is. All Rights Reserved . we get # " 2 Á U2 2 T3 1À r3 2 2 ¼12 g21 1 2 cot a2 þ cot b3 2 T2 a2 r2 But the above value of T3/T2 is very small. W ¼ U 2 2 Substituting all those values in Eq. Inc. jr. V 3 ¼ U 3 cosec b3 . Thus " ( )#21 Á 1 r 3 av 2 À 2 2 2 hts ¼ 1 þ jn cosec a2 þ jr cosec b3 av þ cot b3 av 2 r2 ð7:48Þ !21 ð7:46Þ Copyright 2003 by Marcel Dekker. jn. T ds ¼ dh. and. then " ( )#21 2 Á 1 T3 r3 À 2 2 2 hts ¼ 1 þ jn cosec a2 þ jr cosec b3 þ cot b3 2 T2 r2 ð7:47Þ Taking mean radius. 1 r 3 ¼ ðr 3t þ r 3h Þ 2 Using thermodynamic relation for T3/T2. C 3 ¼ U 3 cot b3 .43) Á 1À hts ¼ 1 þ C 2 þ V 2 jr þ C 2 jn T 3 /T 2 W 3 2 3 Using velocity triangles C 2 ¼ U 2 cosec a2 . (7. Using nondimensional form of specific speed Ns ¼ NQ1/2 3 0 ðDh0 Þ3/4 ð7:52Þ where N is in rev/s. For the 908 inward flow radial turbine. and this change must be taken into account. then 2 ð7:50Þ htt ¼ W ¼ W ts 2 1 C2 2 3 1 1 hts 3 2 2W C2 [ 2 1 1 C2 1 1 r 3 av ¼ 2 3 ¼ 2 2 cot b3 av htt hts 2W hts 2 r2 ð7:51Þ Loss coefficients usually lie in the following range for 908 inward flow turbines jn ¼ 0:063–0:235 and jr ¼ 0:384–0:777 7.12 APPLICATION OF SPECIFIC SPEED We have already discussed the concept of specific speed Ns in Chapter 1 and some applications of it have been made already. volume flow rate changes significantly. Q3 is in m3/s and isentropic total-to-total enthalpy drop (from turbine inlet to outlet) is in J/kg.Axial Flow and Radial Flow Gas Turbines 313 Equation (7. The concept of specific speed was applied almost exclusively to incompressible flow machines as an important parameter in the selection of the optimum type and size of unit.43) as hts ¼ h01 2 h03 ðh01 2 h3 ss Þ 2 ðh03 2 h3 Þ 2 ðh3 2 h3s Þ 2 ðh3s 2 h3 ss Þ ¼ h01 2 h3 ss ðh01 2 h3ss Þ À 2 Á 2 2 2 ð7:49Þ ¼ 1 2 C3 þ jn C 2 þ jr V 3 /C 0 where spouting velocity C0 is given by h À Ág21=g i 1 h01 2 h3 ss ¼ C2 ¼ C p T 01 1 2 p3 /p01 2 0 The relationship between hts and htt can be obtained as follows: W ¼ U 2 ¼ hts W ts ¼ hts ðh01 2 h3ss Þ. All Rights Reserved . According to Balje. Copyright 2003 by Marcel Dekker. Inc. The hts can also be found by rewriting Eq. one suggested value of volume flow rate is that at the outlet Q3.46) is normally used to determine total-to-static efficiency. The volume flow rate through hydraulic machines remains constant. But in radial flow gas turbine. 11 A small inward radial flow gas turbine operates at its design point with a total-to-total efficiency of 0. rad Thus the Ns range is very small and Fig. where it is seen to match the axial flow gas turbine over the limited range of Ns. so that Q3 ¼ A3 C 3 . (7. rev C0 Ad 1 1 C 3 2 A3 2 . (7. 0:3 0:1 . 0:5 Then 0:3 . A3 /Ad . substituting this value 2 in Eq. Inc. rotor disc area Ad ¼ pD2 /4. The flow leaving the turbine is diffused to a pressure of 100 kPa and the velocity of Copyright 2003 by Marcel Dekker. All Rights Reserved .314 Chapter 7 U 2 ¼ pND2 and Dh0s ¼ 1 C 2 .52) 2 0 1/2 1/2 Q U2 U2 N s ¼ À 3 Á3/4 1 2 pD2 pND2 C pffiffiffi3/2 3/2 2 U2 Q3 1/2 ¼ C0 p ND3 2 2 0 ð7:53Þ 1 For 908 inward flow radial turbine. rev ð7:54Þ ND3 2 Equation (7. N s .13 shows the variation of efficiency with Ns. rad ¼ 2:11 C0 Ad ð7:55Þ ð7:56Þ Suggested values for C3/Co and A3/Ad are as follows: 0:04 . factorizing the Eq. Design Example 7. N s ¼ 0:336 1 1 C 3 2 A3 2 . U 2 /C0 ¼ pffiffi ¼ 0:707. 1:1. The stagnation pressure and temperature of the gas at nozzle inlet are 310 kPa and 1145K respectively. then 2 pffiffiffi C0 2 N ¼ U 2 /ð pD2 Þ ¼ 2 pD2 Q3 A3 C3 2 pD2 A3 C3 p2 pffiffiffi ¼ pffiffiffi ¼ Ad C 0 2 2 ND3 2C 0 D 2 2 2 Therefore.54) shows that specific speed is directly proportional to the square root of the volumetric flow coefficient. C 3 /C0 . Q3 1/2 N s ¼ 0:18 . 7. Assuming a uniform axial velocity at rotor exit C3.53).90. flow is negligible at that point. All Rights Reserved . Assume that the gas enters the impeller radially and there is no whirl at the impeller exit. find the impeller tip speed and the flow angle at the nozzle exit.13 Variation of efficiency with dimensionless specific speed. U2 ¼ 539. ðCw3 ¼ 0Þ 2 Now using isentropic p –T relation " g21=g # T 03ss p03 T 01 1 2 ¼ T 01 1 2 T 01 p01 Therefore U2 2 # p03 g21=g ¼ htt C p T 01 1 2 p01 " # 100 0:2498 ¼ 0:9 £ 1147 £ 1145 1 2 310 " [ Impeller tip speed.9. Inc. Take Cp ¼ 1:147 kJ/kg K. g ¼ 1:333: Solution The overall efficiency of turbine from nozzle inlet to diffuser outlet is given by htt ¼ T 01 2 T 03 T 01 2 T 03 ss Turbine work per unit mass flow W ¼ U 2 ¼ Cp ðT 01 2 T 03 Þ.45 m/s Copyright 2003 by Marcel Dekker. Given that the Mach number at exit from the nozzles is 0.Axial Flow and Radial Flow Gas Turbines 315 Figure 7. 316 Chapter 7 The Mach number of the absolute flow velocity at nozzle exit is given by M¼ C1 U1 ¼ a1 a1 sin a1 Since the flow is adiabatic across the nozzle. All Rights Reserved . we have T 01 ¼ T 02 ¼ T 2 þ or C2 U2 2 2 ¼ T2 þ 2Cp 2Cp sin 2 a2 But and T2 U2 gR 2 ¼12 . Inc. Rotor inlet tip diameter Rotor outlet tip diameter Rotor outlet hub diameter Ratio C 3 /C0 92 mm 64 mm 26 mm 0:447 Copyright 2003 by Marcel Dekker. but Cp ¼ T 01 g21 2C p T 01 sin 2 a2 À Á À Á U2 g 2 1 U2 g 2 1 T2 2 2 ¼12 ¼12 2 [ T 2gRT 01 sin 2 a2 2a01 sin 2 a2 01 2 T2 a2 a2 ¼ ¼ since T 01 ¼ T 02 T 01 a01 a02 a2 U2 ¼ a02 M 2 a02 sin a2 À Á 2 U2 g 2 1 U2 2 [ ¼12 2 M 2 a02 sin a2 2a02 sin 2 a2 Á 2 À g21 U2 1 1¼ þ 2 a02 sin a2 2 M2 À Á 2 g21 U2 1 sin 2 a2 ¼ þ 2 a02 2 M2 a2 ¼ gRT 02 ¼ ð1:333Þð287Þð1145Þ ¼ 438043 m 2 / s 2 02 539:452 0:333 1 þ 2 ¼ 0:9311 [ sin 2 a2 ¼ 438043 2 0:9 and or But Therefore nozzle angle a2 ¼ 758 Illustrative Example 7.12 The following particulars relate to a small inward flow radial gas turbine. (3) The power developed by the turbine. 500 rpm 1:75 kg/m3 (1) The dimensionless specific speed of the turbine. Inc. Solution (1) Dimensionless specific speed is 1 1 C 3 2 A3 2 N s ¼ 0:336 .54). rev C0 Ad Now À Á p D2 2 D2 3t 3h A3 ¼ 4 À Á p 0:0642 2 0:0262 ¼ ð2:73Þð1023 Þ m 2 ¼ 4 pD2 p 2 ¼ ð0:0922 Þ ¼ ð6:65Þð1023 Þ m 2 Ad ¼ 4 4 1 ½0:447½2:73 2 N s ¼ 0:336 6:65 ¼ 0:144 rev ¼ 0:904 rad (2) The flow rate at outlet for the ideal turbine is given by Eq. Q3 1=2 N s ¼ 0:18 ND3 2 1=2 ½Q3 ½60 0:144 ¼ 0:18 ½30. 500½0:0923 Dimensionless specific speed Hence Q3 ¼ 0:253 m 3 /s Copyright 2003 by Marcel Dekker. (2) The volume flow rate at impeller outlet.Axial Flow and Radial Flow Gas Turbines 317 Ratio U 2 /C 0 ðidealÞ Blade rotational speed Density at impeller exit Determine 0:707 30. All Rights Reserved . (7. (282 kW. 0. (2) the ratio of the static pressure at the rotor exit to the stagnation pressure at the nozzle inlet.1 A single-stage axial flow gas turbine has the following data: Inlet stagnation temperature 1100K The ratio of static pressure at the nozzle exit to the stagnation pressure at the nozzle inlet 0:53 Nozzle efficiency 0:93 Nozzle angle Mean blade velocity Rotor efficiency Degree of reaction C pg ¼ 1:147 kJ/kgK.214. All Rights Reserved . g ¼ 1:33 Find (1) the work output per kg/s of air flow. 208 454 m/s 0:90 50% Copyright 2003 by Marcel Dekker. For a free-vortex turbine blade with an impulse hub show that degree of reaction r 2 h L¼12 r where rh is the hub radius and r is any radius. 500½0:092 2 60 pND2 60 2 PROBLEMS 7. Inc. 83.78%) 7.318 Chapter 7 (3) The power developed by the turbine is given by _ W t ¼ mU 2 3 ¼ r3 Q3 U 2 3 ¼ 1:75 £ 0:253 £ ¼ 1:75 £ 0:253 £ ¼ 9:565 kW ½ p½30.3 Derive an equation for the degree of reaction for a single-stage axial flow turbine and show that for 50% reaction blading a2 ¼ b3 and a3 ¼ b2. and (3) the total-to-total stage efficiency.2 7. (2) the impeller tip diameter. (644K. The inlet stagnation temperature and pressure for an axial flow gas turbine are 1000K and 8 bar.4 A 50% reaction axial flow gas turbine has a total enthalpy drop of 288 kJ/kg. Calculate (1) the total-to-static efficiency of the turbine. (4) the blade outlet angle at the mean diameter. C ¼ 2.2 bar and isentropic efficiency of turbine is 85%. g ¼ 1:33: (5 stages) Show that for zero degree of reaction. N The flow into the rotor is radial and at exit the flow is axial at all radii. Take Cpg ¼ 1:147 kJ/kgK. T 3 Stagnation temperature at exit from rotor. Calculate the enthalpy drop per row of moving blades and the number of stages required when mean blade speed is 310 m/s. p3 Static temperature at exit from rotor. The axial velocity is constant through the stage.32 m. 0. T 01 Static pressure at exit from nozzles. (3) 0. respectively. p01 Stagnation temperature at inlet to nozzles.399.Axial Flow and Radial Flow Gas Turbines 319 7. T 03 Ratio r2 av r2 7. The nozzle exit angle is 708. (2) 0. find the exhaust stagnation temperature and entropy change of the gas.044 kJ/kgK) The performance date from inward radial flow exhaust gas turbine are as follows: Stagnation pressure at inlet to nozzles.7 705 kPa 1080K 515 kPa 1000K 360 kPa 923K 925K 0:5 25. 500 rpm Rotational speed. 2 0. Assume gas is air. blade-loading coefficient.6 7. [(1) 93%. (4) 72. and (5) the total-to-total efficiency of the turbine. (3) the enthalpy loss coefficient for the nozzle and rotor rows. The exhaust gas pressure is 1. p2 Static temperature at exit from nozzles.5 7. T 2 Static pressure at exit from rotor. Inc. The inlet angle to the rotating blade row is inclined at 208 with the axial direction. (5) 94%] NOTATION A C area absolute velocity Copyright 2003 by Marcel Dekker.019. All Rights Reserved .28. Inc. All Rights Reserved . blade height rotation speed specific speed pressure mean radius temperature rotor speed relative velocity nozzle loss coefficient in terms of pressure angle with absolute velocity angle with relative velocity stagnation temperature drop in the stage static temperature drop in the stage nozzle loss coefficient in radial flow turbine rotor loss coefficient in radial flow turbine flow coefficient isentropic efficiency of stage degree of reaction Copyright 2003 by Marcel Dekker.320 Chapter 7 C0 h N Ns P rm T U V YN a b DT0s DTs 1n 1r f hs L spouting velocity enthalpy. Small bubbles or cavities filled with vapor are formed. In turbines. 4. cavitation is most likely to occur at the downstream outlet end of a blade on the low-pressure leading face. Serious damage can occur from prolonged cavitation erosion. Copyright 2003 by Marcel Dekker. Inc. This cavitation imposes limitations on the rate of discharge and speed of rotation of the pump.8 Cavitation in Hydraulic Machinery 8. when the local static pressure of a liquid falls below the vapor pressure of the liquid. Vibration of machine. which reduces the effective flow areas. These bubbles collapse with tremendous force. noise is also generated in the form of sharp cracking sounds when cavitation takes place. where the fluid is locally accelerated over the vane surfaces. it causes the following undesirable effects: 1. In a centrifugal pump. All Rights Reserved . A drop in efficiency due to vapor formation. 2. Local pitting of the impeller and erosion of the metal surface. 3. giving rise to as high a pressure as 3500 atm. these low-pressure zones are generally at the impeller inlet. When cavitation occurs. The avoidance of cavitation in conventionally designed machines can be regarded as one of the essential tasks of both pump and turbine designers. which suddenly collapse on moving forward with the flow into regions of high pressure.1 INTRODUCTION Cavitation is caused by local vaporization of the fluid. this type of cavitation is often referred to as “tip” cavitation. the important factors are the relative velocities and the absolute pressures.322 Chapter 8 8. the same types of cavitation are found. Traveling cavitation is a type composed of individual transient cavities or bubbles. shrink. the cavities are found in the cores of vortices that form in zones of high shear. Inc. The attached or fixed cavity is stable in a quasi-steady sense. velocity. The discernible bubbles of incipient cavitation are small.1 Cavitation on Moving Bodies There is no essential difference between cavitation in a flowing stream and that on a body moving through a stationary liquid. 8. and then collapse. In both cases. All Rights Reserved . which does not have Copyright 2003 by Marcel Dekker. as it often occurs on the blade tips of ships’ propellers. Hydraulic machinery furnishes a typical example of a combination of the two conditions. the liquid and the guide surfaces are both in motion. With changes in conditions (pressure. The term fixed cavitation refers to the situation that sometimes develops after inception. Tip cavitation occurs not only in open propellers but also in ducted propellers such as those found in propeller pumps at hydrofoil tips. In vortex cavitation. 8. In the casing. the moving liquid flows past stationary guide surfaces. Such traveling transient bubbles may appear at the low-pressure points along a solid boundary or in the liquid interior either at the cores of moving vortices or in the high-turbulence region in a turbulent shear field.2 Cavitation Without Major Flow—Vibratory Cavitation The types of cavitation previously described have one major feature in common. in which the liquid flow detaches from the rigid boundary of an immersed body or a flow passage to form a pocket or cavity attached to the boundary. cavitation grows. When these are similar. Vibratory cavitation is another important type of cavitation. in the runner. The cavitation may appear as traveling cavities or as a fixed cavity. and the zone over which cavitation occurs is limited.2.2. Many cases of cavitation in a flowing stream occur in relatively long flow passages in which the turbulence is fully developed before the liquid reaches the cavitation zone. Fixed cavities sometimes have the appearance of a highly turbulent boiling surface. It is that a particular liquid element passes through the cavitation zone only once. the succeeding stages are distinguished from the incipient stage by the term developed. Vortex cavitation is one of the earliest observed types. temperature) toward promoting increased vaporization rates.2 STAGES AND TYPES OF CAVITATION The term incipient stage describes cavitation that is just barely detectable. In fact. One noticeable difference is that the turbulence level in the stationary liquid is lower. as they expand. which form in the liquid. 4 CAVITATION PARAMETER FOR DYNAMIC SIMILARITY The main variables that affect the inception and subsequent character of cavitation in flowing liquids are the boundary geometry.Cavitation in Hydraulic Machinery 323 this characteristic. As the vibratory pressure field is characteristic of this type of cavitation. Although it is accompanied sometimes by continuous flow. In the field of hydraulic machinery. the name “vibratory cavitation” follows. the flow variables Copyright 2003 by Marcel Dekker. which vibrates normal to its face and sets up pressure waves in the liquid. 8. the velocity is so low that a given element of liquid is exposed to many cycles of cavitation (in a time period of the order of milliseconds) rather than only one. it has been found that all types of turbines. whose performance may be seriously affected by the presence of cavitation. Inc. Although cavitation may be aggravated by poor design. 8. These pressure pulsations are generated by a submerged surface. will serve to emphasize the wide occurrence and the relative importance of this phenomenon. and even the various types of positive-displacement pumps may be troubled by it. with very few exceptions. are undesirable. are susceptible to cavitation. the effects of cavitation. No cavities will be formed unless the amplitude of the pressure variation is great enough to cause the pressure to drop to or below the vapor pressure of the liquid. the forces causing the cavities to form and collapse are due to a continuous series of high-amplitude. it may occur in even the best-designed equipment when the latter is operated under unfavorable condition.3 EFFECTS AND IMPORTANCE OF CAVITATION Cavitation is important as a consequence of its effects. These may be classified into three general categories: 1. high-frequency pressure pulsations in the liquid. Effects that produce damage on the solid-boundary surfaces of the flow 3. structures. Centrifugal and axial-flow pumps suffer from its effects. which form a low-specific-speed Francis to the high-specific-speed Kaplan. Uncontrolled cavitation can produce serious and even catastrophic results. The simple enumeration of some types of equipment. Extraneous effects that may or may not be accompanied by significant hydrodynamic flow modifications or damage to solid boundaries Unfortunately for the field of applied hydrodynamics. The necessity of avoiding or controlling cavitation imposes serious limitations on the design of many types of hydraulic equipment. Effects that modify the hydrodynamics of the flow of the liquid 2. or flow systems. All Rights Reserved . In vibratory cavitation. thus Kb ¼ p0 2 pb rV 2 =2 0 ð8:3Þ Copyright 2003 by Marcel Dekker. gravity effects can be included. namely. At some location on the object. p0 the pressure of undisturbed liquid. which might host undissolved gases. A relative flow between an immersed object and the surrounding liquid results in a variation in pressure at a point on the object. Either procedure will result in lowering of the absolute values of all the local pressures on the surface of the object. This is equivalent to omitting gravity. V0 the velocity of undisturbed liquid relative to body. we can define a cavitation parameter by replacing pmin. These include the properties of the liquid (such as viscosity. not only in affecting cavity dimensions but also in modifying the effects of some of the fluid and boundary flow properties. so that ð2C p Þmin ¼ p0 2 pmin rV 2 =2 0 ð8:2Þ In the absence of cavitation (and if Reynolds-number effects are neglected). and the pressure in the undisturbed liquid at some distance from the object is proportional to the square of the relative velocity. and velocity and in the value of the critical pressure. Denoting this as a bubble pressure pb. including cleanliness and existence of crevices. and ð p0 2 pÞd the pressure differential due to dynamic effects of fluid motion. If surface tension is ignored. the pressure pmin will be the pressure of the contents of the cavitation cavity. and vaporization characteristics). any solid. All Rights Reserved . p will be a minimum. Other variables may cause significant variations in the relation between geometry. Inc. when necessary. pressure. Finally. the pressure gradients due to gravity are important for large cavities whether they be traveling or attached types. p the pressure at a point on object.324 Chapter 8 of absolute pressure and velocity. 2Cp ¼ ð p0 2 pÞd rV 2 =2 0 ð8:1Þ where r is the density of liquid. Let us consider a simple liquid having constant properties and develop the basic cavitation parameter. It is easy to create a set of conditions such that pmin drops to some value at which cavitation exists. pmin. and the condition of the boundary surfaces. This can be accomplished by increasing the relative velocity V0 for a fixed value of the pressure p0 or by continuously lowering p0 with V0 held constant. surface tension. this value will depend only on the shape of the object. However. and the critical pressure pcrit at which a bubble can be formed or a cavity maintained. the physical size of the boundary geometry may be important. In addition to dynamic effects. or gaseous contaminants that may be entrained or dissolved in the liquid. This can be written as the negative of the usual pressure coefficient Cp. (8. Then. pb the absolute pressure in cavity or bubble. For inception. in terms of pressure head (in feet of the liquid). After inception. the cavitation parameter assumes a definite value at each stage of development or “degree” of cavitation on a particular body. the presence of such things as undissolved gas particles. If we now assume that cavitation will occur when the normal stresses at a point in the liquid are reduced to zero. The beginning of cavitation means the appearance of tiny cavities at or near the place on the object where the minimum pressure is obtained. As a consequence. pb will equal the vapor pressure pv. V0 the reference velocity. which tends to remain constant. whereas either V0 increases or p0 decreases. for advanced stages of cavitation. we write Kb ¼ p0 2 pv rV 2 =2 0 ð8:5Þ The value of K at which cavitation inception occurs is designated as Ki. Thus. In considering the behavior of the cavitation parameter during this process. K i : Ki and values of K at subsequent stages of cavitation depend primarily on the shape of the immersed object past which the liquid flows. Copyright 2003 by Marcel Dekker. For the limiting case of parallel flow of an ideal fluid. K ¼ K i . and g the specific weight of liquid.Cavitation in Hydraulic Machinery 325 or. A theoretical value of Ki is the magnitude jð2C p Þmin j for any particular body.5) has been universally adopted as the parameter for comparison of vaporous cavitation events. Ki will be zero since the pressure p0 in the main stream will be the same as the wall pressure (again with gravity omitted and the assumption that cavitation occurs when the normal stresses are zero). Continual increase in V0 (or decrease in p0) means that the pressure at other points along the surface of the object will drop to the critical pressure. Eq. Ki will always be finite. We should note here that for flow past immersed objects and curved boundaries. the value decreases as pmin becomes the cavity pressure. Thus. we again note that if Reynoldsnumber effects are neglected the pressure coefficient (2 Cp)min depends only on the object’s shape and is constant prior to inception. However. the zone of cavitation will spread from the location of original inception. Inc. and turbulence will modify and often mask a departure of the critical pressure pcrit from pv. K . Kb ¼ ð p0 2 pb Þ=g V 2 =2g 0 ð8:4Þ where p0 is the absolute-static pressure at some reference locality. boundary layers. All Rights Reserved . The initiation of cavitation by vaporization of the liquid may require that a negative stress exist because of surface tension and other effects. 1 The Cavitation Parameter as a Flow Index The parameter Kb or K can be used to relate the conditions of flow to the possibility of cavitation occurring as well as to the degree of postinception stages of cavitation. On the other hand. 8. By adjusting the flow conditions so that K is greater than. instead of Eq.5 PHYSICAL SIGNIFICANCE AND USES OF THE CAVITATION PARAMETER A simple physical interpretation follows directly when we consider a cavitation cavity that is being formed and then swept from a low-pressure to a highpressure region.2 The Cavitation Parameter in Gravity Fields As the pressure differences in the preceding relations are due to dynamic effects.4.326 Chapter 8 8. the full range of possibilities. (8. for any degree of cavitation from inception to advanced stages.5). are basically due to changes in the velocity of the flow.5). the relation between dynamic pressure difference ( p0 2 pmin)d and the actual pressure difference ( p0 2 pmin)actual is ð p0 2 pmin Þd ¼ ð p0 2 pmin Þactual þ gðh0 2 hmin Þ where g is the liquid’s specific weight and h is elevation.4. as previously noted. therefore. which tends to collapse the cavity. (8. The variations in pressure. 8. Thus. can be established. Full dynamic similarity Copyright 2003 by Marcel Dekker. For any system where the existing or potential bubble pressure ( pb or pv) is fixed. in terms of actual pressures. All Rights Reserved . which take place on the surface of the body or on any type of guide passage. we have.6) reduces to Eq. Its use. K¼ ð p0 þ gh0 Þ 2 ð pv þ ghmin Þ rV 0 =2 ð8:6Þ For h0 ¼ hmin . the velocity head may be considered to be a measure of the pressure reductions that may occur to cause a cavity to form or expand. The basic importance of cavitation parameter stems from the fact that it is an index of dynamic similarity of flow conditions under which cavitation occurs. Then. (8. Then the numerator is related to the net pressure or head. the parameter has a characteristic value. is subject to a number of limitations. equal to. the parameter (Kb or K) can be computed for the full range of values of the reference velocity V0 and reference pressure p0. For large bodies going through changes in elevation as they move. the cavitation parameter is defined independently of the gravity field. from no cavitation to advanced stages of cavitation. The denominator is the velocity pressure or head of the flow. Eq. Inc. or less than Ki. the work done appears as kinetic Copyright 2003 by Marcel Dekker. etc. Furthermore. 8. Inc. Á 4pp1 À 3 R0 2 R 3 ð8:9Þ 3 If the fluid is inviscid as well as incompressible. has proved interesting to many workers in the field.Cavitation in Hydraulic Machinery 327 between flows in two systems requires that the effects of all physical conditions be reproduced according to unique relations. a particular cavitation condition is accurately reproduced only if Reynolds number.e. the radial flow is irrotational with velocity potential. the radius of the cavity wall. Froude number. and velocity is given by UR 2 u R2 ¼ and ð8:7Þ r U r2 Next. This is especially true with respect to the process by which cavitation produces physical damage on the guiding surfaces. i. gravity. For spherical symmetry. as well as the cavitation parameter K have particular values according to a unique relation among themselves. In other words. U is the cavity-wall velocity at time t. the role played by traveling cavities grows in importance. even if identical thermodynamics and chemical properties and identical boundary geometry are assumed.. the variable effects of contaminants in the liquid-omitted dynamic similarity require that the effects of viscosity. it appears that as more experimental evidence is obtained on the detailed mechanics of the cavitation process. which are the idealized form of the traveling transient cavities. and surface tension be in unique relationship at each cavitation condition. Weber number. the expression for the kinetic energy of the entire body of liquid at time t is developed by integrating kinetic energy of a concentric fluid shell of thickness dr and density r. The result is Z r 1 2 ðKEÞliq ¼ u 4pr 2 dr ¼ 2prU 2 R 3 ð8:8Þ 2 R f¼ The work done on the entire body of fluid as the cavity is collapsing from the initial radius R0 to R is a product of the pressure p1 at infinity and the change in volume of the cavity as no work is done at the cavity wall where the pressure is assumed to be zero. All Rights Reserved .6 THE RAYLEIGH ANALYSIS OF A SPHERICAL CAVITY IN AN INVISCID INCOMPRESSIBLE LIQUID AT REST AT INFINITY The mathematical analysis of the formation and collapse of spherical cavities. where r is greater than R. Rayleigh first set up an expression for the velocity u. Thus. at any radial distance r. the cavity is filled with a gas. This gives sffiffiffiffiffiffiffiffiZ sffiffiffiffiffiffiffiffiZ 3r R 0 R 3=2 dR 3r 1 b 3=2 db t¼ ¼ R0 ð8:11Þ À Á 2p1 R R3 2 R 3 1=2 2p1 b ð1 2 b 3 Þ1=2 0 The new symbol b is R/R0. Rayleigh calculated what would happen if. (8. the external work done on the system as given by Eq. Table 8. Equation (8. (8. which gives 2p1 R3 2 0 U ¼ 21 3r R 3 ð8:10Þ An expression for the time t required for a cavity to collapse from R0 to R can be obtained from Eq. (8.0. (8. (8. positive) value of Q. Therefore. the first movement of the boundary is outward. The time t of complete collapse is obtained if Eq.10) by substituting for the velocity U of the boundary.328 Chapter 8 energy.11) for any other value of b. the integration may be performed by means of functions with the result that t becomes ÀÁ ÀÁ rffiffiffiffiffiffiffiffi rffiffiffiffiffiffi G 3 G 1 r r 6À Á2 t ¼ R0 £ ð8:12Þ ¼ 0:91468R0 6p1 p1 G 4 3 Rayleigh did not integrate Eq. which is compressed isothermally. but U will come to 0 for a finite value of R. (8.11) is evaluated for b ¼ 0: For this special case.10) shows that as R decreases to 0.13). If Q is greater than p1. its equivalent dR/dt and performing the necessary integration. where Q is the initial pressure of the gas. the velocity U increases to infinity.8) and the work of compression of the gas. (8. Eq. which is 4pQR3 lnðR0 =RÞ..8) can be equated to Eq. instead of having zero or constant pressure within the cavity. (8. In such a case. Inc. the cavity will not collapse completely.10) is 0 replaced by 2p1 R3 2Q R3 R0 2 0 ð8:13Þ 21 2 £ 0 ln0 U ¼ 3r R 3 r R3 R For any real (i.9) is equated to the sum of the kinetic energy of the liquid given by Eq. In the detailed study of the time history of the collapse of a cavitation bubble. All Rights Reserved . The limiting size of the cavity can be obtained by setting U ¼ 0 in Eq. (8.1 gives values of the dimensionless time t ¼ t=R0 r=p1 over this range as obtained from a numerical solution of a power series expansion of the integral in Eq. (8. it is convenient to have a solution for all valuespffiffiffiffiffiffiffiffiffiffiffi of b between 0 and 1.11). Thus Eq. which gives z21 2 Q ln z ¼ 0 z ð8:14Þ p1 Copyright 2003 by Marcel Dekker. In order to avoid this.e.9). 95 0.906947 0.856752 0.555078 0. All Rights Reserved .01 0.72 0.93 0.43 0.64 0.914248 0.29 0.908829 0. (8.905885 0.23 0.790068 0.386662 0.76 0.33 0.817993 0.914652 0.39 0.878477 0.84 0.812810 0.58 0.890232 0.770712 0.47 0.62 0.75 0.796068 0.18 0. Inc.48 0.451377 0.909654 0.362507 0.19 0.895956 0.68 0.869969 0.Cavitation in Hydraulic Machinery pffiffiffiffiffiffiffiffiffiffiffi Table 8.212764 0.07 0.016145 0.83 0.845308 0.849277 0.67 0.707625 0.34 0.52 0.907928 0.777398 0.21 0.63 0.00 0.892245 0.523635 0.51 0.31 0.78 0.610515 0.16 0.913130 0.247733 0.912713 0.69 0. the limiting size is a minimum.94 0.894153 0.74 0.14 0.04 0.430965 0.12 0.827798 0.30 0.635059 0.49 0.904738 0.914680 329 in which z denotes the ratio of the volume R3 =R 3 : Equation (8.91 0.597495 0.914604 0.914406 0.57 0.37 0.903505 0.82 0.841181 0.38 0.09 0.783847 0.40 0.24 0.32 0.42 0.28 0. 1.913489 0.756599 0.309297 0.02 0.883528 0.14) indicates that 0 the radius oscillates between the initial value R0 and another.569810 0.03 0.079522 0.914675 0. which is determined by the ratio p1 =Q from this equation.80 0.66 0.15 0.61 0.222110 0.55 0.733436 0.489229 0.911083 0.46 0.899262 0.910404 0.749154 0.89 0.698377 0.174063 0.130400 0.70 0.885876 0.832431 0.822988 0.409433 0.53 0.26 0.860268 0.54 0.866872 0.279736 0.887062 0.801854 0.902182 0.20 0.913793 0.85 0.35 0.96 0.853090 0.678830 0.73 0.836890 0.688784 0.900769 0.06 0.56 0.911692 0.13 0.97 0.741436 0.77 0.623027 0.79 0.92 0.583937 0.59 0.863640 0.41 0.716542 0.36 0.914045 0.506830 0.897658 0.05 0.725142 0.470770 0.539701 0. If p1 =Q .60 0.27 0.763782 0.17 0.65 0. Although Rayleigh presented this example only for isothermal Copyright 2003 by Marcel Dekker.71 0.10 0.657773 0.22 0.50 0.08 0.98 0.872933 0.336793 0.807433 0.646633 0.87 0.88 0.44 0.45 0.99 0.11) (Error Less Than 1026 for 0 # b # 0:96Þ pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi p =r p =r p =r b b b t R1 t R1 t R1 0 0 0 0.25 0.668498 0.86 0.1 Values of the Dimensionless Time t0 ¼ t=R0 r=p1 from Eq.914523 0.90 0.11 0.875768 0.912234 0.81 0. z . Hence.10).330 Chapter 8 compression. ar ¼ 2 du ›u ›u 1 ›p ¼2 2u ¼ ›t ›t r ›r dt ð8:15Þ Expressions for ›u=›t and uð›u=›rÞ as functions of R and r are obtained from Eqs. (8. (8.13) may be formulated. at radius r. As another interesting aspect of the bubble collapse. All Rights Reserved . (8. Substituting these expressions in Eq.7) being taken with respect to r and t. 1.1 shows the distribution of the pressure in the liquid according to Eq.15) yields: ! 1 ›p R ð4z 2 4ÞR 3 ¼ 2 ðz 2 4Þ ð8:16Þ r3 p1 ›r 3r 2 in which z ¼ * ðR0 =RÞ3 and r # R always. He set up the radial acceleration as the total differential of the liquid velocity u. By integration. (8. which gives R0 p ¼ p1 1 2 ð8:19Þ r In Eq. pmax ¼ p1 and occurs at R=r ¼ 0.18). The location rm of the maximum pressure in the liquid may be found by setting dp/dr equal to zero in Eq. This location moves toward the bubble with increasing z and approaches r=R ¼ 1:59 as z approaches infinity. p1 and occurs at finite r/R. the partial differential of Eq. and the partial differential of Eq. It is seen that for 1 .18). this becomes Z r ! Z p Z r 1 R dr dr ð4z 2 4ÞR 3 dp ¼ 2 ðz 2 4Þ ð8:17Þ 2 p1 p 1 3 r5 1 1 r which gives p R R4 2 1 ¼ ðz 2 4Þ 2 4 ðz 2 1Þ 3r p1 3r ð8:18Þ The pressure distribution in the liquid at the instant of release is obtained by substituting R ¼ R0 in Eq. (8. with respect to time.16). (8. equated this to the radial pressure gradient. Inc. 4. (8.18). and equations analogous to Eq. This gives a maximum value for p when r3 4z 2 4 m ¼ z24 R3 ð8:20Þ Copyright 2003 by Marcel Dekker. Figure 8. Rayleigh calculated the pressure field in the liquid surrounding the bubble reverting to the empty cavity of zero pressure. z ¼ 1 at the initiation of the collapse and increases as collapse proceeds.7) and (8. it is obvious that any other thermodynamic process may be assumed for the gas in the cavity.7) with respect to t. z . (8. where r ! 1: For 4 . (8. pmax . and integrated to get the pressure at any point in the liquid. (8.23) taken together show that as the cavity becomes very small.Cavitation in Hydraulic Machinery 331 Figure 8.20) and (8. This would suggest the possibility that in compressing the liquid some energy can be stored.10). Rayleigh himself abandoned this assumption in considering what happens if the cavity collapses on an absolute rigid sphere of radius R.22) and (8. (8. All Rights Reserved .1 Rayleigh analysis: pressure profile near a collapsing bubble. the maximum value of p is obtained as pmax ðz 2 4ÞR ðz 2 4Þ4=3 ¼1þ ¼1þ p1 4r m 44=3 ðz 2 1Þ1=3 ð8:21Þ As cavity approaches complete collapse. the pressure in the liquid near the boundary becomes very great in spite of the fact that the pressure at the boundary is always zero. and Eqs. In this treatment. the assumption of incompressibility is abandoned only at Copyright 2003 by Marcel Dekker. (8. This would invalidate the assumption of incompressibility.18). z becomes great. which would add an additional term to Eq. When rm is substituted for r in Eq. Inc.21) may be approximated by r m ¼ 41=3 R ¼ 1:587R and pmax z R3 ¼ 4=3 ¼ 4=30 3 p1 4 R 4 ð8:23Þ ð8:22Þ Equations (8. especially when it is remembered that there was some variation of pressure p1 during collapse of the actual cavity.12).2 Comparison of measured bubble size with the Rayleigh solution for an empty cavity in an incompressible liquid with a constant pressure field. Copyright 2003 by Marcel Dekker.332 Chapter 8 the instant that the cavity wall comes in contact with the rigid sphere. it is assumed that the kinetic energy of deformation of the same particle is determined by the bulk modulus of elasticity of the fluid. It will be noted that the actual collapse time is greater than that predicted by Eq. All Rights Reserved . From that instant. Figure 8. it is found that ðP0 Þ2 1 p1 R3 p1 2 0 21 ¼ ðz 2 1Þ ¼ rU ¼ 2E 2 3 R3 3 ð8:24Þ where P 0 is the instantaneous pressure on the surface of the rigid sphere and E is the bulk modulus of elasticity. This similarity is very striking. as is common in waterhammer calculations. It is instructive to compare the collapse of the cavity with the predicted collapse based on this simple theory. Inc. On this basis.2 shows this comparison. Both must be expressed in the same units. Figure 8. (8. Cavitation in Hydraulic Machinery 333 8.7 CAVITATION EFFECTS ON PERFORMANCE OF HYDRAULIC MACHINES 8.7.1 Basic Nature of Cavitation Effects on Performance The effects of cavitation on hydraulic performance are many and varied. They depend upon the type of equipment or structure under consideration and the purpose it is designed to fulfill. However, the basic elements, which together make up these effects on performance, are stated as follows: 1. The presence of a cavitation zone can change the friction losses in a fluid flow system, both by altering the skin friction and by varying the form resistance. In general, the effect is to increase the resistance, although this is not always true. 2. The presence of a cavitation zone may result in a change in the local direction of the flow due to a change in the lateral force, which a given element of guiding surface can exert on the flow as it becomes covered by cavitation. 3. With well-developed cavitation the decrease in the effective crosssection of the liquid-flow passages may become great enough to cause partial or complete breakdown of the normal flow. The development of cavitation may seriously affect the operation of all types of hydraulic structures and machines. For example, it may change the rate of discharge of gates or spillways, or it may lead to undesirable or destructive pulsating flows. It may distort the action of control valves and other similar flow devices. However, the most trouble from cavitation effects has been experienced in rotating machinery; hence, more is known about the details of these manifestations. Study of these details not only leads to a better understanding of the phenomenon in this class of equipment but also sheds considerable light on the reason behind the observed effects of cavitation in many types of equipment for which no such studies have been made. Figures 8.3 and 8.4 display the occurrence of cavitation and its effect on the performance of a centrifugal pump. 8.8 THOMA’S SIGMA AND CAVITATION TESTS 8.8.1 Thoma’s Sigma Early in the study of the effects of cavitation on performance of hydraulic machines, a need developed for a satisfactory way of defining the operating conditions with respect to cavitation. For example, for the same machine operating under different heads and at different speeds, it was found desirable Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 334 Chapter 8 Figure 8.3 Cavitation occurs when vapor bubbles form and then subsequently collapse as they move along the flow path on an impeller. to specify the conditions under which the degree of cavitation would be similar. It is sometimes necessary to specify similarity of cavitation conditions between two machines of the same design but of different sizes, e.g., between model and prototype. The cavitation parameter commonly accepted for this purpose was Figure 8.4 Effect of cavitation on the performance of a centrifugal pump. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Cavitation in Hydraulic Machinery 335 proposed by Thoma and is now commonly known as the Thoma sigma, sT. For general use with pumps or turbines, we define sigma as H sv ð8:25Þ H where Hsv, the net positive suction head at some location ¼ total absolute head less vapor-pressure head ¼ ½ð patm =gÞ þ ð p=gÞ þ ðV 2 =2gÞ 2 ð pv =gÞ; H is the head produced (pump) or absorbed (turbine), and g is the specific weight of fluid. For turbines with negative static head on the runner, ssv ¼ H sv ¼ H a 2 H s 2 H v þ V2 e þ Hf 2g ð8:26Þ where Ha is the barometric-pressure head, Hs the static draft head defined as elevation of runner discharge above surface of tail water, Hv the vapor-pressure head, Ve the draft-tube exit average velocity (tailrace velocity), and Hf the drafttube friction loss. If we neglect the draft-tube friction loss and exit-velocity head, we get sigma in Thoma’s original form: sT ¼ Thus Ha 2 Hs 2 Hv H ð8:27Þ V 2 =2g þ H f e ð8:28Þ H Sigma (ssv or sT) has a definite value for each installation, known as the plant sigma. Every machine will cavitate at some critical sigma (ssvc or sTc). Clearly, cavitation will be avoided only if the plant sigma is greater than the critical sigma. The cavitation parameter for the flow passage at the turbine runner discharge is, say, sT ¼ ssv 2 Kd ¼ Hd 2 Hv V 2 =2g d ð8:29Þ where Hd is the absolute-pressure head at the runner discharge and Vd the average velocity at the runner discharge. Equation (8.29) is similar in form to Eq. (8.25) but they do not have exactly the same significance. The numerator of Kd is the actual cavitation-suppression pressure head of the liquid as it discharges from the runner. (This assumes the same pressure to exist at the critical location for cavitation inception.) Its relation to the numerator of sT is V2 ð8:30Þ H d 2 H v ¼ H sv 2 d 2g For a particular machine operating at a particular combination of the operating variables, flow rate, head speed, and wicket-gate setting, Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 336 Chapter 8 V2 d ¼ C1 H 2g ð8:31Þ Using the previous relations, it can be shown that Eq. (8.29) may be written as sT Hf V 2 =2g Kd ¼ 2 12 þ e C1 C1 H V 2 =2g d The term in parenthesis is the efficiency of the draft tube, hdt, as the converter of the entering velocity head to pressure head. Thus the final expression is Kd ¼ sT V2 2 hdt þ e C1 V2 d ð8:32Þ C1 is a function of both design of the machine and the setting of the guide vane; hdt is a function of the design of the draft tube but is also affected by the guidevane setting. If a given machine is tested at constant guide-vane setting and operating specific speed, both C1 and hdt tend to be constant; hence sT and Kd have a linear relationship. However, different designs usually have different values of C1 even for the same specific speed and vane setting, and certainly for different specific speeds. Kd, however, is a direct measure of the tendency of the flow to produce cavitation, so that if two different machines of different designs cavitated at the same value of Kd it would mean that their guiding surfaces in this region had the dame value of Ki. However, sigma values could be quite different. From this point of view, sigma is not a satisfactory parameter for the comparison of machines of different designs. On the other hand, although the determination of the value of Kd for which cavitation is incipient is a good measure of the excellence of the shape of the passages in the discharge region, it sheds no light on whether or not the cross-section is an optimum as well. In this respect, sigma is more informative as it characterizes the discharge conditions by the total head rather than the velocity head alone. Both Kd and sigma implicitly contain one assumption, which should be borne in mind because at times it may be rather misleading. The critical cavitation zone of the turbine runner is in the discharge passage just downstream from the turbine runner. Although this is usually the minimum-pressure point in the system, it is not necessarily the cross-section that may limit the cavitation performance of the machine. The critical cross-section may occur further upstream on the runner blades and may frequently be at the entering edges rather than trailing edges. However, these very real limitations and differences do not alter the fact that Kd and sT are both cavitation parameters and in many respects, they can be used in the same manner. Thus Kd (or K evaluated at any location in the machine) can be used to measure the tendency of the flow to cavitate, the conditions of the flow at which cavitation first begins (Ki), or the conditions of the flow corresponding to a certain degree of development of cavitation. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Cavitation in Hydraulic Machinery 337 Likewise, sT can be used to characterize the tendency of the flow through a machine to cause cavitation, the point of inception of cavitation, the point at which cavitation first affects the performance, or the conditions for complete breakdown of performance. Ki is a very general figure of merit, as its numerical value gives directly the resistance of a given guiding surface to the development of cavitation. Thoma’s sigma can serve the same purpose for the entire machine, but in a much more limited sense. Thus, for example, sT can be used directly to compare the cavitation resistance of a series of different machines, all designed to operate under the same total head. However, the numerical value of sT, which characterizes a very good machine, for one given head may indicate completely unacceptable performance for another. Naturally, there have been empirical relations developed through experience, which show how the sT for acceptable performance varies with the design conditions. Figure 8.5 shows such a relationship. Here, the specific speed has been taken as the characteristic that describes the design type. It is defined for turbines as pffiffiffiffiffi N hp ð8:33Þ N s ¼ 5=4 H where N is the rotating speed, hp the power output, and H the turbine head. The ordinate is plant sigma ðsT ¼ splant Þ: Both sigma and specific speed are based on rated capacity at the design head. In the use of such diagrams, it is always necessary to understand clearly the basis for their construction. Thus, in Fig. 8.5, the solid lines show the minimumplant sigma for each specific speed at which a turbine can reasonably be expected to perform satisfactorily; i.e., cavitation will be absent or so limited as not to cause efficiency loss, output loss, undesirable vibration, unstable flow, or excessive pitting. Another criterion of satisfactory operation might be that cavitation damage should not exceed a specific amount, measured in pounds of metal removed per year. Different bases may be established to meet other needs. A sigma curve might be related to hydraulic performance by showing the limits of operation for a given drop in efficiency or for a specific loss in power output. Although the parameter sigma was developed to characterize the performance of hydraulic turbines, it is equally useful with pumps. For pumps, it is used in the form of Eq. (8.25). In current practice, the evaluation of Hsv varies slightly depending on whether the pump is supplied directly from a forebay with a free surface or forms a part of a closed system. In the former case, Hsv is calculated by neglecting forebay velocity and the friction loss between the forebay and the inlet, just as the tailrace velocity and friction loss between the turbine-runner discharge and tail water are neglected. In the latter case, Hsv is calculated from the pressure measured at the inlet. Velocity is assumed to be the average velocity, Q/A. Because of this difference in meaning, if the same Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 338 Chapter 8 Figure 8.5 Experience limits of plant sigma vs. specific speed for hydraulic turbines. machine was tested under both types of installation, the results would apparently show a slightly poor cavitation performance with the forebay. 8.8.2 Sigma Tests Most of the detailed knowledge of the effect of cavitation on the performance of hydraulic machines has been obtained in the laboratory, because of the difficulty encountered in nearly all field installations in varying the operating conditions over a wide enough range. In the laboratory, the normal procedure is to obtain data for the plotting of a group of sT curves. Turbine cavitation tests are best Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved Cavitation in Hydraulic Machinery 339 run by operating the machine at fixed values of turbine head, speed, and guidevane setting. The absolute-pressure level of the test system is the independent variable, and this is decreased until changes are observed in the machine performance. For a turbine, these changes will appear in the flow rate, the power output, and the efficiency. In some laboratories, however, turbine cavitation tests are made by operating at different heads and speeds, but at the same unit head and unit speed. The results are then shown as changes in unit power, unit flow rate, and efficiency. If the machine is a pump, cavitation tests can be made in two ways. One method is to keep the speed and suction head constant and to increase the discharge up to a cutoff value at which it will no longer pump. The preferable method is to maintain constant speed and flow rate and observe the effect of suction pressure on head, power (or torque), and efficiency as the suction pressure is lowered. In such cases, continual small adjustments in flow rate may be necessary to maintain it at constant value. Figure 8.6 shows curves for a turbine, obtained by operating at constant head, speed, and gate. Figure 8.7 shows curves for a pump, obtained from tests at constant speed and flow rate. These curves are typical in that each performance characteristic shows little or no deviation from its normal value Figure 8.6 Sigma curves for a hydraulic turbine under constant head, speed, and gate opening. (Normal torque, head, and discharge are the values at best efficiency and high sigma.) Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved 340 Chapter 8 Figure 8.7 Sigma curves for a centrifugal pump at constant speed and discharge. (Normal head and discharge are the values at best efficiency and high sigma.) (at high submergence) until low sigmas are reached. Then deviations appear, which may be gradual or abrupt. In nearly all cases, the pressure head across a pump or turbine is so small in comparison with the bulk modulus of the liquid such that change in system pressure during a sigma test produces no measurable change in the density of the liquid. Thus, in principle, until inception is reached, all quantities should remain constant and the s curves horizontal. Figure 8.8 shows some of the experimental sigma curves obtained from tests of different pumps. It will be noted that the first deviation of head H observed for machines A and C is downward but that for machine B is upward. In each case, the total deviation is considerably in excess of the limits of accuracy of measurements. Furthermore, only machine A shows no sign of change in head until a sharp break is reached. The only acceptable conclusion is, therefore, that the inception point occurs at much higher value of sigma than might be assumed and the effects of cavitation on the performance develop very slowly until a certain degree of cavitation has been reached. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved ) 8. Inc. the point where the sigma curve departs from the horizontal may Copyright 2003 by Marcel Dekker. All Rights Reserved .8.3 Interpretation of Sigma Tests The sigma tests described are only one specialized use of the parameter.8 Comparison of sigma curves for different centrifugal pumps at constant speed and discharge.Cavitation in Hydraulic Machinery 341 Figure 8. For example. as already noted. sigma may be used as a coordinate to plot the results of several different types of experience concerning the effect of cavitation of machines. Even though sigma tests are not reliable in indicating the actual inception of cavitation. (Normal head and discharge are the values at best efficiency and high sigma. attempts have often been made to use them for this purpose on the erroneous assumption that the first departure from the noncavitating value of any of the pertinent parameters marks the inception of cavitation. Considering strictly from the effect of cavitation on the operating characteristics. The result of this assumption has frequently been that serious cavitation damage has been observed in machines whose operation had always been limited to the horizontal portion of the sigma curve. Specific speed as used for pumps is defined as pffiffiffiffi N Q N s ¼ 3=4 ð8:34Þ H where N is the rotating speed.8. the indicated siH is at the point where the head has decreased by 0. such points are marked in each curve. Suction specific speed is defined as pffiffiffiffi N Q ð8:35Þ S ¼ 3=4 H sv where Hsv is the total head above vapor at pump inlet. Q the volume rate of flow. the cavitation behavior would be characterized by S. si. The curves of Fig. siH.8. For convenience in operation. 8. For pump B. points could be designated as siP. and H the head differential produced by pump. The knee of this curve. The name “suction specific speed” follows from this concept. where the drop becomes very great. or siQ. It should be equally applicable to pumps and turbines where cavitation depends only on the suction region of the runner. Inc.8 show that at some lower limiting sigma. 8. and Copyright 2003 by Marcel Dekker. suction specific speed. Presumably. which is sensitive only to the factors that affect cavitation. There is remarkable similarity between these sigma curves and the lift curves of hydrofoil cascades.4 Suction Specific Speed It is unfortunate that sigma varies not only with the conditions that affect cavitation but also with the specific speed of the unit. In Fig. All Rights Reserved . for changes in the outlet diameter and head produced by a Francis-type pump runner.342 Chapter 8 be designed as the inception of the effect. the curve of performance finally becomes nearly vertical. This is more likely to be the case in low-specific-speed Francis turbines. It is interesting to note that the knee of the curve for the cascade corresponds roughly to the development of a cavitation zone over about 10% of the length of the profile and the conditions for heavy vibrations do not generally develop until after the knee has been passed. For pumps A and C.5% from its high sigma value. 8. The parameter is widely used for pumping machinery but has not usually been applied to turbines. is called the breakdown point. The suction specific speed represents an attempt to find a parameter. which would indicate the values of si for the specified performance characteristics. siH is shown where the head begins to increase from its high sigma value. The following relation between specific speed (as used for pumps). Runners in which cavitation depends only on the geometry and flow in the suction region will develop cavitation at the same value of S. 33) and (8. (8. All Rights Reserved . Fig. suction specific speed. i. 3=4 N s – pump H sv ¼ s3=4 ¼ sv S H ð8:36Þ A corresponding relation between specific speed as used for turbines. Then 1=2 N s – turb 3=4 ht g ¼ ssv ð8:37Þ S 550 It is possible to obtain empirical evidence to show whether or not S actually possesses the desirable characteristic for which it was developed. Copyright 2003 by Marcel Dekker.Cavitation in Hydraulic Machinery 343 sigma is obtained from Eqs. 8. and propeller pumps. For example.e. Inc. (8. mixed-flow.35).9 hp ¼ Figure 8.34) and (8. to offer a cavitation parameter that varies only with the factors that affect the cavitation performance of hydraulic machines and is independent of other design characteristics such as total head and specific speed.35) together with the expression ht gQH 550 where ht is the turbine efficiency. specific speed for centrifugal..9 Sigma vs. and sigma can be obtained from Eqs. several satisfactory designs have been reported for S in the 20. With special designs. Note that allowable S values for turbines operating with little or no cavitation tend to be higher than those for pumps when compared at their respective design conditions. ssv decreases. as the tendency to cavitate increases. 8. This difference of slope can be taken to indicate that either the parameter S is affected by factors other than those involved in cavitation performance or the different specific-speed designs are not equally close to the optimum as regards cavitation.000 – 20. but S increases.5 shows limits that can be expected for satisfactory performance of turbines. The line for Francis turbines has been added to Fig.e. For cavitating inducers and other special services. cavitation is expected and allowed.000 range. and propeller pumps.000 for standard pumps in general industrial use. Note also that the trend of limiting sigma for turbines is at a steeper slope than the constant S lines. Currently.. the limit for essentially cavitation-free operation is approximately S ¼ 12. It should be noted that ssv and S vary in the opposite direction as the severity of the cavitation condition changes.000 – 35. i. mixed-flow. In cases where the velocities are relatively low (such as condensate pumps). then a curve passing through the lowest point for each given specific speed should be a curve of constant cavitation performance.9 for comparison with pump experience. In the same diagram. Inc. 8. all the best designs represent an equally close approach to the ideal design to resist cavitation. Fig. The latter leads to the conclusion that it is easier to obtain a good design from the cavitation point of view for the lower specific speeds. All Rights Reserved . (8.36)].000 are fairly common. As was explained.344 Chapter 8 shows a logarithmic diagram of sigma vs. specific speed on which are plotted points showing cavitation limits of individual centrifugal. If it is assumed that as the type of machine and therefore the specific speed change. each with a slope of ðlog ssv Þ=ðlog N s Þ ¼ 3=4 [Eq. straight lines of constant S are shown. NOTATION a A Cp E h H K KE N Acceleration Area Pressure coefficient Modulus of elasticity Elevation Head Cavitation parameter Kinetic energy Rotation speed Copyright 2003 by Marcel Dekker. pumps having critical S values in the range of 18. It is based on experience with installed units and presumably represents good average practice rather than the optimum. Cavitation in Hydraulic Machinery 345 Ns p Q r rm R S t u V Z r ht g t0 s Specific speed Pressure Flow rate Radial distance Mean radius Radius of cavity wall Suction specific speed Time Velocity Velocity Dimensionless volume of bubble Density Turbine efficiency Specific weight Dimensionless time Cavitation parameter SUFFIXES 0 atm d e f i min r s v Undisturbed fluid properties Atmospheric values Dynamic effects Exit Friction Inception properties Minimum Radial Static vapor Copyright 2003 by Marcel Dekker. Inc. All Rights Reserved . All Rights Reserved . Inc.Appendix THE INTERNATIONAL SYSTEM OF UNITS (SI) Table 1 Quantity Length Mass Time Electric current Thermodynamic temperature Luminous intensity Amount of a substance SI Base Units Name of unit meter kilogram second ampere kelvin candela mole Symbol m kg s A K cd mol Copyright 2003 by Marcel Dekker. f ohm. J Defining equation 1 F ¼ 1A s/V 1 V ¼ 1V/A 1 N ¼ 1 kg m/s2 1 V ¼ 1 W/A 1 W ¼ 1 J/s 1 Pa ¼ 1 N/m2 K ¼ 8C þ 273. N volt. V watt. All Rights Reserved .15 1 J ¼ 1 Nm Appendix Table 3 Quantity SI Derived Units Name of unit meter per second square square meter kilogram per cubic meter newton-second per square meter newton hertz square meter per second radian watt watt per steradian steradian joule per kilogram-kelvin watt per meter-kelvin meter per second cubic meter Symbol m/s2 m2 kg/m3 N s/m2 N Hz m2/s rad W W/sr sr J/kg K W/m K m/s m3 Acceleration Area Density Dynamic viscosity Force Frequency Kinematic viscosity Plane angle Power Radiant intensity Solid angle Specific heat Thermal conductivity Velocity Volume Copyright 2003 by Marcel Dekker. heat. W pascal. V newton. Pa kelvin. Inc.348 Table 2 Quantity Capacitance Electrical resistance Force Potential difference Power Pressure Temperature Work. energy SI Defined Units Name of unit farad. K joule. Inc.00000 kg m N21 s22 6.6732 £ 10211 N m2 kg22 6.31434 £ 103 J kmol21 K21 6. All Rights Reserved .Appendix Table 4 Quantity — — — Avogadro constant Boltzmann constant First radiation constant Gas constant Gravitational constant Planck constant Second radiation constant Speed of light in a vacuum Stefan-Boltzmann constant Physical Constants in SI Units Symbol e P gc NA k C1 ¼ 2 p hc2 Ru G h C2 ¼ hc/k c s Value 349 2.141592653 1.022169 £ 1026 kmol21 1.997925 £ 108 ms21 5.66961 £ 1028 Wm22 K24 Table 5 Multiplier 10 109 106 103 102 101 1021 12 SI Prefixes Symbol T G M k h da d Prefix tera giga mega kilo hecto deka deci Multiplier 10 1023 1026 1029 10212 10215 10218 22 Symbol c m m n p f a Prefix centi milli micro nano pico femto atto Copyright 2003 by Marcel Dekker.718281828 3.741844 £ 10216 W m2 8.626196 £ 10234 Js 1.380622 £ 10223 J K21 3.438833 £ 1022 m K 2. 685 £ 106 J 1 lbf ¼ 4. All Rights Reserved .581 £ 1025 m2/s 1 (h)(8F)/Btu ¼ 1.343 W/m3 1 Btu/(h)(ft2)(8F) ¼ 5.448 N 1 Btu/h ¼ 0.2 ¼ 6894.0254 m 1 mile ¼ 1.186 J 1 (ft)(lbf) ¼ 1.678 W/m2 K 1 ft ¼ 0.731 W/m K 1 ft2/s ¼ 0.88 N/m2 (Pa) 1 atm ¼ 101.4536 kg/s 1 hp ¼ 745.325 N/m2 (Pa) 1 Btu/lbf ¼ 2326.7 W 1 (ft)(lbf)/s ¼ 1.1 W 1 Btu/(h)(ft2) ¼ 3.1525 W/m2 1 Btu/(h)(ft3) ¼ 10.2931 W 1 Btu/s ¼ 1055.000126 kg/s 1 lbm/s ¼ 0.350 Table 6A Conversion Factors Symbol A r Q 2 Appendix Physical quantity Area Density Energy. heat Conversion factor 1 ft ¼ 0.452 £ 1024 m2 1 lbm/ft3 ¼ 16.4536 kg 1 slug ¼ 14. Inc.8 N/m2 (Pa) 1 lbf/ft2 ¼ 47.0929 m2/s 1 ft2/h ¼ 2.3558 J 1 (hp)(h) ¼ 2.018 kg/m3 1 slug/ft3 ¼ 515.3558 W 1 Btu/s ¼ 1055.8958 K/W (continued) Force Heat flow rate Heat flux Heat generation per unit volume Heat transfer coefficient Length F q q00 qG h L Mass Mass flow rate Power m _ m W Pressure p Specific energy Specific heat capacity Temperature Q/m c T Thermal conductivity Thermal diffusivity Thermal resistance k a Rt Copyright 2003 by Marcel Dekker.2 ¼ 6. ¼ 2.6093 km ¼ 1609.1 J 1 cal ¼ 4.15](9/5) þ 32 1 Btu/(h)(ft)(8F) ¼ 1.0929 m2 1 in.379 kg/m3 1 Btu ¼ 1055.54 cm ¼ 0.293 W 1 lbf/in.1 J/kg 1 Btu/(lbf)(8F) ¼ 4188 J/kg K T(8R) ¼ (9/5) T (K) T(8F) ¼ [T(8C)](9/5) þ 32 T(8F) ¼ [T(K) 2 273.594 kg 1 lbm/h ¼ 0.3048 m 1 in.1 W 1 Btu/h ¼ 0.3 m 1 lbm ¼ 0. 3048 m/s 1 mph ¼ 0.003785 m3 351 Physical quantity Velocity Viscosity.44703 m/s 1 lbm/(ft)(s) ¼ 1.00100 N s/m2 1 ft2/s ¼ 0.3 ¼ 1.488 N s/m2 1 centipoise ¼ 0. liq.Appendix Table 6A Continued Symbol U m n V Conversion factor 1 ft/s ¼ 0.0929 m2/s 1 ft2/h ¼ 2. dynamic Viscosity.S. All Rights Reserved .581 £ 1025 m2/s 1 ft3 ¼ 0.6387 £ 1025 m3 1 gal(U. Inc.) ¼ 0.02832 m3 1 in. kinematic Volume Table 6B Temperature Conversion Table K 220 225 230 235 240 245 250 255 260 265 270 275 280 285 290 295 300 305 310 315 320 325 330 8C 253 248 243 238 233 228 223 218 213 28 23 2 7 12 17 22 27 32 37 42 47 52 57 8F 263 254 245 236 227 218 29 0 9 18 27 36 45 54 63 72 81 90 99 108 117 126 135 K 335 340 345 350 355 360 365 370 375 380 385 390 395 400 405 410 415 420 425 430 435 440 445 8C 62 67 72 77 82 87 92 97 102 107 112 117 122 127 132 137 142 147 152 157 162 167 172 8F 144 153 162 171 180 189 198 207 216 225 234 243 252 261 270 279 288 297 306 315 324 333 342 K 450 455 460 465 470 475 480 485 490 495 500 505 510 515 520 525 530 535 540 545 550 555 560 8C 177 182 187 192 197 202 207 212 217 222 227 232 237 242 247 252 257 262 267 272 277 282 287 8F 351 360 369 378 387 396 405 414 423 432 441 450 459 468 477 486 495 504 513 522 531 540 549 Copyright 2003 by Marcel Dekker. 75 2370.001026 4.71 2262.94 104.98 2304.53 188.05 2683.58 2117.6845 7.33 2382.13021 4.82654 2.0761 0.59 2618.5369 6.001008 19.001000 206.77 146.8721 1.24 2626.96 313.628 7.2571 15.54 188.34 Saturated Saturated vapor liquid (ug) (hf) 2375.55 101.47 2450.87 334.33 2361.99 62.2866 6.15 2396.9548 1.THERMODYNAMIC PROPERTIES OF WATER Table 7 SI Saturated Water Specific volume (m3/kg) Temp C T 0.40612 3.001004 32.1924 1.705 2.6386 0.77 146.3629 9.40 2494.7887 57.02 440.90 2290.0480 5.03 292.8478 7.56835 0.12 2176.131 206.0257 8.12 2205.00 20.6670 6.57 2477.28 2574.2234 7.5219 19.6671 8.001010 15.56734 9.50 2512.35953 2.91 440.2158 0.9912 7.82 376.001001 77.72 2394.88 376.91 2482.98 83.36056 0.19 2592.87 125.81 2443.52 2500.6113 0.7552 7.86 418.85 2321.67071 0.1649 8.4790 7.00 Evap.03 2243.7037 0.001023 5.001020 6. (hg) (sf) (sg) (sfg) 2501.41831 1.16 2333.0164 7.62 2406.9328 9.2148 25.83 0 0.5579 8.1659 6.06 2547.21 2333.384 9.41 209.40 2234.86 2609.34 0.11 2436.54 2519.19 2270.35 2489.001003 43.0752 1.001006 25.13 2676.04114 5.3593 0.54 2191.06 2600.3530 8.4369 0.37 2308.3583 43.593 12.25 2565.84 355. All Rights Reserved .7498 8.98 41.5444 7.1905 8.93 2660.58 2072.66 2651.12 2442.04217 0. Inc.66969 7.8934 0.31 230.8309 7.06 2318.76 2416.0784 6.70 2102.14 84.1510 0.2569 8.00 292.2276 1.3725 7.8922 32.82757 0.5569 8.3706 8.04 2402.91 334.1562 9.62 2162.1342 1.13 Evap.77 2382.40715 0.39 57.4533 8.13123 0.001044 1.169 4.0308 12.56 2506.0762 7.88 355.3 120.6824 7.8311 0.19 2488.03 2147.001012 12. (vfg) Saturated Saturated vapor liquid (vg) (uf) 0 20.377 0.2581 0.86 125.90 397. (ufg) 2375.75 2465.82 397.12 2469.001017 7.35 2510.09 272.4158 7.98082 1.63 2463.2500 1.09 2668.350 15.001000 106.11 272.83 70.97 41.9007 8.924 77.9095 7.70 2087.5261 7.19 38.28 2643.57 2248.05 2283.8932 0.5724 0.001029 3.74 2528.001015 9.8004 6.01 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 Pressure Kpa P 0.001000 147.08 2456.26 2583.0318 0.81 2276.65 167.93 2454.7813 8.117 147.5052 0.94 104.20 251.03 31.19554 6.70 Saturated Saturated Saturated vapor liquid vapor Evap.131 0.66 167.66 2358.94 419.27 2347.3548 7.001047 1.19656 0.118 0.93 331.339 3.4102 6.001032 2.3068 1.77 2296.17 2219.48 2346.1562 8.67185 1.98186 0.376 106.246 5.67290 0.925 0.30 230.91 2538.48 2418.9375 6. (hfg) 2501.58 47.42 209.98 83.41936 Appendix Copyright 2003 by Marcel Dekker.36 2132.0154 1.80 2635.5229 0.001002 57.24 2389.8 Saturated liquid (vf) Internal energy (KJ/kg) Enthalpy (KJ/kg) Entropy (KJ/kg K) 352 Evap.30 2430.9496 8.2966 0.17 2556.7897 0.55 2475.7679 0.3673 0.58 2423.99 62.19 251.941 25.36 2430.001040 1.91 2409.19 2257.758 19.001036 2.6121 7.2245 0.2958 0. 61 632.16 2159.96 2793.0374 353 (continued ) Copyright 2003 by Marcel Dekker.0418 2.20 2216.29 2597.15539 0.2 3648.26 2733.82 675.61 966.001240 0.9298 6.12736 0.70 2773.001114 0.3925 5.69 1080.7078 6.1 198.48 2800.18 1689.19 2583.95 2603.99 1940.46 503.001149 0.70 2586.6663 6.001219 0.64 873.9337 3.001251 0.50 1741.8832 6. All Rights Reserved .8 1553.44524 0.9010 4.58 1617.68 1522.47 2698.89186 0.8075 4.02 2554.4184 1.3612 3.07849 0.24 2534.98 2590.77 990.03 2550.11 2603.001056 0.79 2595.14105 0.5 232.90 1977.17 1594.7934 6.03552 0.03 1821.91 1993.07158 0.58 2778.04887 0.20 741.72 1009.3221 6.46 2580.5275 1.69 524.6256 6.16 631.13 2803.36 850.85 1790.1791 6.90 2545.24283 0.48 524.20909 1.0269 6.96 852.29 567.28 503.44632 0.15654 0.50885 0.1 617.001263 1.84 1764.66 2518.44 2752.17295 0.7501 6.21568 0.7471 2.37 1104.27 2803.0774 7.00 1900.001121 0.9426 1.2 2317.4697 6.43 875.001060 0.53 1766.67 589.34566 0.8202 5.001190 0.4 475.96 1997.73 1716.34 588.44 2055.34676 0.6099 2.06536 0.0 1906.32 719.17 828.1908 5.1863 4.69 1961.26 2746.001065 0.66744 0.37 2572.3 415.2720 4.72 2529.8 2547.46 2720.72 610.19 1056.96 2706.6020 5.12 940.54 2564.001181 0.2835 2.19405 0.6863 3.2386 7.43 2786.4713 2.94 1661.001156 0.05470 0.7 892.001164 0.61 1037.4461 4.3951 6.03658 0.3779 2.7 1254.001052 0.07037 0.02 895.61 2188.50 1813.37 2600.75 1921.05853 0.04 2568.69 1927.2358 2.59 2144.61 943.90 1783.1 3344. Inc.96 2041.7390 1.53 2768.5812 1.27 482.30 2603.17409 0.93 461.001070 0.39 2758.7927 2.23 674.21 1085.30706 0.13990 0.34 1109.10324 0.4733 1.001096 0.08 784.76953 0.0172 3.05976 0.2 1122.5177 2.8 1723.4962 5.1 270.06415 0.77059 0.49 1639.0 1002.001075 0.81 2802.8507 3.85 969.4422 3.76 2009.50 2202.89080 0.4322 6.86 2559.1992 7.11521 0.2 3973.75 1546.1 313.66850 0.001141 0.12620 0.72 986.05346 0.1436 6.96 546.16 762.52 2799.06 2602.8924 1.5347 4.001199 0.9777 6.75 920.09 2523.5233 3.001080 0.76 1959.8419 1.72 546.10 1013.001127 0.49 1703.37 2789.0909 2.95 2804.16 2782.21014 1.51 918.11405 0.52 1875.12 2803.04598 461.31 740.09361 0.89 2604.7906 1.5639 2.1014 4.6343 1.Appendix 110 115 120 125 130 135 140 145 150 155 160 165 170 175 180 185 190 195 200 205 210 215 220 225 230 235 240 245 250 255 143.42 2230.55 2066.6557 2.51 1836.2502 6.8 700.52 2599.2146 6.20 2049.0729 6.24 1570.6869 1.95 2801.5078 6.4 1397.6244 4.30596 0.42 2014.51 2576.9 543.001102 0.11 610.3 169.001134 0.3586 4.8378 6.7100 5.8382 5.5464 6.7153 4.26 2056.97 1858.01 2592.9960 4.80 2691.07729 0.66 1724.9924 2.87 1910.1295 7.91 1858.27269 0.43 1744.05013 0.08619 0.0 361.35 2603.51 1.001173 0.001090 0.62 1802.3584 6.93 1682.50777 0.12 482.37 2602.09 2763.55 718.19292 0.75 2129.00 567.1083 6.7 2794.00 1496.50 2032.08500 0.66 653.10441 0.56 2082.51 2802.63 2539.30 1944.7015 2.3 2104.87 2740.07 1978.04471 1.58110 0.1878 2.5 0.31 1061.39169 0.21 785.2860 6.09497 0.27158 0.001209 0.46 2727.73 1879.16 763.001085 0.44 2601.1395 2.14 1840.3308 2.21680 0.82 1893.36 807.7683 3.9 3060.08 806.39278 0.58217 0.24171 0.03 897.18 653.4247 2.85 963.001229 0.5 791.53 697.2907 5.6047 3.65 2114.2802 3.03 2798.0 4319.50 2174.61 829.5857 6.88 1033.26 2098.18 2796.0926 4.30 2713.1832 7.001108 0. 05 2748.00315 1128.40 2332.7 5941.01267 0.35 1159.0208 3.3903 5.001317 0.01 2596.1812 2.01995 0.04 1316.3633 2.9750 3.65 2784.53 2443.9146 3.97 1359.54 1713.16 1574.64 1191.001348 0.98 1594.6868 0 6.0882 1. (hfg) 1662.69 Saturated Saturated vapor liquid (ug) (hf) 2599.0667 3.2533 3.67 1230.03147 0.37 1121.01 1641.00271 0 Saturated Saturated vapor liquid (vg) (uf) 0.8909 1.02557 0.001447 0.86 964.01690 0.97 2665.26 1605.93 1366.8 6411.27 1344.27 2766.75 0 Saturated Saturated Saturated vapor liquid Evap.00 2421.99 2569.97 1461.97 2779.21 1331.44 2700.08 2683.03279 0.001685 0.02777 0.59 720.00916 0.37 1027.001638 0.58 Evap.39 2388.96 1890.003155 Evap.00398 0.72 1177.55 1510.54 1815.002213 0.47 2302.45 1492.41 1297.26 2.8730 2.5803 5.09 2581. vapor (hg) (sf) (sfg) (sg) 2796.4297 Appendix Copyright 2003 by Marcel Dekker.61 2789.03564 0.1378 5.1129 3.6227 2.9293 2.2111 5.4 7436.56 1086.03 1415.7044 5.16 2546.001528 0.1593 3.00787 0.5361 5.6229 5.001561 0.9 1195.92 2526.2763 5.0018 5.01420 0.57 993.97 1262.33 1387.11 626.9661 5.7821 5.00694 0.11 1080.00599 0.0368 2.87 1416.5375 2.00810 0.00493 0.30 2418.26 Evap.53 2029.41 1252.67 2228.44 441.001302 0.00505 0.0 9201.9551 2.1 Pressure Kpa P 4688.002011 0.4479 3.29 1289.8198 5.0 7992.33 1202.01080 0.01539 0.58 1134.2061 3.4900 5.37 2099.41 1725.91 2482.1379 0.11 1570.7903 2.01186 0.7069 2.88 2464.4336 1.01027 0.01144 0.00978 0.13 1760.00607 0.66 1328.44 1444.03434 0.001384 0. (vfg) 0.001892 0.6040 3.37 1152.48 1238.9487 0.001499 0.53 2773.001597 0.9470 4.97 1283.001276 0.04220 0.27 2714.87 2562.001472 0.8 9856.03877 0.001807 0.9983 4.6 10547 11274 12040 12845 13694 14586 15525 16514 17554 18561 19807 21028 22089 Saturated liquid (vf) 0.97 1477.84 1525.29 526.40 2536.4511 2.72 2727.7972 4.38 2575.94 1159.49 1210.4416 5.03017 0.02 893.4987 3.8937 5.7863 1.8570 5.38 1325.98 1278.01399 0.7776 3.35 2622.5506 3.08 1404.001740 0.6763 1.5594 1.92 1543. (ufg) 1470.7436 5.15 1631.6 5081.3356 5.58 707.3009 3.02420 0.08 1441.13 1843.23 1227.60 2593.66 2590.04093 0.01300 0.13 2758.9913 1.2951 1.8 8581.001366 0.94 1159. Inc.0525 4.34 1605.54 384.55 2525.44 1505.01835 0.17 1670.02642 0.7169 3.001404 0.94 2738.29 1558.29 776.85 2645.354 Table 7 Continued Specific volume (m3/kg) Internal energy (KJ/kg) Enthalpy (KJ/kg) Entropy (KJ/kg K) Temp C T 260 265 270 275 280 285 290 295 300 305 310 315 320 325 330 335 340 345 350 355 360 365 370 374.7 6909.8427 3.19 2563.17 2586.66 945.81 1681.01549 0.26 838.93 1038.89 1305.64 1443.33 1401.6642 5.13 1140.27 1184.52 605.05 1235.6593 3.72 2481.19 1776.01687 0.94 1358.22 1387.38 813.96 2555.3981 3.77 894.01 2498.48 2513.00881 0. All Rights Reserved .02884 0.9301 5.02027 0.12 2099.4297 3.02167 0.8837 2.89 2793.11 1264.001332 0.2737 2.001425 0.54 1634.55 1474.02354 0.01852 0.3 5498.03748 0.1104 4.24 1537.02216 0.52 2351.01 2595.3492 3.00707 0.001289 0.29 1430.01 1372.84 2029.1181 3. 33 2346.0001057 0.37 2561.Appendix Table 8 SI Saturated Water Pressure Entry Specific volume (m3/kg) Internal energy (KJ/kg) Saturated liquid (uf) 0 29.0878 7.24034 2.93 2307.20322 5.89 2721.1271 7.97 60.96 2675.47 88.23674 14.02 2013.43 130.0084 7.8930 0.02 2241.66 2406.30 2514.6700 7.47 384.08 24.68 520.4746 8.13 Saturated vapor (ug) 2375.70 73.29 45.5029 6.99345 3. (ufg) 2375.6492 0.6072 1.50 2437.18 2525.53 2543.5010 7.0001061 0.55 Saturated liquid (vf) 0.49 2533.42 2163.71765 0.35 2181.29 54.77 191.10 75.37 116.64 2524.1059 0.3201 Saturated vapor (sg) 9.69296 1.73 2456.50 21.99243 3.0001020 0.8697 8.0258 1.9187 7.44 444.38 271.88467 0.3 2355.80015 28.0001070 0.37490 1.91 251.18 317.73 2282. Inc.0001005 0.39 2088.70 2618.98013 67. (vfg) 206.33 91.35 2693.4455 5.47 2432.06 2513.49 2430.59 2460.85 355 (continued ) Copyright 2003 by Marcel Dekker.16 466.33 444.5 10 15 20 25 30 40 50 75 100 125 150 175 200 225 250 275 300 Temp T C 0.6407 1.88 40.45 2574.30 Entropy (KJ/kg K) Saturated liquid Evap.3593 7.03 121.12 2025.04 2716.60 133.47 101.25285 45.4335 1.79325 0.0568 5.69400 1.7897 5.10 2002.0001008 0.89 2448.08 548.8319 0.07 2319.22918 3.77 99.0001004 0.2129 1.89 2470.02218 7.00 29.97 504.57 2201.2607 0.01 6.70 73.1501 8.88 289.86 191.71 2463.1562 8.00 2483.93 2423.96 32.21 2179.44 137.74 2246.2434 6.0001001 0.0001030 0.5 2 2.7686 7.79 2584.3025 1.48 2404.63 2326.23931 2.21607 1.0001037 0.2080 87.67355 10.43 121.2536 7.45 535.0001000 0.52 384.54 2700.0001017 0.20702 87.0766 6.00257 0.0520 7.0526 7.20424 5.45 Saturated vapor (hg) 2501.3739 1.9085 7.1956 0. (sf) (sfg) 0 0.29 417.29 2336.56 2444.3311 8.78 504.02117 7.21711 1.19150 19.9918 Pressure Kpa P 0.00385 54.47 101.02 2392.79219 0.87 81.6613 1 1.5 3 4 5 7.17 2420.21 2540.0001022 0.35 271. (hfg) 2501.3 2384.29 54.99 111.65624 0.08 28.6868 5.87 561.71871 0.06 2609.0001026 0.5960 5.67 2506.28 2636. All Rights Reserved .19251 19.37385 1.6441 6.65731 0.30 2484.14 2358.46 2685.62 105.95 1982.9439 1.98 2393.87 2662.48 2519.44 137.63 2599.03 17.14 1991.47 88.4226 0.10 2222.88573 0.40 2408.40 2477.57 561.3545 0.2231 8.19 2625.0910 1.96 2191.6431 8.6751 7.72 2038.00285 54.799915 28.66402 34.0001014 0.32 2052.132 129.55 2172.00363 0.81 53.5939 7.0001064 0.98 13.60582 Evap.02 2315.53 2706.1562 8.49 2540.5300 1.131 129.36 2191.0208 6.85 2496.7548 0.8247 6.49 2143.79 168.21 2159.76 225.0001053 0.34 548.4848 1.22816 3.15828 1.30 467.9383 6.83 2205.05 2226.40 2278.72 2069.82 2373.3120 0.3950 8.02 2451.69 535.0001048 0.29 2725.1717 7.60475 Saturated vapor (vg) 206.23775 14.5173 5.0001000 0.69 2338.5763 0.32 2399.5705 1.4763 0.90 289.19 2305.7236 8.0001003 0.48 2293.03 2545.15933 1.36 417.43 Evap.50 2554.64937 6.56 2537.0001043 0.0001073 Evap.92 186.0001010 0.90 251.8313 7.59 2258.46 2213.55 340.9756 8.30 2533.25385 45.43 2112.21 317.6322 8.47 520.51 2415.64835 6.8278 8.3801 5.9104 5.06 120.06 64.66502 34.6717 9.23 124.08 486.5775 8.63 2712.0001002 0.08 2468.00 127.97913 67.55 Enthalpy (KJ/kg) Saturated liquid (hf) 0.2232 7.60 2529.2843 7.0001067 0.74 2645.67254 10.0001001 0.4563 7.79 168.51 340.4629 8.81 225.70 2261.2514 8.97 69. 70 2570.80 2760.21 655.58 1847.001124 0.39 1693.24043 0.96 175.31457 0.94 761.37489 0.17 1897.3150 2.001216 Evap.7274 1.29268 0.9404 6.001149 0.79 604.77 170.52425 0.98 2048.81 959.34 2548.32 1791.31 2553.37380 0. Inc.4448 6.77 936.76 Saturated vapor (ug) 2546.40 2735.55 1777.8935 3.8972 1.09963 0.19 1972.0001101 0.11349 0.43 2799.97 1806.45 1838.04 2066.67 2752.6421 6.73 Appendix Copyright 2003 by Marcel Dekker.23931 0.63 2773.53 2743.95 2592.26 697.8606 1.48 908.09 187.2646 5.4692 6.25 741.27 813.24 640.001154 0.79 2578.07 198.3408 6.16333 0.93 151.9627 1.1193 5.9922 2.16 1599.22698 0.62 2561.0709 2.08875 0.0001076 0.46 1964.65 1004.42646 0.25560 0.82 2590.5412 Saturated vapor (sg) 6.41 Saturated vapor (hg) 2728. (sf) (sfg) 1.89 1795.34268 0.8206 1.6456 5.04 2039.7600 6.48 162.42 828.001121 0.94 2776.43 2031.1171 2.56093 0.8958 6.31 594.99 191.74 1865.08 2778.86 155.79 2133.3895 6.64 195.30 669.34159 0.0946 2.27126 0.76 212.09 985.15125 0.10 2140.42 933.7766 1.21497 0.41289 0.26 2076.12 2023.15 2796.47 2567.23 2564.86 2804.0001115 0.9311 1.12 233.0044 3.06 2572.45 223.72 1947.04 1621.30 1721.001177 0.88 141.09 982.32 205.08 2781.6190 3.5280 4.82 753.6166 4.6647 4.0199 2.52317 0.2130 5.23 584.2165 2.29 622.2438 4.13177 0.82 2787.19444 0.0001079 0.51 2801.91 184.27286 0.43 172.25449 0.21385 0.17 1821.6040 6.30 138.08 797.0001108 0.3851 2.3067 4.001139 0.6018 2.91 2748.13 2603.81 2120.2514 2.28 1917.0461 2.13062 0.63 147.84 1668.001144 0.2842 2.69 2580.6225 6.57 1909.11 1856.09845 0.67 780.38 604.5034 2.22586 0.04 2086.98 2603.1386 2.20 742.76 2148.356 Table 8 Continued Specific volume (m3/kg) Internal energy (KJ/kg) Saturated liquid (uf) 572.1647 5.62 720.11 2583.0001081 0.67 2108.88 583.13 2771.4953 6.6846 6.87 878.5864 6.5711 4.26 1934.58 2790.20 731.07998 0.49 2574.1850 4.79 781.08756 0.8212 6.8565 6.8289 4.99 2732.14 Entropy (KJ/kg K) Saturated liquid Evap.47 2097. All Rights Reserved .67 2803.001133 0.29 844.94 2756.97 1008.4473 2.2971 6.7158 4.001197 0.7330 6.31567 0.20419 0.2208 6.98 1817.07 2803.11232 0.92 2551.58 2738.90 Saturated liquid (vf) 0.68 2784. (vfg) 0.3744 4.7892 6.29 2000.81 751.50 2766.0001111 0.49029 0.26 2601.30 2056.64 2586.7005 1.53 1764.0001097 0.07154 0.42 218.17753 0.0359 4.46 2582.91 670.55 2557.91 830. (hfg) 2155.88 683.001166 0.93 594.0001093 0.07275 0.001207 0.06668 Evap.17639 0.43 708.7028 3.81 2604.38 177.48 962.29158 0.98 1956.0001084 0.52 1886.56201 0.001187 0.5233 6.00 762.83 2594.66 655.5535 6.55 696.19332 0.95 1890.9606 4.07 1866.19 1840.65 1830.50 2597.20 709.10 732.73 623.69 179.00 2792.14084 0. (ufg) 1973.97 167.08 2015.4478 4.20306 0.001127 0.2574 6.13969 0.46138 0.9174 6.41398 0.68 843.10 Enthalpy (KJ/kg) Saturated liquid (hf) 573.0001104 0.5546 2.67 1959.16220 0.40 2588.15 1751.51 1876.83 2600.54 684.18 1644.6627 6.32 143.87 1921.1869 Pressure Kpa P 325 350 375 400 450 500 550 600 650 700 750 800 850 900 950 1000 1100 1200 1300 1400 1500 1750 2000 2250 2500 2750 3000 Temp T C 136.45 721.06546 Saturated vapor (vg) 0.36 1986.43 2769.64 814.07878 0.0001088 0.99 229.7938 3.8920 4.1298 4.9651 6.44 906.01 164.32 798.15011 0.4869 4.001118 0.7527 1.48 158.75 639.14 876.49137 0.73 2576.7704 4.93 1949.1791 2.34 Evap.30 2763.7080 6. 41 1082.71 2081.8078 3.95 336.88 295.03244 0.01657 0.47 1393.80 675.43 2431.001840 0.15 324.3307 4.54 1350.11 2724.60 2684.80 374.38 2794.9132 0.04978 0.06 303.53 1450.7485 1.94 2742.70 1714.47 1841.9201 3.05 1491.00583 0.9269 4.4294 3.00749 0.28 1151.96 2374.33 1378.5604 3.00072 0 0.78 1205.Appendix 3250 3500 4000 5000 6000 7000 8000 9000 10000 11000 12000 13000 14000 15000 16000 17000 18000 19000 20000 21000 22000 22089 238.001924 0.40 263.40 1096.1044 5.1551 6.04 0 2803.81 369.01426 0.02213 0.06152 0.47 1000.3698 1.1233 1.42 124.58 388.0227 4.76 984.51 1305.26 2.3915 2.4995 1.0700 5.83 2662.56 446.2454 5.001235 0.22 2637.14 1255.43 352.99 928.00473 0.46 1571.34 1384.90 777.001418 0.29 1154.2067 3.00 1433.3097 5.87 1758.00666 0.8625 2.70 2602.33 2784.8718 1.28 2338.4000 3.03112 0.61 1363.6942 0.48 2569.15 2509.001489 0.2330 1.23 869.8132 5.001384 0.001567 0.59 856.04483 0.92 2099.8015 4.85 809.75 2544.14 0.001527 0.001286 0.03944 0.8713 3.4961 3.9733 5.62 1045.21 1213.07 744.74 108.76 1066.37 1753.08 1610.1073 4.99 1449.42 598.06 1040.1917 0 6.92 1511.001611 0.97 2803.27 1322.001226 0.8891 5.01803 0.00868 0.00760 0.58 1029.18 507.79 2557.16 1698.18 1888.74 2513.6847 3.72 2158.04 930.02737 0.0841 0.25 1731.02048 0.00353 0.97 1264.06 318.01278 0.7252 2.001711 0.97 1316.00275 0.70 2404.99 275.08 2476.97 1776.05 2293.45 1650.12 2589.43 1826.01121 0.0266 3.55 1193.3595 3.01149 0.00836 0.33 2772.003155 0.0532 2.04 2603.12 1571.5527 5.63 1660.1252 6.00987 0.16 2029.001252 0.4323 5.001770 0.05707 0.04853 0.76 2455.05 2230.59 1130.7460 3.73 1087.5365 2.4923 5.9962 1.75 342.39 2029.06029 0.25 1207.01274 0.53 1585.23 1407.00931 0.54 365.43 2801.26 1774.4297 3.4685 3.10 1441.6231 3.44 311.09 1548.49 2580.67 2705.41 2529.24 1531.68 1472.6922 2.4297 357 Copyright 2003 by Marcel Dekker.00 1690.6250 1. All Rights Reserved .06 361.6771 5.28 1147.002035 0.58 1622.24 347.60 250.09 1640.86 1739.001452 0.75 330.38 242.00 1505.01907 0.07 2757.11 583.09 2464.55 2610.9387 4.49 2547.29 1519.30 2034.0137 4.00565 0.1776 5.74 2334.02602 0.58 1578.13 688.43 1558.67 2496.002206 0.00315 1025.97 1973.001658 0.00495 0.00380 0.97 2099.6866 2.6140 5.60 1049.7963 2.001319 0.3716 5.44 285.2545 2.89 373.37 357.01599 0. Inc.02352 0.01034 0.2737 3.00659 0.002808 0.5224 4.54 2409.1210 3.24 0 2604.27 2597.32 1266.01450 0.41 1257.001351 0.2857 3.88 1317.69 2580.7431 5.03815 0. 70 (99.05748 8.1271 7.08 2877.58 4891.85 2734.45 3479.6076 9.4479 8.56 2782.98724 72.9400 8.33) 2645.69400 1. Inc.58 P ¼ 200 KPa 2529. All Rights Reserved .97 2811.62) 2675.37198 67.4858 10.99 3075.56 2687.1579 8.4028 10.67896 40.3555 8.4177 9.24034 — 3.61 (120.6281 10.84 3855.45 3663.52 2780.67418 13.51 3279.52 3278.47 4467.73 4159.23) 7.46 2776.87 — 2682.86920 17.27 7.1749 8.10 3928.89 2443.20929 7.88573 0.46 4891.82 4683.05 2706.75669 63.10 4396.03 4053.3593 7.91 4052.6133 7.86 2661.1002 9.4090 11.61 2585.8977 10.5939 — 7.88937 4.1608 10.37 4467.27 2735.8852 10.29498 44.89 3132.90444 10.44 4640.51 3489.81) 2584.14137 58.2795 7.5810 3.60250 2437.49 2576.69 5409.91 4683.8641 9.30 4640.50 2587.52054 s (KJ/kg K) V (m3/kg) u (KJ/kg) P ¼ 50KPa 2483.95 2812.52 2977.08034 8.06252 35.1501 8.75 2658.87 2515.64 2975.17226 2506.94 3302.06 2582.0392 11.70 3854.93636 2.40 3928.51 4158.2287 11.08 5147.13364 8.8382 s (KJ/kg K) Saturated 50 100 150 200 250 300 400 500 600 700 800 900 1000 1100 1200 1300 14.82773 11.19561 19.8342 0.1545 9.92 4396.41833 3.85 — 2511.01 4257.31 3076.13559 26.0967 10.82507 24.6881 8.6662 10.8395 11.9037 9.5066 Copyright 2003 by Marcel Dekker.51251 21.82045 5.2812 9.46 7.98104 9.99 2879.75097 12.78 5409.89 3488.33 2968.2964 10.63 3663.358 Table 9 Temp (C) SI Superheated Vapor Water v (m3/kg) u (KJ/kg) P ¼ 10 KPa h (KJ/kg) (45.61 2659.67355 14.19 5147.63 2592.68 P ¼ 100 KPa Appendix Saturated 150 200 1.22 3479.38 2875.28391 6.6599 9.91052 49.35595 4.80 2870.87 2654.59737 14.06 2968.5372 8.91 4257.6947 7.26 3302.39 h (KJ/kg) (81.62 3705.63 2768.95964 1.05 3705.44508 31.43 3131.52599 54. 02781 4.40604 2.66 4158.11 3488.46246 0.15 3300.85 3131.22 3477.25 4467.8926 8.53 4257.3788 7.95 5147.53 4467.53 4157.70 4683.63876 3.49 4052.36 3478.32 5409.87529 1.28 3274.2157 8.79 3276.8458 10.36 3129.01 4890.79 3478.95174 5.53 3129.41353 5.16 3066.4565 9.96 3927.54 2967.5165 7.27 4395.8319 9.56 5409.56547 4.96 3703.60582 0.0332 8.5182 1.41 2967.16834 3.53 3854.47 P ¼ 300 KPa 2974.69 2806.79 2650.51 2964.63018 2731.38 3662.54930 1.49573 1.72 3928. All Rights Reserved .79636 0.73 2810.0194 9.69 3130.39927 3.83 2726.59512 0.1658 10.49 (133.95 3300.31 4890.12147 1.44 3926.55 2570.75 3301.47084 0.3462 10.63388 0.0329 8.26 (143.3115 7.30 2760.47539 2.79863 7.00555 1.7767 9.2450 9.85 6.01297 2.78139 2.6987 8.41 3484.40 7.88934 1.93740 3.55 3487.33 3074.94 3479.24426 2.33696 6.81 3663.5652 9.78 4257.55 2564.01263 3.98 3485.77 4640.83 8.77 4052.54 3301.3398 9.70643 2.9918 7.77232 0.53 2752.8314 9.5) 2775.95 3662.26030 2733.53422 0.68 5147.7022 8.59 3069.3250 8.1706 7.5892 8. Inc.65 2728.23 P ¼ 400 KPa 2970.0262 10.1982 Saturated 150 200 250 300 400 500 600 700 800 0.19880 1.65484 0.1912 8.09 3704.Appendix 250 300 400 500 600 700 800 900 1000 1100 1200 1300 2.95 2865.23422 2553.24 3663.89 3702.4557 8.10 4157.22 2808.0975 9.48 2646.2217 8.71 4396.9244 (continued ) 359 Copyright 2003 by Marcel Dekker.5661 7.19 3854.87526 6.11 2804.98 3071.71629 0.55 2966.51 6.03 3703.34136 1.46 4683.03151 1.81 2964.5132 8.23 4158.9299 7.8984 8.69 2965.63) 2738.7769 9.9764 10.75 3273.8958 6.20 3927.28 3278.64994 2543.5434 8.48986 4.18669 1.12 4640.31616 1.0575 0.6563 9.7085 7.0778 7.82 2860. 6585 9.5107 8.80406 1.20 4052.00 2881.7059 9.05 h (KJ/kg) (143.95812 2.05594 9.67 3925.77 4467. Inc.23 2642.41 4156.36 4638.61 6.08217 1.17 3853.5463 7.0020 8.5952 8.47424 0.71 4890.41 4890.57012 0.82450 0.03 (151.25 3482.9665 7.59199 0.91 4052.7058 8.5) 4395.42492 0.53 4466.88 6.71093 0.40 2638.0592 7.26718 2561.86 2801.23 4682.3521 8.9485 9.91 2882.0329 9.81511 s (KJ/kg K) V (m3/kg) u (KJ/kg) P ¼ 400 KPa 3853.08 3661.8211 9.12 2962.37 2960.34 4051.15 5146.68 3064.0872 8.71 4639.75 3700.83 3483.64 3477.0109 1.61728 0.99 P ¼ 500 KPa Saturated 200 250 300 350 400 500 600 700 800 900 1000 1100 0.2691 9.83 3853.02 3127.06 4639.35 3299.11214 2.3360 9.80 2850.35288 1.1484 9.41 5147.2328 9.83 5408.7367 8.1361 9.2708 7.07 5409.5255 9.66974 0.42 4639.91 2720.07 3477.52 3662.3381 9.3723 7. All Rights Reserved .27 4256.86) 2748.8212 7.26614 2.11 4889.4224 0.360 Table 9 Temp (C) Continued v (m3/kg) u (KJ/kg) P ¼ 300 KPa h (KJ/kg) (133.80406 0.97883 1.4689 9.90169 0.97 4156.20 3167.37489 0.2673 8.59 2963.96 4394.8389 10.12 2957.98959 1.19 3128.69958 1.76 4256.46847 1.51372 0.51 4256.8780 s (KJ/kg K) 900 1000 1100 1200 1300 1.7937 8.65 3271.63 4051.17469 1.81 4889.89691 0.74720 0.02 4256.63 3156.39383 0.75 P ¼ 600 KPa 2567.29 Appendix Copyright 2003 by Marcel Dekker.7600 6.63) 4395.6528 7.31567 0.16 3061.91 3925.55 3299.43437 0.35202 0.47436 0.67 2855.1816 7.85) 2756.4598 7.99 4682.52256 0.50 2802.82 3701.66 3270.80 (158.52 4394.42013 3854.91 2723.58404 1. 83 2603.38426 0.25 2949.80 3153.65 4638.7467 6.52 4682.13 2839.5864 6.87 (187.61813 0.79 2630.4669 8.5185 9.32411 0.26080 0.41 (195.32 2785.57 (170.0289 8.65 3263.5575 0.90 2709.6627 6.2731 8.16 2959.19235 0.20596 0.2049 9.18228 0.72608 2583.1332 8.74 2704.19444 0.91 3476.09 4465.27 4155.8672 8.16930 0.40109 0.6939 6.19 4050.84974 0.22 2872.77 4051.91 2793.43) 2769.85 5407.88 3478.22 3661.44779 0.94 4637.29314 0.49432 0.4692 6.0384 7.0153 9.67977 0.6028 9.16 2869.61 2715.22 3040.8153 9.2327 7.05 4681.1359 361 (continued ) Copyright 2003 by Marcel Dekker.32 2927.44 3697.59 3051.5233 6.9119 9.0316 7.34 3296.14 3852.86 2942.2120 0.67610 0.52 P ¼ 800 KPa 5164.43 3161.5715 7.09 2698.59 9.82 2815.25794 0.3854 9.46 2797.16 6.76 3475.4542 Saturated 200 250 300 350 0.60 4888.08 5145.33 P ¼ 1400 KPa 6.01 3045.28 P ¼ 1000 KPa 5146.79188 0.58 5408.1228 7.00 4255.34 (179.64 2621.68 3267.14 4154.00 2803.07 3480.23452 2588.38 3924.20 4889.14302 0.14084 0.23268 0.95 3297.82 2612.90758 2576.7118 8.3010 7.13302 1.36 5407.24043 0. All Rights Reserved .21 2875.1016 9.8158 7.21382 0.49 9.16350 0.85 3923.60 3699.8293 7.76 4682.35 3149.2821 9.07) 2790.73401 0.35964 1.50184 0.54075 0.30659 0.Appendix 1200 1300 1.3770 8.97 3056.46 3852.55 5145.9533 7.21009 446.34 5408.66 3125.5898 6.78 4392.91) 2778.63345 0.4975 6.14 2878.44331 0.20026 2592.45210 4464.56007 0.6033 8.4088 7.90 2935.29 3124.7749 1.16333 0. Inc.58 4681.35411 0.20 2789.08 2827.57 4466.35 3660.99) 2784.65 4393.81 P ¼ 1200 KPa 6.18 2957.6906 Saturated 200 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300 0.12 6.9246 7.49 4255.7621 8.15 3157.35439 0.4650 7.58712 0.28247 0. 12497 0.93 3693.41926 0.48919 0.90 4392.45239 0.3773 7.95 2947.28596 0.11042 0.0693 7.90 h (KJ/kg) (195.26 2781.99) 3260.14 4464.66 3117.22199 0.02 2919.4386 7.82 6.18 3250.32 3922.27420 0.3431 8.33772 2595.30859 0.21780 0.3025 7.14021 0.30012 7.05 4049.24998 0.73 4152.87 3657.0099 7.10 3294.15 4390.44 3473.47 Appendix Copyright 2003 by Marcel Dekker.35 3254.56646 0.2808 8.48552 0.12380 0.7558 8.8844 7.00 4888.362 Table 9 Temp (C) Continued v (m3/kg) u (KJ/kg) P ¼ 1200 KPa h (KJ/kg) (187.27 4390.41177 0.15457 0.77 3851.4149 8.1977 9.6732 6.28 3696.6272 8.01 4153.03 2866.84 3292.19550 0.31947 0.48 3659.22029 0.8274 9.02 2776.69 3918.1881 8.5555 8.38606 0.59 4151.39 P ¼ 1800 KPa 2598.0535 8.1793 7.90 3469.60 3474.4934 0. All Rights Reserved .0171 9.9983 8.17456 0.96 3029.27937 0.4217 6.62 4049.09 3119.2983 s (KJ/kg K) 400 500 600 700 800 900 1000 1100 1200 1300 0.33393 0.8080 8.87 4153.75 3691.40) 2794.86 P ¼ 1600 KPa Saturated 250 300 350 400 500 600 700 800 900 0.35281 0.66 3476.7523 7.6026 7.60507 2954.65 4680.2373 7.37294 0.12 4680.9456 9.98 4254.74 3658.02 5144.53 4636.50 3121.25480 0.15) 2797.14148 0.42 3474.21 3141.13 2910.05 2950.5389 7.47 3293.49 (207.16847 0.15862 0.1262 9.20 3034.2258 8.03 4391.3698 0.07) 3257.87 5406.78 3920.1660 8.47 4254.8710 8.61 4465.3793 6.61 3659.6066 6.6758 7.38 2686.45051 0.71 3849.72 3295.11 3694.4824 7.25215 0.29463 0.23 3919.11 6.19005 0.27 3472.9434 8.09 3851.90 3122.38 5406.52783 0.8226 7.23 4637.95 (201.95 2692.41 4887.83 2862.10 3471. Inc.17 3471.83 3145.51864 s (KJ/kg K) V (m3/kg) u (KJ/kg) P ¼ 1400 KPa 2952.49 5144.40 3850.24818 0. 46 8. All Rights Reserved .9562 7.80 P ¼ 3500 KPa 6.0147 7.8435 8.17832 0.Appendix 1000 1100 1200 1300 0.70 5403.6938 8.90) 2804.94 4252.43 2829.32598 0.60 3357.21 3030.28 3350.36678 0.21590 0.40 5405.13857 0.1869 6.19716 0.1817 Saturated 250 300 350 400 450 500 600 700 800 900 1000 1100 1200 1300 0.27185 0. Inc.0096 9.77 3462.18 4463.5452 6.8402 7.58 3690.51 2902.45 4150.57 5141.64 4633.36306 2600.13998 0.2364 0.42) 2799.99) 2803.1748 6.40340 4048.16353 0.60) 2803.65 2737.81 3126.03 3849.05873 0.75 2993.2574 6.06668 0.5960 7.8290 9.27004 0.42482 0.08114 2604.84 2939.35180 0.04 3686.20 4387.2871 6.92 5405.20 3290.56 2851.6437 6.18 4679.1252 6.48 3467.05707 0.67 4251.56 2859.00 2750.15120 0.50 3136.95 5143.5900 8.25322 0.4316 7.7023 7.10976 0.22323 0.08700 0.44 P ¼ 2500 KPa 4635.8569 9.39581 0.3233 7.81 4886.24668 0.89 5142.37761 0.80 3655.07998 0.05 6.26 2679.6390 8.99 3657.13 2662.2844 7.2853 8.97 P ¼ 3000 KPa 6.55 2761.08 4677.7800 8.61 4885.3895 8.0642 9.41 3116.45 4253.07 2880.59 4148.92 P ¼ 2000 KPa 4635.21 4886.1766 8.45382 4048.8837 9.15930 0.06 3008.09890 0.96 3247.99 3468.66 4464.06842 2603.99 6.42 5143.0720 8.52 4462.3408 6.56 (223.9606 9.07058 0.5389 0.4460 363 (continued ) Copyright 2003 by Marcel Dekker.1745 7.93 3470.89 5406.13014 0.33984 0.23458 0.40 4389.7663 6.95 (242.1270 7.71 4679.4860 8.43 3112.09963 0.81 2945.19 2977.12547 0.17568 0.08 3287.33 4047.89 4046.71 4463.14 2855.24 3239.40 4634.10 (233.11144 0.19960 0.96 4253.12 4884.4084 6.31659 0.10 2644.02 (212.9487 8.29046 2603.25 3914.58 2772.14 3917.12010 0.25 4678.29333 0.0921 Saturated 250 300 0.43 8.30 3847.1328 0.03 3025.70 2623.46 3023.6761 8. 21 3674.07661 0.9362 7.48) 2798.04 6. All Rights Reserved .06475 0.5130 6.05884 0.9135 8.23 3439.2827 6.27 2926.00 3456.58 3846.09196 0.1275 8.09053 0.87 3678.65 2919.11095 2602.0900 7.8723 s (KJ/kg K) 350 400 450 500 600 700 800 900 1000 1100 1200 1300 0.08643 0.65 3323.09885 0.03 3466.22652 0.07074 0.27 2725.3288 8.5084 7.9211 7.5911 8.3109 7.33 4460.90) 3115.1571 7.5631 6.11324 0.02 4044.06 3462.24 3337.0833 7.07 3080.38 2960.20749 s (KJ/kg K) V (m3/kg) u (KJ/kg) P ¼ 3500 KPa 2835. Inc.23 3276.7046 6.09918 0.18080 0.75 3020.0700 6.3614 6.05840 0.87 4631.66 (257.25 3230.82 3344.6198 0.59 3653.24206 2843.11 4630.66 2932.04978 0.7689 6.28 5401.84 3845.8745 7.40) 2801.17980 0.15402 0.04406 0.4009 8.76 4675.10787 0.68 3092.81 (250.19415 0.23 3445.47 3903.28 3103.73 3282.92 4676.45 P ¼ 4500 KPa 2600.91 h (KJ/kg) (242.0198 6.80 4384.5820 6.49 3279.44 3905.7571 7.43 3213.37 3015.94 6.16743 0.57 3204.63 P ¼ 4000 KPa Saturated 300 350 400 450 500 600 700 0.88 3010.38 3107.21098 0.48 3682.46 4045.78 2913.14 4881.2337 7.11619 0.14838 0.0300 7.364 Table 9 Temp (C) Continued v (m3/kg) u (KJ/kg) P ¼ 3000 KPa h (KJ/kg) (233.4338 7.14056 0.08765 0.51 3330.99 3222.13 3099.05135 0.34 3911.92 3285.51 3670.12699 0.72 4146.9862 8.08453 0.03 2712.19541 0.33 2826.16414 0.63 4883.29 2943.06 3464.91 3095.08003 0.8404 7.07341 0.40 3908.04 3459.26 5140.60) 3103.94 5139.14 4459.7427 6.00 2817.6837 7.07678 0.06645 0.09936 0.15 Appendix Copyright 2003 by Marcel Dekker.7000 8.1999 8.14 4249.00 4385.3688 7.84 4143.15 3450.6578 6.09847 6.0051 7.9942 0.37 3651.29 3004.13243 0.5191 8.40 4250.49 5402.7719 8. 33 2884.05214 0.03944 0.74 4375.02 5135.12 P ¼ 6000 KPa 4139.14526 2597.78 4457.83 4243.04223 0.28 5397.1217 7.97 3177.17 3301.7942 8.11707 0.07352 0.99) 2794.92 3181.8099 0.15098 0.12 3837.67 3646.06101 0.5407 6.7192 6.22 2789.4492 6.11 3843.07368 0. All Rights Reserved .09749 0.07 5400.04739 0.13 4137.59 4042.45 4673.08958 0.57 3266.0751 8.12106 2589.2108 8.15 4041.76 3422.7440 7.08160 0.5824 8.6458 6.26 4454.00 4669.9733 6.69 2667.58 4627.3334 6.8185 6.30 4671.3519 8.12648 0.89 3453.0673 6.2661 8.6199 (continued ) 365 Copyright 2003 by Marcel Dekker.87 4247.7055 0.15 3433.12 3540.13587 0.96 4458.24 7.12287 0.67 5398.10536 0.14056 0.97 7.06857 0.2083 6.0647 8.4473 8.65 4880.71 4040.6376 8.07879 0.06330 0.64 3316.47 3900.5122 7.10911 0.64 5.38 (275.0287 7.6566 7.0091 8.16139 3648.64) 2784.14645 0.23 3666.8727 8.35 4245.13013 0.15817 0.32 5136.33 2924.37 3842.61 4456.29 P ¼ 5000 KPa 4141.01 3457.4234 7.20 3174.4566 8.34 4328.61 2892.42 5133.1676 7.53 3068.13469 0.58 2999.2661 8.9593 8.17 4378.76 3550.39 3195.05665 0.28 4132.8802 7.05194 0.03244 0.1612 8.67 2906.60 4674.18156 3650.94 2808.05781 0. Inc.62 3840.59 4382.63 5138.82 4625.82 3273.04532 0.84 4037.11321 0.19 3042.Appendix 800 900 1000 1100 1200 1300 0.29 4622.74 4875.12 2697.4014 8.06525 0.81 2988.61 4246.2588 7.16987 0.8891 6.38 4380.52 (263.15 3643.9758 7.87 5399.08849 0.7548 Saturated 300 350 400 450 500 550 600 700 800 900 1000 1100 1200 1300 0.11965 0.03616 0.09811 0.64 3090.8502 8.5330 8.40 3894.90 3082.96 5.10762 0.69 4878.17 4879.62 3658. 7350 7.02641 0.04416 0.40) 2742.4842 s (KJ/kg K) Saturated 300 350 400 450 500 550 600 700 800 900 1000 1100 1200 1300 0.7905 6.88) 2772.60 4449.05195 0.04516 0.07301 0.06) 2724.07669 0. All Rights Reserved .39 4128.9384 8.12 Appendix Copyright 2003 by Marcel Dekker.04845 0.28 3271.43 3444.366 Table 9 Temp (C) Continued v (m3/kg) u (KJ/kg) P ¼ 7000 KPa h (KJ/kg) (285.02352 0.0020 8.03677 2557.02993 0.2119 6.79 2590.02737 0.83 3373.13 3055.4189 6.9442 6.04175 0.81 4238.91 3073.06) 2757.04 3410.5550 6.1933 8.3747 8.27 3521.93 2747.54 5391.09703 0.34 2878.3476 7.6326 6.76 3254.94 2784.83 5130.8132 5.07 2838.00 3636.08 3832.90 5393.9486 7.06283 0.05565 0.2812 7.71 (303.60 3639.02947 0.01803 0.2282 6.30 3138.76 3256.72 4667.02 3158.45 4665.6575 0.67 2923.08 4032.98 2987.03350 0.01 3642.48 2632.06981 0.09080 s (KJ/kg K) V (m3/kg) u (KJ/kg) P ¼ 8000 KPa 2569.55 2977.4477 6.05481 0.25 5128.13 2769.0894 7.1299 8.03 3882.99 3398.46 (311.06097 0.7991 8.08489 0.30 3159.02048 0.08350 0.04814 0.03993 0.02 P ¼ 10000 KPa 2544.0361 6.4846 6.9304 6.03524 0.5965 5.66 3064.02426 0.1300 6.40 3016.69 3448.09027 0.07 3287.03432 0.38 2955.3115 8.84 4368.02975 0.96 4035.2853 6.59 3386.03279 5.32 3045.29 3530.3633 6.8778 7.80 4872.33 3167.5173 7.03817 0.61 3834. Inc.41 2699.0205 7.16 2832.38 2943.26 4616.06702 0.77 4619.07896 0.92 4451.5822 7.10377 2580.75 2966.02580 0.7239 6.02242 0.67 2863.26 3888.55 3117.31 4240.87 4870.39 3096.46 3240.02995 0.77 h (KJ/kg) (295.11 2956.6140 5.30 4371.7431 5.75 2724.38 2848.21 3260.05 5.47 4123.6771 6.87 3650.7974 6.5472 0.33 P ¼ 9000 KPa Saturated 300 350 400 450 0.63 5. 04285 0.0987 0.Appendix 500 550 600 700 800 900 1000 1100 1200 1300 0.38 3632.23 4439.04358 0.6775 6.7117 6.03837 0.82 5118.70 2879.8315 8.51 4119.2965 7.06789 0.1687 7.8141 6.02000 0.03029 0.9218 7.49 2692.56 5112.04523 2455.6782 7.59 4228.89) 2673.05430 2505.15 3039.0236 8.78 5.3442 6.22 3434.18 5389.2221 7.04574 0.12 4092.38 4364.30 4236.44 P ¼ 15000 KPa 3500.4279 7.64 3310.9588 7.5181 7.99 (342.01565 0.02 3633.7561 6.75 3730.0739 8.33 5.77 2826.68 3124.32 3840.0943 7.01845 0.61 3582.41 2975.25 2912.5198 6.7810 6.9028 7.47 2996.43 4343.03248 0.02299 0.27 6.72 3475.03546 0.53 3448.97 4444.20 3248.03210 0.41 4103.84 5386.07544 0.74 4613.05832 0.03869 0.03987 0.04658 0.81 4233.7237 7.13 3604.08072 3152.53 3829.07016 0.2718 6.04101 0.05349 0.02491 0.06482 0.69 5126.94 3225.8821 8.05045 0.52 3104.4597 7.24 4611.02 3819.44 3156.3097 5.14 5123.92 3625.02560 0.68 3338.93 3620.02861 0.30 3199.8282 8.8810 6.22 (327.02801 0.43 3422.28 4447.01350 0.9572 7.0108 (continued ) 367 Copyright 2003 by Marcel Dekker.71 3208. All Rights Reserved .4617 6.8223 6.6289 6.37 3324.4420 5.3783 Saturated 350 400 450 500 550 600 650 700 800 900 1000 1100 1200 0.07265 3144.81 4858.04267 0.6347 7.0397 7.30 3712.54 3241.2054 8.01613 0.41 4222.0416 6.52 4114.15 3308.44 3021.04200 0.4283 0.11 4021.02 6.18 4662.06312 0.94 3610.97 3826.01147 0.89 4015.93 4660.32 4027.48 4603.55 4433.57 2789.43 2520.75 4596.73 3755.09 3343.03875 0.04857 0.02680 0.24) 2610.99 3811.04 4865.1403 6.01034 0.6272 7.9165 8.91 4361.4077 7.32 3876.73 P ¼ 12500 KPa 3511.36 2740.95 4867.69 4352.34 3748.78 3341.4623 5.02080 0.65 3439.02293 0.05409 0. Inc.03460 0.72 3628.2556 8.20 4030.27 3870.08 2624.03564 0.63 4852.37 3410.44 3855.05950 0.0536 7.04859 0.2040 7. 96 5365.34 3174.01 3592.1400 6.05 4589.03391 0.44 4003.37 4582. Inc.12 P ¼ 20000 KPa 2293.57 3675.69 3274.9016 6.1418 5.00994 0.02738 0.84 3191.01818 0.02434 0.25 3083.03031 0.24) 5375.10 4.79 2902.46 3490.8736 7.01270 0.1245 7.51 3296.01929 0.2717 0.95 P ¼ 30000 KPa 1847.52 P ¼ 25000 KPa Appendix 375 0.15 2970.368 Table 9 Temp (C) Continued v (m3/kg) u (KJ/kg) P ¼ 12500 KPa h (KJ/kg) (327.01517 0.81 4637.98 2844.9303 h (KJ/kg) (342.45 3537.73 3797.94 3824.2382 6.60 Copyright 2003 by Marcel Dekker.05 2619.28 4643.90 4428.75) 2528.25 4216.22 2806.32 3809.3347 6. All Rights Reserved .93 4.43 3.0319 0.46 3386.5588 7.00 3281.02113 0.67 3953.03636 8.1093 0.03145 0.89 3804.24 5100.30 4421.45 4840.44 (354.02 3421.37 4846.03876 0.01736 0.7212 6.8706 8.07 3060.001973 1798.0182 6.1839 s (KJ/kg K) 1300 0.67 4009.0026 7.37 5106.02645 0.7530 7.01245 0.18 3393.9269 5.19 2684.37 3560.03316 0.02897 0.13 4335.01656 0.78 3500.02588 0.04 3398.09 3940.50 5.74 2818.03597 0.48 4081.00583 0.59 5370.56 3601.02106 0.5539 5.82 3062.02385 0.7356 6.12 4211.04154 2390.0441 8.9359 8.82 3109.6582 6.76 P ¼ 17500 KPa Saturated 400 450 500 550 600 650 700 750 800 900 1000 1100 1200 1300 0.94 (365.13 3693.04845 s (KJ/kg K) V (m3/kg) u (KJ/kg) P ¼ 15000 KPa 4649.6874 7.81) 2409.80 4326.05813 4654.89) 5381.9308 7.001789 1737.2830 7.00792 0.7993 6.4925 7.01477 0.02274 0.27 4069.16 2942.5866 6.0544 7.5048 6.4229 6.02251 0.75 1791.01969 0.3507 7.06 3238. 26 3782.4057 6.9150 369 (continued ) Copyright 2003 by Marcel Dekker.4631 6.4699 5.8332 7.5229 6.5028 4.05 3346.2330 6.7747 5.31 3100.94 3062.67 2970.7784 6.011124 0.64 3361.3750 6.15 5089.4722 5.023102 0.7604 7.8721 4.015433 0.49 3559.015623 0.63 3517.86 5354.6707 6.7450 6.019196 0.54 4274.61 3158.00 4626.021045 0.04 3283.34 2455.48 3978.003428 0.04 2614.2054 6.36 2994.52 2373.009527 0.5765 7.012724 0.006984 0.6281 5.025119 0.53 3221.Appendix 400 425 450 500 550 600 650 700 800 900 1000 1100 1200 1300 0.06 2619.001908 0.09 4568.01 3395.1418 5.002100 0.4845 7.05 2609.51 3251.3010 6.7904 6.022589 0.96 P ¼ 40000 KPa 2151.2124 4.020903 0.18 4401.004962 0.36 3443.36 3637.009064 0.93 3.88 2920.8283 4.21 2720.81 3753.8432 375 400 425 450 500 550 600 650 700 800 900 0.51 3137.9886 0.67 4024.4423 5.11 2512.46 3777.2867 7.011523 0.26 3149.010168 0.1178 6.03 3275.016647 0.9592 6.29 3978.027115 0.024266 2067.71 2751.52 2096.009942 0.42 1742.91 P ¼ 35000 KPa 2580.014138 0.79 2903.9459 5.010575 0.69 3022.29 3017.029101 2430.4728 5.38 3520.92 4200.44 5.013661 0.04 3335.009162 0.60 3768.09 1854.006004 0.28 5078.86 1914.017448 0.1134 4.89 3739.56 4047.79 3309.0717 7.011446 0.93 3745.62 3491.42 2498.31 4291.93 4554.17 2821.012596 0.35 3081.80 4257.65 2884.16 2949.005304 0.001641 0.005623 0.89 3536.02 2253.1679 7.018913 0.91 3598.37 1987.97 3990.41 2672.007882 0.97 5343.013278 0.08 4309.70 3162.008094 0.3602 6. All Rights Reserved .0342 6.68 4816.0113 6.83 2198.006927 0.6743 5.1503 5.95 4.014883 1702.1764 6.47 4828.16 2806.001700 0.002532 0.012963 1677.58 3681.9342 0.39 3335.008345 0.54 4001.5606 6.36 2869.6691 7.34 3213.87 3.003693 0.03 3189.18 4412.96 1762.29 4615.9025 6.1962 5. Inc.07 2678.002790 0.008679 0.79 4189.39 3574.006735 0.3801 7.49 3713.6662 6.011533 0.83 2365.30 2820.9345 7.71 1930.84 3555. 3440 5.98 3093.26 5333.9320 5.62 7.011317 0.003957 0.28 P ¼ 60000 KPa 1716.63 2159.86 2390.16 3151.07 3019.5290 6.002486 0.7882 7.88 2896. All Rights Reserved .56 3230.61 2941.65 2178.58 2059.16 4748.5224 7.67 4605.016410 0.004835 0.4837 4527.83 3177.08 5057.47 4475.0342 6.1356 7.019360 0.14 3028.59 4793.4110 6.35 2001.72 4359.62 4224.0030 4.97 3906.013561 0.5910 7.006112 0.1194 7.53 2658.38 4380.007459 0.64 4167.012215 1609.017895 0.007727 0.4058 7.5807 0.005118 0.9317 4.55 1788.001633 0.009480 0.25 4390.66 2053.010409 0.2063 7.56 3889.26 3930.77 Appendix Copyright 2003 by Marcel Dekker.04 1959.21 3364.011411 0.25 3441.60 3680.014616 1638.008508 0.4056 7.014324 0.72 5323.34 1892.0146 7.6969 h (KJ/kg) s (KJ/kg K) 1000 1100 1200 1300 0.001559 0.11 4594.55 3553.002007 0.60 2525.45 7.2733 4.6451 5.34 1745.003892 0.15 5303.009076 0.002956 0.370 Table 9 Temp (C) Continued v (m3/kg) u (KJ/kg) P ¼ 35000 KPa h (KJ/kg) s (KJ/kg K) V (m3/kg) u (KJ/kg) P ¼ 40000 KPa 4541.5485 5.5883 5.001731 0. Inc.36 4124.35 1699.53 4145.016940 0.76 2861.7638 4.6805 6.54 3479.001817 0.006272 0.4119 4.09 P ¼ 50000 KPa 375 400 425 450 500 550 600 650 700 800 900 1000 1100 1200 1300 0.0824 6.51 1843.7652 0.3364 7.015643 0.005595 0.1625 4.59 3441.002085 0.18 4551.55 5037.52 1874.07 4338.51 3247.91 2720.006966 0.09 4770.59 5068.8829 6.82 3710.70 4178.012497 0.3082 7.05 4804.28 3.12 4572.19 5284.2189 6.018229 3954.9126 7.96 2567.1725 5.8177 6.61 5017.001503 0.91 3933.12 4191.020815 3966.2183 7.45 2763.98 2283.84 3616.41 4501.7140 3.010283 0.56 3. 00997 0.50 501.5685 0.5128 2.6978 2.00115 0.64 171.11 756.78 0.21 333.00127 1147.85 944.5284 1.1274 2.00099 0.24) 1610.00140 s (KJ/kg K) v (m3/kg) u (KJ/kg) P ¼ 10000 KPa 1393.04 371 (continued ) Copyright 2003 by Marcel Dekker.24 853.06) 1407.00122 0.09 500.00101 0.13 678.83 855.0137 0.61 595.5188 1.58 0.00119 0.00100 0.74 672.90 1328.18 20.20 h (KJ/kg) (311.48 510.47 1134.0001 0.83 422.00110 0.07 1025.62 848.00126 0.000995 0.02 88.00123 0.28 338.97 945.62 176.00104 0.71 507.00 0.00132 0.35 166.49 934.3595 0.7291 1.13 1133.1341 2.00 4.8698 3.36 1037.0003 0.94 1121.68 1234.001452 0.0687 1.00105 0.61 759.Appendix Table 10 Temp (C) SI Compressed Liquid Water v (m3/kg) u (KJ/kg) P ¼ 5000 KPa h (KJ/kg) (263.36 259.00103 0.000993 1585.6872 2.9374 2.07 592.5038 2.5705 0.002035 0.88 1038.9316 2.00129 0.2468 s (KJ/kg K) Saturated 0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 0.00106 0.56 416.15 (342.40 681.7342 1.99) 1154.00118 0.11 1342.45 15.0719 1.9201 0.47 324.10 765.00100 0.95 255.00099 1785.34 1127. Inc.07 584.00110 0.69 417.81 426.2955 0.92 P ¼ 15000 KPa Saturated 0 0.0004 0.3178 2.81) 1826.00112 0.93 250.00101 0.5232 1.53 10.30 2.47 0.00101 0.6847 0.8258 1.03 1220.00115 0. All Rights Reserved .79 586.00107 0.2945 0.00104 0.33 249.63 844.21 5.8829 0.3254 2.001658 0.00108 0.3030 1.34 P ¼ 20000 KPa 3.34 332.0004 3.05 93.03 83.31 (365.67 670.64 166.08 938.00112 0.00102 0.10 83.07 767.2992 1.43 1031.0547 3. 00101 0.5101 1.00117 0.00105 0.3103 2.01 1739.61 185.00136 0.26 514.00100 0.67 665.45 1230.3031 2.04 517.2934 0.000995 0.78 h (KJ/kg) (365.88 0.64 1702.00109 0.03 687.6770 2.27 1040.46 414.07 770.9203 2.1146 2.69 1306.47 949.49 331.97 180.2259 3.000993 0.10 1415.4952 2.23 0.4246 3.2954 1.53 1571.00120 0.0392 3.00121 0. All Rights Reserved .00109 0.00102 0.82 350.62 1333.00100 0.53 1204.47 1316.0655 1.11 773.00105 0.4869 2.00163 83.8575 3.00103 0.5646 0.99 1133.8770 s (KJ/kg K) 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360 0.5144 1.9259 2.52 1038.17 598.58 1431.65 346.00131 0.7192 1.00126 0.7241 1.5665 0.41 1232.66 1539.69 753.79 430.18 860.90 925.75 580.64 667.74 602.13 1591.2917 1.00157 0. Inc.04 1133.00111 0.6074 3.48 858.00114 0.38 413.14 267.89 1020.78 434.00125 0.8205 1.59 1212.00114 0.00144 0.34 750.72 498.00182 s (KJ/kg K) v (m3/kg) u (KJ/kg) P ¼ 20000 KPa 82.18 947.73 248.8459 3.00130 0.70 684.00147 0.0623 1.00107 0.2922 0.23 1453.8231 1.04 929.94 1108.37 496.05 1567.3978 3.75 263.00102 0.372 Table 10 Temp (C) Continued v (m3/kg) u (KJ/kg) P ¼ 15000 KPa h (KJ/kg) (342.42 Appendix Copyright 2003 by Marcel Dekker.00117 0.09 1337.6545 0.82 1114.39 582.6673 2.1209 2.66 330.00138 0.15 247.89 1015.05 165.74 841.00101 0.29 1444.24) 97.00107 0.2071 3.00104 0.81) 102.0248 3.00112 0.75 165.94 837. 5425 3.38 1190.2847 0.16 1746.0793 2.9536 3.001020 0.75 441.43 0.001159 0.2892 2.19 1258.23 1229.0001 0.39 735.001070 0.2844 1.5017 1.37 622.0014 0.26 1322.001230 0.0439 1.5526 0.7860 2.91 608.20 1138.000980 0.4710 2.6915 1.000987 0.17 1522.8242 2.28 410.68 456.001044 0.001082 0.001484 0.36 1837.P ¼ 30000 KPa 0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360 380 0.3538 3.001007 0.71 1626.64 1501.67 990.8051 1.001130 0.1470 3.00 211.001192 0.76 652.97 1667.001330 0. Inc.96 324.9095 2.001105 0.09 1042.2867 3.000995 0.000986 0.000996 0.2703 1.001062 0.16 164.81 745.000989 0.86 487.20 292.7492 4.89 1390.13 49.4556 3.01 246.73 904.7097 1.91 1555.001170 0.69 1078.82 111.69 1287.27 778.4419 2.84 242.57 831.8153 1.46 961.71 1049.29 1228.001004 0.57 1781.2898 0.001016 0.63 569.34 918.001052 0.2634 2.001286 0.1200 3.77 374.20 80.001203 0.001091 0.001588 P ¼ 50000 KPa 0.8890 2.000977 0.47 1675.95 1420.001141 0. All Rights Reserved .9985 3.98 161.03 328.03 130.32 405.32 1006.001400 0.82 193.4857 1.80 1462.07 1630.6158 2.58 576.001242 0.23 1451.0561 1.001869 0.87 539.1024 2.06 1167.54 20.35 29.68 819.8100 Appendix 373 Copyright 2003 by Marcel Dekker.66 1353.87 276.86 660.84 1097.001492 0.96 1327.33 705.24 953.001403 0.001627 0.6489 2.76 493.71 865.91 790.001035 0.0010 0.6290 3.60 1134.5606 0.001275 0.001029 0.001339 0.25 82.63 1546.24 875.63 524.001115 0.73 693.16 358. 80 2385.4 2721.3 2375.3 2497.5 2369.906 940.1 2438.3 Evap.2601 0.0010874 0.662 283.1510 0.7733 11.5 2344.5168 21.824 785.0010905 0.91 2 350.2120 11.0010901 0.7 2836.0 2839.2 2336.753 1986.396 Evap.775 2145.45 2393.0010861 0.93 2381.3928 21.40 2333.9 2714.10 2374.9 Evap.1 2839.3102 0.314 241.7616 9.4 2333.0 2364.0 2490.06 2411.7 2708.776 2145.0 2427.04684 0.4239 21.02 2354.4376 0.0010841 Evap.3772 21.6165 10.45 2 393.2215 10.16 2366.227 3601.752 1986.3 2486.5808 11.05741 0.253 5430.0010864 0.374 Table 11 SI Saturated Solid—Saturated Vapor Water Specific volume (m3/kg) Internal energy (KJ/kg) Saturated solid (ui) 2333.2 2361.1233 11.0010871 0.1 2350.138 394.8 2727.02 2 354.4 2319.0 2839.7 2355.799 334.6 2722.0359 11.14 2370.0625 10.201 2945.6108 0.06 2 411.14 2 362.0 2839.91 2350.7809 10.2526 21.8 2442.112 1357.756 553.0010881 0.3776 10.9704 12.822 4416.3 2501.8 2328.152 206.1 2839.2 Entropy (KJ/kg K) Saturated solid (si) 21.4083 21.14 2362.3 2460.1562 9.8 2835.64 2389.23 2396.08535 0.9 2838.8648 10.663 283.8 2367.0010891 0.3689 0.803 658.01618 0.01286 Saturated solid (vi) 0.0 2339.71 2 404.823 4416.116 6707.6 2838.2939 21.907 940.8 2725.6192 9.7 2838.2995 21.3461 21.6765 11.3 2835.7 2710.3772 10.64 2 389.4 2838.6 2837.1815 0.5 2731.0010854 0.42 2 337.8 2718.0010858 0.9 2353.7 2342.03 2377.09 2358.40 2 408.6 Enthalpy (KJ/kg) Saturated solid (hi) 2 333.98 2 400.863 1639.40 2408.0 2719.1565 9.2176 0.3935 11.42 2337.01 0 22 24 26 28 2 10 2 12 2 14 2 16 2 18 2 20 2 22 2 24 2 26 2 28 2 30 2 32 2 34 2 36 2 38 2 40 Pressure KPa P 0.2 2475.4394 21.7 2834.3 2347.413 466.10 2 374.3150 21.6 2330.5 2712.798 334.4690 10.252 5430.2211 21.0010898 0.3482 9.802 658.2 2449.0010894 0.1 2464.9 2453.2 2836.6 2730.5498 9.4 2434.10355 0.98 2400.6 2837.6898 9.2683 21.78 2345.61 2341.1252 0. Inc.4815 9.0 2837.16 2 366.5 2471.7 2729. All Rights Reserved .2832 9.2 2711.70 Saturated vapor (ug) 2375.757 553.0010908 0.395 Saturated vapor (vg) 2036.021 8366.9 2479.4550 21.115 6707.8 2728.0714 12.2 2322.0010888 0.5014 21.4860 21.228 3601.0010878 0.0010908 0.8 2726. (vfg) 206.70 Saturated vapor (vg) 2501.5546 10.8348 9.5321 Saturated vapor (sg) 9.2 2838.71 2404.6 2456.9852 10.9498 11.200 2945.414 466.315 241.4705 21.61 2 341.0 2325.6447 Temp C T 0.1413 10.0010884 0.6982 10.182 1128.40 2 333.2210 21.113 1357. (vfg) 2834.07012 0.2369 21.1 2839.4143 9.93 2 381.153 206.5368 10.8713 11.139 394.183 1128.0010844 0.8 2468.0010868 0.3028 10.3853 10.022 8366.2 2715. (sig) 10.78 2 345.3 2837.2193 9.4562 10.3306 21.7 2431.0010851 0.9 2839.0010848 0.6 2494.864 1639.4 2358.9093 9.3020 11.23 2 396.3 2372.6113 0.2 2720.02499 0.03 2 377.03090 0.3617 21.5 2445.4864 11.824 785.7 2723.02016 0.6 2482.14 2 370.5177 0.09 2 358.7 2724.80 2 385.1768 Appendix Copyright 2003 by Marcel Dekker.5 2716. (uig) 2708.03810 0. All Rights Reserved .2 997. k (W/m k) Thermal Diffusivity.Appendix THERMODYNAMIC PROPERTIES OF LIQUIDS Table 12 Water at Saturation Pressure Coefficient of thermal expansion.388 £ 1024 ¼ (Btu/lbm 8F) 4266 4206 4195 4187 4182 4178 4176 4175 Thermal conductivity.1 — 3.725 13.6 792. a £ 106 (m2/s) £ 3.624 £ 0.146 0.5 8.577 0. b £ 104 (l/K) Specific heat. Inc.7 999.568 0.7 — 0.146 1.7 994.1 8F 32 41 50 59 68 77 86 95 K 273 278 283 288 293 298 303 308 8C 0 5 10 15 20 25 30 35 £ 0.789 1.540 — Temp T Density r (kg/m3) £ 6. m £ 106 (N s/m2) Kinematic viscosity.874 £ 104 ¼ (ft2/h) 1.874 £ 104 ¼ (ft2/h) 0. Pr gb/v2 £ 1029 (l/K m3) £ 1.1 5.1 7.5556 ¼ (l/R) 2 0.573 £ 1022 ¼ (l/R ft3) — — 0.5777 ¼ (Btu/h ft 8F) 0.615 0.149 0. cp (J/kg K) £ 2.805 0.884 0.585 0.006 0.243 £ 1022 ¼ (lbm/ft3) 999.143 0.95 — 2.6720 ¼ (lbm/ft s) 1794 1535 1296 1136 993 880.9 1000.535 1.131 0. n £ 106 (m2/s) £ 3.4 719.1 998.606 0.035 — 4.1 995.0 6.0 999.0 — £ 0.135 0.4 4.141 0.558 0.551 — 2.7 11.8 Prandtl number.150 Absolute viscosity.300 1.8 375 (continued ) Copyright 2003 by Marcel Dekker.4 9.597 0.137 0. All Rights Reserved .10 1.6 862.01 0.652 0.7 290.244 0.588 0. a £ 106 (m2/s) £ 3.3 907.6 277.333 — 14.5 15.0 139.6 152.23 1.0 23.5556 ¼ (l/R) 3.86 0.160 0.366 0.155 0.95 0.1 974.6 886.18 Appendix Copyright 2003 by Marcel Dekker.611 0.4 201.191 0.141 0.684 0.132 Absolute viscosity.0 712.3 3.156 0.5 926.6 — 7. b £ 104 (l/K) Specific heat.665 0.3 396.0 Temp T Density r (kg/m3) £ 6.874 £ 104 ¼ (ft2/h) 0.9 98.376 Table 12 Continued Coefficient of thermal expansion.8 104.2 671.5 235.671 0.8 12.172 0.162 0.9 3.7 10.5 113.556 0.09 140.170 0.157 0.5 £ 0.564 £ 0.131 0.07 92.90 0.172 0.8 29.59 — 85.0 1766.169 0.658 0.5 517.55 2.874 £ 104 ¼ (ft2/h) 0.0 809. cp (J/kg K) £ 2.147 0.5 1076.173 0. m £ 106 (N s/m2) Kinematic viscosity.682 0.634 0.89 0.388 £ 1024 ¼ (Btu/lbm 8F) 4175 4176 4178 4190 4211 4232 4257 4285 4396 4501 4605 4731 4982 5234 5694 Thermal conductivity.6720 ¼ (lbm/ft s) 658.673 0. Pr gb/v2 £ 1029 (l/K m3) £ 1. k (W/m k) Thermal Diffusivity.0 171.171 0.294 0.243 £ 1022 ¼ (lbm/ft3) 992.212 0.5 9.9 — 4.0 1360.633 0.173 0.685 0.3 124.0 779.1 555.75 1.680 0. Inc.0 211.23 1.640 0.647 0.86 0.2 990.573 £ 1022 ¼ (l/R ft3) 8.0 750.8 837.9 958.151 0.4 848.4 943.149 0.167 0.5777 ¼ (Btu/h ft 8F) 0.1 376.128 4.2 17.5 8.2 988.98 Prandtl number.164 0. n £ 106 (m2/s) £ 3.43 1.1 13.5 8F 104 113 122 167 212 248 284 320 356 396 428 464 500 536 572 K 313 318 323 348 373 393 412 433 453 473 493 513 533 553 573 8C 40 45 50 75 100 120 140 160 180 200 220 240 260 280 300 £ 0.135 0.613 0.0 605.2 20. 4 943.0132 1.0421 0.6 1135.0737 0.45 209.027 15.0424 0.1 675.892 0.0 1345.480 46. Inc.0061 0.4735 0.86 125.1991 0.430 ¼ (Btu/lbm) 20.4 57.430 ¼ (Btu/lbm) 2501 2519 2537 2555 2574 2591 2609 2626 2643 2660 2676 2706 2734 2757 2777 2791 2799 2801 2795 2778 2748 hfg (KJ/kg) £ 0.02 ¼ (ft3/lbm) 206.92 376.6136 6.97 334.1 852.9854 3.4 hg (KJ/kg) £ 0.127 0.1233 0. All Rights Reserved .94 419.362 1.Appendix Table 13 Water at Saturation Temperature Enthalpy Saturation Pressure P £ 1025 (N/m2) 8C 0 10 20 30 40 50 60 70 80 90 100 120 140 160 180 200 220 240 260 280 300 £ 1.673 0.0 1237.7010 1.306 0.430 ¼ (Btu/lbm) £ 0.04 41.09 292.7 589.99 83.0596 0.450 £ 10 ¼ (psi) 0.430 ¼ (Btu/lbm) 2501 2477 2453 2430 2406 2382 2358 2333 2308 2283 2257 2202 2144 2082 2014 1939 1856 1764 1660 1541 1403 K 273 283 293 303 313 323 333 343 353 363 373 393 413 433 453 473 493 513 533 553 573 377 Copyright 2003 by Marcel Dekker.7 1037.940 64.048 7.0122 0.0233 0.0301 0.410 2.3116 0.833 32.917 24 Saturation Temperature T 8F 32 60 68 86 104 122 140 158 176 194 212 248 284 320 356 392 428 464 500 536 572 Specific volume of Vapor vg(m3/kg) £ 16.3 106.929 19.047 3.1804 10.201 33.5 763.06 503.66 167.551 23.193 0.680 5.508 0.191 85.548 12.26 251.0216 hf (KJ/kg) £ 0.0860 0. 5777 ¼ (Btu/h ft 8F) 0.0 900.0 17.0 852.1 864.874 £ 104 ¼ (ft2/h) 911 872 834 800 769 738 710 686 663 Absolute viscosity.2 876.140 0.9 805.0 840.145 0.388 £ 1024 ¼ (Btu/lbm 8F) 1796 1880 1964 2047 2121 2219 2307 2395 2483 Thermal conductivity k (W/m k) Thermal diffusivity. a £ 1010 (m2/s) £ 3.147 0.5 32.1 10. All Rights Reserved . v £ 106 (m2/s) £ 3.3 6.0 799.16 0.0 28. b £ 104 (l/K) Specific heat. m £ 103 (N s/m2) Kinematic viscosity.0 240.4 8.51 Appendix Copyright 2003 by Marcel Dekker.84 Temp T Density r (kg/m3) £ 6.874 £ 104 ¼ (ft2/h) 4280.9 Prandtl number. Inc.9 37.573 £ 1022 ¼ (l/R ft3) 8475 8F 32 68 104 140 176 212 248 284 320 K 273 293 313 333 353 373 393 413 433 8C 0 20 40 60 80 100 120 140 160 £ 0.137 0.0 829. Pr gb/v2 (l/K m3) £ 1.132 £ 0.6 471.0 210.135 0.6720 ¼ (lbm/ft s) 3848.144 0.0 5.76 1.3 12. cp (J/kg K) £ 2.5 20.1 888.0 816.5556 ¼ (l/R) 0.90 2.70 £ 0.7 10.5 4.54 4.0 104.0 83.138 0.133 0.378 Table 14 Unused Engine Oil (Saturated Liquid) Coefficient of thermal expansion.243 £ 1022 ¼ (lbm/ft3) 899.0 72.75 1. 556 ¼ (l/R) 1082 £ 0.628 13.51 11.5777 ¼ (Btu/h ft 8F) 8.Appendix Table 15 Mercury (Saturated Liquid) Coefficient of thermal expansion.20 8.90 15.48 14.506 13.0673 0.0765 0. Pr gb/v2 £ 10210 (l/K m3) 8F 32 68 122 212 302 392 482 600 K 273 293 323 373 423 473 523 588.40 10.124 0. cp (J/kg K) £ 2.6720 ¼ (lbm/ft s) 16.0083 Temp T Density r (kg/m3) £ 6. m £ 104 (N s/m2) Kinematic viscosity.69 9.22 57.0 Thermal conductivity k (W/m k) Thermal diffusivity.0288 0.0 135.0162 0.3 136.0103 0.7 134.0116 0.573 £ 1022 13.49 12.114 0.104 0.06 81.847 Prandtl number.65 £ 1. v £ 106 (m2/s) £ 3.08 74.05 12.5 157.0853 0. All Rights Reserved .5 £ 0.3 139.0207 0.026 12.874 £ 104 ¼ (ft2/h) 0.385 13.243 £ 1022 ¼ (lbm/ft3) 13. Inc.4 138.145 13.579 13.0802 0.73 379 Copyright 2003 by Marcel Dekker.6 137.50 Absolute viscosity.874 £ 104 ¼ (ft2/h) 42.54 69.16 63. b £ 104 (l/K) Specific heat.388 £ 1024 ¼ (Btu/lbm 8F) 140.06 50.7 8C 0 20 50 100 150 200 250 315.99 46.0134 0.264 13. a £ 1010 (m2/s) £ 3.07 14.42 11.0249 0.34 13.54 9.02 £ 0.96 8.31 10.0928 0. 19 6.388 £ 1024 ¼ (Btu/lbm 8F) 1382 1340 1298 1256 1256 Thermal conductivity k (W/m k) Thermal diffusivity.7 8F 220 400 700 1000 1200 K 367 487 644 811 978 8C 94 205 371 538 705 £ 0.26 0.5777 ¼ (Btu/h ft 8F) 86.71 6. cp (J/kg K) £ 2.99 4.4 65.0051 0.79 Appendix Copyright 2003 by Marcel Dekker. a £ 105 (m2/s) Absolute viscosity.243 £ 1022 ¼ (lbm/ft3) 929 902 860 820 778 Prandtl number.3 72.0110 0.36 £ 0. m £ 104 (N s/m2) Kinematic viscosity.4 59.874 £ 104 ¼ (ft2/h) 7.5556 ¼ (l/R) 0. All Rights Reserved .0072 0.6720 ¼ (lbm/ft s) 6.31 4.27 0.2 80.60 3.71 6.0040 0.7 £ 3.16 2. Pr gb/v2 £ 1028 (l/K m3) £ 1.380 Table 16 Sodium Coefficient of thermal expansion.08 1.45 6.83 2.96 16.44 2. v £ 107 (m2/s) £ 3. Inc.874 £ 104 ¼ (ft2/h) 6.32 2.0038 Temp T Density r (kg/m3) £ 6.19 £ 0.573 £ 1022 ¼ (l/R ft3) 4. b £ 104 (l/K) Specific heat. 71 0.508 0.268 8F 32 68 104 140 176 212 392 572 752 932 1832 K 273 293 313 333 353 373 473 573 673 773 1273 8C 0 20 40 60 80 100 200 300 400 500 1000 381 Copyright 2003 by Marcel Dekker.164 0.6 35.388 £ 1024 ¼ (Btu/lbm 8F) 1011 1012 1014 1017 1019 1022 1035 1047 1059 1076 1139 Thermal conductivity k (W/m k) £ 0.6720 ¼ (lbm/ft s) 17.85 1.600 0.164 1.6 33.36 1.790 21.6 19.9 15.71 0.243 £ 1022 ¼ (lbm/ft3) 1.123 19.7 17.4 113.5 23.754 35.445 Kinematic viscosity.874 £ 104 ¼ (ft2/h) 13.907 20.71 0.72 0. v £ 106 (m2/s) £ 3.0193 0.0 24.7 68.41 3.71 0.723 0.11 1.0370 0.240 19.72 0.2 22.5 49.968 0.0237 0. Pr gb/v2 £ 1028 (l/K m3) £ 1.49 1.456 18.442 0.8 27. a £ 106 (m2/s) £ 3.71 0.9 89. m £ 106 (N s/m2) £ 0. Inc.83 2.0307 0.673 35.01 0.Appendix Table 17 Dry Air at Atmospheric Pressure Coefficient of thermal expansion.0485 0.025 0.573 £ 1022 ¼ (l/R ft3) 1.2 64.79 Specific heat.5777 ¼ (Btu/h ft 8F) 0.0265 0.596 0.6 30.322 32.6 81.0 181 0.874 £ 104 ¼ (ft2/h) 19.66 3.00236 Temp T Density r (kg/m3) £ 6. All Rights Reserved .782 0.75 1.0251 0.71 0.0429 0.0709 0.0540 0.68 2.0293 0.0279 0.252 1.71 0.092 1.29 0.00 2.472 0. cp (J/kg K) £ 2.693 29.74 Prandtl number. b £ 104 (l/K) £ 0.794 48.6 49.916 0.71 0.4 21.0350 0.0762 Thermal diffusivity.19 3.2 240 Absolute viscosity.5556 ¼ (l/R) 3. 491 152.174 220.3444 36.288 179.83521 6.57347 34.09071 1.375 649.191 213.25607 7.097 450.0025 68.0518 14.72814 2.335 496.09232 14.20875 7.82905 7.022 207.64448 7. Standard Entropy at 0.93412 6.350 607.360 523.379 185.99515 7.2894 16.02333 5.934 462.306 345.528 670.54188 31.988 286.67573 0.5609 28.39722 1.776 692.15926 7.071 171.120 713.430 298.035 315.1882 87.615 300.05276 7.316 628.257 195.5143 40.56771 13.736 419.390 290.76629 7.18197 11.86926 6.589 441.7479 49.0658 54.80008 7.364 228.1 MPa (1 bar) u KJ/kg 142.54084 7.10735 7.415 465.218 240.323 280.86305 6.30142 7.113 257.60025 20.77044 7.863 381.098 756. All Rights Reserved .726 243.600 h KJ/kg 200.6765 33.57638 7.74010 7.686 565.79998 6.460 800.11226 2.039 359.92569 10.15 300 320 340 360 380 400 420 440 460 480 500 520 540 560 580 600 620 640 660 680 700 720 740 760 780 Ideal Gas Properties of Air.70903 7.6519 17.11458 1.726 389.267 260.060 401.689 404.561 735.51088 0.695 200. Inc.87556 0.473 320.8182 20.55812 6.28916 5.361 182.64535 6.654 99.532 271.37700 0.299 421.16010 25.85061 vr 493.61090 7.42736 7.46260 6.274 256.4358 30.27027 0.704 360.036 214.731 778.584 213.814 503.828 481.871 435.316 574.150 313.939 512.85740 Pr 0.200 113.284 s8 KJ/kg 6.38692 7.50422 7.67717 7.466 389.640 330.34499 7.639 528.76564 16.06119 3.2777 44.474 586.328 560.46642 7.78637 23.340 482.1367 77.63727 4.4088 61.768 157.59801 18.84663 6.982 544.844 374.1553 18.55479 3.98990 1.72562 6.382 Appendix THERMODYNAMIC PROPERTIES OF AIR Table 18 Pressure T K 200 220 240 260 280 290 298.435 544.576 340.487 301.9250 (continued ) Copyright 2003 by Marcel Dekker.7132 23.74188 28.78997 8.728 131.6623 21.0008 25. 974 4046.426 1159.733 2628.68242 0.65077 111.824 716.7426 361.51977 74.86253 1.7245 191.436 1046.18568 3.133 h KJ/kg 822.217 2777.186 2251.29616 8.789 2066.298 1277.43338 0.759 4833.3467 134.211 933.68804 3.939 3364.27308 Pr 38.16913 9.07031 1.83516 8.01581 8.215 4425.907 2116.805 1336.537 3059.630 2565.051 1182.89711 2.790 2017.962 2692.581 2314.789 2516.882 2126.37669 0.28845 5.801 1438.547 1298.69051 8.751 1067.436 1879.19586 9.202 877.202 1205.366 2883.99699 9.079 1344.Appendix Table 18 T K 800 850 900 950 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500 1550 1600 1650 1700 1750 1800 1850 1900 1950 2000 2050 2100 2150 2200 2250 2300 2350 2400 2450 2500 2550 Continued u KJ/kg 592.859 s8 KJ/kg 7.27604 6.970 1486.88514 7.077 2052.498 1485.68035 1.3320 931.478 1161.313 2755.22208 9.95207 8.52891 8.800 1696.059 2946.54020 0.367 977.446 1757.422 674.1554 554.591 1629. Inc.26257 2.9577 634.823 2439.888 1022.445 889.987 2064.1736 226.24781 9.834 1391.38777 48.658 3692.61208 8.518 1725.635 1847.95445 0.7145 310.109 2274.19081 8.40378 0.152 989.936 1113.50124 0.138 2165.6948 9.8560 822.8942 483.77141 4.270 1575.755 1941.52007 1.189 802.25330 1.756 759.46576 0.13493 8.772 1658.714 1774.659 2502.07667 8.30831 2.63027 0.253 1251.329 1818.04294 0.474 1677.76472 8.91692 8. All Rights Reserved .65185 8.57111 8.0192 265.031 vr 13.72811 8.275 2002.873 1581.8972 11.14189 9.96611 8.80410 0.888 1327.74012 0.39402 8.541 1871.398 1635.577 633.93452 8.86908 8.397 933.08573 9.35185 (continued ) 383 Copyright 2003 by Marcel Dekker.46828 60.14204 1.095 845.180 1219.58305 0.912 1968.221 1103.161 1919.057 1822.6192 418.87521 0.58171 2.24449 8.48539 8.44046 8.331 1533.46408 4.9670 723.128 2376.1376 1051.81519 91.429 1515.80039 8.34596 8.951 2189.90219 8.02721 9.2478 160.46770 7.37858 1.11409 9.480 2214.677 1395.782 2819.05678 9.892 1455. 645 Copyright 2003 by Marcel Dekker.188 3460.657 9.990 2664.215 9.21205 0.29790 5269.715 9.359 s8 KJ/kg Appendix Pr vr 0.265 h KJ/kg 3010.868 3203.481 2314.41576 7945.216 9.727 2463.707 9.763 3074.25443 0.797 2613.726 3396.48198 10006.064 3267.39297 7338.733 3525.767 3138.384 Table 18 Continued T K 2600 2650 2700 2750 2800 2850 2900 2950 3000 u KJ/kg 2264.124 9.133 2363.676 9.351 3331.46025 9277.22511 0.36980 6769.27089 0.505 9. Inc.687 2563.883 2413.32228 5736.43818 8590.30805 0.28872 0.19992 9. All Rights Reserved .34625 6236.23291 0.663 2513.32903 0. 1 Temperature-Entropy Diagram for Water Copyright 2003 by Marcel Dekker. Inc. All Rights Reserved .Appendix 385 Figure A. 386 Appendix Figure A.2 Enthalpy-Entropy Diagram for Water Copyright 2003 by Marcel Dekker. All Rights Reserved . Inc. (1970). E. Inc. W. Copyright 2003 by Marcel Dekker. H. (1960). 10. ASME. Harrison. New York: The American Society of Mechanical Engineers.. An axial flow reversing gas turbine for marine propulsion. (1967). Fluid Mechanics for Engineers. Scale Models in Hydraulic Engineering. Reprint. Simons. Caterpillar Tractor Co. C. Ackers. J. B. Centrifugal Compressors for Special Applications. NACA R.. ASME series E. Alleman. (1984). 5. Allen.. 7. Albertson. New York: Scientific American Library. R. M. Jan. D. 1969. P. Barton. Allen. P. R. J. A. 9. (1952). I. L. Dec. J. Anderson. R. 1956. Abbot. 3. (1978). NACA Report. 2. R and M 2974. R. M. NJ: Prentice-Hall. ASME Research Committee on Fluid Meters (1961).. Aerodynamic Design of Axial-Flow Compressors—Volumes. 451.. Atkins. Weirs and Flumes for Flow Measurement.Bibliography 1. Trans. 4. Inc. P. All Rights Reserved . L. W. Butter. Trans. Walther. II and III. The Second Law.. (1951). 6. 37(4). Green and Co.. Air Compressor Handbook. G. The Drag of Two Streamline Bodies as Affected by Protuberances and Appendages.. I. R. 8. (1932). London: Longmans. A. E56B03. J. Axisymmetric one-dimensional compressible flowtheory and application. Ltd. A Method of Performance Estimation for Axial-Flow Turbines. Englewood Cliffs.. Sulzer Technical Review. White. Mathieson. Alley. Journal of Basic Power. G. Flowmeter Computation Handbook.M. 12. A. 11. R. D. Perkins.. New York: John Wiley and Sons. et al. (1978). Batchelor. Cambridge: Cambridge University Press. Wilcox. Baumeister. Came. Bright. Hydrodynamics. Bogdonoff. Prentice-Hall. 21. L. (1931). Hawthorne. ASME 73. J. G. (1967). J. King. Inc. (1964). J. S. W. ed. (1951). T. A. R. Bibliography Bridgman. Markland. L. 20. (1963). P. Aeronaut. M. 24. McGraw-Hill. 22. Mech. A. (1957).. (1967). The Theoretical Determination of the Threedimensional Flow in an Axial Compressor with Special Reference to Constant Reaction Blading. Cocrell. Bradshaw. R. Proc. Dimensional Analysis. G. 25. M.. 7th ed. Feb. Some exact solutions of the flow through annular cascade discs.. K. New York: McGraw Hill Book Co. 33. Advances in Applied Mechanics. L. Church. S. O. T. New York: McGraw Hill. E. application and experimental evaluation of a design procedure for centrifugal compressors. The design and development of centrifugal compressors for aircraft gas turbines. Standard Handbook for Mechanical Engineers. 16. A. P. 15. New York: John Wiley and Sons. H. 29. E. W. Bernoulli. Report. G. Cambridge University.. 14. Sabersky.D... (1943). 36. An Introduction to Fluid Dynamics. Vol. A. Hydraulics. 3184. E. 35. P. (1956).. 28. The development. Thesis. Chemical Rubber Company. R. Conn: Yale University Press. 18. Investigation of axial flow compressors. Aeronaut. 1. Aeronautical Research Council. Inst. Experimental and Theoretical Studies of Surging in Continuous Flow Compressors. 192(5):49 –67. F. Inc. (1958). 17. The turbulent boundary layer. (1961). In: Bernoulli. White.. D. Advanced Fluid Mechanics. L. (1948). F. N. Handbook of Applied Hydraulics. 19. H. H. Church. Inst. Bowen. 17. J. Carmichael. W. Handbook of Tables for Applied Engineering Science. Marks. R and M. T.. Moses. Camody. New York: Dover Publications. 31. S. Eng. D. Mech. Buck. Bolz. 34. (1946). M. G. 23. 26. Aircraft Eng. J. NACA Report No. D. (1944). 153:426 – 440. T. Mechanical Measurements.. Steam Turbines. L. E.388 13. D. 32. New Haven. Proc. H. (1950). 30. NACA cascade data for the blade design of high performance axial flow compressors. (1950). Centrifugal Pumps and Blowers. T. Inc. (1973). Bullock. Mech. (1946).R. Review of incompressible diffuser flow. W. Mass: Addison-Wesley. S. Stall Propagation in Compressors. Copyright 2003 by Marcel Dekker. New York: Pergamon Press.. Brgg. (1945). R. McGraw-Hill. C. J.. Barter. H. Clauser. H. 35(10):286 –292. Rannic. (1968). (1958). Inst.C.. 6th ed. 6842. Eng. Proc. Eng. All Rights Reserved . Sci. 861. Kobus. D. Reading. Binder. H. Experimental Fluid Mechanics. F. R. Trans. W. Ph. Cheshire.. E. J. The development of gas turbine power plants for traction purposes. The Effect of Reynolds Number on the Performance of a Single-Stage Compressor. (1976). Sci. Vol. E. True. J. 27. Beckwith. 4. (1945). New York: Pergamon Press. Csanady. Theory of Turbomachines. Cohen. R. Carter. .O. Mech.. 12. 389 49. P.. The Centrifugal Compressor Stage. Horlock. E.. Constant. S. W. L. (1958). H... 45. Flow regimes in curved subsonic diffusers. Eng. Kline. 1833N Horlock. Standards of the Hydraulic Institute. Van Nostrand. Butterworths Scientific Publication. W. Horlock. A. G. Franzini. G.E. W. (1912).. 57. Proc. H. Harleman. Herbert. Bull. (1915).S. No. H. Horlock. 808. (1968). 42. 58. In: Rouse. (1965). 27:612. M. J. B.. 43. Daily. (1964).R. 14:125. A. New York: McGraw-Hill Book Company. Trans. H. C70/84. Fluid Mechanics with Engineering Applications. 51. M. 41.C.R. (1950). Topps.C.E.. 39. New York: McGraw Hill Book Company. H. D. Pump Operation and Maintenance. V. Marsh. Goldstein. 153. Reading. (1966). Engineering Hydraulics. J. New York: John Wiley and Sons.C. Hydraulic Institute (1975). Mech. K. T. Inst.. 5:18. Eckert. Aeronaut. New York: Hydraulic Institute. (1984-3). (1980). A. No. (1961). 13th ed.S.C.Bibliography 37.C. (1963).O. Freeman. 44. L. New York: McGraw Hill Book Company.P. Inc. Conference Publications. Dixon. The early history of the axial type of gas turbine engine. W. Mass: AddisonWesley Publishing Company. Q. Daily. Gastelow.S. (1965). J. T. J. 40. W. S. Centrifugal Pumps. Potential Flow Through Cascades—A Comparison Between Exact and Appropriate Solutions. R. Fujie. All Rights Reserved . (1963). No 3099. Axial Flow Turbines.E.. ASME 84D:303 – 316. Daugherty. Butterworths Scientific Publishing Company. (1962). Approximate three-dimensional flow theory for axial turbomachines. D. 52. Stow. 3840. (1972). L.. R. R and M. J. J. H. J. Howell. Gastelow. T. 48.M. Daugherty. Eng. J.. Mech. Inst. Fox. Engineers. 47. New York: Dover Publications Inc. Drake. H. The Off Design Performance of Free Vortex Turbine and Compressor Stages. B. R.R.R. Copyright 2003 by Marcel Dekker. S. Proc. 55. R. ed. A Method of Performance Prediction for Centrifugal Compressors. J. (1963). F. 54. (1959) Thermodynamic Data for the Calculation of Gas Turbine Performance. (1957). Modern Development in Fluid Dynamics. 38. 3rd ed. Inst. H.E. Trans. R and M. J. A. 46. Fluid Dynamics. 53. A study of the flow through the rotor of an axial flow compressor. A. Ennis.S. (1966). R. C. Ferguson. P. 50. Hicks. Applied Thermodynamics for Engineers.M. R. Recent Developments in the Aerodynamic Design of Axial Flow Compressors. H. New York: D. Analysis of Heat and Mass Transfer. Fielding. J.. P. 56.M.. New York: McGraw-Hill Book Co. W. (1945). The application of computational fluid mechanics to aero gas turbine compressor design and development.M. J. Hydraulic machinery. 85D(1):55. D.C. Inc. Annulus wall boundary layers in axial flow compressor stages.P. New York: John Wiley and Sons. E. Hydraulic Models. New York: McGraw-Hill.. 75. (1951). 5th ed. (1937). 68. 61. Sorensen. J. Experimental Methods for Engineers. Flow in transonic compressors. J. ed. 2:48– 75. (1981). (1949). (1963).390 59. B. Kramer. 67. 74. (1971). 76. Louis. C. W. Stanitz. Cambridge University. All Rights Reserved .. Steam Turbines Theory and Practice. Engineering Applications of Fluid Mechanics. (1969). McNally. H.D. (1953). In: Davis. (1964). 5th ed. J. New York: McGraw-Hill. Keller. K.E.. 301. W. (1945). J. 16. 77. A. 65. Loeb.M. D. Kerrebrack. Holman. Handbook of Hydraulics. J. C. Moody. A. Hydraulic machinery. Liang hear.A. (1962). T. Ph. C. U. Sockol.. Fluids Eng. 1969. Review of computational methods for internal flows with emphasis on turbomachinery. Lamb.. McGraw-Hill. Rightmire. A. 62. T. Mech. Appl. A study of the supersonic axial flow compressors. 78. Ringros. F. W. Piping Handbook. Impeller Pumps. Pergamon Press.. (1945). Troskolanski. NACA T. G. V. Dimensional Analysis and Theory of Models.. reprint by kreiger. E. Eng.S. V. H. Trans.. L. Horlock. Proc. J. A. Copyright 2003 by Marcel Dekker. Pergamon Press. Reading. 69. J. 70. 72. W.. 26. W. H. A Mathematical Method of Cascade Design. C. (1957). Losses and efficiencies in axial-flow turbines. (1981). Sec. Handbook of Applied Hydraulics. Brater. 26 – 76. Lam. J. pp.A. Hunsaker. 82. New York: Hydraulic Institute. New York: McGrawHill Book Company. (1952). J. S. eds. 12th ed. F. New York: Dover Publications. Handbook of Applied Hydraulics. Performance of Axial Flow Fans. 107:6 –22. Hawthorne. 19:4 – 19. Turbo Blowers and Compressors. S. Lighthill. H. King.. (1960). L. Stalling Phenomena in Axial-Flow Compressors. J. Lazar Kiewics. Sci. (1985). Kearton.S. R. 71. (1967). Inc. Int. D. P. 153:441– 452. W. F.E. A Mean Line Prediction Method for Axial Flow Turbine Efficiency. Mech. Fluid dynamics of axial compressors and design of axial compressors. J.. London: The English Language Book Society. (1945). Howell. 410– 411. A Note on Secondary Flow in Rotating Radial Channels.I. J. Bibliography Hydraulic Institute. S. H. Hydraulic Institute Standards. A. 81. R.N. New York: McGrawHill Book Co. 66.M. R. Zowski. Crocker. J. 64.. 2104. Kacker. H. Thesis. 6th ed. ARC R and M No. Mech. P. 79. H. Okapuu. Principles of Fluid Mechanics. Inst. C. Li. C. London: Sir Issac Pitman and Sons Ltd. G. M. Paper 21-GT-58. Eng. 63. Keratin. L. King. Hickox. S. M. Mass: Addison-Wesley Publishing Company Inc.. Actuator Disc Theory of the Incompressible Flow in Free Vortex. J. W. Hydrodynamics. New York: McGraw-Hill Book Company. In: Davis. (1926). 80. 2nd ed. J. (1965). 60. 73. A. H. London: Anglo American Aeronautical Conference Royal Aeronautical Society. (1934). G. (1954)... H. V.. Norrie. H. J. Shapiro. (1951). E. The Aerodynamics of Compressor Blade Vibration. Investigations of axial flow compressors. Inst. J. (1957). New York: John Wiley and Sons Inc. Naehlizba. W. Inc. 50:154. Rannie.. (1963). Rouse. J.. Marble. W. Nov. Q. S. (1961). Aug. (1956). AIAA Education Series. Cambridge.M. Stepanoff. Research Inst. J. (1944). Robinson. A. Bowon. Aeronaut. H. H. Mechanics of Fluids. F. London: Blacjie and Son. S.E. Bowmer. Gas Turbines. Shames. E.. Eng. 1062. New York: McGraw-Hill Book Company. H. 159:198– 219. Marble. T. E. Soc.: Artia. 15.M. Eng. Aeronaut. M. A. Sci. No. 105. Handbook of Fluid Dynamics. Sabersky. 55. W..M. Rhoden. 97. In: Mayer. Their Design and Equipment. (1962). D. (1948). 89. Q. J. Aircraft Engine Design.. Hydraulic Turbines. D. Spamhake. 3. J. Lewis. 85. Dates. New York: John Wiley and Sons Inc. 103. (1951). Prandtl.A. Prague. Rep. All Rights Reserved . Mass: The Technology Press of the Massachusetts Institute of Technology.G. New York: The Ronald Press Company. A. E. Engineering Hydraulics. F. 87.E 76. Evans. Michelson. 104.. Aeronaut. Advanced Mechanics of Fluids. Experimental investigations of propagating stall in axial flow compressors. 100. 86. (1951)... Essentials of Fluid Dynamics. NACA T. The part load performance of various gas turbine engine schemes. New York: John Wiley and Sons Inc.S. The Dynamics and Thermodynamics of Compressible Fluid Flow. (1934). Pearson. ed. Mattingley. 2nd ed. (1950). R. (1959). Rannie.. New York: McGraw-Hill Book Company. A. Heiser. J. The flow of an incompressible fluid through an axial turbomachine with any number of rows. D. Fundamentals of Hydro and Aeromechanics. C. A Theory of the Secondary Flow in Cascades. A. Incompressible Flow Machines. O. W.. 84. D. R. 91. 2614. 1 & 2. L. Transportation Tech. (1950). P. H. ed. (1928). 99. 95. (1949). Streeter. (1948). 90. Trans. H. Aeronaut. 106. Ostuka. 102. 1006. Copyright 2003 by Marcel Dekker. G. G.. Analytical Investigation of Some Three Dimensional Flow Problems in Turbomachines. Iura. (1951). Aerothermodynamics of Gas Turbine and Rocket Propulsion. D. (1953). 73. American Elseveir. H. L. Prandtl. Report on reheat factors. L. Sorensen. I. W. Mech. Mech. Tietyens. (1953).. D. Pearson.N. Railly. The flow of a perfect fluid through an axial turbomachine with predescribed blade loading. The relative merits of centrifugal and axial compressors for aircraft gas turbines. Surging in axial compressors. et al. Tr. A. New York: McGraw-Hill. New York: The Ronald Press Company. Trans. Rouse. (1987). Vols. ed.. 93.Bibliography 83. (1984). 107. H. Daley. Turbines and Propellers. 98. 391 Mallinson. AIAA Education Series. H. I. 101.S. W. Centrifugal and Axial Flow Pumps. 92. Centrifugal Pumps. Moult. L. C. 88. 108. Investigation of Axial Flow Fans. Pearson. NACA T. H.. 96. E. T. 94. (1951). 110. W. Inc. Copyright 2003 by Marcel Dekker. Heat and Thermodynamics. London: The Macmillan Company. C.. E. J. G. M. 2nd ed. Fundamentals of Classical Thermodynamics. Horlock.. (1956). 118. H. D. 1 & 2. P. 117.N. 111. R. (1953). M. 17280. (1971). Wilson. H. Bibliography Shepherd. D. Experimental Investigations of the Stalled Flow in a Single Stage Axial Compressor. (1958). Inc. A. Theory of Hydraulic Models. Aircraft and Missile Propulsion. H. C. Principles of Turbomachinery. Inc. (1984). (1978). J. Sonntag. J. K. Yalin. 2961. London: The Macmillan Company. 115. 113. NACA T. Compressible Fluid Dynamics. G. Van Wylen. M. MIT Press. M. Wu. Design of High Efficiency Turbomachinery and Gas Turbines. (1937).. Applied Hydrodynamics.392 109. New York: John Wiley and Sons. New York: John Wiley and Sons. Thomson. 112. Zamansky. Vols. New York: McGraw-Hill Book Co. E. Wood. D. 114. New York: John Wiley and Sons Inc. Zucrow. Vallentine. All Rights Reserved . New York: McGraw-Hill. (1953). R. S. (1971). ARC Report No. 116. Subsonic Flow of Air Through a Single Stage and Seven Stage Compressor. (1980). Armstrong.
Report "Turbomachinery Design and Theory, Rama S. R. Gorla & Aijaz a. Khan"