Solution Manual to Principles of Heating Ventilating and Air Conditioning 6th Edition

Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com).This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toront PRINCIPLES OF HEATING VENTILATING AND AIR CONDITIONING SOLUTIONS MANUAL Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto ABOUT THE AUTHORS Ronald H. Howell, PhD, PE, Fellow ASHRAE, retired as professor and chair of mechanical engineering at the University of South Florida and is also professor emeritus of the University of Missouri-Rolla. For 45 years he taught courses in refrigeration, heating and air conditioning, thermal analysis, and related areas. He has been the principal or co-principal investigator of 12 ASHRAE-funded research projects. His industrial and consulting engineering experience ranges from ventilation and condensation problems to the development and implementation of a complete air curtain test program. William J. Coad, PE, Fellow ASHRAE, was ASHRAE president in 2001-2002. He has been with McClure Engineering Associates, St. Louis, Mo., for 45 years and is currently a consulting principal. He is also president of Coad Engineering Enterprises. He has served as a consultant to the Missouri state government and was a lecturer in mechanical engineering for 12 years and an affiliate professor in the graduate program for 17 years at Washington University, St. Louis. He is the author of Energy Engineering and Management for Building Systems (Van Nostrand Reinhold). Harry J. Sauer, Jr., PhD, PE, Fellow ASHRAE, was a professor of mechanical and aerospace engineering at the University of Missouri-Rolla. He taught courses in air conditioning, refrigeration, environmental quality analysis and control, and related areas. His research ranged from experimental boiling/condensing heat transfer and energy recovery equipment for HVAC systems to computer simulations of building energy use and actual monitoring of residential energy use. He served as an advisor to the Missouri state government and has conducted energy auditor training programs for the US Department of Energy. Dr. Sauer passed away in June 2008. Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto PRINCIPLES OF HEATING VENTILATING AND AIR CONDITIONING 6th Edition Ronald H. Howell William J. Coad Harry J. Sauer, Jr. American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto SOLUTIONS MANUAL Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. process. or the like that may be described herein.ashrae. GA 30329 www. or transmitted in any way or by any means—electronic.ashrae. (www.ashrae. process. Requests for permission should be submitted at www. Refrigerating and Air-Conditioning Engineers. and ASHRAE expressly disclaims any duty to investigate. Stephen Comstock This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto ASHRAE STAFF Copyrighted material licensed to University of Toronto by Thomson Scientific. or guaranty by ASHRAE of any product. procedure. No part of this publication may be reproduced without permission in writing from ASHRAE. (www. The appearance of any technical data or editorial material in this publication does not constitute endorsement. 1791 Tullie Circle. Inc. warranty. except by a reviewer who may quote brief passages or reproduce illustrations in a review with appropriate credit. stored in a retrieval system. design.ISBN 978-1-933742-70-0 ©2009 American Society of Heating. distribution. Additional reproduction. Atlanta. N. SPECIAL PUBLICATIONS Mark Owen Editor/Group Manager of Handbook and Special Publications Cindy Sheffield Michaels Managing Editor James Madison Walker Associate Editor Amelia Sanders Assistant Editor Elisabeth Parrish Assistant Editor Michshell Phillips Editorial Coordinator PUBLISHING SERVICES David Soltis Group Manager of Publishing Services and Electronic Communications Jayne Jackson Publication Traffic Administrator PUBLISHER W.org/permissions.org All rights reserved. The entire risk of the use of any information in this publication is assumed by the user. ASHRAE does not warrant that the information in the publication is free of errors. but ASHRAE has not investigated. © (2009). or other—without permission in writing from ASHRAE. recording. Printed in the United States of America ASHRAE has compiled this publication with care.org). design. For personal use only. . and ASHRAE does not necessarily agree with any statement or opinion in this publication. Inc. service. any product. nor may any part of this publication be reproduced. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. American Society of Heating.techstreet.com). service. Inc. or the like. photocopying. procedure.E. Refrigerating and Air-Conditioning Engineers. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Coad Copyrighted material licensed to University of Toronto by Thomson Scientific. Ventilating. with some intermediate computations omitted. The remaining problems are either those requiring discussion or those whose solutions depend on arbitrary assumptions or data selected by the instructor. Inc. Howell W. Additional reproduction. R.Notes to Instructors This manual contains solutions to most of the problems in the textbook. Refrigerating and Air-Conditioning Engineers. distribution. (www. For personal use only. which is based on the 2009 ASHRAE Handbook—Fundamentals.org). or equations in the 2009 Handbook that may not be found in Principles of Heating. Ventilating. Answers and solutions are included for the majority of the problems.com). Some of these problems require the use of tables. Principles of Heating. The solutions in this manual are generally presented in abbreviated form.H. and Air Conditioning.ashrae. and Air Conditioning. Inc. (www. © (2009).J.techstreet. figures. American Society of Heating. American Society of Heating.techstreet. Additional reproduction. Inc. Inc. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.org). (www. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . (www. distribution. For personal use only.com). © (2009).Copyrighted material licensed to University of Toronto by Thomson Scientific.ashrae. Refrigerating and Air-Conditioning Engineers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .CONTENTS Solutions to Chapter 1 . distribution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 Chapter 17 . . . . . . . . . . . . . . . . © (2009). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Chapter 3 . . . . . . . . . . . . . . . . . . 171 Chapter 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Chapter 11 . . . . 177 Chapter 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Chapter 5 . . . . . . . . . . . . . . . . . . . . . . 195 Chapter 20 . . . Inc. . . . . . . . . . . . . . . . . . . . . . . . . . . . .org). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Chapter 19 . . .com). . . . . . . 185 Chapter 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . American Society of Heating. . . . . . . . . . . . . . . . . . . . . . . . . 153 Chapter 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 Chapter 14 . . .ashrae. . . . . . . . . . . . . (www. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Chapter 12 . . 1 Chapter 2 . . . . . . . . . . (www. . . . . . . . 91 Chapter 8 . . . . . . . . . . . . . . . . . . . or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. . . . . . . . . . . . . . . . . . . . . . . . . . . . Refrigerating and Air-Conditioning Engineers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Chapter 9 . . . . . Additional reproduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Inc. . . . . . . . . . . . . . . 203 Copyrighted material licensed to University of Toronto by Thomson Scientific. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 Chapter 4 . . . 61 Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . For personal use only. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .techstreet. . . . . . . . . . . . . . . American Society of Heating. distribution. Additional reproduction. Refrigerating and Air-Conditioning Engineers. Inc. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .com).ashrae.techstreet.org). © (2009).Copyrighted material licensed to University of Toronto by Thomson Scientific. (www. (www. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. For personal use only. Inc. Inc. For personal use only. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.techstreet. (www. Inc. © (2009). Additional reproduction. distribution. (www.Solutions to Chapter 1 BACKGROUND Copyrighted material licensed to University of Toronto by Thomson Scientific. Refrigerating and Air-Conditioning Engineers.ashrae.org).com). American Society of Heating. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . com). © (2009).Copyrighted material licensed to University of Toronto by Thomson Scientific. American Society of Heating. (www.ashrae.techstreet. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .org). Additional reproduction. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Refrigerating and Air-Conditioning Engineers. For personal use only. Inc. Inc. (www. distribution. 000 ft 3 2 From Table 1.7 × 3.5 Btu/h⋅ft2⋅°F (2.5 Estimate the size of heating and cooling equipment that will be needed for a residence in middle America that is 28 × 78 × 8 ft high (8.750 [$54.110.500] This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 1. Assume water is black body.2 cfm/ft × 40 × 150 = 7200 cfm] Costs: Cooling system ($1500/ton) × 17 tons = $25.12 × 12. . Be conservative.500 Heating system ($2. 800 ft .0 Btu/h ⋅ ft ) = 1. For personal use only.000 Btu/h 2 Air movement: 17 tons × 400 cfm/ton = 6900 cfm or [1.= 3.50 ) × 6900 [7200] cfm = $51.1714 ⎛ ---------⎞ → by trial and error ⎝ 100⎠ T w ≈ 410°R = – 50°F ∴ will freeze 1.2°C). Copyrighted material licensed to University of Toronto by Thomson Scientific.50/cfm) × 6900 [ 7200 ] cfm = $17. 40 × 150 × 10 ft high (12. 40 × 150 Cooling unit: ---------------------------= 17 tons 3 350 ft ⁄ ton 3 Heating unit: ( 40 ) ( 150 ) ( 10 ) ( 3 Btu/h ⋅ ft ) = 180.org).1 Estimate whether ice will form on a clear night when ambient air temperature is 45°F (7.800 ft 2 Volume = ( 140 ) ( 220 ) ( 12 ) = 370. cooling.1: 250 ft ⁄ ton and 3.4 m high).8 W/m2⋅K).500 [$97.400 Btu ⁄ h 1.Chapter 1—Background⏐3 1. and air moving) for an office building.000] Total = $94.1: 700 ft ⁄ ton and 3.000 = 37. (www.12 tons or 3.2 × 45.000 ft ) ( 3. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Additional reproduction.0 Btu/h ⋅ ft 3 2 30.com).400 Btu/h ( 700 ) 3 3 Heating: [ ( 28 ) ( 78 ) ( 8 )ft ] ( 3. Inc. distribution.8 × 2.= 123 tons ∴Cooling: ---------------------------2 250 ft ⁄ ton 3 3 Heating: ( 370.techstreet.0 Btu ⁄ h ⋅ ft ( 28 ) ( 78 ) Cooling: ---------------------.1 m high).000 Btu/hr or 1110 Mbh 2 3 From Table 1. Heat in by convection = Heat out by radiation to space (Assume space @ 0°R.7 × 67 × 3.0 Btu/h ⋅ ft ) = 52. Inc. if the water is placed in a shallow pan in a sheltered location where the convective heat transfer coefficient is 0.ashrae. Refrigerating and Air-Conditioning Engineers.000] Fans/ducting ( $7. Floor area = ( 140 ) ( 220 ) = 30. (www.4 Estimate the size of cooling and heating equipment that is needed for a new bank building in middle America that is 140 × 220 × 12 ft high (42. ε = 1 4 hA ( T air – T water ) = σεAT water Tw 4 ( 0. American Society of Heating.6 Estimate the initial cost of the complete HVAC system (heating.5 ) ( 505 – T w ) = 0.250 [$18.5 × 23.7 m high). © (2009). com).6 if it is all-electric. Ventilating. distribution. Copyrighted material licensed to University of Toronto by Thomson Scientific.ashrae.5 ) = 183. 2 From Table 1.4⏐Principles of Heating.2: 30. Inc.640 1. For personal use only.000 kWh Cost = $0. Refrigerating and Air-Conditioning Engineers. Inc.8 Open-ended design problem.org). and Air Conditioning—Solutions Manual 1.7 Estimate the annual operating cost for the building in Problem 1. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . American Society of Heating. Additional reproduction.5 kWh / ft ⋅ yr Energy = ( 40 × 150 ) ( 30. (www.08 ( 183. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.000 ) = $14. (www. © (2009).techstreet. © (2009). distribution. American Society of Heating.Solutions to Chapter 2 THERMODYNAMICS AND PSYCHROMETRICS Copyrighted material licensed to University of Toronto by Thomson Scientific.org). (www. (www.techstreet. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .com). Inc. For personal use only.ashrae. Refrigerating and Air-Conditioning Engineers. Inc. Additional reproduction. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.ashrae. (www. Inc.techstreet. Inc. distribution. © (2009). American Society of Heating. Additional reproduction.com). (www.Copyrighted material licensed to University of Toronto by Thomson Scientific.org). This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . Refrigerating and Air-Conditioning Engineers. For personal use only. 86 ( 192.073 – 0.6 Two pounds of air contained in a cylinder expand without friction against a piston. from R-134a tables: 1 -⎞ 3 ν = 0.44 kJ/kg 6 u = h – Pν = 420.3 3 1 ν 2 – ν 1 = ⎛ -----.0721 ( 10 ) ( 0.2 kJ/kg 1 2. and internal energy? from R-134a tables: S g = 1.8 Saturated R-134a vapor at 42°C is superheated at constant pressure to a final temperature of 72°C.073 ft P2 ( 200 ) ( 144 ) 2 = P ∫ dv = P ( v 2 – v 1 ) = ( 200 ) ( 144 ) ( 2.86 ⎛ ----------= 0. Inc.0049 m ⁄ kg.= ------------------------------------.0123 m ⁄ kg ⎝ 1310⎠ h = 0. 42°C = 315 K: P = 1.0189 ⁄ 1000 ) u = 400.3 kJ/kg s = 0.14 ( 395. The air initially occupies a volume of 0.0.0189⎞ = 0.44 – 1. s 2 = 1.14 ( 0.= 42.083 ) + 0. Additional reproduction. What is the pressure? What are the changes in specific volume.0721 MPa. enthalpy.3 kJ/kg This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 3 1. Refrigerating and Air-Conditioning Engineers. 72°C = 345 K: P = 1.0189 m ⁄ kg . and entropy of 1 kg of R-134a at a saturation temperature of –5°C and a quality of 14%. distribution.082 kJ/kg ⋅ K 2.71 = 0.7 Determine the specific volume. Copyrighted material licensed to University of Toronto by Thomson Scientific.5 – 400.81 u 2 = 608.50 ft3. h 2 – h 1 = 453 – 420.Chapter 2—Thermodynamics and Psychrometrics⏐7 2.6 kJ/kg u 2 – u 1 = 427.= 2. For personal use only. The pressure on the back side of the piston is constant at 200 psia. ⎝ 42 ⎠ s 2 – s 1 = 1.9 ) = 221. © (2009).3 ) ( 560 ) 3 V 2 = -------------.com).5 ) = 45.14 ( 1. Inc. ν2 h 2 = 453. What is the work done by the air in ft-lbf if the expansion continues until the temperature of the air reaches 100°F? PV = mRT W = ∫ P dv mRT 2 ( 2 ) ( 53.300 ft ⋅ lb f 1 2. entropy.7306 ) + 0.44 = 32.976 ) = 1. (www.10 kJ/kg ⋅ k.7108 kJ/kg ⋅ K h g = 420. v g = 0.– 0.81 – 1.techstreet. enthalpy.2 = 27.8 ) + 0.ashrae. (www. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. ----.86 ( 0. American Society of Heating.org). .0037 MPa. techstreet.95 lb m 2. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. How much heat is supplied? kJ Q = mc p ( Δt ) = ( 150 kg ) ⎛⎝ 4. For personal use only.9 – 9. (www.800 kJ Copyrighted material licensed to University of Toronto by Thomson Scientific. Ventilating.39 ) = 351. © (2009).000 kW 0.890 kJ kg ⋅ K 2.ashrae.9 lbm 5.= 33.001004 m ⁄ kg⎠ Q = 350.955 ) ( 1082 ) 29400 ≈ 29700 ∴ m i = 33. Inc.72 – 8.18 kJ ⁄ kg ⋅ K ) ( 30 – 2 ) K 3 ν ⎝ 0.9 ? ( 33. If there is no heat transfer from the tank and the heat capacity of the tank is neglected.9 – 9. and Air Conditioning—Solutions Manual 2.= 9. 400°F flows.955 ) ( 1228 ) = ( 33.28 ) kJ ⁄ kg = 112.180 --------------⎞⎠ ( 85 – 15 ) K = 43. (www. 3 ⎛ ⎞ V· 3 m ⁄s Q = mc p Δt = --. t = 400°F Tank: P 1 = 20 psi. American Society of Heating. sat vapor P 2 = 100 psi 3 200 ν 1 = 20. .000 kW or 3 Q = m Δh = ⎛⎝ ----------------------⎞⎠ ( 125.12 Three cubic meters per second of water are cooled from 30°C to 2°C.org). Additional reproduction.955 = 23.10 Determine the heat required to vaporize 50 kg of water at a saturation temperature of 100°C.09 ft ⁄ lb m ⇒ m 1 = ------------.11 The temperature of 150 kg of water is raised from 15°C to 85°C by the addition of heat. calculate the mass of steam that enters the tank. Attached to this tank is a line in which vapor at 100 psia.8⏐Principles of Heating.09 V2 m 2 = -----ν2 Try and T 2 = 550°F mi hi = m2 u2 – m1 u1 u 2 = 1195 by trial and error 200 m 2 = --------. distribution. Steam from this line enters the vessel until the pressure is 100 psia. Refrigerating and Air-Conditioning Engineers. Steam line: P = 100 psi .955 lb m 20. 2.c p Δt = ⎜ ----------------------------------------⎟ ( 4.9 ) ( 1195 ) – ( 9. Compute the rate of heat transfer in kilojoules per second (kilowatts).9 A tank having a volume of 200 ft3 contains saturated vapor (steam) at a pressure of 20 psia.001004 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Q = mh fg = ( 50 kg ) ( 2256. Inc.com). For personal use only. 100°F. Closed system.171 ) ( 500 – 100 ) = 684 Btu ΔH = mc p Δt = ( 10 ) ( 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Heat is transferred to the air until the temperature reaches 500°F.2 ) ( – 10 ) = ---------------------------------------------------------. Water enters at a pressure of 20 psia and leaves at a pressure of 200 psia.7 psia. and the work done for: a. perfect gas ΔU = 684 Btu . Water enters at a pressure of 138 kPa and leaves at a pressure of 1380 kPa. distribution.013 = – 0.001 ) – ( 1380 ) ( 0. Additional reproduction. W = Q – Δu = 960 – 684 = 276 Btu 2. the heat transfer. Determine the change of internal energy. If there is no heat transfer and no change in kinetic or internal energy. If there is no heat transfer and no change in kinetic or internal energy. (www.+ gz 2⎟ + Q – W = 0 2 2 ⎝ ⎠ ⎝ ⎠ ( 138 ) ( 0.ashrae. a constant-pressure process. ΔH = 960 Btu p = c: Q = mc p Δt = ΔH = 960 Btu Q – W = ΔU (closed system) . Refrigerating and Air-Conditioning Engineers.+ gz 1⎟ – ⎜ u 2 + P 2 ν 2 + ----.techstreet.001 m3/kg. w = 0 since constant volume Q = ΔU = 684 Btu b. The specific volume of the water is 0.--------------J gc J 2. © (2009).org).546 Btu/lb m Copyrighted material licensed to University of Toronto by Thomson Scientific. the change in enthalpy.-------------778 ( 32. what is the work per pound? From 1st Law: P1 ν1 – P2 ν2 g z1 – z2 w = -----------------------------. what is the work per unit mass? 2 2 ν1 ν1 ⎛ ⎞ ⎛ ⎞ m ⎜ u 1 + P 1 ν 1 + ----. a. Inc.240 ) ( 500 – 100 ) = 960 Btu Q – W = ΔU (closed system).016 ft3/lb.15 The discharge of a pump is 3 m above the inlet. Closed system.14 The discharge of a pump is 10 ft above the inlet. Inc.+ ----.Chapter 2—Thermodynamics and Psychrometrics⏐9 2.13 Consider 10 lbm of air that is initially at 14. American Society of Heating.+ --------------. constant pressure process.66 J ( Note 1 J (Joule) = 1 N ⋅ m ) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto ( 0. perfect gas ΔU = mc v Δt = ( 10 ) ( 0. . a constant-volume process b.2 ) ( 778 ) = – 0.001 ) – ( 3 ) ( 9.533 – 0.016 ) ( 20 – 200 ) ( 144 ) ( 32.806 ) – W = 0 W = – 30. (www. constant volume process. The specific volume of the water is 0.com). 38 ft ⁄ lb m P2 100 P1 ν1 Pν = constant ⇒ p = -----------ν w = ∫ P dν 2 ν2 = P 1 ν 1 ∫ dν -----.techstreet. 100°F to 100 psia. steady-flow process from 15 psia. Ventilating. Calculate the work of compression per pound.0674 ft ⁄ lb m ν = νf + χ ( νg – νf ) ν – νf ( 0. and the heat transfer per pound of air compressed.2 = – 101. What is the mass of liquid and the mass of vapor present in the container? Copyrighted material licensed to University of Toronto by Thomson Scientific.21 ) ⎛⎝ ---------⎞⎠ = 1. The volume of the container is 3 ft3 and the mass of nitrogen in the container has been determined as 44.38 w = ( 15 ) ( 144 ) ( 9. Refrigerating and Air-Conditioning Engineers. the change of entropy.5 lb m t = – 240°F = 220°R 3 ⇒ ν f = 0.8445 ( 0.= P 1 ν 1 ln ⎛⎝ -----⎞⎠ ν1 1 ν 1.24 ) ln ⎛ ---------⎞ – ( 53. (www.13 Btu ⁄ lb m ⋅ °R 3 V = 3 ft m = 44. (www. For personal use only.21 ft ⁄ lb m P1 ( 15 ) ( 144 ) P1 15 3 ν 2 = ν 1 ⎛⎝ ------⎞⎠ = ( 9.= 9.0750 – 0.5 = 0.= ---------------------------------------------.5 ) ( 0. distribution.10⏐Principles of Heating.02613 ) νg – νf m v = mx = ( 44.com).0750 ft ⁄ lb m 3 ν = V ⁄ m = 3 ⁄ 44. .8445 ) = 6.16 Air is compressed in a reversible.58 lb m vapor m L = m ( 1 – x ) = ( 44.17 Liquid nitrogen at a temperature of –240°F exists in a container.21 = – 37. t 2 = 100°F RT 1 3 ( 53.21 ) ln ⎛⎝ ----------⎞⎠ 9.5 lbm.92 lb m liquid This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 2.800 ft ⋅ lb f ⁄ lb m P2 T2 100 373 ΔS = C p ln ⎛ -----⎞ – R ln ⎛ ------⎞ = ( 0. and both the liquid and vapor phases are present. P 1 = 15 psi . American Society of Heating.8445 ) = 37.org).5 ) ( 1 – 0.= ----------------------------. t 1 = 100°F P 2 = 100 psi . and Air Conditioning—Solutions Manual 2. isothermal.0674 – 0.= 0. © (2009).02613 ) x = ---------------.3 ) ln ⎛ ---------⎞ ⎝ 15 ⎠ ⎝ 373⎠ ⎝ T 1⎠ ⎝ P 1⎠ = 0 – 101. Additional reproduction. Inc.ashrae.02613 ft ⁄ lb m 3 ν g = 0.3 ) ( 373 ) ν 1 = --------.2 ft lb f ⁄ lb m ⋅ °R = – 0. Inc. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Chapter 2—Thermodynamics and Psychrometrics⏐11 2.7 kJ 2. Additional reproduction. The air is then compressed reversibly according to the relationship pvn = constant until the final pressure is 600 kPa. Inc.674 ) ( 0.= 11.= 0. the tank contains air at 100 kPa.674 ) ( 0. Initially. Determine the work input to the pump.3 + ( 2.-----------------------------.com).= 0.18 A fan in an air-conditioning system is drawing 1. 3 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto ( 287 ) ( 393 ) 3 ( 287 ) ( 294 ) 3 ν 2 = ------------------------------.718 ) ( 120 – 20 ) = 85.ashrae.335 = 0. m = -----. the volume of the tank is 50 m3. .= 1. distribution. The capacity through the fan is 0. the polytropic exponent n b. n = 1. ⎛ -----⎞ = -----.= ------------..31 m in diameter.85 V 1 = --.19 Air is contained in a cylinder. © (2009). What is the temperature rise of the air due to this fan? 2 2 v 1⎞ ⎛ V 2⎞ ⎛ m ⎜ h1 + -----⎟ – ⎜ h 2 + ------⎟ – w = 0 2 2⎠ ⎝ ⎠ ⎝ m ( h1 – h2 ) – w = 0 m Cp ( t1 – t2 ) – w = 0 2 2 πD A = ---------. Copyrighted material licensed to University of Toronto by Thomson Scientific. the work done on the air and the heat transfer V2 V1 1. Refrigerating and Air-Conditioning Engineers.20 Water at 20°C is pumped from ground level to an elevated storage tank above ground level.5 b. the cylinder contains 1. ν 1 0.85 ) m· = --------------------------------------------------. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.188 -------------⎞ = 150 --------a.26 m/s A 0.= 0. For this process determine: a.561 600 P 2 ν 2 = RT 2 .3 ) ( 1000 ) ( 0. ⎝ ν 1⎠ ⎝ ⎠ P2 0. American Society of Heating. Initially.techstreet.= -------------------------.25 ) ( 0. W = ∫ P dv = c ∫ ---= n 1–n 1–n v Q = W + m ( U 2 – U 1 ) = – 106.= 2. The temperature of the air and water remain constant at 20°C. at which point the temperature is 120°C.561 ν2 V 2 = ( 2.561 m ⁄ kg ( 600 ) ( 1000 ) ( 150 ) ( 1000 ) n ν2 n P1 n ⎛ 0.85 m3/s of 24°C air and the inlet and outlet ducts are 0.25 . (www. and the tank is closed so that the air is compressed as the water enters the bottom of the tank.25 hp at 1760 rpm.674 kg = -----.188 ) = 0.746 ) t 1 – t 2 = ---------------------------------.005 ) P 1 V· 1 = m· 1 RT 1 ( 101. Inc.0755 m 4 V· 0.= – 106.503 m P2 V2 – P1 V1 R ( T2 – T1 ) dv.. 20°C. 20°C. (www.27 0.5 m3 of air at 150 kPa.91 K ( 1.= 0.02 kg/s ( 287 ) ( 294 ) 2. The pump is operated until the tank is three-quarters full.= ---------------.1880 m ⁄ kg ν 1 = ------------------------------. For personal use only.org).0755 V 2 = V 1 since small ΔT and ΔP ( 1.3 kJ c.02 ) ( 1. the final volume of the air c. = ------------------------------------.= 19.12⏐Principles of Heating.001002 ) W total from pump = 6930 + 14670 = 21.6 Btu/lb m 2. Additional reproduction. change in specific internal energy d.2 – 1960 ) = – 161. change in specific enthalpy e. change in internal energy per lbm d. and Air Conditioning—Solutions Manual W against air = ∫ Pdv v v = mRT ∫ dv ----. Determine the minimum size motor (in horsepower) to drive this pump. change in enthalpy per lbm e. δq – δw = δu = – 161.769 × 10 ) w· HP = --------------------------------.22 Air undergoes a steady-flow. Δu = c u ΔT = ( 0. The initial state is 1400 kPa. specific work This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 2. (www. American Society of Heating. final temperature b.ashrae. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. reversible adiabatic process. . final specific volume c. 815°C.4 – 1--------------1.25 ) = – 6930 kJ ( 50 ) W to elevate = mg Δz = ( 3 ⁄ 4 ) --------------------------. ν 2 = --------. final specific volume c.7⎠ 8 ( 3.2 ) 3 b. · w· = m· ∫ ν dP = m· ν ( P 2 – P 1 ) 1 8 = ( 100 ) ( 3600 ) ⎛ ----------⎞ ( 500 – 15 ) ( 144 ) = 3.4 Hp ( 778 ) ( 2545 ) P 1 = 200 psi P2 a.171 ) ( 1015.= P 1 v 1 ln ----2v1 v1 v = ( 100 ) ( 1000 ) ( 50 ) ln ( 0.2 – 1960 ) = – 266. Inc. work per lbm Copyrighted material licensed to University of Toronto by Thomson Scientific.7 lbm/ft3.469 ft ⁄ lb m P2 ( 20 ) ( 144 ) c.( 9. ΔH = c p ΔT = ( 0.com).8 ) ( 40 ) = 14670 kJ ( 0.3 ) ( 1051. For personal use only.2 R RT 2 ( 53. reversible adiabatic process. Determine a. 1500°F. (www.600 kJ 2. T 2 = T 1 ⎛⎝ ------⎞⎠ P1 t 1 = 1500°F K – 1-----------K 20 = ( 1960 ) ⎛⎝ ---------⎞⎠ 200 P 2 = 20 psi 1. Inc. Ventilating. and the final pressure is 20 psia.769 × 10 ft ⋅ lb f ⎝ 66. Determine a.4 = 1015.21 A centrifugal pump delivers liquid oxygen to a rocket engine at the rate of 100 lbm/s. and the final pressure is 140 kPa.240 ) ( 1015. distribution.org). Changes in kinetic and potential energy are negligible. The density of liquid oxygen is 66. The oxygen enters the pump as liquid at 15 psia and the discharge pressure is 500 psia.= mRT ln ----2. © (2009).6 Btu/lb m d.= 190.23 Air undergoes a steady-flow. Refrigerating and Air-Conditioning Engineers.8 Btu/lb m e. The initial state is 200 psia. Changes in kinetic and potential energy are negligible.techstreet. final temperature b. The fan takes in outside air at 32.2 ) ( 778 ) = – 13. The fan takes in outside air at 80°F and 14.techstreet. If the process is assumed to be reversible and adiabatic. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 2 v2 ⎞ ⎛ W = m h 1 – ⎜ h 2 + ------------⎟ 2gcJ ⎝ ⎠ Copyrighted material licensed to University of Toronto by Thomson Scientific.= ------------------------------------.= ---------------------------------. Refrigerating and Air-Conditioning Engineers.7 ) ( 144 ) ( 1200 ) Pv· m· = ------. T 2 = T 1 ⎛⎝ ------⎞⎠ P1 k---------– 1k 140 0.25 ) – ------------------------------------( 2 ) ( 32.2 lb m /min RT ( 53. Inc.org). P 2 ν 2 = RT 2 .98 ) V 2 = ---------.= 10.98 ft ⁄ lb m P2 ( 15.7 1. Additional reproduction. In the 0. Inc.5 ⁄ 60 ) = ( 88. Determine the minimum size motor needed to drive the fan.7 ) ( 14. h 2 – h 1 = c p ( T 2 – T 1 ) = ( 1. determine the required power.2°C and 101. w = 528 kJ/kg 2.24 A fan provides fresh air to the welding area in an industrial plant.com). air pressure is 102 kPa.26 A fan provides fresh air to the welding area in an industrial plant. . h 1 – h 2 – w = 0 .64 ) ( 0.= 88.4 ) mν 2 ( 88.78 ) 2.7 = 540 ⎛⎝ ----------⎞⎠ 14.286 = 1088 ⎛⎝ ------------⎞⎠ = 563 K = 290°C 1400 b.3 ) ( 550. [Ans: W = 5.1 hp] ( 14.25°R RT 2 ( 53. (www.ashrae. distribution.= ---------------------------------------------.93 m2 duct leaving the fan.5 ft/min A2 ( 10 ) 2 V2 = m c p ( T 1 – T 2 ) + -----------2gcJ 2 ( 144.005 ) ( 290 – 815 ) = – 528 kJ/kg e.= ------------------------------.3 hp ( η fan ) ( η motor ) ( 0.24 ( 540 – 550.2 ) ( 60 ) 0.4 = 550. (www. R air = 0.Chapter 2—Thermodynamics and Psychrometrics⏐13 P2 a.12 ) HP actual = ------------------------------------.12 hp 2.718 ) ( 290 – 815 ) = – 377 kJ/kg d.= 12. air pressure is 1 psig.25 ) 3 ν 2 = --------.= 144. © (2009).4 – 1--------------1. For personal use only. kW. determine the size motor needed to drive the fan.115 m ⁄ kg c. American Society of Heating.0287 ) ( 563 ) ⇒ ν 2 = 0.7 ) ( 12.3 ) ( 540 ) P2 T 2 = T 1 ⎛⎝ ------⎞⎠ P1 k---------– 1k 15.4 kPa at the rate of 566 L/s with negligible inlet velocity.0287 N ⋅ m ⁄ g ⋅ K 3 ( 140 ) ( 1000 )ν 2 = ( 0.25 If the fan in the previous problem has an efficiency of 64% and is driven by a motor having an efficiency of 78%. u 2 – u 1 = c v ( T 2 – T 1 ) = ( 0.020 Btu/h = 5. In the 10 ft2 duct leaving the fan.7 psia at the rate of 1200 cfm with negligible inlet velocity. HP Ideal ( 5. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. What is the change in entropy of the condensing steam per pound of feedwater heated? [Ans: −0. ⎛ ----⎞ = C p ΔT ⎝ m⎠ FW ⎛Q ----⎞ ⎝ m⎠ F.24 kW 2.4⎠ 1.0932 lb m steam Δh ( 965. Inc.1324 Btu/lb m⋅R] c.= ------------------------------------. a.718 ) ( 305. = ( 1. Refrigerating and Air-Conditioning Engineers. The feedwater is heated from 60 to 150°F at constant pressure. steam condenses at a constant temperature of 220°F. ( m Δh ) steam = m FW ⎛ ----⎞ ⎝ m⎠ F.606 m/s (small.2150 ) – ( 0.com). [Ans: +0. ΔS is positive.= 0. © (2009).1595 Btu ⁄ lbm ⋅ R d.0 [ ( 0.techstreet.ashrae.655 ) ( 0.org).0555 ) ] = 0.0 ) ( 90 ) m steam = -----------------------------------.2 ) ⎛ -------------⎞ ⎝ 101.7 K RT 2 ( 287 ) ( 305. Additional reproduction. Ventilating.655 kg/s RT 1 ( 287 ) ( 305.= 0.27 In an insulated feedwater heater.3 ) ΔS steam = m s s fg = ( 0. American Society of Heating.W.= 0.14⏐Principles of Heating.W.W. Δs T = Δs FW – Δs steam = ( 0.0 ) ( 90 ) = 90 Btu ⁄ lb m F.W.= -------------------------------.93 ) W = m ( h 1 – h 2 ) = mc p ΔT = ( 0.= ----------------------. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto a.860 ) V· 2 = ---------.7 ) 3 ν 2 = --------.4 – 1 ---------------1.0932 ) ( 1. Assuming the specific heat at constant pressure of the feedwater is unity. No] steam = Q again F. Q = m F.4201 ) = 0. distribution.= -----------------------------------------------------. (www.0271 Btu ⁄ lb m ⋅ R Does not violate 2nd Law.4 ) ( 1000 ) ( 0.655 ) ( 0.= 0. Q b. What is the change in entropy of the combined system? Does this violate the second law? Explain.7 ) = 0.1595 Btu/lb m⋅R] d. Assume 1. What is the change in entropy of 1 lb of feedwater as it passes through the heater? [Ans: +0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.1324 Btu ⁄ lb m c.2 – 305. and Air Conditioning—Solutions Manual P 1 V· 1 ( 101. (www. how many Btu are absorbed by each pound in its passage through the heater? [Ans: 90 Btu/lb] b.1595 ) – ( 0.1324 ) = 0.566 ) m = -----------. .4 = 305. ΔS FW = m FW ( s f 150°F – sf 60°F FW ⋅R ) = 1.0271 Btu/lb m⋅R.0 lb m feed water m FW ( Q ⁄ m ) FW ( 1. Inc. neglect kinetic energy) A2 ( 0. C p ΔT .2 ) P2 T 2 = T 1 ⎛ ------⎞ ⎝ P 1⎠ k – 1---------k 102 = ( 305.W. Q loss Copyrighted material licensed to University of Toronto by Thomson Scientific.860 m ⁄ kg P2 ( 102 ) ( 1000 ) mν 2 ( 0. For personal use only. ID pipe. The steam is condensed as it flows through the radiator and leaves as condensate at 88°C.6 ) m = 0. Refrigerating and Air-Conditioning Engineers. (www. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. If the radiator is to have a heating capacity of 1. If 58°F air is desired.28 Steam at 124 kPa and 96% quality enters a radiator. Air at 50 psia and 90°F flows through an insulated turbine at the rate of 1.Chapter 2—Thermodynamics and Psychrometrics⏐15 2.org). American Society of Heating. For personal use only.com). Additional reproduction. 88°C liquid: h 2 = 368.6 lbm/s to an exit pressure of 14. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . The velocity of the air upstream from the restriction is 450 fpm.7 psia. If the air delivers 11.6 lbm/s.00083 kg/s = 6. b.6 lb m ⁄ h 2. at what temperature does the air leave the turbine? c. what must the velocity downstream of the restriction be? Comment on this as a method of cooling. 0.85 = m ( 2595 – 368. What is the lowest temperature attainable at exit? Copyrighted material licensed to University of Toronto by Thomson Scientific. 124 Kpa. how many kilograms per hour of steam must be supplied to the radiator? Steam 1. Inc. Inc.6 Q = m ( h 1 – h 2 ) = 1.ashrae.techstreet. Air at 50 psia and 90°F flows through an insulated turbine at the rate of 1.96 ( 2241 ) = 2595 2.96 quality: h 1 = 444 + 0. Air at 50 psia and 90°F flows through a restriction in a 2 in. (www. distribution.29 Solve the following: a.5 hp to the turbine blades. © (2009).85 kW. 00545 ft Copyrighted material licensed to University of Toronto by Thomson Scientific.2 ) ( 3600 ) ( 778 ) 2gc ⎠ ⎝ = ( 49.78 lb m /min ν1 ( 0. Heat is transferred to the water so that it leaves as saturated vapor at 9 psia.com).= ( 0.43 ] = 95.= -----------.01607⎠ ⎝ v1⎠ U1 A ( 146.78 ) [ 1093. m ( h 1 – h 2 ) – W = 0 mC p ( T 1 – T 2 ) – W = 0 .800 Btu/min 2. Additional reproduction.00545 ) ν2 42.00545 ) m = ---------. .ashrae. b.800 Btu/min] P 1 = 10 psi 3 V· 1 = 0.01607 ft /lb m = 48.7854 in.= ------------------------------------------.techstreet.367 ft /lb m h 2 = 1141. (www.8°F c.8 ft/min A ( 0.8 ft /min t 1 = 80°F P 2 = 9 psi Sat.24 ) ( 90 – 58 ) + ------------------------------------2gcJ 2gcJ ( 2 ) ( 32.16⏐Principles of Heating. © (2009).8 ) U 1 = -----.215 Btu/lb m V· 1 ( 0.024 ft/min ⎝ 0.5 ( 2545 ) t 2 = 68.= -----------------------. diameter tube at the rate of 0. Ventilating. (www.000 Btu/h.7 = ( 550 ) ⎛⎝ ----------⎞⎠ 50 0. [Ans: 95.8 ft3/min. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.= 146.01607 ) 2 2 2 2 U 2 – U 1⎞ ⎛ ( 387024 ) – ( 1468 ) Q· = m· ⎜ h 2 – h 1 + -------------------⎟ = ( 49. and Air Conditioning—Solutions Manual 2 2 V1 ⎞ ⎛ V2 ⎞ ⎛ a.31 A refrigerator uses R-134a as the refrigerant and handles 200 lbm/h. Condensing temperature is 110°F and evaporating temperature is 5°F. distribution. ( 1.6 ) ( 3600 ) ( 0.8 ) ( 0.05 Btu/lb m 2 3 v 2 = 42.286 = 388R = – 72°F 2. Inc.8 ⎛ -------------------⎞ = 387.24 ) ( 90 – T 2 ) = 11.165 + 830. vapor @ 2 2 ν1 ≅ νf h1 ≅ hf 3 80°F 80°F = 0.2 ) ( 778 ) V 2 = 620 fps = 37. For a cooling effect of 11.05 + ------------------------------------------------------( 2 ) ( 32. For personal use only. American Society of Heating. Determine the heat transfer per minute.25 – 48. determine the minimum size motor (hp) required to drive the compressor. = 0.200 fpm Not cooling since temperature will increase as fluid slows down. ⎜ h 1 + ------------⎟ – ⎜ h 2 + ------------⎟ = 0 2gcJ 2gcJ ⎝ ⎠ ⎝ ⎠ 2 2 2 V1 V2 ( 450 ⁄ 60 ) C p + ( T 1 – T 2 ) + -----------. ⇒ A = 0.= 49. Rev. Inc.367 U 2 = U 1 ⎛ -----⎞ = 146.org).30 Liquid water at a pressure of 10 psia and a temperature of 80°F enters a 1 in.78 ) 1141. = 1 in. Adiabatic for Minimum = Isentropic P2 T 2 = T 1 ⎛⎝ ------⎞⎠ P1 K – 1-----------K 14. Refrigerating and Air-Conditioning Engineers. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Dia. Inc.= 1.= 4. Refrigerating and Air-Conditioning Engineers. (www. Inc.= 2483.4286 2. the heat loss from the house is 60. when the outside air temperature is 10°F. distribution. s = 0.4286 TR ⁄ TA – 1 570 ⁄ 465 – 1 11.000 Btu/h if the inside is maintained at 70°F.techstreet.ashrae.22470 @  p = 161.7 ) = 3260 Btu/h 3260 W· = -----------.976 Hp [Minimum for reversed Carnot cycle] 4. s = 0. Copyrighted material licensed to University of Toronto by Thomson Scientific.0 t = 5°F h = 103. For personal use only.745.Chapter 2—Thermodynamics and Psychrometrics⏐17 @  x = 1. © (2009). Determine the minimum electric power required to operate the heat pump (in kW).22470 h ≅ 120 W· = m· ( h 1 – h 4 ) = 200 ( 120 – 103. In winter. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .= ------------------------------. American Society of Heating. Additional reproduction. (www.] 2545 1 1 COP = ------------------------.000 W· = ---------------.org).9 Btu/h = 0.28 Hp [Minimum for mechanical vapor compression cycle.05 psi.32 A heat pump is used in place of a furnace for heating a house. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.com). © (2009).--------------4 ( 144 ) ( 2.30 in.) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto ( 20 ) V πd 5 ⁄ 60 = A --. (www. Additional reproduction.99 kW COP heating ( 8. when the outside air temperature is − 10°C.= ---------.= -------------------. and the mass rate of flow is 5 lbm/min. saturated liquid leaving the condenser is subcooled 6°F.= 2.449 lb/ft .18⏐Principles of Heating.= ------------------------------.35 An R-134a refrigerating system is operating with a condensing temperature of 86°F and evaporating temperature of 25°F. QR QR 1 1 COP heating = ------.33 A heat pump is used in place of a furnace for heating a house. Refrigerating and Air-Conditioning Engineers.techstreet. [Ans: 21. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.48 1 – 263 ⁄ 298 1 – TL ⁄ Th QR 200 W min = --------------------. If the liquid line from the condenser is soldered to the suction line from the evaporator to form a simple heat exchanger. (www. . the heat loss from the house is 200 kW if the inside is maintained at 21°C.449 2 2 d = 1. Inc. 2.ashrae. Inc. Ventilating.34 Refrigerant-134a vapor enters a compressor at 25 psia.= ----------------------------------.com).23 ft /lb 0.= 1.org).70 ⇒ d = 1.833 1 – 470 ⁄ 530 QR ( 60. In winter. Determine the minimum electric power required to operate the heat pump. 40°F ⇒ ρ = 0. and if as a result of this. v = ------------.000 ) W = ----------------------------.= --------------------------.48 2. and Air Conditioning—Solutions Manual For minimum consider reversible Carnot cycle.23 ) v Copyrighted material licensed to University of Toronto by Thomson Scientific. distribution.= 9. What is the smallest diameter tubing that can be used if the velocity of refrigerant must not exceed 20 ft/s? 3 3 1 25 psi. For personal use only.= 8.= ----------------.= -------------------------W QR – QA 1 – QA ⁄ QR 1 – TL ⁄ TH 1 = ------------------------------.1 kW] 1 1 COP h ( max ) = ------------------------.1 kW COP max 9.833 ) ( 3413 ) 2.= 21. how many degrees will the saturated vapor leaving the evaporator be superheated? (Use tables. American Society of Heating. 40°F. 9 Btu/lb m Δh = 614.5 ) – ( 38 ) = h 4 – ( 106. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. calculate per ton of refrigeration a.8 psi ⎭ 4 °SH = 44 – 25 = 19° 2.10 ) ( 614.02446 ) = 0.55 ft3/lb] t 1 = 10°F x 1 = 10% = 0. American Society of Heating.1 Btu/lb m ⎫ ⎬t = 44°F P 4 = 36. Additional reproduction. horsepower required Copyrighted material licensed to University of Toronto by Thomson Scientific. to saturated vapor. Inc.37 For a compressor using R-134a with an evaporator temperature of 20°F and a condensing temperature of 80°F.9 = 505 Btu/lb m This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 2. . mass flow c.Chapter 2—Thermodynamics and Psychrometrics⏐19 h 1 – h 2 = h 4 – h 3 ⇒ ( 40.ashrae.10 ) ( 7. Refrigerating and Air-Conditioning Engineers.7524 ft /lb m 3 Δv = 7.8 h g = 614.com). distribution. For each pound. 10% quality.552 ft /lb m h f1 = 53.02446 ) – ( 0.9 – 53. (www. [Ans: 505 Btu/lb. (www.36 Ammonia is heated in the evaporator of a refrigeration system from inlet conditions of 10°F.10 Sat.304 – 0. determine the changes in enthalpy and volume.6 ) h 4 = 109.techstreet. 6.9 h 1 = 53.org).8 ) = 109. © (2009).9 – 109. The pressure remains constant during the process.7524 = 6.02446 v g1 = 7. vapor @ 2 P 1 = P 2 = constant v f1 = 0.304 3 v 1 = ( 0.8 + ( 0. For personal use only. displacement b. Inc.304 – 0. m· = ----------------. .09 ) ( 0.≈ 12 tons 12000 ( 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. PD = m· v 1 = ( 176.9 = 65. Inc.45 – 39.771 Hp/ton 2. m· = ----------------.20⏐Principles of Heating. Assume a compressor-volumetric efficiency of 70% and frictionless flow. mass flow c.1 Btu/lbm 2 compressor: 1 π3 V· ideal = ( 800 ) ⎛ ---------⎞ ( 4 ) ⎛ ------------⎞ ⎝ 1728⎠ ⎝ 4 ⎠ 3 = 13. Refrigerating and Air-Conditioning Engineers. and Air Conditioning—Solutions Manual qL 12000 b.techstreet.39 An industrial plant has available a 4 cylinder. 3 in.4101 ) = 249 ft /h ⋅ ton = 415 ft /min ton c.978 3 3 a.27 a. the refrigerating effect per kilogram of refrigerant circulated This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 2.12 lb m /min ton h1 – h4 103.ashrae.4 Btu/h ⋅ ton 3 2 3 1 = 1.65 ) ( 60 ) ] ( 65. Ventilating.com). single-acting compressor for use with R-134a. calculate per ton a.40 A mechanical refrigeration system with R-134a is operating under such conditions that the evaporator pressure is 160 kPa and the liquid approaching the refrigerant control valve is at a temperature of 41°C.46 ] = 4018. American Society of Heating. determine: a.= -----------------------------------. w· = m· ( h 2 – h 1 ) = 176.9 ) ( 1. distribution. stroke. Proposed operating conditions for the compressor are 100°F condensing temperature and 40°F evaporating temperature.24 ft /min ton c. Calculate the refrigerating capacity in tons for a system equipped with this compressor. (www. horsepower required qL 12000 b. displacement b.= 186. Plot the cycle on the p-h diagram. It is estimated that the refrigerant will enter the expansion valve as a saturated liquid.38 For a compressor using an R-22 system operating between 100°F condensing temperature and −10°F evaporator temperature.9 lb m ⁄ h ⋅ ton = 3.7 [ 117 – 105.70 ) = 36. that vapor will leave the evaporator at a temperature of 45°F.09 ft ⁄ lbm 3 V· actual = 4 ( 13.65 ft ⁄ min m· q L [ ( 36. (www.80 ) ( 12000 ) 2.org). Inc.0 ft ⁄ lbm q L = 110 – 44.9 ] = 1961 Btu/h ⋅ ton = 0. w· = m· ( h – h ) = 186.= -------------------------------------------------. [Ans: 12 tons] Copyrighted material licensed to University of Toronto by Thomson Scientific. If the system has a capacity of 15 kW. © (2009).58 Hp/ton 3 v ≅ 1.7 ) ( 1.1 ) capacity = --------------.6825 ) = 314 ft /h ⋅ ton = 5. 800 rpm.94 lb m /min ton h1 – h4 105.9 – 37.= -----------------------------------.9 [ 125 – 103. Additional reproduction. and that vapor will enter the compressor at a temperature of 55°F. For personal use only.7 = 2.= 176. PD = m· v 1 = ( 186. bore by 4 in. (www. Refrigerating and Air-Conditioning Engineers. the mass flow rate in kilograms per second per kilowatt the volume flow rate in liters per second per kilowatt at the compressor inlet the total mass flow rate in kilograms per second the total volume flow rate in liters per second at the compressor inlet a) q e = h 1 – h 4 = 389 – 258 = 131 kJ/kg b) 1 m = --------. kg/s c. compressor motor size.0075 kg/s 131 V 1 = mv 1 = ( 0.126 ) ( 1000 ) = 14. e. kW d.Chapter 2—Thermodynamics and Psychrometrics⏐21 b.org).0076 ) ( 0. .com). For personal use only. °C b.4 l/s Copyrighted material licensed to University of Toronto by Thomson Scientific. d.techstreet. American Society of Heating.115 ) ( 0.97 l/s c) d) e) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 2.= 0. refrigerant flow rate. distribution. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. © (2009).ashrae. Determine a. compressor discharge temperature if compression efficiency is 60% 15 m = --------. Inc. ideal compressor discharge temperature.= 0. respectively. The refrigerant leaving the condenser is subcooled 3 degrees and the vapor leaving the evaporator is superheated 5 degrees. c. (www.126 ) ( 1000 ) = 0. Inc. Additional reproduction.115 kg/s 131 v· = mv 1 = ( 0.41 A vapor-compression R-22 refrigeration system is being designed to provide 50 kW of cooling when operating between evaporating and condensing temperatures of 0°C and 34°C. COP for cooling e. Determine a.74 ) ( 28 ) = 0.= 451 ⇒ t 2a ≅ 75°C 0.= 0.( 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. The compressor has 4.4 kW e) Qe 77.= 0.42 For a line of ammonia compressors. Refrigerating and Air-Conditioning Engineers.85% 0.org).1 700 – 525 1 h 2a = 525 + ⎛ ------------------------⎞ = 738. starting with state 1 at the compressor inlet. There is 5°C of subcooling in the condenser and 10°C of superheating in the evaporator. Ventilating. © (2009). v 1 = ------.092 ) PD = --------------------------.= 0.5% clearance and operates at 28 r/s.123 ⎝ 0.85 7.= 5. distribution. s 1 = 10. American Society of Heating.303 Q e = m ( h 1 – h 4 ) = 0.= ---------.76 h 2 = 434 h 3 = h 4 = 237.35 η av = 94 – 6. % The compression efficiency is fairly constant at 82%.= 0.ashrae. each having a 92 mm bore and a 74 mm stroke.4 a) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 2.7 ) P = 0. compressor motor size d.3 1 h 2i = 700.60 c) d) e) 3 1 h 1 = 252.292 ( 434 – 409 ) = 7. refrigerant flow rate b.4 W 14.7 ) ] = 77.4 kW c) W = m ( h 2 – h 1 ) = ( 0.0675 ) [ 738 – 525 ] = 14. v 2i = ------.0275 ) m = ------------------------------------------.0675 kg/s 0. Sketch and label the system.30 kW 50 COP c = ------.0275 m ⁄ s 4 b) ( 0. Inc.0675 [ 525 – ( – 620. the actual volumetric efficiency is given by: ηva = 94 – 6.292 kg/s ( 409 – 237. refrigerating capacity c. Additional reproduction. .5 COP = -----.7 a) t 2i = 55°C b) 50 m = --------------------------------. (www.7485 ) ( 0.= 0.4. compressor discharge temperature e.123 t 2i = 95°C 8.1 d) 1. COPc Copyrighted material licensed to University of Toronto by Thomson Scientific. For personal use only.303 m ⁄ kg 3.1(pd/ps).= 6.43 t 2a ≈ 110°C 2 3 2π ( 0. including appropriate values for the thermodynamic properties. A compressor in this line has two cylinders.3 ( 434 – 409 ) h 2a = 409 + ---------------------------. Inc. and Air Conditioning—Solutions Manual h 1 = 409 s 1 = 1.techstreet.com). The system is being selected for an air-conditioning unit and will therefore operate between an evaporating temperature of 0°C and a condensing temperature of 35°C. v 2a = ------.22⏐Principles of Heating.82 ⎠ 8.1 ⎛⎝ ----------⎞⎠ = 74. (www. 4 ) = 1115. (www. © (2009). h 1 = h g @ – 4°C = 494 h 2 = 650 h 3 = h 4 = – 621.99 Hp This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 2. s 1 = 1.org).03 ) ⁄ 0.1 ) = 9. and operates at 1725 rpm.ashrae. Inc. h 2 ≅ 432. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. COP Sketch and label a p-h diagram showing values. For the ideal standard vapor compression cycle. refrigerating effect b.⎛⎝ ---------⎞⎠ ⎛⎝ ------------⎞⎠ = 0.0147 ⎠ 2 4 1725 3 π(5) Piston Displacement = ------------------2.4 ) COP = -----.0147 68 h 3 = 256.– 1⎞ = 100 – 4 ⎛ ------------------.7349.062 ( 432 – 408.Chapter 2—Thermodynamics and Psychrometrics⏐23 2. and the required motor size. v 2 = -----.746 = 1. Additional reproduction.– 1⎞ = 94. 1 s 2 = 1.= ----------------.4 a) b) Q e = ( h 1 – h 4 ) = 494 – ( – 621. determine a. Clearance volume is 4%.15 h2 – h1 W ( 650 – 494 ) 1. .002258 ) Flow Rate = ---------------------------------------------. Copyrighted material licensed to University of Toronto by Thomson Scientific. h 1 = 408.= 7.002258 m ⁄ s 100 60 4 ( 100 ) ( 0. (www.42 kW W = 0.com).062 ( 408. v 1 = 0.6% ⎝ v2 ⎠ ⎝ 0.3.4 kJ/kg h1 – h4 Qe ( 1115. American Society of Heating.03462 3.03462 ) Q e = 0.1 2. Refrigerating and Air-Conditioning Engineers.03 – 256.946 ) ( 0. Estimated volumetric efficienty based on re-expansion of TRAPPED GAS.7349. Inc. h 4 = 256. Determine as close as possible the actual refrigerating capacity.techstreet. if the compressor is used in a system operating between 10°C and 40°C. distribution.062 kg/s ( 0. a 40 mm stroke. in hp. respectively.= 0. kW.1 4. evaporating and condensing temperatures.= 0. v1 0.43 An ammonia refrigerating system is operating with a condensing temperature of 30°C and an evaporating temperature of −4°C.03462 η v = 100 – cv ⎛ ----.44 A single-cylinder R-22 compressor has a 50 mm bore. For personal use only.= ---------------------------. 10 mm 52 0 1084.65 4.com).45 For the lithium-bromide/water absorption refrigeration system shown below.24⏐Principles of Heating.techstreet.197 0.= 0.98m 1 0.50 –70 9.0 ) = 0. heat rejection ratio.ashrae.= (---------------------------------------= 3.98m 1 .65 ( – 41 ) – 0.5 – 68.85 m 3 = 9.5 0. 50 mm 150 0 1128 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. American Society of Heating.96 0.65 ) ( – 41 ) – ( 9. heat required at generator per ton of cooling b.197 ) ( 1128 ) + ( 9.85 2. Inc.96 – 1128 ) = – 12520 Btu/h = – 209 Btu/m QR 30450 + 12520 )------.org).197 3.51 m 1 = 0. © (2009).51 –41 9. 50 mm 150 0. (Qabsorber + Qcondenser)/Qevaporator State P t x h m 1.58 Qe ( 12000 ) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 12000 m 2 = m 4 = m 5 = -------------------------------------------60 ( 1084. m 1 = 9.5 ) ] c) = 30450 Btu/h = – 508 Btu/min Q c = 60 ( 0. For personal use only.197 Copyrighted material licensed to University of Toronto by Thomson Scientific.65 a) Q g = 60 [ ( 0. and Air Conditioning—Solutions Manual 2. 50 mm 101 0 68.197 ) ( 68.197 + 0. Ventilating.85 ( – 70 ) – 9.197 5. Additional reproduction. Inc.50 m 3 = ⎛⎝ ----------⎞⎠ mi = 0. .197 ( 1084. distribution. Refrigerating and Air-Conditioning Engineers.85 ) ( – 70 ) ] b) = 30964 Btu/h = 516 Btu/min 12000 COP = --------------. 10 mm 90 0. determine a. (www.39 30964 Q a = 60 [ 9. (www. COP c. 62 ) m 1 = 4. American Society of Heating.74 m 1 x 1 = m 2 x 2 + ( m 1 – m 2 )x 3 m 1 ( 0.0 ) ( 0.Chapter 2—Thermodynamics and Psychrometrics⏐25 2.9 mm 170°F 0.16.74 ) – ( 56. m 3 = 3. © (2009). .2 mm 75°F 0.47 For the aqua-ammonia absorption refrigeration system shown in the sketch below.7 lbm/h 1025.34 Btu/lb Q a = – 14800 Btu/hr Q g = ( 1 ) ( 1138 ) + ( 3.16 ) ( – 38 ) – ( 4. distribution.7 Q c = ( 1 ) [ ( 56. (www.47 ) = ( m 1 – 1.16 ) ( – 38 ) – ( 1 ) ( 1081.7 Btu/lb Q c = – 12659 Btu/hr Q a = ( 4.46 In the basic lithium-bromide water absorption system.62 –38 4.6 Btu/lb Q e = 15460 Btu/hr Qa + Qc ------------------. 8. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Per ton of refrigeration 12000 m 2 = m 4 = m 5 = ---------------. the generator operates at 170°F while the evaporator is at 47°F.47 –73 2.techstreet.28 Qe 2.04 5. Inc.com). State P t x h 1.74 ) = – 1265. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Inc.16 ) ( – 73 ) = 1321.96 Btu/lb Copyrighted material licensed to University of Toronto by Thomson Scientific. 33.04 ) ] = 1025. -- 88°F 0 56. (www.16 ) ( – 73 ) – ( 3.= 2. Additional reproduction. 8.org).2 mm 47°F 0 1081. Refrigerating and Air-Conditioning Engineers. Calculate the heat rejection ratio for these conditions. The absorbing temperature is 75°F and the condensing temperature is 88°F.9 mm 170°F 0 1138 3. complete the table of properties.04 ) – ( 1138 ) ] = – 1081. For personal use only. 33.= 11.16 Q e = ( 1 ) [ ( 1081.ashrae. 999 550 2.26⏐Principles of Heating.= 0. Point p. psia t. Inc.24 ) = 36.com). Btu/lb 1 25 80 0. Experiments indicate that about 200 Btu/h·ft2 of energy can be collected when the plate is operating at 190°F. °F x. and the engine will reject energy as heat to the atmosphere. dew-point temperature mass of water vapor contained in the room This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Q R = ToΔs = ( 530 ) ( 0. . be transferred as heat to a fluid within a heat engine. Inc. distribution.= 92.26 165 For minimum area. in turn.= --------. For personal use only.7°C). This energy will. The barometric pressure is standard and the partial pressure of the water vapor is measured to be 0.74 ft ⁄ kW 36. © (2009).4 m) room contains an air-water vapor mixture at 80°F (26. when the atmospheric temperature is 70°F. relative humidity b. lb NH3/lb mix h. (www. and Air Conditioning—Solutions Manual 3 200 260 0. Ventilating.ashrae. Refrigerating and Air-Conditioning Engineers.992 606 6 200 97 0. USE CARNOT CYCLE QA 200 Δs = ------.8 Btu/h ⋅ ft 2 3413 2 area = -----------. Calculate a.2 psia (1.techstreet. (www.85 735 4 200 160 0.38 kPa).8 2. Additional reproduction. maximum efficiency.org).992 75 7 25 20 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. d.24 Copyrighted material licensed to University of Toronto by Thomson Scientific. Estimate the minimum collector area that will be required for a plant producing 1 kW of useful shaft power.54 40 5 200 160 0. American Society of Heating.6 by 2. humidity ratio c.308 TH 650 W = Q A – Q R = ( 200 ) – ( 163.48 Solar energy is to be used to warm a large collector plate.39 –48 2 200 260 0.50 A 20 by 12 by 8 ft (6.1 by 3.308 ) = 163. 2 Btu/lb m c) t d = t sat @ 0.= 0.= ---------------.258 ) ( 144 ) e) 0.5073 b) pw 0. enthalpy c. For personal use only.0111 ) [ ( 1061 ) + 0.51 Given room conditions of 75°F (23.43 w s = 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.= ( 0.43 ) = 0.0187 14.62198 ) ⎛⎝ ------------------------------⎞⎠ = 0.= 0.72 ft /lbm Pa ( 14.258 w = 0. enthalpy [Ans: 30.4% p ws 0. © (2009). distribution.2°F b) h = 30.6 ) ( 0.3 ) ( 535 ) 3 v = ---------.62198 --------------.7 – 12⎠ p – pw c) t d = t sat @ 0.2 ) ( 144 ) ( 1920 ) m w = ----------------------. degree of saturation a) pw 0.19 lb m ( R ⁄ m w )T ( 1545 ⁄ 18 ) ( 80 + 460 ) 2.com).0086 lb m /lb air ⎝ 14.258 = 60.45t ) = ( 0.394 = 39.62198 ⎛⎝ ---------------------------⎞⎠ = 0.0111 μ = -----. (www.2°F (18.24t + w ( 1061 + 0.593 ws 0. using the ASHRAE Psychrometric Chart.2 Btu/lbm (70.2 φ = -------.4°C)] b. dew-point temperature d.7 – 0.258 ) 2.0112 kg/kg)] a) t wb = 65. find a.= ( 0. humidity ratio [Ans: 0.= ------------------------------------------------.Chapter 2—Thermodynamics and Psychrometrics⏐27 a) pw 0. humidity ratio b.62198 -----.7 – 0.45 ( 75 ) ] = 30.= ----------------------------------------------------.9°C) dry bulb and 60% RH.2 w = 0.0112 lb/lb (0.2 PSI = 53. .2 J/g)] c.43 ( p w = φp s = ( 0. wet-bulb temperature [Ans: 65. Inc.org).52 For the conditions of Problem 2. (www.0112 lb m /lb air This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto w 0. determine for the air vapor mixture without using the psychrometric charts a.= ---------------.0111 lb m /lb air pa 14. Inc.24 ) ( 75 ) + ( 0.7 – 0.2 Btu/lb m c) w = 0. Refrigerating and Air-Conditioning Engineers.15°F d) Pw V ( 0.techstreet.= 1.0187 Copyrighted material licensed to University of Toronto by Thomson Scientific.185°F d) RaT ( 53. American Society of Heating.258 b) h = 0. specific volume e.= 13.62198 ) ⎛ ----------------------⎞ = 0. Additional reproduction.ashrae.51 (above). 8 28. .801 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 100 Copyrighted material licensed to University of Toronto by Thomson Scientific.75 89. Btu/lbair Specific Volume. °C Humidity Ratio.5 0.0088 40 28.88 2.7 85 70 62 0.86 40 29 11 0.54 Using the ASHRAE Psychrometric Chart complete the following table: Dry Bulb.7 29.6 2. Inc.1 13.902 66 90.8 70 55 43 0. Relative Humidity.2 0. Ventilating. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.2 0.0115 26 70 0.0207 40 95 0.28⏐Principles of Heating.005 18 44.875 27 27 27 0. For personal use only.56 Complete the following table.4 21 0.0076 26.8 86 60 40 0.4 13.8 13.87 4 –2 –10 0.5 0.0058 38 23.012 47 33. kJ/kg Specific Volume.2 25.91 30 72.org). and Air Conditioning—Solutions Manual 2. °F Wet Bulb.011 60 30 13.7 0. ft3/lbair 78 70 0.006 40 36. complete the following table. °F Dew Point. °F Dew Point. Relative Humidity.5 53.0227 100 85.53 Using the ASHRAE Psychrometric Chart.001 20 6.ashrae. kg/kg % 32 24 20.85 32 16 4 0.55 Complete the following table using the Psychrometric Chart.techstreet.01 60 48 0.7 0. Enthalpy h.0054 75 60 50 0. distribution. Dry Bulb.886 34 81 0. m3/kg 2.0130 33.4 0.47 Enthalpy.3 12.92 22.5 0.1 22 13.022 7 7 7 0.9 100 23 0. Refrigerating and Air-Conditioning Engineers.5 65 59.84 38 25.78 39. Relative Humidity.5 70 61 0. Dry Bulb.8 0.0155 40 26.5 0.7 0. kg/kg % 26.4 42 13. Inc. °C Dew Point.2 18 0.5 17. (www. Relative Humidity φ. lbv/lba % 80 63.1 99 85.2 0.0063 Enthalpy.001 20 10.3 79 65 57 0.2 17 14.905 41. m3/kg 52 72.1 26.904 30 21 17 0. Additional reproduction. °F Humidity W. Btu/lbair % ft3/lbair 26. °C Wet Bulb.5 0.012 45 61 0.85 Dry Bulb.8 23. °F Wet Bulb.8 14. °C Humidity Ratio.3 21. American Society of Heating.65 74.8 24. (www.47 97 77 68 0.0 80 80 80 0.8 14.2 0.com).6 0.0052 20 26.0157 40 40 14. lb/lbair 85 60 41 0.4 16 0.0087 41 49 0.01 46 30 13. © (2009). Specific Volume v.3 12. °C Wet Bulb.0155 38 78. °C Dew Point. kJ/kg Specific Volume. °F Humidity Ratio.0224 100 43.016 39 42 14.86 21 13 7.01143 34 38 14.011 30 60 13.2 0.6 74 65 60 0.1 Enthalpy.0160 38.0238 50 58 14.2 0.5 82 0. = 0.Chapter 2—Thermodynamics and Psychrometrics⏐29 2. relative humidity φ = 59%] @ 90F P g = 0.7 – 0.622 -----------------.018 lb/lb.475 PSI P ws 0. [Ans: W = 0.4119 psia ( 0.622 + 0.= 0.com).6989 Copyrighted material licensed to University of Toronto by Thomson Scientific.018 lb m /lb air (P )( W) P w = -----------------------.59 or 59% 0.622 ) ⎛ ---------------------------------⎞ = 0.02077) 1093 – 0.0179 )---------------------------------------= 0. The barometric pressure is 14.622 + W Pw φ = --------. Additional reproduction. Inc.7 psia.24 ( t – t* ) W = -----------------------------------------------------------------------------------(t∗ = 78. © (2009).org). determine the humidity ratio and relative humidity of an air-water vapor mixture with a dry-bulb temperature of 90°F and thermodynamic wet-bulb temperature of 78°F. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.= ( 0. Refrigerating and Air-Conditioning Engineers. American Society of Heating. For personal use only. Check your result using the psychrometric chart.= P ws ( 14.4748 W* = 0. W∗ = 0. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .44t – t* W = 0.4119 ---------------. (www.ashrae. (www.556t* )W* – 0.0179 ) 0.6989 PSI @ 78F P g = 0.7 ) ( 0.techstreet. Inc.57 Without using the psychrometric chart.4748⎠ P – P ws ( 1093 – 0.02077 ⎝ 14. distribution. © (2009). Additional reproduction.Copyrighted material licensed to University of Toronto by Thomson Scientific. (www.com). Inc. For personal use only.ashrae. distribution. Refrigerating and Air-Conditioning Engineers.techstreet. Inc. American Society of Heating.org). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . (www. Additional reproduction. Refrigerating and Air-Conditioning Engineers.Solutions to Chapter 3 BASIC HVAC SYSTEM CALCULATIONS Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc. For personal use only. (www. Inc.org).com). distribution. (www.ashrae. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.techstreet. American Society of Heating. © (2009). This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . (www. American Society of Heating. distribution. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Additional reproduction.org). Inc. Inc.techstreet. (www.com).Copyrighted material licensed to University of Toronto by Thomson Scientific. Refrigerating and Air-Conditioning Engineers. For personal use only.ashrae. © (2009). 4 ) = 0.4 ) a) W = 0. dew point c.3 A room of dimensions 4 m by 6 m by 2.( P – Pw ) P – Pw 0.( 14. Calculate: a.4 m contains an air-water vapor mixture at a total pressure of 100 kPa and a temperature of 25°C. total mass of water vapor in the room 3 V = ( 4 ) ( 6 ) ( 2.000017 kg/kg)? To what temperature must this air be cooled if its pressure is 10 atm? Pw W W = 0.7 – Pw ) 0.0088 kg ⁄ kg air P – Pw ( 100 – 1.0000161 kPa : T < – 60°C 3.1 One of the many methods used for drying air is to cool the air below the dew point so that condensation or freezing of the moisture takes place.4 kg m w = Wm a = ( 0.0001 = 0.4 ) b) @ Pw = 1.000017 10 ATM = 147 PSI ⇒ Pw = ---------------------.techstreet. Refrigerating and Air-Conditioning Engineers.000402 – 0.0000273Pw Pw = 4. Chap 1 HF] 0.= ------------------------------. Suppose an experiment requires a humidity ratio of 0.6 ) ( 57.287 ) ( 298.1 kPa in order to achieve this humidity? Copyrighted material licensed to University of Toronto by Thomson Scientific.4 kPa Pw ( 0.com).2 One method of removing moisture from atmospheric air is to cool the air so that the moisture condenses or freezes out.Chapter 3—Basic HVAC System Calculations⏐33 3.62198 0. Inc.02 × 10 –3 PSI ⇒ t sat = – 27°F Pg φ = 100% → Pw = Pg : W = 0.02 × 10 –4 PSI ⇒ t sat = – 63°F [Table 3. To what temperature must atmospheric air be cooled in order to have a humidity ratio of 0.000017 Pw = ---------------------.0001.62198 ⎛⎝ -----------------⎞⎠ ⇒ Pw = ------------------. Pw = 1.= 0. To what temperature must the air be cooled at a pressure of 0.62198 Pw = 4.2 ) = 66.6 ) ⁄ ( 0. © (2009). humidity ratio b. The partial pressure of the water vapor is 1.6 m .8°C c) m a = PaV ⁄ RaT = ( 98.4 kPa t sat = Dew point = 11.( 147 – Pw ) 0.622 ⎛⎝ --------------------⎞⎠ 0.ashrae. American Society of Heating. Additional reproduction. (www.1 – Pg Pg = 0.org). (www.4 ) = 57. distribution.584 kg This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 3. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.000017 lb/lb (0. P = 100 kPa . . For personal use only.62198 Pw = 0.4 kPa. Inc.622 ) ( 1.0088 ) ( 66.622 ----------------. 005 )287 1 2 m· 3 = 3. The rate of flow of dry air is 287 lbm/min. (www.02 – 0.009 ) ( 22 ) ] Q° = – 5860 Btu ⁄ min This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 3. and the vapor at exit has an enthalpy of 1085 Btu/lbm. the amount of moisture removed from the air (lbm/ min) b.7. The dry air at exit has an enthalpy of 13.com).7 – 0. φ = 50% Pws 1 = 0. © (2009).02 lbm of vapor per pound of dry air at entrance and 0.6 Btu/lbm of dry air and 1100 Btu/lbm of water vapor. Ventilating.009 lb m ⁄ lb air m· 3 = m· w – m· w = ( W 1 – W 2 )m· a = ( 0.org).2 Btu/lbm.009 lbm of vapor per pound of dry air at exit. Determine: a. Inc. the rate of heat removal required Copyrighted material licensed to University of Toronto by Thomson Scientific.618 psia 50 – x φ 3 = Pw 3 ⁄ Pws 3 = 100 ⇒ Pws 3 = Pw 3 = 0. Additional reproduction.16 lb m ⁄ min ° [ ( h – h ) + ( W h – W h ) + ( W – W )h ] b) Q° = m a a a 2 w 1 w 1 2 w 2 1 2 1 3 = 287 [ ( 13.1°C). Refrigerating and Air-Conditioning Engineers.6 ) + ( 0.1816 2 P 2 = 50 psi 3 P 3 = 50 psi.1816 W 1 = 0.618 → t 3 = 86°F a) m· a = 287 lb m ⁄ min hw 3 = 22 Btu ⁄ lb min ha 1 = 21. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.622 ------------.02 – 0. There are 0. t 1 = 70°F.techstreet.00778 lb/lb Pws 1 14.34⏐Principles of Heating.622 ⎛⎝ ---------------------------------⎞⎠ = 0. and 14. . Condensate leaves with an enthalpy of 22 Btu/lbm. The air is compressed to 50 psia (344.2 – 21. If condensation of water vapor from the air is to be prevented.00778 → x = 0. then sent to an intercooler.2 Btu ⁄ lb m hw 1 = 1100 Btu ⁄ lb m hw 2 = 1085 Btu ⁄ lb m W 1 = 0.= 0.1816 x W 2 = W 1 = 0.009 ) ( 1085 ) – ( 0.4 The air conditions at the intake of an air compressor are 70°F (21. and Air Conditioning—Solutions Manual 3. American Society of Heating. t 3 = ? Pw 1 0.6 Btu ⁄ lb m ha 2 = 13. distribution. For personal use only.02 ) ( 1100 ) + ( 0.622 ⎛⎝ --------------⎞⎠ = 0. φ 3 = 100%.3 kPa). (www.3632 Pw 1 = φPws = 0. 50% RH.ashrae. what is the lowest temperature to which the air can be cooled in the intercooler? @ 1 P 1 = 14.7 psia (101.7 kPa).02 lb m ⁄ lb air W 2 = 0.5 Humid air enters a dehumidifier with an enthalpy of 21. Inc. Chapter 3—Basic HVAC System Calculations⏐35 3.6 Air is supplied to a room from the outside, where the temperature is 20°F (−6.7°C) and the relative humidity is 60%. The room is to be maintained at 70°F (21.1°C) and 50% RH. How many pounds of water must be supplied per pound of air supplied to the room? t 1 = 20°F t 2 = 70°F φ 1 = 60% φ 2 = 50% W 2 = 0.0078 lb m ⁄ lb air Pw 1 = φPg 1 = ( 0.6 ) ( 0.0505 ) = 0.0303 P1 ( 0.0303 ) W 1 = 0.62198 --------- = ( 0.62198 ) ------------------------------------- = 0.00128 Pg 1 ( 14.7 – 0.0303 ) mw m a w 1 + m w = m a w 2 → ------- = w 2 – w 1 1 2 ma m ------w- = 0.0078 – 0.0013 = 0.0065 lb m ⁄ ( lb air ) ma a) φ 1 = 49% b) t i * = 41°F c) W = 0.0054 lb m ⁄ lb air d) h i = 20.3 Btu ⁄ lb m, e) h e = 25.2 Btu ⁄ lb m f) q = 4.9 Btu ⁄ lb g) φ f = 25% This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 3.7 Air is heated to 80°F (26.7°C) without adding water, from 60°F (15.6°C) dry-bulb and 50°F (10°C) wet-bulb temperature. Use the psychrometric chart to find: a. relative humidity of the original mixture b. original dew-point temperature c. original humidity ratio d. initial enthalpy e. final enthalpy f. the heat added g. final relative humidity Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 36⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual 3.8 Saturated air at 40°F (4.4°C) is first preheated and then saturated adiabatically. This saturated air is then heated to a final condition of 105°F (40.6°C) and 28% RH. To what temperature must the air initially be heated in the preheat coil? t 2 = 101°F 3.9 Atmospheric air at 100°F (37.8°C) dry-bulb and 65°F (18.3°C) wet-bulb temperature is humidified adiabatically with steam. The supply steam contains 10% moisture and is at 16 psia (110.3 kPa). What is the dry-bulb temperature of the humidified air if enough steam is added to bring the air to 70% RH? 1. Air t = 100F t* = 65F h = 29.8 W = 0.0052 2 3. Air φ = 70% h f = 184.5 m· s = m· a ( W 3 – W 1 ) 2 m· a ( h 3 – h 1 ) = m· s ( h s ) = m· a ( W 3 – W 1 ) ( h s ) 2 2 h3 – h1 Δh h s = --------------------- = --------- = ( 184.5 ) + ( 0.9 ) ( 1152.1 – 184.5 ) W3 – W1 ΔW Δh-------= 1055 Btu ⁄ lb m ΔW Using Psychrometric Chart Δh through Pt. 1. along --------- = 1055 and 70% ΔW t db = 96°F 3.10 The conditions on a day in New Orleans, Louisiana, are 95°F (35°C) dry-bulb and 80°F (26.7°C) wet-bulb temperature. In Tucson, Arizona, the air conditions are 105°F (40.6°C) dry-bulb and 72°F (22.2°C) wet-bulb temperature. What is the lowest air temperature that could theoretically be attained in an evaporative cooler at these conditions in these two cities? New Orleans 80°Fwb Tucson 72°Fwb This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto · ma 1 h 1 + m· s h s = m· a h 3 2. Steam x = 0.90 P = 16 psia h g = 1152.1 Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Chapter 3—Basic HVAC System Calculations⏐37 3.11 Air at 29.92 in. Hg enters an adiabatic saturator at 80°F dry-bulb and 66°F wet-bulb temperature. Water is supplied at 66°F. Find (without using the psychrometric chart) the humidity ratio, degree of saturation, enthalpy, and specific volume of entering air. Pw 1 s * W s = 0.62198 ⎛⎝ --------------------⎞⎠ P – Pws Pw 1 s @ t = 66F = 0.31636 PSI ( 0.62198 ) ( 0.31636 ) * W s = ------------------------------------------------- = 0.01368 lb m ⁄ lb aire ( 14.696 – 0.31636 ) * ( 1093 + 0.556t* )W s – 0.24 ( t – t* ) W = ------------------------------------------------------------------------------------1093 + 0.444t – t* (------------------------------------------------------------------------------------------------------------------------1093 + ( 0.556 ) ( 66 ) ) ( 0.01368 ) – ( 0.24 ) ( 80 – 66 -) W = = 0.01044 1093 + ( 0.444 ) ( 80 ) – 66 W 0.01044 μ = ------- = ------------------- = 0.467 Ws 0.01368 h = 0.240t + W ( 1061 + 0.444t ) = ( 0.24 ) ( 80 ) + ( 0.01044 ) ( 1061 + ( 0.444 ) ( 80 ) ) = 30.65 Btu ⁄ lb RaT ( 53.3 ) ( 540 ) v = ---------- ( 1 + 1.6078W ) = ----------------------------- ( 1 + ( 1.6078 ) ( 0.01044 ) ) P ( 14.7 ) ( 144 ) 3 v = 13.832 ft ⁄ lb m 3.12 An air-water vapor mixture enters an air-conditioning unit at a pressure of 150 kPa, a temperature of 30°C, and a relative humidity of 80%. The mass flow of dry air entering is 1 kg/s. The air-vapor mixture leaves the airconditioning unit at 125 kPa, 10°C, 100% RH. The moisture condensed leaves at 10°C. Determine the heat transfer rate for the process. 3.4 W 1 = 0.622 ⎛ ----------------------⎞ = 0.0144 ⎝ 150 – 3.4⎠ 1.228 W 2 = 0.622 ⎛ ----------------------------⎞ = 0.0062 ⎝ 125 – 1.228⎠ m a ha 1 + m a W 1 h w – m a ha 2 – m a W 2 h w – m a ( W 1 – W 2 )h f + Q = 0 1 2 3 m a [ ( ha 1 – ha 2 ) + W 1 hg 1 – W 2 hg 2 – ( W 1 – W 2 )h f ] + Q = 0 3 Q = – 1 [ ( 1.0035 ) ( 20 ) + ( 0.0144 ) ( 2556.3 ) – ( 0.0062 ) ( 2519.8 ) – ( 0.0144 – 0.0062 ) ( 42.01 ) ] Q = – 41.0 kJ/s = – 41.0 W This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Pw 1 = φPg 1 = ( 0.8 ) ( 4.25 ) = 3.4 kPa Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 38⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual 3.13 Air at 40°C, 300 kPa, with a relative humidity of 35% is to be expanded in a reversible adiabatic nozzle. How low a pressure can the gas be expanded to if no condensation is to take place? What is the exit velocity at this condition? S1 = S2 Isentreopic Pw 1 = φ 1 Pg 1 = ( 0.35 ) ( 7.384 ) = 2.584 kPa 2.584 w 1 = 0.622 ⎛ ----------------------------⎞ = 0.0054 ⎝ 300 – 2.584⎠ Pg 2 W 1 = W 2 if no condensate. ⇒ 0.0054 = 0.622 ⎛ ---------------------⎞ ⎝ P 2 – Pg 2⎠ P 2 = 116.18Pg 2 P2 S 2 = S 1 ; T 2 = T 1 ⎛ ------⎞ ⎝ P 1⎠ k---------– 1k 116.18Pg 2 0.286 = 313.2 ⎛ --------------------------⎞ ⎝ 300 ⎠ Trial and Error → Try T 2 = 17.2°C = 290.4 K ; Pg 2 = 1.984 ? ( 116.18 ) ( 1.984 ) 0.286 290.4 = 313.2 ⎛⎝ ----------------------------------------⎞⎠ = 290.5 300 P 2 = ( 116.18 ) ( 1.984 ) = 230.5kPa 2 2 2 V2 V1 ⎞ ⎛ V2 ⎞ ⎛ ⎜ h 1 + --------⎟ – ⎜ h 2 + --------⎟ = 0 ; -------------------- = 1.0035 ( 313.2 – 290.4 ) 2g 2g 2 ( 1000 ) ⎝ c⎠ ⎝ c⎠ V 2 = 214 m/s v = va + μ ( vs – va ) v = ( 1 – μ )v a + μv s W W = ⎛⎝ 1 – -------⎞⎠ v a + ⎛⎝ -------⎞⎠ ( v s ) Ws Ws Ra T ⎧ ⎫ v a = --------⎪ ⎪ P ⎪ ⎪ ⎪ ⎪ Ra T ⎪ v s = -----------------⎪ P – P ws ⎪ ⎪ ⎨ ⎬ W⎪ ⎪ -----= μ ⎪ ⎪ Ws ⎪ ⎪ ⎪ ⎪ P ⎪ w = 0.622 ---------------- ⎪ P – P ⎩ w⎭ P – Pw Ra T Ra T P – P ws P – P ws Pw Pw W Ra T W = 1 – ⎛ ----------------⎞ ⎛ ----------------⎞ --------- + ------- ------------------ = ------------------ 1 – ⎛ ----------------⎞ ⎛ ------------------⎞ ------------------ + ------⎝ P – P w⎠ ⎝ P ws ⎠ P ⎝ P – P w⎠ ⎝ P ws ⎠ W s P – P ws P – P ws P Ws Pw P – P ws P – P ws Pw R a T P – P ws = ------------------ ------------------ – ⎛⎝ -------------⎞⎠ ⎛⎝ ------------------⎞⎠ ( P – P ws ) + ⎛⎝ ----------------⎞⎠ ⎛⎝ ------------------⎞⎠ P – Pws P P ws P ws P P – P w P – Pw 2 R a T ( P – Pws ) ( P ws ) ( P – P w ) – P w ( P – P ws ) + P w ( P – P ws )P = ------------------ --------------------------------------------------------------------------------------------------------------------------------------------P – P ws P ( P ws ) ( P – P w ) Ra T R a T PP ws ( P – Pws ) = ------------------ ------------------------------------- = ---------------P – P ws PP ws ( P – P w ) P – Pw This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 3.14 By using basic definitions and Dalton’s Law of partial pressure, show that v = RaT/(p – pw) Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 0 tons td = 54.380. © (2009). Determine a. the dew point of the air leaving the conditioner 1.3 tons .7°C) dry bulb and 50% RH are mixed with one pound of air at 50°F (15. Inc.= 143.900 Btu/h = 143. American Society of Heating.0 3 13.16 Four pounds of air at 80°F (26.2 tons 12000 Q L = m· w ( h fg ) = ( 1287 ) ( 1076 ) = 1.244 ) ( 80 – 57 ) s a p 1 2 = 1. Additional reproduction. Inc. h ≈ 25 φ = 60% φ = 90% v = 13.33 SCFM = 71000 ⎛⎝ -------------⎞⎠ = 67900 h c = 29.1 ( SCFM ) ( t 1 – t 2 ) = --------------------------------------------. and standard atmospheric pressure. Calculate the following: a. latent heat load on the conditioner.1 ) ( 67.Chapter 3—Basic HVAC System Calculations⏐39 3.900 ) ( 23 ) = 1.385. 2.4 tons Copyrighted material licensed to University of Toronto by Thomson Scientific. in Btu/h b.475 ) ( 0.5 ) – ( 0.techstreet. .8 tons · m w = m· a ( W 1 – W 2 ) = ( 306.or - e) . enter the unit.org).009 ) = 1287 lb/h Q = m· c ( t – t ) = ( 306. 71.6°C) and 50% RH. rate of water removal from the unit c.009 )25 ] Q = – 3.2°F 3. Refrigerating and Air-Conditioning Engineers.719.000 = 115.5°F This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto d) ( 1.9 V· ( 71000 ) ( 60 ) m· a = --.com). (www. relative humidity of the mixture b.475 ) ( 0.009 W = 0.000 cfm at 80°F dry bulb.009 ) = 1. For personal use only.9 h = 23.0132 – 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. t = 80F t = 51F 3.2 ( h f @ 57°F ) 71000 CFM v = 13.or · = 4840 ( SCFM ) ( Δw ) = ( 4840 ) ( 67.000 Btu/h = 115.093.0132 – 0.= ------------------------------. sensible heat load on the conditioner.ashrae.15 In an air-conditioning unit. The air leaves the unit at 57°F dry bulb and 90% relative humidity. cooling capacity of the air-conditioning unit.7 W = 0.900 ) ( 0.9 m· a [ ( h 1 – h 2 ) – ( W 1 – W 2 )h 3 ] = – Q – Q = ( 306475 ) [ ( 33.5 h = 33. in Btu/h e.0132 a) b) c) 13.900 Btu/h = 257. in Btu/h d. (www. distribution.7 – 23. 60% RH.0132 – 0. dew-point temperature of the mixture by graphical solution φ mix = 52% t dp = 55.= 306475 lbm air ⁄ h v 13. 5°F 3.3576 ( 0.97 ) = 204700 Btu/h = 17. Inc.01266 W = 0. distribution.7°C).1°C) wet-bulb temperature. American Society of Heating.17 Air is compressed from 85°F.62198 ⎛ ------------------⎞ ⇒ P s = 1.696 – 0.62198 -------------------P – Pw 2 P w = φP g = ( 0.ashrae.18 An air-water vapor mixture flowing at a rate of 4000 cfm (1890 L/s) enters a perfect refrigeration coil at 84°F (28. For personal use only.00857 ) ( 21.[ ( 34 – 22 ) – ( 0. (www. 14.0155 lb/lb ( 14.00857 2 ( 4000 ) ( 60 ) Q = ---------------------------. © (2009).techstreet. Additional reproduction.7 psia to 60 psia and then cooled in an intercooler before entering a second stage of compression. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Inc.org). The air leaves the coil at 53°F (11. and Air Conditioning—Solutions Manual 3.40⏐Principles of Heating.9°C) and 70°F (21.97 34 0. 60% RH.3576 ) W = ---------------------------------------------. What is the minimum temperature to which the air can be cooled so that condensation does not take place? Pw 2 W 1 = W 2 = 0.6 ) ( 0.1 ) ] ( 13.0155 = 0.62198 ) ( 0. . (www. How many Btu/h of refrigeration are required? 1.596 ) = 0.01266 – 0.1 84F 70F 13. Ventilating. t = 53F saturated h = 22 h f = 21.3576 ) Ps 0. t = t* = v = h = W = 2.= 0.459 psia ⎝ 60 – Ps⎠ ⇒ t d = 114.com).1 tons This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto m· a [ – ( h 1 – h 2 ) – ( W 1 – W 2 ) h f ] = Q Copyrighted material licensed to University of Toronto by Thomson Scientific. Refrigerating and Air-Conditioning Engineers. 8°F This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 3.and wetbulb temperatures for the mixed airstream? Copyrighted material licensed to University of Toronto by Thomson Scientific.= 0.techstreet.= 30.3°F .1 h 2 = 40. American Society of Heating.0031 W 2 = 0.0096 ) w mix = -------------------------------------------------------------------------------.65 Btu/lb 6000 · · m o W o + m R W R = ( m· o + m· R )W mix ( 1000 ) ( 0. distribution. .19 Air at 40°F dry bulb and 35°F wet bulb is mixed with air at 100°F dry bulb and 77°F wet bulb in the ratio of 2 lb of cool air to 1 lb of warm air.5 m· 1 h 1 + m· 2 h 2 = m· h 3 → 2m· h 1 + m· h 2 ( 13.23 Btu/lb h 3 = ----------------------------3m· 3 · m· 1 W 1 + m· 2 W 2 = m 3 W 3 → 2m· W 1 + m· W 2 ( 0.9°C) and 52% RH. For personal use only.0148 h 1 = 13.org).20 Outdoor air at 90°F (32.018 ) + ( 5000 ) ( 0.0070 lb/lb air 3m· 3 m· o h o + m· R h R = ( m· o + m· R )h mix ( 1000 ) ( 41.ashrae.com). Inc. What are the dry.5. t mix = 65. t 1 = 40F * t 2 = 100F * t 1 = 35F m· 1 = 2m· t 2 = 77F m· 2 = m· W 1 = 0.4 ) + ( 5000 ) ( 28. Chart → t mix = 78.1 ) + 40. There are 1000 lb (454 kg) of outdoor air for every 5000 lb (2265 kg) of return air.6°C) wet bulb is mixed with return air at 75°F (23. (www. Additional reproduction. Compute the resultant humidity ratio and enthalpy of the mixed air.Chapter 3—Basic HVAC System Calculations⏐41 3. (www.0031 ) + ( 0.0148 -) .2°C) and 78°F (25. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.= 2--------------------------------------------------w 3 = --------------------------------= 0. Inc. © (2009).= 2---------------------------------= 22. Refrigerating and Air-Conditioning Engineers.5 ) h mix = ----------------------------------------------------------------------.011 lb/lb air 6000 * From Psych. 86 kPa 1 100 – P w 1 Pw 2. (www.169 kPa a) ma [ ( h1 – h2 ) + w1 hw 1 0.ashrae. and a pressure of 100 kPa.86 φ = --------1 = ------------. and Air Conditioning—Solutions Manual 3.246 1 b) φ = 67% .= 0. Additional reproduction.21 In a mixing process of two streams of air. dew-point temperature From Psych.= 0. © (2009). an adiabatic saturation temperature of 25°C. Determine the humidity ratio and relative humidity of an air-water vapor mixture that has a dry-bulb temperature of 30°C. Refrigerating and Air-Conditioning Engineers.3 kPa ) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 3.0 Btu/lb e) t DP = 60. American Society of Heating.018 kg/kg ( for 101.169 ) W 2 = --------------------------------. Calculate the following conditions after mixing at atmospheric pressure: a.0183 kg/kg 2556.622 ( 3.0% d) h = 32.67 ⇒ 67% Pg 4.= 0. dry-bulb temperature b.0035 ) ( 25 – 30 ) + 0. Inc.0204 ( 100 – 3.62189 ----------------------⇒ P w = 2. .42⏐Principles of Heating. Chart as graphical solution a) t dB = 81.techstreet. Use the psychrometric chart to determine the humidity ratio and relative humidity of an air-water vapor mixture that has a dry-bulb temperature of 30°C. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.com). a wet-bulb temperature of 25°C. enthalpy e.0112 lb/lb c) φ = 48. Inc.5°F b) W = 0. Ventilating. W = 0.000 cfm of air at 75°F and 50% RH mix with 4000 cfm of air at 98°F dry-bulb and 78°F wet-bulb temperature. and a pressure of 100 kPa.3 ) w 1 = ----------------------------------------------------------------------------------------.5°F Pw 2 = Pg 2 = 3. Copyrighted material licensed to University of Toronto by Thomson Scientific.89 Pw 1 0. relative humidity d.3 – 104.0204 ( 2442.169 ) – w 2 h w + ( W 2 – W 1 )h f ] = 0 2 2 c p ( T 2 – T 1 ) + W 2 ( hg 2 – h f ) 2 W 1 = ------------------------------------------------------------------hg 1 – h f 2 ( 1. b.0183 = 0. For personal use only. distribution.22 Solve the following: a. (www. humidity ratio c.org). 10. = 0.5 ) m a = ---------. (www.0255 ( 2414. © (2009).2 – 2423.7 ( 5. Supply air at 39°F is to absorb 100.techstreet.4 ) ] = – 277 kJ Qs + Σm· w h w Δh100000 ) + 35 ( 1100 -) ------.81 T 2 = 30°C : 3.94 kPa 1 1 @ dew point : P w = P g . For personal use only.169 ≠ 0. Determine the temperature at which condensation begins and the heat transfer for the process.6 – 14.94 ) Pa V ( 100 – 3.= (--------------------------------------------------= -----------------------------Δw 35 Σm· w = 3957 Btu/lb m From Psychrometric Chart φ supply = 90% t dp = 36°F m· a h 1 + Q s + m· w h w = m· a h 2 Qs + mw hw ( 100000 ) + ( 35 ) ( 1100 ) m· a = --------------------------. distribution.= 0.88 By Interpolation T 2 = 28.= ------------------------------------------. Inc.7165 ( 28.94 ) ( 0. Additional reproduction. How many pounds of dry air per hour are required? What should the dew-point temperature and relative humidity of the supply air be? Copyrighted material licensed to University of Toronto by Thomson Scientific.0255 kg/kg ( 100 – 3. American Society of Heating.org).24 A room is to be maintained at 76°F and 40% RH. W = constant . Inc.543 kg Ra T ( 0.com). and 70% RH is contained in a 0. Refrigerating and Air-Conditioning Engineers.0128 ( 303 ) = 3.Chapter 3—Basic HVAC System Calculations⏐43 3.000 Btu sensible heat and 35 lb of moisture per hour.0128 ( 298 ) = 3.2°C 0.2 ) Q = m ( u2 – u1 ) = ma cv ( T2 – T1 ) + ma W2 ug – ma W1 ug 2 1 = ( 0. (www. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.287 ) ( 308. P w = P g = ---------------2 2 m w RT 1 308.= 11260 lb/hr ( 26.ashrae.622 ( 3.0128T 2 Trial and Error Try T 2 = 25°C : 4.2 – 35 ) + 0.5 m3 closed tank.246 ≠ 0.628 ) = 3.23 An air-water vapor mixture at 100 kPa.543 ) [ 0. Assume the moisture has an enthalpy of 1100 Btu/lb.2 P w1 v = 0. . v = constant 2 2 Pw v m w RT 2 3.3 ) h2 – h1 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 3.94T 2 2 ----------. P w = φP g = 0.= ----------------.= --------------------------------------------------------. The tank is cooled until the water just begins to condense. 35°C.94 ) W 2 = W 1 = -----------------------------. For personal use only. Refrigerating and Air-Conditioning Engineers.000.000 ) ( SHR ) = -------------------------------------------------. Inc.0099 lb/lb ma ( 7000 ) ( 143. Determine the dry-bulb and wet-bulb temperatures of the leaving air. Air is supplied to the auditorium at 67°F.= 236.000 ) Q L = m· w h w = ---------------------------. in lb/h? b. a.000.443 lb/h ( 0. .= 143. Inc. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.com). and Air Conditioning—Solutions Manual 3.25 Moist air enters a chamber at 40°F dry-bulb and 36°F wet-bulb temperature at a rate of 3000 cfm. How much latent heat is picked up in the auditorium? d.000 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 3. What is the dew-point temperature of the entering air.= -----------.= (--------------------------------------------------------= -------------------------m· w 83 Δw m· a 1 = 2555 Btu/lbm V· 1 3000 = -----. the air absorbs sensible heat at a rate of 116.( 1100 ) ≈ h g ( 7000 ) d) 90°F (people) = 157. a sensible-heat load of 350. In passing through the chamber. (www. Chart t wb = 64°F a) b) Q s = m· a c p ( t R – t s ) 350.26 In an auditorium maintained at a temperature not to exceed 77°F.= 0. Ventilating.443 ) φ = 70% .244 ) ( 77 – 67 ) m· w 1.000 W s = W R – ------.7 83 m· a = m a + m w = 236. American Society of Heating. What is the sensible heat ratio? Copyrighted material licensed to University of Toronto by Thomson Scientific. © (2009). (www.6 ) ( 60 ) t 2 = 40 + 34 = 74°F From Psych.000.000 ) + 157. How much air must be supplied.6 lb/min 2 1 1 60 116000 Qs = m· c p ΔT ⇒ Δt = -----------------------------------------------. and a relative humidity not to exceed 55%.44⏐Principles of Heating.= 237.000 m· a = ----------------------------------------.= 34°F ( 0.69 ( 350.000 Btu/h and picks up 83 lb/h of saturated steam at 230°F.org). V· a = 3000 cfm Q s + m· w h w 116000 ) + 83 ( 1157 -) Δh.techstreet.ashrae.000 Btu and 1. distribution.0109 – ----------------------------------------.2 + -----.100 Btu/h ( 350.= 0.000 grains of moisture per hour must be removed. Additional reproduction.2 lb/min v1 12.240 ) ( 257. t dp = 57°F c) ( 1. and what is its relative humidity? c.= 0. 967 lb/h c p ( Δt ) ( 0.com).500 Btu/h = 25. distribution.4 ) + ( 29.92 in. Refrigerating and Air-Conditioning Engineers. American Society of Heating.Chapter 3—Basic HVAC System Calculations⏐45 3.= 35. in lb/h? b.000 ) ( 0. The space has a load of 200.000 Btu/h. what is the refrigeration load? Q L = m· ( w R – w s ) ( 1100 ) 120.000 Btu/h latent.and wet-bulb temperatures in the space? a) Q s = m· c p ( t R – t s ) b.techstreet. The barometric pressure is 29.000 Btu/h sensible. For personal use only. (www. Additional reproduction.000 Btu/h. If a mixture of 50% return air and 50% outdoor air at 98°F dry bulb and 77°F wet bulb enters the air conditioner.05 Btu/lb m 2 ≈0 m· a ( h m – h s ) – m· a ( w m – w s )h f + Q = 0 h s = 24. t* = 58°F ( 200. (www.7Btu/lb t wb = 64.05 – 24.= 0.000 ) . 50% RH. How much air must be supplied.28 A structure to be air conditioned has a sensible heat load of 20. What is the sensible heat ratio? a) b) Qs 200.000 Not possible to meet conditions with supply since SHR doesn't pass through both room and supply conditions.= 0.= 81.7 ) = – 310.000 ) ( w R – 0.24 ) ( t R – 60 ) t R = 76.6°F 30. a.7 Btu/lb m 60°F Q = – ( 30.7 ) h m = ------------------------------------. The temperature of the supply air to the space cannot be lower than 65°F dry bulb.000 = ( 30. and 200.967 ) a c) From Psych.50 ( 200. © (2009). Inc. is it possible to meet the load conditions by supplying air to the room at 100°F and 60% RH? If not.2 100.000 ) SHR = -------------------------------------------------.ashrae.01031 ⇒ h R = 29.000 Btu/h and a latent load of 30. What are the dry.29 A flow rate of 30. Copyrighted material licensed to University of Toronto by Thomson Scientific.244 ) ( 75 – 65 ) h v ≅ h g @ 85-95°F = 1100 Btu/lb m· w ( 200. What is the required wet-bulb temperature of the supply air? c.000 m· a = ---------------.000 Btu/h at a time when the total load is 100. discuss the direction in which the inside state would be expected to move if such air were supplied.= 0.0094 ) ( 1100 ) W R = 0.5°F m· a h a + m· R h R – ( m· a + m· R )h m ( 40. Hg.27 A meeting hall is maintained at 75°F dry bulb and 65°F wet bulb.011 – ------------------------------------= 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.= ----------------------------------------. 3.000 lb/h of conditioned air at 60°F and 85% RH is added to a space that has a sensible load of 120.000 SHR = ------------------.000 = ( 30. Chart → t = 65°F . If used room becomes warmer and/or less humid. If the inside state is to be at 80°F. Inc.000 ) 3. . a.9 tons This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 20.0088 lb/lb W s = W R – -----m· ( 1100 ) ( 81.000 + 200.org).000 ) ( 35. 00524 lb/lb m· R W R + m· OA W OA = ( m· R + m· OA )W m ⇒ Wm From Psychrometric Chart t m = 54. © (2009). ( 0. φ m = 58% This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Return Outside Air t = 75°F t = 40°F φ = 50% φ = 50% h = 28.0022 W 2 = W 1 + ------. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.4 ) ( 28.0027 + ---------------.5 ) h m = -------------------------------------------------------------(1) h m = 18. .0 kJ/kg 3 3.0092 ) + ( 0.5 W = 0.0247 m a ( 5°C. 100 kPa. The total airflow to the room is 60% outdoor and 40% return air by mass. the relative humidity at the outlet b.6 ) ( 12.31 A room is being maintained at 75°F and 50% RH.78Btu/lb ( 0.7 kJ/kg m a [ h 1 – h 2 + ( W 2 – W 1 )h f ] + Q = 0 3 h 2 = 93.1 kg/s.46⏐Principles of Heating. 100 kPa.2 ) + ( 0. the rate of heat transfer to the unit a) b) mw 0. For personal use only. The mixture leaves the unit at 30°C.1 [ 12.0092 W = 0.30 An air-water vapor mixture enters a heater-humidifier unit at 5°C.94 kJ/s = 7. Liquid water at 10°C is sprayed into the mixture at the rate of 0.6 ) ( 0.= 0.4 ) ( 0. Refrigerating and Air-Conditioning Engineers.5°F . Ventilating.94 kW h f = 42. 50% RH. Inc. Return air from the room is cooled and dehumidified by mixing it with fresh ventilation air from the outside.0026 m· R h R + m· OA h OA = ( m· R + m· OA )h m Copyrighted material licensed to University of Toronto by Thomson Scientific.org). and humidity content of the mixed air going to the room.7 – 93 + ( 0.com).0022 )42 ] = 7. The flow rate of dry air is 0.2 h = 12.= 0.1 and 30°C φ 2 = 91% h 1 = 12. The outside air conditions are 40°F and 50% RH at this time. American Society of Heating.50% ) 0. Calculate: a.0026 ) W m = -------------------------------------------------------------------------(1) = 0. (www.0022 kg/s. Inc. Additional reproduction.techstreet.0 kJ/kg Q = – 0.ashrae. Determine the temperature. (www. relative humidity. and Air Conditioning—Solutions Manual 3. distribution. = 5455 lb m /h ( 28. (www. Find the following: a.Chapter 3—Basic HVAC System Calculations⏐47 3. (www.m R = 0. Inc. American Society of Heating.405 m R h2 – hR 20. which is on the condition line for the room.m R = --------------------------.32 A room with a sensible load of 20.5% of m R This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Q s – 0.78Q s 20. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.000 Q L = ----------------------------.78 QL 5640 ------. the air-conditioning processes on the psychrometric chart b. The air is then mixed with some room return air so that the temperature of the air entering the room is at 60°F. For personal use only.000 Btu/h is maintained at 75°F and 50% RH.= ---------------.2 m 1 = 40. . Inc.org).3 – 28. airflow rate d. which is mixed.= 0.282 Qs 20.2 – 23. The outdoor air.5 ) 2 to 3: ( m R – m 1 )h 2 + ( m 1 )h R = ( m R )h 3 h2 – h3 20. ratio of latent to sensible load c. Refrigerating and Air-Conditioning Engineers.78 0.= --------------= 0.– 20.com).78 Qs + QL c) d) Room: mR h3 + QT = mR hR 20000 + 5640 m R = --------------------------------.techstreet.000 = 5640 0.3 – 23. This air is then cooled and dehumidified by a coil and leaves the coil saturated at 50°F.5 m 1 = ----------------. distribution.ashrae. is 25% by mass of the total flow going to the conditioner. the percent by mass of room return air mixed with air leaving the cooling coil b) Qs from Load line ------------------. © (2009). Outdoor air at 95°F and 80°F wet bulb was mixed with the room return air.000 Copyrighted material licensed to University of Toronto by Thomson Scientific. Additional reproduction. = ---------------.org).4 ) ( 144 ) ( 15 ) 1 m a = -----------= ---------------------------------------.⇒ m w1 = m w2 m a1 m a2 pg ps 0. and 60% RH passes over a coil with a mean surface temperature of 40°F.425 m3) tank.= 14. Ventilating.3 ) ( 545 ) m cond = 0.techstreet. © (2009).ashrae.4°C).7 psia. At what temperature will condensation begin? If the tank and mixture are cooled an additional 15°F (8.2 p a = p a ----.= 70. how much water will condense from the mixture? v 1 = 15 ft 3 v 2 = 15ft 3 t 1 = 85°F t2 = ? φ 1 = 50% φ 2 = 100% P 1 = 14. and 50% RH is contained within a 15 ft3 (0.42⎠ pa v1 ( 14. Refrigerating and Air-Conditioning Engineers. What is the required cooling capacity of the coil? Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc.42 psia 3 1T 545 1 0. distribution.0167 psia s 0.287 psia 2 t 3 = 63. (www. .596 ) = 0.298 T3 508.08 ) ( 60 ) [ [ 0.0 = 48. 85°F (29.000547 = -------2 T1 T2 545 Trial and error produces t 2 = 63.288psi p s2 = p g 2 m w2 m w1 W 1 = W 2 ⇒ ---------.00552 lb m V1 1000 m· a = -----.2°F ⇒ p g = 0.298 p s v = m w R w T ⇒ ----.167 W 3 = 0.4 ⎛⎝ -------------⎞⎠ = 13.0129 lb/lb 14.com).08 lb/min v1 14.2°F .= ---------.0052 – 0.0183 ) ( 8.070 lb T1 ( 53.00774 lb/lb ⎝ 13.24 ( 40 – 90 ) ] – [ 0.7 psia (101.5 kPa).33 An air-water vapor mixture at 14. 90°F.622 ⎛ -------------⎞ = 0.2 – 15. Inc.268 · Q· = m a [ ( ha 2 – ha 1 ) + ( W 2 hw 2 – W 1 hw 1 ) + ( W 1 – W 2 )h ws ] = ( 70.7 – 0.3 tons This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 3. A spray on the coil assures that the leaving air is saturated at the coil temperature.622 ⎛⎝ ------------------------------⎞⎠ = 0.840 Btu/h = 9.0183 ( 1100 ) – 0.= 1. For personal use only. American Society of Heating.298 W 1 = 0.34 Air flowing at 1000 cfm and at 14. (www.48⏐Principles of Heating. p g = 0.= 0.7 psia P2 = ? p s1 = ( 50 ) ( 0.0052 ( 1074 ) ] + [ ( 0. Additional reproduction. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.04 ) ] ] = 111.1 = ------------.3°C). and Air Conditioning—Solutions Manual 3. = 45.004 ) c) Supply conditions cannot maintain design for given Q s and Q L This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto a) Copyrighted material licensed to University of Toronto by Thomson Scientific.435 φ = ------. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. distribution.ashrae. The conditions to be maintained inside the space are 50°F and 75% RH.techstreet. For personal use only.= ------------.35 An air-vapor mixture at 100°F (37. W s = 0.= 0.02 lb water vapor per pound of dry air (20 g/kg). (www.2 ⇒ p w = 0.949 s @ p w = 0. sensible heat is added at the rate of 45.435 psi ⎝ p – p w⎠ From Chapter 1 p w = 0.com). The barometric pressure is 28.000 G L = -----------------------------.8% pw 0. Calculate the relative humidity. What must the air exhaust rate (lb/h) from the space be to maintain a 50°F temperature? What must the air exhaust rate (lb/h) from the space be to maintain a 75% RH? Discuss the difference.7 kPa). Qs = Gs [ cp ( t 0 – t1 ) ] b) 45.622 ⎛ ---------------⎞ = 0.561 in.= 6250 lb/h 0.8°C) dry bulb contains 0. Refrigerating and Air-Conditioning Engineers.000 Btu/h.0432 s pw 0.463 0.000 Btu/h and latent heat is added at the rate of 20.0432 Ws 3.949 psi . Hg (96.org). Within the space. Inc.36 Air enters a space at 20°F and 80% RH.245 ( 30 ) Q L = G L ( ΔW ) ( 1060 ) 20. © (2009). and degree of saturation. pw W = 0. Additional reproduction. American Society of Heating.000 G s = -----------------------.= ---------------. dew-point temperature.3°F 0. (www.Chapter 3—Basic HVAC System Calculations⏐49 3.= 4740 lb/hr 1060 ( 0. Inc. .435 t dp = 75.02 W μ = ------. © (2009). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. For personal use only.50⏐Principles of Heating.= 32% p ws 1. (www. °F c.org). Hg 11 – p w From Table 2-1: c) t dp = 52°F pw 0.techstreet.24 ( t – t* ) W = --------------------------------------------------------------------------------.37 Moist air at a low pressure of 11 psia is flowing through a duct at a low velocity of 200 fpm. The duct is 1 ft in diameter and has negligible heat transfer to the surroundings. would you expect some condensation to occur? Why? If yes.= 0.491 ) p – pw b) pw W = 0.39 φ = -------.739 ( 0. . % a) t = 85 t* = 70 ( 1093 – 0.0177 ⇒ p w = 0. the Compression Process would lower t dewpoint .= -------------------------------------------. Copyrighted material licensed to University of Toronto by Thomson Scientific.1915 psi = 0. where would the condensation form? How would you remove it? condensation occurs in discharge pipe when exiting discharge valve. distribution. humidity ratio.39 Does a sling psychrometer give an accurate reading of the adiabatic saturation temperature? Explain. lb/lb b.213 3. and Air Conditioning—Solutions Manual 3. relative humidity.38 If an air compressor takes in moist air (at about 90% RH) at room temperature and pressure and compresses this to 120 psig (827 kPa) (and slightly higher temperature). Additional reproduction. (www. The air is heavily saturated @ Inlet.739 ) ( 0. Refrigerating and Air-Conditioning Engineers.556t* )W s – 0.39 in.= 0.0212 psia 11 – 0. dew-point temperature. The dry-bulb temperature is 85°F and the wet-bulb temperature is 70°F.622 --------------*. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Yes. Calculate the following: a.= 0. Ventilating.= ------------. Remove moisture in an aftercooler.622 -----------------.com).ashrae. 3. Inc.0177 lb/lb 1093 + 0. Normally within 1°F with a shielded thermometer in air-water vapor mixtures.444t – t* pw ( 0. Inc. because Lewis Relation is approximately equal to one. American Society of Heating.491 ) * W s = 0. 0038 ) = 2.244 ) ( 15 ) 7430 = 69. If the room conditions are to be 78°F and 50% RH. Also.27 lb/h of moisture. The air is heated with a finned heat exchanger with 78 ft2 of heat transfer surface area and a UA value of 210 Btu/h·°F.= 155 lb/min v 12.= -----------. Calculate the heat added by the coil. lb/min.5 ( 60 ) ( 1100 ) ( 0. d. a sling psychrometer reads 80°F (26.7°C) and 67°F (19. Do the following: a.4°C) wet bulb. Compare these to the corresponding values for the same readings at sea level.5 lb/min ( 0.41 At an altitude of 5000 ft (1500 m). American Society of Heating. Estimate the sensible and latent load for a room with 25 people in it (the lights give off 9000 Btu/h).techstreet.0208 – 0. Show the processes on the psychrometric chart.org). a steam spray system adds moisture to the air from saturated steam at 16 psia. . Copyrighted material licensed to University of Toronto by Thomson Scientific.Chapter 3—Basic HVAC System Calculations⏐51 3. The outlet air is at 100°F and 50% RH.27 ) ( 1100 ) ( 25 ) = 7430 Btu/h QL total = 7430 Q s = m· c p Δt Q L = m· h fg ( W R – W s ) 15250 m· = ----------------------------. Determine correct values of relative humidity and enthalpy from the chart.00850 lb/lb @ 63°F φ = RH = 70% This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 3. Refrigerating and Air-Conditioning Engineers. © (2009).= 69.9 m a ( w 1 ) + m w = m a w 2 ⇒ m w = m a ( ΔW ) b) c) m w = 155 ( 0. (www. Inc. b.9 Btu/lb h = 31.ashrae.4 ) – 2.40 An air processor handles 2000 cfm of air with initial conditions of 50°F and 50% RH. Btu/min. what flow rate of air would be required if the supply air came in at 63°F? What would be the supply air relative humidity? Qs people = ( 250 ) ( 25 ) = 6250 Btu/h Qs total = 6250 + 9000 = 15250 QL people = ( 0. c. Calculate the mass flow rate. V· 2000 m· = --.65 Btu/lb 3.65 ( 1152 ) Q s = 1900 Btu/min @ 5000 ft @ Sea Level %RH = 53 %RH = 51. Additional reproduction. Calculate the pounds per minute of steam required.5 h = 34. (www.65 lb/min d) Q s = m a ( h 2 – h 1 ) – m w ( h w ) = 155 ( 475 – 16.com). distribution.0102 – W s ) w s = 0.42 The average person gives off sensible heat at the rate of 250 Btu/h and perspires and respires about 0. Inc. For personal use only. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Btu/h heat supplied to reheat coil. (www.1 ( CFM )Δt ⇒ t 4 = -----------------------.com). (www. c.44 Using the SI psychrometric chart at standard atmospheric pressure.1 ( CFM ) ( Δt ) = ( 1. 100% outside air is required for ventilation.7 kJ/kg t dp = 19. dew point and humidity ratio for air at 28°C dry bulb and 22°C wet bulb b.+ 75 = 101°F 1. heat supplied to preheat coil.872 m /kg This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto b) 200.33 3. For personal use only.700 Btu/h d) 7000 0. Inc.5°C 3 v = 0.52⏐Principles of Heating.000 Btu/h and a negligible latent heat load (latent losses to outside are made up by latent gains within the space). Determine the following: a. and then reheated. Btu/h amount of water required for humidification.1 ( CFM ) ( t 2 – t 1 ) = ( 1. .8 – 20 ) = 552.1 ( 7000 ) Q PH = 1. °F b. © (2009). gpm Qs = 200.0093 – 0. The space is to be maintained precisely at 75°F and 50% RH.01403 kg/kg b) h = 64. Refrigerating and Air-Conditioning Engineers. American Society of Heating. Due to the nature of the process.8°F c) Q RH = 1. humidified with an adiabatic saturator to the desired humidity. d.1 ) ( 7000 ) ( 101 – 60 ) = 315.techstreet. The temperature out of the adiabatic saturator is to be maintained at 60°F dry bulb. distribution.org). and Air Conditioning—Solutions Manual 3.1 ) ( 7000 ) ( 91.43 A space in an industrial building has a winter sensible heat loss of 200.000 t3 = 60°F W3 = w4 = 0. temperature of air entering the space to be heated. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.002152 m w = m a ΔW = ⎛ -------------⎞ ⎛ ---------------------------------------------⎞ = 0.ashrae. find a.45 gal/min ⎝ 13. The amount of ventilation air is 7000 scfm and the air is to be preheated.900 Btu/hr a) Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc.0093 t2 = 91. Additional reproduction. The outdoor air conditions can be taken as saturated air at 20°F. Ventilating. enthalpy and specific volume a) W = 0.33⎠ ⎝ ⎠ 8.000 Q s = 1. ashrae.8 kJ/kg a.com).8 = 24.2 kJ/kg q s = h a – h 2 = 48 – 36. W 1 = 0. and latent heat removal for the process. sensible. American Society of Heating. Inc. For personal use only. © (2009).= ( W 1 – W 2 ) = ( 0. find: a.00903 kg/kg h 2 = 36.0145 – 0.2 kJ/kg q L = h 1 – h a = 61 – 48 = 13 kJ/kg Copyrighted material licensed to University of Toronto by Thomson Scientific. 24°C db 21°C wb 2.8 = 11. Refrigerating and Air-Conditioning Engineers. (www.45 Using the SI chart.org). distribution. 21°C wet bulb to 13°C dry bulb.00903 ) = 0. 1. moisture that must be removed in cooling air from 24°C dry bulb.techstreet. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . Additional reproduction. total. saturated b. Wa = W2 ta = t1 h a = 48 kJ/kg mm ------. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. (www.Chapter 3—Basic HVAC System Calculations⏐53 3.0145 kg/kg h 1 = 61 kJ/kg Assume Saturated 13°C db W 2 = 0. Inc.00547 kg/kg ma q t = h 1 – h 2 = 61 – 36. 6 0. % Enthalpy h. Conditioned air leaves the apparatus and enters the room at 60°F dry bulb.0130 0.0130 – 0. distribution. Outside air is at 95°F dry bulb and 76°F wet bulb.000 ⁄ 1100 W s = W r – ------.4 ) = 34.244 ) ( 84. e.8 tons 45540 ( 0.7 0.0123 – -------------------------------.2 ) + 0.6 . a latent load of 50. and fan motor rating. © (2009).0 0.015 – 0. d.5 ) d) w f = --------------------------.6 45540 e) q = 45540 [ 34.3 mw 50.5% 357. Inc. On a mass basis.0123 ) + 0.9 f 84.4% 357.techstreet.0113 )28 ] = 357.= 34. What percent of the required refrigeration is due to the outside air load? Dry Bulb t. t f = 84. lb/h scfm acfm vact OA 95 42 39.0130 .35 b) 0. (www.000 Btu/h.0113 ma 45540 200. (Hint: See Fig.8 m 82. Ventilating.0150 11380 2528 2710 14. c.244 ) ( 78 – 60 ) 10120 ( 3.000 Btu/h.5 in.0123 34160 7590 7860 13. 3-1) b.600 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Point Copyrighted material licensed to University of Toronto by Thomson Scientific.0 0. Refrigerating and Air-Conditioning Engineers.600 Btuh = 29. Inc.000 m a = ----------------------------------------. Specify the fan size. Fan efficiency is estimated as 55%.600 2528 [ 1.4 0. scfm.25 ( 95 ) = 82.25 ( 39.540 lb/h ⇒ 10120 cfm ( 0.0 s 60 100 26.× 100 = 22. Btu/lb W. American Society of Heating.org).75 ( 32.0113 45540 10120 10130 13. and is maintained at 78°F dry bulb and 60% RH. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.0130 45540 10120 10630 14.25 ( 0.6 – 60 ) f) % sensible = ------------------------------------------------------------.1 ( 2545 ) h f = 34 + --------------------------.3 55 34.75 ( 0.54⏐Principles of Heating. The fan must produce a pressure increase of 3.= 0. HP. Draw and label the schematic flow diagram for the complete system. For personal use only. Plot and draw all processes on a psychrometric chart. Determine the size refrigeration unit needed.= 0. °F φ. and Air Conditioning—Solutions Manual 3. 25% outside air is mixed with return air.= 10. lb/lb ma.46 An air-conditioned space has a sensible heat load of 200.10 ( 95 – 78 ) + 4840 ( 0.6 50 34. W f = 0.0123 ) ] g) % due to OA = -------------------------------------------------------------------------------------------------------------.× 100 = 76.0150 ) = 0.7 – ( 0. in Btu/h and tons.6 – 26.1 HP 0. a. What percent of the required refrigeration is for (1) sensible cooling and (2) for dehumidification? g.75 ( 78 ) + 0.ashrae.3 r 78 60 32.55 ( 6350 ) 10. Complete the table below. water to overcome the system pressure loss. f.com).= 45. (www. Additional reproduction.0130 45540 10120 10550 13. .2 0. (www. distribution. Additional reproduction.org). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.techstreet.ashrae. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . © (2009). Inc. (www.com). For personal use only. American Society of Heating. Refrigerating and Air-Conditioning Engineers. Inc.Solutions to Chapter 4 DESIGN CONDITIONS Copyrighted material licensed to University of Toronto by Thomson Scientific. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .com).ashrae. © (2009). (www. American Society of Heating. Inc. Additional reproduction. For personal use only. Inc.Copyrighted material licensed to University of Toronto by Thomson Scientific. Refrigerating and Air-Conditioning Engineers.org). (www. distribution.techstreet. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. what air temperature must the heating system maintain for comfort of the workers if the air movement is 40 fpm? 700 Btu/h= 2 met Activity level = ----------------------2 19.5t a + 0.108 ( 120 ) 0.2 ) = 71°F Assuming light clothing (0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. determine the air temperature necessary for comfort. (www.( 20 ) + -----------.4 ( 1 + 0. Assume occupants have equal view (i.( 60 ) + -----------.( 54 ) + -----------. Inc. and 540 ft2 of floor with a surface temperature of 70°F. 55: t o = at a + ( 1 – a )t r ) a = 0. 55: t r = ∑ Fρ – i t i . 5 in.5 t air = 2 ( 75 ) – 64 = 86°F = 30°C 4.org).ashrae. Inc. (4-1): W = 120 lb 0.12 The living room in a home is occupied by adults at rest wearing medium clothing. Refrigerating and Air-Conditioning Engineers.425 0. all angle factors are identical) of all surfaces.techstreet. compute the body surface area (ft2). 45 ft2 of glass with a surface temperature of 20°F.. If the air movement is 20 fpm and light clothing is being worn.1 ) t a = 86°F 4.108m Eq. distribution.1 2070 2070 2070 2070 2070 t o = 75°F = 0.75 clo): Correcting for activity level: Fig.com).e. When the MRT for the area is 69°F. 670 ft2 of partitions with a surface temperature of 70°F.7 (5 ft.Chapter 4—Design Conditions⏐57 4.5t a + 0.5t r = 0. The mean radiant temperature is 18°C (64°F).70 + -----------.13 A room has a net outside wall area of 275 ft2 with a surface temperature of 54°F. © (2009).1 ft 2 4. 120 lb). Determine the air temperature necessary for comfort. 540 ft2 of ceiling with a surface temperature of 60°F.425 ( 65 ) 0. Additional reproduction.5 ft 89 + 69 t o = -----------------. American Society of Heating.( 70 ) = 64.= 79°F 2 t a = 79 – 5. (www. 4-3..725 = 17.725 l l = 65 in. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto N Std. of which 310 Btu/h is latent heat. t a = 89°F.14 Workers on an assembly line making electronic equipment dissipate 700 Btu/h. Copyrighted material licensed to University of Toronto by Thomson Scientific. For personal use only. .5 ( 64. A = 0. [Ans: 30°C (86°F)] t air + MRT t o ≈ 75°F ≅ --------------------------2 ( Std.75 ) ( 2 – 1. ∑ A i = 2070 i =1 275 45 540 670 540 MRT = t r = -----------.8 For the person in Problem 4. A D = 0. 7 ] CFM = 2417 Supply air must have sufficiently low humidity ratio since latent load actually represents moisture added. .17 In an auditorium. [Ans: 243°F? unfeasible] Assume uniform view of surfaces: 880 120 t r = -----------.org). For personal use only. distribution. How much air (cfm) should be supplied to remove the sensible heat? b. Inc. (www. © (2009). The average temperature of other surfaces in the room remains at 60°F and the air temperature is still 68°F. a. The room is occupied by light clothed adults at rest. Refrigerating and Air-Conditioning Engineers. Explain what must be done to remove the latent heat.( 13. What panel area will be required if the room is occupied by adults at rest? ( 1000 – P ) ( 60 ) + P ( 120 ) t r = 82 = -------------------------------------------------------------1000 P = 367 ft 2 a. the maximum allowable panel temperature is 120°F. floor. Air is supplied at 60°F.475 · m· = 11 . Ventilating. Additional reproduction.150 students are watching slides.( 60 ) + -----------. Determine the required flow rate (lb/h) to handle the heat gain from the occupants if the return air temperature is not to exceed 75°F.58⏐Principles of Heating. [Ans: 11. Air enters the room at 57°F. QL M H O ≅ -----------2 1100 4. Assume the lights are out and no heat gain or loss occurs through the walls.t p 1000 1000 t p = 243°F 4.techstreet. of which 120 ft2 is to be heated.475 lb/h] q s = 210 Btu/person (Chap.( 60 ) + -----------.5t a + 0.1 ( CFM ) ( 72 – 57 ) 330 [ Chap.475 lb/h = --V = ---------------. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.ashrae.( t p ) 1000 1000 t o = 75 = 0. and Air Conditioning—Solutions Manual 4. (www.33 ) = 2550 cfm ν 60 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 4. Copyrighted material licensed to University of Toronto by Thomson Scientific.⎞⎠ = 1. and ceiling.com). 17. American Society of Heating.16 Assume that in Problem 4.15 A room has 1000 ft2 of surface. Inc. b.18 Two hundred people attend a theater matinee. 7) q s = m· C p ( t r – t s ) = 200 ( 210 ) = m· ( 0. and the balance has an average surface temperature of 60°F.244 ) ( 75 – 60 ) · V 11 . The MRT is 80°F and the average room air temperature is 72°F.15. 390 Q s = 150 ( 225 ) ⎛⎝ --------.5t r 880 120 t r = 150 – 68 = 82°F = -----------. Determine the surface temperature of the heated panel necessary to produce comfort if the air velocity is 20 fpm. The air temperature in the room is 68°F. com). [Ans: 92°F] From Fig. 4-3 t a ≅ 28°C MRT ≈ 22°C 4. Outside: to = 4. Φo = 100% b. Rochester. 1 clo ⇒ ti = 75 – 5. Dry bulb = 25°C.1 db/76. Winter: Summer: Dry bulb = 22°C. St.1°F. summer.3°C φ o = 100% φ a = 50% This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Assuming Icl = 0. For passengers seated without coats.2 ) ]22 t r = ---------------------------------.600 M w = ---------------.= 23.2 ) ( 1. winter. Summer: Dry bulb = ____°C. distribution. W = _______ kg/kg b. Inside: ti = 72°F.075 ) 4. The room is 6 m by 4 by 3 m high and the other surfaces are all at 22°C. .0065 lb/lb air Ma ( 1350 ) ( 60 ) ( 0.2 by 1. Assume uniform view for each surface: ( ∑ A = 108 ) 4 ( 1.2) = 66°F. St. Assume very little activity and light clothing. Missouri.1°F wb c.techstreet. W = _______ kg/kg a.2 )49 [ 108 – 4 ( 1.5 with ta = MRT + 6 Copyrighted material licensed to University of Toronto by Thomson Scientific. Georgia. for sedentary activity and light clothing.4°C 108 108 Fig.2 ) ( 1. 43 . (www.2°F. For personal use only.64 ΔW = -------. MRT = 23.64 lb/h 1100 Q L = 80 ( 545 ) = 43 .22 The mean radiant temperature in a bus is 6°C lower in winter than the air temperature. Φo = 100% 4. American Society of Heating. b.23 For Atlanta.64 m3/s (1350 cfm). Inside: Assume 2 met.= 0. (www. © (2009). having a 1. Trial and Error with Fig. Refrigerating and Air-Conditioning Engineers. Minnesota a. 4-3: MRT = 92°F [Assume medium activity is sedentary] 4. Winter: Dry bulb = ____°C.2 m/s. Inside: ti = 78°F. Louis.20 Specify the MRT for comfort in a space where the air temperature is 68°F and the relative velocity is 20 fpm. Missouri b. Missouri c. specify the normal indoor design conditions listed below for a. 30% RH Outside: to = –15. factory (medium activity).600 Btu/h Mw 39. W = 0.004 kg/kg W = 0. 30% RH.ashrae. determine the desired air temperature if the relative air velocity is 0. Inc. winter. apartment building. 4-3 SI: for v = 0. Inc. 60% RH. Outside: 93.4(1 + 1)(2 – 1.org).012 kg/kg 4. Additional reproduction.21 Specify completely a suitable set of indoor and outdoor design conditions for each of the following cases: a.+ -----------------------------------------------------.2 m/s. apartment building. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.2 m radiant panel at 49°C on each of the four walls. Louis.Chapter 4—Design Conditions⏐59 4.= 39.24 Specify completely indoor and outdoor design conditions for winter for a clean room in Kansas City.= ---------------------------------------------.19 Determine the increase in humidity ratio due to 80 people in a dance hall if air is circulated at the rate of 0.4 ⇒ t a = 28°C t o = – 16. American Society of Heating. Additional reproduction. © (2009). distribution.org).ashrae. (www. (www. Refrigerating and Air-Conditioning Engineers. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .techstreet. Inc. For personal use only. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc.com). This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . American Society of Heating.org). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. (www. Refrigerating and Air-Conditioning Engineers. distribution. © (2009). Inc.techstreet. Inc.ashrae.com). Additional reproduction. (www.Solutions to Chapter 5 LOAD ESTIMATING FUNDAMENTALS Copyrighted material licensed to University of Toronto by Thomson Scientific. For personal use only. (www.techstreet. Refrigerating and Air-Conditioning Engineers.ashrae. Additional reproduction.com). For personal use only. (www. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . distribution. American Society of Heating.org). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Inc. Inc.Copyrighted material licensed to University of Toronto by Thomson Scientific. © (2009). 1 ( SCFM ) Δt = 1. For personal use only.1 ⎛ -----------⎞ ( ΔW ) = 80. or Q s = 1.18 in.23 ( 236 ) ( 24 – 9 ) = 4350 watts Copyrighted material licensed to University of Toronto by Thomson Scientific.302 in. and (b) the latent load due to infiltration. What would be the leakage rate for the building of Problem 5.30 ⎠ 0. Inc. Additional reproduction. ΔP = 0. Fig.2 m/s wind blowing uniformly against one face of a building.3 A double door has a 1/8 in. Btu/h ⎝ 60 ⎠ This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 236 L/s at 9°C with 24°C inside 500 cfm 48°F with 75°F Q s = 1. Chap. (www.3 in. assume 6 × 7 ft doors.1 ( SCFM ) Δt = 1. H 2 O = 453 cfm 5. © (2009).techstreet.000482 ) ( 25 ) = 0. Sensible CFH Q s = 1. H 2 O Average crack width is 1/8 in. distribution.000482 at Standard ρ V2 2 P v = ( 0.ashrae. Latent CFH Q L = 4840 ( SCFM ) ( ΔW ) = 1.com).302 ) = 0.18 CFM = 600 ⎛ ---------.6 ( 0. Refrigerating and Air-Conditioning Engineers. 5. HBF → 300 cfm/door 0. 14.1. what pressure differential would be used to calculate the air leakage into the building? 11. H2 O C s = 0. .447 = 25 mph 11. (www. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. crack.1 With an 11.018 CFH ( t i – t o ). 5.⎞ ⎝ 0. a.1? From Problem 5.850 Btu/h *There is also a latent load. 27.18 in. which has a 1/4 in. Btu/h ⎝ 60 ⎠ b.55 Two doors at 0. H 2 O 5.1 ( 500 ) ( 75 – 48 ) = 14 .6 [ Fig.2 ⁄ 0.2 m/s wind P v = 0.1 ⎛ -----------⎞ Δt = 0. Inc.5 Give an expression for (a) the sensible load due to infiltration.4 Determine the heat loss due to infiltration of 236 L/s of outdoor air at 9°C when the indoor air is 24°C. American Society of Heating.Chapter 5—Load Estimating Fundamentals⏐63 NOTE: For the problems in this chapter the answers may vary depending on which tables are used for the R and k values as well as the assumptions in the selection of the tabulated listings.org).6 (CFH ) ( W i – W o ). 5-5 at 0° ] ΔP in = 0. crack on all sides except between the two doors. 890cfm 3 a.1 ( 937.000 ft @ in W i = 0.org).1 CFM ( Δt ) = 1. There are adequate openings in the roof for the passage of exhaust air. (www.5 ) ( 0. 3 × 3 ft opening Q = C 4 C v AU 25 mph 2005 HBF. Specify the necessary humidifier size (lb/h).300 Btu/h Q L = 4840 ( CFM ) ( ΔW ) = 4840 ( 937. 30% RH minimum. latent.9 A building 20 by 40 by 9 ft has an anticipated infiltration rate of 0. © (2009). Equations (27) – (29) Q = 88 ( 0. The infiltration rate is estimated to be 0.0007875 3 NV ( 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Indoor design conditions are 75°F. distribution.75 air changes per hour. Refrigerating and Air-Conditioning Engineers. Ventilating. American Society of Heating. Volume = ( 20 ) ( 40 ) ( 9 ) = 7200 ft ( 7200 ) ( 3 ⁄ 4 ) SCFM = NV = -------------------------------. Outdoor design temperature is 5°F. Volume = 75 × 100 × 10 = 75 .ashrae.0046 – 0. m w = Q L ⁄ 1100 = 1960 ⁄ 1100 = 1.techstreet.001 ) = 1960 Btu/h Q t = 6390 + 1960 = 8890 Btu/h b.0055 – 0.8 lb/h This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 5. Additional reproduction. a. Inc.55 ) ( 9 ) ( 25 ) = 10 .= ------------------------------------. and total heat loads (Btu/h) due to infiltration.8 A 3 by 3 ft ventilation opening is in a wall facing in the prevalent wind direction.000 ) CFM = -------. (www.75 ) ( 75 . .5 ) ( 75 – 0 ) Q s = 77 . and the outside conditions are 0°F and saturated. The indoor conditions are 75°F and 24% RH.64⏐Principles of Heating. Determine sensible.0046 @ out W o = 0. Estimate the ventilation rate for a 25 mph wind. For personal use only. b. Copyrighted material licensed to University of Toronto by Thomson Scientific.= 937.0007875 ) = 17 . Inc. and Air Conditioning—Solutions Manual 5.000 Btu/h 5.1 ( 90 ) ( 70 ) = 6930 Btu/h Q L = 4840 ( 90 ) ( 0.= 90 60 Q s = 1.1 ( CFM ) Δt = 1. Calculate the sensible and latent heat loss.5 60 60 Q s = 1.6 A building is 75 ft wide by 100 ft long and 10 ft high.com).75 ach. 00079 ) = 1070 Btu/h = 0 no humidifier 5.21 = max.000 ⁄ 60 = 725 cfm b.10 × 58 × ( 75 – 0 ) = 4785 Btu/h = 1.06 V bz = 5 × 4 + 0. INFILTRATION Office: 1/2 ach × 7000 ⁄ 60 = 58 cfm (more if traffic) Shop: 1 1/2 ach × 29 . Determine for each area a. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.= 0.000 ) = 4650 cfm 1000 2 2 0°F W o = 0.9 q s = 1. Copyrighted material licensed to University of Toronto by Thomson Scientific. = 4840 × 58 × ( 0.12 ( 20 . cfm b. VENTILATION ( Std. Btu/h 2 Office: 700 ft . 12 by 12 ft private office with 8 ft high walls b. On a winter day when the outside temperature is 0°F.1-2004 ) V bz = R p R z + R a A z Office: Rp = 5 R a = 0.× 15 + 0. Z p ⇒ E V = 0.10 × CFM × Δt qs office qs shop = 1.08 Z P = -----------.6 cfm b. 62. d. American Society of Heating. Inc.0046 – 0. For personal use only. .= --------.11 Specify an acceptable amount of outside air for ventilation of the following: a.00079 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto V ou = Σ = 62 + 742 = 804 cfm.10 × 725 × ( 68 – 0 ) = 54 .= 893 cfm EV 0. There are 22 employees normally in the shop area and 4 employees in the office area.06 ( 12 × 12 ) = 5 + 8.⎞ = --------. 29 .5 × ---------------.0046.000 ft2 of floor area Table 5-9 2 a.06 ( 700 ) = 62 cfm Shop: R P = 10 R a = 0.5 cfm/person & 15 p/1000 ft + 0. Inc.= 0.06 cfm/ft ∴ OA = 5 × 1 + 0.18 V bz = 10 × 22 + 0. minimum required outside air. V oz shop = 742 Assume single system for both spaces: V oz 62 742 Z P = ⎛ -------.techstreet. distribution. 25% } W i = 0. infiltration.org). sensible heat loss due to infiltration. 7000 ft 2 3 Shop: 2900 ft .Chapter 5—Load Estimating Fundamentals⏐65 5. Additional reproduction.000 ft CFM supply = 700 = V pz 3 o CFM supply = 3600 = V pz s a. the office is maintained at 75°F.9 ⎝ V pz ⎠o o s 700 3600 c.230 Btu/h q L = 4840 × CFM × ΔW qL offic qL shop @ 75°F.000 ∴ OA = 7. Refrigerating and Air-Conditioning Engineers.18 ( 2900 ) = 742 cfm V oz = V bz ⁄ E z Assuming ceiling supply (floor return): E z = 1. cfm c.com).6 = 13.10 A small factory with a 10 ft high ceiling is shown. Btu/h d. (www. department store with 20. (www.12 cfm/ft 20 . © (2009). V ou 804 V ot = -------. Use 7. and the shop is kept at 68°F with no humidity control.0 V oz office = 62. latent heat loss due to infiltration.ashrae. Use 5 cfm/person + 0. 25% RH. Inside air ΣR = Copyrighted material licensed to University of Toronto by Thomson Scientific.44 0.215 Winter Summer 0. 8 in. 4 in.66⏐Principles of Heating.44 0.211 Btu/h/ft2 ·°F 5. Refrigerating and Air-Conditioning Engineers. 3/4 in. 4 in.55 Air gap (top) 4. 1/2 in. 4 in. Inc. and 3/4 in. Stone.95 0.47 0.15 Find the overall coefficient of heat transmission U for a wall consisting of 4 in. and Air Conditioning—Solutions Manual 5. 1/2 in.17 0. The outside air velocity is 15 mph and the inside air is still. and 1/2 in. 1/2 in.10 0. of face brick. Inc.com). R Outside air Face brick.17 0.68 2. Inside air ΣR = 0.13 The ceiling of a house is 3/4 in. of face brick.68 4.techstreet. Airspace.50 (assume lightweight aggregate) U = 1/ΣR = 0. Find the U-factor for both winter and summer.14 Calculate the winter U-factor for a wall consisting of 4 in.10 1. 4 in.459 Btu/h/ft2 ·°F 5. Metal lathe and plaster.85 0.89 Room air 0. For personal use only.55 Air gap (bottom) 0. (13 mm) of gypsum plaster (sand aggregate). (100 mm) face brick.44 0.74 0. Determine the U-factor for cooling load calculations. of cement mortar.10 1.80 0.47 0. Additional reproduction.18 U = 1/ΣR = 0. (19 mm) acoustical tile on furring strips with highly reflective aluminum foil across the top of the ceiling joists. (www. Hollow clay tile.83 U = 1/ΣR = 0.92 Acoustical tile 1. R Outside air Face brick.68 2.ashrae. Inside air ΣR = U = 1/ΣR – 0. an airspace 1 5/8 in.92 ΣR = 12.95 0. 4 in. of stone. wide. (www. . of cement mortar. Cement.68 4. Common brick. American Society of Heating. 3/4 in. 0. (1/R = 0.47 0. 8 in. hollow clay tile. Cement mortar.64 0. 8 in. 1 5/8 in. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 1/2 in.12 A wall consists of 4 in. Gypsum (sand) plaster.08) Gypsum plaster. Ventilating. of gypsum plaster. 1/2 in.org).40 Btu/h/ft2 ·°F This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto R Outside air Face brick. © (2009). distribution.25 0.09 0.17 0.85 0.66 0. 8 in.078 Btu/h/ft2 ·°F 5. (100 mm) common brick. and wood lath and plaster totaling 3/4 in.44 0. R Attic air 4. 120 ( m ⋅ K ) ⁄ W 2 R con = ( 0.88 U = 0.81 0. 2 R o = 0.138 ( m ⋅ K ) ⁄ W 1 2 2 1 U = --------.⎞ ( 100 ) = 9. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.072 Btu/h· ft ·°F 5.ashrae. (www.35 Neglecting Studs U = 1/4.03 + 0.+ ---.63 ( W/(m ⋅ K ) 5.17.19 Rework Problem 5.Chapter 5—Load Estimating Fundamentals⏐67 5.067 ) 2 U av = 0.7 m/s and the inside air is still? 11 1 1 --= ∑ R = ---.55 ) = 0.17 0.7 m/s.094 ) ( 0.10 0. For personal use only.47 W ⁄ ( m ⋅ K ) = 3.= ----------------------------------------------.12 ) + ( 0. 1. 5.067.18 including the effect of the studs.com).206 Btu/h·ft2 ·°F 5.18 For the wall of Problem 5.06 2.+ ----U hi C ho 11 1 1 --------= ---------.31 8.16 A wall has an overall coefficient U = 1. permeable felt building paper. U i = 0. metal lath and sand plaster.97 0.97 + 4.08 2 C = 1. American Society of Heating. R Outside air Siding Felt paper Sheathing Airspace Lathe and plaster Inside air ΣR = 0. 25/32 in. wood fiber sheathing.030 ( m ⋅ K ) ⁄ W 2 R i = 0. R s = 4. 2 by 4 studding on 16 in.5 S = ⎛ ------. Inc.29 C 34. Inc. .85 – 0.+ ---. (www.12 U av = ( 0.+ ------------1. centers. Additional reproduction.35 = 8.85 – 0.17 Compute the U-factor for a wall of frame construction consisting of 1/2 by 8 bevel siding.138 Q = U ( A ) ( Δt ) = 3. What is the conductance of the wall when its outside surface is exposed to a wind velocity of 6. © (2009).97 + 11.12 + 0.= 3. Outside wind velocity is 15 mph. Neglect the effect of the studs. Refrigerating and Air-Conditioning Engineers.85 = 0.31 W/(m2 ·K).47 W ⁄ ( m ⋅ °C ) ∑ R 0.23. ⎝ 16 ⎠ U s = 0.org).6°C.techstreet.9 m2 of this wall.067 Btu/h·ft2 ·°F Copyrighted material licensed to University of Toronto by Thomson Scientific.906 ) ( 0.6 – ( – 15 ) ] = 1540 W This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto R corrected for insulation 4.47 ( 14. Inside air temperature is 15.68 4. and 3/4 in.20 A concrete wall 250 mm thick is exposed to outside air at −15°C with a velocity of 6. determine U if the space between the studs is filled with fiberglass blanket insulation. distribution.9 ) [ 15.4%.06 0. Determine the heat flow through 14.25 ) ( 0.0 = 14. org).61 R4 0.3 + ---------.23 A composite wall structure experiences a –10°F air temperature on the outside and a 75°F air temperature on the inside.9°F RT 8.21 Find the overall coefficient of heat transfer and the total thermal resistance for the following exterior wall exposed to a 25 mph wind: face brick veneer. 25/32 in.32 0. 3 in.com).0 T 4 = T 3 + -----.68 ΣR = 15. 1. and a 3/8 in.( ΔT ) = – 3.ashrae.00 0.( 85 ) = 68. Additional reproduction.064 Btu/h·ft2 ·°F 5. per ft2.2°F RT 8.06 Airspace.) thick precast concrete (stone aggregate.2 + ---------. Determine the U-factor and the heat flow rate. Inc.61 R2 0.1 cm (4 3/4 in. (www.68 RT = ΣR = 8. Ventilating.( ΔT ) = – 65. sheet of gypsum board.116)(85) = 9. thick outer facebrick.( 85 ) = 65.44 T 3 = T 2 + -----.87 Btu/h·ft2 T 1 = – 10°F R1 0. (R3) Gypsum board (R4) Inside air (R5) 0.( ΔT ) = – 8. walnut veneer plywood panels for the interior. and 1/4 in. The wall consists of a 4 in.( ΔT ) = – 10 + ---------. insulating board sheathing.76 (m· K)/W ] = 0.3°F RT 8. Copyrighted material licensed to University of Toronto by Thomson Scientific.32 T 5 = T 4 + -----. a 2 in.techstreet. oven dried)? 2 R = ( 0.75 ) ( 0. (www.3°F RT 8. 1 1/4 in.9 + ---------. For personal use only.116 Ti = 75°F.44 Insulating board sheathing 2.61 R3 7. Inc.22 What is the thermal resistance of 12. © (2009).31 Inside air 0. distribution.68⏐Principles of Heating. To = –10°F q = UAΔT = (0. Plot the steady-state temperature profile across the wall. 3/4 in.17 0. and Air Conditioning—Solutions Manual 5.( 85 ) = – 3. 0.09 Face brick 0.59 ⇒ HBF U = 1/ΣR = 0.61 T 6 = 75°F This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 5. batt of fiberglass insulation.61 = 0.092 ( m ⋅ K ) ⁄ W or 2 R = ( 4.01 Fiberglass. 11.52 h· ft ·°F/Btu R Outside air (R1) Face brick (R2) Fiberglass.44 7. American Society of Heating.61 U = 1/ΣR = 1/8. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Refrigerating and Air-Conditioning Engineers.11 ) = 0.121 m ) [ 0. fiberglass insulation in stud space.17 T 2 = T 1 + -----. .0 Plywood.( 85 ) = – 8. 3 in. Neglecting Studs: R Outside air 0. 27 The top floor ceiling of a building 30 by 36 ft is constructed of 2 by 4 in.Chapter 5—Load Estimating Fundamentals⏐69 5.9 – 5. The heat loss through the outside walls is 28% of this total when the overall coefficient for the outside walls is 1.89 0.6 kW 5.0.61 4.com). Inc. Airspace.20 0.05 ( 27.4 with fiberglass = 0.9 kW.70 = 14. Additional reproduction. lower Acoustical tile Inside air ΣR = 0. thick. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.01 = 12.22 Btu/h·ft2 ·°F 5.01 kW.ashrae. © (2009).53 U = 0.1 q total = 12.89 + 1. Refrigerating and Air-Conditioning Engineers.13 Rock wool 11.0 Attic air 0.1 ∑ R with fiberglass 0. upper Airspace. ∑ R = -----1. but the space between the joists is filled with rock wool. .61 ΣR = 12. For personal use only. Inc. centers.24 Find the overall heat transmission coefficient for a floor-ceiling sandwich (heat flow up) having the following construction. determine the new total heat loss for the house. Find the overall coefficient of heat transfer for the ceiling. the total heat loss is calculated as 17.714 = ------------.( 5. U = 0. On the underside is metal lath with plaster.89 Copyrighted material licensed to University of Toronto by Thomson Scientific. 3/4 in.714. If 50 mm organic bonded fiberglass is added to the wall in the stud space.4 W/(m2 ·K).61 Lathe and plaster 0. 2 1/2 in.61 1.01 ) = 1.70 kW 2. airspace and have metal sashes.26 In designing a house.9 ) = 5.61 0. The air temperature at the ceiling in the room is 78°F and the attic temperature is 25°F. American Society of Heating.25 The exterior windows of a house are of double insulating glass with 1/4 in. (www. R Room air 0.61 0.87 Btu/h·ft2 ·°F (Table 5-16) 5. Determine the design U-factor for heating.714 + 0.= 0.81 Btu/h·ft2 ·°F This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto q walls q other = 17.org).76 ) = 2. joists on 18 in. distribution. R Upper air Concrete.35 U = 1/ΣR = 0. (www.28 ( 17. On top of the joists there are only scattered walking planks.techstreet. 1 . q walls = 0. 172 ( 100 ) ( 70 – 10 ) = 1030 Btu/h 5.( 80 ) = 59. 2 in. Refrigerating and Air-Conditioning Engineers.17 0.91 ) ( 9.32 1. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.172 q = UA ( t i – t o ) = 0. Inc. 3/8 in.2°C).82 A = 100 ft2 to = 10°F ti = 70°F U = 1/ΣR = 0.04 ( 100 ) ( 70 – 10 ) = 6240 Bth/h This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 5. concrete slab. Cellular glass insulation. Inc. Roof insulation.com).17 0. 1 in. For personal use only. Omit correction for framing.17 0. and Air Conditioning—Solutions Manual 5. The roof has 3/8 in. 1 1/2 in.( 80 ) = 0.17 6. with a metal lath and 3/4 in.2 ) ] = 1833 W q = 1. Ventilating.61 0. Concrete (80)#.61 7. .5 mm) plate glass with inside and outside air temperatures of 70 and 10°F (21.6°F 7.17 5. Additional reproduction.29 ) [ 21. distribution.17 2 U = 0. (www. 2 in.13 0. built-up roofing.33 4. Acoustical tile. outside air 10°F. thick.30 Calculate the heat loss through 100 ft2 (9.82 T 5 = – 5 + ---------.29 Calculate the heat loss through a roof of 100 ft2 area where the inside air temperature is to be 70°F. built-up roofing.1°F 7.17 0.28 Determine the U-factor and the temperature at each point of change of material for the flat roof shown below.9°F 7. 3/8 in. 4 in. cellular glass insulation.techstreet. and the composition from outside to inside: 3/8 in. © (2009). acoustical tile. respectively. American Society of Heating.17 0. Inside air ΣR = 0. roof insulation.17 T 1 = – 5 + ---------. Concrete.61 5.29 m2) of 1/4 in. Airspace (top) Airspace (bottom) Lather and plaster Inside air T1 T2 T3 T4 T5 T6 T7 ΣR = 0.33 2. 1 in.( 80 ) = – 3.54 0. (6.50 0.( 80 ) = 68. and 3/4 in.61 0.04 Btu/h· ft ·°F q = UA ( Δt ) = ( 5. 80 lb/ft3 lightweight aggregate concrete on corrugated metal over steel joists. (sand) plaster ceiling.5 T 2 = – 5 + ---------.1 and –12. 2 2 U glass = 5.50 T 7 = – 5 + ---------.89 0.139 Btu/h· ft ·°F R Outside air Built-up roofing. R Outside air Built-up roofing.2 – ( – 12.91 W/ ( m ⋅ K ) { Table 5-15 } or 1. Copyrighted material licensed to University of Toronto by Thomson Scientific.ashrae.2°F 7. (www. 4 in. 1 1/2 in. 3/4 in.org).70⏐Principles of Heating. 44). expanded polystyrene insulation (k = 0. Additional reproduction. dry bulb = 75°F ⇒ φ = 61% max.techstreet. lightweight gypsum plaster on 1/2 in.05 Plywood 0.17 Face brick 0.0 in. still air inside. airspace.com). and 1/2 in.33 A wall is constructed of 4 in.31 A building has single glass windows and an indoor temperature of 75°F.org).77 = ----------------------------------( 1.32 Inside air 0. Inc.3 + x/0.ashrae. airspace.81 2 --q.46 ( 75 – t s ) = 1.44 32 – ( – 15 ) -------------------------------------------------= -------------------------3.68 ΣR = 3 + x/0. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Outside air 0. 5. 5.24 Airspace 0. and after sundown. a vapor seal having negligible thermal resistance. Refrigerating and Air-Conditioning Engineers. distribution. Inside: 55% rh ⇒ 55°F dew point E = 0.24 ) A x = 1. (www. airspace. For personal use only.44 Copyrighted material licensed to University of Toronto by Thomson Scientific.12).34 The roof of a rapid transit car is constructed of 3/8 in. The outside air temperature is 40°F. .Chapter 5—Load Estimating Fundamentals⏐71 5.6 h i = 0. face brick.32 Repeat Problem 5. plasterboard.59 ( 75 – 40 ) t s = 60.81 ( 72 – 55 ) = 13. © (2009).24).44 70 – ( – 15 ) x = 1. steel with welded joints and aluminum paint.78 Inside air 0 ΣR = 1. what can the maximum relative humidity of the inside air be without condensation forming on the glass? h i ( 75 – t s ) = U ( 75 – 40 ).44 Sheathing x/0.0 + x ⁄ 0.24 55 – ( – 20 ) q --.= 13.44 Airspace 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.9°F = dew point. 1. American Society of Heating. 1/16 in.46 ( 75 – t s ) = 0. 3 1/2 in.04 ( 75 – 40 ) t s = 50.44 + x ⁄ 0.94 Plaster 0.3 + x ⁄ 0. pressed fiber board sheathing (k = 0. plywood (C = 2.17 + 0. how thick must the sheathing be to prevent water pipes in the stud space from freezing? R 0. what thickness of insulation is necessary to prevent condensation when the inside conditions in the car are 72°F dry bulb and 55% RH? 72°F db.= 0.31 for a double glass window with a 1/2 in. With a 15 mph outside wind. (www. dry bulb = 75°F ∴φ = 40% max.1°F = dew point. h i ( 75 – t s ) = U ( 75 – 40 ).0 Btu/h·ft2 ·°F) when the ambient temperature is –20°F.47 Polystyrene insulation x/0. When the inside air temperature is 70°F and the outside temperature is –15°F.0 in. 3/4 in. Inc. 1. 5.45 Plasterboard 0.77 Btu/h· ft ·°F A R Outside air 0. If the car is traveling at 60 mph (film coefficient is 20. 024 = 0.44 0.259.17 Rins R ins = 2.25 0. Inc.44 Sheathing.20 0. a 2 in. U = 0. sliding patio door with insulating glass (double) having a 0.25 0. 2 by 6 studs on 24 in. .org). The roof area is 2717 ft2 while the ceiling area is 1980 ft2. and the outside temperature is 20°F.com). 0.85(0.93 0.17 + R ins 0.61 2.68 7.24 Btu/h·ft2 ·°F (1.92 Outside air Shingles Plywood Attic Air ΣR = 2.25 Btu//h·ft2 ·°F This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 5.420 c.437 ΣR = 0.23 U = 1/ΣR = 0.48 a. Assuming that the insulation forms a perfect vapor barrier.68 ΣR = 3.5°F t o = 20°F R Inside air Wood deck Outside air Insulation – 44.techstreet.129 Ceiling Attic air Plasterboard Inside air Rs Outside air 0. 3/4 in. plywood sheathing.50 in.35 A roof is constructed of 2 in.50 -------------------------------------------------------. and 3/8 in. insulation on top of deck. (www. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. built-up roofing.68 Paneling 0. American Society of Heating.36 Determine the summer U-factor for each of the following: a.46 Btu/h·ft2 ·°F (Table 5-17) Ro.= 70 --------------------70 – 20 0. ceiling/roof where the ceiling is composed of 1/2 in.38/(2717/1980) = 4.92 0. R film Δt film ------------= --------------R total Δt total t i = 70°F 40% RH dew point = 44.25 Face brick 0. Refrigerating and Air-Conditioning Engineers. © (2009).20 Inside air 0. distribution.81 Btu/h·ft2 ·°F d. U = 0.38 u = 0.44 0. (www. Additional reproduction. centers. Ri 0.5 0.129) + 0.259 0. Us = 0.ashrae. and Air Conditioning—Solutions Manual 5.36 W/m2 ·K) — 0.c = 2.024 Uo.50. plasterboard nailed to 2 by 6 joists on 16 in. determine the required resistance of the insulation to prevent condensation from occurring at the deck insulation interface when indoor conditions are 70°F and 40% RH. Ventilating. no insulation.45 0.259) = 0. b. centers and the roof consists of asphalt shingles on 3/4 in. airspace in a metal frame d. plywood paneling b.61 + 2. c.68 0. For personal use only.15(0.61 + 2.29 + 2.29 U = 0. Inc.73 Roof 0.72⏐Principles of Heating. centers.c = 1/4. 3/4 in. It has no ceiling.93 Studs — Air { 0.107 Uav = 0. solid wood door with a wood storm door Copyrighted material licensed to University of Toronto by Thomson Scientific.76 2. plywood on 2 by 4 rafters on 16 in.86 Ui = 0. building wall consisting of face brick veneer.93 5. wood decking. and 5/32 in.5 + 0. ashrae. and 7.techstreet.44 Insulating board sheathing 2.45 Inside air 0. (www.5 mph wind. For personal use only. R U = 1/ΣR = 0. q = 0.40 Btu/h· ft 2 5.01 Fiber batt 7.Chapter 5—Load Estimating Fundamentals⏐73 5. 25/32 in. –0 Polyurethane (2 × 6. (www.50 Plywood. © (2009).81 Copyrighted material licensed to University of Toronto by Thomson Scientific.17 + 0. and 1/2 in. b.0728 Btu/h· ft ·°F Rw c. 1/8 in. ∑ R = 11.38. face brick. c.38 An outside wall consists of 4 in. Inc.06 Airspace 1. 1/4 in.81 – 0. 0.68 ΣR = 13. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.81 – 0. Inc.0 = 14. American Society of Heating. value of U c. plasterboard. Compute the summer U-factor for the wall of Problem 5. Determine the winter U-factor. . heat gain per ft2 R Outside air 0.25) 12.38. Additional reproduction.89.37 A prefabricated commercial building has exterior walls constructed of 2 in. 1 2 U w = ------.084 U = 0.067 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Outside air 0. 72°F indoor air. mineral fiber batt between the 2 by 4 studs.25 = 11. ∑ R = 11. Refrigerating and Air-Conditioning Engineers. If full wall insulation is used.74 b. aluminum sheet and 1/4 in.39 Solve the following: a. Determine: a. veneer plywood.31 Inside air 0.68 ΣR = 11. U = 0. overall thermal resistance b.com).= 0. expanded polyurethane bonded between 1/8 in.0 + 11.38 if the wind velocity is 30 mph. Compute the winter U-factor for the wall of Problem 5. distribution.085 U = 0.81 – 1.01 – 7.085 Btu/h·ft2 ·°F 5.17 + 1 ⁄ 0.org).88.0728 ( 105 – 72 ) = 2.17 Face brick 0.0 Plasterboard 0. c. b. compute the summer U-value for the wall of Problem 5. insulating board sheathing 2 in.72. a. Design conditions include 105°F outside air temperature. ∑ R = 11.25 Aluminum.26 = 11. . Door: solid wood. Determine the summer U-factor for each door. © (2009).0706 Btu/h ft 2 °F ∑ R i 0. Ventilating.40 An exterior wall contains a 3 by 7 ft solid wood door. 2 by 4 studs (16 in. and a 6 by 7 ft sliding patio door with double insulating glass having a 1/2 in.31 + 0.68 b.1 + 0. nail-base insulating board sheathing. 1 1 U = --------.60 5. oc) with full wall fiberglass insulation. and Air Conditioning—Solutions Manual 5. 3/4 in.= ----------------------------------------------------------------------------------------------..68 = 2. thick. Inc.25 + 0.76 Btu/h· ft ·°F 2 5. (www.org).79 + 1..72 + 2.31 + 0.= ----------------------------------------------------------------------------------------------. 1 3/8 in.68 R ∑ 2 2 U = 0.31 + 0.99 Sliding patio door: U = 1 ⁄ R = 0.25 + 0.74⏐Principles of Heating. R w = 3.68 = 1. 1 1 U i = ---------.784 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 5.39 – 0. 4 in.50 + 0. U s = 0. 1/4 in.com).25 + 0. 1 3 ⁄ 8 in.41 If the doors of Problem 5. determine the U-factor for each door. wood door: R inside = 1 ⁄ 0. 1 1/2 in.960 W ⁄ ( m ·K ) ] a. paneling b.14 + 4. 1. For personal use only. distribution.79 + 1. cellular glass board insulation.46 Btu/h· ft ·°F Sliding patio door: U = 0. wood dome: U = 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 9 in.25 = 3. American Society of Heating..43 Determine the summer U-factor for the following building components a. 2 1 3 ⁄ 8 in.44 + 1.5 S = ------.= -----------------------------------------------------------------------------------------------------------0.703.ashrae. Inc. thick.5% 16 U w = 0. .( 100 ) = 9. (www. airspace and metal frame..264 R s = 3.0 + 0. with wood and glass storm door Copyrighted material licensed to University of Toronto by Thomson Scientific.35 + 0.17 + 0.33 R inside = 1 ⁄ 0. concrete. plywood paneling.68 1 1 U s = ----------.= 0.81 – 0. Additional reproduction.= 0. 1 in.66 U = 1 ⁄ R = 0. 1 by 8 in.40 are between the residence and a completely enclosed swimming pool area. 1/2 in.techstreet.14 + 11.25 + 0.17 + 0. 1/4 in.703 – 0.133 Btu/h ft 2 °F ∑ R s 0. airspace.42 Determine the winter U-factor in W/(m2 ·K) for the wall of a building which has the following construction: face brick.169 Btu/h· ft ·°F [ 0.27. Refrigerating and Air-Conditioning Engineers. Wall: wood drop siding. org).46 Copyrighted material licensed to University of Toronto by Thomson Scientific.61 + 19.44 Determine the combined ceiling and roof winter U-value for the following construction: The ceiling consists of 3/8 in. ceiling joists.17 + 0. and one 1 3/4 in. Six inches of fiberglass (mineral/glass wool) insulation fills the space between the joists.16 R t = -----.= 0.12 ΣR = 0.Chapter 5—Load Estimating Fundamentals⏐75 area to ceiling area is 1:3.c = 0.46 A wall is 20 by 3 m.ashrae. The wall itself has the construction of Problem 5. U c nU R 1.+ ---------.= ---------------------------------------------------------------.6 ) ( 37 ) + ( 4. Additional reproduction.techstreet. The wall proper consists of one layer of face brick backed by 250 mm of concrete with 12 mm of gypsum plaster on the inside. respectively.463 0. kW. (www.6 m2 Awindow = (0. one double-glazed (1/2 in.89 A w = 4 × 3 × 5 = 60 ft 2 A p 5. Inc.0487 0. Window: (Assume A1 frame operable) U = 4.02 + 0.064 5. © (2009). 5 1/2 by 10 ft.44 + 0. distribution.20. 5.4 ) ( 37 ) q = 5800 W = 5.446 .08 0.934 + 0.93 W/(m2 ·K) Awall = (0. For indoor and outdoor design temperatures of 22°C and –15°C. R Wall Outside air Face brick Concrete.61 R ∑ ceiling Ceiling: 1 1 2.5 × 10 = 55 ft w A wallproper = ( 70 × 8 ) – 60 – 55 – 21 = 424 ft 2 2 A d = 3 × 7 = 21 ft U wall = 0.24 W/(m2 ·K) q = UA Δt = ( 2.= 0. Specify the U-factor and corresponding area for each of the various parts of the wall with normal winter air velocities.4 m2 U = 2. 3 by 7 ft. Refrigerating and Air-Conditioning Engineers. The attic is unvented in winter. 4 Windows: U sin gle pane = 0.138 0.54 + ---------.14)(20 × 3) = 8. airspace) picture window. 12 mm Inside air 0.com). determine the heat loss through this wall. For personal use only. . The ratio of roof Roof: 1 1 U roof = -----------------. The pitched roof has asphalt shingles on 25/32 in.= 22.86)(20 × 3) = 51.24 ) ( 51.= ---------------------------------------------------------------.62 R ∑ roof 1 1 U ceiling = -----------------------.078 0.50 Door: U = 0. Inc.03 0. American Society of Heating.32 + 0. 250 mm Plaster. gypsum board on 2 by 6 in. thick solid wood door. The wall contains four 3 by 5 ft wood sash 80% glass single pane windows each with a storm window. solid wood sheathing with no insulation between the rafters.93 ) ( 8.21.8 kW 2 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Picture window: U = 0.45 The west wall of a residence is 70 ft long by 8 ft high.= 20.3 2 U o . which includes 14% doubleinsulating glass windows with a 6 mm airspace. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. (www.045 Btu/h· ft ·°F 5. distribution. Inc. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. For personal use only.Copyrighted material licensed to University of Toronto by Thomson Scientific. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . © (2009). Additional reproduction. (www.ashrae. Inc. American Society of Heating.techstreet.com).org). (www. Refrigerating and Air-Conditioning Engineers. (www.ashrae.org). distribution.Solutions to Chapter 6 RESIDENTIAL COOLING AND HEATING LOAD CALCULATIONS Copyrighted material licensed to University of Toronto by Thomson Scientific. For personal use only. American Society of Heating. Inc. Refrigerating and Air-Conditioning Engineers. Additional reproduction. © (2009).techstreet.com). Inc. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. (www. ashrae.Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc. distribution. Inc. (www. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.org). This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . For personal use only. Refrigerating and Air-Conditioning Engineers. © (2009).techstreet. American Society of Heating. (www.com). Additional reproduction. 167 + 0.13 ( 0.5 ΔT c.= 0.25 Btu/h·ft2 ·°F) Wall of 10% single pane glass and 90% of 6 in.org). Refrigerating and Air-Conditioning Engineers. Additional reproduction. q a = 0. American Society of Heating. poured concrete with ho = 6. Inc.6 ) Copyrighted material licensed to University of Toronto by Thomson Scientific.6 W/(m2 ·K).+ --------------------------------------.90 0.25 ) ( 150 ) ΔT q a = 70. q b = 0.5 ΔT b. 6.345 W/(m 3 ·K) R 2. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.42 ( 159 ⁄ 133 ) ( 1.345 ) ( 133 ) ( 41 ) = 1880 W ( 6420 Btu/h ) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 1 1 R = ---------.2 A house has a pitched roof with an area of 159 m2 and a U of 1.48 + 0.9 ) ( 150 ) ΔT 0.52 = 2.5 ) ( 150 ) ΔT + 0.techstreet.25 ( 0.25 ( 0. . The ceiling beneath the roof has an area of 133 m2 and a U of 0.62 q c = 123 ΔT Wall (c) has greatest heat loss.42 W/(m2 ·K). For personal use only.Chapter 6—Residential Cooling and Heating Load Calculations⏐79 6.1 Determine which of the following walls of 150 ft2 gross area will have the greatest heat loss: a. © (2009).75 ) ( 150 ) ΔT + 1. distribution. Determine the heat loss through the ceiling. The attic is unvented in winter for which the design conditions are –19°C outside and 22°C inside.25 Btu/h·ft2 ·°F) Wall of 50% double-glazed windows with the remainder of the wall brick veneer (U = 0.88 kW (6400 Btu/h)] 1 1 q = UA ΔT U = --.5 ) ( 150 ) ( ΔT ) q b = 70.0 and hi = 1.= ------. 1 q c = 1.com).38 + 0. b. (www. Inc.( 0. (www.9 q = ( 0.= 2. Wall of 25% single glass and the remainder brick veneer (U = 0.13 ( 0.6 Btu/h·ft2 a.ashrae. c.1 ) ( 150 ) ΔT + ----------------------------------------------.69 ( 0. [Ans: 1. 433 ) With storm window. 3/4 in.4 ) = 2254 Btu/h U C/R = 0.0525 ) ( 2185 ) ( 85. American Society of Heating. 3/4 in.61 6. 0.08 6. on centers) ceiling joists. distribution.68 10. 4 in. Insulation.17 0.55. and Air Conditioning—Solutions Manual 6.44 1. The ceiling consists of 1/2 in.85 ) ( 0.61 19. .433 Uav (0.1)(0. 1/2 in.0 1.4 ) = 9796 Btu/h Window loss = ( 0.01 – 0.047.17 ER Attic air Insulation Acoustical tile.4 ) = 5533 Btu/h Ceiling loss = ( 0.ashrae. t i = 72°F (selected) . Additional reproduction.19 0. plasterboard interior wall. 2 1/2 in.095 ) = 0. Two 4 by 6 ft single pane glass windows with storm windows. Assume wood frame.7 1. glass fiber insulation in 2 by 4 stud space (16 in. acoustical tile with R-19 insulation between the 2 by 6 (16 in.31 UR = 0. Centers U avg = ( 0.08 – – 4.16 Ui = Us = 0. The roof has asphalt shingles on 3/4 in.095.4°F ( Fig. 1/2 in.55 ) ( 48 ) ( 85.08 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Refrigerating and Air-Conditioning Engineers. t c = – 13.047) = 0.9)(0. Plywood sheathing.19 0.139 ) + ( 0. face brick. A 2185 ft2 ceiling topped by a 2622 ft2 hip roof. c. plywood sheathing on the roof rafters. on centers).25 1.054 + ( 2622 ⁄ 2195 ) ( 0.45 Inside air 0.62 2.45 Rs 0.53 7.6 % Value Wall loss = ( 0.10 Ri 0.122) + (0. Inc. 2-1/2 in. (www. Inside air ER w Rs 0.15 ) ( 0. Wall having 648 ft2 of area and construction of 4 in. Ventilating. treat as double glazing Wwindow = 0. (www.054 1 U C/R = -----------------------------------------------------------------------------1 ⁄ 0.com). © (2009). The attic is unvented in winter.122 S = 10% Ui = 0.1 ) ( 648 ) ( 85.44 1. 4-4 ) Wall Ceiling Ri 0.35 0.44 1. plywood sheathing.17 0. ΔT design = [ 72 – ( – 13.61 21.techstreet.3 Determine the design winter heat loss through each of the following components of a building located in Minneapolis.80⏐Principles of Heating.17 0. Airspace Studs Plasterboard Roof Copyrighted material licensed to University of Toronto by Thomson Scientific. Minnesota: a. Inc.4°F Selected ⇑ Outside air Shingles Plywood Attic Air ER ⇑ 99. Us = 0.0525 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Outside air Face brick.41 R 0.org).68 0.4 ) ] = 85. For personal use only.139 S ≅ 15% for 2x on 16 in.61 8. b. 6-1) Δt = t i – t g = 31 q = ( UA Δt ) walls + ( UA Δt ) floor q = ( 0.0042 – 0. For personal use only.75 AC/h ) ⎜ -------------------------⎟ = 219 cfm ⎝ 60 min/h ⎠ q s = 1.480 ft 3⎞ Infiltration ≈ ( 0. Wall area = ( 2 ) ( 8 + 12 ) ( 2. determine a.1 ) = 84 m 2 .av = 0.100 Btu/h (Table 6-18) b.5 For a frame building with design conditions of 72°F indoor and 12°F outdoor.4 If the building of Problem 6. on grade without perimeter insulation [Ans: 12. 3 by 5 ft. 3 by 7 ft.Chapter 6—Residential Cooling and Heating Load Calculations⏐81 6.100 Btu/h] Single glass double-hung window.81 ) ( 6 × 7 ) ( 72 – 12 ) = 2041 Btu/h (Table 5-16) 6.164 W/(m 2 ·°C) t i = 21°C ( selected ) . and entirely below grade.980 W/(m 2 ·°C) U f .6 Determine the heat loss for a basement in Chicago. of standard concrete construction. metal frame with double insulating glass having 1/4 in. Refrigerating and Air-Conditioning Engineers.164 ) ( 96 ) ( 31 ) = 3040 W This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto a.58 ) ( 3 × 7 ) ( 72 – 12 ) = 731 Btu/h (Table 5-17) d. determine the heat loss through each of the following components: a. b.av – A = 2 – 12 = – 10°C (Table 4-8) (Fig. Sensible heat load due to infiltration Latent heat load due to infiltration ⇓ Estimated ⎛ 17 .av = 0. d. Additional reproduction.200 Btu/h ( ΔW ) = 4840 ( 219 ) ( 0. (www.3 is a residence having a volume of 17.1 ( 219 ) ( 88 ) = 21 . q = UA ( t i – t o ) = ( 0. thick solid wood door.com). (www. Inc.10 (cfm) q l = 4840 (cfm) ( Δt ) = 1. c. which is 8 by 12 by 2. Slab floor. 100% 6. q = UA ( t i – t o ) = ( 0. t g = t w .ashrae. 56 by 28 ft.00038 ) = 4049 Btu/h 72°F.techstreet.20 ) [ 2 ( 56 + 28 ) ] ( 72 – 12 ) = 12 . Copyrighted material licensed to University of Toronto by Thomson Scientific. with storm window in common metal frame [Ans: 780 Btu/h] 1 3/8 in. distribution. Illinois. air space [Ans: 2192 Btu/h] q = F p P ( t i – t o ) = ( 1. Inc.4°F. with 25% single glazing [Ans: 490 Btu/h] Sliding patio door. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. © (2009).org). 25% ⇑ ⇑ –13. q = UA ( t i – t o ) = ( 0.81 ) ( 3 × 5 ) ( 72 – 12 ) = 729 Btu/h (Table 5-16) c.980 ) ( 84 ) ( 31 ) + ( 0.1 m high. 6 by 7 ft.480 ft3 and is equipped with a humidifier set for 25% RH. . American Society of Heating. Floor area = 8 × 12 = 96 m 2 U w . b. Inc.43 0.techstreet. Fig.900 Btu/h Q floor = U f A ( t i – t o ) = ( 0. Determine the design heat loss from the uninsulated below grade concrete walls and floor.200 Btu/h This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 6. The ratio of roof area to ceiling area is 1.5 – 23 ) ] = 10 . °F Appropriate temperature difference.17 + 0. including a 72°F inside dry bulb at the 5 ft line.62 1 1 1 1 R t = -----.com).049 ( 1. located in Des Moines. 30 by 100 ft.82⏐Principles of Heating.44 + 1. has a total ceiling area of 1960 ft2 and consists of 3/8 in.6% value b. d.3 ) ( 0.9°F [ Table 4-7 ] .157 ) ( 260 × 9 ) [ 72 – ( 35. Outside design temperature.026 (conservative) [Table 6-17] Q walls = U av A ( t i – t o ) = ( 0.= 21.61 + 19 + 0. 1 U T = ---------.7 A residence located in Chicago. gypsum board on 2 by 6 ceiling joists. °F Appropriate overall coefficient U. t a = 35.= 0. 72°F inside – ( – 4°F ) outside = 76°F 1 U c = -------------------------------------------------------. Iowa.org). © (2009).046 21. Additional reproduction.08 + 0.+ ---------.= 0.34 + 0. determine a.157 (use conservative 8 ft value) [Table 6-16] Floor: U f = 0.3. solid wood sheathing with no insulation between the rafters. d.43 ) c.61 1 U R = ------------------------------------------------------------. (www.049 0. The attic is unvented in winter. Ventilating.046 ) ( 1960 ) ( 76 ) = 6852 Btu/h t o = – 6. Btu/h·ft2 ·°F Ceiling heat loss q. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Six inches of fiberglass (mineral/glass wool) insulation fills the space between the joists. Btu/h t o = – 4°F. b. Chicago CO a. The pitched roof has asphalt shingles on 25/32 in.= ------------. (www. has a conditioned space which extends 9 ft below grade level. distribution. Inc.300 Btu/h -------------------------------Total = 32 .5°F .ashrae.8 q = U T A Δt = ( 0. 6-1] Walls: U av = 0. American Society of Heating.= 0.8 A residential building. For winter design conditions.026 ) ( 3000 ) [ 72 – ( 35. Illinois. Refrigerating and Air-Conditioning Engineers. A = 23 [Table 4-8.5 – 23 ) ] = 21 . 99. For personal use only.8 U c nU R 0. c. and Air Conditioning—Solutions Manual 6.+ ---------------------------. . Copyrighted material licensed to University of Toronto by Thomson Scientific. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. (www. American Society of Heating. 1/2 in. no flooring above. solid with glass storm door D2: Sliding glass door.org). two section. double glazed. 16 in. concrete floor over 4 in.techstreet. 1 in. airspace W3: 5 by 3 ft wood sash casement. no insulation between rafters. 1 ft overhang on eaves. 1/2 in. fiberglass ceiling insulation Diagram for Problem 6. 3/8 in. double glazed. Missouri) with a.ashrae. gravel Fireplaces: One in living room on first floor Garage: Attached but unheated Windows: W1: 3 by 5 ft singles glazed. 1 3/4 in. Louis. 1/2 in. 1/2 in. 2 by 6 rafters. 25/32 in. 4 in.Chapter 6—Residential Cooling and Heating Load Calculations⏐83 6. centers. each 3 by 6 ft. Refrigerating and Air-Conditioning Engineers. 10 in.9 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Basic Plan Wall construction: Face brick. Full wall fiberglass insulation and 4 in. double glazed. double-hung wood sash. fiberglass wall insulation and 2 in. gypsum board ceiling Roof: Asphalt shingles on solid wood sheathing.com). gypsum board interior Ceiling: 2 by 6 ceiling joists. 1:4 pitch. (www. Inc. 8 in. © (2009). double glazed. insulating board sheathing. fiberglass ceiling insulation b. airspace. For personal use only.000 Btu/h (15 kW)] Copyrighted material licensed to University of Toronto by Thomson Scientific. all below grade. distribution. Additional reproduction. on center. airspace Doors: D1: 3 by 6 ft-8 in. 2 by 4 studs on 16 in. 3/8 in. concrete walls. weather stripped with storm window W2: 10 by 5 1/2 ft picture window. . aluminum frame [Ans: (b) 52. airspace W4: 3 by 3 ft wood sash casement. Inc.9 Determine the heating load and specify the furnace for the following residence (located in St. no ceiling applied to rafters. no overhang on gables Full basement: Heated. 26 0.81 0.026 ) ( 70 × 28 ) ( 50 ) = 2550 Btu/h Infiltration: (Fig.10 ) ( 72 – 4. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.1°F W i = 0. 5-7) ACH ≅ 0. Additional reproduction. 010 Btu/h This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto NOTE: U or R values may differ depending on which table is used.0042 Surface W1 windows (6) W2 windows W3 windows (2) W4 windows (2) D1 door (West) D1 door (South) D2 door Wall (N.9 67.0042 – 0.9 67. Ventilating. distribution.50 0. (www.00097 ) = 2036 Btu/h q T = 24. . Copyrighted material licensed to University of Toronto by Thomson Scientific.366 + 12 .com).org).5 ( 70 × 28 × 8 ) ⁄ 60 ] ( 1.300 + 2550 + 9758 + 2036 = 51.9 67. E) Wall (S) Ceiling/Roof A 90 55 30 18 20 20 40 1091 204 1960 W o = 0.157 Btu/h· ft 2 ·°F (Table 6-16) Basement floor: q = ( 0.ashrae. Refrigerating and Air-Conditioning Engineers.techstreet.9 67.067 Δt = 67.9 67. and Air Conditioning—Solutions Manual Problem 6.067 0.9 continued t a ≅ 44°F A ≅ 22°F t i = 72°F t o = 4.51 0. W.00097 U 0.26 0. For personal use only.5 ( 70 × 28 × 8 ) ⁄ 60 ] ( 4840 ) ( 0. (www. American Society of Heating.157 ) [ ( 2 ) ( 70 + 28 ) ( 8 ) ] ( 50 ) = 12 . Inc.84⏐Principles of Heating.9 t g = 44 – 22 = 22°F q 3114 1872 1038 621 349 349 2202 4963 890 8916 24366 Btu/h Δt = 72 – 22 = 50°F Basement wall: U av = 0.068 0.9 67.026 (Table 6-17) q = ( 0.1 ) = 9758 Btu/h q L = [ 0.300 Btu/h U f = 0. Inc.9 67.9 67.51 0.9 67.51 0.5 q s = [ 0. © (2009). = 0.ashrae.77 kW)] q = UAΔt Wall: q = (0.85 Floor: F P = 0. the heat losses from a room of a building as shown in the diagram. The window has a wooden sill and the plate glass (U = 1.480 ) ⁄ 60 = 104 Heat Losses: qglass = (0.Chapter 6—Residential Cooling and Heating Load Calculations⏐85 6. 260 Btu/h qwalls = (0.89 U ceiling = -------------------------------------------------------------------------------------------------------------.5 ⁄ 5 ) + ( 1 ⁄ 7. Additional reproduction.7 kW) NOTE: U or R values may differ depending on which tables are used.11 Calculate. . Inc.techstreet. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Diagram for Problem 6. Refrigerating and Air-Conditioning Engineers.39)(520 – 192– 21)(70 – 40) Door: q = (0.2 ) 1 U glass = 0.65 ) + ( 0.org).68)(64)(70 – 0) = 3.5 ⁄ 5 ) + ( 1 ⁄ 0. P = 64 ft Infiltration: Assume 1/2 ACH ⇒ CFM = 1.381 ( 1 ⁄ 1.06)(192)(70 – 40) = = = 3592 403 6105 10. if the outside ambient is 0°F.10 Determine the total conductance loss through the wall panel as shown below.800 Btu/h qfloor = (0.89)(245)(70 – 0) = 15.65 ) + ( 16 ⁄ 9 ) + ( 0.75 ) ( 9 ) + ( 13.26 ) + 0.06) covers 85% of the window area.98 + 0.4 kW)] Volume = ( 24 ) ( 40 ) ( 13 ) = 12 . American Society of Heating.381)(587)(70 – 0) = 15.820 Btu/h (18.10)(104)(70 – 0) = 8. (www. [Ans: 9640 Btu/h (2. distribution. for design purposes.com).100 Btu/h 6.050 Btu/h qinfil = (1. (www. 660 Btu/h qceiling = (0.800 Btu/h (13.5 ) ( 9 ) = 245 ft 2 Net wall = ( 24 ) ( 13 ) + ( 40 ) ( 13 ) – 245 = 587 ft 2 Floor/Ceiling = ( 24 ) ( 40 ) = 960 ft 2 1 U wall = -------------------------------------------------------------------------------------------------. Inc.11 Copyrighted material licensed to University of Toronto by Thomson Scientific.143)(960)(70 – 0) = 4. For personal use only.480 ft 3 Glass = ( 13.2 ( 12 . © (2009).000 Btu/h Total Loss = 46. [Ans: 62.64)(21)(70 – 40) Window: q = (1.143 ( 2 ⁄ 1.= 0.68 . 3 . U c = 0.9 ) ( 0. determine: a. Ventilating. q tran = AU ( Δt ) = ( 6.com).08Vc ) + A R U R Copyrighted material licensed to University of Toronto by Thomson Scientific. (www.61 ( 1960 ) ( 0.1 ) ( 0. A c = 1960 . distribution. The effect of the joists themselves can be neglected.14 Estimate the heat loss from the uninsulated slab floor of a frame house having dimensions of 18 by 38 m.362 ) ] t a = -----------------------------------------------------------------------------------------------------------------------------------------------.= 0.6°F 1960 [ 0.61 + 0.61 + 0. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto t o = 7. The attic contains louvers which remain open all year. Assume single story frame house.62 a.6 kW] q = FP P ( t i – t o ) = ( 2.08 ( 0.055 ) ( 1960 ) ( 67.115 ) + ( 0.362 0.362 ) Δt = 75 – 7. Kentucky.055 ) ( 75 ) + 6 [ ( 1. The residence is located in Louisville.org). For personal use only. Six inches of fiberglass (mineral/glass wool) insulation fills the space between the joists.= 9. q s = ( 1.ashrae. Inc. Leakage area = ( 3 + 0. The house is maintained at 22°C. (a) appropriate temperature difference Δt overall coefficient U ceiling heat loss.= 0.32 + 19 + 0.08A c Vc + A R U R ) · t a = --------------------------------------------------------------------------Vc ≅ 0.5 ) ] + ( 2548 ) ( 0. © (2009).3 ) ( 3 ) ( 0.5 ) + ( 2548 ) ( 0.32 + 7. Additional reproduction. Refrigerating and Air-Conditioning Engineers. calculate: a.12 A room has three 760 by 1520 mm well-fitted double hung windows. For winter design conditions.055 Btu/h· ft 2 ·°F c. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.08 )1960 ( 0.techstreet.00413 ) ( 6.7 ) 2 ] 11.4 W b.7 m/s wind.115 ⎪ ⎭ 0. Louisville: 1 U R = ------------------------------------------------------------.61 ⎪ ⎬ U av = ( 0.52 ) = 11.055 1 U cs = ------------------------------------------------------------.= 0.14 + 0. b. 1 U ci = -------------------------------------------------------. The ratio of roof area to ceiling area is 1:3. t i = 75°F . c. (www. American Society of Heating. and Air Conditioning—Solutions Manual 6.15 kW ) 6. b.1 = 67. gypsum board on 2 by 6 in. plywood with no insulation between the rafters.4 cm 2 = L 1/4 2 1/2 · V = [( A ) ( Δt ) + ( B ) (V )] L = [ ( 0.17 + 0.5 cfm/ft 2 · A c ( U c + 1. q = U c A c ( t c – t a ) = ( 0. ceiling joists.77 + 0.232 ) ( 5800 ⁄ 3600 ) ( 39 ) = 73.055 + 1.52 ) ( 39 ) = 838 W 6.76 ) ( 1. Inc.00188 ) ( 39 ) + ( 0.1°F . . A R = 2548 · A c U c t c + t o ( 1.9°F b.76 ) ( 1.8 m 3 /h a.13 A residence has a total ceiling area of 1960 ft2 and consists of 3/8 in.049 ⎫ 0. A R ⁄ A c = 1.2 ) ( 3 ) ( 0. For design conditions of –1°C and 21°C.9 ) = 7319 Btu/h ( 2. 6.049 ) = 0. Outdoor design temperature is –15°C in a region with 5400 degree kelvin days. heating load from air leakage heating load from transmission through the windows. The pitched roof has asphalt shingles on 5/8 in. [Ans: 8.4 = 5.86⏐Principles of Heating.07 ) ( 112 ) [ 22 – ( – 15 ) ] = 8580 W NOTE: U or R values may differ depending on which tables are used.44 + 0. 7 W/(m2 ·K). (www.17 For a residence in Roanoke.7 ) ] = -------------------------------------------------------------------------------------------------------------------------------------------------203 ( 0. Refrigerating and Air-Conditioning Engineers.9 ) = 2125 W ( 7250 Btu/h ) 1 U o . Inc.9°C .649 ) ( 72 ) + ( 14.2 ) ( 1.543 ) NOTE: U or R values may differ depending on which tables are used.649 0. plywood has an area of 2950 ft2.= ---------.08 ) ( 325 ) + ( 2950 ) ( 0.059 ) + ( 244 ) ( 2. · A c U c t c + t o ( 1200A c Vc + A R U R ) t a = ----------------------------------------------------------------------------· A c ( U c + 1200Vc ) + A R U R ( Eq. Virginia. Determine the ceiling heat loss W with ventilation and compare to the loss if there had been no ventilation.c = --------------------------------------------------------------------------.543 0.3 ) ( 203 ) ( 22 + 12. q = ( 0. Additional reproduction.61 + 0.com). plasterboard on 2 by 6 joists on 24 in.32 + 0.649 ) + ( 1.Chapter 6—Residential Cooling and Heating Load Calculations⏐87 6. distribution.1°F ( 2300 ) ( 0. The attic has forced ventilation at the rate of 325 cfm. Inc.7 ) t a = – 12. (www. . For personal use only.62 + 0.2°F Roof: 1 1 U R = ------------------------------------------------------------.84 ( 2300 ) ( 0.44 + 0. 4-6 ) ( 203 ) ( 0.30 W/(m2 ·K). an attic ventilation rate of 59 L/s is provided with outside air at –13°C.3 ) + [ ( 203 ) ⁄ ( 244 ) ( 2. Determine the attic air temperature at winter design conditions.techstreet.275 ( 1 ⁄ 0. The ceiling area is 203 m 2 and Uclg = 0. The 2300 ft2 ceiling consists of 3/8 in.54 t o = 14.059 ) ) + ( 244 ) ( 2. Inside design temperature is 22°C.= 0. Copyrighted material licensed to University of Toronto by Thomson Scientific.61 1.275 ) ( 203 ) ( 22 + 13 ) = 1954 W ( 6670 Btu/h ) If uninsulated: t i = 72°F Ceiling: 1 1 U c = -------------------------------------------.3 + 1200 ( 0.08 ) ( 325 ) + ( 2950 ) ( 0.15 Repeat Problem 6.14 for the case where insulation [R = 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 6. © (2009).92 ) ( 112 ) ( 22 + 15 ) = 3800 W 6.ashrae.16 To preclude attic condensation.9 (m2 ·K)/W] is applied to the slab edge and extended below grade to the frost line. American Society of Heating.7 ) ] q = ( 0.17 + 0.= 33. centers.543 ) t a = -------------------------------------------------------------------------------------------------------------------------------------------. q = FP P ( t i – t o ) = ( 0.= 0.= ---------. the hip roof consisting of asphalt shingles on 1/2 in.3 ) ( 22 ) + ( – 13 ) [ ( 1200 ) ( 203 ) ( 0.= 0. The roof area is 244 m 2 and Uroof = 2.org).61 1. Cinder block. Copyrighted material licensed to University of Toronto by Thomson Scientific. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . to = –13.11 ) ( 85. If the wall of Part (a) is converted to 60% single glazed glass. cinder aggregate concrete blocks.com). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.4°F R 0.18 Solve the following: a. Q = ( 0. Btu/h. American Society of Heating.= 0. clay tile interior. b. Inc.60 ) ( 115 ) ( 10 ) ( 1. (www.44 ~ 1. a 3/4 in. 8 in. Btu/h? ti = 72°F. © (2009). Minnesota.72 4. Ventilating. Insulation. 8 in. For personal use only.13 ) ( 85.techstreet. organic bonded glass fiber insulation. Inc. (www. what is the winter design heat loss through the total wall.12 ΣR Outside air Face brick Air gap.0 1. 4 in. 803 Btu/h Q = ( 0. Refrigerating and Air-Conditioning Engineers. Clay tile.11 ) ( 115 ) ( 10 ) ( 72 + 13. and 4 in.40 ) ( 115 ) ( 10 ) ( 0.4 ) = 70.110 Btu/h· ft 2 ·°F ∑R b. distribution. 1 in. and Air Conditioning—Solutions Manual 6. Determine the design heat loss through the wall in winter. a. Inside air 1 U = --------.4 ) = 10. consists of face brick. 1 in. air gap. 911 Btu/h U glass ¡ NOTE: U or R values may differ depending on which tables are used. A 115 by 10 ft high wall in Minneapolis.org). Additional reproduction.ashrae.17 0.88⏐Principles of Heating.4 ) + ( 0. 3/4 in.11 068 = 9.0 1. 20.61 1.700 Btu/h Problems 6.17 0. 3 in.75 in. 3/4 in. some 18 in. (www.44 0. metal-framed glass (1/4 in.0763)(3000)(69) = 15.070 Btu/h = (1. nonoperable.975 Btu/h = (4840)(250)(0.com). Additional reproduction. plywood sheathing.84 1 30 × 100 × 10 Infiltration: @ 1/2 ACH --. of rigid.33 0.techstreet.19 Determine the design heating load for a residence. Cellular glass insulation. thick. 2 in.46)(168)(69) = 5. (www. Inc.21. 4 in.0676)(1262)(69) = 5. Air Air Acoustical tiles.ashrae. cellular glass insulation.25 0. φ o = 100% rh . W i = 0.6.10)(250)(69) = 18.10 0. lightweight concrete deck.17 0. 2 in.330 Btu/h = (0.93 12.⎞⎠ = 250 CFM 2 60 Design Values: t i = 72°F .68 14. 4 in. Inside air ΣR = Uw = R 0. plasterboard Ceiling/roof: 3 in.12 0.84)(260)(69) = 15. American Society of Heating. 1/2 in. acoustical tiles.79 0.61 0. and a drop ceiling of 1/2 in. Inside air ~ ΣR = Uc = Windows: U g = 0.940 Btu/h Total Loss = 121. Plasterboard. For personal use only. © (2009). 3/4 in.Chapter 6—Residential Cooling and Heating Load Calculations⏐89 6. Plywood sheathing. below the roof.0763 Copyrighted material licensed to University of Toronto by Thomson Scientific. which has an uninsulated slab on grade concrete floor. distribution. built-up roofing.00092 Heat Losses: Walls: Roof: Doors: Windows: Floor: Infiltration: Q Q Q Q Q Qs QL = (0.880 Btu/h = (0.005 – 0.69)(1170)(69) = 55. Refrigerating and Air-Conditioning Engineers. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Doors: U d = 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. expanded rubber insulation.69 Floor: F P ≈ 0.61 13. 6. 30 by 100 by 10 ft. and 1/2 in. Walls Outside air Face brick.org). and 6.42 9. Doors: Two 3 by 7 ft.0676 Ceiling/Roof Outside air Built-up roofing Lightweight concrete. . Inc.790 Btu/h = (0. face brick. The construction consists of Walls: 4 in. to be located in Windsor Locks.00092) = 4.45 0.46 R 0. Windows: 45% of each wall is double pane.10 0. NOTE: U or R values may differ depending on which tables are used.700 Btu/h = (0. solid wood doors are located in each wall. 1/2 in. illustrated in Section 6. Rubber insulation. 1. W o = 0.⎛⎝--------------------------------. air gap). 4 in. Connecticut.22 are essentially openended but should be solved using the RLF method.005 t o = 3°F . φ i = 30% rh . © (2009).techstreet. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .ashrae. (www. (www. Inc. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.com). Additional reproduction.Copyrighted material licensed to University of Toronto by Thomson Scientific. Refrigerating and Air-Conditioning Engineers. For personal use only. distribution.org). Inc. American Society of Heating. ashrae. For personal use only.org). This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .techstreet. © (2009). Inc. Refrigerating and Air-Conditioning Engineers. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Additional reproduction. (www.Solutions to Chapter 7 NONRESIDENTIAL COOLING AND HEATING LOAD CALCULATIONS Copyrighted material licensed to University of Toronto by Thomson Scientific. (www. American Society of Heating.com). Inc. distribution. American Society of Heating. © (2009). Additional reproduction.Copyrighted material licensed to University of Toronto by Thomson Scientific. For personal use only. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. (www. distribution. (www.com). Inc. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . Inc.ashrae. Refrigerating and Air-Conditioning Engineers.org).techstreet. 3 9.327 ) = 9. b.81 Btu/h· ft 2 ·°F.1 The exterior windows are of double insulating glass with 0. a. For personal use only.: Ω = 50° Copyrighted material licensed to University of Toronto by Thomson Scientific. Sept.9 ft P west = 7 × cot 37° = 7 ( 1.: Ω = 37° P south = 7 × cot 50° = 7 ( 0.com). Inc. on the other face of the building? What is the elevation of the top of the awnings above the sidewalk? P = S H cot Ω West.25 in.3 = SH ( 0. Determine the design U-factor for cooling for the window. if fixed 7. (www. Refrigerating and Air-Conditioning Engineers.M.68 Btu/h· ft 2 ·°F. c. ID #4: U = 0. the show windows are 7 ft high. distribution.ashrae. © (2009). (www.org). An aluminum awning with a 3 in. sun time? Which face of the building governs the awning dimensions? Where will the shade line be at 3 P.839 ) = 5. From Table 5-6. The bottom of the show windows are 2 ft 6 in. Sept. or U = 0. . Both south and west awnings are to have the same dimensions. What minimum distance should the strut extend from the building to keep the shade line on the windows at 3 P. (6 mm) airspace and have metal sashes. 9.2 A store in Lafayette.3 ft b. if operable. above the sidewalk.1 ft.8 ft above ground ⎝ 12 ⎠ This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto a. shadow to ground d. Inc. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.5 = 11.839 ) ⇒ S H = 11. West c. rise per horizontal foot is to be hung with the bottom strut at the window header.⎞ + 9. Indiana. d. South.M.techstreet.Chapter 7—Nonresidential Cooling and Heating Load Calculations⏐93 7. Additional reproduction. American Society of Heating.3 ⎛ -----. is on the northeast corner of an intersection with one street running due north. 67 Table 7-4. and Air Conditioning—Solutions Manual 7.27 ) ( 92 ) ( 0.27 . SHGCD ( 71 ) = 0.67 ) ( 1 ) + ( 12 ) ( 28 + 31 ) ( 0. A = 1093 W/m 2 exp ( E ⁄ sin β ) B = 0. . American Society of Heating.tan 37 = 1. Additional reproduction.ashrae.1 – 0. Table 7-4.186 ⁄ sin 66 ) E D = E DN cos θ = 283 ⋅ cos 71 = 92 Btu/h· ft 2 2 E d = CYE DN .138 + sin 71 ) ( 0.2 (typical) = 31 Btu/h· ft 2 ·°F q = ( 2. (www. M . For personal use only. Y = 0.9 + 552.94⏐Principles of Heating. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.3 Calculate the solar radiation entering through clear glass as shown below. © (2009). 1 P . Refrigerating and Air-Conditioning Engineers. Inc.186 .2 = 692 Btu/h This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Table 7-13.313 cos θ Y = 0. July 21.1 ft T = -----.725 ) ( 283 ) = 28 Btu/h· ft 2 E r = E DN ( C + sin β )ρg ( 1 – cos Σ ) ⁄ 2 .83 ) ] = 2.437 cos θ + 0.725 E d = ( 0.tan 66 ⁄ cos 37 = 4 ft 12 12 Area in Sun = ( width – M ) ( height – T ) A s = [ 3 – ( 1.com).27 ft 2 Total glass area = 3 × 4 = 12 ft 2 q = As ( E D )SHGC ( θ )IAC + AT ( E d + E r )SHGCD IAC A s = 2. cos Σ = 0 for vertical = 283 ( 0. C = 0.org). distribution. (www. θ = 71° .. [Ans 692 Btu/h] Mullion: M = P tan γ Tansom: T = P tan β⁄ cos γ 40° N latitude.55 + 0. Inc.138 1093 E DN = ---------------------------------------------.138 ) ( 0. south γ ( ψ = 0° from south ) = φ = 37°.2 ) ⁄ 2 ρ g = 0. β = 66° 17 17 M = -----.78 Copyrighted material licensed to University of Toronto by Thomson Scientific. SHGC ( 71 ) = 0. A T = 12 A E DN = ---------------------------------.= 892 W/m 2 = 283 Btu/h· ft 2 exp ( 0.78 ) ( 1 ) q = 139. IAC = 1 . Ventilating..83 ) ] [ 4 – ( 4 – 0.techstreet. 413 ) = ( 50 ) ( 4 ) ( 40 ) ( 1 ) ( 1.000 Btu/h (Table 7-14) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 1.503 ⇒ φ = 60° cos θ v = cos β cos γ = 0. (www. Determine the solar angle of incidence for a vertical wall facing 15° west of south when the sun has an azimuth of 79.8° δ ≅ – 10.5° sin β = cos L cos δ cos H + sin L sin δ ⇒ sin β = 0.8872 ⇒ θ v = 28. [Ans: 83. b.5 What environmental factors affect the solar intensity reaching the earth’s surface? Angle through which the solar radiation passes through the atmosphere increases or decreases the quantity of air mass. Copyrighted material licensed to University of Toronto by Thomson Scientific. . distribution. 7. Inc. before noon ) H = 0. ozone.ashrae.4° b.7 γ = φ – ψ = 64. For personal use only. Gas molecules. β = 75.M.6 Determine the heat being dissipated by 50 pendant mounted fluorescent luminaires with four 40 W lamps in each luminaire.2 .7°.2 –1 θ v = cos [ ( cos 75. cos θ v = cos β cos γ .7 ) ( cos 64.4°] a. Additional reproduction. © (2009). (www. Height of ground upon which the structure is to be built where the elevation is substantial. Refrigerating and Air-Conditioning Engineers.8° LCT = 8:30 + 4 ( 90 – 95 ) = 8:10 A.396 ⇒ β = 23. CST on October 22 at 32° N latitude and 95° W longitude.Chapter 7—Nonresidential Cooling and Heating Load Calculations⏐95 7. Dust and other contaminants 4. Equation of time = 15. 4-40 Watt lamps in each of 50 luminares. φ = 79. ψ = 15 .8°] Find the solar incident angle (for direct solar radiation) for a vertical surface facing southeast at 8:30 A.25 ( 215 ) = 53.7 How much sensible. 2.2° west of south and an altitude of 75.4 AST = 8:10 + :15 = 8:25 A. ( 215 min. factor ) ( 3.M. allow.15 ) ( 3. Inc.500 Btu/h 25.techstreet.400 Btu/h 7.3° cos φ = ( sin β sin L – sin δ ) ⁄ ( cos β cos L ) ⇒ cos φ = 0. and water vapor 3.500 Btu/h 12. q = ( wattage ) ( use factor ) ( spec. [Ans: 28. and total heat is contributed by 50 customers shopping in a drugstore? Sensible: Latent: (50)(250) = (50)(250) = Qtotal = 12. American Society of Heating.M. latent. 7.2 ) ] = 83.com).org).413 ) = 31 .4 Solve the following: a. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Inc. distribution. Inc.com). and Air Conditioning—Solutions Manual 7. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Determine heat dissipated to the space from the motor and pump. . This area should have separate system.61 + 0.96⏐Principles of Heating.ashrae. American Society of Heating. The air temperature at the underside of the floor is 100°F.( 10 × 25 )5 CFM/person + 0. The floor is 4 in.= ------------------------------------------------------------. concrete with vinyl tile finish. [Ans: 18.7 ) ( 1 ) ( 100 – 70 ) = 21 Btu/h· ft 2 Description General office Director’s room Conference room 5 private offices Size 25 by 50 ft 25 by 25 ft 10 by 25 ft 10 by 10 ft Occupancy 75 ft2 per person 16 people Plush furnishings Smoking permitted Values from Table 5-9: General Office: Directors Room: Conference Room: Private Offices: 25 × 50 ft cfm. and the room air temperature desired close to the floor is 70°F.70 1 1 0.+ 0.+ R con + R tile + ---hi hi q = ( U ) ( A ) ( Δt ) = ( 0.61 ---.06 ( 10 × 25 ) = 78 cfm 1000 ft 2 Smoking areas not covered by ASHRAE Standards.06 -------5 CFM/person × ------------------------------( 25 × 50 ) = 158 cfm 75 ft 2 /person ft 2 16 people × 5 + 0.31 + 0.10 Calculate the maximum heat gain through the floor for a room directly over a boiler room.9 W/m2)] 1 1 U = --------------------------------------------------. Ventilating.9 Calculate the heat gain to a room from an open deep fat fryer if (a) hooded and (b) nonhooded.11 An air-conditioning unit serves an office having the following areas: What quantity of outdoor air must be brought into the air-conditioning unit for ventilation? Copyrighted material licensed to University of Toronto by Thomson Scientific. Table 7-18 (a) Hooded: (b) q s = 47.06 ( 25 × 25 ) = 118 cfm 50 people ----------------------.techstreet. Total = 354 cfm This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 7. Additional reproduction.51 + 0. 800 Btu/h Nonhooded: NOT RECOMMENDED 7. © (2009). (www.= 0. For personal use only. [Ans: 3390 Btu/h] Table 7-16 1 Hp Motor Assume both in the space 3390 Btu/h 7.87 Btu/h·ft2 (18.8 A 1 hp motor driving a pump is located in a space to be air conditioned.org). (www. Refrigerating and Air-Conditioning Engineers. c. Copyrighted material licensed to University of Toronto by Thomson Scientific. b. Inc.4 × 9.100 ft3/60 min/h = 1910 cfm 158 + 118 + 78 % OA = -----------------------------------.1 m with a 3 m ceiling. what will be the percentage of outdoor air? [Ans: 10. 11%. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. supply air.ashrae.( Problem 7.4 by 9.11 supplies 3200 cfm to the ductwork.000 158 + 118 + 78 % OA = -----------------------------------. b.techstreet.= -----------------------------. Total volume of spaces: [(25 × 50) + (25 × 25) + (10 × 25)]9 = 19. (www.= 10 ach (high) 19 . 3200 ( 60 ) Supply air changes/h = ----------------------. Smoking is not allowed. d. b. a. .12 Suppose the fan of the air-conditioning unit in Problem 7. a “typical” number of 6 ach is sometimes used. Determine Sensible heat load from the occupants Latent heat load from the occupants Moisture added from the occupants The minimum volume of outdoor air for ventilation Suitable summer design inside dry-bulb temperature a.9 ( 27.Chapter 7—Nonresidential Cooling and Heating Load Calculations⏐97 7.11 ) × 100 = 11% 3200 For proper air distribution. Refrigerating and Air-Conditioning Engineers. cfm = 6 × 19.× 100 = 15% 1910 7.) What is the percentage of outdoor air? Suppose the minimum recommended quantities of total air and outside air are used. 15%] Assume smoking spaces are on separate dedicated system.3 lb/h ( 0. (www.( 235 ) = 2554 W 230 b.100 ft3 a. c.1 ) = 349 L/s 20 to 22°C This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto a. © (2009). e. q s = 25 100 --------. q L = 25 130 --------.13 A small parts assembly area in a factory has a working force of 25 men and occupies a space 27. e. If so. For personal use only.= 10. How many air changes per hour are being used? (Assume a ceiling height of 9 ft. q L ( Btu/h ) 3320 ( 3.00128 kg/s ) 1100 Btu/h 1100 Occupancy (Table 5-9 for metal shop) ( 5 × 25 ) + 0.com). Inc. distribution.( 235 ) = 3320 W 230 c. American Society of Heating. c. Additional reproduction. d.413 ) Moisture added = --------------------------.org). 17 0.110 Btu/h· ft 2 ·°F ∑R b. 8 in.4 ) = 10 .60 ) ( 115 ) ( 10 ) ( 1.4 ) + ( 0. © (2009). 4 in.0 1. a 3/4 in.4°F a. and Air Conditioning—Solutions Manual Problems 7. Refrigerating and Air-Conditioning Engineers. Minnesota consists of face brick. Inside air ΣR = Copyrighted material licensed to University of Toronto by Thomson Scientific. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.40 ) ( 115 ) ( 10 ) ( 0. distribution. (www.72 4. (www. Inc.= 0. A 115 by 10 ft high wall in Minneapolis. organic bonded glass fiber insulation.org). R 0.11 ) ( 115 ) ( 10 ) ( 72 + 13.” 7. American Society of Heating.ashrae. to = –13.4 ) U glass ¡ q = 70. If the wall of Part (a) is converted to 60% single glazed glass. Determine the design heat loss through the wall in winter in Btu/h.12 ) ( 85.0 1. cinder aggregate concrete blocks. clay tile interior. b. 3/4 in. Insulation.11 ) ( 85.14 through 7. 8 in.11 0.44 1. 1 in. No manual (hand) cooling load method is currently recommended by ASHRAE.18 Solve the following: a. For personal use only.com). Cooling loads for each of these buildings/spaces should be determined using the RTS software available on the CD provided with “Principles.12 1 U = --------.98⏐Principles of Heating. what is the winter design heat loss through the total wall in Btu/h? ti = 72°F.techstreet. Clay tile. . Additional reproduction. and 4 in. Inc. ~ Cinder block.17. 1 in.68 9. q = ( 0. Ventilating.800 Btu/h q = ( 0. 300 Btu/h This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Outside air Face brick Air gap. air gap. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. (www. Inc. (www. American Society of Heating. distribution. © (2009). For personal use only.com).org). Additional reproduction.techstreet. Inc.ashrae. Refrigerating and Air-Conditioning Engineers.Solutions to Chapter 8 ENERGY ESTIMATING METHODS Copyrighted material licensed to University of Toronto by Thomson Scientific. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Refrigerating and Air-Conditioning Engineers.ashrae. distribution. © (2009). (www. For personal use only. American Society of Heating.Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc. Additional reproduction. (www.com). Inc.techstreet. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .org). the annual heating cost if electric heat is used with the single rate of 16¢/kWh. av) to.1 The total design heating load on a residence in New York City is 32. and 6 A. The home has an oil-fired furnace. Find the savings in gallons of fuel oil if the owner lowers the temperature in the home to 68°F between 10 P.6 ) – ( 70.03 ) = 13.⎜ ---------------------------.000 Btu/h) for an indoor temperature of 72°F. Additional reproduction. and 6 A.0% ( 72 – 27.M. every day during January.( 100 ) = ---------------------------------------------------------------------. the maximum savings effected if the thermostat is set back to 65°F between 10 P. Estimate: a.techstreet.M. V = 1 a.67 – 27. Refrigerating and Air-Conditioning Engineers. © (2009). (www.( 72 ) = 70.av = -----. and has a design heat loss of 112.av = 27.( 100 ) = 3.M.8 ) Cleveland: ti = 72°F. (www.M. Cost = ( 49 .16 ) = $7969 00 c.3 A home is located in Cleveland.Chapter 8—Energy Estimating Methods⏐101 8.800 kWh ( 72 – 13 ) ⎝ ( 1.⎟ ( 0.67°F 24 24 ( 72 – 27. The furnace is off from June through September. Inc.000 Btu/h at an inside design temperature of 72°F and an outside design temperature of 0°F. Inc.6°F for January Without Setback: ( q L ) ( DD ) ( 24 ) ( 112 .org). 816 ⎛ ----⎞ -----( t i – t o ) with ⎝ 24 ( 65 ) + 24 ( 72 ) ⎠ – 42.5 % with Setback ---------------------------------.800 ) ( 0. Ohio.av) = 1159 = 31(65 – to.4 gal ( 50 L ) With Setback: This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 8.6 ) Fuel Savings = ( 448 ) ( 0.0 ) ( 1 ) ⎠ b.8 ⎛ ( 4848 ) ( 24 ) ⎞ E = ---------------------. 32. American Society of Heating.77 ) = 49 .000 ) ( Δt ) ( η ) (V ) E = 448 gal 8 16 t i . η = 1 CD = 0.7 ) ( 140 . January 1159 DD DD = (Days in Period)(65 – to..77.( C D ) = --------------------------------------------------------.6 ) % Savings = -------------------------------------------------------------------. For personal use only. ti = 72. Copyrighted material licensed to University of Toronto by Thomson Scientific.com). the annual energy requirement for heating b.000 ) ( 1159 ) ( 24 ) E = ----------------------------------. distribution.= 92% or 8% Savings ( t i – t o ) without ( 72 – 42. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.8 kW (112. to = 1°F.ashrae. $/yr c. .( 68 ) + -----. $/yr New York City: DD = 4848 to = 13.( 1 ) ( 72 – 1 ) ( 0. 8 kW (112.77 ) = 1.= 1300 gal ( 4950 L ) ( 0.org). Refrigerating and Air-Conditioning Engineers. © (2009). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. American Society of Heating.46 ×108 Btu ( Δt ) ( 73 ) ( 1.102⏐Principles of Heating.77 ( q L ) ( DD ) ( 24 ) ( 112 .9°F 8 16 t i .3°C (72 to 65°F) between 10 P.9 ) Cost Savings = ( 0. to = –1°F.46 ×108 ) E F = ----------.000 ) ( 5161 ) ( 24 ) E = ----------------------------------. Ventilating. b. The furnace is off during June through September. c. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .( 65 ) + -----.80 ) ( 140 . is 32. Inc.com).av = -----.1% Savings ( 72 – 43.9 ) – ( 70 – 43. in $/yr Kansas City: HL = 32. CD = 0.9 ) % of Savings = -----------------------------------------------------------. distribution. Annual heating cost if No. (www. Additional reproduction.M. Annual energy requirement for heating b. For personal use only.4 The total design heating load on a residence in Kansas City.2 to 18.= ---------------------------------------. Estimate: a. Missouri.000 Btu/h).000 ) η⋅V Cost = ( 4950 ) ( 0.( 100 ) = 7. Maximum savings effected if the thermostat is set back from 22. and 6 A. and Air Conditioning—Solutions Manual 8.( C D ) = ----------------------------------------------------.( 72 ) = 70°F 24 24 ( 72 – 43.av = 43.techstreet.071 ) ( 2300 ) = $165 Copyrighted material licensed to University of Toronto by Thomson Scientific. 2 fuel oil is used in a furnace with an efficiency of 80% (assume fuel oil costs 68¢/L) c.68 ) = $3366 t o .8 kW (112.000 Btu/h). ti = 72°F DD = 5161 a. (www.M. Inc.( 0.ashrae. 2 fuel oil per season if used as heating fuel c.77 ) = 7.2 ) (V ) + ( 54.( 0. Determine the following: a. Total steam flow in kg/s if a steam system is used Tulsa.48 ×106 L ) F = -------------------------------------------------( 1050 Btu/ft 3 ) ( 0.2 ) ( V ) ( t supply – t return ) .4 kW.Chapter 8—Energy Estimating Methods⏐103 8. Typical · · Q s = ( 1. Inc.8 ) c. . 20 kJ/s = m· ( 232. © (2009).37 ×107 Btu ( 21 590 kWh ) ( Δt ) ( η ) (V ) ( 72 – 9 ) b.2 ) · V = 518 L/s Q s = m· ( h fg ) . ¡ f. Inc.⎜ ----------------------------⎟ = ---------------------------------------. OK: DD = 3680. 20 . (www. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. distribution. Litres of natural gas per season if used as heating fuel d.4 f.ashrae.org). (www. to = 97. 7. Additional reproduction. Copyrighted material licensed to University of Toronto by Thomson Scientific.= 21 600 kW ⋅ h ( 3413 ) ( 1 ) e.000 = ( 1. American Society of Heating.= 8770 ft 3 ( 2.80 ) d.0086 kg/s This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto a.8 kJ/kg ). to = 9°F. Heating energy requirements. Total airflow rate in L/s if a warm air system is used g.37 ×107 F = ------------------------------------.techstreet.( C D ) = -------------------------------------------------------.= 6800 kWh COP ⎠ ( 97 – 78 ) ( 3. 7.37 ×107 . Q c ⎛ ( CDD ) ( 24 )⎞ ( 9.5 A residence located in Tulsa. kWh b.com).000 ) ( 0.4 ) Δt ⎝ g. 7. For personal use only.4 – 22. ti = 72°F CDD = 1949. m· = 0.= 658 gal ( 2500 L ) ( 140 .4 ) ( 1949 ) ( 24 ) E c = -----. Refrigerating and Air-Conditioning Engineers. kWh of electric energy if used for air-conditioning system having COPseasonal = 3. ti = 78 ( H L ) ( DD ) ( 24 ) ( 20 ) ( 3413 ) ( 3680 ) ( 24 ) E = ------------------------------------. kWh of electric energy if used as heating fuel with baseboard units e.37 ×107 F = ------------------------. Litres of No. Oklahoma has a design heating load of 20 kW and a design cooling load of 9. Inc.900 .95 ) 68 .000 kWh.000 Btu with Setback 83 . Additional reproduction. American Society of Heating. e. Gallons of No. b.----------------------------. c. The furnace is off from June through September.000 ( lb/h ) steam = ---------------.= 966 gal/yr ( 140 . Ohio. distribution.900 .6 Estimate the annual energy costs for heating and cooling a residence located in Cleveland. av = -----. c.= 6.000 ( cfm ) air = ----------------------------------------.77 Design Temperatures: 1°F Winter .= 68 lb/h 1000 68 . (www.3 ) ( 1000 ) Cost: ( 7560 ) ( 0.= 7560 kWh ( 8 – 78 ) ( 7.000 613 ( 24 ) Summer: ER = ------------------. For personal use only.000 ) ( 0.2 ) – ( 68. c. In winter the thermostat is set back to 60°F for 10 h each night.900 .≅ 20 kW 3413 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 8. Cubic feet of natural gas/yr kWh Total airflow rate in cfm if a warm air system is used Total steam flow in lb/h if a steam system is used Total water flow rate in gpm if a hydronic system is used Total electric power in kW if electric heating is used q 68 . having design loads of 65.ashrae.70 ) 94 .3 Average Winter Temperature = 37.( 60 ) = 68. g.8°F Savings = ----------------------------------------------------------------.= 29 .000 ER = -------------------. g.( 0.000 Btu/yr ( ti – to )d ( 75 – 3 ) 94 .800 ft 3 /yr ( 1050 ) ( 0.7 A residence in St.= 120 . Natural Gas = -------------------------------. h. Cleveland: HDD = 6154. respectively. © (2009). Missouri.2 ) 14 10 Setback Savings: t i .000 Winter: ER = ---------------.000 Cost: Heating → --------------------------.org).900 . CDD = 613. SEER = 7.( DD ) ( 24 )C D = ------------------.000 Btu without Setback ( 75 ) ( 75 – 37. e.( 75 ) ÷ -----. d. h.10 ) ( 130 – 75 ) 68 .000 ( gpm ) water = -----------------------------------------.75 ) 94 . Joseph.com). Refrigerating and Air-Conditioning Engineers. 86 °F Summer 65 .0725 ) = $1774 3413 30 . Electricity costs 0.= 1120 cfm ( 1.3 (Btu/h)/W.104⏐Principles of Heating. Inc.300 kWh/yr ( 3413 ) ( 0.000 Btu/h (cooling) based on a 75°F indoor temperature.900 . Electricity = -------------------------------.2 ) ER = ( 99 .0725 $/kWh year round. CD = 0. has a design heating load of 68. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.000 kW = ---------------.8 gal/min ( 1 ) ( 20 ) ( 60 ) ( 8.000 Btu/h (heating) and 30.4% 24 24 ( 75 – 37.( 5435 ) ( 24 ) ( 0.0725 ) = $548 Annual Cost: 1774 + 548 = $2322 a.164 ) = 83 . Ventilating.techstreet. f.8 – 37.500 . The furnace is on from October 1 through May 31.3 ) 68 . Fuel Oil = ---------------------------------------. 2 fuel oil/yr Copyrighted material licensed to University of Toronto by Thomson Scientific. f.( 100 ) = 16. and Air Conditioning—Solutions Manual 8. Determine the fuel and energy requirements for heating in: a.000 gal. Btu b. (www.77 ) = 99 .( 6154 ) ( 24 ) ( 0.000 Btu/h when design indoor and outdoor temperatures are 75°F and 3°F.77 ) = 94 . .900 .000 ) ( 1 – 0.2°F.500 .000 ft 3 . The air conditioner has an SEER of 7. Electric baseboard heat is used. 5°F Copyrighted material licensed to University of Toronto by Thomson Scientific.77 a. distribution. Inc. 2.600 ) = $951 8.000 Btu/gal and costing $2. © (2009). Missouri. 2 .065 ) ( 14 . .Chapter 8—Energy Estimating Methods⏐105 8.= 14 . Annual energy usage for heating b.50/ gal Springfield: DD = 4570.000 Btu/h for design condition of 75°F inside and 10°F outside.5¢/kWh New Orleans: q c = 12 kW ( 41 . Cost of this energy if the electric rate is 6.77 ) = 2.81 ×109 Btu ( 75 – 10 ) b. the design cooling load is 12 kW (41. For personal use only. American Society of Heating.com).70 ) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Average Winter Temperature = 44. 800/yr ( 140 .ashrae.000 Btu ) Summer Design: 92 ⁄ 78°F ( Inside = 78°F ) CDD = 2706 Assume Air Conditioner SEER = 13 41 . The heating system is operational between October 1 and April 30.000 Btu/h). Louisiana.( 4570 ) ( 24 ) ( 0.50 ) = $71. CD = 0.8 For a residence located in New Orleans. Annual energy requirements for cooling. Estimated fuel cost if No. ER = ---------------------. Cost = ( 0.techstreet.9 An office building located in Springfield. Additional reproduction. Refrigerating and Air-Conditioning Engineers. kWh b.000 ( 2706 ) ( 24 ) a. (www. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.600 kWh ( 92 – 78 ) 13 ( 1000 ) b.000 ) ( 0. 2 fuel oil is used having a heating value of 140.160.81 ×109 Fuel Cost ≈ ---------------------------------------. Determine: a.( 2.000 ER = -----------------------.160 . Inc.org).---------------------------. has a heat loss of 2. Determine: a. (www. 7 1 41.( $1.Heat Pump Electric Operating Supplied Elec.54 0 1139 47 18 355 27. A B Calculation of Annual Heating Energy Consumption House C D Heat Pump E F G H Supplemental I Jd Ke Lf Mg Nh Heat Seasonal SuppleTotal Heat Pump Cycling Pump Heat Pump mental Electric Weather Rated Heat Integrated Capacity Adjusted Temp.000 ) ( 0.85 56.techstreet.-------------------------------.59 81.03 0 418 57 8 692 12.500 .000 Btu ( 72 – 32 ) Chapter 19: AFUE = 0. Cycling capacity = 1 at the balance point and below.49 9. respectively. b. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.81 0. American Society of Heating. © (2009).01 1071 10.( 0.30 0.4 6.89 71.20 ) = $487 ( 100 .0 1 51.40 85 206 22 43 2 65.org).3 5.3 1 46.19 436 4. 106⏐Principles of Heating.0 5. a.400 cooling need Winter: ( Energy = 5946 kWh see BIN Sheet ) Cost = 0.56 694 6.52 1139 10. ConsumpHeating.94 52. Con. Time Rate. and Air Conditioning—Solutions Manual 8.05 0.200 ER = ---------------------.47 0 908 52 13 530 19.84 56. Compare the heating energy cost for the heat pump to that for a condensing gas furnace with natural gas costing $1.13 0 1071 37 28 154 42.79 60.19 0 436 27 38 24 58.13 12 22 17 12 7 2 –3 TOTALS: a Cycling Capacity Adjustment Factor = 1 − Cd(1 − x).6% and 1% outdoor design dry bulb temperatures.78 0 1052 42 23 288 35. (www. CDD = 2596 a.3 5.34 10.Copyrighted material licensed to University of Toronto by Thomson Scientific.05 1 1. The design heating and cooling loads are 61.500 .81 1 0.925 ) b. Select an appropriate heat pump from the XYZ Corporation models listed on the next page and estimate the energy costs for summer and winter if electricity is 8¢/kWh. Refrigerating and Air-Conditioning Engineers. Capacity.400 ( 2596 ) ( 24 ) Summer: Cost = ---------------------.64 10.5 ) ( 1000 ) 61 .000 Cost = ------------------------------------------.1 6.08 × 5946 = $476 55 .com).0 0. Jacksonville: Winter: 32°F. Load.76 1052 9.19 61.8 6.3 6. Heating Energy Data Loss Heating Adjust.08 ) = $1753 ( 93 – 78 ) ( 10. ment °F tbal − tbin hours 1000 Btu/h 1000 Btu/h Factora 1000 Btu/hb 106 Btuf kWhe kWhg tionh kW Fractionc 106 Btud 62 3 879 4.400 Btu/h. sumption.99 4.3 0.82 58.0 0.11 121 1. based on 99.54 66. Inc.60 0 694 32 33 83 50.49 51.56 0. For personal use only.6 5. Balance point has been estimated as 65°F.ashrae.( 1327 ) ( 24 ) ( 0.89 54.79 0.07 3. Required.925 = η 37 . b Col G = Col E × Col F c Operating Time Factor equals smaller of 1 or Col D/Col G 5849 d Col J = (Col I × Col G × Col C)/1000 e Col K = Col I × Col H × Col C f Col L = Col C × Col D/1000 g Col M = (Col L – Col J) × 106/3413 h Col N = Col K + Col M 97 5946 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Problem 8-10 Climate .56 0. distribution. HDD = 1327 Summer: 93/77°F. where Cd = degradation coefficient (default = 0.3 0.14 46. (www. Ventilating.200 and 55. Temp.2 0.80 418 4.20 per therm.35 908 8.0 0.08 10 0.10 A small football promotion office is being designed for Jacksonville.20 8.76 61. Diff. Florida. Inc. Input. Additional reproduction.77 ) = 37 .32 0.24 76. Heat Pump A060 for 55.Space Bin.7 4.25 unless part load factor is known) and x = building heat loss per unit capacity at temperature bin. Capacity.79 41.81 6. Bin. 61 0.77 a.= 30 . distribution. Con.000 Btu/h.000 ⁄ 1. kWh of electricity for cooling for air conditioner with SEER of 8. Heating Bin.44 0.0 1.8 0.0 1. respectively. where Cd = degradation coefficient (default = 0.000 ) ( 0. Determine a.903 kWh → See table below.= 1220 cfm ( 1.2 3. sumption. for warm air systems f.org).903 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 103.30 27 22 43 249 36. °F Heat Pump Output.88 ) h.5 34.25 unless part load factor is known) and x = building heat loss per unit capacity at temperature bin. b Col G = Col E × Col F c Operating Time Factor equals smaller of 1 or Col D/Col G 261 562 867 1081 1530 2123 2499 1656 872 432 218 136 52 22 12.1 4.5 4. Time ment a b c 6 d e 6 f g kW 10 Btu kWh kWh tionh Fraction 10 Btu Factor 1000 Btu/h 0.= 8900 kWh ( 90 – 78 ) ( 8.765 0. (www.75 0.60 Heat Pump Weather Heat Integrated Data Loss Temp.80 22 12 53 68 45.55 15 –3 68 8 57.80 13 G H Supplemental Jd I Lf Mg Nh Heat Seasonal SuppleTotal Cycling Rated Pump Heat Pump mental Electric Capacity Adjusted Heating Energy Adjust.8 31.3 4. CD = 0.885 0.2 3.27 0.10 ) ( cfm ) ( Δt ) e.1 4.06 2. Required airflow.86 9.818 0.0 1.8 ) (assumed 30% latent) f. 1000 Btu kW 62 57 52 47 42 37 32 27 22 17 12 7 2 −3 44 43 41 39 36 33 30 27 24 22 19 17 15 13 4. (www.98 2.27 0. Input.20 0.4 4.5 ×106 F = ---------------------------------------.54 0. for air conditioning Performance Data for Model WA-36 Heat Pump Air Temperature. Ohio.848 0. Bin.80 33 32 33 711 28.0 3. Heating energy requirements. cfm.311 d Col J = (Col I × Col G × Col C)/1000 e Col K = Col I × Col H × Col C f Col L = Col C × Col D/1000 g Col M = (Col L – Col J) × 106/3413 h Col N = Col K + Col M 14.7 29. 103. Capacity.78 0.80 43 52 13 611 11.930 0. Rate. °F tbal − tbin hours 1000 Btu/h 1000 Btu/h × 62 3 726 2.= 1160 cfm ( 1.42 5.6 3.05 30 27 38 460 32.5 3.9 2.10 TOTALS: a Cycling Ke Capacity Adjustment Factor = 1 − Cd(1 − x).88 1.9 3. 2 fuel oil if 75% efficient oil-fired warm air system is used c. Inc. For personal use only.( 5070 ) ( 24 ) ( 0.000 = ------------------.0 33.9 3.1 2.Chapter 8—Energy Estimating Methods⏐107 8.0 33. Diff.5 ×106 F = -------------------------------.5 ) ( 1000 ) ( 35 .34 3. cfm.5 27 24 22 19 17 15 13 4. Therms of natural gas if 88% efficient gas-fired warm air system is used d. Cycling capacity = 1 at the balance point and below.( DD ) ( 24 ) ( C D ) ( ti – to )d Copyrighted material licensed to University of Toronto by Thomson Scientific. .000 ( 1080 ) ( 24 ) kWh = ---------------------------------------------------.29 0.7 3.30 39 42 23 627 19.16 0.5 using degree-day estimation h.55 24 17 48 131 40.ashrae. Load.983 1.0 3.5 ×106 F = ---------------------------------------.Space Capacity. ConsumpHeating.6 3. Btu b.96 0.95 1 1 1 1 1 1 1 12.85) 1 1 1 1 1 1 1 0.77 = 850 (0.000 ) ( 0. Heat Pump Input.3 3.000 Btu/h and 35.0 1.55 44 57 8 639 6. c.3 ) cfm = ------------------------------------.08 0. American Society of Heating.000/(72 – 5) × 0. Summer: 90°F.com).46 715 914 721 519 416 202 105 3592 15. d.0 1.9 30. Climate Q s = ( 1.11 A 1980 ft2 residence located in Cincinnati. kWh of electricity if 98% efficient baseboard units are used e.5 ×106 Btu ( 72 – 5 ) g.10 5. 103.3 4.55 36 37 28 698 23. Temp. 35 . Required.77 ) = 103. Required airflow.3 3.30 17 2 63 18 53. Additional reproduction. CDD = 1080.7 3.75 ) Problem 8-11 74 .93 0.05 19 7 58 44 49. Inc.7 34.790 0.5 3. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.39 0. © (2009).900 kWh ( 3413 ) ( 0.techstreet.17 0.Heat Pump Electric Operating Supplied Elec.33 0.000 cfm = -------------------------------------. DD = 5070. Refrigerating and Air-Conditioning Engineers.= 985 gal ( 100 .0 1.1 ) ( 78 – 5. Gallons of No. qL ER = -------------------.5 4. has design heating and cooling loads of 74.8 Cincinnati: Winter: 5°F.9 2.98 ) A B Calculation of Annual Heating Energy Consumption House C D Heat Pump E F 74. kWh of electricity if heat pump system (WA-36 specifications follow) including supplementary electric resistance heat is used g. b.1 ) ( 130 – 72 ) kWh = 15 . 78 .1 2.72 0.4 4.05 41 47 18 599 15.= 1175 therms ( 100 . 69 ×108 C f = ------------------------------------.0 ) t i .75 ) c.. Inc.12 A small commercial building located in Oklahoma City.067 ) = $5280 ( 3413 ) ( 1.5 ) ( 1000 ) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto d.M.07 = $2467 ( 96 – 78 ) ( 11.7°F Copyrighted material licensed to University of Toronto by Thomson Scientific.3 e. heating. 245 .000 ) ( 0.77 Average: 48.108⏐Principles of Heating.com). and 162. Determine: a. and all day Sunday.× 0. (www. and 6 A. distribution. Monday through Saturday.-------------------------------. and Air Conditioning—Solutions Manual 8. CDD = 1876 a. Ventilating. 2.50/gal. 2.= 0. Cooling season energy cost using cooling degreedays if conditioner has SEER of 11. $ c. . 162 . or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.( $2. CD = 0. Annual energy requirements for heating.000 ( 1876 ) ( 24 ) C e = ---------------------.000 Btu/h. For personal use only. Inc.7¢/kWh.ashrae.3°F. Btu b.av = [ ( 6 ) ( 16 ) ( 72 ) + ( 168 – ( 6 ) ( 16 ) ( 55 ) ] ⁄ 168 = 64.( 0. cooling.M. Fuel cost using LPG at $2. American Society of Heating. % e.( 3695 ) ( 24 ) ( 0.5 and electricity is 7¢/kWh Oklahoma City: Winter: 10°F Summer: 96°F HDD = 3695.69 ×108 C f = ----------------------------. © (2009). Refrigerating and Air-Conditioning Engineers.963 ( 90 .77 ) = 2.3 % Savings = 1 – --------------------------. Balance point for the building has been estimated at 65°F.50 ) = $9. Savings if setback to 55°F is effected between 10 P. has design loads of 245. Additional reproduction. Fuel cost using electric baseboard units with electricity at 6.7 – 48.000 Btu/h.69 ×108 Btu ( 72 – 10 ) b. $ d.31 or 31% 72 – 48. Oklahoma.org).techstreet.000 ER = ---------------------. (www. 64. Chapter 8—Energy Estimating Methods⏐109 8.13 A small commercial building in Indianapolis, Indiana, has design heating and cooling loads of 98,000 Btu/ h and 48,000 Btu/h, respectively. Internal heat gains throughout the winter are relatively steady at 4.5 kW. Electricity costs 7.1¢/kWh. Estimate: a. Annual heating cost if baseboard electric resistance units are used. b. Annual cooling cost with a conventional vapor compression air-cooled unit, using your choice of method. Select a heat pump system for the building from the XYZ Corporation models. Determine the a. b. Annual heating cost and Annual cooling cost. Indianapolis: Winter: –3°F; Average: 39.6°F; DD = 5577; CD = 0.77 Summer: 88°F; CDD = 974 a. 98 ,000 ER = --------------------------- ( 5577 ) ( 24 ) ( 0.77 ) = 135 ×106 Btu [ 72 – ( – 3 ) ] b. 135 ×106 Cost = -------------------------------- ( 0.071 ) = $2970 ( 3413 ) ( 0.95 ) Cooling Unit: Assume SEER = 11 Btu/Wh 48 ,000 CDD = ---------------------- ( 974 ) ( 24 ) = 112 ×106 Btu ( 88 – 78 ) 112 ×106 Cost = ---------------------------- ( 0.07 ) = $714 ( 11 ) ( 1000 ) ( 4.5 ) ( 3413 ) t bal = 72 – ------------------------------------------------ = 60°F 98 ,000 ⁄ [ 72 – ( – 3 ) ] Heat Pump: Model A-048; SEER = 10.5 ; Watts = 5270 a. Energy Input = 13 ,220 + 5976 = 19 ,196 kWh → See table below. Balance Point: Cost = 19196 (0.071) = $1363 Problem 8-13 Climate A B 112 ×106 ------------------------------- ( 0.071 ) = $757 ( 10.5 ) ( 1000 ) CDD Method: Calculation of Annual Heating Energy Consumption House C D Heat Pump E F G H Supplemental I Jd Ke Lf Mg Nh 98,000/[72 – (–3)] = 1306 (1.3) Heat Seasonal SuppleTotal Heat Pump Cycling Weather Heat Rated Pump Heat Pump mental Electric Integrated Capacity Adjusted Data Loss Temp. Heating Energy Heating Adjust- Heat Pump Electric Operating Supplied Elec. Con- Space Bin, Rate, Capacity, Input, Bin, Load, Required, ConsumpHeating, sumption, Capacity, Time ment 60 a b c 6 d e 6 f g kW °F tbal − tbin hours 1000 Btu/h 1000 Btu/h Factor 1000 Btu/h 10 Btu kWh kWh tionh Fraction 10 Btu 62 — 57 3 585 3.9 59.2 0.766 45.3 4.99 0.09 263 52 8 586 10.4 55.4 0.797 44.2 4.81 0.24 676 47 13 579 16.9 51.0 0.833 42.5 4.60 0.40 1065 42 18 605 23.4 48.0 0.872 41.9 4.46 0.56 1511 37 23 712 29.9 44.4 0.918 40.8 4.28 0.73 2225 32 28 791 36.4 40.8 0.973 39.7 4.10 0.92 2984 27 33 551 42.9 37.3 1 37.3 3.93 1 20.55 2165 23.64 905 22 38 293 49.4 33.8 1 33.8 3.76 1 9.90 1102 14.47 1339 17 43 152 55.9 30.0 1 30.0 3.58 1 4.56 544 8.50 1154 12 48 97 62.4 27.3 1 27.3 3.45 1 2.65 335 6.05 996 7 53 60 68.9 24.2 1 24.2 3.31 1 1.45 199 4.13 785 2 58 35 75.4 21.2 1 21.2 3.18 1 0.74 111 2.64 557 –3 63 13 81.9 18.4 1 18.4 3.06 1 0.24 40 1.06 240 Temp. Diff. TOTALS: a Cycling Capacity Adjustment Factor = 1 − Cd(1 − x), where Cd = degradation coefficient (default = 0.25 unless part load factor is known) and x = building heat loss per unit capacity at temperature bin. Cycling capacity = 1 at the balance point and below. b Col G = Col E × Col F c Operating Time Factor equals smaller of 1 or Col D/Col G 13,220 d Col J = (Col I × Col G × Col C)/1000 e Col K = Col I × Col H × Col C f Col L = Col C × Col D/1000 g Col M = (Col L – Col J) × 106/3413 h Col N = Col K + Col M 5976 19,196 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto b. Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 110⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual 8.14 A small (2200 ft2) food mart store located in Charlotte, NC, has design heating and cooling loads of 94,500 Btu/h and 57,400 Btu/h, respectively, based on inside design temperatures of 72°F (winter) and 78°F (summer). The store is open 24 h a day and has a relatively constant internal load due to lights, food cases, people, etc., of 3.3 W/ft2. Select a suitable heat pump for the XYZ Corporation and estimate its operating energy costs for both summer and winter if the price of electricity is 7.4¢/kWh. Charlotte: Winter: to = 18°F Summer: to = 91°F CDD = 1596 HL 94 ,500 ------- = ---------------------- = 1480 Δt ( 72 – 18 ) Q int = ( 3.3 ) ( 2200 ) ( 3.413 ) = 24 ,780 ⎛ 24 ,780⎞ t balance = 72 – ⎜ ----------------⎟ = 55°F ⎝ 1480 ⎠ Select Model 060JA Heat Pump: Total Watts = 6250 SEER = 10.5 Winter: 7187 kWh × 0.074 = $532 Summer: Climate A B ( 1596 ) ( 24 ) -------------------------------- ( 0.074 ) = $1192 ( 10.5 ) ( 1000 ) Calculation of Annual Heating Energy Consumption House C D Heat Pump E F G H Supplemental I Jd Ke Lf Mg Nh 98,000/(72 – 18) = 1480 (1.48) Heat Seasonal SuppleTotal Heat Pump Cycling Weather Heat Rated Pump Heat Pump mental Electric Integrated Capacity Adjusted Temp. Data Loss Heating Energy Heating Adjust- Heat Pump Electric Operating Supplied Elec. Con- Space Bin, Load, Required, ConsumpBin, Rate, Capacity, Input, Time Heating, sumption, Capacity, ment 55 a b c 6 d e 6 f g °F tbal − tbin hours 1000 Btu/h 1000 Btu/h Factor 1000 Btu/h 10 Btu kW Fraction 10 Btu kWh kWh tionh 62 57 52 3 730 4.4 71.2 0.765 54.5 6.32 0.08 369 47 8 684 11.8 66.0 0.795 52.5 6.05 0.22 910 42 13 634 19.2 61.0 0.829 50.6 5.81 0.38 1400 37 18 515 26.6 56.0 0.869 48.7 5.56 0.55 1575 32 23 360 34.0 51.0 0.917 46.8 5.30 0.73 1393 27 28 166 41.4 46.3 0.974 45.1 5.05 0.92 771 22 33 64 48.8 41.9 1 41.9 4.81 1 2.68 308 3.12 129 17 38 23 56.2 37.0 1 37.0 4.61 1 0.85 106 1.29 129 12 43 5 63.6 33.2 1 33.2 4.35 1 0.17 22 0.32 44 7 48 2 71.0 29.4 1 29.4 4.13 1 0.06 8 0.14 23 2 –3 Temp. Diff. TOTALS: a Cycling Capacity Adjustment Factor = 1 − Cd(1 − x), where Cd = degradation coefficient (default = 0.25 unless part load factor is known) and x = building heat loss per unit capacity at temperature bin. Cycling capacity = 1 at the balance point and below. b Col G = Col E × Col F c Operating Time Factor equals smaller of 1 or Col D/Col G 6862 d Col J = (Col I × Col G × Col C)/1000 e Col K = Col I × Col H × Col C f Col L = Col C × Col D/1000 g Col M = (Col L – Col J) × 106/3413 h Col N = Col K + Col M 325 7187 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Problem 8-14 57 ,400 ---------------------( 91 – 78 ) Solutions to Chapter 9 DUCT AND PIPE SIZING Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto (c) 40 to 64 ft/min. Group E: These outlets would be satisfactory if properly designed and selected. Group B: These outlets would be satisfactory but probably not as good as Group A. 9. .com). (www. H 2 O V = 4005 P v = 4005 0.2 in. What rate of airflow is expected? Copyrighted material licensed to University of Toronto by Thomson Scientific. b.org).5 in. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. a. Discuss your selection of outlets and locations for each of the following combinations: (a) Group A or Group C.54 in. c.33 in. (305 mm).15 m/s ) a. distribution. P v = P T – P S = 0. Inc. by 24 in. 40 to 60 ft/min 9. (www. Refrigerating and Air-Conditioning Engineers. Group C: These outlets would not be completely satisfactory where cooling is predominant.( 100 ) = 0. of water (125 Pa) and total pressure is measured at 0. (b) Group B or Group E. H 2 O ⁄ 100 ft 100 ft 60 From Fig. American Society of Heating. a.2 ------------= ------. Inc. The duct has an ID of 12 in.2 You are to select the type of outlets for a home to be constructed in Houston. © (2009).1 The air velocity in a human occupied zone should not exceed (a) 10 to 25 ft/min.4°C. Group A: These outlets would be satisfactory for Houston where cooling is predominant and heating is minimum. (305 mm by 610 mm) duct if the static pressure is measured at 0.4 Solve the following problems. (b) 25 to 40 ft/min. Additional reproduction.ashrae. Q = AV = 801 ( 1 ) ( 2 ) = 1600 cfm ΔP 0. Texas. of water (50 Pa). of water (135 Pa) b.04 = 801 fpm b.04 in.techstreet. 9-2 V = 1650 cfm ⇒ 1300 cfm This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 9. (d) None of the above.3 m) length of circular duct is 0.Chapter 9—Duct and Pipe Sizing⏐113 9. For personal use only.07 ( V x – 30 ) for t x = 76°F and Δt = 0 V x = 30 fpm ( 0.54 – 0. Find the airflow through a 12 in.3 What velocity of air is necessary at a location in a room such that most people will feel neither cool nor warm? Assume that the local temperature is equal to the control temperature of 24.4°C = 76°F Δt = ( t x – 76 ) – 0. 24. The pressure difference available to a 60 ft (18.50 = 0. © (2009). Δps. 9.08 ⎛ --------.1 m/s)? For 20. plot pv. Inc.6 How large of a duct is required to carry 20. one branch duct must supply 207 cfm to one of the rooms.5 For a residential air-conditioning system. Refrigerating and Air-Conditioning Engineers. H 2 O 100 ft \ = 0. (a) Determine the branch duct size and the pressure drop from the main duct to the room. Fig. and (b) specify the supply and return grille sizes for the room. or 5 × 11 in.2 9.techstreet.08 in. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . and Air Conditioning—Solutions Manual 9. American Society of Heating. 9-2 @ 600 fpm. a. For personal use only. (www. H 2 O b. Additional reproduction. 9.⎞ ⎝ 100 ⎠ ΔP ------------= 0. use 48 in.000 cfm (9400 L/s) of air if the velocity is not to exceed 1600 fpm (8. 207 cfm – 16 ft branch run V recomm = 600 fpm From Fig. 207 cfm ⇒ D = 8 in.0128 in. distribution. or larger. and Δpt for the flow through the system. Inc. 16 ΔP = 0. The branch duct has a run of 16 ft.114⏐Principles of Heating. (www.com).000 cfm @ 1600 fpm.org). Ventilating.7 Given the duct system shown below. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. return grille – 600 fpm supply grille ≈ 6 × 9 in.ashrae. Copyrighted material licensed to University of Toronto by Thomson Scientific. 093 m2) duct to a 2 ft2 (0. loss Increase cross-sectional area 9.32 – 0. distribution. one side is 26 in.g.= 1000 fpm 1 2 V1 ⎛ A 1 ⎞ H L = ------.ashrae.9 Determine the dynamic loss of total pressure that occurs in an abrupt expansion from a 1 ft2 (0.11 Find the equivalent rectangular duct for equal friction and capacity for the duct in Problem 9. w.org). 2 = ( 0.⎟ 2g ⎝ A 2 ⎠ 2 2 ⎛ V1 ⎞ 1 2 = ⎜ -----------. (www. Inc.g. (0.05 ) + ( 0.6. find the frictional pressure loss between points (1) and (2).com). H2O/100 ft (150 ft) = 0.Chapter 9—Duct and Pipe Sizing⏐115 9. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 9. b. Additional reproduction.7. a. 2 ⎝ 4005 ⎠ From Fig.914 m) diameter galvanized steel duct. H loss = H 1 – H 2 = P1 – P2 + HV – HV 1 b. © (2009). H2O 9. 9-2 Loss = 0. w.186 m2) duct carrying 1000 cfm (470 L/s) of air. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. For Problem 9. For personal use only. .10 Determine the friction loss when circulating 20.0156 in. Copyrighted material licensed to University of Toronto by Thomson Scientific.000 cfm (9430 L/s) of air at 75°F (23.⎟ ⎛⎝ 1 – ---.0 ) = 0. From Table 9-1 80 in.⎞⎠ = 0. (www.375 in. How can the static pressure be increased in a duct system as the air moves away from the fan? a. American Society of Heating. Inc.25 in.7 m) of 36 in.techstreet.8 Solve the following. Refrigerating and Air-Conditioning Engineers. A 1 = 1 ft 2 A 2 = 2 ft 2 Q V = ---A Q = AV 1000 V 1 = -----------.39 in.9°C) through 150 ft (45.12 – 0.⎜1 – -----. Copyrighted material licensed to University of Toronto by Thomson Scientific.12 Find the pressure loss between points A and D for the 12 by 12 in. . For personal use only.116⏐Principles of Heating. respectively.059 = 0. Inc.08 W 12 H 12 ----. H 2 O ⎝ 4005 ⎠ Loss straight = 0.249 in.03 ) ( 0.21 V 13 ----.03 V = 2000 fpm V 2 P v = ⎛ -----------.= 1. ⇒ C D = ( 0. duct shown below.ashrae.g. 1 and No.= 1.g. distribution. American Society of Heating.27 + 0.33 in. Additional reproduction. H 2 O⎞ ΔP loss = ( 10 + 20 + 30 ) ⎜ -------------------------------⎟ + ( 0. Elbows No.1 in.6 ) = 0.= -----. (www. Refrigerating and Air-Conditioning Engineers. Elbow #1 Elbow #2 H12 ---= -----.techstreet..⎞ = 0.= 1.= -----. Ventilating.21 + 0.org).05 ) ( 0. w.0 W 12 V24 ---= -----.= 2.0 From Table 9-4 W 12 ⇒ C D = 0.0 W 12 Table 9-1 Fig. (www. Air at standard conditions is being supplied at the rate of 2000 cfm in galvanized duct of average construction.249 ) 100 ft ⎠ ⎝ This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto = 0. Inc. 9-2 D c = 13. ⁄ 100 ft ⎛ 0.45 in. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. and Air Conditioning—Solutions Manual 9. © (2009). w.com).45 in. 2 have center line radii of 13 and 24 in. ⎞ ( 0. H 2 O C D = 1. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.0351 in.= 500 --------. H 2 O This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 9. H 2 O V -----b.28 Table 9-4 ΔP branch = ( 0.2 ( 0.= 2550 -----------. Refrigerating and Air-Conditioning Engineers.0 D Section d → 2 elbows C D = 0. H 2 O ⎝1⎠ ΔP loss – ΔP loss = 0 d No damper needed. H 2 O u b d V--= 1.22 Vd -----.037 + 2 ( 0.407 ) = 1. H 2 O Q b = 1000 cfm V b = 1275 fpm H v = 0.0277 ) = 0.102 ) = 1. If a damper is needed.02 ρ = 0. Find the actual static pressure regain and the total pressure loss in the straight through section if the static regain coefficient is 0.8 ( 0. H 2 O Q d = 2000 cfm V d = 2550 fpm H v = 0. w. .415 ) = 1. H2 O c. Additional reproduction.8 Table 9-4 Q1 2000 V 1 = -----. find the static pressure loss in this section. For personal use only.g.02 ⎛ -----. (www.= 667 fpm A1 3×1 1500 V 2 = -----------.org). ΔP branch = ( 1 – R ) ( P v – P v ) = 0.04 ΔP loss = 0. (R/D)elbows = 1 Q u = 3000 cfm V u = 3825 fpm H v = 0.= 0.com). 2 a. b R = 0.33 Vu Section b → Diverted Flow fitting ΔP fitting = ( 1.= 500 fpm 1×1 P v = R ( P v – P v ) = 0.915 in.03 in.80.006 ≈ 0 in. and if so. (www.01 + 0. H 2 O ≅ 0 in.407 ) + ( 50 + 40 + 10 ) ( 0.ashrae.= 750 fpm 2×1 P v = 0.14 A 1 ft high by 3 ft wide main duct carries 2000 cfm of air to a branch where 1500 cfm continues in the 1 ft by 2 ft straight through section and 500 cfm goes into the branch.0015 ≅ 0 in.techstreet.01 in. in what section.13 Analyze the air-handling system shown in the following diagram.666 Vu 3825 Straight Section Table 9-4 C D = 0. what is the pressure loss across the damper? f = 0.75 V1 667 Assume Branch 1 × 1 ft 1 2 1 2 C = 0.g.0277 in.407 in.0351 ) = – 0. Determine if a damper is needed in either section (d) or (b). Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc. V -----L. w.0277 – 0.0351 ) ≅ – 0.0277 – 0.008 in.03 in.075 lb/ft3 D = 12 in. H2 O b.1 Table 9-4 10 ΔP loss = 1.1 ) ( 0. © (2009).102 in.Chapter 9—Duct and Pipe Sizing⏐117 9.= 0. There is a damper located in section (u) so that the proper static pressure can be maintained in section (u).22 ) ( 0.= 0. 500 V b = -----------.28 ) ( 0.02 ) ( 0. 1 P v = 0. distribution.= -----------. American Society of Heating. If the branch take off is a 45° cylindrical Y. Inc. 014 0.25 0. (www.18 Run G Radius Ratio = 1.ashrae.5 in.054 in.86 H12 ----= -----. Loss C = 0. in. H 2 O Section cfm ΔP/100 ft Deq. Inc.122 ) = 0. Table 9-1 ΔP --------.065 0.183 ⎛ 1130⎞2 = 0. American Society of Heating.34 Longest Run: ΔP elbow ≅ 0.211 in. 12 in. Length ΔP C = 0.techstreet.= -----------. For personal use only. H 2 O each ΔP div.35 + 0.0 ( 0.4 0. of water.15 The supply ductwork for an office space is shown in the following diagram. Each outlet grille has a pressure loss of 0.13 0.5 Run D Table 9-4 Rectangular Size.35 in. H O 2 Damper in A-E and A-F or reduce size appropriately.= W 14 Elbows Radius Ratio = 1.09 2 = 0.0 ⎛⎝ ------------⎞⎠ = 0. and each outlet is to handle 2000 cfm. Size the ductwork by the equalfriction method and calculate the pressure drop. Additional reproduction. H 2 O Straight through loss ≅ 0 in. V AB 12 × 30 in.0754 0. © (2009).4 0.118⏐Principles of Heating.054 + 0.105 in.013 + 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.25 0. (www. flow ≅ 0.20 0.167 0. Total cfm = 800 + 900 + 1200 = 2900 cfm Deq = 20 in. Refrigerating and Air-Conditioning Engineers.078 + 0. H O 2 ⎝ 4005⎠ Table 9-4 ⎛ 1010⎞ = 0.122 in.= 0. H 2 O This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto loss Loss 0.105 0. H O 2 100 ( Table 9-4 ) V b 1060 -----.12 = 1.776 V u 1300 Divided Flow Fitting Cum.5 0.com). galvanized duct.27 0.211 0.0 1300 2 ΔP loss = 1. Inc. H O 2 Table 9-3 C = 1.= 0. 14 in.0 C = 1. Ventilating.065 0.005 in.25 30 ft 50 ft 90 ft Table 9-4 r ⁄ D = 1.105 + 0. H 2 O ΔP loss = 0. The velocity in the ducts is to be 2000 fpm.76 V 1400 u First branch take-off = 1400 fpm C = 1.= 0. Estimate the required pressure increase of the fan.13 0. distribution.13 20 in.12 in.= -----------.16 The following duct system contains circular.0 ΔP loss = 1.36 0.org).02 in. and that all duct take-offs are straight rectangular take-offs.2 13.36 + 0.054 + 0.09 ⎜ ------------⎟ ⎝ 4005⎠ V b 1010 -----. cfm fpm D ΔP/100 ft ΔP Pv Length 4000 2000 2000 2000 2000 2000 19.026 + 0. and Air Conditioning—Solutions Manual 9. 17 in. 9.5 loss 0.014 0. .18 ⎜ ------------⎟ = 0.5 13.13 0. Assume a maximum duct depth of 12 in.081 0. 1400 1300 1130 1010 1060 12 × 30 12 × 21 12 × 14 12 × 10 12 × 11 H12 ----= -----. Velocity A-B C-D C-E C-H B-F 2900 2000 1200 800 900 0. H 2 O 4005 Through loss ≅ 0 Total Pressure Drop ( No Outlet Grille ) = 0.13 0.0054 0.014 in.13 in.= 1 W 12 ΔP 50 ft 8 ft 60 ft 20 ft 10 ft Copyrighted material licensed to University of Toronto by Thomson Scientific.081 + 0. 12. Calculate the frictional pressure loss for the system and the total capacity required by the fan.283 in. Q total = 300 cfm 100 cfm branch ΔP through ≅ 0.46 --------.138 in.07 Straight ΔP = 0.283 in.08 Elbows (Table 9-3) ΔP ≅ 0. Refrigerating and Air-Conditioning Engineers.Chapter 9—Duct and Pipe Sizing⏐119 9. (www. © (2009). H 2 O static for 300 cfm Damper the 100 cfm branch Copyrighted material licensed to University of Toronto by Thomson Scientific. The pipe is circular. 100 20 100 cfm at 1000 fpm → ΔP = 0. For personal use only.10 200 cfm branch 20 300 cfm at 1000 fpm → ΔP = 0.046 in. American Society of Heating.092 in.087 ----------------------------------------------------------ΔP loss = 0. The radius ratio of the elbows is 1. Additional reproduction. distribution.17 Select a fan for the following system. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .org). Inc. Inc.com).0 and the elbows are 3 piece. straight 300 cfm at 1000 fpm → ΔP = 0.24 --------.techstreet.= 0.= 0. H 2 O Fan must supply 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.ashrae.046 Branch Loss (Table 9-3) ΔP = 0. (www. 100 ΔP loss = 0. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . Estimate the total pressure loss between points (1) and (2) and between (1) and (3) in the following take-off. Additional reproduction. of commercial fabrication. (www. P S = 1. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.ashrae.75 V1 1 1600 ΔP loss = 0.= 1.16 ) = 0.techstreet. distribution. © (2009).04 P V = 0.08 A1 2 = 0.375 V1 C = 0. V 1 = 1600 fpm Q 1 = 3200 cfm V 2 = 1200 fpm Q 2 = 2600 cfm V 3 = 600 fpm Q 3 = 600 cfm Table 9-4 a.= 0.1 m/s Q3 = 283 L/s V3 = 3. H 2 O C = 0.54 –­ P V = 0. American Society of Heating.120⏐Principles of Heating. and has mastic tape joints.05 m/s The duct is rectangular.12 m/s Q2 = 1227 L/s V2 = 6.086 in. Inc. H 2 O 1 3 P S + P V – P S – P V = 0. of water.54 ⎛⎝ ------------⎞⎠ 4000 A2 -----.086 in. Refrigerating and Air-Conditioning Engineers. b.16 V -----2.023 – 0.  –  P T – P T = 0.16 – 0. Ventilating.org).09 2 ΔP loss = ( 0. Inc.= 0.  –  V3 -----.086 3 Copyrighted material licensed to University of Toronto by Thomson Scientific.04 ) ( 0.0064 in.0 in.086 1 1 3 P S = 1. For personal use only. and Air Conditioning—Solutions Manual 9.05 in. H 2 O b. Estimate the static pressure at (3) if the static pressure at (1) is 1.0 + 0. H 2 O 3 3 .18 a. (www.com). when: Q1 = 1510 L/s V1 = 8. . (www. H 2 O b. what friction drop will be required of a damper at 68°F? What size duct would be required (for ducts C and D) if the damper is eliminated? What is the velocity in the line? Assume a temperature of 68°F. If the static pressure at (3) is 0.+ 0. distribution.1 in w. What is the expected approximate frictional pressure from (1) to (2) in the below length of duct. (www.5 in.35 – 0.003 in w. Section D = 0.13 ⎛ ------------⎞ ⎝ 4003⎠ 2 ΔP V 2 ΔP L = 0.2 ⎛⎝ ------------⎞⎠ 4000 2 Copyrighted material licensed to University of Toronto by Thomson Scientific. ΔP damper = 0.028 + 0.techstreet.Chapter 9—Duct and Pipe Sizing⏐121 9. of water and that the R/D of the elbow is 2.031 ) ( 30 ⁄ 100 ) = 0.071 ) = 0.ashrae.g. and air at standard temperature and pressure.0093 + 0.35 in.350 in. H2 O c.13 6000 2 = 0.35 = ( 30 + 3 ) --------.122 in w. Assume that the static pressure at (3) is still 0.org).0093 in.071 in.075 in w. Inc.225 in.5 = 2. © (2009).0093 + 0. using 1000 cfm Find a D. ΔP = ( 0.g.0093 + 0. H 2 O/100 ft ΔP = ( 40 ⁄ 100 ) ( 0. H 2 O Total loss = 0. ⎝ 4000⎠ ΔP = ( 0. Elbow Radius = 36 in. Also assume that the grille loss is linear with velocity.6 Section C Elbow D = 17.031 ) ( 30 ⁄ 100 ) = 0.13 ⎛ ------------⎞ = 0.g. Inc.228 in.003 + 0.003 + 0.10 = 0. at  P S = 0. H 2 O R ⁄ D = 36 ⁄ 17.35 in. Section A D = 19 in. of water.35 in. Grille Loss = 0. For personal use only.05 ΔP loss C = 1. fpm Length.028 in w.0093 + 0.122 = 0. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 1600 ΔP loss = 1. ft A B C D 2000 1000 1000 1000 1000 40 — 30 30 600 600 600 Grille loss = 0. H 2 O and P v = P v 2 3 ΔP friction = 0. H 2 O R⁄D = 2 C = 0.13 V ΔP = 0. ΔP = 0.10 = 0. Branch cylindrical tee Table 9-4 V c ⁄ V a = 0. Refrigerating and Air-Conditioning Engineers. c.g. at 600 fpm Duct cfm Velocity.13 ⎛⎝ ------------⎞⎠ + ΔP grille 100 4000 Trial and Error Solution: Assume a velocity. Additional reproduction.g. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.075 + 0. thus → V = 1300 fpm D = 12 in.com). P S at  = 0. clean sheet metal.0093 in. b. Assume round ducts.10 in H20 at 600 fpm a.2 C = 0.19 Solve the following problems: a. American Society of Heating. Size the ducts between A and E. and Air Conditioning—Solutions Manual 9. Additional reproduction. A-C. The static pressure required in the duct at points B.10 in. Inc.org).20 The pressure (energy) loss between A-B. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto This is an open-ended design problem with a number of satisfactory solutions. 5. Ventilating. Refrigerating and Air-Conditioning Engineers. and E to produce the proper flow from the air diffusing terminal units (ceiling diffusers) is assumed to be uniform and to be 0. What would the static pressure be at F? Note: In practical application. Ignoring interference losses due to terminal unit take off: 1. For greater imbalances.ashrae. Inc. A-D.122⏐Principles of Heating. minor static pressure imbalances up to 0. Size the ducts between A and C.05 in. Calculate the total pressure loss between A and D. (www. duct sizing must be modified and/or butterfly dampers installed in the branch ducts. of water can be absorbed by adjustment of dampers installed in air-diffusing terminal units and further minor adjustments of diverting dampers. American Society of Heating. and A-E must be equal if a proper air balance is to be achieved. D. distribution. C. Copyrighted material licensed to University of Toronto by Thomson Scientific.com). of water. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 4. © (2009). 2. Size the ducts between A and B. 3. For personal use only.techstreet. . in branch fittings. (www. 4 gpm Size. 1 1/2 1 1 3/4 3/4 5/8 1 1/4 1 1 1 5/8 5/8 3/4 Return Side V-VI 4 gpm VI-VII 5. 4. B at Conv.000 Btu/h (1.000 gpm = ------------------------.3 ft/100 ft) 1100 milli-inch (9.000 Btu/h (3 gpm) 36. (See Table 11 in Chap. at 5 fps – 2 in.95 ft b. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 27 .5 ft/100 ft Multiply heat required at convectors by ( 20 ⁄ 10 ) = 2 and use Fig.= 9.8 ft 9.000 Btu/h (2.ashrae. American Society of Heating.113 L/s) and has a 3.8 gpm) 14. at 5 fps – 2 in.23 Size the system shown in Example 9. From Fig. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.4 gpm) Size.4 gpm) 30.techstreet. 2.7 gpm.3 ⎞ -⎟ = 3. For personal use only.9 ft open globe valve → 12 L eq = 12 ( 5. Refrigerating and Air-Conditioning Engineers.2 ft/100 ft) 600 milli-inch (5 ft/100 ft) 2400 milli-inch (20 ft/100 ft) 1100 milli-inch (9.4 gpm) 54. See Notes in Example for this design. E at Conv.5 ( 5.2 ft/100 ft) 1500 milli-inch (12.000 Btu/h (1. The head loss for this straight pipe in this run from Fig.8 ⎠ 2 = 5. pipe → L eq = 5.1 ft/100 ft Convector Losses = 11. G at 14.6 gpm) 24. ⎛W ⎞ H 2 = H 1 ⎜ ------2. 10° ΔT Copyrighted material licensed to University of Toronto by Thomson Scientific. 2009 HBF) a.4 gpm XI-XII 3 gpm XII-XIII 6.000 Btu/h (2.8 gpm (0.21 Determine the equivalent feet of pipe for a 2 in.⎟ ⎝ W1 ⎠ 2 ⎛2. 2 friction loss for elbow is 35 ft/100 ft.9 ft open gate valve → 0. This should be verified later. pipe → L eq 1 in.3 ft/100 ft) 1960 milli-inch (16.3 ft/100 ft) Total = 212. © (2009).5 ft/100 ft) 4750 milli-inch (39. Estimate the pressure loss with a flow of 2.2 gpm) 18. A at Conv.000 Btu/h (3. (See Chap.2 gpm) 22. Fig.000 Btu/h (9 gpm) 54.000 Btu/h (3. 1 1 1 1 1/4 1 1/4 1 1/2 Assume run containing Conv.18 gpm ( 480 ) ( 10 ) Assume design friction loss = 2.= 5. Inc.-X XI-XII XII-XIII XIII-VII VII-I (280)(2) = (280)(16) = (280)(7) = (100)(11) = (200)(3) = (200)(12) = (100)(11) = (150)(10) = (250)(14) = (280)(25) = 560 milli-inch (4.22 A convector unit is rated at 1.5 elbow L eq = 0. C at V-VIII at VIII-IX at IX-X at Conv.000 Btu/h (1.6 ft/100 ft) 7000 milli-inch (58.Chapter 9—Duct and Pipe Sizing⏐123 9. Additional reproduction.9 ft ) = 70. 2 the friction loss for Tees is 158 ft/100 ft.com).4 ⎜-----⎝1. in.400 Btu/h (14. 2009 HBF) A-B-C Supply Side I-II at II-III at III-IV at Conv. Inc.000 Btu/h (1.5 m/s).6 ft 9.51 gpm ( 480 ) ( 10 ) 45 .org). distribution. (www. F at Conv. 22.0 gpm VII-I 14. (50 mm) open gate valve and a 2 in.3 gpm through the convector. elbow = 5.6 gpm) 14. .4 gpm) 90. (50 mm) open globe valve at a flow velocity of 5 fps (1. D at Conv.4 gpm) 32.5 for a 10°F temperature drop.4 ft (10 kPa) pressure loss at rated flow. (www. Conv.000 Btu/h (5.67 ft/100 ft) 4480 milli-inch (37. Total friction loss for this run = 4. 22.6 gpm XIII-VII 9.4 gpm) 16. in.000 Btu/h (5.9 ft ) = 2.000 D-E-F-G gpm = ------------------------. 2 I-II II-VIII VIII-IX IX-X X-Conv.17 ft Pump must supply at least this head at 14. E is longest run.9 ft/100 ft From Table 4 and Fig. 6 gpm) 4670 Btu/h (0.techstreet.8 gpm) Pipe Size.6 ft/100 ft.000 Btu/h (1.8 ft/100 ft) 5500 milli-inch (45.0 gpm) 12.2 gpm) 4670 Btu/h (0.8 gpm) 10. 2. 2009 ASHRAE Handbook—Fundamentals) 27 . 4. 30° ΔT A-B-C Supply Side I-II at II-III at III-IV at Conv. For personal use only. Additional reproduction. Inc. American Society of Heating. Inc.000 D-E-F-G gpm = ------------------------. © (2009). 1 3/4 5/8 1/2 1/2 1/2 1 3/4 5/8 5/8 1/2 1/2 1/2 Return Side V-VI 1.8 ft/100 ft) Total = 142 ft/100 ft Convector Losses = 11.670 Btu/h (1.124⏐Principles of Heating. 2 (240)(2) = (100)(16) = (170)(7) = (130)(11) = (85)(3) = (85)(12) = (130)(11) = (220)(10) = (100)(19) = (220)(25) = 480 milli-inch (4 ft/100 ft) 1600 milli-inch (13.000 gpm = ------------------------.3 ft/100 ft) 1190 milli-inch (9.7 gpm) 6000 Btu/h (0. Convector loss = 11.000 Btu/h (1.ashrae.24 Size the system shown in Example 9. C at V-VIII at VIII-IX at IX-X at Conv. 2 the total friction loss for Tees = 79.2 gpm XIII-VII 3. See Notes in Example 4 in Chap. A at Conv.0 gpm XII-XIII 2. B at Conv.8 gpm) 18.com). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 2 to get pipe size.1 ft/100 ft) 1020 milli-inch (8. The head loss for this straight pipe from Fig.33 gpm VI-VII 1. (www.06 gpm ( 490 ) ( 30 ) Assume design friction loss = 2. Conv.9 ft/100 ft) 2200 milli-inch (18.000 Btu/h (1.0 gpm) 18.3 ft/100 ft) 1900 milli-inch (15. distribution. G at 48. Refrigerating and Air-Conditioning Engineers.8 gpm Pipe Size.5 gpm) 30. Verify after various calculations.5 ft/100 ft) 1430 milli-inch (11.5 for 30°F temperature drop.0 gpm VII-I 4. D at Conv. E is longest run.9 ft/100 ft Total friction loss for this run = 2.9 gpm of flow.8 gpm) 10. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto I-II II-VIII VIII-IX IX-X X-Conv. 5/8 3/4 5/8 3/4 1 1 Assume Conv.000 Btu/h (3.= 3.1 gpm) 7340 Btu/h (0.5 gpm) 5340 Btu/h (0. loss for elbows = 10. Fig.5 ft/100 ft Multiply heat required at convectors by ( 20 ⁄ 30 ) = 0. 9 PHVAC.5ft/ 100 ft. in.-XI XI-XII XII-XIII XIII-VII VII-I Copyrighted material licensed to University of Toronto by Thomson Scientific.43 ft Pump must supply at least this head at 4.84 gpm ( 490 ) ( 30 ) 45 . F at Conv.000 Btu/h (4.org). (See chapter 22.9 ft/100 ft From Table 4. E at Conv.667 and Fig.9 ft/100 ft) 225 milli-inch (2.9 ft/100 ft) 1430 milli-inch (11. Ventilating. in.8 gpm XI-XII 1.000 Btu/h (1. and Air Conditioning—Solutions Manual 9. Fig.5 gpm) 8000 Btu/h (0. . (www. From Fig.= 1. supply tee Loss = ( 3.025 milli-inch Table 11. The water leaves the boiler at 200°F and has a 20°F temperature drop. .025 milli-inches for Loop B.25 Size the system shown for iron pipe. What head must be developed by the pump and what flow rate (gpm) is required? (See chapter 22. For personal use only. Additional reproduction. Loop B gpm = ------------------------. pipe Loss = ( 180 ) ( 15 ) = 2700 milli-inch Table 11. Loop A Copyrighted material licensed to University of Toronto by Thomson Scientific.8 ) ( 2.techstreet. 1 Loss = ( 225 ) ( 49 ) = 11 .500 ) = 8250 milli-inch Fig.7 ) ( 225 ) ( 1.000 27 .7 ) ( 225 ) ( 1.org).8 ) = 1500 milli-inch return tee -----------------------------------------------------------------------------------------------------Total loss = 23 . (www. HBF. HBF.25 ) = 810 milli-inch Convectors Loss = ( 0.8 ) = 1500 milli-inch Loss = ( 3. Inc.ashrae. For Loop B Measured length = 49 ft From Fig. Refrigerating and Air-Conditioning Engineers. Assume 300 milli-inch/ft friction loss.3 ) ( 27 . Note: 1 milliinch = 0.3 (Btu/h output).Chapter 9—Duct and Pipe Sizing⏐125 9. © (2009).500 gpm = ------------------------.45 . HBF.= 2.085 milli-inch Boiler Circuit: L = 15 ft – Fig. boilers -------------------------------------------------------------------------------------------------Total = 5940 milli-inch Total friction loss = 29 .4 ) ( 180 ) = 864 milli-inch 6 ft riser and return Loss = ( 6 ) ( 180 ) = 1080 milli-inch Loss = ( 3 ) ( 2. 2009 ASHRAE Handbook—Fundamentals) A B Iron pipe 20°Δt This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 24.81 ( 490 ) ( 70 ) ( 490 ) ( 70 ) Total gpm = 5. 1 – 1 in. HBF. of water. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 7. distribution. Inc.= 2.com). American Society of Heating. elbows Loss = ( 2 ) ( 2.25 Assume Loop B will have longest pipe. elbows Loss = ( 2 ) ( 1. Assume a 3 ft rise is needed to get to the convectors and then a 3 ft drop to return to the boiler. The convectors have a loss given by the equation: Loss (milli-inches) = 0. (www.4 ) ( 180 ) = 1296 milli-inch Table 11.001 in. chapter 22. Inc. Ventilating.25 gpm.5 ) ( 1.41 ) = 0.28 Determine the pipe sizes for the refrigeration systems shown in the following figure. Determine the size of schedule 40 pipe required and the velocity in the steam pipe.000 lb/h Pinitial = 150 psig (Fig.8 ) ( 2. 2.8 gpm.5 ft Total friction = 2. Additional reproduction.09 ft Convectors ( 0.33 ft Tees ---------------------------------------------------------------Total = 2. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 9.techstreet. © (2009).= 1. For personal use only. gpm = 5. American Society of Heating. Use copper tubes. 15.ashrae. distribution. and Air Conditioning—Solutions Manual 9.org). chapter 22.5 ) = 0.5 × --------.126⏐Principles of Heating.41 ( 21 ) + 2 ( 2.3 ) ( 2. there are many solutions. 2009 ASHRAE Handbook—Fundamentals) 6 psi/100 ft – Schedule 40 Select 3 1/2 in.7 ) ( 2.26 Rework Problem 9. 13D. (www.27 A steam system requires 15. 2009 ASHRAE Handbook—Fundamentals.25 using Type L copper tubing. Inc.69 ft 2 ( 3. The design pressure drop is to be 6 psi per 100 ft. Refrigerating and Air-Conditioning Engineers. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.29 ) = 0. See Figure 5.5 ft/100 ft. 3/4 in.22 ft 100 Elbows 2 × ( 1. a. Refrigerant lines using R-22 b. .83 ft for Loop A including Boiler 9.41 ) + 4 ( 2.000 lb/h of steam at an initial pressure of 150 psig. Loop A = 2. tube 44 Pipe loss = 2. pipe Velocity at 150 psig = 10.com). Condenser water lines Copyrighted material licensed to University of Toronto by Thomson Scientific.4 ) ( 1. (www.33 ft Boiler circuit 1.000 fpm This is an open ended design problem.4 ) ( 1.8 ) = 0. Chapter 9—Duct and Pipe Sizing⏐127 9.29 A gas appliance has an input rating of 80,000 Btu/h and is operated with natural gas (specific gravity = 0.60) having a heating value of 1050 Btu/ft3. What size of supply pipe is necessary when the equivalent length is 70 ft and a pressure loss of 0.3 in. of water is allowable? (See chapter 22, Table 26, 2009 ASHRAE Handbook—Fundamentals.) Leq = 70 ft; 0.3 in. H2O 80,000/1050 = 76.19 ft3/h 3/4 in. pipe 9.30 A fan operating at 1200 rpm has been delivering 6500 cfm against a static head of 3.25 in. of water and a total head of 5.25 in. of water. The air temperature is 130°F, gage temperature, 90°F, and the input power is 6.1 kW. During a time of power difficulty, the operator notices that the static head is now 2.36 in. of water. There has been no change in the system. Find: a. New capacity in cfm b. New power input in kW c. Original efficiency of the fan (%) a. n 2⎞ 2 h 2 2.38 ⎛ ---- = ----- = ---------- ; ⎝ n 1⎠ h 1 3.25 h 2 = 0.855 ( 1200 ) = 1028 rpm b. c. 3 kW 2 = ( 6.10 ) ( 0.855 ) = 3.82 kW QP t ( 144 ) ( 0.746 ) ( 6500 ) ( 5.25 ) ( 0.036 ) ( 144 ) ( 0.746 ) AHP Eff = ------------ = ------------------------------------------ = -----------------------------------------------------------------------------------( 33 ,000 ) ( 6.1 ) ( 33 ,000 ) ( 6.1 ) BHP EFF = 65.5% 9.31 A certain damper design introduces a head loss of 0.5 velocity heads when wide open. A damper of this design is to be installed in a 12 by 30 in. duct that handles 3000 cfm. The pressure drop in the undampered system is 1.5 in. of water. If the pressure drop through the damper when wide open is to be 5% of the total system resistance, how much cross-sectional area in the duct should the damper occupy? ΔP T = ΔP duct + ΔP damper = 1.5 + 0.05 ΔP T ; ΔP T = 1.58 ΔP damper = 1.58 – 1.5 = 0.08 in. w.g. wide open 12 ( 0.075 )V 2 12 Pa V 2 CFM 2 HV ( in. w.g. ) = --------------------- = ---------------------------------- = ⎛ ----------------⎞ ⎝ 4005A⎠ 2 g ρw 2 ( 32.2 ) ( 62.4 ) ΔP damper open 3000 = 0.08 = 0.5 ⎛ ----------------⎞ ⎝ 4005A⎠ 2 A = 1.87 ft 2 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Q 2 = ( 6500 ) ( 0.855 ) = 5570 cfm Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 128⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual 9.32 What effect on the following parameters does a variation in air density have for a fan operating in a system: a. flow rate b. developed head c. horsepower Air Density Variation: a. Flow rate; Volume – Remains same; Mass – Increases with increases. b. Developed head – Increases directly with density increase. c. Horsepower – Increases directly with density increase. 9.33 A fan delivers 1500 cfm (708 L/s) of dry air at 65°F (18.3°C) against a static pressure of 0.20 in. of water (50 Pa) and requires 0.10 BHP. Find the volume circulated, the static pressure, and the BHP required to deliver the same weight of air when the air temperature is increased to 165°F (73.9°C). (Note: Atmospheric pressure is constant.) 1500 cfm, 65°F, P s = 0.20 in. H 2 O, 0.10 bph for air temperature of 165°F T ρ2 460 + 65----- = -----2 = ----------------------= 0.81 ρ1 T1 460 + 165 CFM 2 = CFM 1 = 1500 cfm ρ2 H P = H P ⎛ ----- ⎞ = 0.10 ( 0.84 ) = 0.084 BHP 2 1⎝ ρ ⎠ 1 9.34 Should fans be placed before or after air heaters? Why? For the same mass flow rate through the air heater, the fan law gives: ρ BHP 1 -------------- = -----2 BHP 2 ρ1 ρ2 BHP 2 = ⎛⎝ ----- ⎞⎠ BHP 1 ρ1 if ρ 1 is cold, ρ 2 is warm then BHP 2 > BHP 1 – Place fan before heater. 9.35 A 40 in. by 24 in. rectangular duct conveying 12,000 cfm of standard air divides into 3 branches (see This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto ρ2 P S = P S ⎛ ----- ⎞ = 0.20 ( 0.84 ) = 0.168 in. H 2 O 2 1⎝ ρ ⎠ 1 Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Chapter 9—Duct and Pipe Sizing⏐129 figure). Branch A carries 6000 cfm for 100 ft, B carries 4000 cfm for 150 ft, and C carries 2000 for 35 ft. (a) Size each branch for equal total friction of 0.15 in. of water. Do not exceed upper velocity limit of 2000 fpm. (b) What is the total friction loss if the same quantity of air, 12,000 cfm of air at 150°F and 14.0 psia, is passed through the same system as part (a)? (c) For a fan selected for part (a), at what percentage of the speed in part (a) must the fan run to satisfy part (b)? Branch A Total length = 100 ft + 6 ft. Assume branch similar to elbow. with r ⁄ w = 1.5 ; H ⁄ W = 1.0 → C = 0.09 ΔP ΔP = 0.15 in. H 2 O = ( 100 + 6 ) --------- + 0.0225 100 ΔP ⁄ 100 ft = 0.12 in. w.g. ( Chap. 21, 2009 HBF ) Assume W = duct width = 24 in. for 600 cfm and D eg = 33.6 V = 1400 fpm Size = 24 in. × 24 in. ΔP = 0.15 in. w.g. = ( 6 + 150 ) ΔP′ ⁄ 100 ΔP′ = 0.096 in. w.g./100 ft with W = 24 in. V = 1350 fpm Size = 24 in. × 19.5 in. Branch C ΔP = 0.15 in. w.g. = ( 35 + 6 ) ΔP′ ⁄ 100 + 0.0225 ΔP′ = 0.31 in. H 2 O for 2000 cfm, D eg = 11 in. Branch B b. V = 1800 fpm ( Chap. 21, HBF ) Size = 24 in. × 7.5 in. K = 0.89 Q actual = 12 ,000 cfm H o = ( 0.89 ) ( 0.15 ) = 0.133 in. H2 O c. For Part a. ΔP static, fan = 0.15 + 25 Δp = 0.183 in. H 2 O Q fan = 12 ,000 cfm Using fan lows, the required speed for 150°F with 12,000 cfm is rpm2 = rpm1 × 1 × 1. 9.36 A centrifugal fan operating at 2400 rpm delivers 20,000 cfm of air through a 32 in. diameter duct against a static pressure of 4.8 in. of water. The air is 40°F. The barometer is 29.0 in. Hg. Determine the horsepower input This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 2000 2 ΔP = 0.04 ⎛⎝ ------------⎞⎠ = 0.0225 in. w.g.; 4000 Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 130⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual if the efficiency is 70%. If the fan size, gas density, and duct system remain the same, calculate the horsepower required if operated at 3200 rpm. π 2 π 32 2 2 A = --- D = --- ⎛ ------⎞ = 5.6 ft ; 4 4 ⎝ 12⎠ Q 20 ,000 V = ---- = ---------------- = 3570 fpm A 5.6 2 12ρ u V cfm 2 20 ,000 h v = ------------------ = ⎛ ---------------- ⎞ = ⎛ ----------------------------- ⎞ = 0.8 in. w.g. ⎝ 4005A ⎠ ⎝ ( 4005 ) ( 5.6 ) ⎠ 2gρ w h t = h s + h v = 4.8 + 0.8 = ( 5.6 in. w.g. ) ( 0.0361 ) = 0.202 psi ( 144 ) AHP = ( 0.202 ) ---------------- ( 20 ,000 ) = 17.6 hp 33 ,000 17.6 a. BHP = ---------- = 25.2 hp 0.70 b. 3200 2 BHP 2 = ( 25.2 ) ⎛ ------------ ⎞ = 59.7 hp ⎝ 2400 ⎠ 9.37 Compute the efficiency of Fan 303 (Fig. 9-11b) when delivering 15,500 cfm at 4 in. static pressure (SP). From Fig. 9-116, fan BHP ≅ 13.4 CFM × ΔP ( in. w.g. ) 15 ,500 ( 4 ) Ideal BHP = ------------------------------------------------- = ------------------------- = 9.8 6350 6350 Wi 9.8 η f = ------- × 100 = ---------- × 100 = 74% Wa 13.4 a. CFM × ΔP ( in. w.g. ) CFM × ΔP W i = m· ∫ ν dp ≅ m· ν ΔP = ------------------------------------------------- = -------------------------conversion factors 6350 Wi CFM × ΔP W a = ------ = -------------------------ηf 6350η f b. c. W ( kWh ) = HP × 0.746 × kW ⁄ HP × time Motor input ( kWh ) = Fan input ⁄ η m CFM × 60 CFM × ΔP × 2545 W = ( 0.240 ) ( m· ) ( Δt ) = m ∫ ν dp ( 0.240 ) ------------------------ ( Δt ) = -------------------------------------------13.33 6350η f Δt = 0.371 ΔP ⁄ h f d. W = – m ∫ ν dp = – m ν dp ∼ CFM ⋅ ΔP V 2 CFM 2 L ΔP = ⎛⎝ f ---- + C o⎞⎠ ------ ∼ ⎛⎝ ------------ ⎞⎠ 2 A D 2 W ∼ CFM ( CFM ) ∼ ( CFM ) 3 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 9.38 Develop and explain the following relations for fan performance: (a) HP = CFM × ΔP/6350ηf (b) kWH = HP (0.746) Hours/ηm (c) Δtf = ΔP(0.371)/ηf (d) HP ~ CFM3 Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. reads 180 psi.= 385 hp ( 33 .= ---------------------------.2 BHP = ------------------------------. (www. The pump efficiency is 80% and the motor efficiency is 87.= 133 fps.8 ) ( 0.5 hp ) ( 2545 ) ( 1 ⁄ 3413 ) ( 0.= 35.2 fps 12.39 A water pump develops a total head of 200 ft.Chapter 9—Duct and Pipe Sizing⏐131 9. what is the power cost for pumping 1000 gal? Hm ( 200 ) ( 1000 ) ( 231 ⁄ 1728 ) ( 62.= 1.000 50.= 72. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.355 ( 3. Copyrighted material licensed to University of Toronto by Thomson Scientific. calculate the necessary input to the motor in kilowatts.= 257 ft 2g c H t = 437 – 40 + 257 = 654 ft This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 9. ( 180 ) ( 144 ) H d = ---------------------------. 3. The inlet pipe is 4 in. .11 ft ⁄ s ⎝ 1728 ⎠ ⎝ 60 ⎠ ( 3.4 ) WHP = ---------------.4 H S = +40 ft 231 1 3 Q = 1400 ⎛ -----------. American Society of Heating.60 ) kW = ( 385 ) ( 0.41 A pump delivers 1400 gpm of water.5¢ per kilowatt-hour.11 ) ( 60 ) ( 62.techstreet.11 ) ( 144 ) V s = ----------------------------.= 1540 η s = -----------3⁄4 3⁄4 H ( 360 ) Use a centrifugal pump. If the pump and motor combined efficiency is 60%.2 hp 33 . distribution. © (2009).35¢ for 1000 gal 60 9.40 For a certain system it is required to select a pump that will deliver 2400 gpm (150 L/s) at a total head of 360 ft (110 m). which is 22 ft above the pump centerline.81/h 0.81 ---------. Inc. Inc. (www. and a pump shaft speed of 2400 rpm. For personal use only.com). If the power rate is 1.875 ) Cost = ( 72.746 ) = 287 kW 2 H vd – H vs 2 Vd – Vs = -----------------.+ 22 = 437 ft. nominal standard pipe.73 ( 3. The discharge gage.= 50. Additional reproduction. The surface of the inlet supply is 40 ft higher than the pump centerline.11 ) ( 144 ) V d = ----------------------------. nominal and the outlet pipe is 2 in. Refrigerating and Air-Conditioning Engineers.⎞ ⎛ -----. What type of pump would you suggest? n Q 2400 2400 .ashrae.000 33 .000 ) ( 0. 62.⎞ = 3.015 ) = $0.= ----------------------------------------------------------------------------. The water temperature is 40°F.org).5 hp ( 0.5%.4 ) ( 654 ) H p = -------------------------------------------------------. 000 ( 0. A certain system is found to have losses due to frictional effects according to the equation H = 0.000 ) ( 0. Inc.741 psi ( 10 ) ( 144 ) H s = ------------------------.org).16 L/s) against 60 ft (18.techstreet. Refrigerating and Air-Conditioning Engineers. ( 300 ) ( 231 ⁄ 1728 ) ( 61.43 How many horsepower are required to pump 66 gpm (4.= 23. what would be the speed ratio n2/n1 for the same pump. BHP = 8. If the mechanical efficiency of the pump is 65%.9 lb/ft ν 165° 0. Should a backward.74 psi c.= 0. . If a capacity of 400 gpm is desired. The system is handling water at 160°F. yes P s = 10 psi η2 -----.000 ) 9.01 ) ( 90 ) BHP = -------------------------------------------------------------------------( 33 . and system? d.5 kPa).300 milli-inches of water (20.33 hp η 33 . For personal use only.= 400 --------.= ---------------.0164 –3 Qρh m· h ( 9250 ) ( 82 . Ventilating.80 ) 2 a. distribution. density of fluid.= ------. backward. what is the head developed by the pump and the BHP if the pump efficiency is 80%? b. (www.6 ft H2 O ( 61. American Society of Heating. What would be the theoretical maximum length of suction in order to prevent cavitation if the level of the supply tank is below the centerline of the pump? Assume atmospheric pressure to be 14. © (2009). (www.3 m) of head assuming 75% efficiency? Qρh ( 66 ) ( 231 ⁄ 1728 ) ( 62.132⏐Principles of Heating.ashrae.001 (gpm)2 where H is in ft of water. find the required horsepower input.= 1.01 ) η2 Q 1 = 300 Q 2 = Q 1 -----η1 Q 2 = 400 d. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.75 ) h = 0. c. Inc. 1 1 3 ρ = -----------.7 psi.05 hp η η ( 60 ) ( 12 ) ( 0.= 60. Additional reproduction. and Air Conditioning—Solutions Manual 9.42 A pump is required to force 9250 lb/h (4200 kg/h) of water at 165°F (74°C) through a heating system against a total resistance of 82.001 ( 300 ) = 90 ft H 2 O b.65 ) ( 33 .44 Solve the following problems: a.= 1.com).= -------------------------------------------------------------------.35 hp P v = 4. For a design capacity of 300 gpm.300 ) ( 10 ) BHP = ----------.= ----------------------------------------------------------. P b = 14.or forward-curved blade pump be chosen? Would you make arrangements for a priming system for the pump? Copyrighted material licensed to University of Toronto by Thomson Scientific.4 ) ( 60 ) BHP = ----------.33 η1 300 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 9. distribution. Inc. Additional reproduction.ashrae.Solutions to Chapter 10 LIFE-CYCLE COSTS Copyrighted material licensed to University of Toronto by Thomson Scientific. Refrigerating and Air-Conditioning Engineers. American Society of Heating. For personal use only. Inc. © (2009).com). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. (www. (www.org).techstreet. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . © (2009). American Society of Heating.Copyrighted material licensed to University of Toronto by Thomson Scientific. (www.com).techstreet.ashrae.org). Inc. (www. distribution. Refrigerating and Air-Conditioning Engineers. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . For personal use only. Inc. Additional reproduction. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 08 ) – 1$3000 = R --------------------------------------20 0. Find the amount accumulated.08)10 Sum = $2159 10. (www.Chapter 10—Life-Cycle Costs⏐135 10.11 ) – 1 S = 1000 -------------------------------------0. Series present worth n 1 ( 1 + i ) – 1P = R ⎛⎝ ------------⎞⎠ = R -------------------------n CRF i( 1 + i) 20 ( 1 + 0. Find the annual cost if the salvage value is $0 and the interest rate is 8%. find the yearly withdrawal that will use up the money in 20 years.com).09)3 Amount = $27. Neglect energy and maintenance costs. determine the value of this money in 10 years.08 ) R = $306 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto (1 + i) – 1 1 -⎞ .000 in 3 years with an interest rate of 9%.722 10.4 If $100. Sum = Amount (1 + i)n $35.= R ( CAF ) i 10 ( 1 + 0. Refrigerating and Air-Conditioning Engineers. Additional reproduction.000 = R -------------------------------------------20 ( 0.08 ) ( 1 + 0.08 ( 1 + 0.3 $1000 is invested at the end of each year for 10 years.2 Find the present worth of money that will have a value of $35.000 is invested at 8% interest.185 10.ashrae.1 If $1000 is invested at 8% interest. Interest is 11%.08 ) – 1 $100.5 The cost of a new heat pump system is $3000 with an expected lifetime of 20 years.org). Sum = Amount (1 + i)n = $1000 (1 + 0.000 = Amount (1 + 0.techstreet.11 S = $16. Inc. Copyrighted material licensed to University of Toronto by Thomson Scientific. For personal use only. n 20 ( 1 + 0. distribution. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Inc.= R ⎛ ----------P = R -------------------------n ⎝ CRF⎠ i( + i ) . American Society of Heating. n (1 + i) – 1 S = R --------------------------. (www.08 ) R = $10. © (2009).026 10. 20 yr. 9%. Ventilating.10185 110 × 1/0.000 ( CRF.org).10185 $1080 100 × 1/0. and Air Conditioning—Solutions Manual 10.85 $1200 × 0.com).500 trial and error solution for x% yields rate of return = 11. American Society of Heating. Present Worth System B $1000 $1200 Initial Cost CRF (8%. x%. 9%.22 $110 $100 $211. Calculate the break-even point if i = 9%.10185 Operating Cost Uniform Annual O&O Cost $122.10185 Present Worth $982 $2080 $2182 System A is least costly to own. N years ) – $5. 20 ) – $10.000 in 20 years. Initial cost – Salvage = Savings $15. distribution.10185 $101. 20 ) = $7.) Operating Cost = 0. x%. For personal use only. Compute the rate of return. (www.400 trial and error solution for N years yields N = 34 years (break even) (reasonable?) 10.ashrae.25% 10.8 on the basis of uniform annual costs. The new system saves $1400 per year in fuel cost. Inc. Costs System A System B Owning Cost Initial Cost × CRF $1000 × 0.8 The costs of two small heat pump units A and B are $1000 and $1200 and the annual operating costs are $110 and $100. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. N years ) = $1.000 ⁄ ( CAF. Initial cost – Salvage = Savings $60.000 and a salvage value of $5000. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto System A Copyrighted material licensed to University of Toronto by Thomson Scientific. Compare the systems on the basis of present worth. Neglect maintenance costs. The system has a salvage value of $10. Neglect maintenance costs.22 System A is least costly to operate. Refrigerating and Air-Conditioning Engineers. The interest rate is 8% and the amortization is selected as 20 years. . independent of age.9 Compare the units in Problem 10.85 $222.136⏐Principles of Heating.000 and saves $7500 in energy costs each year.6 A new heating system has a cost of $15.000 ( CRF.000 ⁄ ( CAF. Inc. © (2009). 10.techstreet. respectively. Additional reproduction.7 A new high-efficiency cooling system costs $60. (www. Refrigerating and Air-Conditioning Engineers.000 1. 101.000. Average Chiller Efficiency Chiller A Chiller B 0.500 $19.10 An installation is going to require a 500 ton chiller.73 kW/ton 0.323.100.423 Copyrighted material licensed to University of Toronto by Thomson Scientific.org). .6 × 13423 = $250.000 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 259.= ------------------. (www.000 1.000 kWh 0.techstreet.500 $240.480 1. The economic data is given below.6 yrs. Inc.500 – 240.500 $240. American Society of Heating. An annual energy analysis for this office building application shows that the required ton-hours over the year will be 2.63 × 2.500 $10.com).059) +10000 $88.63 kW/ton Initial Cost $221.) savings in operating cost would be 18.000 $19.057 $13.100.000 (0. A B Energy Required: 0. For personal use only.ashrae.000 Electricity Cost 6¢/kWh 5.000 $240. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.Chapter 10—Life-Cycle Costs⏐137 10.533.9¢/kWh Maintenance Costs $9. (www.323. © (2009). distribution.480 – 88.100.500 For the remaining time (18. Inc.533.000 (0.06) +9500 $101.000 kWh Annual Operating Cost: 1. Additional reproduction.41 yrs.500 + $19.73 × 2.500 Installation Cost $19.500 + $19.057 Initial Cost: $221.000 Estimated Life 20 years 20 years Perform a simple payback analysis for this option.000 $259.= 1.000 Y pb = --------------------------------------------. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . © (2009). Inc. American Society of Heating. Additional reproduction.techstreet. (www. Inc.Copyrighted material licensed to University of Toronto by Thomson Scientific. distribution. (www. For personal use only. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Refrigerating and Air-Conditioning Engineers.ashrae.org).com). Additional reproduction. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .Solutions to Chapter 11 AIR-CONDITIONING SYSTEM CONCEPTS Copyrighted material licensed to University of Toronto by Thomson Scientific. (www.techstreet. Refrigerating and Air-Conditioning Engineers. (www. © (2009).com). For personal use only.org).ashrae. Inc. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. American Society of Heating. distribution. Inc. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. distribution. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .ashrae.org). (www. Inc. American Society of Heating. © (2009). Additional reproduction. Refrigerating and Air-Conditioning Engineers. (www.techstreet. For personal use only.com).Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc. 12°F 11.cool .5 ( 0. how much air must be circulated? q s = q T ( SHR ) = 20 ( 12000 ) ( 0.90.techstreet.1 ( Δt ) 1.com). If the conditioned air is to be supplied at 20°F less than the room temperature.org).87567 ) = 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. what is the highest temperature at which the conditioned air can be supplied (ts)? Why? ts max = dew-point temperature of the space conditions P s at 75°F = 0. If there is negligible latent load within the space.43784 t s at 0.9 ) q s = 216000 Btu/h q s = 1. Hg = 55.1 A room is to be cooled to a temperature of 75°F and a relative humidity of 50%. (www.1 ( CFM ) ( t r – t s ) qs 216000 CFM = -----------------.Chapter 11—Air-Conditioning System Concepts⏐141 11. Additional reproduction. distribution.87567 in. © (2009).43784 in. Inc.3 What are the four generic types of air systems expressed by thermodynamic methods? Copyrighted material licensed to University of Toronto by Thomson Scientific. Hg P s at 75°F/50% = 0.4 What are the 18 fundamental parameters that must be addressed in the selection and design of an HVAC system? Load dynamics Performance requirements Availability of equipment Capacity Spatial requirements First cost Energy Consumption Operating cost Simplicity Flexibility Operations requirements Service ability Maintainability Availability of service Availability of replacement components Environmental requirements of space Environmental requirements of community Reliability This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 11. For personal use only.ashrae. (www. Inc.1 ( 20 ) 3 CFM = 9820 ft ⁄ min Heat . Refrigerating and Air-Conditioning Engineers.2 A room has a total space cooling load of 20 tons and a sensible heat ratio of 0.off Dual stream Reheat Variable Air Volume 11. . American Society of Heating.= -----------------1. 000 cfm of air at a total fan pressure rise of 6 in. the cooling and heating load for each room in a building must be calculated? Why? Because the system must be designed to add heat or remove it in each room at the same rate at which the load occurs in order to maintain thermal equilibrium. How much power (hp) is required to drive the fan? b.g. .142⏐Principles of Heating. and Air Conditioning—Solutions Manual 11. If the minimum outdoor air dampers are sized for 6000 cfm of ventilation air.8 A constant-flow air-handling system is designed to circulate 60. Refrigerating and Air-Conditioning Engineers.9 ) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto CFMO t m = t i + -----------------. w. American Society of Heating.6 If outdoor air at 95°F dry bulb and 78°F wet bulb is cooled to 75°F dry bulb without any dehumidification.ashrae.7 ) Hp = 81 CFM ( ΔP f )θ kWh = ------------------------------( 8512 )η f η m 60000 ( 6 ) ( 8760 ) kWh = ---------------------------------------.5 Before designing a system.7 In the air handling unit of Figure 11-1.6 Btu/lb Copyrighted material licensed to University of Toronto by Thomson Scientific. Additional reproduction. under design conditions the outdoor air temperature is 95°F dry bulb and 78°F wet bulb and the space temperature is 75°F and 50% RH.org). For personal use only. (www. ) = ------------------------------6350 ( 0. what will the relative humidity be? 90% RH 11. What will be the annual fan energy consumption? a) b) CFM ( DP t ) Hp = ---------------------------6350η f 60000 ( 6 in. The fan efficiency is 70% and the motor efficiency is 90%.techstreet.( t o – t i ) CFMT 6000 = 75 + --------------.= 588100 kWh 8512 ( 0. what is the statepoint (dry-bulb and wet-bulb temperatures) of the mixed air? From Psych. The system is designed to operate continuously. distribution.7 ) ( 0. 11. chart 11. Inc. The supply fan handles 60. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. a.( 95 – 75 ) 60000 = 77°F db 64°F wb h m = 29.000 cfm of air at 55°F saturated (entering the fan). an understanding of the nature of the load is fundamental to the selection of a system type. (www. © (2009). Inc. Also. Ventilating.com). Inc.09 = 65 –7 Hp w 6. the sensible space load were reduced by 25% by using a more energy effective building envelope and improved lighting system. (www. American Society of Heating.09 × 10 q 6350 ( 1.Chapter 11—Air-Conditioning System Concepts⏐143 11. of water total pressure or by a water system with a pump head of 40 ft.ashrae. Calculate the ratio of fan power required for an air system to pump power required for a water system with the following system variables: Fan efficiency 70% Pump efficiency 80% Air Δt 20°F Water Δt 40°F q = 1.1 ) ( 20 ) ( 0.8 ) 3960 ( 500 )Δt w ( η p ) –5 Hp air × 10 q------------------------------------------.= 6.31 × 10 q 3960 ( 500 ) ( 40 ) ( 0.7 ) GPM ( ΔH ) Hp w = --------------------------3960 ( η p ) q = GPM ( 500 ) ( Δt w ) GPM = 8 ⁄ ( 500 ) ( Δt w ) q ( 40 ) q ( ΔH ) –7 Hp w = ----------------------------------------------. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Additional reproduction. . in the above problem.1 ) ( Δt ) q s ≈ CFM and kWh ≈ CFM kWh ≈ q s ∴Δ ( kWh ) = kWh (% reduction in load) = 588100 ( 0.1 ) ( Δt air )η f 6350 ( 1.= -------------------------------------------------.31 × 10 q This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto CFM ( ΔP t ) Hp air = --------------------------6350η f Copyrighted material licensed to University of Toronto by Thomson Scientific. and this change were accommodated by reducing the air flow rate at the same fan pressure and efficiencies.10 It is desired to transfer a given quantity of heat energy from one location to another location in a building.= -----------------------------------------------. Two methods being considered are either by an air system operating at 4 in.com).= 4. Inc.1 ( Δt air ) q ( ΔP t ) q(Δ) –5 Hp air = ----------------------------------------------.9 If.techstreet.1 ( CFM ) ( Δt air ) q CFM = -----------------------1.= 4. distribution. For personal use only. (www.org). © (2009). Refrigerating and Air-Conditioning Engineers.25 ) = 147000 kWh 11. what would be the reduction in annual fan energy? q s = CFM ( 1. e. Hot water entering the convectors (radiators) of a hydronic system f. d. Air leaving a gas-fired warm air furnace b.046 TR = -------------------------2 15 ⎛ ----------⎞ – 1 ⎝ 3.g--2π y 1 386 f n = -----.42 N min f = ----------60 = 60f = 60 ( 5. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.42 ) N min = 325 rpm 11.12 Specify typical temperatures for the following: a. and it is installed on a spring isolator mount with 1 in. (www. and Air Conditioning—Solutions Manual 11. static deflection.= ---------------. Transmissibility of the isolator b. . 135°F 105°F 58°F 55°F 190°F 170°F (57°C) (41°C) (14°C) (13°C) (88°C) (77°C) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto N min 3f n Copyrighted material licensed to University of Toronto by Thomson Scientific. c. Air leaving the cooling coil of a commercial air conditioner e. © (2009). (www.50 1 TR = ------------------------2 ( f ⁄ fn ) – 1 a) N 900 f = -----.5 f = f = 3 ( 3. Refrigerating and Air-Conditioning Engineers. Hot water returning to the boiler from the convectors a.org). For personal use only. Inc.techstreet. b. Ventilating. distribution. Air leaving a heat pump condenser c.--------.= 0.com). Air leaving the cooling coil of a residential air conditioner d.= 15 Hz 60 60 1 f n = -----. Determine a.= 3.= 3 TR 0. Inc.13⎠ 1 TR = ------------------------2 ( f ⁄ fn ) – 1 b) 2 ( f ⁄ fn ) TR – TR = 1 2 TR + 1 0.11 A fan with a variable speed drive is selected to operate at 900 rpm.144⏐Principles of Heating. Additional reproduction.13 Hz 2π 1 1 .= --------. American Society of Heating.13 ) = 5. Minimum speed that the unit can be operated at before the transmissibility is 0.ashrae. f.5 + 1 ( f ⁄ f n ) = ---------------. 13 An air-conditioned room has a sensible heat load of 200.2 31.25 ) ( 39.010 – ----------------------------------.01125 D 60 56. © (2009). 64°F wb.015 ) + ( 0. W A = 0. ( 0. h B = 29. an occupancy of 20 people.3 b) SHR = q s ⁄ ( q s + q L ) = 200000 ⁄ 250000 = 0. Use the following letters to designate state points: A Outside design conditions B Inside design conditions C Air entering apparatus (mixed air) D Air entering room (supply air) a. v D = 13.6 67. distribution.3 ft ⁄ lb.010 C 80.0091 A: 95°F db. . What is the apparatus load in tons? e.75.25 ( 51230 ) ( 0.244 )(76 – 60) M da = 51230 lb/h MW ( 50000 ⁄ 1054 ) D: W D = W B – ---------. Outside air is assumed to be at design conditions of 95°F dry bulb and 76°F wet bulb. Does the room load plus the outside air load equal the coil load? a) Point Dry Bulb Wet Bulb h W A 95 76 39. (www. V· = --------------. Complete the table provided. and is maintained at 76°F dry bulb and 64°F wet bulb.010 ) ( 1054 ) = 67495 Btu/h 126872 Btu/h f) 250000 + 126872 = 376872 ≈ 378600 YES This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto C: 80.com). 67. Conditioned air leaves the apparatus and enters the room at 60°F dry bulb. American Society of Heating.2 0.01125 – 0. Calculate the room SHR.01125 q s = 200000 = M da C p ( t B – t D ) = M da ( 0.2 ) = h c . h D = 24.25 ) ( 0. b.010 M A h A + M B h B = ( M A + M B )h c . Twenty-five percent of the air entering the room leaves through cracks and hoods.4.ashrae. Inc.org). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. c.= 0.000 Btu/h.000 Btu/h. h A = 39. M A W A + M e W e = ( M A + M e )W c ( 0.0091 )28 – 24.015 – 0.75 – ( 0.244 ) ( 95 – 76 ) = 59376 Btu/h c) q L = M OA ( W A – W B )1054 = 0. (www.5 tons O. What is the load of the outside air? In lb per hour? In cfm? f.015 B 76 64 29.3 ] = – q q = – 378600 Btuh = 31. Load: q s = M OA C p ( t A – t B ) = 0.4 ) + ( 0. a latent heat load of 50.3 0.8°F wb.Chapter 11—Air-Conditioning System Concepts⏐145 11.015 B: 76°F db.A. Inc. Additional reproduction. 56. W c = 0. W B = 0.6°F db.( 13.techstreet.= 0.4 0.0091 M da 51230 D: 60°F db.25 ( 51230 ) ( 0. For personal use only.2°F wb Assuming occupancy is included in q s and q L as should be: Copyrighted material licensed to University of Toronto by Thomson Scientific. 76°F wb. Refrigerating and Air-Conditioning Engineers.75 0.8 d) 3 51230 51230 lb/h. What air quantity must enter the room? d.75 ) ( 29.2.8 24.75 ) ( 0.010 ) = W c C: h c = 31.3 ) = 11356 cfm 60 M da h c – M da ( W c – W D )h f60°F – M da h D + q = 0 e) 51230 [ 31. θ r = 40%.242 ( 102 – 70 ) Copyrighted material licensed to University of Toronto by Thomson Scientific. American Society of Heating. v = 13. Refrigerating and Air-Conditioning Engineers.001.0053 ) = 20.0057 m: 0.0053 )33 ] b) = 30300 Btu/h = 2.7 ) = h m = 22.001 ) + 0. Estimate the temperature and humidity ratio of the air entering the conditioned space. 10% outside air 0.0062 ) = W m = 0. AS = M ( W s – W m ) = 7748 ( 0.000 Btu/h and a latent loss of 20.7 ) + 0.7 m: 90% recirc.0088 60000 SHR (space) = --------------. Ventilating. Rate of dehumidification 75°Fdb 65°Fwb h 1 = 30 65°F. The space is to be maintained at 70°F and 40% RH. The return air from the rooms has average dry. W s = 0. The conditioner consists of an adiabatic saturator and a heating coil. W x = 0. distribution.0109 – 0.7 ) = 90563 Btu/h M W. Inc.0062. (www. © (2009).9 ( 0.7 40°Fdb.= 3600 lb/h 13. 20% RH.7°F.15 Air at the 800 ft3/min leaves a residential air conditioner at 65°F with 40% RH.0088. t m = 67°F t wb.4.7°F = t x x: t = 53.000 Btu/h.242 ) ( 102 – 53. Size of the unit in tons (12.org). m = 53. v s = 14.52 tons · M w = M da ( W 1 – W 2 ) = ( 3600 ) ( 0. W OA = 0. Determine a. .com). chart: at intersection of W = 0. and Air Conditioning—Solutions Manual 11. Inc. h 2 = 21. W 2 = 0.0053. 40% RH.0057 ) = 24 lb/h 11. W r = 0. respectively.146⏐Principles of Heating.0109 a) q = M da [ h 1 – h 2 – ( W 1 – W 2 )h f ] = 3600 [ 30 – 21.= 7748 lb/h Cp ( ts – tr ) 0.335 w 1 = 0.4 – ( 0. h OA = 10.1 ( 10.and wet-bulb temperatures of 75°F and 65°F.ashrae. h r = 23.0088 and space condition line (slope of SHR = 0.9 ( 23.techstreet.14 A space has a sensible heat loss of 60.0088 – 0. θ = 100%.000 Btu/h = 1 ton) b.335 ( 800 ) ( 60 ) M da = ------------------------.= 0..35 or 1853 cfm q HC = mCp ( t s – t x ) = 7748 ( 0. For personal use only.75 ): t s = 102°Fdb.4 .1 ( 0. What is the flow rate in lb/h and cfm? How much heat is added by the coil to the air in Btu/h? How much water is added to the air by the adiabatic saturator (lb/h)? q L = 20000 Btuh q s = 60000 Btuh t r = 70°F t OA = 40°F r: OA: t r = 70°F. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. (www.2 lb/h This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto qs 60000 M da = ------------------------.0109 – 0.75 80000 From Psych.= --------------------------------------. The air that passes through the conditioner is 90% recirculated and 10% outdoor air at 40°F and 20% RH. Additional reproduction. WB = 57°F ∴t = 91°Fdb a) Entering preheater. and then reheated. The temperature out of the adiabatic saturator is to be maintained at 60°F dry bulb.002152. d. Temperature of the air entering the preheater Temperature of the air entering the space to be heated Heat supplied to preheat coil.8 M da h 1 – M da h 2 – M da ( W 1 – W 2 )h 3 + q c = 0 26100 [ 33. Cooling capacity of the air-conditioning unit.0093 – 0. (www. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. c. © (2009). American Society of Heating.0093. b. 100% RH. lb/h Sensible heat load on the conditioner.009 ) = 109.8 tons e) Dew Point = 54°F r: OA: a. d. gpm Temperature of the spray water Show the processes and label points on the psychrometric diagram 75°Fdb. W OA = 0. The outdoor air conditions can be taken as saturated air at 20°F.16 In an air-conditioning unit 6000 cfm at 80°F dry bulb.5 – 60 ) = 313740 Btu/h (reheat) e) 700 lb gpm = ---------. °F 6000 × 60 M da = -----------------------.com).Chapter 11—Air-Conditioning System Concepts⏐147 11. The space is to be maintained at 75°F and 50% RH. tons Rate of water removal from the unit. Btu/h Latent heat load on the conditioner. distribution. W r = 0.( 0. Inc. Refrigerating and Air-Conditioning Engineers.0132 – 0. Calculate: Copyrighted material licensed to University of Toronto by Thomson Scientific. h r = 28. humidified with an adiabatic saturator.4 – ( 0. For personal use only. .1 cfm ( t s – t r ) = 1.5 gal⎠ f) 57°F g) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 11. 60% RH.106 Leaving adiabatic saturator: t = 60°Fdb. enter the unit.244 ) ( 80 – 57 ) = 146470 Btu/h d) = 12.17 A space in an industrial building has a winter sensible heat loss of 200. Due to the nature of the process.2 tons q L = M c ( 1076 ) = 109.1 ( 7000 ) ( 101.0022 ) ⁄ ⎛ 8.000 Btu/h and a negligible latent heat load (latent losses to outside are made up by latent gains within the space). e. Btu/h Heat supplied to reheat coil. e. 50% RH. (www.org).ashrae. f.0093 ∴wb = 57°F Leaving preheater: W = 0. Additional reproduction. Inc.6 lb/h c) q s = M da C p ( t 1 – t 2 ) = 26100 ( 0.44 gpm ⎝ 13.2 20°Fdb. and standard atmospheric pressure. g. t = 20°Fdb b) q s = 200000 = 1. h OA = 7.6 ( 1076 ) = 117930 Btu/h = 9. Calculate a) a. b. c.33 -------⎞ = 0.1 ( 7000 ) ( 91 – 20 ) = 536760 Btu/h (preheat) q ≅ 1.1 ( 7000 ) ( t s – 75 ) t s = 101.techstreet. 100% outdoor air is required for ventilation.4 tons b) M c = M da ( W 1 – W 2 ) = 26100 ( 0.5°F c) d) q ≅ 1. W = 0. Btu/h Dew point of the air leaving the conditioner. The leaving condition of the air is 57°F dry bulb and 90% RH.002152. The amount of ventilation air required is 7000 cfm and the air is to be preheated.009 )25 ] = – q c q c = – 268700 Btu/h = 22.8 – 23. Btu/h Quantity of make-up water added to adiabatic saturator.0132 – 0.= 26100 lb/h 13. W s = 0. M da = 10246 lb/h c) r: 77°F db. (www.023. chart: t = 49°F db. b.5 ( 0.ashrae. Additional reproduction. what is the refrigeration load? Copyrighted material licensed to University of Toronto by Thomson Scientific. and floor is 33. distribution. determine the required airflow (cfm) and the maximum relative humidity permissible in the incoming air. 60% RH. and Air Conditioning—Solutions Manual 11.70 70000 + 30000 Using protractor on Psych. Specify the desired interior dry-bulb temperature and relative humidity. a.148⏐Principles of Heating.08 cfm ( 95 – 75 ).8 kW (30. h m = 0.8 M da [ h m – h s – ( W m – W s )h f49° ] = – q q = – 10246 [ 39. a) q s ( loss ) = 33600 – 640 ( 3.5 kW (70.= 0.5 ( 0.3 kW This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 11.600 Btu/h.0052 – -----------------------------------------.000 Btu/h). If the heating system provides air at 95°F. W r = 0.0046 M da 5497 φ s = 14% max b) 70000 SHR = -----------------------------------. ∴φ = 28%.6 loss Mw ( 18 ) ( 200 ) ( 1070 ) W s = W r – ---------.010. The room is to accommodate 18 adult males [250 Btu/h (sensible) and 200 Btu/h (latent) per person].8 – ( 0.0074. W = 0. The inside glass temperature on the design day is 40°F. b.7 m: W m = 0.18 In winter. For personal use only. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. The zone is to be maintained at 25°C (77°F) and 50% RH.= 0.5 ( 48. a. American Society of Heating.5 ( 29. .7 ) W m = 0.org).= 5497 lb/h 13.= 0. Calculate the conditions (t and W) of the entering air to the zone if the air leaves the coil saturated. © (2009). h s = 19. ceiling.000 Btu/h) and a latent load of 8.0074 )17 ] = – 195650 Btu/h = – 57. 50% RH. h m = 39.0165 . h OA = 48. Inc.4 a) OA: 97°F db. The heat loss through the walls. (www.0165 – 0.19 A zone in a building has a sensible load of 20.023 ) .P. a meeting room with a large window is to be maintained at comfort conditions. Ventilating. Refrigerating and Air-Conditioning Engineers. W r = 0.05 – 19. What flow rate is required in order to maintain the space temperatures? c.0052 c) 60 26916 = 1.0074 at 100% RH q s = M da ( 0. Condensation on the window is highly undesirable. cfm = 1246 × ---------. 100% RH.com). If a mixture of 50% return air and 50% outdoor air at 36.05 s: 49°F db.413 ) – 18 ( 250 ) = 26916 Btu/h b) t db = 75°F (selected). c. Inc. W OA = 0. Determine the sensible heat loss or gain. h r = 29. = 40°F.techstreet. There are 640 watts of lights in the room.244 ) ( 77 – 49 ) = 70000.01 ) + 0. D.4 ) + 0.1°C (97°F) and 60% RH enters the air conditioner. org). The heating system operates with 25% outside air mixed with return air. a) H L × DD × 24 9 1570000 × 4900 × 24 E = ---------------------------------.385 × 10 Btu Δt × k × v ( 75 – ( – 5 ) ) × 1 × 1 = 405800 kWh 9 b) 1. a. and typical uses of the following systems: a.7.× C D = --------------------------------------------------.000 Btu/h = 460 kW 11. Annual fuel cost using No. including temperatures and flow rates at each location.60/gal. .21 A general office building in St.6 = 1. For personal use only.570. Missouri.21.3. Louis.ashrae. Note: different outside design conditions may be selected.4 Article 12.techstreet. Inc. 2 fuel oil at $1.5 Article 12.1 Article 12.10)(130 – 55) = 1. Btu b. determine: a. disadvantages.Chapter 11—Air-Conditioning System Concepts⏐149 11. American Society of Heating. b. Inc. Schematically draw the flow diagram and label.2 Article 12. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Double-duct system e. Specify the necessary furnace size. Annual energy requirements for heating. Terminal reheat system c.150. has a winter sensible space heating load of 1.= 19000 scfm 1.com). Additional reproduction.1 ( 130 – 75 ) b) Qf = 19000(1.1 Article 12.000 Btu/h for design conditions of 75 and −5°F.× $1. (www.× 0. © (2009).2 11.22 For the building of Problem 11.8. Fan-coil units b.385 × 10 Btu Cost = ------------------------------------. distribution. Refrigerating and Air-Conditioning Engineers. Induction system a) b) c) d) e) f) Article 12.60/gal = $17600 126000 Btu/gal This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto a) Copyrighted material licensed to University of Toronto by Thomson Scientific. Variable volume system f. 1150000 CFM s = --------------------------------. Multizone system d.20 Sketch (with line diagrams) and list the advantages.3. (www. 3°F M a [ h m – h s – ( W m – W s )h f ] + Q c = 0 4600 -----------.0033 ) = 20000 Btu/h c) Furnace size = 307400 Btu/h ( say 300000 ) Summer s: T s = 58°F. Winter: Sensible space heat loss is 189.3 8.150⏐Principles of Heating.T m = 83. Fan operation is constant all year long. The minimum supply air temperature from the cooling coil is 58°F.0119 – 0.10 ( 4600 ) ( T s – 72 ) ⇒ T s.2 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto use 4600 cfm Q s. Size the humidifier.001074.0033 ) = 2.3 m: 1350 ( 0. W rmin = 0.10 ( scfm ) ( 78 – 58 ) ⇒ SCFM = 4600 b) Winter SCFM on T OA + SCFM r T r = SCFM T T m 1350 ( 6 ) + 3250 ( 72 ) = 4600 T m . or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.001074 ) + 3250 ( 0. d) Humidifier h m = 33.4 – 52.= 0.0104 + -----------------------------------.000 Btu/h at design conditions.0124 Outside: 96°Fdb. distribution.0042 – 0.0042 Winter design Outside: 6°F.013 Inside: 72°F.ashrae. Refrigerating and Air-Conditioning Engineers. For personal use only.3 Q c = – 165200 Btu/h A/C size 4600 60 1 M c = -----------. 25% RH (min). Inc. Missouri. 74°Fwb. Determine the fan size (scfm) needed to provide sufficient air b. American Society of Heating. φ s = 100%.org). 20° range.com).583 Ventilation air a) General office: 15 cfm/person × 90 = 1350 cfm Q s.( 0. w = 109. Ninety people are normally employed doing light work while seated. a.0104. .2 Ms 90 ( 255 ) ⁄ 1100 r: T r = 78°F. 38 ft by 80 ft by 8 ft. W m = 0. Inc.10 ) ( 109. w = 189000 = 1. W O = 0. The building is in Kansas City. Size the heating unit needed. an air-treating unit consisting of cooling coil.6 ) = 287400 Btu/h Q L = 4600 ( 4840 ) ( 0.24 or 2 ------------13. and humidifier is provided for this space with the flow diagram as shown. gal/h Summer design Inside: 78°F.0119 1350 ( 96 ) + 3250 ( 78 ) = 4600 T m . h O = 2. and Air Conditioning—Solutions Manual 11.23 To provide comfort conditions for a general office building. © (2009). (www.0114 ) = 4600 W m .0042 ) = 4600 W m . h s = 25. Btu/h d.6°F SCFM O W O + SCFM ⋅ W r = SCFM T W m 1350 ( 0.= 0.0033 Q s = 4600 ( 1.W m = 0.0042 – 0.2 – 25.( 60 ) [ 33.0114 60 Ma 46000 × ---------13.4°F Copyrighted material licensed to University of Toronto by Thomson Scientific. Btu/h c. s = 101200 = 1.10 ( scfm ) ( 155 – 72 ) ⇒ SCFM = 2070 Q s. w = 189000 = 1. 60% RH (max). W O = 0. W r = W s + ------. W rmax = 0.2 – ( 0. W s = 0. (www. Maximum supply air temperature is 155°F. Summer: Sensible space heat gain is 101. T m = 52.techstreet. heating coil. Additional reproduction. Ventilating.0104 )26 ] = – Q c 13. latent load is negligible.200 Btu/h at design conditions.× ---------.013 ) + 3250 ( 0. Size the cooling coil needed. Latent load is due entirely to the occupancy.33 4 gph * Note: different outside design conditions may be selected. 23 – 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. % 4.33 = 16230 cfm @ s: P w = P w.47511 Copyrighted material licensed to University of Toronto by Thomson Scientific.2141 60 0.0112 ) + ( 73060 – 11250 ) ( 0. (www.8°F c Q s ≅ 1. s = 0.000 Btu/h Latent = 220.= 0.8 – 55 ) = 443500 Btu/h c or Q s = Q T – Q L = 622200 – 184600 = 437600 Btu/h c 7. distribution.47511 psia Ma 73060 h r = 0. Q L ≅ 4840 × CFM × ΔW = 4840 × 16230 × ( 0.× 110 = 55.h o = 0. Latent component of (4) 6.ashrae.0110 + ---------------------------------.80 – 25.74 4.80 Btu/lb a W m = [ ( 11250 ) ( 0.240 ( t ) + 0.1335 lb v ⁄ lb a m a [ h m – h s – ( W m – W s )h c ] + Q c = 0 73060 [ 33. P w 1 s0 = 0.01335 – 0.23 psi).= 0.23 3.444 – 59 ) = 25.26432 w = 0. cfm 3.30 ( 0. 1. Inc.01374 ) ] ⁄ 73060 = 0. Size of cooling unit.= 0.01335 – 0. P w 10 = 0.23 – ( 0.0110 12. determine 1. Refrigerating and Air-Conditioning Engineers. Sensible cooling load due to outside air.622 ---------------.2141 W s = 0.622 --------------------------------------.6% P – Pw 12.444 ⋅ t ) ⇒ t m = 79.⇒ P w = 0. American Society of Heating.= 73060 lb/h 0.techstreet.80 – 0.01335 ( 1061 + 0.011 ( 1061 + 0.21634 0.= 0.01374 = 0.0112 . m O = 2500 × 60 ⁄ 13. Btu/h 5.01874 lb v ⁄ lb a .28 – 2. 33.444 × 78 ) = 33.74 ) ] ⁄ 73060 = 33. Colorado (elevation = 5000 ft. Relative humidity at return. (www.0110 ) ( 23. Sensible component of (4) 7. Outside air at the rate of 2500 cfm is required for ventilation.41 h s = 0.240 ( 91 ) + 0.07 ) ] = – Q c = 622200 Btu/h 5. s = 0.18 ) + ( 73060 – 11250 ) ( 33.0110 ) = 184600 Btu/h 6.24 A view of the air-conditioning system for a building in Denver.24 ( 58 ) + 0.× 13.33 = 11250 lb/h .622 -------------------------.23 – P w 0. Supply airflow.26432 psia .72113 ) = 0.444 × 91 ) = 34.21634 W O = 0. Q s ≅ 1. Other conditions at summer design are Space Loads Sensible = 410. is given.com).21634 h m = [ ( 11250 ) ( 34. .org).72113 . © (2009). Supply airflow. lb/h 2.01374 ( 1061 + 0.Chapter 11—Air-Conditioning System Concepts⏐151 11. For personal use only. Btu/h 410000 Q s = m a c p ( t r – t s ) ⇒ m a = -----------------------------------. 30% RH For an indoor design temperature of 78°F. Inc.000 Btu/h Outside Air: 91°F. Q L ⁄ 1100 220000 ⁄ 1100 W r = W s + ----------------------.0112 ( 1061 + 0. P w.240 ( 78 ) + 0.10 × 16230 × ( 79.622 -----------------------------. 2.244 ( 78 – 55 ) 73060 v· = m a v = --------------.10 × 2500 × ( 91 – 78 ) = 35750 Btu/h o This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Pw Pw 0.10 × CFM × Δt = 1.18 12. φ = ------------------. Additional reproduction. barometric pressure = 12.= 0. Inc.Copyrighted material licensed to University of Toronto by Thomson Scientific. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.techstreet. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . (www. Refrigerating and Air-Conditioning Engineers. distribution. (www. Inc. American Society of Heating. © (2009). For personal use only.org). Additional reproduction.ashrae.com). This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . For personal use only. Inc.ashrae.techstreet. American Society of Heating. Inc.com). (www. Refrigerating and Air-Conditioning Engineers.Solutions to Chapter 12 SYSTEM CONFIGURATIONS Copyrighted material licensed to University of Toronto by Thomson Scientific. © (2009). Additional reproduction.org). distribution. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. (www. Refrigerating and Air-Conditioning Engineers. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.techstreet. For personal use only. Additional reproduction. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . (www.Copyrighted material licensed to University of Toronto by Thomson Scientific.org).ashrae.com). © (2009). Inc. (www. Inc. American Society of Heating. distribution. com).org). describe the operating sequence of the zone or terminal control of 1. the reheat valve opens and reheats the minimum volume flow air to prevent overcooling and maintain room temperature control at all times. they will turn in the wrong direction.techstreet. why must all of the terminal fans be running prior to turning on the system fan? If the main fan is blowing air through the terminal fans. thus running backwards. 1. All as the primary supply air throttles down. distribution.off 3. list the four fundamental psychrometric system types from least consumption to most consumption. Heat . Reheat 12. They.cool . which keeps the airflow high enough to assure adequate ventilation and air circulation rate. reducing the flow of conditioned air to the space. American Society of Heating. Most systems have a lower limit setting to assume adequate ventilation and air circulation. below which there is no room temperature control.2 In a VAV system with series fan powered terminals. A VAV reheat system 3. 12. (www. will run in the direction they are turning when turned on. For personal use only.Chapter 12—System Configurations⏐155 12. . (www. 1) VAV system: As the space cooling load decreases. Inc.5 a. because the fan does not have to run continuously.ashrae. Additional reproduction. 2) VAV reheat system: Similar to (1) above except that when the load falls below the minimum airflow setting. A dual-duct VAV system Copyrighted material licensed to University of Toronto by Thomson Scientific. being single phase motors.1 From an energy consumption perspective. the space thermostat closes a supply air damper. 2) A VAV system alone cannot handle a heating load. In your own words. the warm duct damper starts modulating open mixing warm air with the conditioned air in increasing amounts to prevent overcooling and maintaining temperature control at all times. 3) Dual-duct VAV system: Similar to (1) above except that when the load falls below the minimum airflow setting. 12. Why do some VAV systems also use dual-duct or reheat features? b. a) b) 1) The dual duct or reheat provides false loading. Refrigerating and Air-Conditioning Engineers. © (2009). Dual stream 4. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 12.4 What is the purpose of using a fan powered terminal in a variable air volume system? Provides better mixing. A VAV system 2. Prevents dumping.air . Variable . Inc. ambient air circulation.3 What is the advantage of a parallel fan powered terminal over a series fan-powered terminal? Less energy consumption. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. and air distribution.volume 2. 1 is a variable volume. and IAQ concerns. 12.6 What is the primary advantage of a separate outdoor makeup air conditioning unit? The outdoor air make-up air-conditoning unit (primary air unit [12. causing discomfort and sometimes freeze damage or microbial growth.11 Size the basic components and sketch the equipment arrangement if the HVAC system now under consideration for the building of Example 12.7 Why is a high-pressure primary system fan required with an induction system? Because there is a high pressure required to create the high velocity flow through the induction nozzles. For personal use only. Also. leakage of pans causing damage to ceilings. . Additional reproduction. Vertical units can be mounted under windows in extremely cold climates. (www. and neutral decks).1]) provides a constant volume of outdoor ventilating air and is used in conjunction with VAV systems. difficulty of access for maintenance and service. and the cooling coil usually operates dry. This is a design problem and many solutions can be found. Ventilating. dual fan. Refrigerating and Air-Conditioning Engineers. untreated ventilation air can enter the sapce. Inc. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 12. cold. distribution.com). (www. Inc. providing better heating performance. no return air fan. American Society of Heating.10 Size the basic components and sketch the equipment arrangement if the HVAC system now under consideration for the building of Example 12.1 is a triple deck multizone (hot. the quantitiy of ventilation air is not controllable because of varying pressure differentials resulting from chimney effects and wind variations.9 Why are vertical floor mounted fan coil units recommended in some applications in preference to horizontal ceiling mounted units? Explain. Because when the room thermostat turns the fan coil unit off. Copyrighted material licensed to University of Toronto by Thomson Scientific. This is a design problem and many solutions can be found. 12.ashrae.8 Are fan coil units with connections to the outdoors recommended as an acceptable method for providing ventilation air? Why? No. 12.techstreet. no ventilation control dampers. reducing the likelihood of microbial growth in the system. and Air Conditioning—Solutions Manual 12. mixing of return air from other rooms. © (2009). It also simplifies the design of the system for the space with no outside air entering the space conditioning unit there is danger of freezing.156⏐Principles of Heating. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 12.6. dual duct. and horizontal models overhead and above ceilings can create problems related to condensate collection and disposal.org). Minimum outside air for meeting the ventilation requirements of the anticipated 550 occupants will be maintained throughout the year.54 7.3 – 16. P w.00092 37200 550×15=8250 * .0 – 20.44 – 100.042 . % OA 3 100 1.039 14.Chapter 12—System Configurations⏐157 12.0017 45600 10100 s 130 5.g.8 17. Btu/h Humidifier. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Enthalpy h.88 – 78. °F φ.005 = 0.7 – P w 45600 0. Pw 0.005 8400 1850 m 16 94 5.4 21.46 – 12.005 – 0.505 17.57 – ( 0. w.500 ≤ 6. to use the HVAC system shown in the sketch.30 34.0017 – 0.2256 h s = 0. the air is heated to 130°F at which temperature it is supplied to the conditioned space.4 Max Unit Wt.00092 ) + 8400 ( 0.3 36. s = 2.79 Q s = 645000 = 0.820 6.244M a ( 130 – 72 ) .005 45600 10100 Design cfm Face Area*. Q h = 1428000 Btu/h M h = ( 45600 ( 0. d.770 4. Tennessee.86 5. Btu/lb lb/lb Point Unit Coil Unit Size Copyrighted material licensed to University of Toronto by Thomson Scientific.2 12.699 ) + 8400 ( 22.31 – 5.75 73. At winter design conditions. Fan speed will be changed between summer and winter. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.039 37200 ( 3 ) + 8400 ( 72 ) t m = ---------------------------------------------------. Additional reproduction. W.3 10.240 ( 130 ) + 0.005 ) w m = ---------------------------------------------------------------------------.005 [ 1061 + 0.000 Btu/h latent (gain) Winter: 645. Inc.100 = 94% 0.000 Btu/h sensible (gain) 139.22 – 49.7°F.org).715 32.57 45600 Pw 37200 ( 0.119 .01 – 9.72 – 34.815 39.0 26.8 – 24. P w = 0.57 . (www. American Society of Heating. lb/h SCFM .375 50.ashrae. Btu/h and ft2 of face area Boiler.64 9.12 continued on next page. e.2256 .600 ≤ 4.. Inc. * Other outdoor design conditions could be used.. the cooling coil supplied air to the conditioned space at 58°F.444 ( 130 ) ] = 36.2 – 14.650 24.7 ) h m = ------------------------------------------------------------------. P w = 0. Select an appropriate air handler from the following data.= 15. The duct system will be designed so that at summer air flow rate the pressure drop does not exceed 3. b. During summer operation.= 0.180 2. s = 0.119 W s = 0. distribution. ft2 3 6 8 10 12 14 17 21 25 30 35 40 50 66 80 100 1. (www. φ = ---------------. The winter conditioning unit includes both a heating coil and a humidifier supplied with city water at 60°F.0017 ) ) ⁄ 8.8 – 5.com).000 Btu/h sensible (loss) negligible latent Unit Physical Data (Approximate) Size the following system components: 2 Cooling coil.techstreet.3 14. P w. For personal use only. M a = 45600 lb/h = 10100 cfm 37200 ( 1.= 5.63 56.005 ) ( 60 – 32 ) ] + Q h = 0 . gph a.8 15.43 48.10 30.00167 = 0.78 – 39.110 8.622 ----------------------.33 = 18 gph Problem 12.390 12. Btu/h Heating coil. φ = ------------. The humidistat in the return air steam maintains the design relative humidity of 30% in winter. The space design loads are Summer: 423.400 10.13 – 65.622 ----------------------.050 19.× 100 = 5.2 – 29.000 (all modules) * Actual face area varies with unit coil type.8 .75 in.699 r 72 30 22. . c. © (2009).190 14.7 – P w 2. Refrigerating and Air-Conditioning Engineers.34 – 3. Winter Dry Bulb.7 ma.150 7.. lb ≤ 3.12 A small single-zone classroom building is being designed for Knoxville. Btu/h and ft of face area Chiller unit.042 45600 45600 [ 36.32 4.660 2.49 – 7.930 3. (www. (www.0 .56 = 35.10 ( 75 – 58 ) 49000 CFM 2 = --------------------------------.6 Δt = 2.= 3580 1. At summer design outdoor conditions of 95°F db and 75°F wb.com).= 58.10 ( 3580 – x 1 ) ( 75 – 58 ) .4 – 25.158⏐Principles of Heating.0105 )26 ] + Q c = 0 . lb/h Point Dry Bulb.1 ) ( 62.2 . Duct pressure drop is 3. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.= 32 ft 600 Unit Size 40 a) b) 112000 Winter: CFM 1 = -----------------------------------. W.24 )6200 ( 60 ) ⁄ 13.0133 – 0. Hg ( 12 )778 ( 2545 ) 1.ashrae.3°F 6200 ( 0.000 Btu/h while SPACE 2 experiences a design sensible heat gain of 49. American Society of Heating. Summer Enthalpy h.0120 86700 W m = [ 37200 ( 0.10 ⋅ x 1 ( 130 – 75 ) = 112000 + 1.Q c = – 887000 Btu/h Select 600 fpm face velocity 2 19300 A = --------------.10 ( 130 – 75 ) 23500 CFM 2 = --------------------------------.5 ) + 49500 ( 32 ) ] ⁄ 86700 = 34.8 .33 Q Hc = 1. motor horsepower) b. Calculate the size of a.to 60.5 .= 3 hp 6200 SCFM 3. At winter design outdoor temperature of 0°F. Heating coil (Btu/h).0105 86700 19300 SCFM Q s = 423000 = 0. Refrigerating and Air-Conditioning Engineers.56 Btu/lb 0.000 Btu/h while interior SPACE 2 has a net sensible heat gain of 23.9 49 35.0105 + -------------------------------------.3°F h f = 34. Ventilating.4 . and Air Conditioning—Solutions Manual Problem 12.4 86700 [ 35.10 × 2581 ⋅ ( 130 – 58.0133 86700 19300 s 58 100 25. For personal use only.1 in.1 .4 ) ( 60 ) FAN: hp = v· ΔP = --------------------------------------------------. pressure.8 Btu/lb t m = [ 37200 ( 91 ) + 49500 ( 78 ) ] ⁄ 86700 = 93. Inc. . Additional reproduction.= 48300 Btu/h = 19 hp or 0.to 60.5x 2 = – 23500 + 48994 – 18.0151 37200 8250 r 78 58 32.= 1850 1.= 1260 1.10 ⋅ x 2 ( 130 – 75 ) = – 23500 + 1. Copyrighted material licensed to University of Toronto by Thomson Scientific.000 Btu/h.org).3 ) = 200000 Btu/h This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 12.13 A double duct system is to be used for air conditioning of a two-zone building.= 2620 1. exterior SPACE 1 has a design sensible heat loss of 112.0151 ) + 49500 ( 0.to M a = 86700 lb/h ( 139000 ⁄ 1100 ) W r = 0. Fan (scfm. water.244M a ( 78 – 58 ) .0120 ) ] ⁄ 86700 = 0.techstreet. % OA 91 48 38. Inc. all year long. °F φ.5x 1 = 112000 + 66950 – 18. © (2009). Btu/lb lb/lb ma.7x 1 x 1 = 2259 cfm 1.0120 49500 11050 m 83. Outside air requirement is 1400 cfm.10 ( 75 – 58 ) 6200 ( 3.12 continued. Interior design temperatures of both spaces is 75°F.2 + -------------------------------------------------------.10 ( 75 – 58 ) 67000 Summer: CFM 1 = --------------------------------.0133 lb/lb h m = [ 37200 ( 38.6°F W fan = 19300 ( 3.0361 ) ( 144 )60 ⁄ 778 = 29000 Btu/h 29000 Assume R f = 60% W fan = --------------.= 0.= 58.10 ( 2620 – x 2 ) ( 75 – 58 ) .75 ) ( 0.7x 2 x 2 = 322 cfm CFM Hc = 2259 + 322 = 2581 4800 ( 75 ) + 1400 ( 0 ) 3 ( 2545 ) t m = ------------------------------------------------. t f = 58.6 53 34.500 Btu/h.0133 86700 19300 f 85.1 in.8 + 0. SPACE 1 has a design sensible heat gain of 67. distribution.1 – ( 0. 0114 lb v ⁄ lb a Copyrighted material licensed to University of Toronto by Thomson Scientific.7 Btu/lb CONDITION OF AIR LEAVING COOLING COIL: 50°F. After mixing of the outside ventilation air with return air.ashrae.7 – 0. The thermostat controls the temperature leaving the reheater. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. RH = 100%.4 ) = 31. The return air conditions from the space must be exactly 50% relative humidity and 78°F.178 psia W cc = 0. select an appropriate chilled water cooling coil.91 Btu/lb [ Note: Psychrometric properties could also have been obtained from the Psychrometric chart. s = p v ⁄ 0. . © (2009).622p v ⁄ ( p – p v ) = 0. Additional reproduction.5 = p v ⁄ p v. select an appropriate electric resistance reheater coil. From a manufacturer's catalog.com). (www.244 Btu/lb °F for moist air 78 – 85000 ⁄ ( 0.5 ) = 29.6°F COIL SIZES (RATINGS) Q cc = – 16300 [ 31.org).7 – 0. s @ 78°F = 0.Chapter 12—System Configurations⏐159 12.3375 ) = 0.240 ( 50 ) + 0. Size the reheater (kW) and the cooling coil (Btu/h).244 ) ( 56.14 To maintain necessary close control of humidity and temperature required for a computer room. the reheat air-conditioning system shown in the sketch is used.0102 ( 1095. American Society of Heating.3375 ) ⁄ ( 14.3 ) = 20. distribution.000 Btu/h and a moisture load of 42 lb/h. p v = p v. (www.00762 ) ( 18 ) ] = – 185500 Btu/h Q rh = 16300 ( 0.25 Btu/lb SUPPLY AIR CONDITION: W s = W f = W cc = 0.240 + 0.6 – 52 ) = 18300 Btu/hr = 5.0102 lb v ⁄ lb a h r = 0.7 – 20.178 ) ⁄ ( 14. RETURN AIR CONDITION: p v.0102 ( 1096.00762 lb v ⁄ lb a h cc = 0.3375 W r = 0.25 – ( 0. the mixed air is at 80°F dry bulb with a relative humidity of 0. The air is then cooled to saturation at 50°F by the cooling coil.0102 – 0.622 ( 0.00762 lb v ⁄ lb a m a = m s ⁄ ( W r – W s ) = 42 ⁄ ( 0. Inc.475 p v = 0. There is a 2°F temperature rise across the fan.178 ) = 0. W m = 0.4 kW (can probably get by with 5 kW heater) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto MIXED AIR CONDITION: 80°F.00762 ) = 16300 lb/h t s = t r – Q ⁄ [ m a c p ].622 ( 0. For personal use only. where c p = 0. Space loads for the computer room include a heat load of 85. s = 0. From a manufacturer’s catalog.0114 – 0.24t + Wh g = 0.00762 ( 1083.240 ( 80 ) + 0. Air flow is controlled by a humidistat in the return air duct.244 ) ( 16300 ) = 56. Refrigerating and Air-Conditioning Engineers.0114 lb/lb. Inc.475 psia RH/100 = 0. ] h m = 0.techstreet. 6°F 1.techstreet. 3.000 Btu/h (a loss) Summer inside temperature = 78°F Summer design heat gains = 124.10 ( 80.10 ( 8696 × 0.com).000 Btu/h (sensible) and 52.15 A small commercial building located in St.= 4901 1. could reset t s to higher value] t m = [ 4000 ( 94 ) + ( 27431 – 4000 ) ( 78 ) ] ⁄ 27431 = 80.= 40.= 111. t r = 78°F ⇒ φ r = 48% [Plenty low.10 ( 78 – 55 ) Zone 2 (an interior space) Winter inside temperature = 78°F Winter design heat loss = 40.10 ( 8300 × 0.org). Relative humidity off the coil is approximately 90% in both cases.5 ) ( 92.0083 ) ] = 27431 [ 27.000 Btu/h (latent) Zone 4 Winter inside temperature = 72°F Winter design heat loss = −180.5 ) REHEATERS ( Q m ) 1 = 1. Louis.10 ( 78 – 55 ) 1. There are four zones (separately thermostated spaces) in the building. The VAV boxes are not to be cut back beyond 50% of rated flow. Supply air from the cooling coil is maintained at 55°F during the summer and 58°F during the winter. The design conditions and calculated design load for each zone are as follow: Zone 1 Winter inside temperature = 72°F Winter design heat loss = −55. © (2009).0097 ) ] ⁄ 27431 = 0.10 ( 8696 × 0.000 Btu/h (latent) 1.4°F 1.0104 Q cc = 27431 [ 1.000 Btu/h (a loss) Summer inside temperature = 78°F Summer design heat gains = 210. 90%) Σm w ( 31000 + 71000 + 42000 + 51500 ) ⁄ 1100 W r = W s + ----------. Additional reproduction.4 – 58 ) = 93000 Btu/h ( Q m ) 2 = 1.83 + 10.6 – 58 ) = 55700 Btu/h ( Qm )3 = 0 none needed ( Q m ) 4 = 1. (www.5 ) – 40000 ( t m ) 2 = 78 + ----------------------------------------.= 8696 1.5 ) ( 111.10 ( 4901 × 0. (www.10 ( 4901 × 0.= 0.= 0.160⏐Principles of Heating.3 W m = [ 4000 ( 0. Ventilating. and of the fan (scfm) is to take place. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.5 ) 180000 ( t m ) 4 = 72 + ----------------------------------------. preliminary sizing of the central cooling unit.0083 + ---------------------------------------------------------------------------------------------------.10 × CFM s × Δt ] 140000 115000 CFM 3 = --------------------------------. as shown in the following sketch.= 8300 1.10 ( 8300 × 0.10 ( 78 – 55 ) 220000 CFM 2 = --------------------------------. . Minimum outside air of 4000 scfm is maintained at all times (just don’t ask how).= 5288] 1.000 Btu/h (sensible) and 71.4°F 1.0097 ma 123750 Copyrighted material licensed to University of Toronto by Thomson Scientific.5 ) – 115000 ( t m ) 3 = 78 + ----------------------------------------.4 – 58 ) = 244000 Btu/h This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 210000 CFM 4 = --------------------------------.10 ( 78 – 58 ) 2. American Society of Heating.= 5534 [check CFM 3 = --------------------------------.16 ] = 1042000 Btu/h Cooling coil → 1042 Mbh * Note:Other outdoor design conditions could be selected. Determine required airflow rates 124000 CFM 1 = --------------------------------.2°F < 58 ∴ no reheating 1. Inc. For personal use only.0104 – 0.ashrae.0144 ) + ( 27431 – 4000 ) ( 0.= 69.000 Btu/h (latent) Zone 3 (an interior space) Winter inside temperature = 78°F Winter design heat loss = 115.3 – 55 ) + 4840 ( 0.000 Btu/h (a gain) Summer inside temperature = 78°F Summer design heat gains = 220.000 Btu/h (sensible) and 42. Inc.000 Btu/h (a gain) Summer inside temperature = 78°F Summer design heat gains = 140. distribution. Missouri is to be conditioned using a variable air volume (VAV) system with reheat. At this stage of the process. Winter reheater discharge temperature 55000 ( t m ) 1 = 72 + ----------------------------------------. and Air Conditioning—Solutions Manual 12.5 ) ( 69. Refrigerating and Air-Conditioning Engineers. of the reheaters.500 Btu/h (latent) [ Q s = 1.000 Btu/h (sensible) and 31.= 92.10 ( 5534 × 0.10 ( 78 – 55 ) FAN → 27431 SCFM ⇒ 123750 lb/h Summer (55°F. 10 ( 82. negligible latent Conduct the preliminary sizing of the fan (scfm and horsepower).= 5000 1.10 ( 78 – 58 ) Copyrighted material licensed to University of Toronto by Thomson Scientific.000 Btu/h sensible (loss).Chapter 12—System Configurations⏐161 12.6 – 58 ) + 4840 ( 0.0115 ) t m = ------------------------------------------------------. cooling coil (scfm and Btu/h). Missouri where summer outdoor design conditions are 94°F db and 75°F wb and winter outdoor design conditions are 3°F and 100% RH.1 ( 2545 ) W f = 0.= 13. distribution. A blow-thru multizone unit will be used with cold deck temperature maintained at 58°F all year long and with hot deck temperature varying from a maximum of 130°F at winter design to 85°F during the summer.0361 )144 ( 60 ) W = ----------------------------------------------------------------.000 Btu/h sensible (loss).6°F 122000 ( 0.000 Btu/h latent (gains) Zone 2: 290.= 0.to t s = 58. (www. all spaces will experience their peak loads at the same time.16 continued on next page. .000 Btu/h latent (gains) Winter Zone 1: −215.g.1 Hp 778 ( 0.techstreet. Additional reproduction. w.10 ( 78 – 58 ) 190000 = --------------------------------. the control humidistat in the common return air duct is set at 30% RH. 59.0 in.000 Btu/h sensible.000 Btu/h sensible.33 27090 ( 2 ) ( 0.to W o = 0.000 sq ft of floor space. For personal use only. In winter. The space design loads at indoor design temperatures of 78°F summer and 72°F winter are Summer Zone 1: 116. negligible latent Zone 2: 110. 39.ashrae. φ s = 100% (max) ⇒ W s = 0.= 5270 1. Design occupancy is to be 10 people per 1000 ft2 of floor area.000 Btu/h Latent (gains) Zone 3: 190.5 + -----------------------------------.0115 and 78° ⇒ φ r ≅ 55% ok 122000 6000 ( 94 ) + 21090 ( 78 ) 6000 ( 0.= 2620 1. heating coil (scfm and Btu/h).10 ( 72 – 58 ) 171000 CFM 3w = -----------------------------------.0104 + -----------------------------------------------------------------------------. CFM 1s CFM 2s CFM 3s 215000 CFM 1w = -----------------------------------. Louis. © (2009). negligible latent Zone 3: −171. and humidifier (gal/h). Provide a completely labeled sketch of the system.10 ( 130 – 72 ) CFM = 5270 CFM = 13180 CFM = 8640 Fan CFM = 27090 Summer 27090 M a = --------------.10 ( 78 – 58 ) 290000 = --------------------------------.com).10 ( 130 – 72 ) 110000 CFM 2w = --------------------------------.= 81. Inc.16 A commercial three zone office building is being designed for St.= 0. The amount of outside air is to equal the recommended 20 cfm per person in accordance with ASHRAE Standard 62-1989.0121 t f = 81.0144 ) + 21090 ( 0. (www. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto USE 116000 = --------------------------------.0121 – 0.000 Btu/h Sensible.0121 27090 27090 13.244 ) CFM cc ( max ) = 27090 ( since all zones peak at same time ) Q cc = 27090 [ 1.0144 .60 = 122000 lb/h 13.000 Btu/h sensible (gain).65 )2545 t o = 94 .to t r = 78 . Each zone is to contain 10.= 13180 1.org). Inc. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 43. The duct system will be designed so that the pressure drop does not exceed 2.= 8640 1. Refrigerating and Air-Conditioning Engineers. Fan efficiency is estimated at 65%.0104 ) ] = 956000 Btu/h Problem 12. American Society of Heating.= 82.= 3370 1.0104 ( 43000 + 59000 + 39000 ) ⁄ 1100 W r = 0. Due to the building orientation and internal zoning.5°F W m = ---------------------------------------------------------------------------. h = 39.3 .10CFMh3 ( 130 – 72 ) = 171000 + 1.4 tons b) q rh = m da c p ( t rh – t s ) = 34479 ( 0.3. For personal use only. 50% RH. © (2009).6.00092 ( 3°.org).5 – 52 ) = 52160 Btu/h = 15.005 – 0.3 ( 0.8°F 27090 Humidification: W o = 0.162⏐Principles of Heating. .00092 ) = 100 lb/h ÷ 8 --. (www. v = 13. CFM h2 = 1174 zone 3 1.3 kW Cost = 15.5°F dry bulb with electric heaters.10CFMh1 ( 130 – 72 ) = 215000 + 1. h = 20. W m = 0.0. 12.7 . m· = 34479 lb/h m: 8392 ( 0.33 3 Q h = 110 ( 1076 ) + 1.16 continued. W r = 0. Ventilating. (www.07 . 76°Fwb.17 An air-conditioning unit takes in 2000 cfm of outside air at 95°F dry bulb and 76°F wet bulb. and 6000 cfm of return air at 78°F dry bulb and 50% RH. h = 30.0102. Winter zone 1 1. and Air Conditioning—Solutions Manual Problem 12. 6000 cfm. American Society of Heating. What would be the operating cost of these heaters at 2.00745. 90% RH.10 ( 13180 – CFM h2 ) ( 72 – 58 ) – 110000 .10 ( 5270 – CFM h1 ) ( 72 – 58 ) . Inc.com). CFMh3 = 3839 CFM h = 8752 6000 ( 3 ) + 21090 ( 72 ) t m = ---------------------------------------------------.4.( 0. 100% ) .techstreet.3 – 20.244 ) ( 58. v = 14.8 .1 = 57.to 30% ) 6000 ( 60 ) 1 M c = ----------------------.= 13gph 13. t m = 82°F a) m da [ h m – h s – ( W m – W s )h f ] = q c c 34479 [ 32.5 cents per kWh? h m = 32.10 ( 8752 ) ( 130 – 57. m· = 8392 lb/h r: 78°Fdb.6 – ( 0. 2000 cfm. W = 0. W = 0. Refrigerating and Air-Conditioning Engineers.ashrae. CFMh1 = 3739 zone 2 1. v = 13.= 56.005 ( 72 . Additional reproduction. t f = 56.10CFMh2 ( 130 – 72 ) = 1. a.025 ) = 38 cents/hour This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto OA: 95°Fdb.8 ) = 118400 + 695100 = 813500 Btu/h c * Note: Other outdoor design conditions could be selected.0114 – 0. Assume the conditioned air were reheated to 58.00745 ) ( 20 ) ] = q c q c = 400680 Btu/h = 33. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.10 ( 8640 – CFM h3 ) ( 72 – 58 ) . m· = 26087 lb/h s: 52°Fdb.0102 ) = 34479 W m .7 + 1.015. The conditioned air leaves the chilled water coil at 52°F dry bulb and 90% RH. distribution.0114 lb/h Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc. What is the refrigeration load on the chiller in tons? b.015 ) + 26087 ( 0. W = 0. to Q c = 228000 Btu/h d. Inc.9 ( 31.10 ) ( 122 – 58 ) .0119 . s = 31.= 2500 SCFM 1 = --------------------------------.97 8696 ( 0.06 rh: 8696 ( 30 ) + 25783 ( 20. Sensible design heating loads Space 1: 162. Return air relative humidity at summer design conditions.10 ( 5432 ) 24444 [ 33.0124 – 0. (www.10 ( 2932 ) 1. © (2009).001.000 Btu/h 64500 55000 SCFM 2 = --------------------------------.0093 ) + 38 + 26 = 24444 W r .00745 for 2.000 Btu/h Copyrighted material licensed to University of Toronto by Thomson Scientific.0081 t rh = 58°F W = 0. W rh = 0. 24444 ( 0. scfm b.= 124°F 1. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. indoor 72°F.19 For the building and reheat system shown below. % c. W r = 0. 90% RH.8 m: 8392 ( 0.000 Btu/h Space 2: 143.0119 ) + 0.= 122F.com).4 ) = 32.0093 )26 ] + Q c = 0 . h f = 33. s ≅ 58%. 162000 143000 t s1 = 72 + --------------------------.Chapter 12—System Configurations⏐163 12.ashrae. φ r. Fan rating. American Society of Heating.4 1. ∴ can absorb less latent load in conditioned space. Size reheat coils. Comment on the ability of the leaving air to absorb latent load in the conditioned space. t s2 = 72 + --------------------------.015 ) + 17391 ( 0.8 av c. 78°F wet bulb. Sensible design cooling loads Space 1: 64. (--------------------------4000 ) ( 60 )= 17391 lb/h 13.= 2932 1.0118 8392 ( 39. Btu/h d.18 In Problem 12.10 ( 2500 ) Q R = 2500 ( 1.0102 ) + 25783 ( 0. assume 2000 cfm of return air bypasses the chilled water coil and is used for reheat. W = 0. For personal use only. h m = 33.techstreet. How does the final condition of the air compare with the reheated air in part (b) of 12.0169 ) = 0. Btu/h and scfm for each Latent design loads (moisture produced) Space 1: 38 lb/h Space 2: 26 lb/h Year-round: 10% by mass outside air required for ventilation. distribution.6 ) = 34479 h rh .1 ( 0. Summer: Outdoor 95°F dry bulb.9 ( 0. Fan SCFM = 2932 + 2500 = 5432 cfm ( = 24444 lb/h ) b.4 – 24 – ( 0. Conditions of cooling coil: 58°F. Refrigerating and Air-Conditioning Engineers. Inc.10 ) ( 124 – 58 ) Q R = 2932 ( 1.11(b). . av h m = 0. no humidity control.0081 compared to 0.0124 t m = 80°F 5 ( 2545 ) t f = 80 + --------------------------. determine a. Size cooling coil.4 ) + 17391 ( 30 ) = 25783 h m . a. indoor 78°F. 12.0102 ) = 25783 W m . Additional reproduction.8 (--------------------------2000 ) ( 60 -) = 8696 lb/h 13. 1 2 = 206400 Btu/h = 181500 Btu/h This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Winter: Outside 6°F. (www.10 ( 78 – 58 ) a.1°F.17.10 ( 78 – 58 ) 1.h r.1? b.= 82.00745 ) = 34479 W rh .8 ) + 0. W m = 0.org).1 ( 41.8 w m = 0.500 Btu/h Space 2: 55. h rh = 22. distribution.ashrae. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . Inc. t s = 59. and Air Conditioning—Solutions Manual 12.0°F 1 2 3 ∴ Coil discharge temperature = 59.org). Ventilating. (www.1°F #2 149600 = 8500 × 1. (www.10 × ( 78 – t s ) . Inc.20 A basic reheat system has been retrofitted with an improved control system.com).techstreet. Refrigerating and Air-Conditioning Engineers. For personal use only. For the operating conditions shown in the sketch below and with all thermostats set at 78°F. American Society of Heating.0°F #3 30100 = 2100 × 1. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.10 × ( 78 – t s ) . Additional reproduction.10 × ( 78 – t s ) . t s = 65. for what cooling coil discharge temperature T should the logic system of the controller be calling if there is no humidity override? #1 81000 = 3900 × 1.164⏐Principles of Heating.1°F Copyrighted material licensed to University of Toronto by Thomson Scientific. t s = 62. © (2009). American Society of Heating. For personal use only.ashrae. distribution. Refrigerating and Air-Conditioning Engineers. (www. Inc.techstreet.Solutions to Chapter 13 HYDRONIC HEATING AND COOLING SYSTEM DESIGN Copyrighted material licensed to University of Toronto by Thomson Scientific. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .org). © (2009). Inc. (www.com). Additional reproduction. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. (www. Inc. Additional reproduction. American Society of Heating.com). © (2009). For personal use only. distribution. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.Copyrighted material licensed to University of Toronto by Thomson Scientific. Refrigerating and Air-Conditioning Engineers. (www.ashrae.techstreet.org). Inc. 3 Calculate the size of the expansion tank for a hot water heating system of 1. (www. What is the water circulation rate (GPM) required if the temperature range of the water is 12°F.7 psia [ ( v 2 ⁄ v 1 ) – 1 ] – 3αΔt V t = V s ---------------------------------------------------Pa ⁄ P 1 – Pa ⁄ P2 (from Equation 13.7 = 49. Additional reproduction.79 hp 3960 η r 3960 ( 0.org)./in.7 ⁄ 44.746 -------.000 Btu/h heating capacity if the tank is a closed tank with an air/water interface and the following system parameters are known: Supply water temperature Ambient temperature Fill pressure (at tank) Max.746 ) ( 1 ⁄ 0.79 × ⎛⎝ ------------⎞⎠ ( 0.9 ) = 9173 kWh 3 Copyrighted material licensed to University of Toronto by Thomson Scientific. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.7 ) – ( 14.200.= -----------------------. b.1 What is the maximum temperature at which a heating water system can be operated if the boiler (hot water generator) is rated as low pressure by the ASME Boiler and Pressure Vessel Code? From 13.Chapter 13—Hydronic Heating and Cooling System Design⏐167 13.12) –6 [ 0. . For personal use only.01604 – 1 ] – 3 × 6. Refrigerating and Air-Conditioning Engineers.7 ⁄ 49.7 = 44. distribution. what is the pump horsepower? Motor size? c.7 ) V t = 6930 gallons 210°F 60°F 30 psig 35 psig 6.8 ) Motor Hp = 5 Hp c) kW 1 Annual Energy (kWh) = Hp × hours × 0.01670 ⁄ 0.ashrae.2. If the head loss in the system is 60 feet.× ------Hp η m 13. maximum temperature is 250°F.5 × 10 in.01670 ft ⁄ lb 3 v 1 = v f @ 60°F = 0. (www.2 Sketch the fundamental components for a chilledwater system with a single load and source. American Society of Heating.7 psia P 1 = 30 + 14. and the pump is 80% efficient. If the motor is 90% efficient and it operates for onethird of the total hours in the year.°F Δt = 210 – 60 = 150°F Pa = 14.5 × 10 × ( 150 ) V t = 6000 --------------------------------------------------------------------------------------------------------------------( 14.techstreet. Inc. what is the annual energy consumption of the pump? a) b) q = GPM ( 500 ) ( Δt ) GPM = q ⁄ 500 ( Δt ) GPM = 100 ( 12000 ) ⁄ ( 500 × 12 ) = 200 gpm GPM ( ΔH ) 200 ( 60 ) Hp = -------------------------.= 3. and a capacity of 100 tons of cooling. a.000 gallons This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 8760 kWh = 3.1. operating pressure (at tank) System water volume Steel piping system material 3 v 2 = v f @ 210°F = 0. 13.com).01604 ft ⁄ lb –6 α = 6. © (2009). Inc.7 psia P 2 = 35 + 14. 77 150 4. distribution.01 300 17.techstreet.168⏐Principles of Heating.19 650 82.01 200 7. a.7 V t = 2280 gallons 13.80 600 70.= ---------.21 50 . the pump head required at 640 gpm is 80 ft.50 100 1.16 Q Copyrighted material licensed to University of Toronto by Thomson Scientific. . Q = C s ΔH Q 640 gpm C s = -----------.= 71. Refrigerating and Air-Conditioning Engineers. Cs? b.95 500 48.01670 ⁄ 0. Ventilating.4 What tank size diaphragm tank would be required for the above system? From Equation 13.7 ⁄ 49.org).ashrae.84 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto From Equation 13.39 550 59. Inc. © (2009).5 × 10 × 150 V t = --------------------------------------------------------------------------------------------------------------------1 – 44. and Air Conditioning—Solutions Manual 13. (www.6 ) 2 ΔH Q ΔH 0 0 400 31. American Society of Heating. Plot the system curve from 0 to 800 gpm. For personal use only.6 ---------ΔH 80 ft ΔH = ( Q ⁄ 71.5 In a given chilled water system.49 450 39. What is the system constant. Inc. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.90 750 108.14 [ ( v 2 ⁄ v 1 ) – 1 ] – 3αΔt V t = V s ---------------------------------------------------1 – P1 ⁄ P2 –6 [ ( 0. (www.01604 ) – 1 ] – 3 × 6. Additional reproduction.com).72 800 124.56 700 95.22 250 12.58 350 23. © (2009). The dynamic head losses in the system are Piping and fittings 30 ft Chiller 20 ft Control valve 10 ft Cooling coil 10 ft When the system is filled (at 95°F ambient temperature) it is desired to have a pressure of 10 psig at the high- est point in the system which will reduce to 5 psig when the water temperature reduces to 45°F.= 54. explain the difference between a three-way control valve and a two-way control valve as they affect the hydraulics of the system. Refrigerating and Air-Conditioning Engineers. the pump is located in a basement equipment room with the expansion tank connected to the pump suction. (www. which is 115 feet above the pump. Additional reproduction.3 2 85 psig 13. Inc.techstreet. (www. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 62. distribution. the three-way valve provides a constant flow variable Δt in the system as it modulates and the two-way valve provides a variable flow as it modulates. a. .85 psig ≈ 55 psig 144 b) H = ΣH = 30 + 20 + 10 + 10 = 70 ft c) Assume pressure at pump inlet equals the tank pressure. what is the pressure at the pump suction? The pump discharge? P i = 10 psig @ 95°F P f = 5 psig @ 45°F P 2 = P t @ 95°F w = 62.ashrae. American Society of Heating.7 In your own words. p2) should the expansion tank be designed for? b. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.5 psig ≈ 60 psig Copyrighted material licensed to University of Toronto by Thomson Scientific. For personal use only. What pump head is required? c. The pump is the lowest point in the system and the highest point is a pipe in the penthouse.6 In a chilled-water system.Chapter 13—Hydronic Heating and Cooling System Design⏐169 13. What operating pressures (p1. With the pump off and a cold (45°F) system.org).05 lb/ft 3 P 1 = P t @ 45°F w = 62. Inc.com).4 ⎛⎝ ---------⎞⎠ 2 1 1 144 P x = 55 + 30. from the system perspective. With the pump on and a cold (45°F) system.05 ( 115 ) P 2 = 10 + w ( 115 ) = 10 + --------------------------144 P 2 = 59. what is the pressure at the pump suction? The pump discharge? d. (Pump off) P x = P x = P 1 = 55 psig 2 d) 1 P x = P 1 = 55 psig 1 70 P x = P x + wH = P x + 62.42 lb/ft 3 62. However. Both vary the flow (gpm) through the controlled load as they modulate.42 ( 115 ) P 1 = 5 + --------------------------. 28 ΔP @ 50°Δt = 1380 psi P 2 = P 1 + ΔP P 2 = 50 + 1380 P 2 = 1430 psig Copyrighted material licensed to University of Toronto by Thomson Scientific. If the chase is at a temperature of 95°F and the pipe reaches thermal equilibrium with the chase. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. What is the required Cv of the valve? GPM = C v ΔP C v = GPM ⁄ ΔP q = GPM ( 500 ) ( Δt ) 8 360000 GPM = ------------------. American Society of Heating. © (2009). Inc. steel pipe from Figure 13. steel pipe in a 45°F chilled-water system at 50 psig is in a pipe chase and is isolated between two service valves. Additional reproduction. The water temperature entering the coil is at 44°F with a 12°F Δt.9 A section of 1 in. what will the final pressure in the pipe be? ( β – 3α )Δt ΔP = -----------------------------------------------5 ⁄ 4 ( D ⁄ ( EΔr ) ) + α solving for 1 in. Inc.com). This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .170⏐Principles of Heating.= 60 gpm 500 ( Δt ) 500 ( 12 ) C v = 60 ⁄ Δ5 = 26. and Air Conditioning—Solutions Manual 13.org). For personal use only. (www.= -------------------. Ventilating. distribution.8 A control valve is to be sized for a cooling coil with a capacity of 30 tons of cooling.techstreet. It is determined that the valve should have a pressure drop of 5 psi.83 13.ashrae. Refrigerating and Air-Conditioning Engineers. (www. © (2009). Inc. (www. Additional reproduction.ashrae. American Society of Heating. Refrigerating and Air-Conditioning Engineers.techstreet.com). Inc.Solutions to Chapter 14 UNITARY AND ROOM AIR CONDITIONERS Copyrighted material licensed to University of Toronto by Thomson Scientific. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.org). This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . (www. distribution. For personal use only. Additional reproduction.ashrae. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . (www.org).techstreet. distribution. For personal use only.com). Refrigerating and Air-Conditioning Engineers. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc. American Society of Heating. (www. © (2009). Inc. ashrae.) In passing through the condenser coil. distribution.241 ) ( 115 – 95 ) m· = 319.95ft 2 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 14.239 m· = ------------------------------------------------------( 60 ) ( 0.2 If the ductwork supplying the air to and from the condenser section in Problem 14.12 ft 3 /min a. .com).000 + ( 3 ) ( 3413 ) q rej = 46.85 ft3/lb. The unit has a total cooling capacity of three tons of refrigeration.18 ft 3 /min A o = CFM o /v = 4758.77 ) ( 14. From the outdoors to the condenser? b. American Society of Heating.Chapter 14—Unitary and Room Air Conditioners⏐173 14. From the condenser back to the outdoors? Copyrighted material licensed to University of Toronto by Thomson Scientific.37 ) CFM = 4595. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. (www. © (2009). Inc. and the power requirement to the compressor is 1 kW per ton of cooling.77 ) ( 14. How many cfm of air must be brought into the condenser from an ambient outdoor temperature of 95°F db and 78°F wb if the condensing temperature is to be 115°F with a 10°F approach to the leaving air temperature? Total heat rejected = q rej q rej = q ref + q motor = 36. CFM o = m· 2 = ( 319.88 ) CFM o = 4758.74 ft 2 b.1 An air-cooled packaged air conditioning unit with a hot water heating coil is to be used to condition a small office suite in a high-rise office building. (www.) A i = CFM i /v = 4595. Refrigerating and Air-Conditioning Engineers.49 gr/lb) heated to 115°F db has a final specific volume (v) of 14. Inc.12/800 A i = 5.77 lb/min CFM = m × v v = 14.org).241 ) ( 115 – 95 ) 46.18/800 A o = 5. Additional reproduction.239Btu/h q rej = m· ( 0. what would be the cross-sectional area of the ductwork? a.techstreet. For personal use only.37 ft 3 /lb ( at 95°F db and 78°F wb ) CFM = ( 319.1 were sized for a velocity of 800 ft/min. the air would be heated a a constant humidity ratio. Air at 95°F db and 78°F wb (w = 117. 1 and 14. . Ventilating. Inc. (www.3? Copyrighted material licensed to University of Toronto by Thomson Scientific.69 (water-cooled) b.ashrae. For personal use only.) To COP a = --------------T – To T o = 40 + 460 = 500°R T = 105 + 460 = 565°R 500 COP a = -----------------------565 – 500 COP a = 7.techstreet.174⏐Principles of Heating.3 If the packaged air-conditioning unit of Problem 14. and 1) the water was supplied at 85°F. what would be a. distribution. (www. American Society of Heating.1 were provided with a water-cooled condenser instead of an air-cooled unit.com).69 = 0. The Carnot COP between the 40°F suction temperature and the 105°F condensing temperature of the Problem 14. and 3) the condensing temperature was 105°F. Inc.67 = ---------7. © (2009). and Air Conditioning—Solutions Manual 14. The Carnot COP between 40°F suction temperature and the 90°F condensing temperature? b. 2) the leaving water temperature was 95°F. what would be the kW per ton for the water-cooled unit of Problem 14.org).867kW/Jon This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 14.) T o = 40 + 460 = 500°R T = 105 + 460 = 575°R 500 COP a = -----------------------575 – 500 COP a = 6.3 were proportioned in the same relationship as the Carnot COPs of Problem 14. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.1 air-cooled unit? a. Refrigerating and Air-Conditioning Engineers.67 (air-cooled) Short Solution: ( 1 )COP a ( air-cooled ) kW/J on = ----------------------------------------------------COP a ( water-cooled ) kW/J on 6.4 Assuming that the actual power requirement for the cooling cycles of Problems 14.3. Additional reproduction. Inc. For personal use only.Chapter 14—Unitary and Room Air Conditioners⏐175 14.000 + 3 ( 0. distribution. Refrigerating and Air-Conditioning Engineers. American Society of Heating.techstreet.48 ) GPm = ---------------------------------------------( 1 ) ( 10 ) ( 60 ) ( 62.867 ) ( 3413 ) = 44. Inc.4 ) GPm = 8. (www.97 gal/min Copyrighted material licensed to University of Toronto by Thomson Scientific.org). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Additional reproduction.881 q r = m w ( 1 ) ( 95 – 85 ) ( 60 ) Btu/h ( 44.3? q r = Total heat rejection q r = Refrigeration capacity and compressor heat = 36. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . © (2009).881 ) ( 7. (www.5 How many gallons per minute of water would be required for the water-cooled unit of Problem 14.ashrae.com). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Inc.ashrae.org). American Society of Heating.Copyrighted material licensed to University of Toronto by Thomson Scientific. © (2009).com). Additional reproduction. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . For personal use only. (www. distribution. Inc.techstreet. Refrigerating and Air-Conditioning Engineers. (www. Refrigerating and Air-Conditioning Engineers. Additional reproduction. distribution. For personal use only. (www. (www. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. © (2009).techstreet.Solutions to Chapter 15 PANEL HEATING AND COOLING SYSTEMS Copyrighted material licensed to University of Toronto by Thomson Scientific. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .ashrae. Inc. American Society of Heating.com). Inc.org). For personal use only. Inc. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. © (2009).com). distribution.ashrae. Additional reproduction. Refrigerating and Air-Conditioning Engineers. (www. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .org).techstreet.Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc. (www. American Society of Heating. radiant output for a 100 ft2 heating panel with a panel surface temperature of 120°F b. 560 ft2 of ceiling with a surface temperature of 70°F. © (2009). Inc. (www.org). chapter 4. PHVAC @ 20 fpm.radiant 55 Btu/h ⋅ ft 2 Btu 55 ------------2. estimate the following: a. Additional reproduction.= 65. Inc. natural convection output for the ceiling panel when the air temperature is 70°F a) From Figure 1. The air temperature in the room is 75°F. chapter 6.6 = -----------------------------------------------------------1500 t = 131°F This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto From Figure 3. Estimate the average unheated surface temperature or the area-weighted mean radiant temperature.1 A room has a net outside wall area of 300 ft2 that has a surface temperature of 55°F. American Society of Heating. 50 ft2 of glass with a surface temperature of 30°F. 75°F air temp. The average unheated surface temperature in the room is 67°F. The room is occupied by adults in light clothing at a sedentary activity.3 A room has 1500 ft2 of surface area and 320 ft2 is to be heated. MRT = 80.techstreet. and 560 ft2 with a surface temperature of 70°F.ashrae. 2008 Systems and Equipment Handbook 2 Btu 17.. 2008 Systems and Equipment Handbook b) . From Figure 4-3. . or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.2 For the room in Problem 15.× 100 ft = 1750 Btu/h h ⋅ ft Copyrighted material licensed to University of Toronto by Thomson Scientific. 300 ( 55 ) + 50 ( 30 ) + 560 ( 70 ) + 560 ( 70 ) MRT = AUST = ------------------------------------------------------------------------------------------------300 + 50 + 560 + 560 96400 MRT = --------------.Chapter 15—Panel Heating and Cooling Systems⏐179 15.6°F 1470 15. Refrigerating and Air-Conditioning Engineers. (www. chapter 6.5 ------------2.× 100 ft = 5500 Btu/h h ⋅ ft t p – t a = 120 – 70 = 50°F 2 15.6°F ( 1500 – 320 )67 + 320 ( t ) 80. Determine the surface temperature of the heated panel necessary to produce comfort if the air velocity is 20 fpm.1.com). For personal use only. distribution. and Air Conditioning—Solutions Manual 15. © (2009).org). Additional reproduction.3. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . 2008 Systems and Equipment Handbook Copyrighted material licensed to University of Toronto by Thomson Scientific. chapter 6. (www. For personal use only.180⏐Principles of Heating.4 For Problem 15. Ventilating. Inc. 2008 Systems and Equipment Handbook Q R = 20160 Btu/h Q C = 20 Btu/h ⋅ ft 2 2 Q C = 20 ( 320 ft ) = 6400 Btu/h Q T = Q C + Q R = 20160 + 6400 Q T = 26560 Btu/h Figure 3. t p = 131°F. (www.techstreet. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. chapter 6. American Society of Heating.ashrae. AUST = 67°F t p – t a = 131 – 75 = 56°F Q R = 63 Btu/h ⋅ ft 2 2 Q R = 63 ( 320 ft ) Figure 1. Refrigerating and Air-Conditioning Engineers. determine the total heat transferred by the ceiling heating panel. distribution.com). Inc. org). American Society of Heating.ashrae. (www. © (2009). For personal use only.techstreet. COGENERATION. Refrigerating and Air-Conditioning Engineers. AND HEAT RECOVERY SYSTEMS Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc. Inc. distribution.Solutions to Chapter 16 HEAT PUMP. (www. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . Additional reproduction. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.com). American Society of Heating.org). Refrigerating and Air-Conditioning Engineers.Copyrighted material licensed to University of Toronto by Thomson Scientific.com). Inc. (www.techstreet. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Additional reproduction. (www. distribution.ashrae. © (2009). Inc. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . For personal use only. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto From the table in Problem 8. A reciprocating natural gas engine cogeneration plant is to serve the building. Refrigerating and Air-Conditioning Engineers. with salvaged heat being used for heating and for driving a single-effect absorption chiller. Select a heat pump from the table in Problem 8.13.com). At 2°F outdoor. the absorber requires 20. . © (2009). this heat pump has an output of 15100 Btu/h. Qh 100000 Btu/h W = -----------. Any shortfall in cooling by the absorber with recovered heat must be made up by the boiler as input to the absorber. Qh 100000 Btu/h W Carnot = -------------------------------.0 kW h b. Supplemental heat = 52000 Btu/h – 15100 Btu/h = 36900 Btu/h = 10. (a) Determine the minimum electric power (Carnot COP) required to operate the heat pump.org).10/kWh for purchased electricity cost. and Heat Recovery Systems⏐183 16. Cogeneration. Additional reproduction. Compare design operating costs with hourly design operating costs using conventional equipment (purchased electricity for the building and for cooling with an electric chiller at 1.13. The engine-generator is sized for the electrical load.= --------------------------------COP 3 = 33333 Btu/h = 9. As a heat pump. the heat loss from the house is 100. it is to maintain 70°F in winter with an outside air temperature of 2°F and a heating load of 52.64 Btu = 10373 -------.000 Btu/h. boiler efficiency of 80% for fuel cost. purchased gas for a boiler for heating).00 per therm.8 kW Copyrighted material licensed to University of Toronto by Thomson Scientific. What size resistance heater is required at the winter design condition? 16. Any shortfall in heating from recovered heat must be made up by a boiler. Use $1. American Society of Heating. The design cooling load is 250 tons.8 kW 16.3 A 100.0 kW/ton.000 Btu/h. (b) Determine the actual electric power to operate the heat pump with a heating COP of 3. The heating load is 52000 Btu/h.000 ft2 building design has a design electrical load of 5 W/ft2.2 An air-source heat pump is to be used for both air conditioning and heating of a residence. Inc.1 A heat pump is used in place of a furnace for heating a house. when the outside air temperature is 15°F. Solution on following page. Inc.= ------------------------------------. In winter. a cooling load of 36000 Btu/h at 95°F outdoor requires an A036 heat pump.ashrae. sized for cooling.= 9. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.= 3. distribution.000.000 Btu/ton⋅h input. maintaining the interior at 80°F in summer with an outside air temperature of 95°F and a cooling load of 36.Chapter 16—Heat Pump. The design heating load is 3. (www. $0. Carnot 9. (www.64 TH – TL 530°R – 475°R a.000 Btu/h.techstreet. For personal use only. TH 530°R COP h. Calculate hourly design operating costs for heating and cooling.= --------------------------------COP h.000 Btu/h if the inside is maintained at 70°F. Carnot = ------------------. = 5000000 -------ton-h h Btu Btu Supplemental fuel input = ⎛⎝ 5000000 -------.org).25/h h h $62.ashrae. Additional reproduction. Refrigerating and Air-Conditioning Engineers.00/therm ) = $30.10/kWh ) = $50.input h Salvaged heat = 70% (30% + 30%) = 42% of input Btu = 2172000 -------h Design heating = 3000000 Btu/h Btu Btu Supplemental fuel input = ⎛ 3000000 -------. . For personal use only.35 ---------------.06/h Conventional Plant Building electricity Chiller electricity ⎛ 500 kWh -----------⎞ ( $0.10/kWh ) = $50.184⏐Principles of Heating.= $10.71/h h h Btu therms 1035000 -------.⎟ ⎝ therm ⎠ $80.00/h ⎝ h ⎠ Electricity Heating ⎛ 3000000 Btu --------⎞ ⎜ h ⎟ ⎜ ------------------------------⎟ ( $1.71 ---------------.0 -------------⎞⎠ ( $0.techstreet.00/h This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Btu therms 5171000 -------.06/h Copyrighted material licensed to University of Toronto by Thomson Scientific.00/h ⎝ h ⎠ kWh ( 250T ) ⎛⎝ 1. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.71/h Btu ⎛ 3535000 --------⎞ ⎜ h ⎟ ⎜ ------------------------------⎟ ( $1.10/kWh ) = $25. distribution. © (2009).00/h ton-h $75.– 2172000 --------⎞ ⁄ 80% ⎝ h h ⎠ Btu = 1035000 -------h Design cooling = 250T Btu Btu Absorption chiller input = ( 250T )20000 ------------. and Air Conditioning—Solutions Manual W 2 Engine-generator output = ⎛ 5 -----2-⎞ ( 100000 ft ) = 500000 W ⎝ ⎠ ft = 500 kW = 1706500 Btu/h which is 33% of the fuel input Btu Fuel input = 1706500 Btu/hr ⁄ 0.com).35/h Btu . Ventilating. American Society of Heating.= 51.= 10.00/h 5 Btu ⎜ 10 -------------. Inc.00/h Cooling Design Condition Cogeneration Plant Engine input Supplemental fuel input $51. (www. (www.33 = 5171000 -------.⎟ ⎜ 10 5 ------------⎝ therm ⎠ $87.= $51.– 2172000 --------⎞⎠ ⁄ 80% h h Btu = 3535000 -------h Heating Design Condition Cogeneration Plant Engine input Supplemental fuel input Conventional Plant ⎛ 500 kWh -----------⎞ ( $0.00/therm ) = $35. Inc. Solutions to Chapter 17 AIR-PROCESSING EQUIPMENT Copyrighted material licensed to University of Toronto by Thomson Scientific. (www. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . distribution. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Inc. Refrigerating and Air-Conditioning Engineers. Additional reproduction.techstreet. (www.com).ashrae. American Society of Heating. © (2009).org). Inc. For personal use only. org). (www. For personal use only. Inc. Refrigerating and Air-Conditioning Engineers. (www. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. American Society of Heating. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . © (2009). Inc. Additional reproduction.com).techstreet. distribution.ashrae.Copyrighted material licensed to University of Toronto by Thomson Scientific. 60 wb : w 2 = 0.83 B/lb b) m c = ( 0.1°C wet bulb and leaves at 16. distribution. Find the total. © (2009). moisture condensed from air c. 95 db w = 0. 1.2 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 1. 62 db. How much condensate drains off the coil? 29. h 1 = 41. (www.0184 – 0.0124 – 0.0107 )1076 = 1. heat removed from air b. 2.0124 w 2 = 0.0107 : h c = 34.5 – ( 0. 90 db.0105 q = 42 – 25. 56°F 4.= 0.0184 h 2 = 25. Inc.1°C = 70 wb.6 Btu/lb 17.1 or q s ≈ 0.0168 – 0. 95% RH : : h 1 = 42. and sensible cooling loads on the coil with air at 14.53 16.0124 – 0. How much sensible heat and how much latent heat is removed from the air by the coil? b. (www.1 Air enters a coil at 95°F dry-bulb and 78°F wet-bulb temperature and leaves at 62°F dry-bulb and 60°F wetbulb temperature.com).6°C).75 Btu/lb q s ≅ h c – h 2 = 34.244 ( 95 – 62 ) = 8.4 – 34. American Society of Heating.0107 2.05 Btu/lb q L ≅ h 1 – h c = 41.0168. For personal use only. . °F 85 db. 95 db. w 2 = 0.2°C) 60% RH and leaves the coil at 60°F dry bulb (15.Chapter 17—Air-Processing Equipment⏐187 17.8.6 = 6.5 = 8.5 3.3 Air enters a direct-expansion coil at 90°F dry bulb (32. latent.0168 – 0.7°C dry bulb and 90% RH.4°C = 85°F 16. w 1 = 0.6 h s = h f = 24 B/lb [ h 1 – h 2 – ( w 1 – w 2 )h 3 ] = q c 41. 70 wb : w 1 = 0. a.0107.0017 lb/lb air 17. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 62 db.7°C = 62°F 21. Inc.0107 )1076 = 6. 78 wb : w 1 = 0.2 Btu/lb Δw = 0.7 psia. SHR for the condition line 1.8 or q L ≈ ( 0.0105 = 0. 90% RH : a) q s = 0. Refrigerating and Air-Conditioning Engineers. h 2 = 26. The condensate is assumed to be at a temperature of 56°F.61 B/lb q L = ( 0.4°C dry bulb and 21.0079 lb/lb air 42 – 33. Find a.4 SHR = q s ⁄ q T = ---------------------.6 – 26.techstreet. 95% RH.8 = 16.107 ) = 0.0107 )24 = q c q c = 14.4 – 26.2 Air enters a direct expansion coil at 29.4 2.244 ( 85 – 62 ) = 5.ashrae.org). 60% RH 60 db. Additional reproduction. Copyrighted material licensed to University of Toronto by Thomson Scientific. 9 – 33. Inc. The direct-expansion evaporator uses R-12 and operates at 35°F.4 Assume humidifier is steam coil heated pan type and process is pure humidification. (www. Ventilating. American Society of Heating.0128.687 51 – 35 t in – t out. w = 0.= 0. 40% RH. 70% RH : w oA = 0.2 ) = ( 38725 + 84312 )h m ⇒ h m = 19. should be supplied to the heating coil? b.6 ) = 1673300 Btuh 1673300 q HC = 1673300 ≅ m s hf s : m s = --------------------. For personal use only.4 Water flowing at 60 lb/min and at 51°F is chilled in an evaporator to 40°F. Additional reproduction.005 h: 38725 ( 11. © (2009). distribution. Find the evaporator effectiveness. h r = 23.org).2 = v r = 13. and Air Conditioning—Solutions Manual 17. Recirculated air is returned from the plant at 69°F dry bulb and 40% RH. v OA = 12. 20% RH.2 ) – ( 0. a) Leaving heating coil: t = 115°F. OA 35 db. max 17.= -----------------.= 1673 lb/h 1000 b) q Hum = m da ( h out – h in ) – m da ( w out – w in )h f h f = 28 water @ 60°F = 123037 [ ( 41.6 at w m and h m → t m = 59 db s : 115 db.techstreet. .2 – 14. h = 33. w s = 0. in pounds per hour. Refrigerating and Air-Conditioning Engineers.45 m r = ( 18900 ( 60 ) ) ⁄ 13.6 ) q HC = 123037 ( 13. The mixture is heated by a steam coil and humidified by a pan humidifier to final conditions of 115°F dry bulb and 20% RH. h s = 41.005 )28 ] 1043500 = – 1043500 Btuh = m s h fg ⇒ m s = --------------------1000 m s = 1043 lb/h This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto m OA = ( 8100 ( 60 ) ) ⁄ 12.ashrae.006.003. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. a.005. The heat transfer area is 20 ft2 and the heat exchanger has an overall heat transfer coefficient of 60 Btu/h·ft2 ·°F. h OA = 11.2 q HC = m da ( h out – h m ) = ( 38725 + 84312 ) ( 33. t in – t out 51 – 40 Effectiveness = -------------------------------. w r = 0.900 cfm of recirculated air.com).45 = 84132 lb/h m : W: 38725 ( 0.8. 8100 cfm of outside air mixes with 18.003 ) + 84312 ( 0. What steam flow.55 r : 65 db. Estimate the steam consumption of the humidifier. Inc.188⏐Principles of Heating.006 ) = ( 38725 + 84312 )w m w m = 0.58 = 38725 lb/h Copyrighted material licensed to University of Toronto by Thomson Scientific.8 ) + 84312 ( 23. (www.0128 – 0.5 Outside air at 35°F and 70% RH is supplied to an air-conditioning apparatus. org).8 ( 23. h 2 = 14 14 – h 3 h2 – h3 E = -------------------------. = 40% RH (b) h steam = 1056 Btu/lb.com). distribution. h = 23.8 ( 14 – 3 ) = 7445 Btu/min × -----------. How many pounds of steam per hour should be supplied to each? : 55°F. Additional reproduction.. a.6 Outdoor air (8000 cfm) at 10°F dry bulb and 50% RH enters the central apparatus of a split heating system.Chapter 17—Air-Processing Equipment⏐189 17. v 1 = 11. the air flows through a steam heating coil and is heated to 70°F dry bulb. W 1 = 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.80 = -----------------.00628 (a) : 70°F.82 Copyrighted material licensed to University of Toronto by Thomson Scientific. Assume steam at 2 psig and 90% quality is supplied to the tempering coil.techstreet.= 257. the sump water heat exchanger. It is tempered to 55°F dry bulb. . it flows through a spray humidifier where the leaving sump water is maintained at 50°F. After leaving the humidifier.00628.00064. The spray humidifier has a performance factor of 0. W 2 = 0.6. For personal use only.4 lb/h 1038 60 Heat Coil: 676.00064.= 430 lb/h 1038 60 Sump: 676. © (2009).9 lb/h 1038 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto at :10°F.= 0. Inc. Refrigerating and Air-Conditioning Engineers. h 1 = 3. What is the final relative humidity and humidity ratio of the air as it leaves the heating coil? b. and the heating coil.6 – 17 ) = 4462 × -----------. h 3 = 17 h 2 – h f50°F 14 – 18 : W 3 = 0.80. Then. (www. W = 0. American Society of Heating.ashrae. 50%.8 ( 17 – 14 ) = 2030 Btu/min × -----------.= 117. Δh H O = 1056 – 18 = 1038 Btu/lb 2 60 Tempering Coil: 676. (www. Inc. (www.5 55 – 37..2 – 14 ) = 4873 Btu/min × -----------.75 ( 1.0063 ) = 0. © (2009). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. a. 75% RH.7 – P v 2.com).6 grains/lb da This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 17.8 = ---------------------.= 267 lb/h 1038 1. What is the steam rate (lb/h) for the tempering coils and for the heating coil? h1 = 3 w 1 = 0. and Air Conditioning—Solutions Manual 17.0371 – 0. t 4 = 70°F. For personal use only. 100% RH.ashrae.2°C) saturated. Air leaves the spray chamber at 45°F dry bulb (7.622 ---------------------.6 is out of service for maintenance.190⏐Principles of Heating. w 2 = 0. American Society of Heating.. P v = 0.0308 lb s /lb da × 7000 = 215.8 60 Heating Coil: 676. Inc.00386 .8 ( 21. 45°F db.0371 lb/lb 14.techstreet. distribution. How many grains of moisture per pound of entering air are condensed? Copyrighted material licensed to University of Toronto by Thomson Scientific. t sat = 37. The split heating system is operating as specified except that the sump water is recirculated.8 Air at 105°F dry bulb (40. .00064 55 – t 3 E = 0. Additional reproduction. φ = ----.1021 ) = 0.8266 Ps Pv w 1 = 0. RH 4 = 25% (b) Tempering Coils:to 430 lb/h from Problem 17. (www.org).6°C) and 75% RH passes through a chilled water spray.5 t 3 = 41°F w 3 = w 4 = 0. Refrigerating and Air-Conditioning Engineers. What is the final relative humidity and humidity ratio of the air leaving the heating coil? b. Inc. Assume make-up water to the sump is 37°F and saturating effectiveness is equal to the performance factor.00064 h 2 = 14 w 2 = 0.= 0. Ventilating.0063 lb/lb Δw = ( 0.7 The heat exchanger for the spray water in Problem 17. Pv 105°F db. Refrigerating and Air-Conditioning Engineers.5 ) = 342000 Btuh (c) H = m ( w 3 – w 2 ) = 20492 ( 0.= 0.org). . t 2 = 135. latent.0074 – 0. t s = 120 .0074 – 0.= 4545 cfm m a = --------------------------------------.244 ( 120 – 70 ) 1500 ( 3. 13.techstreet. and 45.000 Btu/h.8 ) 1500 ( 0.1 – ( 0. 1. 26°C wb : W 1 = 0.0040 . 17°C.2°F (b)2.000 Btu/h. lb/h. Btu/h.= 20492 lb/h 1.= 0.7 = 34.0110 ) = 16.1 – ( 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. w 1 = ------------------------------------------------------------------------------. Ventilation requires that 1500 cfm of outdoor air be used.5°C. the capacity of the humidifier. Additional reproduction. Find the total.0054 ) h 1 = ---------------------------------------------------------------.0110.803 ) + 3045 ( 22.= 16. h 2 = 37. 16°C wb : W 2 = 0.Chapter 17—Air-Processing Equipment⏐191 17. Inc.0074 – 0.1 ma 20492 h 2 + ( w 3 – w 2 )h h – h 3 = 0 (b)1. h 1 = 80. Inc.7 q T = h 1 – h 2 – ( w 1 – w 2 )h 3 = 80.ashrae. Supply air is to be at 120°F. and cfm b. © (2009). h s = 37.001315 ) + 3045 ( 0. the amount of supply air required.com).2 2.9 Air enters a coil at 35°C dry-bulb and 26°C wetbulb temperatures and leaves at 17°C dry-bulb and 16°C wet-bulb temperatures.2 – 16.2 – 45 – ( 0.0 Copyrighted material licensed to University of Toronto by Thomson Scientific. Q = m ( h 2 – h 1 ) = 20492 ( 33.0177 – 0.5 . and sensible cooling loads on the coil with air at 101 kPa. American Society of Heating.10 A building space is to be maintained at 70°F and 35% RH when outdoor design temperature is 10°F.0074 .02 ( 35 – 17 ) = 18.10 ( 120 – 70 ) 0. 35°C.4 J q w ≅ ( h g – h f )Δw = 2500 ( 0. The conditioning equipment and nomenclature are shown in the following sketch.5 ) = 420000 Btuh (b)2. Determine: a. h 2 = 45 3.6 lb/h This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 250000 250000 CFM s = -----------------------------------. Q = m ( h 2 – h 1 ) = 20492 ( 37 – 16. (www. if 1. saturated steam at 17. h 2 = 37.8 J q s ≅ c p Δt = 1.8 J 17.00405 ) ( 1153.0110 )56.0177.= 0.5°C : h 3 = h f = 56. For personal use only.2 psia c.4 ) = 33. lb/h. (www.0054 + ------------------------------. the capacity of the heating coil.2 t 2 = 119°F (b)1. The condensate is assumed to be at a temperature of 13. latent. Design heat losses from the space are 250. distribution. sensible.to the humidifier is a steam humidifier using dry.00405 ) = 68.00405 ) ( 28 ) = 37. t 1 ≅ 51°F 4545 4545 mw 45000 ⁄ 1100 w s = w r + ------.0177 – 0.to the humidifier is a spray washer using recirculated spray water with makeup water provided at 60°F 2. . w s = 0. Initially. Desired humidity ratio at outlet is 0.techstreet.x100 = 65.005 – 0. Inlet conditions to the washer are 75°F db and 48 F wb.x100 = 57% 72 – ( – 2 ) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto (a) Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc. American Society of Heating.005 .005 lbw/lba. Refrigerating and Air-Conditioning Engineers. (www. (a) w i = 0. (b) Q = 1.org).007 0. Inc.33 17. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.001 E w = --------------------------------. Additional reproduction.007 – 0. For personal use only.( 0. a separate preheater was planned for bringing the outside air from its –2°F design ambient outdoor temperature to 40°F. Ventilating.ashrae.001 ) = 162 lb/h 13. © (2009). and Air Conditioning—Solutions Manual 17.005 – 0.12 A heat pipe air-to-air energy recovery device is being considered for a system requiring 9000 SCFM of outside air.com). Determine: (a) the necessary humidification efficiency of the washer.11 A spray-type air washer is to be used for humidification as well as cleaning of 9000 SCFM of air.001 9000 ( 60 ) (b) m w = ----------------------.6% 0. % (b) the make-up water requirements (humidifying capacity) of the unit. distribution.192⏐Principles of Heating.10 ( 9000 ) ( 40 – ( – 2 ) ) = 415800 Btu/h 40 – ( – 2 ) E s = ----------------------. Determine: (a) the rating (Btu/h) and (b) the sensible effectiveness (%) to specify for the heat pipe unit if it is to eliminate the need for the air preheater. (www. lbw/h. both with and without the energy recovery unit.6 + --------------------------------.00102 t r = 72°F. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.0093 ) = 183500 Btu/h (d) with: Q T = 188700 + 183500 = 372200 Btu/h without: Q T = 279400 + 183500 = 462400 Btu/h 19.org).10 ) ( 83.244m a ( 130 – 72 ). American Society of Heating. (b) sensible coil load.= 96. . Refrigerating and Air-Conditioning Engineers. Include fan effects. φ s = 90% ⇒ w s = 0.005 – 0.Chapter 17—Air-Processing Equipment⏐193 17.10 ) ( 96. Btu/h.8 – 58 ) = 188700 Btu/h (c) with/without :w 2 = 0.2°F = t 2 29900 ( 0. Determine: (a) fan size (hp & scfm). Btu/h. all year long. (www. 1.⇒ 6650 scfm 60 ( 3.(a) 146000 = 0.244 ) Q s = 6650 ( 1.0244 ) w 1 = w f = w 2 = w o = 0.005 t s = 58°F. (www. Btu/h. At summer design conditions (Indoor: 78°F. t f = 95 + -----------------------------------. At winter design conditions (Indoor: 72°F & 30% RH.8°F 29900 ( 0. water.00102 ) = 66 lb/h ≈ 190 gal/day 13. Deter- 2.= 3. m a = 16600 lb/h m a [ 0. m a = 29900 lb/h × ------------. Additional reproduction.25 ) ( 0.73 ( 95 – 78 ) = 82.000 Btu/h (sensible) and 79.25 in.5% savings This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 1600 [ 0. Q h = 292000 Btu/h same with heat pipe since cooling coil discharge is same (b) m h = m a ( w r – w o ) = 16600 ( 0.244 ( t s – t c ) ] + Q h = 0 Q s = 6650 ( 1. Neglect fan effects. Inc.015 – 0.ashrae.015 Q L = 4840 ( 6650 ) ( 0.6 .4 ( 2545 ) t 1 = 95 – 0. Outdoor: 95°F db/76°F wb). φ r = 30% ⇒ w r = 0. t 2 = 82. Inc. t o = 5°F. the space cooling loads are 146. Design duct system pressure drop (summer) is 3.036 ) ( 144 )60 W = V· ΔP = 6650 -----------------------------------------------------. mine: (a) the necessary size of heating unit (Btu/h) both with and without the energy recovery unit and (b) the humidifier size (gallons/ day). φ o = 100% ⇒ w o = 0.015 Copyrighted material licensed to University of Toronto by Thomson Scientific. distribution.000 Btu/h (latent).(a) without heat pipe unit: 235000 = 0. Outdoor: 5°F & 100% RH) the space load is 235. (c) latent coil load.224m a ( 78 – 58 ).244 ( 130 – 58 ) ] + Q h = 0. air leaves the heater at 130°F.13 The HVAC system for a hospital operating room which requires 100% outside air is shown in the following figure and includes an air-to-air heat pipe energy recovery unit having a sensible effectiveness of 73%.2 – 58 ) = 279400 Btu/h with 3.com).4 ( 2545 ) t 1 = t o = 95°F. For personal use only.4 HP 778 ( 2545 ) (b) without heat pipe unit: 3.0093 1. and (d) necessary size of cooling unit.33 2.techstreet. © (2009). Fan speed is changed between summer and winter operation.000 Btu/h (sensible) with negligible latent load. 90% RH. During winter operation.= 83. The air leaving the cooling coil is maintained at 58°F. techstreet.com). This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Refrigerating and Air-Conditioning Engineers. Inc. American Society of Heating.ashrae. (www. (www. © (2009).Copyrighted material licensed to University of Toronto by Thomson Scientific. Additional reproduction. Inc.org). For personal use only. distribution. (www. Refrigerating and Air-Conditioning Engineers.com).ashrae. (www. © (2009). Inc.org). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. distribution. Additional reproduction.Solutions to Chapter 18 REFRIGERATION EQUIPMENT Copyrighted material licensed to University of Toronto by Thomson Scientific. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . American Society of Heating. Inc. For personal use only.techstreet. © (2009).Copyrighted material licensed to University of Toronto by Thomson Scientific. For personal use only. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .com). (www. distribution. Additional reproduction. Inc. American Society of Heating.org). Inc. Refrigerating and Air-Conditioning Engineers.ashrae. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.techstreet. (www. 7°C) and evaporating at 20°F (−6.2 Given a compressor using R-22 condensing at 80°F (26.2°F 20 ( 8.2°F 90 – 75 ln ---------------------90 – 88.ashrae.53 Btu/lb a) b) c) 158.3 ⁄ 12000 ) h1 – h2 – w = 0 w = 106. Additional reproduction. find the enthalpy of the refrigerant when it enters the a.– 1⎞ = 10 ⎝ TR ⎠ ⎝ 500 ⎠ QA 10 tons COP = 10 = -------------------.1 A condenser used in a refrigeration system has a capacity of 10 tons at a 40°F evaporating temperature.53 – 118 = – 11.74 Ton ( 73.0. P 4 = P 1 = 57.+ 75 = 88. with a heat transfer area of 83 ft2.795 psia 20°F. (www. s = 0.= 6.org).= ------------------QR – QA Q R – 10 Q R = 11 tons ⇒ Q R = 11 × 12000 = 132000 Btuh 132000 t w.53 – 33.33 psia.o = ---------------------------. When 20 gpm of cooling water enters at 75°F.techstreet.33 )60 ( 90 – 75 ) – ( 90 – 88.com). Inc.2 ) Δt m = -------------------------------------------------------.22415 h 1 = 106.7°C). .= 0.34 q e = 73. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. © (2009).342 = h 4 HP( 11. (www. American Society of Heating. Inc. Are these claims reasonable? Why? Q = UA Δt m = m w C p ( tw o – tw i ) To 550 Ideally: COP = 1 ⁄ ⎛ -----.5 ⁄ 2545 ) -------= ---------------------------------.33 psia Copyrighted material licensed to University of Toronto by Thomson Scientific. condenser c.2 ) More than 95 inconsistent 18.22415i h 2 = 118 Btu/lb 80°F. s = 0. h 3 = 33. compressor b.– 1⎞ = 1 ⁄ ⎛ --------. evaporator Find the power required for the compressor. Refrigerating and Air-Conditioning Engineers.2 Btu/lb This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto P 2 = P 3 = 158.= 257(needed) 83 ( 6.2 132000 U = -----------------. the condensing temperature is 90°F. x = 1. For personal use only. The manufacturer claims a U-factor of 95 Btu/h· ft2· °F. distribution.5 B/lb q e = h 1 – h 4 = 106.Chapter 18—Refrigeration Equipment⏐197 18. 14 TH – TL 75 18.94 ) = 2018 lb/h = 33.1 lb/h/ton = 2. x = 0. h g = 93.47 – 44.7357 83. Why is this theoretical limit difficult to obtain? TL 460 Ideal COP = ------------------. Is this correct? Substantiate your answer with calculations based on knowledge of R-22 for these conditions.588 ft ⁄ lb ⎝ 83.328 + x ( 104. v g = 1.198⏐Principles of Heating.1 ( 18.547 = 90.6 cfm 18. © (2009).334 ( 1. Determine the volume of the refrigerant leaving the expansion valve in cubic feet per minute. h 3 = 45.5 An R-134a refrigerating system develops 10 tons of refrigeration when operating at 100°F condensing and +10°F evaporating.328 ).= --------. Refrigerating and Air-Conditioning Engineers.334 1) 3 1 v 4 = 0.43 3) – 30°F.0.com). (www.. – 100°F.7357 ) + 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. verify the mass flow rate in lb per min. h g = 104.588 ) = 19.5 cfm This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 1) Copyrighted material licensed to University of Toronto by Thomson Scientific.82 = 132.43 ) = 2435 cfh = 40.org).37. Ventilating.8°C and 23. → x = 0. with no liquid subcooling or vapor superheating. (www.3 What is the maximum theoretical COP of a refrigeration device operating between 0°F and 75°F (−17. v g = 18. x = 1.6 lb/min v· = 33. h = 2.29 100°F.ashrae.= 6.2 lb/min v· = 132. v f = ------------.techstreet.37 – 2.4 A reference book on refrigeration indicates that a compressor using R-22 requires a displacement of 40. . Inc.617.29⎠ m = 120000 ⁄ ( 104. x = 0.155 4) h 4 = h 3 = 45. Inc. Also.6 ( 0.82 Btu/lb 12000 ⁄ 90. For personal use only. distribution. h f = 15. Additional reproduction.155 = 15.617 – 15.666 ⎛ -------------⎞ = 0. American Society of Heating. and Air Conditioning—Solutions Manual 18. 3) 1 + 10°F.9°C).328.547 q e = 93.59 cfm per ton for evaporation at −100°F and condensing at −30°F. 83⎠ v1 2. Inc. (www.25 ( 138.org).94 – h 3 ) 1 h 3 = 40.83 1 ( 112 – 107.= 29.39 ft ⁄ lb m ⎝ 138.819 h 1 = 107. For an evaporator temperature of 0°F and a condenser temperature of 100°F.4 lb/min 1 --- n p1 v1 = p2 v2 n P1 n ⇒ v 2 = v 1 ⎛⎝ ------⎞⎠ P2 1------ 21.39 ⎠ ⎝ v 2⎠ n v = 0.4 ( 2. What is the flow rate of the refrigerant? Copyrighted material licensed to University of Toronto by Thomson Scientific. For personal use only.94 P H = 138.04 – 0. distribution. Inc.6 An expansion device has a mass flow rate for R-134a given by m = 60 + 0.16 ) = 89.1587 = 0. P 1 = 21.com).ashrae.7 cfm nv 0.32 – 40. Additional reproduction.32 1 h1 h3 = hf P L = 40.83 psia v 1 = 2. Liquid leaves the condenser saturated at 100°F (37.16 1.16 psia P 2 = 138. estimate the piston displacement required for a compressor if C = 0.= -------------------------------.26 )60 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 18.1587 m = 60 + 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.1 for the compression process.32 ) = ( 44.04 ⎛ ----------------⎞ ⎝ 0. © (2009).04 and the polytropic compression coefficient n = 1.techstreet.76 psia = 112 ( 40.7 A liquid-to-suction heat exchanger is installed in an R-134a system to cool liquid that comes from the condenser with vapor that flows from the evaporator. and vapor leaves the heat exchanger at a temperature of 50°F (10°C). vapor leaves the evaporator saturated.8 – 21.1587 n v = 1 + C – C ⎛ -----⎞ = 1 + 0. 50°F ) 100F = 44.1 3 v 2 = ⎛ ----------------⎞ 2.Chapter 18—Refrigeration Equipment⏐199 18.26 Btu/lb 10 ( 12000 ) m = ----------------------------------------------.25Δp where m = flow rate in lb/min Δp = pressure drop across the valve in psi.76 psia.17 kW) of refrigeration at 30°F (−1.8°C).1°C).819 mv 1 89. The evaporator generates 10 tons (35.= 235. Refrigerating and Air-Conditioning Engineers. (www.82 lb/min ( 107. American Society of Heating.1587 ) PD = --------. . Copyrighted material licensed to University of Toronto by Thomson Scientific. The vapor enters the compressor at 30°F.600 = mCp ( t o – t i ) 436600 t o = 85 + ---------------------------.7 ft ⁄ lb 3) h 3 = h f = 155.3°C) and is cooled with 90 gpm (5. Inc. If the average piston speed is to be 600 ft/min and the actual volumetric efficiency at this condition is 83%. The ammonia leaves the condenser as saturated liquid.2 30 ( 12000 ) m = --------------------------------------.2 ) v· = 12. calculate the condensing area required.83 PH = 211.7 ) Δt m = -------------------------------------------------------------. Assume an ideal Carnot cycle QA QA 1 1 COP = ------. If the expected U-factor of the condenser is 130 Btu/h⋅ft2·°F [738 W/(m2·K)].6 ⇒ A = 171 ft 2 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 18.= 8 ( 800 ) ⎛ ------------------⎞ ( 0.7 = -------------------.0 in bore x = vt 1 2L = 600 --------800 L = 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.= --------------------------------------.7 570 W QR – QA TH --------. The condenser is to operate at 110°F (43.⇒ Q R = 436.600 = 130 × A × 19.= 94.7 ( 7. find the bore of the compressor.= ---------------.375 ft = 4.9 A condenser is to be selected for a system that generates 30 tons (105.375 ) ⎝ 4 × 144⎠ 0.= 12.= -------------------. . (www. For personal use only. © (2009).8 cfm 2 97. (www.ashrae.68 L/s) of water at 85°F (29.org).= 19. Assuming superheating to 30°F takes place in the evaporator 3 1) 30°F. Inc.200⏐Principles of Heating.5 in.7 lb/min 60 ( 627 – 155.com).33 )60 ( 110 – 85 ) – ( 110 – 94.5 psi D = 3.8 An eight-cylinder ammonia compressor is designed to operate at 800 rpm and deliver 30 tons of refrigeration. distribution.8 πD ---------.5 psi.= 4.6°F 25 ln ---------15. v 1 = 7. PL = 38.5 kW) of refrigeration at 10°F (−12.9 psi.3 436. h 1 = 627. Refrigerating and Air-Conditioning Engineers.techstreet. Additional reproduction.2°C). and Air Conditioning—Solutions Manual 18.– 1 470 TL QA 30 ( 12000 ) 4. 38. Ventilating.7 90 ( 8.– 1 -----.= -----------------.4°C). American Society of Heating.600 Btuh QR – QA Q R – 30 ( 12000 ) Assuming 85°F inlet → Q R = 436. The evaporator is to operate at 10°F with a condensing temperature of 100°F.7 ) = 97. The heat transfer area is 1. 11 – 4. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Assuming refrigerant flow is not limiting. distribution. to what temperature can this air cool water that enters at 110°F with a flow of 80 lb/min? What is the makeup water rate? m da ( h out – h in ) air = mCp ( t ou – t off ) air: in water: out on 1000 Cfm. 78°F wb.35 84°F db. h = 48.0168 1000 m da = ------------.org).2 – 41. Find the evaporator effectiveness.9 m2 and the heat exchanger has an overall heat transfer coefficient of 341 W/(m2· K).techstreet.4.0 69. 95°F db. American Society of Heating.Chapter 18—Refrigeration Equipment⏐201 18. (www.7 [ 48.61 ( 1046 ) – 0. Refrigerating and Air-Conditioning Engineers. . Inc.10 A cooling tower cools water by passing it through a stream of air.61 lb/min 0.7 lb/min 14. m = 80 lb/min.0256 – 0.× 100 = 72% 11 – 2 Copyrighted material licensed to University of Toronto by Thomson Scientific. w = 0.2.35. Inc. The direct-expansion evaporator uses R-12 and operates at 2°C.0256 t = 110°F. v = 14.4 ] = 80 ( 1 ) ( 110 – t off ) t off = 104°F make-up = m da ( w out – w on ) check: = 69. C p = 1.11 Water flowing at 28 kg/min at 11°C is chilled in an evaporator to 4. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.com). © (2009). 100% RH. (www.7 ( 0. For personal use only.5°C.244 ( 95 – 84 ) = 635 = 80 ( 110 – t off ) t off = 102°F 18. w = 0.0168 ) = 0. Additional reproduction.= 69.ashrae. h = 41. If 1000 cfm of air at 95°F dry bulb and 78°F wet bulb enters the tower and leaves saturated at 84°F.5 Effectiveness = ------------------. (www. distribution. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.techstreet. © (2009). Inc. For personal use only. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . Inc. Additional reproduction.org).ashrae. American Society of Heating. (www.com). Refrigerating and Air-Conditioning Engineers.Copyrighted material licensed to University of Toronto by Thomson Scientific. org). For personal use only. © (2009).ashrae.com). This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . American Society of Heating. distribution. (www. (www. Inc.techstreet. Refrigerating and Air-Conditioning Engineers.Solutions to Chapter 19 HEATING EQUIPMENT Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc. Additional reproduction. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . (www. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. American Society of Heating.Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc. © (2009). distribution.com). Refrigerating and Air-Conditioning Engineers. For personal use only. Inc.techstreet. (www.org). Additional reproduction.ashrae. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .5995 59.32 ÷ 44 = 0. lb/lbmix lb/mol mol/molmix CO2 0. Inc.6 kg air ⎝ 56 44 ⎠ 19.com).00411 0.2 The gravimetric analysis of a gaseous mixture is: CO2 = 32%. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.00727 O2 0.7665 psia (5.97% ⎛ n CO ⎞ P CO = P mix ⎜ -----------2⎟ = 3 ( 0. (www. The mixture is at a pressure of 20.115 ÷ 28 = 0. of air = ⎛ ------------.5995 ) = 1.techstreet. O2 = 54.47% molmix/lbmix 99.4341 psia (2.org).76 ) ( 29 ) = 3. a. and N2 = 11.2555 ) = 0.02843 = 0.642 wt.99 kPa) 2 Copyrighted material licensed to University of Toronto by Thomson Scientific.01705 N2 0.02843 molmix/lbmix mol/molmix % Volume ÷ 0.02843 = 0.ashrae.+ ------⎞ O 2 → -----.CO + CO 2 ⎝ 28 44⎠ ⎝ 56 44⎠ 28 x.5%.642 kg CO. Refrigerating and Air-Conditioning Engineers.----x x x ⎛ ----+ -⎞ C + ⎛ -----.1447 ) = 0. x = 0. (www.642 0.45 kg C.7985 psia (12.1447 14. © (2009).45 kg (1 lb) of pure carbon to equal masses of CO and CO2.28 kPa) 2 ⎝ n mix ⎠ P O = 3 ( 0. American Society of Heating. Inc.5%.1 Set up the necessary combustion equations and determine the mass of air required to burn 0. x x.95% ÷ 0.2555 ~ 25. CO 2 0.Chapter 19—Heating Equipment⏐205 19.55% ÷ 0. For personal use only.545 ÷ 32 = 0.+ -------------⎞ ( 4. distribution.02843 = 0. Additional reproduction.7 kPa (3 psia).----x-⎞ ⎛ ----⎝ 28 + 44⎠ 12 = 0. b.39 kPa) 2 P N = 3 ( 0. Determine (a) the volumetric analysis and (b) the partial pressure of each component. 8 Theorical Mols Air ( 7.76 )N 2 Copyrighted material licensed to University of Toronto by Thomson Scientific.3 A liquid petroleum fuel. C2H6OH is burned in a space heater at atmospheric pressure.= 11.N 2 11. 83.5 + 14.1 Air Mol H 2 O 3 ----------------------= ---------------------------------------------.0 ) ( 3.1 P v = 0.5 × 3.com).0% Theroretical): C 2 H 6 OH + ( 1.8 ) ( 3.25 ) ( 3.5 ) ( 18 ) m H O = ----------------------.2 + 18.ashrae.5H 2 O + ( 3.7 ) = 2.÷ 11.5O 2 + ( 7.org).25 )O 2 + ( 0.25 ) ( 3.= 53.76 )…35.7 ÷ 11.5 ) ( 32 ) + ( 7.7 ---------18. Ventilating.94% CO 2 CO 1.= ------------------------------------------------------. 11.600 Btu/lb (45 600 kJ/kg).797 psia.= 0. 133.054 Mol Exh. For combustion with 20% excess air.45 lb air ⁄ lb fuel F 2 × 12 + 6 + 16 + 1 ( 3.78 = 5.76 )N 2 → 0. For personal use only.65O 2 b. Refrigerating and Air-Conditioning Engineers.02% -----------------. and the dew point of the combustion products.5 ) ( 3.76 )N 2 → 2CO 2 + 3.25 ----------Mol Fuel × lb fuel ⁄ mol fuel lb fuel lb fuel ( 1 ) ( 78 ) Mol Air × 29Mol Air 78 × 20 --------------------------------= 20.25 ) ( 3. Inc.= ----------. American Society of Heating. © (2009).4 Find the air/fuel ratio by mass when benzene (C6H6) burns with theoretical air and determine the dew point at atmospheric pressure of the combustion products if an air/fuel ratio of 20:1 by mass is used. [Ans: 5.04% CO 9.65 DP = 133.8 g/s) when the brake output is 75 hp.5 A diesel engine uses 30 lbm of fuel per hour (3.9°F 80% Theoretical Air: C 2 H 6 OH + ( 0.5 P w = ----------------------------------------------------. For combustion with 80% theoretical air. determine the dry analysis of the exhaust gases in percentage by volume. (www. (www.02% N2] C 2 H 6 OH + 3. and Air Conditioning—Solutions Manual 19. [Ans: 11.5% 30 lb/h × 19 .66 + 0.9°F] a. the mass of water formed by combustion per pound of fuel.25 ) ( 3.600 Btu/lb This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto C 6 H 6 + 7.3 ÷ 11.= --------------------------------------------------------------------.2 ) ( 3.25 ) ( 4. 6 + 3 + 28.78 = 11.206⏐Principles of Heating.= 13.34 lb H O ⁄ lb fuel 2 2 47 3.76 ) ( 28 ) -----------------------------------------------------------.3CO + 3.5H 2 O + ( 8.= -----------------.5 ) ( 3.25 )O 2 + ( 1.5 + 7.7 ) = 0. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.8 Mol Fuel × 78 Mol Fuel 29 Actual Mols Air ………………………53.25 ) ( 3.2 ) ( 3. If the heating value of the fuel is 19. Inc.76 )N 2 CO 2 0.8 ) ( 3.78 100.1 mols excess air C 6 H 6 + 53.25O 2 + ( 3.76 )N 2 20% Excess (12.45.5 ) ( 3.2N 2 + 18.054 ( 14.04% CO.76 ) ( 29 ) A --.22 )N 2 + 0.7CO 2 + 1. 1. distribution.( 14.78 N2 ------------.94% C02.techstreet.= 1. determine the air/fuel ratio by mass.5H 2 O + ( 1. ΔP = 94°F 19. Theoretical: b. what is the brake thermal efficiency of the engine? 75 hp × 2545 Btu/hp·h η 6 = --------------------------------------------------------.76 )N 2 → 6CO 2 + 3H 2 O + ( 7.× 100 = 32.0% 19. a.47 psia 2 + 3.76 )N 2 → 2CO 2 + 3.78 = 83.34. . ( 1.2 ) ( 12. Additional reproduction.8 Air → 6CO 2 + 3H 2 O + 28.2 ) ( 3. ∴---------------------. lb air lb air Mol Air × lb air ⁄ mol air ( 7. 16 a.52CO + 2.76 )N 2 → CO 2 + 2H 2 O + 7.48 (vol.52 = 10.07N 2 Since nitrogen is from air: moles air ft 3 air A ⁄ F ( by volume ) = ( 23.52CO + 2.24 ) ( 1544 ) ( 960 ) 2 3 PV = nRT. .16 + 87. PH O 16 2 -----------= -------------------------------------. 18. V = -------------------------------------------------------------------------. 16 % CO 2 (vol.89 (mass).76 )N 2 → 16CO 2 + 16H 2 O + 90.1%] aCH 4 + bO 2 + cN 2 → 10.52 ) → d = 21.24 moles 2527 lb = 0. b = ------------.ashrae.48 ---------------------. For personal use only. distribution. PT 16 + 16 + 90. Refrigerating and Air-Conditioning Engineers. the percent theoretical air.52%. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.1%.89 – 17.6 Methane (CH4) is burned with air at atmospheric pressure.1% 17.16 18. (www.× 100 = 10. Inc.= 0.76 C: a = 10 + 0.52 = 10. Moisture formed per kg of fuel.16 lb a ⁄ lb f 16 18.1% 17.16 ( 32 ) + 87.= 379. American Society of Heating.7 Fuel oil composed of C16H32 is burned with the chemically correct air-fuel ratio.16 b 21 3. moisture formed per lb of fuel Partial pressure of the water vapor. (www.7 m /kg) ( 14.Chapter 19—Heating Equipment⏐207 19.52 ) ( 16 ) = 18.89 % Theoretical Air = ------------.24 ( 16 + 16 + 90.24 moles 16 moles 224 lb or kg 768 lb 2527 lb 704 lb a.41O 2 + 21..00%.= 3. O2 = 2.07%. c.924 psia ( 133 kPa ) [dew point = 125°F(52°C)] c.24 PH 2O 16 moles 288 lb 90.= 79 -----.13 ( 14. and the percent excess air.16O 2 + 87.76. b. if the gas is at 260°C (500°F) and 102 kPa (14.1% 16 + 90.= 17. d.).16 % Excess Air = --------------------------------.8 ) = 1.8 ) ( 144 ) ( 224 ) This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 19.04 10.52CH 4 + 23. 10.com).07 82.8 psia).5 ft /lb (23.× 100 = 15. and N2 = 87. CO = 0.07 ) ⁄ 10.41O 2 + dH 2 O + 87. Find Copyrighted material licensed to University of Toronto by Thomson Scientific.286 lb m /lb fuel also 1.13.0CO 2 + 0.41%.52 H: 4a = 2d = 4 ( 10.24N 2 1 mol 24 moles 90. water vapor psia Percentage of CO2 in the stack gases on an Orsat basis Volume of exhaust gases per unit mass of oil. [Ans: 10. kPa. dry) = ------------------------. © (2009). M ⁄ F = 288 ⁄ 224 = 1.04H 2 O + 87.0CO 2 + 0. Additional reproduction.07 --c. Balance the combustion equation and determine the airfuel ratio.or ---------------mole fuel ft 3 fuel lb air A ⁄ F ( by mass ) = [ 23. The Orsat analysis of the flue gas gives: CO2 = 10.07N 2 N2 : c = 87.89 -----------lb fuel Theoretical: CH 4 + 2O 2 + 2 ( 3.org).= 23.techstreet.07N 2 → 10.52 N 2 2 ( 32 ) + 7. Inc.07 ( 28 ) ] ⁄ ( 10. C 16 H 32 + 24O 2 + ( 24 ) ( 3. d.× 100 = 110.286 kg/kg b.52 ( 28 ) A ⁄ F ( by mass ) = -----------------------------------------. 110. 50O 2 + 12.0%.28 lb O ⁄ lb fuel 2 0.× 100 = 7. 47.3N 2 N2 : c = 82.= 3.7 + --.com).5 )O 2 + ( 0. Additional reproduction.× 1 = 8.0H 2 O + 39.75 CO remaining C 8 H 18 + ( 0.0CO 2 + 1.0 + ------. Refrigerating and Air-Conditioning Engineers.49] H2 : C: 0.org).0H 2 O + 47.8 or C 45 H 94 19.3%. Inc. 2 2 A ⁄ F = 1.20 ) = 17..25 for CO → CO2 8.52 lb O ⁄ lb fuel 2 2H 2 + O 2 → 2H 2 O 4 32 36 1 8 9 C + O 2 → CO 2 12 32 44 1 32/12 44/12 A = 3. and Air Conditioning—Solutions Manual 19.0 )N 2 → 4.4 2 2 C: a = 8. theor.8%.9%.9 39.0 + 1.24 lb O ⁄ lb fuel 2 3.20 lb air ⁄ lb fuel . and N2 = 82.10 Compute the compositions of the flue gases on a percent by volume on dry basis (same as Orsat) resulting from the combustion of C8H18 with 85% theoretical air.ashrae.5 ) ( 2 ) = 21.+ 8.8 ∴ Composition of fuel: C 9 H 18.7O 2 + eH 2 O + 82.9 3. Inc.9N 2 4.9 1.0N 2 0.49 lb air ⁄ lb fuel C 8 H 18 + 12.85 ) ( 47.25 – 8.9 Determine the air/fuel ratio by mass when a liquid fuel of 16% hydrogen and 84% carbon by mass is burned with 15% excess air.0CO + 8. CO = 1.00 for C → CO 4.00 for H 2 → H 2 O 12. [Ans: 17.208⏐Principles of Heating.9 N 2 = ---------.2315 lbO ⁄ lb air = 15.9 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 19. distribution.25 CO 2 = ---------.52 lb O ⁄ lb fuel ⁄ 0. © (2009).× 100 = 83.8 Determine the composition of a hydrocarbon fuel if the Orsat analysis gives: CO2 = 8. Ventilating.0 = 9.75CO + 9.00 – 4. 47.9 = 8.16 × 8 = 1.84 × 32 ⁄ 12 = 2.7%.0 H: b = 2e = 18.25 = 2H 2 + O 2 → 2H 2 O C + 1 ⁄ 2O 2 → CO 2C + O 2 → 2CO C + O2 → CO 2 3. For personal use only. Copyrighted material licensed to University of Toronto by Thomson Scientific.25 oxygen atoms available – 9.techstreet.5 ( 3. e = 9.25CO 2 + 3. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. (www.75 CO = ---------.0CO 2 + 9.0%.76. C a H b + dO 2 + cN 2 → 8. O2 = 8.85 ) ( 12.0 e O 2 : 21.15 ( 15.3 --c. . (www. d From composition of air: d = 21.85 ( 12.76 )N 2 → 8. American Society of Heating.3% 47. Chapter 19—Heating Equipment⏐209 19.11 A liquid petroleum fuel having a hydrogen to carbon ratio of 0.169 by mass is burned in a heater with an air/fuel ratio of 17 by mass. Determine: a. b. the volumetric analysis on both wet and dry bases of the exhaust gases the dew point of the exhaust gas. H H ---- = 0.169 ---- = 0.169 × 12 = 2.028 = 2.03 C C CH 2.03 + 1.5075O 2 + 5.67N 2 → CO 2 + 1.015H 2 O + 5.670N 2 x ( 1.5075 + 5.6700 )29 ----------------------------------------------------- = 17; x = 1.145 or 14.5% excess 1 ( 12 + 2.03 ) CH 2.03 + 1.725O 2 + 6.49N 2 → CO 2 + 1.015H 2 O + 0.2175O 2 + 6.59N 2 a. b. 1 wet CO 2 = ---------------- = 8.7225 1.015 H 2 O = ---------------- = 8.7225 0.2175 O 2 = ---------------- = 8.7225 6.4900 N 2 = ---------------- = 8.7225 P w = 0.1165 ( 14.7 ) = 11.50% 1 dry CO 2 = ---------------- = 12.95% 7.7075 11.65% 0.2175 O 2 = ---------------- = 2.82% 7.7075 6.490 74.40% N 2 = ---------------- = 24.20% 7.7075 1.71 psia ( 11.8 kPa ); Dew Point = 120°F ( 49°C ) 2.49% 19.12 Compare the heating value for semianthracite coal as given in Table 8, chapter 28, 2009 ASHRAE Handbook—Fundamentals with the value predicted using the Dulong Formula. [Ans: 1.24% difference] HV = 13 ,600 Btu/lb O2 = 5 ; H 2 = 3.9; C = 80.4; N 2 = 1.1; S = 1.1; Ash = 8.5 Dulong: Difference: HHV = 14 ,544C + 62 ,028 [ H – ( 0 ⁄ 8 ) ] + 4050S = 14 ,544 ( 0.804 ) + 62 ,028 [ 0.39 – ( 0.05 ⁄ 8 ) ] + 4050 ( 0.011 ) = 13 ,769 Btu/lb 13 ,769 – 13 ,600 --------------------------------------- × 100 = 1.24% 13 ,600 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto From Table 8, pg. 18.8, 2009 HBF Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 210⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual 19.13 Natural gas with a volumetric composition of 93.32% methane, 4.17% ethane, 0.69% propane, 0.19% butane, 0.05% pentane, 0.98% carbon dioxide and 0.61% nitrogen burns with 30% excess air. Calculate the volume of dry air at 60°F, 30 in. Hg (15.6°C, 101.5 kPa) used to burn 1000 ft3 (28.3 mL) of gas at 68°F and 29.92 in. Hg (20°C and 101.4 kPa) and find the dew point of the combustion products. Table 1, pg. 28.2, 2009 HBF Natural Gas; 30% Excess Air Methane (CH4) 93.32; Ethane (C2H6) 4.17; Propane (C3H7) 0.69 Butanes (C4H10) 0.19; Pentanes (C5H12) 0.05; CO2 0.98; N2 0.61 A ⁄ F = ( 9.57 ) ( 0.9332 ) + ( 16.75 ) ( 0.0417 ) + ( 23.95 ) ( 0.0069 ) + ( 31.14 ) ( 0.0019 ) + 38.29 ( 0.0005 ) 3 3 3 3 = 9.88 ft air /ftgas at 68°F, 29.92 in. Hg = 9880 ft air /1000 ft gas 3 3 3 3 A ⁄ F at 30% excess air → A ⁄ Fa = 1.3 × 9.88 = 12.89ft air /ft gas or 12 ,890 ft air /1000 ftgas T1 P2 ( RT 1 ) ⁄ ( MP 1 ) V1 - = ---------------- = ---------------------------------V2 ( RT 2 ) ⁄ ( MP 2 ) T2 P1 3 3 520 29.92 V 60°F, 30 in. Hg = 12 ,890 ⎛ ---------⎞ ⎛ -------------⎞ = 12 ,660 ft air /1000 ft gas ⎝ 528⎠ ⎝ 30 ⎠ Methane: → ( 0.9332 )CO 2 + ( 0.9332 ) ( 2 )H 2 O + ( 0.9332 ) ( 2 ) ( 3.76 )N 2 Ethane: → ( 0.0417 ) ( 2 )CO 2 + ( 0.0417 ) ( 3 )H 2 O + ( 0.0417 ) ( 3.5 ) ( 3.76 )N 2 Propane: → ( 0.0069 ) ( 3 )CO 2 + ( 0.0069 ) ( 4 )H 2 O + ( 0.0069 ) ( 5 ) ( 3.76 )N 2 Butane, Pentane: small, neglect 0.0098CO 2 CO 2 : N2 : 2.018 1.047CO 2 ------------------------2.019H 2 O 1.8% 2.019H 2 O P w = 15.11 ( 30 in. Hg ) = 4.0 in. Hg; 0.0061N 2 ----------------------- = 10.71 Total 7.702N 2 71.5% 10.0N 2 + 2.047O 2 = 15.11 Total Dew Point = 126°F 19.14 The proximate analysis of a coal is: moisture = 4.33%, volatile matter = 40.21%, fixed carbon = 45.07%, and ash = 10.39%. The heating value was determined as 29 000 kJ/kg (12,490 Btu/lb). Find the ASTM rank of the coal. Prox.: M = 4.33 ; VM = 40.21 ; FC = 45.07 ; A = 10.39 ; HV = 12 ,490 Btu/lb Change to Dry: VM = 42% ; FC = 47.1% ; A = 10.9% Class II. Bituminous, Group 5 High-Volatile C Bituminous Coal From Table 7, pg. 28.8, 2009 HBF This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto With 30% excess air -----------------------1.047CO 2 9.7% Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Chapter 19—Heating Equipment⏐211 19.15 A fuel oil shows an API gravity of 36. Calculate the specific gravity at 60/60°F and the pounds per gallon of fuel. Estimate the ASTM Grade. [Ans: 0.845, 7.05, No. 2] Eq. (2), pg. 28.7, 2009 HBF 141.5 Degrees API = -------------------------------------- – 131.5 S.G. at 60/60°F 141.5 36 = -------------------------- – 131.5 ; S.G. 60/60°F = 0.845 S.G. 60/60°F at 60°F: ν H ρ fuel oil 2O = 0.01603 ft 2 /lb ; ρ = 62.383 lb/ft 3 = 0.845 ( 62.383 ) = 52.714 lb/ft 3 M = 52.714 lb/ft 3 × 0.13368 ft 3 /gal = 7.047 lb/gal Grade No. 2 19.16 A representative No. 4 fuel oil has a gravity of 25° API and the following composition: carbon = 87.4%, hydrogen = 10.7%, sulfur = 1.2%, nitrogen = 0.2, moisture = 0, and solids = 0.5%. a. b. Estimate its higher heating value. Compute the mass of air required to burn, theoretically, 1 gallon of the fuel. lb a ⁄ gal fuel = ( 13.82 ) ( 7.5 ) = 103.6lb a ⁄ galf This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 141.5 25 = ------------------------ – 131.5 S.G./60°F Eq. (2), pg. 28.7, 2009 HBF S.G./60°F = 0.904 No. 4 fuel oil, 25°API C = 87.4; H = 10.7; S = 1.2; N = 0.2; Solids = 0.5 a. HHV, Btu/lb = 22 ,320 – 37 ( S.G. ) = 18 ,903 Btu/lb Eq. (3), pg. 28.7, 2009 HBF Table 6: No. 4 1 gal = 7.5 lb ; HV = 145 ,000 Btu/gal b. A ⁄ F = 0.0144 ( 8C + 24H + 3S – 30 ) Eq. (6), pg. 28.10, 2009 HBF = 0.0144 [ 8 ( 87.4 ) + 24 ( 10.7 ) + 3 ( 1.2 ) ] = 13.82 lb a /lb f Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. 212⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual 19.17 The following data was taken from a test on an oilfired furnace: Fuel rate = 20 gal oil/h Specific gravity of fuel oil = 0.89% by mass Hydrogen in fuel = 14.7% Temperature of fuel for combustion = 80°F Temperature of entering combustion air = 80°F Relative humidity of entering air = 45% Temperature of flue gases leaving furnace = 550°F a. b. Calculate the heat loss in water vapor in products formed by combustion. Calculate the heat loss in water vapor in the combustion air. [Ans: 1672.5 Btu/lb (3888 kJ/kg), 29.4 Btu/lb (68.3 kJ/kg)] a. b. 9H 2 q 3 = ---------- ( h ) tg – ( h f ) ta Eq. (19), Chap. 28, 2009 HBF 100 ( 9 ) ( 14.7 ) = ----------------------- ( 1312.2 – 48.05 ) = 1672.5 Btu/lb f 100 q 4 = Mw a [ ( h ) tg – ( h f ) ta ] Eq. (20), Chap. 28, 2009 HBF = ( 0.0098 ) ( 13.89 ) ( 1312.2 – 1096.12 ) = 29.4 Btu/lb 19.18 An office building requires 2901 MJ (2.75 × 109 Btu) of heat for the winter season. Compute the seasonal heating costs, if the following fuel is used: a. Assume that the conversion efficiency is 75% for the oil and 61% for the coal. a. 2 ,750 ,000 ,000 Btu ----------------------------------------------------------------------------------- × $70.00/ton = $11, 680 coal 13 ,500 Btu/lb × 2000 lb/ton × 0.61 b. 2 ,750 ,000 ,000----------------------------------× $2.75/gal = $73 ,067 oil 138 ,000 × 0.75 This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto b. Bituminous coal; 31 380 kJ/kg (13,500 Btu/lb); $70.00 per ton No. 2 fuel oil, 38 500 kJ/L (138,000 Btu/gal); $2.75 per gallon. Copyrighted material licensed to University of Toronto by Thomson Scientific, Inc. (www.techstreet.com). © (2009), American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. (www.ashrae.org). For personal use only. Additional reproduction, distribution, or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. a.009.= 0.0067 – 0.2°C).0004 lb ν /lb a 14. 50% RH. Inc. w = 0. b. Pv = 0.242 .0067 – 0.0094 – ---------------. .6 r: 0.24 ( 110 – 65 ) = 10. American Society of Heating. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.127 cfm.6 Btu/lb gain by air q L = ( 0. and has a low-pressure steam heating system.240 [ 70 – ( – 1 ) ] + 1060 ( 0. Outside design conditions are −1°F dry bulb.5(1. A makeup air system is being added to the plant and it has been decided that the input air should be 10.380 lb da /h a.7 – 0.127 ( 60 ) ⁄ 11.org).009 70°F db.Chapter 19—Heating Equipment⏐213 19.754 × 10–2) = 0. it leaves the furnace at 110°F dry bulb (43. What are the total steam requirements for the heating coil and the humidifier? What capacity should the humidifier have in pounds of water per hour? Copyrighted material licensed to University of Toronto by Thomson Scientific.0004 ) } = 1 .622 -----------------------------. Refrigerating and Air-Conditioning Engineers.300 Btu/h m H = 52 . 60% RH. For personal use only. a.3°C) and 0.20 A plant is maintained at 70°F dry bulb.= 0.com).00543 lbv /lba (0.0054 )h f = 0 g 19.19 Saturated air at 41°F dry bulb (5°C) enters a furnace.3 kJ/kg)] a. [Ans: qs = +6 Btu/lb (14.00543 kg/kg) and circulates through a factory.0094 m w = 325 ( 475 ) ⁄ 1100 = 140. 50% RH.techstreet.380 q = 52 . Inc. (www.24 ( 110 – 41 ) = 16. see (a) Across furnace: q s = 0.0054 )1060 = 7 Btu/lb gain g b. 140.380 ( 0. ql = +7 Btu/lb (16.0067 lb v /lb a 52 .3°C) and 63°F wet bulb (17.ashrae.012 – 0. Air leaves the factory at 65°F dry bulb (18. q s = mc p Δt = 0.6 = 52 .380 { 0.0 kJ/kg). Additional reproduction. distribution. b.009 w = 0. What is the sensible and latent heat change for the air passing through the factory? State whether the air gains or loses sensible and latent heat during each process. gains latent. 60% RH.0054 – 0. The plant is 250 by 560 ft and normally has 325 people working per shift.8 Btu/lb loss q L = m ( Δw )h f = ( 0.0004 ) = 330 lb/h This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto b. © (2009). Across space: loses sensible. Makeup air system for ventilation: OA: –1°F db.3 w s = 0. (www. ν = 11.3 lb w /h lightbench work m a = 10 . from catalog: ∼ 250 W/ft = 852 Btu/h/ft 55 .23 For the residence of Problem 19-22.000 = 1. 170°F outlet q s = 490 ( gpm ) Δt . b.org).22 A residence with a design heating load of 16 kW (55. Return air to the furnace is at 22.= 16. (www. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.10 ( cfm ) ( 135 – 72 ) Airflow = 1280 cfm 19. Copyrighted material licensed to University of Toronto by Thomson Scientific. electric baseboard units replace the hot water system. a. The baseboard units house copper tubing with aluminum fins and operate with the inlet air temperature at 18.techstreet. b.10 ( cfm ) ( t s – t r ) . American Society of Heating. (www.com). 55 . This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto a. 89 . kW Total length of baseboard units 55 . Inc.ashrae. e. c.5 ft 852 a. distribution.6 L ≅ 90 ft from catalog for t av = 180°F 19. Ventilating. Specify the following: This problem requires catalog data. 135°F q s = 1. Total rating of electric system. c. For personal use only. Specify This problem requires catalog data. Refrigerating and Air-Conditioning Engineers. a.000 Btu/h) is to use an oil-fired warm air system with forced circulation. © (2009).2°C (72°F). Inc.1 = 16 kW 3413 b. and Air Conditioning—Solutions Manual 19. b. b. Supply air temperature Airflow rate Make and catalog number of suitable furnace a. a.000 = 490 ( gpm )20 c. b.000 Btu/h) is to use a forced circulation hot-water baseboard radiator system. Specify the following: This problem requires catalog data.214⏐Principles of Heating.000 L ≅ ---------------. Additional reproduction. Hot water inlet temperature and outlet temperature Total water flow rate Total length of radiator panel for house Location of panels Make and catalog number of suitable hot water heater 190°F inlet. gpm = 5.21 A residence with a design heating load of 26 kW (89. d.3°C (65°F). .= 64.000 P = ---------------. 000 Btu/h).org). Supply air temperature Airflow rate q s = 1. Copyrighted material licensed to University of Toronto by Thomson Scientific.Chapter 19—Heating Equipment⏐215 19. Specify a. Specify the following: This problem requires catalog data. b. Inc.3°C (74°F). Hot water inlet temperature and outlet temperature Water flow rate Length of radiator panel a.000 = 490 ( gpm )20 gpm = 6. American Society of Heating. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.25 A large classroom has a winter design load of 26 kW (89. A forced circulation warm air system is to be used with return air at 23.3°C (65°F). (www. distribution. b.ashrae. (www.9 kW (68. 170°F outlet q s = 490 ( gpm ) Δt .com).000 Btu/h) with installed forced circulation hot-water baseboard radiators. b.000 = 1. Inc.24 A large classroom has a winter design heat loss of 19. The baseboard units house copper tubing with aluminum fins and operate with the inlet air temperature at 18.10 ( cfm ) ( 135 – 74 ) Airflow rate = 1330 cfm This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto a. 89 . © (2009).techstreet. For personal use only. Refrigerating and Air-Conditioning Engineers. a. c. . ∼ 600 Btu/h/ft 68 .95 ∼ 7 c. from catalog: at t av = 180°F.000 L = ---------------. 190°F inlet.10 ( cfm ) ( t s – t r ) t s = 135°F (selected) b. Additional reproduction. 68 .= 113 ft 600 19. Copyrighted material licensed to University of Toronto by Thomson Scientific. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Amount of air L/s (cfm). m st = -----g hf d. and Air Conditioning—Solutions Manual 19.5 ×10 ( 93 ) ( 235 – 60 ) = 0. Steam: qs q s = m st h f . Ventilating.26 For a heat loss from the space to be conditioned of Q. 4. boiler. Refrigerating and Air-Conditioning Engineers. 3. . Additional reproduction. distribution. –6 ΔL = αL Δt = 6. Hot air: qs · · q s = 1. (www. 5. b. 6. in watts. Select pipe sizes required.28 Compute the increase in length of 28. write the expression for determining a.5 ×10 –6 in.techstreet.6°C (60°F).org). Inc. which must be supplied if a steam heating system is used Size. which must be supplied if a hot air system is used Amount of hot water L/s (gpm). This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 1. d. Inc.. α = 6. P = 1000 q s 1000 g 19. Find ΔP/ft capacity of pump. Electric: P ( watt ) q s ( kW ) = ------------------./in.106 ft = 1. Select pump. Compute the heat required for each room or space Sketch runs. 19. Compute gpm for each circuit. of electric heaters required if electric heat is used a. (www.2 ( t s – t r ) b.3 m (93 ft) of steel steam pipe when the average steam temperature is 113°C (235°F) and the air is 21°C (70°F).ashrae. For personal use only.2V ( t r – t s ). 2. and convector locations. © (2009).com). c.27 List the steps taken when designing a forcedcirculation hot water heating system.216⏐Principles of Heating. The pipe was installed during a period when the temperature was 15. V ( l ⁄ s ) = -------------------------1.27 in. which must be supplied if a hydronic system is used Amount of steam kg/h (lb/h). American Society of Heating. Hydronic: · q s = m w c p ( t in – t out ) w = ρ w V w c p ( t in – t out ) w w · Vw w qs = ------------------------------------------ρ w c p ( t in – t out ) w w c. American Society of Heating. determine the required furnace size.000 Btu/h) output. The operating temperature is 93°C (200°F).000 × 1 + 20 .000 lb) of water. 2 2 2O + m s C s ) Δt q ( 1 ) = ( 15 .M. Inc.techstreet. when should the furnace be started to be up to the operating temperature of 93°C (200°F) by 7:30 A. γ = 5.com). Monday morning? [Ans: 2:16 A. The piping and boiler also contain 6810 kg (15. 7:30 A.M. Additional reproduction. For personal use only.12 ) ( 200 – 50 ) = 2 . Copyrighted material licensed to University of Toronto by Thomson Scientific.M. distribution. Assuming the system should be warmed up in one hour.org).610.29 The total mass of steel in the boiler and piping of a school’s heating system is 9080 kg (20. [Ans: 764 kW (2. Inc.000. the temperature of the system is 10°C (50°F). or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. b.22 h = 5 h – 13 min. © (2009).000 × 0.000 Btu/h 500 . Sunday] q a γ = m H O C H O Δt + m s C s Δt = ( m H O C H 2 a.000 Btu/h)] For a furnace size of 146 kW (500. After a weekend shut-down.M.Chapter 19—Heating Equipment⏐217 19. (www. – 5:13 = 2:17 A. Refrigerating and Air-Conditioning Engineers.610 .000 lb).610 . a.000 ( γ ) = 2 . b.ashrae. (www. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . Additional reproduction. (www. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.ashrae.techstreet. Refrigerating and Air-Conditioning Engineers. distribution. (www. American Society of Heating. Inc. © (2009).org).Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc.com). For personal use only. (www.Solutions to Chapter 20 HEAT EXCHANGE EQUIPMENT Copyrighted material licensed to University of Toronto by Thomson Scientific. Inc. Additional reproduction. distribution. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .ashrae. For personal use only.org).techstreet.com). © (2009). Inc. American Society of Heating. Refrigerating and Air-Conditioning Engineers. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. (www. For personal use only. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Additional reproduction. © (2009). Refrigerating and Air-Conditioning Engineers. American Society of Heating.ashrae. distribution.Copyrighted material licensed to University of Toronto by Thomson Scientific. (www. Inc. Inc.com).org).techstreet. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto . (www. Inc. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission. Additional reproduction.+ --------.AR di AR fo AR c A fi --------+ ---------. The shell-and-tube heat exchanger will use standard size copper tubes with a steel pipe as the shell. (www. distribution.Chapter 20—Heat Exchange Equipment⏐221 NOTE: All of the problems in this chapter are open-ended design problems and require the reader to make certain design assumptions.5 is provided below to illustrate the open-ended nature of these design problems. © (2009). (www. cf Q = UAFΔt m. 20.130 kg/s is to be condensed from saturated vapor to saturated liquid at a pressure of 0. Space limites the length of the exchanger to 2 m.ashrae. Refrigerating and Air-Conditioning Engineers. R-410A surrounds the tubes and evaporates under pool boiling conditinos at a pressure of 1.1 MPa.+ ----------. American Society of Heating.org).91 MPa as it flows through the tubes. There are a number of possible solutions.com). cf = ---------------------Δta ln ⎛ ---------⎞ ⎝ Δtb⎠ Saturated vapor → saturated liquid at 0.techstreet. For personal use only.5 Design the evaporator/condenser for a cascade lowtemperature refrigeration system using R-410A in the high temperature loop and R-22 in the low temperature loop.91 MPa R-22 at 293 K condensing This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto Governing Relations: Copyrighted material licensed to University of Toronto by Thomson Scientific. The exterior of the heat exchanger shell is to be well insulated. . cf = ----------------ΣR where: 1 U = -----------------------------------------------------------------------------------------------------------------------------------------do A ln ----A .+ --------------------------------------------2πkl Ai Ao A o h o ( A plane + φA fin ) hi Ai Δt a – Δt b Δt m. R-22 at the rate of 0.+ ---------------. A possible solution for 20. Q = m c ( h co – h ci ) = m h ( h hi – h ho ) FΔ m. Inc. 6 m/s 0.9 ( 175 ) ) ]10 – 6 = 159 × 10 – 6 Inside: 40.com).0034 1200 μ = [ 0.9 -----------.org).6 ) N T = 1.6 kJ/kg) = 24.or Δt avg = 293 – 283 = 10°C Δta ln ⎛ ---------⎞ ⎝ Δtb⎠ m i = N T ρA T V Select: 5--in.13 kg/s (187. .97 Two tubes per pass.000 × 159 × 10 – 6 ( 0.000 kW Δta – Δtb Δt m = -----------------------.00014 ( 1. For personal use only.00014 m 2 4 at x = 0. distribution. (www.1 ( 12.4 kW = 24. Refrigerating and Air-Conditioning Engineers.9 mm ) πD i2 A T = ---------.= 0. and Air Conditioning—Solutions Manual R-410a at 1.222⏐Principles of Heating. Inc. OD 8 ( D i = 13.000 = ------------μ Copyrighted material licensed to University of Toronto by Thomson Scientific. Ventilating. © (2009).ashrae. American Society of Heating.0034 ) N T = -------------------------------------------------0. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto D i VP Rev = 40.techstreet.= 0. (www.1 0. Inc.026 + 0.4 mm ) ( D o = 15. Additional reproduction. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.1 MPa evaporating at 283 K Q = m· h fg R – 22 = 0.4 + 0.0034 ) V = -----------------------------------------------------------------------0.0134 V = 1.13 kg/s × ( 0.1 Select: 1 v x = 0. 0159 ) × L T ( 10 )ΔT m ) L T = 3. Additional reproduction.7: h i = 26.00035 dirty 1 U clean = --------------------------------------.00018 + --------------30. © (2009).000 W/ ( m 2 ⋅ K ) Table 20.ashrae.Chapter 20—Heat Exchange Equipment⏐223 From Equation 20. (www.techstreet.000 = 1660 w/m 2 °C U Copyrighted material licensed to University of Toronto by Thomson Scientific. (www.6 m.= 0.26Pr l1 / 3 Re l + Rev ⎛ -----l ⎞ ⎝ ρ v⎠ ⎝ K 0.org).000 = 7800 ( 2 )π ( 0.8 ⎞ ⎠ h i = 30.7°C From Table 20. Inc. distribution.000 W/ ( m 2 ⋅ K ) R fo = 0.000 ( 20 – T s ) = 14. American Society of Heating. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto 1 U dirty = ------------------------------------------------------------------------------------------1 1 ---------------. . Inc.00018 Assume: ΔT = 5 h o = 14.7°C ΔT = 6. For personal use only.000 U avg = 7800 W/ ( m 2 ⋅ K ) ( Q = U o A o ΔT m = 24.000 26.00035 + 0. use two tubes for each pass where a tube is 5/8 in.000 26. OD.com).+ --------------30.6 2 Use two passes where a pass equals 1.7 → Rfi = 0.2 L max pass = ------. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.500 ( T s – 10 ) ) → T s = 16. 500 W/ ( m 2 ⋅ K ) ( 30.= 1. Refrigerating and Air-Conditioning Engineers.2m 3.000 W/ ( m 2 ⋅ K ) 1 1 ---------------.= 14.+ 0.33: ρ 1/2 i Di ⎛ h--------. Additional reproduction. distribution. For personal use only. Inc. This copy downloaded on 2016-03-20 16:05:39 -0500 by authorized user University of Toronto .org). (www.ashrae. American Society of Heating. Inc. or transmission in either print or digital form is not permitted without ASHRAE’s prior written permission.com). © (2009). Refrigerating and Air-Conditioning Engineers. (www.Copyrighted material licensed to University of Toronto by Thomson Scientific.techstreet.