Introduction to Transport Phenomena Momentum, Heat and Mass.pdf

Introduction to Transport Phenomena Introduction to Transport Phenomena Momentum, Heat and Mass Bodh Raj Formerly Professor Department of Chemical Engineering Seth Jai Parkash Mukand Lal Institute of Engineering and Technology (JMIT) Haryana New Delhi-110001 2012 INTRODUCTION TO TRANSPORT PHENOMENA—Momentum, Heat and Mass Bodh Raj © 2012 by PHI Learning Private Limited, Delhi. All rights reserved. No part of this book may be reproduced in any form, by mimeograph or any other means, without permission in writing from the publisher. ISBN-978-81-203-4518-8 The export rights of this book are vested solely with the publisher. Published by Asoke K. Ghosh, PHI Learning Private Limited, Rimjhim House, 111, Patparganj Industrial Estate, Delhi-110092 and Printed by Raj Press, New Delhi-110012. To my beloved students . 2 Flow through a Circular Tube (Gravity Flow) 18 2.1 Flow of a Liquid Falling Film 13 2.5 Flow of Two Immiscible Fluids 30 Problems 35 3 Equations of Change for Isothermal System 36–45 3.4 Flow through an Annulus 26 2.2 Convective Momentum Transfer 7 1.3 Shell Momentum Balances and Boundary Conditions 8 1.3 Prandtl Mixing Length Model 53 5 Unsteady-State Flow of Newtonian Fluids 55–58 5.1 Equation of Continuity 36 3.2 Boundary-Layer Thickness for Flow Near a Solid Surface 50 4.2 Equation of Motion (Navier–Stokes Equation) 38 3.1.2.3.1 Time-Smoothed Equation of Change for Incompressible Fluids for Turbulent Flow 48 4.3 Laminar Flow in a Narrow Slit 22 2.1 Flow Near a Wall Suddenly Set in Motion 55 Section B: Heat Transfer .1 Time-Smoothed Equation of Change for Turbulent Flow (Incompressible Fluid) 47 4.CONTENTS Preface xi Section A: Momentum Transfer 1 Introduction to Momentum Transfer 3–12 1.1 Newton’s Law of Viscosity (Molecular Momentum Transfer) 5 1.1 Navier–Stokes Equations 42 Solved Examples 43 Problems 45 4 Momentum Transfer in Turbulent Flow 46–54 4.1 Boundary Conditions 9 Solved Examples 10 Problems 11 2 Shell Momentum Balance and Velocity Distributionin Laminar Flow (Typical Cases) 13–35 2. 3 Shell Mass Balances and Boundary Conditions 122 Solved Examples 124 Problems 127 12 Shell Mass Balances and Concentration Distribution forLaminar Flow 128–146 12.6 Heat Transfer 61–67 6.3 Diffusion with a Heterogeneous Chemical Reaction (Slow Reaction) 136 .3 Heat Conduction through Composit Walls 76 7.2 Convective Mass Transport 120 11.1 Fick’s Law of Binary Diffusion (Molecular Mass Transport) 117 11.1.2.4 Heat Conduction in a Cooling Fin 79 7.2 Diffusion with a Heterogeneous Chemical Reaction (Instantaneous Reaction) 133 12.3 Shell Energy Balances and Boundary Conditions 65 Solved Examples 66 Problems 67 7 Shell Energy Balances and Temperature Distributionin Heat Conduction in Solids (Typical Cases) 68–98 7.3 Prandtl Mixing Length Model in Heat Transfer 109 10 Unsteady-State Heat Conduction in aSemi-Infinite Slab 111–113 Section C: Mass Transfer 11 Mass Transfer 117–127 11.1 Diffusion through a Stagnant Gas Film 129 12.1 Mass and Molar Fluxes 121 11.2 Boundary-Layer Thickness for Heat Transfer Near the Solid Surface 105 9.1 Time-Smoothed Equations for Energy for Incompressible Fluids in Turbulent Flow 103 9.2 Heat Conduction with a Nuclear Heat Source 72 7.1 Heat Conduction with an Electrical of Heat Source 69 7.6 Heat Conduction with a Viscous Heat Source 85 7.1 Fourier’s Law of Heat Conduction (Molecular Energy Transport) 61 6.7 Heat Conduction with a Chemical Reaction Heat Source 88 Solved Examples 94 Problems 97 8 The General Energy Equation 99–102 8.2 Convective Energy Transport 63 6.1 Some Features of Fick’s Law of Diffusion 119 11.1 Special Cases for Energy Equation 102 9 Temperature Distribution in Turbulent Flow 103–110 9.5 Heat Conduction from a Sphere to a Stagnant Fluid 82 7. 4 Von Kármán Analogy 182 16.2 Reynolds Analogy 173 16. Heat andMass Transfer 165–190 16.1 Time-Smoothed Concentration in Turbulent Flow 150 14.1 Analogy among Momentum.5 Diffusion through a Spherical Stagnant Gas Film Surrounding a Droplet of Liquid 141 Solved Examples 143 Problems 145 13 The General Equation of Diffusion 147–149 14 Concentration Distribution in Turbulent Flow 150–158 14.2 Boundary-Layer Thickness for Mass Transfer 152 14.5 Chilton–Colburn Analogy 184 Solved Examples 187 Problems 190 Appendices Appendix A: Conversion Factors and Fundamental Units 193 Appendix B: Gas Law Constant R 198 Appendix C: Properties of Water (Liquid) 199 Appendix D: Properties of Liquids 202 Appendix E: Properties of Gases 204 Appendix F: Properties of Solids 207 Appendix G: The Equation of Continuity 209 Appendix H: Equation of Motion for a Newtonian Fluid withConstants r and m 210 Appendix I: The Equation of Energy for Pure Newtonian Fluidswith Constants r and k 212 Appendix J: Fick’s (First) Law of Binary Diffusion 213 Appendix K: The Equation of Continuity for Species a in Terms of ja 215 Index 217–218 .3 Prandtl Analogy 177 16.12. Heat and Mass Transfer 16 Analogies among Momentum.4 Diffusion with a Homogeneous Chemical Reaction 138 12.3 Prandtl Mixing Length Model in Mass Transfer 156 15 Unsteady-State Evaporation of a Liquid 159–161 Section D: Analogies among Momentum. Heat and Mass Transfer 166 16. In other words. we can understand mass transfer or heat transfer from momentum transfer. By analogies. But there are only a few books on the market which cover all the topics together. heat transfer and mass transfer by flux expressions. sedimentation. So. The subject has grown rapidly in the recent past. The one-dimensional equations evolved are taken up in all the examples.e. if the reaction in the reactor takes place at high temperatures (generally that happens). This subject. liquid–liquid extraction. The partial differential equations so obtained are solved using the boundary conditions. Another approach adopted is to write shell balances for these transfer processes. Heat transfer occurs in conduction and in convection transfer of heat during evaporation and drying. The mass transfer operations such as distillation. care has been taken to use simple mathematical expressions and also to provide complete solutions of the problems. i. biotechnology. fluid flow and filtration. This book caters to the needs of the introductory level. After the reaction. crystallization and adsorption. The subject of transport phenomena covers mainly three aspects: momentum transfer (also known as fluid dynamics). Moreover. the separation processes take place with a view to isolating the product in pure form. the subject of transport phenomena should be studied together on account of the following reasons: 1. heat and mass are also developed. Then. all these three transport phenomena take place. we can understand one transport phenomenon from another. 2. absorption. biology. There are many books and literature available which cover these three topics separately. in many universities and engineering colleges including Indian Institutes of Technology. the governing mathematical equations are similar in nature. this subject is taught at two levels. absorption or crystallization may also take place. In the field of chemical engineering. whereas mass transfer takes place in operations such as distillation. chemical reactions. the fluid flow phenomenon comes into the picture.Preface “Transport Phenomena” is a subject of interest to many scientists and engineers in their respective fields of study. why do we need to study these three phenomena together? There are reasons for it. For example. An attempt has been made in this book to explain the phenomena of momentum transfer. So. Let us take a simple process where the raw materials are transported from the storage vessel to the reactor. Students are expected to have a good knowledge of . Also. apart from being of prime interest to chemical engineers. Generally a question is asked. nanotechnology and micro-electronics. The mechanisms of these three transport phenomena are closely related. The expressions for the conservation of momentum. Since it is an introductory book. It is not possible to cover all the aspects of the subject in one book of this size. this book covers only the introductory part to the subject. heat transfer and mass transfer. So. the fluid dynamics or momentum transfer occurs in industrial operations such as mixing. introductory level and advanced level. is also increasingly gaining popularity and application in the fields of agriculture. then the raw materials have to be heated. In this chapter. The general equation of continuity and the equation of motion are explained in Chapter 3. namely the Newton’s law of viscosity. von Kármán and Chilton–Colburn are explained in this chapter.differential equations and methods of integration. which covers the Fick’s law of diffusion. . These problems are solved by taking shell energy balances. Chapter 2 deals with the typical cases for shell momentum balance in order to obtain the velocity distribution and momentum flux for laminar flow conditions. Simple problems for mass transfer are discussed in Chapter 12 by taking shell mass balances for laminar flow conditions. It explains the Fourier’s law of heat conduction. Section C: Mass Transfer (Chapters 11 to 15) The governing law of molecular mass transfer is explained in Chapter 11. Constructive suggestions from readers. In this chapter. convective energy transport and shell energy balances are also discussed. Section A: Momentum Transfer (Chapters 1 to 5) In Chapter 1. Since there are many similarities in molecular transport of momentum. Head and Mass Transfer. The various models like the Prandtl mixing length are also covered in this chapter. convective mass transfer and shell mass balances are also covered. The temperature distribution in turbulent flow conditions is dealt with in Chapter 9. This book is an outgrowth of my interactions with the students during class work and tutorials. teachers. A specific case of diffusion for the unsteadystate condition is explained in Chapter 15. Simple illustrations and mathematical tools have been adopted to explain the transport phenomena in order to fulfil the needs of the students. Momentum transfer in turbulent flow conditions is explained in Chapter 4. This book is divided into four sections: Section A: Momentum Transfer Section B: Heat Transfer Section C: Mass Transfer Section D: Analogies among Momentum. Section D: Analogies among Momentum. The mechanism of momentum transfer is explained. The unsteady-state flow of Newtonian fluids is explained in Chapter 5. The general energy equation is developed in Chapter 8. students and well wishers to improve this first edition of the book will be gratefully acknowledged. Different analogies like Reynolds. Section B: Heat Transfer (Chapters 6 to 10) The mechanism of molecular heat conduction is discussed in Chapter 6. Specific problems of heat conduction in solids are highlighted in Chapter 7. This introductory book on transport phenomena will go a long way to cater to the needs of the students. the basic law of momentum transfer. analogies among these three transport processes are also demonstrated to understand them better. heat or mass. heat and mass transfer are discussed in Chapter 16. Heat and Mass Transfer (Chapters 16) The mechanism and governing equations for momentum. The unsteady-state case of heat transfer in conduction is discussed in Chapter 10. The shell momentum balance for momentum transfer is also covered. Prandtl. is covered. Mass transfer in turbulent flow condition is discussed in Chapter 14. The general equation of diffusion is explained in Chapter 13. com [email protected] RAJ E-mail: dr. Section A MOMENTUM TRANSFER . how does a particular phenomenon take place? Second.e.1 INTRODUCTION TOMOMENTUM TRANSFER The matter consists of a multitude of extremely small particles called molecules.e. solids. i. the unit of momentum is given by momentum = mass velocity = kg · m/s whereas that for momentum flux is given by· momentum flux = rate of momentum per unit area Similarly. The heat and mass transfers will be discussed in detail in subsequent chapters. therefore. In gases. write a general statement such as rate of transfer process = . the driving force is the difference in velocities. . the transfer of heat energy takes place because of temperature gradient and mass transfer takes place because of concentration gradient. the molecules are arranged in a packed fashion. it carries some momentum with it as the mass of the molecule moves with a certain velocity in a particular direction. When the molecule of a fluid moves. The scientist studies the properties of matter at the molecular level. we generally ask two questions. Now. One. molecules move because of velocity gradient. Momentum. liquids and gases. heat and mass transfer phenomena take place by two means: . why does it take place? The same is the case when we deal with momentum transfer. In solids. whereas in engineering applications we are concerned with the study of the bulk or macroscopic behaviour of solids. Why does the molecule move? It moves because of velocity gradient in the system. i. While studying engineering applications. liquids and gases rather than their molecular behaviour. . We can. the molecules have more free space between them compared to the molecules in liquids. Thus. The behaviour or the movement of the matter can be studied at the molecular level.1) In momentum transfer. (1. microscopically or it can be studied macroscopically at the bulk level. and matter may be divided into three different classes of substances or states. In Mode II. A similar type of mechanism is simulated in a molecular transfer process. each man is allowed to fill up the bucket with water from the pond. In Mode I. Convective transfer Now. Figure 1.e. i. it is applicable to momentum. Let us also assume that there is a water pond in the vicinity. This mechanism is called molecular transfer. heat and mass transfer. and so on. suppose there is a fire to be extinguished at a particular place. Molecular transfer 2. Such a mechanism is called convective transfer.1 Mechanism of momentum transfer. men are stationary but the bucket of water is moving from one man to another. Here all men remain stationary but the bucket of water keeps moving. Mode I: Allow all men to stand in a queue and let the first person fill up the bucket of water and pass it on to the next person. In this way.1. In Mode III. In molecular transfer. It is similar to each molecule moving with the energy. Men are moving carrying the buckets of water. each man is moving with a bucket of water. Mode II: In this case. It is applicable to phenomena of momentum. heat and mass transfer. a medium is required to transport . where water is sprayed from the pond to the fire place. water will be transported to the fire place and used to extinguish the fire. let us discuss these two mechanisms in detail. As shown in Figure 1. Both the mechanisms of Modes I and II will be discussed further in subsequent sections. Here both men and buckets of water remain moving. Mode III: We install a pump to spray water from the pond onto the fire. the molecules remain stationary but the energy is transferred from one molecule to another. Each man carries the water bucket to the fire place.1. Here again. There are three different modes available to us to extinguish the fire. here the molecules and the energy move together. Let us analyse the above mechanisms in a scientific way. Let F be the force r equired to maintain the motion of the lower plate and y be the distance between the plates. Figure 1. We select the x–y coordinates as shown in Figure 1. The viscosity of fluids. Under laminar flow conditions. Let us simulate this situation.2. the mechanism of transport is accomplished by two means: 1. At steady-state condition. In summary. the velocity distribution becomes linear as shown in Figure 1. 1. the velocity of the fluid settles down. Convective transfer We will now discuss the phenomenon of molecular momentum transfer and subsequently that of the convective momentum transfer. The Newton’s law of viscosity will tell us the definition of viscosity. This will introduce “Newton’s law of viscosity”. both the plates are at rest. It is one of the modes used for heat transfer but not much important for momentum and mass transfer.energy from one place to another. It has been experimentally found that the force F can be expressed as F A F v F Combining these factors. The viscosity is a physical property of a fluid. The lower plate is set in motion suddenly with a constant velocity v in the x-direction. when the fluid is flowing between two parallel plates. gases and liquids varies with temperature and pressure. We know that oils are more viscous than water.2.1 NEWTON’S LAW OF VISCOSITY (MOLECULAR MOMENTUM TRANSFER) Let us discuss the basic molecular momentum transfer process. As the time passes. Molecular transfer 2. we get .2 Laminar velocity distribution: flow between two plates. Consider the lower plate to be moving with a velocity v and suppose the area of the plates is A. Initially. molecules of a fluid move in layers that slide past one another in a orderly fashion. Such a mechanism is called radiation. we can designate v/y in the differential form as –dvx/dy. (1. . At the neighbourhood of a solid surface (i. Let us summarize the molecular fluxes in Cartesian coordinates as shown in Table 1. . (1. SI units of viscosity n can be arrived at as follows.or . (1. which do not obey this law be called non-Newtonian fluids.2) in symbol form as . which is defined as the force per unit area required perpendicular to the ydirection. . especially polymers. It is called shear stress. . Newton suggested that those fluids which obey the law of viscosity be called Newtonian fluids. . It is a fluid property. the fluid adjacent to the solid acquires x-momentum in the positive direction of y. Now. we write F/A = xyx.4) where = o = kinematic viscosity.1 Molecular Fluxes in Cartesian Coordinates .e. (1. y = 0).1. etc. So xyx is also called xmomentum in the positive direction of y. Table 1.3) where xyx = shear stress and = velocity gradient Equation (1. Let us interpret the behaviour of this concept. Further. The negative sign indicates that velocity decreases in the positive direction of y. And those fluids. . We have xyx = [N/m2 = Pa] vx = [m/s] y = [m] We can write another form of Newton’s law of viscosity as . We can now write Eq.2) where n is a proportionality constant called the viscosity of the fluid. this law is called “Newton’s law of viscosity”. Since the constant of proportionality is called viscosity.3) states that the shear stress is proportional to the negative of the velocity gradient. we get momentum fluxes vy t v and vz t v. we can easily apply the momentum balances. z with vector velocity v. we will not be discussing the estimation of viscosity. pressure. we have to consider three mutually perpendicular planes through x. average velocities. The summary of the convective momentum flux components is given in Table 1. However. When a fluid enters the space of coordinates x. z. Let the fluid enter with a velocity vx. Viscosity of fluids. we will study only the steady flow conditions in subsequent chapters. can be easily worked out. It can be estimated by the following methods: 1. Leonnard–Jones potentials 3. Similarly for the velocity in y. the unsteady flow problems can be solved separately.and z-directions. is affected by temperature and pressure. whereas the viscosity of a gas at low pressure increases with increasing temperature. y. For liquids. Once we know these fluxes. then multiplying by the area perpendicular.x-component Flux y-component z-component y xxx xyx xxy xyy xxz xyz z xzx xzy xzz Direction of Velocity x The viscosity is a fluid property. the viscosity decreases. therefore. y. then the momentum flux per unit volume is vx t v. This is well discussed in other chemical engineering books and handbooks.3 SHELL MOMENTUM BALANCES AND BOUNDARY CONDITIONS Once we know the molecular and convective momentum fluxes. obtained from critical values 2. the other engineering problems such as shear stress. Once the velocity profiles are known. density and velocity do not change with time. we can calculate the forces. In this book.1. 1. Under steady flow conditions. liquids or gases. Table 1. Let us. Since momentum is . as the temperature increases. momentum can also be transported by the bulk flow of the fluid. Reduced temperature–pressure and reduced viscosity.2 Convective Momentum Flux Components Velocity Direction x y z Flux of Momentum Convective Momentum Flux Components Perpendicular to x-component y-component z-component the Surface t vx v t vx vx t vx vy t vx vz t vy v t vy vx t vy vy t vy vz t vz v t vz vx t vz vy t vz vz 1. we discussed momentum transfer by the molecular way. Each of these planes is taken as unit area. Here.2 CONVECTIVE MOMENTUM TRANSFER In Section 1. In addition. write the general conservation of momentum for steady flow. etc. This process is called convective transfer. we develop the velocity distribution in a particular case. Molecular theory of liquids In this book.2. whether (x. Apply a momentum balance over the shell and to the area perpendicular to the variable velocity selected. for example. Integrate the differential equation to get the velocity distribution. 3. There are two methods to evaluate it. i. 9.3. then vy = 0. 5. Get the velocity distribution equation. Cartesian. 4. 7. vz = 0 but vx is considered. Either solve the constant here itself from physical concepts or carry it forward to solve it later.being transferred. the term momentum transport is also usually used. as shown below. Identify the coordinate system. Apply the Newton’s law of viscosity and obtain a differential equation for the velocity distribution. the other parameters such as average velocity. z. etc. Apply the boundary conditions to evaluate the constant values. Integrate the differential equation formed to get the momentum flux distribution. Thus. It will appear with a constant of integration. maximum velocity. The following concepts are commonly used in momentum transport: . Formulate the differential equation for the momentum flux by considering the shell thickness approaching zero. gravity) = 0 (1. i) or (r. if the flow is in the x-direction. Identify the non-vanishing velocity component.5) The following procedure is adopted for the shell momentum balance: 1. z) or (r. For example. for example. 8. Rate of “momentum in” by molecular transport – Rate of “momentum out” by molecular transport + Rate of “momentum in” by convective transport – Rate of “momentum out” by convective transport + External forces acting (e. z). 2. Select the thin shell over which the momentum balance is to be applied. for example. shear stress. we have to use the boundary conditions from physical concepts. y. 11. can be easily evaluated. 10. From the velocity distribution. we will get the first differential equation for the momentum flux as 6. cylindrical or spherical depending upon the system of the problem. 1.g.1 Boundary Conditions In order to solve the differential equations formed for velocity and momentum flux. There will be a constant of integration appearing now. Liquid–liquid interface: In such cases. How far apart should the two plates be placed so that the shear stress x is 0. 3. xyx = 0. so the shear stress at the gas–liquid interface will be zero. 2. Gas–liquid interface: Since the viscosity of the gases is low compared to that of the liquids. The velocities of both the liquids will be the same at the interface. vx = 0.3 N/m2? Also calculate the shear rate. Shear stress. shear rate. Solution: The relative velocity of two plates. the velocity of the liquid will be equal to the velocity of the solid at the interface. This situation will be analysed later while solving flow problems in viscometer.9142 10–3 kg/m· s Let y be the distance between the two plates. the liquid velocity will be zero.4 m/s Viscosity of water at 24°C.00122 m = 0.3 N/m2 Applying the Newton’s law of viscosity.9142 cP = 0.1 There are two parallel plates some distance apart. SOLVED EXAMPLES EXAMPLE 1. . The lower plate is being pulled at a constant velocity 0.122 cm Thus. Solid–liquid interface: If the solid surface over which the liquid flows is stationary. there is no slip between the two liquids at the interface. water is used at 24°C. n = 0. Also the momentum flux will be the same for both the liquids at the interface. Between the plates.1. or or or = 0. In other cases.4 m/s faster relative to the top plate. then at the solid–liquid interface. 3. Given: f/k = 190 K v = 3. The lower plate is stationary and the upper plate requires a force of 100 N to keep it moving with a velocity of 2.00914 m and the lower plate is being pulled at a constant velocity 0.069 kg/m · s. or shear rate = PROBLEMS 1. Assuming a linear velocity distribution.EXAMPLE 1. R = 7 cm Wall shear stress = 16.069 kg/m· s Applying the Newton’s law of viscosity. determine the dynamic viscosity of oil in poise. The oil film between the plates has the same viscosity as that of the oil at the surface of contact.00914 m The relative velocity. Calculate the shear stress and the shear rate. vx = 0. Determine the ratio of eddy viscosity to molecular viscosity at a distance 0.996 n = 1.548 where f and n are Leonnard–Jones parameters and K is temperature in kelvin.2 The distance between two parallel plates is 0. y = 0. The fluid filled between the plates is glycerol at 293 K having a viscosity 1.5 m/s. 2.366 m/s faster relative to the top plate.2 10–5 kPa . Predict the viscosity of CO2 at 200 K and 1 atm.5 mm apart. Two square plates with each side 60 cm are spaced 12.366 m/s The viscosity of glycerol at 293 K = 1. Solution: The relative distance between the plates.6R from the wall for water flowing at a steady rate in a long smooth round tube under the following conditions: Tube radius. Two parallel plates are 0. A Newtonian fluid flows between two parallel plates at rest initially.77 cP. The lower plate moves at a velocity 10 cm/s and the upper plate is stationary.001 m. Compute the steady-state momentum flux when the lower plate velocity is 1 m/s in the positive x-direction and the plate separation is 0. The fluid between the plates is ethyl alcohol at 273 K having a viscosity of 1. Calculate the shear stress and the velocity gradient.7 cP. Fluid viscosity is 0.5 cm apart. 5. Assume linear velocity distribution.Density = 1000 kg/m3 Kinematic viscosity = 10–6 m2/s 4. . then the practical problems can be solved. Next. the other parameters like maximum velocity.1 FLOW OF A LIQUID FALLING FILM Let us consider that a liquid is flowing over a vertical flat plate of length L and width W as shown in Figure 2. we will apply the Newton’s law of viscosity to molecular momentum flux.2 SHELL MOMENTUM BALANCEAND VELOCITY DISTRIBUTIONIN LAMINAR FLOW (TYPICAL CASES In this chapter. Flow in a narrow slit is discussed in Section 2. The unsteady-state flow equations are solved separately in Chapter 5.3. Once the simple problems are understood. will be evaluated. The examples in this chapter are considered “idealized problems”.2 deals with the flow through a circular tube. we will obtain the velocity profile. 2. These are basically force balances over the shell considered. The flow in an annulus (Section 2. In Section 2. Then.1. the problem of liquid falling film has been solved.4) and the flow of two immiscible fluids (Section 2. average velocity. In all the cases. we will learn how to make shell momentum balances for laminar flow conditions.5) give an idea of solving the problems with different boundary conditions. The tools to solve these problems should be understood thoroughly. steadystate conditions are assumed. Here. only the gravity flow is considered. Section 2. An attempt has been made to solve the viscous flow problems first. Only the one-dimensional system flow has been selected purposely to make the treatment easy.1. Later. Such cases are seen across many chemical processes such as: (a) wetted wall columns (b) evaporation of liquids (c) gas absorption in liquids (d) surface coatings . 9. 8. 5. 2. viscosity n. The thickness of the film of the liquid is d and is very small compared to the length and width of the plate.2.Figure 2. Laminar flow of the liquid. 6. remain constant. The fluid is Newtonian. The flow of liquid is under gravity only. d L and d W. xzz = 0 and xyz = 0. vy = 0. 3. The z-momentum considered will be as follows: Molecular momentum flux = xxz i. liquid flow is in the z-direction only. i. (2. . Let us consider the liquid film of thickness d as shown in Figure 2. No end effects. Liquid properties like density t. The assumptions made in the analysis of flow of a liquid over a vertical flat plate are as follows: 1. Steady-state conditions. we obtain: Rate of molecular “momentum in” across the surface at x – Rate of molecular “momentum out” across the surfaceat at x + x + Gravity force acting on the fluid in the z-direction = 0 . Let the density t and the viscosity of the fluid n be constant.e. Unidirectional flow.e.1 Flat plate.1) . .e. No convective momentum is considered. The disturbances of the flow of liquid at the edges (i.e. Let us consider that a liquid is forming a film over the flat plate. 4. z = 0. Applying a momentum balance at the shell. Velocity = vz Let us consider a shell of the liquid at a distance x and of thickness x. etc. Consider that the flow of the liquid is in the z-direction only. z = L) are neglected. i. vx = 0. so the velocity vz is considered. 7. we get . (2.2).3) and (2.3) Gravity force acting on the fluid in the z-direction = WL tx g .5) by WL x and taking the limit x 0. (2. .6) We know from calculus that Hence we get from Eq. (2. (2. .5) Dividing Eq. . (2. .2 Liquid falling film. (2.4) in Eq. .Figure 2. . . . .2) Rate of molecular “momentum out” across the surface at x + x = (W L) xxz|x+x . (2.1). .6). (2. (2. we get .4) Substituting the values of Eqs. Rate of molecular “momentum in” across the surface at x = (W L) xxz|x . . (2. 9). Integrating it. x = 0 and the shear stress is zero.8).7). (2. Boundary condition 1: At x = 0. . . . . . hence the liquid is also stationary at the surface. xxz = 0. . becomes xxz = t g x . . C1 = 0 Equation (2. i. . . we have xxz = t gx + C1 .8) where C1 is a constant of integration. since the plate is stationary. .13) Substituting this value in Eq. vz = 0 . (2. . We know that at the solid surface of the flat plate. . (2. (2. Physically we know that at x = 0.7) Integrating both sides of Eq. which is evaluated by the boundary condition. . Therefore. we obtain . we get velocity distribution as . we have . Boundary condition 2: At x = d.12). the fluid velocity is zero. i. at the solid surface of the plate. (2. (2.10) where n is the viscosity of the liquid.11) This equation is the differential equation for the velocity distribution. the fluid velocity is maximum and at x = d. .9) Applying the Newton’s law of viscosity to the left side of Eq. at the gas–liquid interface. we will get the velocity.14) Substituting the value of C2 in Eq.12) where C2 is a constant.12).. (2. (2.e. . (2. . (2. (2.e. . . there is no slip between the liquid and the solid. (2. therefore. av = (c) Mass flow rate . (2. (2. The shear stress is given by Eq. .16) (b) Average velocity vz. (2. It is clear that this velocity distribution is a parabola as shown in Figure 2. From the velocity distribution. . vz.av Thus. . the other quantities can be easily calculated as follows: (a) Maximum velocity At x = 0.2.15) This is the parabolic velocity distribution solution. .av vz.17) .2. Hence . (2. the velocity becomes maximum. .. .9) and is a straight line as shown in Figure 2. (2. .18). (2. . . the film thickness can be calculated as . .18) (d) Film thickness From Eqs. (2. .20) = weight of the entire liquid film For falling film. the Reynolds number is defined as .. (2. (e) Force of the liquid on the solid surface Fz = LWdtg . . .17) and (2. For turbulent flow. NRe < 20. . (2. .19) where is the mass flow rate. NRe > 1500.21) For laminar flow. The disturbances of the flow of liquid at the edges (i. i. xzz = 0 Let us consider a shell at a radius r and of thickness r as shown inFigure 2.e. Laminar flow. Steady-state conditions. i. NRe 1800. The flow pattern can be understood by applying momentum balance. cylindrical coordinates are considered. density t is constant and viscosity n is constant. No slip between the liquid and the solid surface of the wall. i.e. .3. at z = 0. Incompressible fluid. 5.2. Here.22) . Hence.e. Applying a momentum balance at the cylindrical shell in the z-direction. Hence only the velocity vz is considered here. i.2 FLOW THROUGH A CIRCULAR TUBE (GRAVITY FLOW) In many chemical engineering processes. 6.e. Under these assumptions. liquid flow is only in the z-direction. The flow of liquid is under gravity only. 2. Rate of molecular “momentum in” across the cylindrical surface at r – Rate of molecular “momentum out” across the cylindrical surface at r + r + Gravity force acting on the fluid = 0 . z = L) are neglected. 7.e. we get Figure 2.3 Axial flow in a circular tube. 4. xiz = 0. 8. The assumptions made in the analysis of flow of a liquid in a circular tube are as follows: 1. Let us consider that the fluid is flowing in the z-direction. 3. No end effects. Unidirectional flow. (2. Newtonian fluid. one comes across a fluid flowing in a circular tube. Let a fluid of constant density t and constant viscosity n be flowing in a circular tube of radius R and length L. the z-momentum considered will be as follows: Molecular momentum flux = xrz. v(i) = 0 and v(r) = 0. . otherwise the momentum flux would become infinite at the centre of the tube. the molecular momentum flux. . C1 must be zero. . (2. .22).25) Substituting the values of Eqs. centre of the tube. (2. . Therefore. . (2.26) Dividing Eq.27) Integrating both sides of Eq. . . we get the differential equation. the momentum flux would become . (2. . . .26) by 2pLr and taking the limit r 0. (2. we have .25) in Eq. So physically. . (2.30). (2.31) Substituting Eq. . . . (2.30) Applying the Newton’s law of viscosity.24) Gravity force acting in the z-direction of the liquid on the cylindrical shell = (2rr rL)tg . (2. we get .24) and (2.28) where C1 is a constant of integration. we obtain . (2. .32) Integrating this separable differential equation. will be finite at the centre of the tube. . At r = 0. (2. (2.27).31) into Eq. . (2. Boundary condition 1: At r = 0. .23) Rate of molecular “momentum out” across the cylindrical surface at r + r = (2rrL)xrz|r+r . . (2. (2. the velocity of liquid will be zero. (2. (2.29) So. the differential equation for the velocity is given by . xrz. .23). .Rate of molecular “momentum in” across the cylindrical surface at r = (2rrL)xrz|r . we get . xrz = finite . the liquid velocity will be maximum and at the wall of the tube. (2. (2. we obtain velocity distribution as . the velocity becomes maximum. . . vz = 0 or .34) Substituting the value of C2 in Eq. The velocity profile is parabolic as shown in Figure 2. The various other quantities can be evaluated as follows: (a) Maximum velocity At r = 0.4 Momentum flux distribution and velocity distribution in a circular tube. Figure 2.4. Hence . . The shear stress xrz is also shown in Figure 2. .4 which is linear.33) The constant C2 can be evaluated from the boundary condition. Boundary condition 2: At r = R. .33). (2. incompressible flow of a Newtonian fluid. (2.35) This velocity profile is for laminar.. . i.35) and integrating. the width of the slit in very small compared to the length and width of the plate.av = (c) Mass flow rate = (average velocity)(area)(density) = . we get . (2.37) Thus.39) The flow must be checked before using these equations for the velocity profile. .e. . Here. B << W << L. . (2. . There are no end effects. vz.36) (b) Average velocity vz. .. Also. NRe 2100. . i. we consider that the flow is in the z-direction only. 2. . (2. length L and width W as shown inFigure 2. . (2. The flow is laminar and the fluid is incompressible of density t and viscosity n.av Substituting vz from Eq.5.38) (d) Force exerted at the wall of the tube or .e. .3 LAMINAR FLOW IN A NARROW SLIT Let us consider that a Newtonian fluid is flowing in a narrow slit formed by two parallel walls at a distance 2B. (2. vy = 0. vx = 0. Here.Let us consider the coordinates to be (x. The flow of the fluid is under pressure and gravity. p0 is the pressure at z = 0. The velocity here is vz. . Select a shell at a distance x and of thickness x as shown in Figure 2. pL is the pressure at z = L. Figure 2. z).5 Flow through a slit.6. (2. .46) Integrating both sides of Eq. . we get .45) by WL x and taking the limit x 0. We make the following simplifications.46).44) Substituting the values of Eqs.42) Rate of molecular “momentum out” across the surface at x + x = xxz|x+x (W L) . .44) in Eq. . Applying a momentum balance at the shell.41).42). . (2. z) coordinates with x = 0 at the centre of the slit. . (2. .Figure 2. . we get (W L)xxz |x – (W L)xxz |x+x + (P0 – PL)Wx = 0 . . (2. . . . .45) Dividing Eq. (2. (2.43) External forces = (P0 – PL)W x . we have . (2. (2.43) and (2.41) Rate of molecular “momemtum in” across the surface at x = xxz|x (W L) . (2. we get Rate of molecular “momentum in” across the surface at x – Rate of molecular “momentum out” at the surface x + x + External forces acting on the fluid = 0 . (2.6 Velocity profile and momentum flux distribution. (2. . .40) ( P = p + tgz) The molecular momentum flux will be xxz. Let us select the (x. (2. . (2. (2.48) Applying the Newton’s law of viscosity. vz = 0 Boundary condition 1: At x = 0.47) now becomes . . xxz = 0 C1 = 0 Equation (2.50) Integrating this equation by the variable separable method. . Physically we know that the velocity at the centre. at x = B or at x = –B. . .e. .e. we obtain the velocity distribution as .52) This is the velocity profile solution in a slit. (2. i. (2. . the other quantities can be easily calculated as follows: (a) Maximum velocity .51) Boundary condition 2: At x = B. .51). It is clear that this velocity distribution is a parabola as shown in Figure 2.47) where C1 is a constant of integration. we get . The shear stress is given by Eq. The velocity of the fluid at both the plates will be zero. we have . vz = 0 Substituting the value of C2 in Eq. x = 0. (2.. .6. (2. will be maximum. (2. From the velocity distribution.48) and is a straight line as shown in Figure 2. .49) .6. i. . (2. . the velocity beccomes maximum. . Hence. i. (2. (2.54) Comparing Eqs. we get (c) Mass flow rate .53) and (2.55) A very important relation can be obtained from this analysis. .av . . (2.e. . slit analog of the Hagen–Poiseuille equation is . . (2.54).53) (b) Average velocity vz. . .At x = 0. Figure 2. we consider the fluid to be flowing upwards in the annular space. p0 is the pressure at z = 0.e. . The fluid is Newtonian. liquid is flowing in the z-direction only. A typical example is shell and tube heat exchanger where we want to know the fluid flow behaviour through the annulus. Laminar flow of the fluid.e. i.2. The assumptions made in the analysis of flow of a liquid through an annulus are as follows: 1. Unidirectional flow.4 FLOW THROUGH AN ANNULUS Many a time in chemical engineering. 5. The length of the cylinder is L as shown in Figure 2. Let us consider that an incompressible fluid of density t and viscosity n is flowing in an annular space between two coaxial cylinders of radii R and kR. Hence vi = 0. where k becomes the ratio of the radius of the inner cylinder to that of the outer cylinder. i. we come across flow through an annulus. Incompressible fluid. vr = 0. Steady-state conditions. 2. 4. opposite to gravity. 3.7. Here.7 Cylindrical annulus. 58) External forces acting in the z-direction of the liquid on the cylindrical shell = 2rrr (P0 – PL) . (2. Applying a momentum balance at the cylindrical shell.8. (2. we get .59) in Eq. Let us consider a cylindrical shell of radius r and thickness r as shown in Figure 2. .56) Rate of molecular “momentum in” across the cylindrical surface at r = (2p rL) xrz | r . . . the momentum flux becomes xrz.58) and (2. (2. Rate of molecular “momentum out” across the cylindrical surface at r + r = (2rrL)xrz | r + r . . (2. (2. we obtain Rate of molecular “momentum in” across the cylindrical surface at r – Rate of molecular “momentum out” across the cylindrical surface at r + r + External forces acting on the fluid = 0 .pL is the pressure at z = L.57) Figure 2. .59) Substituting the values of Eqs. (2.56).57). .8 Velocity distribution and shear stress distribution. . (2. We make the following simplification: ( P = p + tg) For the cylindrical coordinate. . . But we know from the concept that wherever the velocity is maximum.61) Integrating both sides of Eq. (2.65). (2. . . we obtain .67) Integrating this first order differential equation by the separable method. (2. Let that plane (unknown) be at r = mR at which the shear stress will be zero. . . we get . (2.65) Now applying the Newton’s law of viscosity. . since we have no knowledge about the momentum flux xrz at the solid surfaces r = kR and r = R. . . Now Eq. we get .60) Dividing Eq.63) . . we have .66) Substituting the value of xrz in Eq. Hence. (2. . . (2.(2rLr)xrz | r – (2rLr)xr+r + 2rrr(P0 – PL) = 0 .62).62) becomes . (2. (2.62) where C1 is a constant. xrz = 0.68) where C2 is a constant. we get . (2. . (2. . at r = mR. . (2. .64) Substituting the value of C1 in Eq. we obtain a differential equation for vz as . there will definitely be a plane where the velocity will be maximum. (2. C1 cannot be determined. . (2. . In the space between the annulus. . (2. the shear stress will be zero at that plane. At this stage.61) by the variable separable method. .60) by 2rLr and taking the limit r 0. we need two boundary conditions to obtain their values. (2. vz = 0 . (2. The other quantities here.75) (b) Average velocity vz. .68). we obtain the momentum flux distribution as .71) and 0 = 1 + C2 .In Eq. (2. Boundary condition 1: At r = kR. .72) C2 = –1 Substituting the value of C2 in Eq. (2. Hence. (2. . (2. . we get the velocity profile distribution as . . there are two unknowns C2 and m.65). . .71). .av . (2. . . . vz = 0 .8. (2. . can easily be calculated as follows: (a) Maximum velocity . (2.74) The velocity profile distribution and momentum flux distribution are shown in Figure 2. we get Substituting the values of C2 and 2m2 in Eq.69) Boundary condition 2: At r = R. (2. .73) Substituting the value of m2 in Eq.70) 0 = k2 – 2m2 ln k + C2 . (2. .68). The fluids are flowing under pressure gradient only.78) We must check for the Reynolds number since we have assumed laminar flow condition. the pressure is pL.79) For laminar flow. . At z = 0 the pressure is p0 and at z = L. .av rR2(1 – k2)t or .9. . (2. .av . The flow rates of the fluids are adjusted in such a way that the lower half of the slit is filled with fluid I and the upper half is filled with fluid II. Then the fluid I will be flowing in the lower portion of the plate and the fluid II will be flowing above the fluid I. NRe 2100. . . (2.5 FLOW OF TWO IMMISCIBLE FLUIDS Let us consider that two immiscible. (2. incompressible liquids are flowing in the z-direction in a horizontal narrow slit of length L and width W. The fluids are flowing sufficiently slowly so that the interface is not disturbed. 2. We can encounter such a situation with the transportation of petroleum products and water in a chemical industry. . .76) (c) Mass flow rate = (average velocity)(area)(density) = vz. The fluid I is more dense and more viscous than the fluid II as shown in Figure 2.or vz.77) (d) Force exerted on the solid surfaces The force Fz is the sum of the forces exerted on the inner cylinder wall and the outer cylinder wall. . (2. . or Fz = pR2(1 – k2)(P0 – PL) . 83) .Figure 2. The distance between the plates is 2b. . 2. . Applying a momentum balance at the shell.80) Rate of molecular “momentum in” across the surface at x = (W L)xxz | x . The assumptions made in the analysis of flow of two immiscible fluids flowing between two horizontal plates are as follows: 1. 3.10. . becomes the zaxis. (2.9 Two immiscible fluids flowing between two horizontal plates. .82) External force acting on the fluid = (p0 – pL)Wx .81) Rate of molecular “momentum out” across the surface at x + x = (W L)xxz | x+x . . 4. (2. (2. . . 5. we obtain Rate of molecular “momentum in” across the surface at x – Rate of molecular “momentum out” at the surface at x + x + External pressure force acting on the fluid = 0 . . z) are selected in such a manner that the interface in Figure 2. (2. Laminar flow Incompressible liquids Steady-state conditions Only the pressure gradient occurs Liquids are Newtonian The coordinates (x.9. The shear stress considered will be: Molecular momentum flux = xxz Let us consider a shell at a distance x and of thickness x as shown inFigure 2. at x = 0). (2. (2. (2.10 Velocity distribution and shear stress distribution for two immiscible fluids flowing between two horizontal plates.85) Integrating this equation for two phases I and II. . Let us now apply the boundary conditions. .80). .88) . (2. .Figure 2. (2.90) . Boundary condition 1: At x = 0.87) We know that at the interface (i.83) in Eq. .89) Fluid I: . .e. (2. (2. . . .86) Fluid II: . .82) and (2. the momentum flux is continous. . we have Fluid I: . we obtain . . (2. we get . . we have . (2. Substituting the values of Eqs. and the constants are Now applying the Newton’s law of viscosity for both the fluids.84) by WLx and taking the limit x 0. (2.81). (2.84) Dividing Eq. . (2. .95) Boundary condition 4: At x = –b. .96) From boundary condition 2: . (2.98).90) and (2. . (2. we get . (2. we get Fluid I: . (2. (2. .94) (there is no slip of fluid I and fluid II at the interface) Boundary condition 3: At x = b.99) Subtracting Eq. .99) from Eq. .100) Substituting the value of C1 in Eq. . .91) by the variable separable method. . .93) The three constants in Eqs.92) and (2. . (2. we have or or . . .97) From boundary condition 4: . (2. . (2. (2. . .98). . . (2.91) Integrating Eqs. Boundary condition 2: At x = 0.92) Fluid II: .98) From boundary condition 3: . . (2.93) can be determined from the boundary conditions. (2. . .Fluid II: . (2. . (2. . (2. let us calculate the plane of zero shear stress. . PROBLEMS 1. we get the velocity profile distribution for fluid I and fluid II as follows: Fluid I: . (2. (2.or or . Equating Eq. we get the value of shear stress as follows: .86). . Now. we can easily calculate the average velocity and maximum velocity.104) These velocity profiles and the shear stress profile are shown in Figure 2. .104) to zero. .102) Fluid II: . .103) Substituting the value of C1 in Eq.93). . (2. . (2.92) and (2.101) Substituting these values of C1 and C2 in Eq. Show that the flow of a liquid along an inclined plate is . (2. we get the plane of zero shear stress as Once we know the velocity profiles of both the fluids.10. (2. . 8 103 kg/m3. 2.5 mm on a vertical wall per metre of vertical plate? [Ans. What should be the mass flow rate of the liquid for a falling film thickness of 2. (a) Derive an expression for the velocity distribution. (a) . (b) ] 3. (b) What is the ratio of the average velocity to the maximum velocity? [Ans. 0.2 kg/s] . A Newtonian fluid is in laminar flow in a narrow slit formed by two parallel walls at a distance 2B apart. It is assumed that B << W so that the “edge effects” are unimportant. An oil has a kimematic viscosity of 2 10–4 m2/s and density of0.where d = liquid film thickness b = angle of inclination of the plate with the vertical axis w = mass flow rate per unit width of the plate. The mass balance can be written over the element x y z as follows: (Rate of “mass in”) – (Rate of “mass out”) = (Rate of increase in mass) . and vx.2) Rate of “mass out” at the surface x + x is Rate of increase in mass = Volume Rate of change of density . solved it with the help of boundary conditions and derived the velocity profile.3 EQUATIONS OF CHANGEFOR ISOTHERMAL SYSTEM In Chapter 2.1 EQUATION OF CONTINUITY Let us consider a fluid element of volume x y z through which a fluid is flowing. Such problems can easily be solved by two general equations: 1.3) . respectively. But it is not always possible to apply the mass balance to any problem or situation. Further. z-direction. Let t be the density of the fluid. To derive all these equations we used the mass and force balance notation. . the motions of fluid are nonlinear and it is not always easy to solve these problems by shell momentum balances. vy. y-. these general equations are simplified for constant density and viscosity and are called Navier–Stokes equations which are more famous for tackling engineering problems. We then integrated this equation. . Equation of motion (momentum balance) These equations are derived in this chapter. can easily be found out.1) Rate of “mass in” at t`he surface x is . Equation of continuity (mass balance) 2. Once the velocity profile is known. We often encounter complicated problems in chemical engineering. We applied a shell momentum balance and formed the differential equation for shear stress and applied the Newton’s law of viscosity to form a differential equation for the velocity profile. . vz be the velocity components in x-. . maximum velocity. we learnt about the velocity distribution in simple situations of common interest. 3. In such cases. The mass of fluid entering at x (see the left shaded face) and the mass of fluid leaving at x + x (see the right shaded face) are shown in Figure 3. (3.1. (3. other quantities like average velocity. . etc. These equations are used as the starting point for studying any problem involving the isothermal flow of a pure liquid. . (3. and t changes with respect to space and time.6) The rate of change in fluid density described by Eq. (3. ..1 Fluid element of volume x y z through which a fluid is flowing.and z-directions. . . .7) where is the divergence of mass velocity vector tv.1) becomes . we can write the expressions for the mass entering and leaving iny. Similarly.8) . (3. In vector notation. .6) as follows: . . (3. y 0 and z 0.5) by xyz and taking the limit as x 0. we obtain . (3. (3. We can expand the term ∂(tvx) and rearrange Eq. we can write . .6) is known as the equation of continuity. (3. (3. . . (3. . (3.5) Dividing both sides of Eq. Hence the mass balance from Eq.4) Figure 3. (3. The mass of fluid entering at x surface and the mass of fluid leaving at x + x surface are shown in Figure 3. and vx.2 as the shaded faces. (3. the external forces acting on the system are pressure and gravity. Here. respectively. (3.e. Similarly. . First. Figure 3. Case 1: For a compressible fluid (where t is constant). convective and molecular. Eq.7) becomes (·tv) = 0 3. Convective momentum = (tv)v Molecular momentum also called shear stress = x Let us write the conservation-of-momentum equation: (Rate of convective “momentum in”) – (Rate of convective “momentum out”) + (Rate of molecular “momentum in”) – (Rate of molecular “momentum out”) . Similarly.9) is called the substantial derivative. we consider the momentum transfer by two mechanisms. vy. . Also.2 EQUATION OF MOTION (NAVIER–STOKES EQUATION) Let us consider a fluid element of volume xyz through which a fluid of density t is flowing. the fluid is entering and leaving at y and y + y and z and z + z. we can write such expressions for the y-component and z-component.9) where D/Dt on the left-hand side of Eq. we write the momentum balance for the x-component of momentum and external forces. i. y-. through which a fluid is flowing.2 Fluid element of volume x y z. respectively. vz be the velocity components in x-. z-direction.or . we must be clear that the y-direction momentum and the z-direction momentum contribute to the x-direction momentum as well.+ (Sum of external forces acting on the system) = (Rate of momentum accumulation) .10) Before writing the individual terms of Eq. Rate of x-component of convective momentum entering at x in the x-direction Rate of x-component of convective momentum leaving at x + x in the x-direction Rate of x-component of convective momentum entering at y Rate of x-component of convective momentum leaving at y + y Rate of x-component of convective momentum entering at z Rate of x-component of convective momentum leaving at z + z The net convective x-momentum flow into the element xyz is given by . (3.11) Now let us write the molecular momentum terms: Rate of x-component of momentum entering at x by molecular transfer Rate of x-component of momentum leaving at x + x by molecular transfer Rate of x-component of momentum entering at y by molecular transfer* * xyx is the flux of x-momentum at the face perpendicular to the y-axis. . .2. Let us consider the volume xyz of fluid as shown in Figure 3.10). (3. (3. . . . we write in this fashion. (3. the y-component and z-component of the differential equation of motion can be written as follows: y-component: . and z 0. y 0. . . .13) Gravitational force acting in the x-direction = tgxxyz . (3. (3.10) and dividing by xyz and taking the limits x 0. in the x-direction.16) Similarly. (3. we obtain the x-component of the differential equation of motion as follows: . Here we assign the symbol gx to the x-component of the gravitation vector g. Net fluid pressure acting in the x-direction . . .15) in Eq. . . . . let us consider the external forces. pressure and gravity.15) Substituting Eqs. but to make it more clear that we are considering the x-direction forces. (3. Let p |x be the pressure exerted at the face at x and p |x+x be the pressure exerted at x + x. as a matter of convenience. Rate of accumulation of x-momentum . (3.11) to (3.Rate of x-component of momentum leaving at y + y by molecular transfer Rate of x-component of momentum entering at z by molecular transfer Rate of x-component of momentum leaving at z + z by molecular transfer So. the net x-component of momentum by molecular transfer is given by .14) There is no meaning of writing the gravitational force by the symbol gx. (3. .12) Now. (3. . (3. we obtain the equation of motion as: x-component: . .18) Simplifying these equations by using the equation of continuity (3.. . . . (3. . (3. .6).17) z-component: . i.19) y-component: . . (3. . .20) z-component: .21) .e. The general form of the equation of motion for pure liquids can be written as . i. we obtain the following equations of motion for the Newtonian fluids.24) z-component: . (3. . . .e. .1 Navier–Stokes Equations The general equation of motion as discussed above can be simplified to a more useful equation assuming constant values of density t and viscosity n. In such a case ( · v) = 0 Applying the Newton’s law of viscosity for the Newtonian fluids. (3.22) where D/Dt = substantial derivative = divergence 3. .23) y-component: .2. . (3. x-component: . we obtain v1t1A1 = v2t2A2 . Solution: Applying the mass balance over the flow system.26) where . Figure 3. The fluid enters a pipe of area A1 at a velocity v1 and density t1. The fluid leaves a pipe of area A2 at a velocity v2 and density t2. we arrive at the general form of the equation as .25) By combining these three components.. the rate of mass accumulation is zero.3. . . (3.1 A fluid enters a flow system as shown in Figure 3. . . SOLVED EXAMPLES EXAMPLE 3. These equations are called Navier–Stokes equations and are most commonly used for solving engineering problems. Rate of mass entering = v1t1A1 Rate of mass leaving = v2t2A2 Applying the mass balance. we have Rate of mass entering – Rate of mass leaving = Rate of mass accumulation At the steady-state conditions. Find an expresion for the velocity v2 of the fluid leaving the system. (3.3 Mass balance over a flow system. 3 m Let v3 be the velocity in nozzle 3 of diameter.Let the pipes be circular of D1 and D2 diameters. Solution: Let us consider the nozzles’ configuration as shown in Figure 3. The diameter of nozzle 2 is 30 cm and the flow rate is 2 m/s.3)2 + v3(0. Find the velocity in the nozzle 3. The diameter of the nozzle 3 is 20 cm.2 A fluid enters a nozzle 1 of 40 cm diameter at a velocity 3 m/s.4. 3/(0.2 m Applying the mass balance. Let the fluid enter nozzle 1 at a velocity.2. v2 = 2 m/s Diameter of nozzle 2 = 30 cm = 0. Then the fluid splits into two nozzles 2 and 3 connected in a Y shape. we get Rate of mass of fluid entering = Rate of mass of fluid leaving or where t is the density of the fluid.4 m Fluid velocity in nozzle 2.4)2 = 2(0. D3 = 20 cm = 0. v1 = 3 m/s Diameter of nozzle 1 = 40 cm = 0.2)2 . Figure 3. or EXAMPLE 3.4 Nozzle configuration—Example 3. 3.892 m3/h. An oil of density 892 kg/m3 flows through a pipe of diameter 52.0321 m/s] .388 10–3 m3/s entering the pipe. (a) Calculate the velocity at which the fluid will leave the system in a pipe of diameter of 40.4 m/s. If the diameter of the pipe in which the oil leaves the system is 77.5 mm at a flow rate of 1.3. 0.9 mm. (b) Also calculate the mass velocity of the fluid. then at what velocity will the oil leave the system? [Ans. (a) 0.or v3 = 7. (b) 2032 kg/h] 2.9 mm. The fluid enters a pipe of diameter 52. The total flow rate is 1.5 mm as shown in Figure 3. [Ans. A 20% (wt) sucrose solution having a density of 1074 kg/m3 flows through a system as shown in Figure 3.5 m/s PROBLEMS 1. . In order to improve the efficiency of the process. mathematical formulation becomes an easier way to understand the laminar flow phenomenon. e. But as the number of variables becomes large. experimental or models. There are no exact solutions of flow problems similar to those which exist in the laminar flow conditions. It has been postulated that fluctuations of velocity in turbulent flow form eddies.1 TIME-SMOOTHED EQUATION OF CHANGEFOR TURBULENT FLOW (INCOMPRESSIBLE FLUID) The flow pattern and behaviour of the fluid are different in turbulent flow conditions than in laminar flow conditions. Boundary-layer formation and boundary-layer separation also take place. Then we solved the equations formed with the help of boundary conditions. The molecules move randomly with different velocities in all directions. Some empirical or experimental approaches have been developed to understand the turbulent flow problems. Let us discuss fluctuations of velocities in the turbulent flow. it has been investigated extensively. the situation for turbulent flow is different. Turbulent flow is important in many areas of engineering.g. The mathematics becomes more complex and solutions are not feasible. An attempt has been made to understand the flow behaviour under turbulent flow by time–smoothed velocities. 4. In Chapter 2. These fluctuations take place in all directions. These methods are empirical. The mathematical formulation of the differential equations is easy and straightforward. turbulence is created. the mathematics becomes difficult. The mechanism is difficult to understand. walls or baffles. Nowadays with the help of the computer. On the other hand.1–1 mm. The velocities of the fluid changes in direction and magnitude when the fluid faces some solid obstruction. Since the mechanism is easy. the numerical methods of solving such problems are available. In laminar flow. Molecules of the fluid behave orderly and move in straight lines and stream-wise. boundary-layer thickness near the solid surface and Prandtl mixing length models. we discussed typical cases by shell momentum balance for laminar flow conditions. the fluid moves in a particular fashion. Eddy formation takes place.4 MOMENTUM TRANSFERIN TURBULENT FLOW Laminar flow or streamline flow is easier to understand. The turbulence is created when two streams of fluids meet with different velocities or the flowing stream comes in contact with a solid. High-speed photography has evidence of measuring these eddies to as small a size as 0. Scientists have taken different approaches to understand the fluid behaviour under turbulent flow conditions. 2) Some of the properties of these time-averaged velocities and fluctuations of the velocities are now considered.e. (v¢z) 0 but have positive values. vx = + vx . . . let us first consider these fluctuations in the x-direction. . . The level of intensity of the . and z-directions are zero. The timeaveraged fluctuations of velocity vanish over a short period of time.In order to understand the process. Let vx = instantaneous velocity of main stream in the x-direction. We can measure the instantaneous velocity at a given point w. (v¢y)2 0. Figure 4. These plots of variations of the instantaneous velocity vx in the x-direction are shown in Figure 4. (v¢x)2 0. i.and z-directions.t. the data have been analysed by statistical techniques. (4. The time-averaged values of these fluctuations of velocities in the x-. i. the squares of these fluctuation velocities are not zero. Since the fluctuations are random.e. (4. time with the help of a Pitot tube.1 Velocity fluctuations in turbulent flow.1. = mean velocity (time averaged) in the x-direction of flow stream. However. y-. vx = deviation of velocity from the mean velocity in the x-direction.1) Similarly for the y. we can write The time-averaged velocity can be calculated as .r. . .3) The intensity of the tubulence. .5) and the x-component of equation of motion as . . . .23) Substituting. Eq. Experimentally. . this parameter I can be measured by a hot-wire anemometer. . (4.1 Time-Smoothed Equation of Change for Incompressible Fluids for Turbulent Flow After defining the time-smoothed velocities and some of the properties of these fluctuations. i. (3. . (4. (3. is an important parameter and is helpful in understanding heat and mass transfer processes under turbulent conditions.6) Equation of motion: x-component: . we can now discuss the equation of motion and the continuity equation. (3. we get the equation of continuity as . I. 4. For simplicity.turbulence may be related to the square root of the mean squares of the fluctuating velocity components. in the equation of continuity and the equation of motion. (4. .6) and Eq. (3.e. (4.4) Let us reproduce here the equation of continuity and the equation of motion. Equation of continuity with constant density: . . That is. I.3). let us consider only the case of incompressible fluids with constant density and viscosity. Similar equations hold good for pressure as well. .1.23). The intensity of the turbulence. can be defined mathematically as in Eq. we observe: (a) The equation of continuity (4. (4. Similar is the case for vy and vz.6) and (3. (3. . .9) These components are generally known as Reynolds stresses. The three terms of this expression are described by the momentum transport associated with the turbulent fluctuations.2 BOUNDARY-LAYER THICKNESS FOR FLOW NEAR A SOLID SURFACE The behaviour of flow pattern near a solid surface is different than that of the bulk of fluid away from the solid surface.7) is the same as the previous one except that vx is replaced by (time-smoothed average velocity). we obtain . . (b) The equation of motion (4. The solutions are approximate.7) . . . (4. . Comparing these equations with Eqs. 4. For study of velocity in a boundary layer near the wall.8) Similar relations can be obtained for the y-component and the z-component.6) Now applying the simplification procedure noting that the time-averaged product as discussed above and is not zero. viscous flow is considered. .8) also has the same changes in terms of time-smoothed velocity and pressure. These equations are timesmoothed equation of continuity and equation of motion for constant density and viscosity. The only change is that a third expression on the right-hand side is added..23). Let us consider a flat plate over which the fluid is entering at a uniform velocity v. we consider the boundary-layer region near the solid surface. (4. (4. . The thickness of the boundary layer d is arbitrarily selected at some distance away from the surface of the flat plate . Here. Let these components be defined as: . (4. OL is the boundary-layer thickness line as shown in Figure 4. With the above assumptions. x-component: .10) becomes . . . (4. which is . hence t is constant. No gravity force is acting. (4. Hence.2 Boundary layer for flow over a flat plate. (4.10).2. the term ∂2vx/∂x2 is negligible compared to the other terms in the equation. 3. Incompressible fluid.10) y-component: .11) The continuity equation may be used to simplify the above equations. 2. Some simplifications of the Navier–Stokes equations are done to the concept of a thin boundary layer: Figure 4. . the general Navier–Stokes equation can be simplified. . . .12) In Eq. 4. The terms containing vy and its derivative are small.where the velocity reaches 99% of the main stream velocity. The assumptions made are: 1. (4. Flow is steady state. The flow is in the x and y-directions only.Eq. (4. Blasius reduced these two equations to a single ordinary differential equation which is nonlinear and obtained a series solution. . (4.15) where NRe. . . . only the skin friction plays an important role. Boundary condition 1: vx = 0 at y = 0 vy = 0 at y = 0 Boundary condition 2: At some distance away from the flat plate..16) From the relation of vx as a function of x and y obtained from the series solution. hence Eq. (4. (4. .12) and (4.13) becomes . is given approximately by . (4.x = It is clear that the boundary-layer thickness d is proportional to .17) Let L be the length of the plate and b be the width.99v . one should be clear that for calculating the drag force for flow over a flat plate. Before proceeding further. . Since the velocity at the entrance v is constant. Eq. vx = v at y Equations (4.12) are solved with the following boundary conditions.14) are nonlinear equations and hence the solutions become complex. (4. The total drag force can be calculated as .14) Equation (4. .13) Further simplification can be done. where vx 0. The boundary-layer thickness d. Let us calculate the shear stress at the surface at y = 0 for any value of x. . (4.16) becomes .14) and the equation of continuity (4. . . . dp/dx is zero. The results of Blasius are outlined below. More simplifications and semi-empirical correlations are available in the literature.3 PRANDTL MIXING LENGTH MODEL In order to solve for velocity profile in turbulent flow. the Reynolds stresses have to be evaluated.18) and integrating.21) Substituting the value of FD from Eq.19) in Eq.23) or where The Fanning friction factor f for laminar flow in circular pipes. (4. (4.20) But A = b L . we know. . let us compare Eq. .18) Substituting Eq. .19) The drag coefficient is defined from the drag force as . (4. .24). 4. Boussinesq used the eddy-diffusivity model to solve these Reynolds stresses. (4. (4. . . . . (4. (4. These are similar equations for different situations. Now. Also. .21).23) has been experimentally verified. . is given by .. . (4. (4. Eq. (4.22) . L < 5 105. . (4. we obtain .17) in Eq. The only constraint is that CD applies to the laminar boundary layer only for NRe. . . (4.23) with Eq. (4. we get . (4. .24) where . The Newton’s law of viscosity can be written for laminar flow as . . . (4.25) By analogy, we can write the turbulent shear stress as . . . (4.26) where ht = turbulent viscosity or eddy viscosity. The turbulent viscosity is a function of position and flow, whereas viscosity is a fluid property. Equation (4.26) can be rewritten as . . . (4.27) where ft = ht/t is called eddy diffusivity of momentum. This is similar to the momentum diffusivity, n/t, for laminar flow. Prandtl assumed that the eddies formed in the turbulent flow behave like a gas and eddies move a little distance L, before losing their identity. This distance L is called Prandtl mixing length. Prandtl also assumed that the fluctuation of velocity vx is due to a lump of fluid moving a short distance L in the y-direction. After travelling the distance L, the lump of fluid will have a deviation of velocity from the mean velocity, which is shown in Figure 4.3. Let be the velocity of the fluid at y and be the velocity of the fluid at y + L, where L is the small distance which the eddies travel. We can write the fluctuation of velocity as . . . (4.28) Figure 4.3 Eddy velocity in the y-direction. Dividing both sides of Eq. (4.28) by L and taking the limit L 0, we get . . . (4.29) Similarly, we can write . . . (4.30) Prandtl assumed that these deviations of velocities are small and vx vy, then the time-averaged, , is given by . . . (4.31) In Eq. (4.31) the minus sign and the absolute value were used to make the quantity the experimental data. Substituting this value of agree with in the Reynolds stresses defined by Eq. (4.9), we get . . . (4.32) We can compare Eq. (4.27) with Eq. (4.32) and find ft as . . . (4.33) which we know is the eddy diffusivity of momentum. In this chapter, we have understood momentum transfer under turbulent flow for three models, namely, time-smoothed equation of change, boundary-layer thickness and Prandtl mixing length. We now move on to unsteady-state problems in the next chapter. 5 UNSTEADY-STATE FLOWOF NEWTONIAN FLUIDS In Chapter 2, we solved many cases of streamline flow. Those problems can easily be solved by applying shell momentum balances. It is easy to formulate such simple problems and their solutions are also simple. Chapter 3 deals with the general equations of motion. We can start the problems from such general equations, and simplifications are done based upon some assumptions. Here also the mathematical solutions are easy. But in actual situations, one encounters unsteady-state problems. The partial differential equations formed involve analytical methods to solve these problems. The unsteady-state behaviour creates more than one direction. Such problems are extensively solved in fluid dynamics. These mathematical equations formed are solved by analytical and numerical methods. With the help of computers, these complicated practical problems can be solved by numerical methods. In this chapter, a very simple case is taken up for unsteady state. 5.1 FLOW NEAR A WALL SUDDENLY SET IN MOTION Let us consider a semi-infinite plate over which a liquid of constant density t and viscosity n is bounded over the solid surface as shown in Figure 5.1. Initially, the liquid and the solid plate are at rest. The plate suddenly starts moving with a velocity v0. Let us consider Cartesian coordinates (x, y) for the system under the following assumptions: 1. 2. 3. 4. Flow is laminar No gravity force No pressure gradient Incompressible fluid Figure 5.1 Laminar flow of a liquid near a flat plate suddenly set in motion. We can write the equation of motion (3.22) as . . . [refer to Eq. (3.26)] Here with the above assumptions made, this equation becomes . . . (5.1) or . . . (5.2) where n = n/t. In order to solve Eq. (5.2), we need the initial and boundary conditions. Initial condition: At t 0, vx = 0 for all y (5.3) Boundary condition 1: At y = 0, vx = v0 for all t > 0 (5.4) Boundary condition 2: At y = , vx = 0 for all t > 0 (5.5) Equation (5.2) can be solved with the help of the above initial and boundary conditions by numerical methods. Let us introduce the following dimensionless velocity and time parameters to solve this equation: Equation (5.2) becomes . . . (5.6) By the introduction of dimensionless time as shown above, Eq. (5.6) will become These functions are well known in mathematics. z = 0 .6) becomes . one can get the first-order separable equations from Eq.8) Boundary condition 2: h = . . (5. (5. the boundary conditions become Boundary condition 1: h = 0.7). . z = 1 . (5.Equation (5.7). = 1 – erf(h) = erfc(h) . (5. . . . . (5. .9) By introducing another variable for ∂z / ∂h. One can plot the dimensionless velocity vs.7) In order to solve Eq. dimensionless time as shown in Figure 5. .9) where the error function can be defined as 1 – erf(x) = erfc(x) where erfc(x) is called the complementary error function. Now the original variable becomes . . . With the help of the above boundary conditions. (5.10) The complementary error function is a decreasing function which varies from 1 to 0.2. (5. . one can easily get the solution by the definition of error functions. 2 Velocity distribution (dimensionless velocity vs. numerical methods too. .Figure 5. dimensionless time) for a flow near the wall when the wall is suddenly set in motion. In this chapter. in practice. we have explained a simple situation of unsteady-state flow near a wall solving it analytically. However. are extensively used to solve such problems with the help of computers. Section B HEAT TRANSFER . Radiation energy mechanism is different and will not be discussed in this book. 6. its temperature initially is T0.6 HEAT TRANSFER Some solid materials conduct heat easily. Heat conduction in fluids is slightly different than that in solids. Molecular energy transfer 2. there is a third mechanism. Y distance apart. the temperature profile is shown in Figure 6. In fluids (such as liquids and gases). At steady state. namely radiation. At steady-state conditions. In radiation.1 Steady-state temperature profile for a solid slab.1 FOURIER’S LAW OF HEAT CONDUCTION (MOLECULAR ENERGY TRANSFER) Let a solid slab of area A be placed between two parallel plates. After a small period of time. In heat transfer. we shall define and explain the thermal conductivity by Fourier’s law of heat conduction.1. Metals conduct heat. the temperature of the solid slab starts increasing. Wood insulates heat.1. As time passes. Suddenly. energy is interdiffusing and does not require any media. energy is being transferred by two mechanisms: 1. whereas others do not. Convective energy transfer These two modes of energy transfer have been explained well in Chapter 1 with the help of an example. the lower plate is raised to a higher temperature T1. whereas wood does not conduct heat. the temperature of the slab changes. the temperature profile becomes linear. The physical property that describes the rate at which heat is conducted is called the thermal conductivity k. Let . Now. the temperature attains a linear profile as shown in Figure 6. When the slab is placed. Figure 6. As time elapses. (6. T = T1 – T0 It is obvious from the above that Q A Q T and Q Combining all these quantities. (6.2) in x.3) . . . (6. a thermal property of the solid. Similarly. Equation (6.1) where k is a constant called thermal conductivity. Then.1) in the usual differential form when the thickness of the slab approaches zero. we get or . . . Equation (6.1) now becomes . The negative sign indicates that heat flow decreases in the positive direction of y. (6. .1) is also valid if we replace the solid slab by a fluid (liquid or gas).Q be the rate of heat flow in the slab. .and z-directions as well. Let us write Eq.5) There is another way of defining thermal conductivity. .4) In the vector notation. J/kg · K a = thermal diffusivity. . . we can write Eq. . . known as thermal diffusivity a.6) where cp = specific heat capacity at constant pressure. . It is defined as . (6. In other words. we can say that the heat flux by conduction is proportional to the temperature gradient. The only precaution to be taken is that there is no convection or radiation of heat taking place. . The rate of heat flow per unit area is written as qy which is called heat flux in the positive direction of y.2) This equation is called “Fourier’s law” of heat conduction in one dimension. (6. (6. all these equations can be combined and we can write . temperature gradient. . (6. vz n x-. . New York. Convective energy transfer. 1 M. is similar to the momentum diffusivity as discussed in Chapter 1. It is discussed in detail by Jacob1. vy. It has three components. (b) Chapman–Enskog kinetic theory of gases. . In the first step. Momentum and thermal diffusvity are correlated by a dimensionless number called Prendtl number. The thermal conductivities for liquids and gases can be estimated by the (a) corresponding state chart. In solids. y-.Units are: k = W/m· K a = m2/s qy = W/m2 Here. if a molecule of higher thermal energy migrates to a region of lower energy level. Radiation: This mode will not be discussed in this book.2 CONVECTIVE ENERGY TRANSFER So far we have discussed that energy can e transported by three modes: 1. (6. reduced pressure and reduced thermal conductivity. z-directions. Heat Transfer. In the second step. This is discussed in other heat transfer and chemical engineering handbooks. We assume the concentration of thermal energy to be constant in the given volume. It is dfined as . Vol. In this book. 3. 2. we are not discussing the estimation of thermal conductivity. Wiley. onsderthat dS is the area perpendicular to the x-axis as shown in Figure 6. it must distribute its excess energy among the low energy molecules. whereas the thermal conductivities of most liquids decrease with increasing temperature. Chapter 6. the thermal diffusivity. Molecular energy transfer (conductive heat transfer). Jacob. The mechanism of transport of thermal energy is done in two steps. .7) The thermal conductivity has to be measured expeimentally because it depends upon so many parameter. vx.using reduced temperature. 1. molecules migrate by random motion in a given volume. the thermal conductivity and electrical conductivity are closely related. The iternal energy is associated with the random translational motion of molecules and internal motion of molecules. The thermal conductivities of gases at low density increase with increasing temperature.2. Let v be the velocity vector with which the fluid s moving. a. 6. The energy of molecules consists of kinetic energy and internal energy. (6. . do not change with time. However. Under steady-flow conditions. Here. (6. we develop the temperature distribution in a particular case.10) where U is the internal energy per unit mass.Figure 6.11) Similar equations can be written for the y-direction and z-direction. (6. (6. . density. The other parameters like average temperature and surface heat fluxes can easily be calculated. we can have the convective energy flux. .12) By using Eq. Total energy associated by convective transfer . . Total convective energy = .12). etc.3 SHELL ENERGY BALANCES AND BOUNDARY CONDITIONS After knowing the convective and conductive heat fluxes. Volumetric flow rat across the area dS ^ to the x-axis = vx × dS . pressure. only heat transfer under steady-state conditions is discussed. . . . . (6.2 Energy transfr by convective mode with velocity vx across the surfae d perpendicular to the x-axis.9) Energy associated with internal energy per unit volume = tU . temperature. 6. Let us write the general equation of conservation of energy for the steady flow: (Rate of “energy in” by molecular transport) – (Rate of “energy out” by molecular transport) + (Rate of “energy in” by convective transport) – (Rate of “energy out” by convective transport) . .8) Energy asociated with kinetic enery per unt volume . In the subsequent chapters. we can easily apply the energy balances. the unsteady-flow problems can be solved separately. . (6. Identify the non-vanishing temperature component. (6. air temperature is given. Identify the coordinate system depending upon the system of the problem. From the temperature distribution.13) The molecular and convective fluxes are discussed in Sections 1. Apply the energy balance over the shell and to the area perpendicular to the variable heat transfer being transferred. It will appear with the constant of integration. 9. 5. T0 the ambient temperature and Tb the bulk fluid temperature. 3. Sometimes heat flux normal to the surface is given. At the solid–fluid interface. Select the thin shell over which the heat/energy balance is to be applied. 4. Thus. i. other quantities can be calculated such as average temperature. q = h(T0 – Tb) .1 The thermal conductivity of an insulating material was measured. Solution: . . Integrate this differential equation to get the temperature distribution. The temperature of a flat slab of 25 mm of the material was measured to be 318. 10. Integrate the differential equation formed to get the heat flux distribution.1 and 1. SOLVED EXAMPLES EXAMPLE 6.13). This constant can be evaluated from the boundary conditions. 7. Get the temperature distribution equation. we will get the first differential equation for the heat flux as . 8. (6. the heat flux at the interface for both the fluids is the same.14) 6. Formulate the differential equation for the heat flux by considering the fact that the shell thickness approaches zero. maximum temperature. etc. At the interface of two solids or fluids. Substitute in Eq.e.1 W/m2. . 4. Ambient temperature at the outer surface may be specified. The following procedure is followed for the shell heat transfer: 1.2. 11. 2. Apply the boundary conditions to evaluate the constants. Calculate the thermal conductivity of the material. Boundary conditions: 1. The heat flux was measured to be 35. i.e. . 2. the Newton’s law of cooling may be applied. At the same time.2 K. the temperature at the interface is also the same for these two phases. . 3.+ (Rate of work done on the system) + (Rate of work done on the system by external forces) + (Rate of energy production) = 0 (6. (6.15) where h is the heat transfer coefficient. A constant of integration will appear in the equation.4 K and 303. Apply the Fourier’s law of heat conduction and obtain the differential equation for the temperature distribution. 5 m2 and thickness 0. Calculate the heat loss per m2 of the surface area of an insulating wall composed of 25.1 m was found to conduct heat at a rate of 3 W at steady state with temperature T0 = 24°C and T1 = 26°C on the two main surfaces. Solution: Applying Fourier’s law of heat conduction.4 mm of corkboard having thermal conductivity k of 0. . . 0.2 Calculate the heat loss per m2 of surface area for an insulating wall of a cold storage room where the outside temperature is 299.025 m EXAMPLE 6.4 – 303.2 = 15. What is its thermal conductivity? . . [Ans.5 K.0433 = 39. .1 W/m2 T = 318. where the inside temperature is 352.4 mm = 0. A plastic panel of area 0.7 K and the outside temperature is 297. .9 K and the inside room temperature is 276.9 – 276.1 K.1 W/m2] .048 W/m · K.5 = 23. The wall is made of 25. 105.0254 m Heat loss per m2 = 0.4 mm thick fibre insulating board.3 W/m· K] 2.0433 W/m· K.2 K x = 25 mm = 0. [Ans. The thermal conductivity of the board k is 0.Using Fourier’s law of heat conduction = heat flux = 35.4 K x = 25.0433 W/m· K T = 299. where k = 0.9 W/m2 PROBLEMS 1. Then the Newton’s law of viscosity was applied. In this way.7 SHELL ENERGY BALANCESAND TEMPERATURE DISTRIBUTION IN HEAT CONDUCTION IN SOLIDS (Typical Cases) In Chapter 2. In Sections 7. the Fourier’s law of heat conduction is applied.2 to understand the continuity of heat flux and temperature at the interfaces. we applied momentum balance over a shell and solved for momentum flux.6 and 7. we learn how to apply the boundary conditions.1 HEAT CONDUCTION WITH AN ELECTRICALSOURCE OF HEAT When an electrical current passes through a wire. But one should be very careful while solving energy problems. the wordings used are similar and the methods to solve momentum and heat problems are also similar. After integration. In order to design the material of construction of the insulation.1 with an electrical heat source. in this chapter too. we obtained the differential equation for the momentum flux. In both these chapters. But the constants of integrations are evaluated from the boundary conditions. Thus. we obtain the temperature distribution profile. In these problems. this electrical energy is converted into heat energy.7 illustrate the heat conduction problems with a viscous heat source and chemical reactions.3 and 7. Heat conduction in a nuclear heat source is discussed in Section 7. steady-state conditions are assumed. these differential equations for momentum flux and velocity were obtained after evaluating the constants of integration from physical concepts. we select the shell over which heat balance is applied. After integration of the differential equation. We start with a very simple heat conduction problem in Section 7.4. Heat conduction from a spherical ball to a stagnant fluid is solved in Section 7. the differential equation was obtained. The differential equation for heat flux is obtained. Sections 7. we consider heat conduction in one dimension only to understand the solution very clearly. This is different than the thermal conductivity of the material of the wire. One thing should be kept in mind that the area should be taken perpendicular to the flow of heat flux. Let us consider a cylindrical wire of radius R and length L through which electrical current is passed . Here.5. Afterwards. the constants evolved were evaluated from the boundary conditions. Let the current density through the electrical wire be I (A/cm2) and the electrical conductivity of the material of electrical wire be ke (–1 cm–1). heat conduction through composite walls and cooling fins are dealt with. In all these problems. As discussed earlier in Chapter 2. we need to know the temperature distribution produced by electrical current. Similarly. These problems are of practical nature. 7. 1. is constant.2) .1 Cylindrical electric wire through which current is flowing. . we obtain Rate of “heat in” – Rate of “heat out” + Rate of heat produced by electrical energy = 0 . Let us consider a cylindrical shell of radius r and thickness r as shown in Figure 7.1(b). Electrical conductivity ke is constant. . A-/cm3 I = current density. –1 cm–1 The following assumptions are made in the analysis of heat conduction with an electrical source: 1. Figure 7. 4.as shown in Figure 7. Heat energy flows only in one direction. Applying energy balance at the cylindrical shell. Let qr be the heat flux in the r-direction. in the r-direction. Heat energy flows by conduction only. i. Steady-state conditions prevail.1) where Se = energy produced per unit volume. Thermal conductivity k of the wire. The heat produced per unit volume by the electrical wire is given by (7. (7.e. 2. 3. A/cm2 ke = electrical conductivity. 5. 11) It is clear that the heat flux is a linear relationship which is shown in Figure 7. .Rate of “heat in” at the cylindrical surface at r = (2rrL)qr |r . the heat flux distribution is . . (7.1(b).6) by 2rL r. .5) Substituting these values in Eq. i.10) Boundary condition 1: At r = 0 (centre of wire). . we get . . .7) Taking the limit r 0. Now.4) Rate of heat produced by electrical energy = (2rrL r)Se .8) Integrating both sides of Eq. (7. (7.e.2). (7. . (7. (7. we obtain . we get (2rrL)qr |r – (2rrL)qr |r+r + (2rrL r)Se = 0 . (7. . (7. . (7.12) Substituting this value in Eq. we get . qr is finite. C1 = 0 Therefore. (7. (7. (7.3) Rate of “heat out” at the cylindrical surface at r + r = (2rrL)qr |r+r . . . . . (7. . . . .8). . we apply the Fourier’s law of heat conduction. . .6) Dividing both sides of Eq.9) where C1 is the constant of integration. . . (7.11) and integrating the first-order differential equation. . . . . (7. T = T0 (surface temperature). . (7. (7. (7.15) is parabolic as shown in Figure 7. we get . C2 = T0 + .we get .13) where C2 is the constant of integration. .14) Substituting the value of C2 in Eq.13).1(b).16) (b) Average temperature rise . . . (7.15) The temperature distribution given by Eq. (7. . Boundary condition 2: At r = R. which can be solved by using the boundary condition. Other temperatures can now easily be calculated from this temperature distribution as follows: (a) Maximum temperature at r = 0 (Tmax – T0) = . we get Tav = Tmax . It is surrounded by aluminium cladding of radius R2 as shown in Figure 7. As the fissionable material is consumed.11). .16) with Eq. . (7. .e.. . The kinetic energy is produced by this nuclear fuel in the reactor. i. we obtain Q |r = R = 2rRL = rR2LSe . . . We know that this energy is minimum at the centre. . (7. (7. (7.21) where Sn0 = volume rate of heat produced at the centre of the sphere b = constant . the temperature may not be uniform.2 HEAT CONDUCTION WITH A NUCLEAR HEAT SOURCE Let us consider that heat is produced by a nuclear fissionable material. (7. Let the fissionable material be a spherical ball of radius R1. .17). . Q |r = R = 2rRL qr |r = R . Let us assume a simple parabolic model.17) Comparing Eq. (7. .18) (c) Heat flow at the surface It can be calculated for the whole length of the wire. (7. .20) All the heat produced by the electrical wire at the surface must be allowed to dissipate.2. Let us assume that the heat energy resulting from the nuclear heat source is Sn (cal/cm3· s). 7. Aluminium cladding is done to prevent the leakage of nuclear radiation from the reactor. (7.19) Substituting for qr |r = R from Eq. (7. Applying the heat balance at the shell at the steady-state condition. There is no convection energy. (7. .28) Similarly.23). .Figure 7. Let us also consider a shell element of radius r and thickness r through which the heat flux qrF is passing in the fissionable material. Let us consider that heat is being conducted away only by conduction. (7. . we have . . .21) in Eq. . the heat flux qrC in the aluminium cladding can be evaluated by heat conduction balance but there is no heat source term here. we get Rate of “heat in” at r – Rate of “heat out” at r + r + Rate of heat produced by the nuclear fission = 0 . (7. (7. (7. .27) Substituting the value of Sn from Eq. .2 A spherical nuclear fuel heat source. .25) in Eq. (7.26) by 4rr and taking the limit as r 0.24) Rate of heat produced by the nuclear fission = Sn(4rr2r) . Hence .22). (7. . we get . (7.24) and (7. . (7. (7. we obtain (4rr2qrF)|r – (4rr2qrF)|r+r + Sn(4rr2r) = 0 .26) Dividing both sides of Eq.22) Rate of “heat in” at r by conduction = (4rr2)qrF |r .25) Substituting the values of Eqs.27). .23) Rate of “heat out” at r + r by conduction = (4rr2qrF) |r+r . . . (7. (7. . . (7. . Boundary condition 1: At r = 0.35) and (7. .31) where C1 and C2 are the constants. . (7. we have . (7. (7. . . . qrF = finite .33) By boundary condition 1.31). we obtain . putting r = R1.38) . qrF = qrC . . the heat flux is the same.30) and . (7. . (7. We know that at the interface of the fissionable material and cladding. . . Hence . . (7. (7.29). we get . C1 = 0. These are calculated by boundary conditions.32) Boundary condition 2: At r = R1.34) By boundary condition 2. we get .36). . (7. .28) and (7.36) By equating Eqs. (7.. (7. . (7. .37) Substituting this value of C2 in Eq.29) Integrating Eqs. (7. . . .35) and qrC becomes . . .Now applying the Fourier’s law of heat conduction for both these materials of different thermal conductivity. (7. .46) By boundary condition 3. which can be evaluated by boundary conditions. .45) Substituting the value of C4 in Eq.43) Boundary condition 4: At r = R2. . .44) By boundary condition 4. . .45) in Eq.48).39) and . TF = TC (temperature at the interface) . (7.47) . at r = R1 and TF = TC. . we have . .44). (7. (7. . . .48) Substituting the value of C4 from Eq. (7.34) and (7. . (7.41) . (7. .38) and integrating. . (7. (7. we obtain . (7. (7. we get . . we get . . C4 can be evaluated as . .42) where C3 and C4 are constants of integration.40) Substituting these values of qrF and qrC in Eqs. (7. . (7. . (7. Boundary condition 3: At r = R1. TC = T0 (ambient temperature) . 7. . at the interface. . . . Another example is of refrigration chambers.41).51) Substituting the value of C3 in Eq.52) We have learnt two very important notions in this problem. (7. the heat flux is the same. These materials in multilayer walls have different thermal conductivities. At the interface of the two solid materials.49) or . we come across the problem of heat conduction through various materials.50) At the interface TC = TF. . we need to know the thicknesses of different layers of materials and the temperature distribution across each of them. 2. . (7. Let us consider that the furnace consists of three layers of slabs of width W and height H as shown in Figure 7. we get or . .3 HEAT CONDUCTION THROUGH COMPOSITE WALLS Many times in the industry. . we get . A typical example is that of furnace walls which are made up of different refractory materials. .. (7. 1. Also. If we were to construct a furnace.3. (7. (7. the temperatures of the two materials are the same. i. Steady-state conditions prevail. . 3. The thicknesses of these slabs are (x1 – x0). The following assumptions are made in the analysis of heat conduction through composite walls: 1. (7. 2. respectively.e. . In Figure 7. The slab surfaces are at places x0. Rate of “heat in” at the surface x – Rate of “heat out” at the surface x + x = 0 . and x = x3) is given by the “Newton’s law of cooling” with heat transfer coefficients h0 and h3. k1.54) Dividing both sides of Eq. let qx be the heat flux flowing in the x-direction. we have .Figure 7. . (7.3 Heat conduction through composite walls. Thermal conductivities of the slabs k1. The temperature Ta is higher than the temperature Tb and thus the heat will flow from fluid A towards fluid B passing through these three slabs. Heat flows in one direction only. i.3. The third slab comes in contact with fluid B which is at a temperature Tb. k2. With the above assumptions. x2 and x3. k3 are constant and not changing with temperature. the three slabs are of different thicknesses and of different materials of thermal conductivities. We are interested in finding the temperature profiles in the slabs and also the heat flux. 4. x1. . at x = x0.53) Region 1: qx |x WH – qx |x+x WH = 0 . k3. (7. There is no heat generation. Let us consider a shell at x and of thickness x for applying heat balance in the region 1. The surface of the first slab comes in contact with fluid A which is at a temperature Ta. k2. (x2 – x1) and (x3 – x2).54) by WH x and taking the limit x 0.e. Heat transfer (at boundaries. in the x-direction. . (7. . (7. .56).63) Region 3: . . Integrating these equations over the thickness of each slab.58) Now applying the Fourier’s law of heat conduction for all the three slabs. . we get for: Region 1: . k2 and k3 are the respective thermal conductivities of these three slabs and are constant. . (7. .56) Integrating Eq. (7. (7.57) = heat flux at surface x = x0 This equation is valid for all the three slabs. . (7. we get qx = constant = q0 .61) where k1. (7. . .65) . (7. (7.64) Applying the Newton’s law of cooling at the surfaces x = x0 and x = x3. . . . . . (7. qx = q0 .62) Region 2: .. (7.55) or . we will get: At surface x = x0: q0 = h0(Ta – T0) or .59) Region 2: . we obtain Region 1: . . . .60) Region 3: . . . (7. . . 70).4. length L and thickness 2B as shown in Figure 7.68) Equation (7. A typical example is heat transfer in a motorbike.62) to (7. Adding these five Eqs. .71) We can now generalize Eq. It is a very useful relation for designing furnaces. . (7.69) and (7.71) for n number of slabs as .68) can be written either in heat flux (J/m2 · s) form or in heat flow Q(J/s) form. (7. (7. Let us consider a rectangular fin of width W. . If we know the heat transfer coefficients. . . (7. . Let us calculate the efficiency of a cooling fin. . . .70) Comparing Eqs. .At surface x = x3: . 7. (7. respectively.67) .4 HEAT CONDUCTION IN A COOLING FIN Fins are practically used to enhance the surface area between the hot body and the poorly conducting fluids such as gases.69) or Q = U(W H)(Ta – Tb) . . (7. then the thickness of the slabs can be found out. . (7. (7. (7.66) where h0 and h3 are the heat transfer coefficients of the fluids A and B. we get . the fins are of small thickness in comparison to their length and width. we get . Generally. . . These fins may be rectangular or simple circular plates.66). (7.72) This equation is very useful for calculating heat transfer through composite walls separating two fluid streams. Let us now define the overall heat transfer coefficient U as q0 = U(Ta – Tb) . . e. we get . 4.4 A rectangular cooling fin (B << L and B << W). B << L and B << W. . No heat is lost by the edges of the fin. .4. i.75) Dividing both sides of Eq. . We make certain assumptions and consider Cartesian coordinates. the thickness of the fin is very small in comparison to its length and width. Steady-state conditions prevail. i. . (7. the heat balance at the shell. . .e. Heat flux at the surface of the fin is given by the Newton’s law of cooling. 5.75) by 2BWz and taking the limit z 0. Heat is lost by conduction in the fin.76) Now we apply the Fourier’s law of heat conduction and get .e. we select the shell as shown in Figure 7. (7. . qz = h(T – Ta) . . 7.73) where h = heat transfer coefficient (constant) Ta = ambient air temperature at the surface of the fin. the heat flux will be qz only. (7. Here. Heat flows only in the z-direction. Now applying. 2. we have Rate of “heat in” by conduction at z – Rate of “heat out” by conduction at z + z – Heat lost at the surface of the fin = 0 . qx and qy are zero. i.Figure 7. The rectangular fin is attached to a hot wall at a temperature Tw which is known. (7. (7. Temperature is a function of the z-direction. 6. With the above assumptions. Thermal conductivity k is constant. 8. . Assumptions: 1.e.74) i. 3. .76).79) Boundary condition 2: At z = L.82) . . . . .77) where k is the thermal conductivity of the material of fin which is assumed constant.85) Boundary condition 2: . . Boundary condition 1: At z = 0. . .86) The solution of the differential Eq. (7. . . . (7. we introduce the following dimensionless quantities: . (7.77) in Eq. . (7.81) . (no heat loss from the edges) . (7.84) is well known in mathematics.78) This differential equation can be solved with the help of boundary conditions. . (7.78) and the boundary conditions become: . We now get by substituting Eq. dimensionless distance .84) Boundary condition 1: i |Z=0 = 1 . Eq. . (7.. . (7.83) ** . . (7. (7. . . (7. dimensionless heat transfer coefficient* . . dimensionless temperature . T = Tw (wall temperature) . (7. (7. .80) In order to solve the above differential equations. . . With the introduction of the above dimensionless quantities. where NBi is called “Biot” number. (7. . we get .81) and Eq. (7.89). . (7. . . (7.96) The effectiveness of the fin can now be calculated as .or (D2 – N2)i = 0 . . Now applying boundary conditions C1 + C2 = 1 . (7.94) . we obtain i = cosh NZ – tanh N sinh NZ .87) The roots are D = N. . . (7. (7.88) where C1 and C2 are constants. (7. . .82) we get . . . (7.97) . (7. (7.91) Substituting C1 = 1 – C2 from Eq.90) Applying the second boundary condition. .92) .95) the values of i and Z from Eq. .95) Substituting in Eq.93) Substituting the values of C1 and C2 in Eq. (7. . (7.88) and simplifying. . (7. . . (7. . . (7. . 0 = –NC1e–N + NC2eN . thus the solution becomes i = C1e–NZ + C2eNZ . .89) . . (7. (7. . Heat flows only in the radial r-direction.99) or . Let us also consider a shell of radius r and thickness r. The thermal conductivity k of the fluid is constant. (7.101) 7.5. Steady-state conditions prevail. . The surface temperature of the sphere is TR and the temperature of the fluid far away from the sphere is TC. 3. 5. The following assumptions are made: 1.5 HEAT CONDUCTION FROM A SPHERETO A STAGNANT FLUID Let us consider that a heated sphere of radius R is suspended in a large. This hot sphere conducts heat to the surrounding fluid in the absence of convection. 4. .98) (7. . With the above assumptions. 2. Heat flux = qr . let us apply the heat balance to the shell.or . motionless fluid. Heat flux at the surface can be evaluated by the Newton’s law of cooling.100) . . . No convection heat flows in the fluid. Let us consider that the heated sphere is suspended in a pool of fluid as shown in Figure 7. (7. . (7.108) where C2 is constant. . we get r2qr = C1 . . . Rate of “heat in” at the spherical surface at r – Rate of “heat out” at the spherical surface at r + r = 0 . . Boundary condition 1: At r = R. (7.e. . . i. .103) Dividing both sides of Eq. (7.109) Boundary condition 2: At r = .111). . (7. . .104).104) Integrating Eq.112) Substituting the value of C2 in Eq. .5 Heated sphere in a stagnant fluid. (7. we get C1 = (TR – TC)kR . . (7.108). we have . (7. . .113) Substituting the values of C1 and C2 in Eq. T = TR . (7. . Now C1 and C2 are evaluated from the boundary conditions. Now applying the Fourier’s law of heat conduction. .e. . (7. (7. .Figure 7. (4rr2qr)|r – (4rr2qr)|r+r = 0 .110) Now.111) and TC = C2 . . . (7. (7.102) i. (7.106) we get . (7. .114) . .107). we obtain .105) where C1 is the integration constant. (7. . . . . (7. (7.107) Integrating Eq. (7.103) by 4rr and taking the limit r 0. we obtain . T = TC . . we ignore the curvature of the cylinders and select the Cartesian coordinates.6.118) where h is the heat transfer coefficient.121) This is a very important relationship. it is simplified as shown in Figure 7. . . . (7. (7. Equating Eqs. (7.Now let us calculate the heat flux at the surface of the sphere. (7. . i.117) We can write the Newton’s law of cooling at the surface of the sphere as qr |r = R = h(TR – TC) . .6 HEAT CONDUCTION WITH A VISCOUS HEAT SOURCE High speed viscometers are used to find out the viscosity of highly viscous fluids. the heat flux at the surface of the sphere can be calculated. (7. Here. The system is shown in Figure 7.118). Nusselt number = 2 . The mechanical energy is converted into heat energy.115) By differentiating Eq. .120) where D is the diameter of the sphere . . . (7. .114) and substituting r = R. It is also assumed that the velocity distribution is linear. . (7. and the fluid is filled in the gap of the cylinders. . The inner cylinder is stationary. A highly viscous fluid is filled between these cylinders. . . The width of the gap between these cylinders is b. In these viscometers. . Once the temperature profile is known. . In order to solve this complicated problem. (7. we get . we get . Let the volume heat source from this “viscous dissipation” be called SV. . its adjacent layers rub against each other.e.117) and (7.6(b). when the fluid is moving with a high speed. . 7.119) or . Here. Let us consider two cylinders in which the outer cylinder is of radius R and is moving with an angular velocity ~. the inner cylinder is stationary and the outer cylinder is moving. . (7.116) . (7.122) where vb = ~R . (7. 1. Newtonian fluid between the cylinders. 6. The gap b between the cylinders is small in comparison to the length L and width W. The curvature of the cylinder is neglected. (7.124) Integrating Eq. Convective energy flux 2. . (7. Thermal conductivity is constant.125) The combined flux e has three components.124). 2. 5. WLex |x – WLex |x+x = 0 . . Here. Rate of work done by molecular transfer mechanism . we have ex = C1 . we apply the heat balance at the shell. (7.123) by WLx and taking the limit x 0. The temperature of the fluid is a function of R only. . i. 3. we get . Steady state conditions prevail.b = width of the gap between the cylinders.6 Heat conduction with a viscous heat source. 4. Incompressible. . . The other assumptions made are: 1. we use the concept of combined flux e.123) Dividing both sides of Eq. (7. . Figure 7. With the above assumptions.e. 131) We have assumed the linear relationship of the velocity. . we have . The x-component of the second term can be written as [x · V] = xxxvx + xxyvy + xxzvz . . . .131). (7. (7. . Rate of heat transfer by molecular mechanism .133) Integrating this equation. (7. we obtain . . . .3.128) . (7. . . (7.129) The third term q can be written after applying the Fourier’s law of heat conduction. (7. Boundary condition 1: . (7. can be neglected since vx = 0. . These constants can be evaluated by the boundary conditions. The first term.125).130) Substituting these values of qx and x · v in Eq.134) where C1 and C2 are constants. . . .127) Since the velocities vx = 0. (7. (7. we have .132) Substituting the value of dvz/dx in Eq. (7. .126) where is enthalpy per unit volume.e. . i. . vy = 0. The x-component of q can be written as . convective energty flux. . vy = 0 and hence applying the Newton’s law of viscosity. (7. or . . . . (7. . . SV. . T = Tb . (7. (7. . .137) Applying boundary condition 2.T = T0 . . we have .139) On simplification.136) Here we assume T0 Tb. (7. Rate at which the work is done by molecular momentum flux . . (7. C2 = T0 . (7. . .135) Boundary condition 2: At x = b.138) Substituting the values of C1 and C2 in Eq. we get or or . (7.140) where NBr = Brinkman number The maximum temperature can easily be evaluated at Now let us calculate the “viscous heat dissipation”. . By boundary condition 1. . .At x = 0.134). the heat energy is either produced or consumed depending upon whether the reaction is exothermic or endothermic. . . Thus. (7. . (7. Sv .143) The reaction generally takes place in a reactor. . . Let us consider a tabular reactor of radius R which extends from z = – to z = + as shown in Figure 7.7.= xxz area velocity . . .141) We know from the above discussions that vz = vb Rate at which the work is done by molecular momentum Rate of viscous energy dissipation.142) 7. . (7. The reactors are of various types depending upon the flow conditions.7 HEAT CONDUCTION WITH A CHEMICAL REACTIONHEAT SOURCE When a chemical reaction of the reactants takes place and the product is produced. Let us formulate the simple form in which the chemical energy is a function of temperature only. the temperature is independent of the r-direction but is a function of the z-direction only. The chemical energy is converted into heat energy. Let Sc be the chemical energy per unit volume. Zone III is packed with inert spherical particles which are again non-catalytic. z = L to z = + . z = 0 to z = L. . Zone III: Exit zone. composition of the reactants and catalyst. Zone II: Reaction zone. superficial velocity and mass flow rate of the fluid are independent of r. the heat of reaction depends upon the operating conditions of the reactor such as pressure. The reactor wall is insulated. i.7 Axial flow fixed bed reactor.and z- . The reactor is insulated.e. is given by . . . Here. temperature.144) where SC1 is a constant and i. Zone II is packed with catalyst particles where the reaction takes place. 2. z = – to z = 0. (7. 3. The fluid enters with uniform axial velocity in the “plug flow reactor”. Steady-state conditions prevail. Zone I is packed with inert spherical particles which are non-catalytic. The reactants enter the reactor with a uniform velocity. . 4. (7. Sc = SC1 F(i) . The following assumptions are made in the analysis of heat conduction with a chemical reaction heat source: 1. It is a very complicated parameter to be evaluated. Density. The reactor is divided into three zones: Zone I: Entrance zone.Figure 7. the dimensionless temperature.145) where T1 is the reactant temperature at which it enters. Rate of work done by molecular momentum transfer. The superficial velocity v0 is based upon the empty reactor. Rate of heat flux due to conduction. kg/m3. 3. . (7. (7. (7.147) by rR2z and taking the limit z 0. and applying the Newton’s law of viscosity. . (7.151) Here vx = 0. in kJ/kg · K.e. . . kg/s t = density of fluid. The second term for z-component: [x ·v] = [vxxxz + vyxyz + vzxzz]vz .directions. . Let us consider a disk of radius R at a distance of z and of thickness z. i. we get . . and [x · v] = 0 . . Convective energy flux. (7. vy = 0.148) This combined flux e has three components: 1.147) Dividing both sides of Eq. we can now apply heat balance in Zone II.153) .150) where cp = specific heat. rR2ez |z – rR2ez |z+z + (rR2z)Sc = 0 . . Using the concept of combined flux e. . . (7. . . we have . (7. . 2. . (7.e. . .152) where vz = constant.149) The first term can be re-written by neglecting the pressure term which is constant in the z-direction: i. . . .146) where w = mass flow rate. (7. liquid to liquid. liquid to solid catalyst particle. . Substituting these various values in Eq. .156) Here vz is constant and we have assumed vz = v0. (7. (7. .160) These differential equations can be solved by using boundary conditions. . (7. . since the heat is being transferred from the catalyst particle to the other particles. . (7. . (7. Boundary condition 1: At z = –.157) Similar equations hold good for all the three zones. .162) Boundary condition 3: . . (7.The third term is of the z-component and applying the Fourier’s law of heat conduction. . . .154) Here. we obtain . . TI = TII (temperature at the interface) . .158) Zone II: 0 < z < L . Zone I: z < 0 . . so we assume the thermal conductivity in all the three zones to be constant. so we get . except that the Sc term will not appear in Zones I and III.161) Boundary condition 2: At z = 0. (7. we consider thermal conductivity keff (effective). (7.148). . . (7. we get . . TI = T1 .159) Zone III: z > L .155) or . (7. . (7.166) The equations for all the three zones can be solved with the help of these boundary conditions. (7. So the conductive heat transfer terms can be neglected. (7. . .168) . TIII = finite .160). .At z = 0. . In this case. (heat flux at the interface) . . (7. . (7. (7. .164) Boundary condition 5: At z = L. But let us take a practical case of interest to engineers of limiting solution.158) to (7. the convective heat transfer is more significant than the conductive heat transfer. . . . . . .172) Zone III: z > L. (7. . (7.165) Boundary condition 6: At z = . . . we get Zone I: z < 0. . . . Let us introduce the dimensionless terms: . . With the introduction of the above dimensionless terms. we can transfer the differential Eqs. .163) Boundary condition 4: At z = L.167) . TII = TIII . . (7.169) where N is the dimensionless thermal heat source. (7.171) Zone II: 0 < z < 1. . and neglecting the conductive heat transfer terms.170) where is a constant. . (7. iI = iII . .180) Now. . . (7.174) Boundary condition 2: At Z = 0. . (7.175) Boundary condition 3: At Z = 1. (7. iIII = eN .8.173) The above equations are the first-order differential equations which are solved with the separable variable methods.179) Zone III: z > L. .8 Dimensionless temperature vs. The boundary conditions become: Boundary condition 1: At Z = – . (7. (7. .178) Zone II: 0 < z < 1. . . . . (7. (7. (7. Figure 7. . dimensionless distance as shown in Figure 7. iI = 1 . . a limiting solution for the case will be discussed: Let F(i) = i . iII = eNZ . . iII = iIII . let us plot these dimensionless temperature vs. Dimensionless distance. iI = 1 . . .177) The solutions are: Zone I: z < 0..176) With these boundary conditions. . . The thermal conductivities are given below.1 K at the outside surface of the concrete. kB = 0. The outside temperature. C = thermal conductivity. 2.0433 W/m· K Concrete.2 mm of concrete. T4 = 297. T1 = 255.9 Example 7.1 K.151 W/m· K Cork board. W/m · K 1. B = thermal conductivity.762 W/m· K Thickness: xA = 0. Area = 1 m2.762 Solution: Let us consider the cold storage room to be made up of three materials A. kA = 0.0127 m xB = 0. The cold storage room temperature. A = thermal conductivity. Pinewood Cork board Concrete 0.6 mm of cork board and an outer layer of 76.151 0.1 A cold storage room is constructed of an inner layer of 12.1. Material Thermal Conductivity.0762 m Let us calculate the resistances: .4 K.1016 m xC = 0. B and C as pinewood. Data: The thermal conductives of the materials used are as follows: S.SOLVED EXAMPLES EXAMPLE 7. 3.4 K inside the cold room and 297. cork board and concrete. Pinewood.No. kC = 0. Figure 7. respectively.0433 0. The wall surface temperature is 255. a middle layer of 101. Calculate the heat loss in watts for 1 m2 and the temperature at the interface between the pinewood and cork board.7 mm of pinewood. 032 Btu/Ft·h·F.The heat flow q0 is given by The heat flows from outside to inside the cold storage room. What is the rate of heat flow through the wall in Btu/h? Solution: The thermal conductivity k1 of the pulverised cork at 32°F = 0. T1 = 40°F. The area of the wall is 25 ft2.8 K EXAMPLE 7.10 Example 7. we get T2 = 256.021 Btu/ft · h · F at 32°F and is 0. T2 = 180°F.2. The thermal conductivity of the cork is 0. Applying the heat balance. The temperature on the cold side of the cork is 40°F and that on the warm side is 180°F. x = 6 in. thick is used as a layer of thermal insulation in a flat wall. Figure 7. The temperature of the hot side. The thickness of pulverised cork.032 at 200°F. the thermal conductivity k2 of the pulverised cork = 0. A = 25 ft2 The temperature of the cold side.021 Btu/ft · h · F at 32°F. Area.2 A layer of pulverized cork of 6 in. Let T2 be the temperature at the interface between the pinewood and the cork board. we have Substituting the values. At 200°F. . we get the values of a and b as a = –0.3 An aluminium fin 3 mm thick and 7. . The data given is: (a) The thermal conductivity of aluminium. Substituting the values of k1. . (iv) By solving (iii) and (iv). . T1 and T2 in (i) and (ii).0112 b = 6. Solution: We consider the case when the end of the fin is insulated. . Calculate the heat loss from the fin per unit depth.5 kJ) EXAMPLE 7. .Let the thermal conductivity vary linearly. (iii) 0. (ii) where a and b are constants. then k1 = a + bT1 .96 Btu/h (= 29. Here. Let the width of the fin be 1 cm. . . we obtain q = 27.3.5 cm long is extending from a wall. k = 300 W/m· C (b) Heat transfer coefficient.032 = a + b(200 + 460) .55 10–5 We can calculate heat flow as . The base is maintained at 300°C and the ambient temperature is 50°C.021 = a + b(32 + 460) . . Tw = 300°C Ta = 50°C . (v) Substituting the values in (v). . k2. (i) k2 = a + bT2 . h = 10 W/m2 · h · C Figure 7.11 Example 7. we get 0. . (i) k = thermal conductivity = 300 W/m· C P = perimeter of the fin L = 7.8(300 – 50) tanh (3.8 Substituting the values in Eq. At the wall. . we get Q = 300 2(0. Develop the overall heat transfer coefficient for the composite cylindrical pipes as where the terms used have their usual meaning.5 cm = 0. .01 0. i. 3. Consider a heat conduction with electrical heat source from an electrical wire of radius R and length L. Develop the temperature profile for the above system. .2 m fire brick. surrounded by 0.003) 3. at r = R qr |r=R = h(T – Tair) where h is the heat transfer coefficient and Tair is the ambient temperature. Assume the heat transfer coefficient and thermal conductivity to be both constant.8 0. .e. T0 is unknown but the heat flux at the wall is given by the Newton’s law of cooling.075) Q = 4. The inside of the oven is made up of 0. 2.746 W or PROBLEMS 1. The wall of an oven consists of three layers of bricks. . (i).1 m thickness of insulating brick and the outer side layer is made .075 m h = 10 W/m2 · h · C or m = 3.The heat transfer from the fin: Q = kAc m(Tw – Ta) tanh mL . layer of common brick of thermal conductivity of 0.04 m and the thickness is 2 mm. 5. layer of sil-o-cel brick with a thermal conductivity of 0.2K. (b) 190.08 Btu/ft · h · F backed by a 9 in.6°C] 4. 237.15 m of building material. Heat is flowing through an annular wall of inside radius r0 and outside radius r1.9 W/m · K) is attached to a copper tube having an outside radius of 0. The temperature of the inner face of the wall is 1400°F and that of the outer face is 170°F. A flat furnace wall is constructed of a 4. The oven operates at 870°C and the outside temperature is 50°C.up of 0.5 in. 7.5 W/ m2. 2. Find the expression for the heat flow if the solid is in the shape of a pipe and the heat flow is in the radial direction only.04 m.5°C or 463. A circular stainless steel fin (R = 17.5 0. (a) 689.225 0. Calculate the fin efficiency and the rate of heat loss from the fin. (a) What is the heat loss through the wall? (b) What is the temperature of the interface between the refractory brick and the common brick? [Ans. The outside wall or tube base is at 523. Material Thermal Conductivity.No. The length of the fin is 0. (a) (= 179 kJ/m2)] (b) Interface temperatures: 570. and the external surrounding air at 343.45°C. Fire brick Insulating brick Building brick 0. (a) How much heat is lost per m2 of the surface and (b) what are the temperatures at the interfaces of the layers? Data: The thermal conductivities of the materials of the bricks are as follows: S.8 Btu/ft · h · F. [Ans. k (kcal/h · m · C) 1.60 [Ans. The thermal conductivity varies linearly with the temperature from k0 at T0 and k1 at T1. (i) 89% (ii) 150 W] .5 K] 6. Develop an expression for the heat flow through the wall at r = r0. 3.2K has a convective coefficient of 30 W/m2 · K. Let the thermal conductivity of a certain solid be expressed as a linear function of temperature k = a + bT. it is necessary to develop general energy equations and simplify them to solve particular problems. In order to understand the general energy equation. (a) Energy associated by convection This constitutes the internal energy and the kinetic energy. So.1. .1) Figure 8.1 Fluid element of volume xyz. (8. let us take the x-component. It enters at the left-hand shaded portion as shown in the figure and leaves at the right-hand shaded portion. The fluid carries different forms of energies. It is not always easy to solve the complex problems in such a way. similar equations can be written for the y. These aspects are discussed in physics. Applying the energy balance to the volume of the fluid xyz. Let us identify the different energies associated with the fluid. Let us consider a fluid element of volume xyz as shown in Figure 8. .8 THE GENERAL ENERGY EQUATION In Chapter 7. steady-state heat transfer problems by shell energy balance.and z-components. The transfer of energy takes place by two modes. we solved the simple. convection and molecular transport or conduction. . we obtain Rate of “energy in” – Rate of “energy out” – Rate of external work done by the system on the surrounding = Rate of energy accumulation . This is the energy associated with the random movement of molecules and molecular interactions. x-component: yz y-component: zx z-component: xy (b) Energy transport by conduction x-component: yz y-component: zx z-component: xy (c) Work done against the gravitational force x-direction: –txyz(vxgx) y-direction: –txyz(vygy) z-direction: –txyz(vzgz) (d) Work done against the static pressure x-direction: yz y-direction: zx z-direction: xy (e) Net work done against the viscous forces x-component: (f) Rate of energy accumulated It consists of an expression of kinetic energy and internal energy as follows: .But we will discuss only the engineering aspects. We can write the equation of energy for Newtonian fluids and constant thermal conductivity as . .1 SPECIAL CASES OF ENERGY EQUATION Let us now consider the special cases of energy equations which are commonly used in the field of . (8. (8. So it is simplified by using the equation of continuity and the equation of motion derived earlier in Chapter 3. . we get . z 0. (8.3) This general equation of energy is not easy to handle in this complicated way. .2) We can write the general equation of energy in the following form: . . . (8. . y 0. 8.1) and dividing by xyz and taking the limits x 0.4) where This is called Fourier’s second law of heat conduction.Substituting the preceding values in Eq. (a) Fluid at constant pressure In Eq.6) (c) Heat generation If there is a heat produced in the fluid by chemical reaction or electrical means. .engineering. (8. (8. . W/m3. General energy equations are useful tools for solving energy problems. (8. . (8. . In order to solve the heat transfer problems. . The simplifications are done with a view to discarding the terms which are not required.4). . Simplifications are done for particular situations in the field of engineering applications.5) (b) For solids In this case t is constant and v = 0. then SV is added to the energy equation. we can use the above general equation of energy in a similar manner as we do for the equation of motion and the equation of continuity. .7) where SV = rate of heat generation. Thus. . . . conditions are different. we solved some of the heat conduction problems in laminar flow conditions. . The time-averaged temperature is defined as . Some of the concepts discussed are: time-smoothed temperature. The thermal problems can be understood in a similar pattern as that for the flow problems in a system. In order to solve the heat transfer problems. many empirical methods are available in the literature.9 TEMPERATURE DISTRIBUTIONIN TURBULENT FLOW In Chapter 7. 9.1) The instantaneous temperature and fluctuation in temperature for turbulent flow are shown in Figure 9. and Prandtl mixing length models.1). . To find the temperature profiles in turbulent flow. we create turbulence in the system to enhance the heat transfer rates. Let us introduce the temperature terms: = time-smoothed temperature T = temperature fluctuation Similar to Eq. Mixing takes place in the fluid. (4. (9.1 TIME-SMOOTHED EQUATIONS FOR ENERGY FOR INCOMPRESSIBLE FLUIDS IN TURBULENT FLOW Time-smoothed quantities and turbulent fluctuations have been discussed in Chapter 4 for the flow conditions.1. The differential equations formed by shell balance have been solved by using the appropriate boundary conditions and initial conditions. Sometimes. for temperature we can write as . thermal boundary-layer thickness. A similar concept holds good for the temperature profiles. 2) All the properties of the velocity fluctuation hold good for the temperature fluctuation as well.1 Temperature fluctuations in turbulent flow. (9. . for example will not be zero. (9. and k as follows: .7) and (4. t. (4.Figure 9. (8. . let us substitute the terms the time-smoothed equation becomes: and . The energy Eq. .4) can be written for a fluid with constant n. cp.3) Now for turbulent flow. .8). Then . The time-smoothed equation of continuity and the equation of motion for a fluid with constant density and viscosity were given in Eqs. . and so on. Let us consider a flat plate over which the fluid is entering at a uniform temperature T. (9. The surface temperature of the plate is Ts. . all the three equations of motion. Figure 9. (9. . . (9.4) We can observe that the time-smoothed equation has the same original form of the terms.2 Thermal boundary layer for a flow past a flat plate.2 BOUNDARY-LAYER THICKNESS FOR HEAT TRANSFER NEAR THE SOLID SURFACE We discussed the boundary layer for a laminar flow over a flat plate in Chapter 4. continuity and energy can be written as the same equations as discussed earlier but we have to change the vx and T terms by and . (9.. A similar situation also arises for the heat transfer problems.6) . . The thermal boundary layer dT is arbitrarily selected at some distance away from the surface of the plate where the temperature reaches 99% of the main stream. OL is the boundary layer as shown in Figure 9. . 9. . . .2. . except those indicated with dashed terms.5) . The turbulent heat flux components can be defined as follows: .7) To summarize. viscosity. (4.14) and . . (9.10) can be solved with the help of boundary conditions. T = Ts .e. 2.8) . . 3. (9. (4. . T = T . . 4. . . . These equations are .12) Boundary condition 3: At x = 0. (9. .10) The differential Eq. specific heat and thermal conductivity are constant. (9. . . The following assumptions are made: 1. . (9. . vx = v . T/t = 0. T = T .13) Let us see the similarity between the above energy equation and the equation of motion earlier derived in Chapter 4. (9. . i. Boundary condition 1: At y = 0. vx = 0 Boundary condition 2: At y = 0. . (9.11) Boundary condition 2: At y = . The heat conduction in the x-direction is negligible in comparison to the heat conducted in the y-direction.12) These equations were solved by using the following boundary conditions: Boundary condition 1: At y = 0. No heat is conducted in the z-direction. With the above assumptions.We write the energy equation as .9) where cp = heat capacity in kJ/kg mol · K. . Steady-state conditions prevail. vy = 0 Boundary condition 3: At y = 0. . The fluid properties such as density. the energy equation becomes . .10) and (4. Boundary condition 3: At y = .e. . The boundary conditions become: Boundary condition 1: At y = 0. . (9. .17).S. .16) We conclude that these equations and boundary conditions are identical for velocity profile and temperature profile for the Prandtl number being equal to 1. replacement of v by T is feasible. Blausius discussed the solution in a similar way as done for momentum transfer. .18) . . Boundary condition 2: At x = 0. Also for any x and y point.16) . the solution of the velocity profile is applicable to the temperature profile solution. .H.14) . (9. (9. of Eqs. . (9. the momentum transfer and heat transfer are analogous in the boundary-layer thickness. Therefore. (4. Blasius assumed the Prandtl number cp n/k = 1.17) where As discussed above.14) is clear. . . Hence.The similarity of Eqs. It is concluded that the thermal boundary-layer thickness dT being equal to the hydrodynamic boundarylayer thickness d is valid only for the Prandtl number being equal to 1. . we get .17) The velocity gradient at the surface is given by . . the coefficients on the R.14) and Eq.16) and (4. (4. . . the dimensionless velocity vx /v and the dimensionless temperature (T – Ts)/(T – Ts) are equal. (9.15) . . Now let us see the boundary conditions when we introduce the dimensionless temperature as (T – Ts)/(T – T) and the dimensionless velocity as vx /v.10) are the same. In such a case. The above heat transfer equations are nonlinear and hence the solution becomes complex. . (4. (9. i. (9. By combining Eqs. . (4. . 17) and (9.23) Let us calculate the mean heat transfer coefficient between x = 0 andx = L for the whole length of the plate. (9. (9. (9. Combining Eqs. we get .21) where = Nusselt number.24) Integrating Eq. .25) where .13). (9. (9. (9. (9. Pohlhausen gave a relationship between the hydrodynamic boundary layer and the thermal boundary layer for fluids. . This is valid only for the Prandtl number more than 0.18).23). . . we get . .20) where hx is the heat transfer coefficient at a point x and k is the thermal conductivity. we get . (9. we get the temperature gradient at the surface as . (9. (9.24).6. . . A = area = bL Substituting the value of hx from Eq.19) and (9. (9.22) Then. Thus. . .By combining Eqs. . the local heat transfer coefficient becomes . . . b = width of the plate.19) The convective heat transfer qy (J/s) can be written from the Fourier’s law as . . . . (9. . Substituting T from Eq. . This situation is exactly similar to the case of velocity. After travelling this distance L.3 Eddy temperature in the y-direction. the lump of the fluid will have a deviation of temperature from the mean temperature as shown in Figure 9. Similarly.26) by L and taking the limit L 0. .3. Prandtl also assumed that the fluctuation of temperature is due to the “lump” of the fluid moving a short distance L in the y-direction.28) where cp = specific heat capacity in kJ/kg · K. . = (Mass)(Heat capacity)(Temperature difference) = –(tvy)cpT . . (9. we get T . we have . In a similar way. In fluid transport. energy is transported to a distance L with a velocity v¢y in the y-direction. We can write the fluctuation of temperature as: T . (9.27). it is assumed that the eddies formed in the turbulent flow conditions move a little distance L before they lose their identity. where L is small. And then they get absorbed in the bulk of the fluid.9. Prandtl proposed a model for heat transfer in the turbulent flow conditions.3 PRANDTL MIXING LENGTH MODEL IN HEAT TRANSFER In Chapter 4.26) Dividing both sides of Eq. (9.27) Let the rate of heat transferred per unit area be qy /A. Figure 9. Now. (9. we discussed the Prandtl mixing length model for the turbulent flow conditions. . Suppose T = deviation of temperature from the mean = mean temperature T = instantaneous temperature. we get . .31) becomes . .29).32). . it is called thermal eddy diffusivity at.3. .31) As per the earlier discussion in Section 4. . . Three models are applied in this chapter to understand temperature distribution in turbulent flow: time-smoothed equation of energy.33) When we consider heat transfer. (9. Eq. . (9. Also. we arrived at eddy diffusivity ft as . . (9. boundary-layer thickness and Prandtl mixing length. (9. . . Then. (9. . (9.30) Substituting the value of vy in Eq. . (4.29) We assumed in Section 4. Combining the Fourier’s equation and Eq. .32) The Fourier’s law of heat conduction for the molecular heat transfer can be written as .33) where a is known as the molecular thermal diffusivity. . . (9. (9.. .3. (9.34) If can be seen from the above that the energy problems are different for turbulent flow conditions. we get . Sometimes we have to handle more than one variable. Initially. Let us consider the x-y coordinate system as shown in Figure 10. The following assumptions are made: 1. T(y. the surface temperature is raised to T1. There is no external heat source. are constant. these differential equations were solved with the help of boundary conditions. 4. We are interested to know the temperature profile. one variable is space and the other variable becomes time. k. t). Heat does not flow in the x-direction. No heat flows by convection. the energy Eq. n. In Chapter 8.10 UNSTEADY-STATE HEAT CONDUCTION IN A SEMI-INFINITE SLAB Heat conduction problems for laminar flow conduction have been solved for the steady-state conditions in Chapter 7. etc. 3. With the above assumptions. A typical problem is solved in this section.1. These equations are useful to tackle complex problems. We used the shell energy balance to formulate the differential equations. Generally.5) can be used to solve this problem. Let us consider a semi-infinite plate. 2. Suddenly. we developed the general energy equations. Figure 10. Then. Such a situation is called unsteady state. cp. The fluid properties such as t.1 Flat plate whose surface temperature is suddenly raised. . (8. its surface is at temperature T0 at time t = 0. (10. . T = T1 for t > 0 ..6) The boundary and initial conditions become Initial condition: At t 0. (8.7) Boundary condition 1: At y = 0. Initial condition: At t 0. . . . This differential equation can be solved with the help of the initial and boundary conditions. (10. T = T0 for t > 0 . . i = 0 for all y .3) Boundary condition 2: At y = . (10.9) Just as we solved the velocity profile in Chapter 5.2) Boundary condition 1: At y = 0. defined as .14) . . .6) then becomes: .1) becomes . (10. . . we get 100(10. i = 0 for t > 0 . (10. . . . we also introduce the dimensionless distance y here as: . (10. (10. .4) We now introduce the dimensionless temperature i. (10. (10. . .1) where a is the thermal diffusivity.11) The solution obtained is: i= .12) or i = 1 – erf h . . . . i = 1 for t > 0 . (8. . . (10. . . .5) by tcp. . T = T0 for all y . .5) Dividing both sides of Eq.10) The differential equation (10. (10. . . (10.5) The above differential Eq.13) or = . (10. (10.8) Boundary condition 2: At y = . . The erf is called the error function. A simple case of heat conduction in a semi-infinite slab is solved in this chapter by analytical methods. . The number of variables becomes more and the mathematical tools to solve the unsteady-state heat transfer problems. These functions can be solved using any mathematics handbook. In real situations. become complicated. we generally come across unsteady-state heat transfer problems. Section C MASS TRANSFER . we will consider the third fundamental transport process. The molecular transfer of mass takes place because of the concentration difference.e. the mass fraction wA is defined for species A as: Similarly. Figure 11. the silica plate is exposed to air and then to helium gas. concen-tration diffusion or mass diffusion.1 Helium (A) diffusing into silica plate (B). wA = 0. we will now study the Fick’s law for molecular diffusion of mass. we consider helium as A and fused silica as B as shown in Figure 11. i. the silica plate is exposed to helium gas. Just as we have seen Newton’s equation for momentum transfer and Fourier’s law of heat conduction. mass transfer.11 MASS TRANSFER In this Section C of the book.1. the silica plate is exposed to air and there is no helium gas.1 FICK’S LAW OF BINARY DIFFUSION (MOLECULAR MASS TRANSFER) Let us consider a binary system in which “species” A is diffusing into “species” B. Suddenly. The molecular transport of one substance to another is called diffusion. As the time passes.2. the concentration of helium is built up and later the steady-state concentration of helium is achieved. Here. 11. . This process is shown in Figure 11. Helium gas penetrates into the silica plate. that is. In this case. First. the mass fraction of B is given by At time t = 0 or t < 0. wAy. . . when A is diffusing in B Here wAy/A is expressed as molar mass flux when species A is diffusing in the y-direction and is denoted by jAy. . (11. The mass flow of helium in the y-direction.2) becomes . . we can write the Fick’s law as . The concentration of A at the top of the silica slab is zero and as time passes.1) or . (11. (11. Let A be the area of the silica slab and Y be the thickness. . (6. is given as wAy A wAy wAy where is the initial concentration of A.5) for heat conduction.3) Equation (11. (11. The differential form of Eq. (11. .4) . Equation (11.3) is called the Fick’s law of diffusion. .2 Steady-state concentration profile for diffusion of helium A into fused silica B. Helium is moving slowly and its concentration is very small. jAy is the molecular mass flux of helium in the positive direction of y. Here.Figure 11. the concentration profile becomes straight. . we get . The x-y coordinates are selected.2) where t = density of the silica–helium system DAB = diffusivity of silica–helium system = proportionality constant called diffusivity.2) for molecular momentum transfer and similar to Eq. Combining these.3) is similar to Eq. (1. Similarly in other directions too. . m2/s. . we can write . Similarly. the velocities of A and B are taken as weighted velocities according to the mass fractions and not instantaneous velocities. jAy + jBy = 0. The Fourier’s law of heat conduction has been discussed in heat transfer in Section 6. we can say flux gradient momentum flux velocity gradient heat flux temperature gradient mass flux concentration gradient The constant of proportionality can be written as: n = molecular momentum diffusivity a = thermal diffusivity DAB = mass diffusivity All these diffusivities have the same units. (1. . All these equations are basically similar.6). (11. . (11.5) Combining all these equations. 3.1.6) where = divergence = . we can write the equation as jB = –tDBAwB . . 2. In brief. (11. (11.5) and (11. then the diffusivity of A in B and the diffusivity B in A are not equal.e. (11. . In a system when species A and B are moving. Let us look into Eqs. . i.9) where vAy = weighted velocity of A according to mass fraction of A in the y-direction.1 Some Features of Fick’s Law of Diffusion Let us discuss some of the features of the Fick’s law of diffusion. These fluxes are measured with respect to the motion of centre of mass of A and B. . (6. 1.2). when B is diffusing into A.1. Normally when A and B are diffusing in a system. The Newtons’s law of viscosity has been discussed in momentum transfer in Section 1.e.1. . i. vBy = weighted velocity of B according to mass fraction of B in the y-direction. . Then the mass fluxes can be defined as jAy = twA(vAy – vy) .and . . 4.7) 11.8) and jBy = twB(vBy – vy) . So. These are similar equations used for the estimation of viscosity as discussed earlier. If t is the density of the fluid and dS is the area of cross section perpendicular to the direction of flow. . In mass transfer process. In such cases. When a chemical reaction takes place. (c) Hydrodynamic theory as stated by the Nernst–Einstein equation is used for diffusion in binary liquids. vy.1 Mass and Molar Fluxes The diffusion equations are written along with the equation of motion when no chemical reaction takes place. (11.DAB DBA The diffusivity of a system can be estimated by a number of methods. In this book. z-directions. 11. mass may also be transferred by the motion of molecules of the fluid. increases with increasing temperature and is almost independent of the composition for a given pair of gases. For binary gas mixtures at low pressures. (b) Chapman–Enskog theory is applied for the estimation of diffusivity for gases at low density. one can express fluxes in either mass flux or molar flux. Some of them are: (a) Reduced temperature and pressure and the corresponding reduced diffusivity. dy and dz are the unit vectors in the x-. There are many empirical equations available in the literature for the estimation of diffusivity for gases and liquids. respectively. We express mass concentration. vz. Let us define the terms: Mass concentration of species A . the convective mass flux for component A can be written as tAv = tAvxdx + tAvydy + tAvzdz . convective mass flux = kg/m2 · s convective molar flux = kmol/m2 · s 11. . DAB is inversely proportional to pressure. then: Volumetric flow rate = vx dS Rate of mass flow in the x-direction = t vx dS Rate of mass flow in the y-direction = t vy dS Rate of mass flow in the z-direction = t vz dS Multiplying each term by a unit vector and dividing by dS. z-directions as vx.10) where dx. mass units are expressed. y-.2. This velocity will have three components in the x-. This is well discussed in other chemical engineering books. then molar concentration and molar fluxes are expressed. obtained from critical values. Each molecule is having some velocity v of the fluid in the bulk.2 CONVECTIVE MASS TRANSFER In addition to molecular mass transfer. we will not describe the estimation of diffusivity. y-. mass average velocity and mass flux for a particular species. We can summarize the gradient and Fick’s law of diffusion for mass and molar terms as shown in Table 11.3 SHELL MASS BALANCES AND BOUNDARY CONDITIONS .1A and Table 11. Let us say that the combined flux is the sum of the fluxes due to molecular diffusion and convective diffusion. (11. .1A Mass Fluxes Mass Units Mode Mass flux Gradient Fick’s law Molecular diffusion jA wA jA = –tDABwA Convective diffusion tv vA – v — Combined diffusion nA wA nA = tAv – tDABwA or nA = wA(nA + nB) – tDABwA Table 11. cA = number of moles of A per unit volume of solution The Fick’s law of diffusion can be written in mass and molar units as Mass units: jA = tA(vA – v) = –tDABwA . .1B respectively. vA – v* = diffusion velocity of species A with respect to the molar average velocity v*. (11.1B Molar Fluxes Molar Units Mode Molar flux Molecular diffusion Convective diffusion Combined diffusion cv NA Gradient Fick’s law xA = cDABxA * v– v — xA * NA = cv – cDABxA = xA(NA + NB) – cDABxA 11. Table 11. . .11) Molar units: = cA(vA – v*) = cDABxA .wA = mass of A per unit volume of the solution Molar concentration of species A.12) where vA – v = diffusion velocity of species A with respect to the mass average velocity v. Let NA be the number of moles of A diffusing per unit area per unit time in a binary system. Integrate the differential equation formed to get the concentration equation. The steps involved are as follows: 1. The reactions may take place by two mechanisms: 1. Homogeneous reactions 2. express the terms in molar form for molar balance. After formulation of the steadystate differential equations. A is diffusing but B is not diffusing. . (11. Otherwise. 5. Generally.14) We may come across the problem of solving for NBz. 9. One may say that in a system. Under steady-state conditions. we shall see that the steady-state diffusion problems may be formulated by shell mass balances. we select the combined flux and simplify the other terms. The molar flux of A in the z-direction can be written as: . . In other cases. Select the coordinate system of the problem. if mass balance is to be applied. NBz = 0. there is no source term in the differential equations in the shell mass balance.Steady-state momentum balances have been made in shell momentum balances in Chapter 1. 8. Heterogeneous reactions In homogeneous reactions. let us discuss the molar flux and mass flux produced by chemical reactions. Also. 11. where mass balance has to be applied. This can be solved by physical or chemical reasoning. Express the terms on the basis of mass. Similarly. Formulate the differential equation in terms of fluxes. 2. apply the shell balance as: Rate of “mass in” of A – Rate of “mass out” of A + Rate of mass of A produced by chemical reaction = 0 (11. For heterogeneous reactions. These differential equations can be solved with the help of boundary conditions. This is similar to the heat source problem for energy shell balances. Apply the boundary conditions to solve for the constants of integration. Before discussing the boundary conditions. we have made shell energy balances for heat conduction in Chapter 6. Select the shell. these equations have been solved by setting boundary conditions.13) 4. . 6. Integrate the differential equation formed and either keep the constant of integration or solve its value from the physical concepts. 10. Let the reaction be taking place as: A products The reaction rate equations can be written for homogeneous and heterogeneous reactions. convective or combined as the case may be. We have selected the fluxes as molecular. 3. NAz/NBz may be known from the physical or chemical concepts. Get the concentration profile and solve other important items. 7. the source term may appear in the differential equations. In such a case. Apply the Fick’s law of diffusion and formulate the differential equation for concentration. in mol/cm2 · s kn ≤ = reaction rate constant based on the surface area Boundary conditions: 1. . NAz = or = given or NBz = 0 for B is not diffusing. . The rate of chemical reaction may be specified. then we can write: molar flux concentration gradient or molar flux = kc(concentration gradient) or . SOLVED EXAMPLES .g. (11. in mole of A/cm3 n = order of reaction.16) where RA = rate of reaction. xA = . . . similar to heat transfer coefficient 4.g. The mass flux at the surface may be specified. . If the solid substance A is lost to the surroundings. . (11. for first order n = 1 NAz = combined molar flux.15) Heterogeneous reaction: NAz |surface = kn ≤ cAn |surface .17) where = molar flux of A at the surface = concentration of A at the surface cAb = concentration of A in the bulk fluid stream kc = mass transfer coefficient. . e. e. e. in mol/m3 · s kn = reaction rate constant cA = concentration of A. .g.18) where k1 ≤ = rate constant for the first-order reaction based on the surface area. 3. 2.Homogeneous reaction: RA = kncAn . (11. The concentration at the surface may be specified. . (11. or Now the molar flux of A becomes . The concentration gradient is constant over this distance and diffusivity is 1. .5 10–5 m2/s. Mole fraction of A (CO2) at z2 = 0. it is 0. .3 Example 11.1 + 15 = 288.2. .1 K z2 – z1 = 3 m Let us write the Fick’s law: .1 CO2 gas is diffusing through N2 in one direction at atmospheric pressure and temperature 15°C. The gas phase as a whole is stationary. (i) where c = concentration of the CO2–N2 system. at point B. .EXAMPLE 11. The mole fraction of CO2 at point A is 0. T = 15°C = 273. i.e. but in the opposite direction.0195.5 10–5 m2/s Figure 11. 3 m away in the direction of diffusion.1. (ii) where xA = mole fraction of A. DAB = 1.0195 Mole fraction of A (CO2) at z1 = 0. . N2 is diffusing at the same rate as the CO2.2 Total pressure = 1 atm = 1.0132 105 Pa Temperature. (a) What is the molar flux of CO2 in kmol/m2 · s? (b) What is the net mass flux in kg/m2 · s? Solution: Let A be the CO2 species and B be the N2 species. (iv) R = 8314 Substituting the above values in Eq.3 10–5 m2/s.32 kPa is 2. The diffusivity at 298 K and 101. (ii).38 10–7 kg mol of A/m2 · s For the component B. The partial pressure of NH3 is constant at 20. .61 m long with an inside diameter 24. = 1 – 0. i. Solution: Let A(NH3) be diffusing in the chamber as shown in Figure 11. (iv).Integrating Eq.e. .0195 = 0.38 10–7 kmol of B/m2 · s EXAMPLE 11. we get = 0.4 mm at 298 K and 101. . we get = –0. we have . (iii) or or .666 kPa in another. Both ends of the tube are connected to a large mixed chamber at 101.32 kPa. (v).4 and B(N2) be diffusing in the opposite direction.9805 The mole flux for B.2 Ammonia gas (A) and nitrogen gas (B) are diffusing in counter diffusion through a straight glass tube 0. (v) Substituting the values in Eq. . .32 kPa. N2.0 kPa in one chamber and 6. mole fraction of B.8 mole fraction of B. Calculate the diffusion of NH3 in kg mol/s and also the diffusion of N2. . = 1 – 0.2 = 0. . can be written as . . 4 mm = 0.03 10–5 kmol of A/m2 · s Diffusion of A (NH3): = 9. D = 24. T = 298 K Diameter of the pipe.32 103 – 20 103) = 81. (i) Substituting the values in Eq. (i). .32 103 Pa .Figure 11.666 103 Pa z2 – z1 = 0.61 m Total pressure. we get = 2.4 Example 11. .0244 m Let us write the molar flux of A: . DAB = 2. P = 101.2.3 10–5 m2/s = 20 103 Pa = 6.32 103 Pa Temperature.49 10–11 kmol of A/s Diffusion of B (N2): = (101. . (ii). the partial pressure the other end at 0. The data given is: Thickness of the silica plate = 10–2 mm Density of silica = 2. At one end of the pipe at point A (1).49 10–11 kmol of B/s PROBLEMS 1. DAB = 0. Is this relation true for a diffusion in a binary liquid system? 2. The partial pressure of helium is 1 atm at y = 0 and zero at the upper surface of the plate. Calculate the flux of He at steady state if DAB of theHe– N2 mixture is 0.2 10–7 cm2/s Make suitable assumptions.2 atm.= (101. .654 103 Pa .2 m.687 10–4 m2/s.0084 m3/m3 Diffusivity of helium–silica. A mixture of He and N2 gases is contained in a pipe at 298 K and 1 atm total pressure which is constant. Calculate the steady-state mass flux jAy of helium for the fusedsilica–helium system at 500 K. [Ans. Prove that DAB is equal to DBA for an ideal gas.666 103) = 94.03 10–5 (0. . of He is 0.0244)2 = –9.63 10–6 kg mol A/m2 · s] 3. we have = –2.6 g/cm3 Solubility of helium in silica = 0.6 atm and at = 0. 5. (ii) Substituting the values in Eq.32 103 – 6.03 10–5 kmol of B/m2 · s Diffusion of B = –2. All these problems are solved for the steady-state conditions.5. These boundary conditions are taken from the knowledge of physical concepts of the problem. in this chapter. This will result in the differential equation for concentration distribution. we can express the combined flux of A(NAz) in the z-direction as . benzene) is being evaporated into a gas B (e. The reaction may be instantaneous reaction or slow reaction. It will enable us to understand the mass balance easily. Next. we formulated the shell energy balance for heat conduction. The steps involved will be the same as those previously defined for momentum and heat balances. Section 12.4 deals with diffusion with homogeneous reaction. the mass flux. Let NA. These constants are evaluated with the help of boundary conditions. we will develop shell mass balances for laminar flow.1. .1 deals with the simple diffusion problem. This will lead to the solution for mass flux distribution. Diffusion from a spherical droplet through a stagnant gas film is discussed in Section 12. All these problems are solved for steady-state conditions. then NB = 0.1 DIFFUSION THROUGH A STAGNANT GAS FILM Let us consider that a liquid A (e. air). This diffusion system is shown in Figure 12. Mass balance is done over the shell to formulate the first-order differential equations. In the same manner.g. be defined as the number of moles of A diffusing per unit area per unit time. Section 12.g.3 discuss the diffusion problems with heterogeneous reaction. In Chapter 7.2 and 12.14) where if the B component is non-diffusing.12 SHELL MASS BALANCES AND CONCENTRATION DISTRIBUTION FOR LAMINAR FLOW In Chapter 2. (11. . 12. On integration. . we developed the shell momentum balance for viscous flow. Normally. we get the constants. or where mostly the NBz/NAz ratio is specified. Sections 12. we will introduce the relation between the mass flux and the concentration gradient. Several mass transfer problems are solved by considering the mass flux. (12. As soon as A evaporates. 8. Applying a mass balance at the shell.1. so diffusivity DAB is independent of composition. . .e. The solubility of B in A is negligible. Vapours of A and B form an ideal gas mixture. Let xA1 be the concentration of A at the liquid–gas interface. air is not soluble in benzene. in K R = gas constant.1) where pA = vapour pressure of A p = total pressure Also. Steady-state conditions prevail.e. the flows at the centre and near the wall. i. Let us consider a shell at a distance z and of thickness z as shown in Figure 12. The following assumptions are made: The level of liquid A is maintained throughout during evaporation. p = cRT where c = concentration. affect the diffusion process. No radial flow of A. 7. assumed constant T = temperature. Then. 6.e. 5. it is carried away by the B gas stream. . 1. unidirectional flow). Let the combined flux be NAz. 3. it flows only in the z-direction (i. 4. As the pressure and temperature are assumed constant. 2. As assumed above. we have .Figure 12. diffusion of A takes place in the z-direction only. After the evaporation of A. i.1 Steady-state diffusion of liquid A into stagnant gas B. . . (12.3) by S and z and taking the limit z 0. . .10) and in order to evaluate C1 and C2. Substituting the rate of mass in Eq.2) Let S be the area of cross section of the vessel in which evaporation takes place.4).Rate of “mass in” of A at z – Rate of “mass out” of B at z + z = 0 . (12. (12. . also we have assumed DAB to be constant). we get . . (12.5) or . we assume: C1 = –ln K1 . (12.14) When B is not diffusing into A.2). . which is . we get SNAz |z – SNAz |z+z = 0 .9) where C1 is a constant of integration. NBz = 0. (12. (12.6) Substituting this value of NAz in Eq. .7) For ideal gas We know that p = cRT where c is the concentration (assumed constant.7) becomes . . . (12. Integrating this equation again.14). we get . we get . .10) where C2 is a constant. . (12.3) Dividing both sides of Eq.4) We can write the equation for the combined flux NAz from the previously developed Eq. (12. . . (11. (12. . Eq. . . (12. we get –ln (1 – xA) = C1z + C2 . Looking into the pattern of Eq. .11) . . . .8). Therefore. (12.8) Integrating Eq. (12. (12. (11. . . (12. We obtain Eq. (12.C2 = –ln K2 . . xA = xA1 . (12. . . (12.18) Substituting this value of K1 in Eq.19) Substituting the values of K1 and K2 in Eq. . (12. (12.20) .14) Boundary condition 2: At z = z2. (12. (12. (12.16) .13) Boundary condition 1: At z = z1. (12. . .15) . xA = xA2 . . (12.16). . . . (12. . .17) Dividing Eq.17) by Eq. (12. we obtain or .12) where K1 and K2 are constants.16). . we get (12.13). we get or . .10) as: ln (1 – xA) = ln (K1z · K2) or 1 – xA = K1z · K2 . . (12. . (12. (12. . .1. . (12. (12.24) Boundary condition 1: At z = z1. (12. Let us write the average concentration for B: . . . . Once the concentration profiles are known. (12.27) or . . .25) Boundary condition 2: At z = z2. Z = 0 .28) . . The flux can be obtained from Eq. . . The slope dxA/dz is not constant. Average concentration Let . .29) . (12.or .21) For component B xB = 1 – xA . . . (12.6). then the average concentration and mass flux can be easily calculated.26) Now Eq. . (12. (12.24) becomes . . (12.23) where Z is the dimensionless distance. . Z = 1 .22) The concentration profile is shown in Figure 12. . Reactant A is diffusing in the positive direction of z.3 Idealized model for a catalytic reaction. of the terminal concentrations. There are spherical balls coated with a catalytic material. As the reaction takes place at the surface of the catalyst. we assume that 2 moles of A (reactant) produce 1 mole of B (product). diffusing through the gas film and the reaction takes place at the surface of the catalyst.2. Two moles of A are Figure 12. the reaction takes place instantaneously. concentration profile and a shell for mass balance. The following assumptions are made: . the gas film and the surface of the catalyst are as shown in Figure 12. Although some heat is produced in the catalytic reaction. We again assume an “idealized model” for the system. Reactant A diffuses through the gas film surrounded by the catalyst. (xB)ln. Then.2 Catalytic reactor and a catalyst. Also. we assume it to be very small and negligible. At the catalyst surface. while the product B is diffusing in the negative direction of z. the product B formed diffuses back to the main stream. After the reaction. Figure 12.Thus. the product B diffuses back in the mainstream. the average value of xB is the logarithmic mean.3. 12.2 DIFFUSION WITH A HETEROGENEOUS CHEMICAL REACTION (INSTANTANEOUS REACTION) Let us consider that a reaction takes place in a catalytic reactor as shown in Figure 12. we get or . . Therefore. i. . and B is moving in the negative direction of z. .35) . (12. DAB is constant. (12. (12. 2 moles of A penetrate the gas film and produce 1 mole of B. Gas film surrounding the catalyst is isothermal.31) The general Fick’s law can be written as .31). . Now applying a mass balance on the shell. .32). the temperature remains constant. the flux for B. can be written as .e. NAz is the combined flux operating in the system. (12. W as the width and L as the length of the shell for consideration. (12. . . . we obtain or .1. (12. .30) by WLz and taking the limit as z 0. we have W LNAz |z – W LNAz |z+z = 0 . . Steady-state conditions prevail. So. . we get . NBz.32) As discussed above. Let us consider a shell at a distance z and of thickness z. Let us take. (12.34) Substituting this value of NAz in Eq. . Diffusion of A and B takes place only in the z-direction. 3. (12. (12.30) Dividing both sides of Eq.33) Substituting this value of NBz in Eq. 2. xA = . (12.40) Let d be the gas film thickness as shown in Figure 12. (12. .37) where C2 is a constant.41) Boundary condition 2: At z = d. (12. . Now. For mathematical similarity. (12. we obtain . (12. we get . K2 are constants.42) Substituting these values and solving for the concentration of A. Boundary condition 1: At z = 0. .36) where C1 is a constant of integration. . . . (12. .Integrating this equation. . (12. (12.39) where K1.45) .36). we get . . . we may assume: C1 = –2 ln K1 .43) or . . .44) or or . or . . . . xA = 0 . (12. .38) C2 = –2 ln K2 . (12. .3. . . Integrating again Eq. . (12. r.e. (12. we obtain .t. we will get the same relation as that of Eq. we get or . Let us assume that after diffusion of A through the gas film. It is a series process. It is assumed that the rate at which A disappears at the catalyst surface is proportional to the concentration of A at the interface. reactions take place at the catalyst surface slowly. The whole process is explained as illustrated in Figure 12.49) where k1 ≤ is the rate constant for the first-order surface reaction. one comes across catalytic chemical reactions which are not instantaneous.e.Substituting the values of K1 and K2 in Eq. . We can write the flux as NAz = k1 ≤ cA = ck1 ≤ xA . it appears as if no reaction term appears and only diffusion takes place. . (12. we can conclude: 1. the reaction takes place.48) From the above.31). (12. .46) This is the concentration of A which is shown in Figure 12.46) w. (12. 12. . . we can say. is a “diffusion” controlled one for the case of an instantaneous reaction. (12. . (12. i. At the catalyst surface at z = d.47) Differentiating Eq.3. 2. Although chemical reaction occurs instantaneously at the surface of the catalyst. .3 DIFFUSION WITH A HETEROGENEOUS CHEMICAL REACTION (SLOW REACTION) Many a time. the reaction of A takes place at the catalyst surface. . i. . (12. z and taking the value at z = d. Mostly. The molar flux at the catalyst surface can be calculated as . First.2. After applying mass balance at the shell. 3. diffusion takes place and then the chemical reaction takes place. A to B formation.40). we obtain or . Substituting the value of boundary condition 2 in Eq.34). (12. (12.52) At steady state.40). (12. (12. we get or . . (12. we get the same relation as that of Eq. Differentiating Eq. (12.54) and evaluating at z = d and substituting in Eq. .53) Substituting the values of K1 and K2 in Eq. (12. one can find out the flux at the catalyst surface. (12. NAz = K1cxA .51) . .Integrating this equation and putting proper constants of integration. . i. . the flux NAz becomes constant. . (12. . Boundary condition 1: At z = 0. . . we get .40). . (12.50) Boundary condition 2: At z = d.40). .54) From this equation.e. The system is shown in Figure 12. .56) Two limiting cases may arise: 1.Na2CO3 + H2O In general. we say that when A is absorbed in B. (12. then the reaction can be written as CO2 + 2NaOH --. If CO2 gas is absorbed in a concentrated aqueous solution of NaOH. . The surface reaction kinetics is described by the term DAB/k1 ≤ d in Eq. Let us take liquid B in a vessel and allow gas A to be absorbed in it. neglecting the higher-order terms. where diffusion is negligible.55). . We have obtained the combined effect of diffusion and chemical reaction. Here.4. 12. We obtain the diffusion controlling mechanism. A typical example of such reactions is absorption of a gas in a liquid. a chemical reaction takes place. If k1 ≤ is large. 2.4 DIFFUSION WITH A HOMOGENEOUS CHEMICAL REACTION It is very common in chemical engineering that homogeneous reactions take place to get the desired product. .. a chemical reaction takes place to produce AB. (12. The other limiting case is the reaction kinetics controlling mechanism. (12. It is related to the Damköhler number which describes the reaction kinetics as . the term can be expanded.55) Let us analyse the result. . i. we assume that A and B form a binary mixture. Let NAz be the combined molar flux. . in s–1 3. Applying a mass balance at the shell.58) Let S be the area of cross section of the vessel. . we obtain . The concentration of AB formed is very small compared to the concentration of A or B. The following assumptions are made: 1. In other words. (12. SNAz |z – SNAz |z+z – k1cASz = 0 . Steady-state conditions prevail. . . (12. we have Rate of molar “mass in” of A – Rate of molar “mass out” of A – Depletion of A per unit volume = 0 .59) by Sz and taking the limit z 0.60) Applying the Fick’s law of diffusion. (12. Then. Let us consider a shell at a distance z and of thickness z as shown in Figure 12. we get .4. RA = k1cA . 2. . . the diffusivity DAB is constant.60). (12. The diffusion of gas A in liquid B takes place isothermally. (12. i. (12.Figure 12. 4.4 Absorption of gas A in a liquid B with a homogeneous reaction.e.e.57) where RA = rate of reaction k1 = rate constant based on volume. Therefore.59) Dividing both sides of Eq. . .61) From Eq. . (12. An irreversible first-order reaction takes place. . . . . .71) where C1 and C2 are constants and can be solved with the help of boundary conditions. . Let us write the boundary conditions: Boundary condition 1: At z = 0. i.62) can be solved with the help of definitions of the following dimensionless numbers. (12. .68) By using the dimensionless variables. . .e. i.. . . .e. (12. We know the concentration of A at the surface of the liquid. .67) where z is called Tiele modulus.69) This equation can easily be solved mathematically as (D2 – z2)b = 0 . . . NAz = 0.64) This second-order differential Eq.62) This differential equation can be solved with the help of boundary conditions. the differential Eq. (12. which is a dimensionless number. . b = C1e–za + C2eza . . . . It is the ratio of the chemical reaction rate to the diffusion rate. (12. (12. (12. From the physical concept. NAz = 0 or . (12. (12. . (12. i. . (12. . . (12. it is understood that no diffusion of A takes place at the bottom of the vessel.70) where D2 = z2 or D = z.e.63) Boundary condition 2: At z = L.62) becomes .66) . (12. .65) . . . . (12.76) The concentration profile is shown in Figure 12.75) Substituting the values of a.5. . (12.71) becomes b = C1 cosh za + C2 sinh za . b and z in Eq. . (12. .74) Solving for C1 and C2 values from the boundary conditions. (12. .5 DIFFUSION THROUGH A SPHERICAL STAGNANT GAS FILM SURROUNDING A DROPLET OF LIQUID Let us consider a spherical droplet of liquid A. . we get . (12. .75) from the above.73) The solution of Eq. . we get or . . (12. . Liquid A is being evaporated. Surrounding the liquid is a stagnant gas film B as shown in Figure 12.72) Boundary conditon 2: At a = 1. b = 1 . 12. (12. .Boundary conditon 1: At a = 0.4. . . we get Rate of “mass in” of A at r – Rate of “mass out” of A at r + r = 0 . we have . (12. we get r2NAr = C1 . DAB is constant. (12. NAr = NAr1 . It is assumed that the diffusion takes place only in the radial direction. we have . Boundary condition 1: At r = r1.79) Integrating Eq. The radius of the liquid droplet is r1 and the radius of the stagnant gas film is r2. (12. (12.78) by 4pr and taking the limit r 0.79). (12.78) Dividing both sides of Eq.80) where C1 is a constant. Let us take the spherical coordinate system for mass balance. It is also assumed that during diffusion. . the temperature remains constant.e. We consider that NAr is the mass flux operating in the system.Figure 12. .80).77) 4p(r2NAr)|r – 4p(r2NAr)|r+r = 0 . Let us consider a shell of radius r and thickness r as shown in Figure 12. . . (12. (12. .5. . These will be used as boundary conditions for evaluating the constants of integration. . . (12.5 Diffusion through a spherical stagnant gas film surrounding a droplet of liquid A. It is also assumed that the concentration of A in the gas phase is xA1 at r = r1 and xA2 at r = r2 at the outer edge of the film. Applying a mass balance over the shell. . i.81) Substituting this value in Eq. . 01325 105 Pa). . (12.675 10–4 m2/s. (12.01325 105 Pa Partial pressure of methane at point (1). . we get .83) between the limits at r1.85) This equation gives the concentration profile. .83) Substituting this value in Eq. . Solution: Let methane = A and helium = B z2 – z1 = 0.4 104 Pa at one end and 1. T = 298 K Total pressure. .79). . cDAB is constant. If we integrate Eq.84) At constant temperature..1 m Temperature. (12.1 Methane gas is diffusing in a straight tube 0. . xA = . . . p = 1. (12. The partial pressure of methane is 1. Integrating this equation.82) But by the Fick’s law. we get . NAr is given as . (12. (12. Helium gas is insoluble in methane and it is non-diffusing. .1 m long containing helium at 298 K and the total pressure is 1 atm (=1. The diffusivity of methane–helium is 0.333 103 Pa at the other end. xA = and at r2. we get . Calculate the flux of methane at steady-state conditions. (12.86) where SOLVED EXAMPLES EXAMPLE 12. . 29 cm2 exposed for evaporation. Calculate the evaporation rate in g/h of the liquid chloropicrin. we get = 1. The data given is: Total pressure = 770 mm Hg Diffusivity (CCl3NO2–Air) = 0.4 104 Pa Partial pressure of methane at point (2).675 10–4 m2/s The flux of A is given by the following equation where B is non-diffusing: Substituting the values.14 cm Density of CCl3NO2 = 1.333 103 Pa DAB = 0.2 Liquid chloropicrin (CCl3NO2) evaporates into “pure air” from a long tube of surface area 2. pA2 = 1.29 cm2 Solution: Let us consider CCl3NO2 = A Air = B .088 cm2/s Vapour pressure of CCl3NO2 = 23.pA1 = 1. The temperature is 25°C.81 mm Hg Distance from the liquid level to top of the tube = 11.028 10–5 kg mol A/s · m2 EXAMPLE 12.65 g/cm3 Surface area of the liquid exposed for evaporation = 2. 5 Substituting the values in Eq. . (i) Total pressure = 770 mm Hg Temperature = 25°C = 298 K = 770 mm Hg = 770 – 23. CCl3NO2 (liquid) is taken in a tube as shown in Figure 12.06 Molecular weight of CCl3NO2 = 164. (i).Figure 12. we get = 1. The evaporation rate is given by . .19 mm Hg z2 – z1 = 11.81 = 746.6.088 cm2/s R = 82.027 10–8 g mol/cm2 · s Evaporation rate in g/h . We assume steady-state conditions.6 Example 12.14 cm DAB = 0.2. [Ans. CO2 is diffusing through N2 in one direction at atmospheric pressure at 15°C.555 mm Hg.*. . The reaction is not instantaneous at the catalytic surface at z = d.8 kg/m3. Assume that the rate at which A disappears at the catalyst-coated . An ethanol (A)–water (B) solution in the form of a stagnant film of 2 mm thickness at 293 K is in surface contact with an organic solvent in which ethanol is soluble and water is insoluble. Its vapour pressure at 318 K is 0. (a) 3.0139 g/h PROBLEMS 1.5)(2. N2 is diffusing at the same rate as the CO2. The surface temperature of the naphthalene can be assumed to be at 318 K. [Ans. [Ans. the concentration of ethanol is 6. Deduce an expression to find the rate of diffusion of gas A through a non-diffusing gas B. . The diffusivity of ethanol is 0. .1 kg/m3. The concentration gradient is constant over this distance and diffusivity is 1. 3 m away in the direction of diffusion. The diffusivity of the naphthalene–air system at 318 K is 6. but in the opposite direction.74 10–9 m2/s. A sphere of naphthalene of radius 2 mm is suspended in a large volume of still air at 318 K and 1 atm pressure. Calculate the steady-state flux NA.92 10–6 m2/s. * 2.29) 3600 = 0.0195. . it is 0. Specises A is assumed to be a spherical particle of radius r1. The gas phase as a whole is stationary. At point 2.8% (wt) and the solution density is 972. and at point B.027 10–8 (164. 9. Derive an equation for the molar flux for a situation of diffusion of A in a stagnant gas B at steady-state conditions. NA = 9 10–7 kmol/m2·s] 4. (c) 1.5 10–5 m2/s.8 10–8 kg mol/m2 · s. At point 1 the concentration of ethanol is 16. .8% and density is 988. i. .68 10–8 kg mol/s · m2] 3.45 10–5 m/s] 5. . The mole fraction of CO2 at point A is 0.2. (a) What is the molar flux of CO2 in kmol/m2 · s? (b) What is the net mass flux in kg/m2 · s? (c) At what speed (m/s) would an observer have to move from one point to antoher so that the net mass flux relative to him would be zero? . .5 10–6 kg/m2 · s.e. Calculate the rate of evaporation of the naphthalene from the surface.= 1. (b) 1. Derive a concentration profile for a reaction 2A B. [Ans.000 and 6500 N/m2. with CO nondiffusing. The diffusivity of the mixture is 1.0208 cm3 CCl4 evaporates in a 10-hour period after steady state has been attained. The total pressure is 1 105 N/m2 and the temperature is 0°C. That is. . What is the diffusivity of the system CCl4–O2? . . respectively. It is found that 0. The partial pressures of oxygen at two planes 2 mm apart are 13. .67 10–5 kg mol/m2 · s] 7. The vapour pressure of CCl4 at that temperature is 33 mm Hg. 6.87 10–5 m2/s. 0. The length of the tube containing CCl4 is 17.82 cm2. 2. [Ans. The cross-sectional area of the tube is 0. NAz = k1CA where k1 is the rate constant for the pseudo-first-order surface reaction.0636 cm2/s] . The total pressure of the system is 755 mm Hg and the temperature is 0°C.1 cm above the gas–liquid interface. . The diffusivity of the pair O2–CCl4 is determined by observing the steady-state evaporation of liquid CCl4 contained in a vertical tube and in contact with O2 on top. Oxygen A is diffusing through carbon monoxide B under steady-state conditions. .surface is proportional to the concentration of A in the fluid at the interface. Calculate the rate of diffusion of oxygen. No chemical reaction takes place. 3. we have to start from the general mass balance equations and simplify these general equations with suitable assumptions. we applied the mass balance on the shell and then formulated the differential equation. Let NAx be the mass flux operation in the system. There is no generation of mass of the fluid. Now applying a mass balance over the element. In Chapter 3. these diffusion equations can be used to solve any problem.1. in kg mol/m2 · s. the diffusivity DAB remains constant. Figure 13. 2.1 A fluid element of xyz through which a fluid is flowing. Let cA be the concentration of the component A of the fluid. we shall develop the general diffusion equations in this chapter. The mass flux equations are set up from these general mass balances. it is not possible to formulate the problems by the shell balance. In Chapter 8. the general energy equations have been developed. Similarly. The fluid A is entering the shaded area at x as shown in Figure 13. Sometimes. Then.1 and leaving at x + x.13 THE GENERAL EQUATIONOF DIFFUSION In Chapter 12. Let us consider an element of x y z through which a fluid is flowing as shown in Figure 13. we developed the general equations of motion. For such complicated problems. We consider a binary mixture of the fluid. As the temperature of the fluid remains constant. we have Rate of “mass in” of A at the surface x – Rate of “mass out” of A at the surface x + x . The following assumptions are made: 1. (13. (13. (13. . (13. (13.2) Rate of “mass out” of A at the surface x + x = (yz)NAx|x+x . . . .5) by xyz and taking the limit as x 0.3) Rate of mass of A accumulated = (xyz) . let us substitute the NAx value from the Fick’s law of diffusion. . . we can also obtain the diffusion equations for the y. . . . .1) Rate of “mass in” of A at the surface x = (yz)NAx |x . .10) Hence we can write the general equation as . (13.5) Dividing both sides of Eq. . .12) with the heat transfer equation.4) Substituting these values in Eq. . . (13. (13.and z-directions as . . (13. we get .11) or . . . (13. . (13. . which is . (13.8) Similarly. . we obtain .12) where When we compare Eq.1). (13. .7) Then. (13. we get . .6) Now. .9) and . (13.= Rate of mass of A accumulated . . . Equation (13. (13. (13. . The molar diffusion equation can be written as . also called the second Fick’s law of diffusion. we can note a similarity between these two equations. (13. (13. If we include the convective mass transfer and the rate of reaction terms.12) is. . (13.13) where the second term v · cA contributes towards the convective mass transfer term. (13. the general equations of diffusion have been developed.where a is the thermal diffusivity. . . and RA is the rate of reaction.12) becomes . . therefore.14) for the binary system results in: (a) When tDAB is constant: .15) (b) When the velocity is zero: . The simplifications can be done for particular situations of mass transfer.14) The simplification of Eq. .16) In this chapter. . . then the general diffusion Eq. The instantaneous concentration of A and time are shown in Figure 14. one can obtain the concentration distribution. boundary-layer thickness and Prandtl mixing length model. There are many empirical correlations available in the literature for obtaining concentration distribution in turbulent flow conditions. Also. in Chapter 9. no turbulence. we discussed the time-smoothed velocity and time-smoothed temperature in turbulent flow. Similarly. Then. we can explain the concentration distribution in turbulent flow. Instead. the Fick’s law of diffusion has been incorporated. we assumed the laminar flow conditions.e. Let us consider here the fluctuations of concentration in turbulent flow. we will understand the concentration distribution in turbulent flow conditions by timesmoothed concentration. These fluctuations take place in all directions and in all species. In arriving at the concentration equation. . we assumed the steady-state conditions. 14. we discussed momentum transfer in turbulent flow. In Chapter 12.1.1 TIME-SMOOTHED CONCENTRATION IN TURBULENT FLOW In Chapters 4 and 9. we discussed heat transfer for turbulent flow conditions. i. By this method.14 CONCENTRATION DISTRIBUTION IN TURBULENT FLOW In Chapter 4. In this chapter. the equations of mass transfer or diffusion have been derived by considering the shell mass balances. Let us consider the concentration of only one component and in only one direction for understanding the problem. by analogy. It is not required to repeat the physical mass transfer phenomenon taking place in turbulent flow conditions. 3) . Let cA = molar concentration of A in the turbulent stream cA = concentration fluctuation of A = time-smoothed average concentration of A Then cA = + cA .1) The time-smoothed concentration of A can be measured experimentally by taking fluid samples at various points and at various times. .5) where is the average of the concentration fluctuations and is the average of the product of concentration fluctuations and velocity fluctuations in the x-direction. By analogy. The will vary slightly with position in the turbulent core region. Let us write the equation of continuity for rectangular coordinates for the first-order chemical reaction (the order of reactions can be of any order): . (14. .Figure 14. . (14. i. (14.e. . . . (14. as done for the velocity fluctuations. .2) . .4) . whereas it varies to a large extent near the solid surface. .1 Concentration fluctuations in turbulent flow condition. . the properties of the concentration fluctuations will be similar in nature to those of velocity fluctuations. (14. . No diffusion takes place in the x. So. .e. The following assumptions are made: 1. We used the Blasius solution in the case of hydrodynamic boundary layer and thermal boundary layer. we can write the fluxes for the turbulent flow conditions as follows: Turbulent diffusion molar flux: Turbulent momentum flux: Turbulent heat flux: 14. vz = 0. the only difference being that cA is replaced by which is the time-smoothed concentration. . RA = 0. We may thus conclude that the concentration terms in the diffusion equations may be replaced by the time-smoothed average concentration for turbulent flow conditions. Let us consider a flat plate over which the fluid of constant concentration cA is entering as shown in Figure 14. Similar is the situation for the temperature distribution equations. it appears to be the same. the Fick’s law of diffusion is applicable. In an analogous manner. vx with . Laminar flow conditions exist near the surface of the flat plate.2.. we obtain .e. . Similar concepts are applicable for the mass transfer too. i. .. Steady-state conditions prevail. No chemical reaction takes place. Here.e.7) When we compare this equation with the general equation of concentration. i.2 BOUNDARY-LAYER THICKNESS FOR MASS TRANSFER Earlier in Chapters 4 and 9. (14. 3. (14. We have also seen earlier that the velocity terms are replaced by the time-smoothed velocity in the equation of continuity and equation of motion for turbulent flow conditions. 2. we discussed the boundary-layer thickness for flow and heat transfer conditions. Now replacing cA with . . etc.6) where k1 is the rate constant for the first-order reaction. No flow occurs in the z-direction i. 4. 5. It may also be noted that the reaction terms may differ for the higher-order reactions. again we consider thex–y coordinate only. we can use the Blasius solution for convective mass transfer for the laminar flow over a flat plate.and z-directions. Figure 14.2 Concentration on boundary layer of a fluid past a flat plate. Let us define the following terms: cA = concentration of the fluid approaching the plate cAS = concentration of the fluid adjacent to the surface of the flat plate. cA = concentration of A in the boundary layer The general diffusion Eq. (13.14) is written as: . . . (14.8) Considering the above assumptions, we can simplify and write the diffusion equation as: . . . (14.9) This equation can be solved with the help of boundary conditions: Boundary condition 1: At y = 0, cA = cAS . . . (14.10) Boundary condition 2: At y = , cA = cA . . . (14.11) The similar boundary-layer equation for momentum transfer can be written as: . . . (14.12) The thermal boundary equation can also be similarly written as: . . . (14.13) where o = momentum diffusivity = a = thermal diffusivity = . All these equations are solved with the help of boundary conditions and the equation of continuity, i.e. . . . (14.14) Equation (14.9) can be solved with the help of the dimensionless concentration boundary conditions, i.e. Boundary condition 1: At y = 0, Boundary condition 2: At y = , . . . (14.15) . . . (14.16) For the momentum boundary-layer Eq. (14.12), the dimensionless boundary conditions are: Boundary condition 1: At y = 0, Boundary condition 2: At y = , . . . (14.17) . . . (14.18) Similarly, the dimensionless boundary conditions for the thermal boundary-layer Eq. (14.13) are: At y = 0, At y = , . . . (14.19) . . . (14.20) All the boundary-layer Eqs. (14.9), (14.12) and (14.13) for mass, momentum and heat transfer are similar in nature. Also, when we look into the dimensionless boundary conditions for mass, momentum and heat, the boundary-layer equations are similar in nature. Blasius has formulated the solution for the convective mass transfer when the Schmidt number is 1.0. The Schmidt number is defined as: . . . (14.21) i.e. o = DAB The solutions for the momentum boundary-layer equations are valid for the mass boundary-layer equations for Schmidt number 1.0. The Blasius solution for the momentum boundary-layer equation is given as . . . (14.22) where Comparing Eqs. (14.15) and (14.17), we get . . . (14.23) Differentiating Eq. (14.23), we have . . . (14.24) or . . . (14.25) Substituting this value in Eq. (14.22), we get . . . (14.26) The convective mass transfer equation can be expressed in terms of mass transfer coefficient as . . . (14.27) where k¢c is the mass transfer coefficient. The mass flux can also be written as . . . (14.28) Now combining Eqs. (14.26) to (14.28), we get . . . (14.29) where = Sherwood number = NSh. Equation (14.29) holds good only when the Schmidt number is one. For the Schmidt number not equal to 1.0, Pohlhausen gave a relationship between the hydrodynamic boundary layer d and the concentration boundary-layer thickness dc as: . . . (14.30) As discussed above, the local convective mass transfer coefficient is given as . . . (14.31) Let L be the length of the plate and B its width. The mean mass transfer coefficient can be calculated as: . . . (14.32) Substituting the value of k¢c from Eq. (14.31) and integrating, we get . . . (14.33) This equation is similar to Eq. (9.25) for heat transfer, i.e. where 14.3 PRANDTL MIXING LENGTH MODEL IN MASS TRANSFER In Chapter 4, the Prandtl mixing length in the turbulent flow condition has been discussed. Similarly, the Prandtl mixing length model has been dealt with in Chapter 9 for heat transfer. In the mass transfer operation, mostly the flow pattern is turbulent. The turbulent condition enhances the mass transfer rates tenfolds than that in the laminar flow. As discussed in the momentum transfer, the turbulent flow is very complex in nature. It is not easy to understand the mass transfer in the turbulent flow conditions. It was explained in momentum transfer that eddy formations occur in turbulent flow conditions. Similarly, the fluid undergoes random eddy movements throughout the turbulent core region. When the mass transfer takes place, we call this eddy diffusion. Prandtl postulated that eddy diffusion takes place in turbulent flow. After eddies are formed, they travel a small distance L, called the Prandtl mixing length, before their entities are lost. This process is similar to the momentum transfer and heat transfer. Let cA |y be the concentration of A at y and cA |y+L be the concentration of A at y + L. Here, L is the small distance which the eddies travel. (14. . .39) The term is given as: is called eddy mass diffusivity. .36). . Let cA be the deviation of concentration of A from the mean concentration and let concentration of A.35) The rate of mass transferred per unit area (mass flux) is velocity . (14. (14. . . (14. we get . (14. . For the molar mass diffusion. .34) be the mean Dividing both sides of Eq.3 Eddy concentration in the y-direction. . . . (14.37) Earlier in momentum transfer in Chapter 4. we arrived at Eq. .38) Substituting the value of from Eq.36) Combining Eqs. (14.38) in Eq.35) and (14. the Fick’s law . fm. (14. The rate of mass transfer of A at the for a distance L in the y-direction is given as: . we get . (14.38): . (14. (14. we get .34) by L and taking the limit when L 0.37). The fluctuation of concentration can be written as: .Figure 14. (14.41) Similar equations have been derived earlier for the momentum and heat transfer. . In this chapter we have explained the mechanism of mass diffusion by three models: time-smoothed concentration. .43) where ft = eddy momentum diffusivity at = eddy thermal diffusivity. (14. . .42) Heat transfer: . . we obtain the total mass flux as: . boundary-layer thickness. (14.39) and (14. .. . .40). (14. . The fluxes are given as: Momentum transfer: .40) Combining Eqs. and Prandtl mixing length. (14. the steady-state diffusion problems have been solved.1) where xA0 = interfacial gas-phase concentration. . Temperature and pressure are constant. Let us define the dimensionless terms: . The molar average velocity in the gas phase does not depend on the radial coordinates. xA = xA0. This equation can be solved with the help of boundary conditions: Initial condition: At t = 0. 4.1). 2. . Many diffusion problems can be solved by looking into the solutions to the analogous of heat and momentum transfer. As the chemical reactions take place along with diffusion. these problems become more and more difficult. the same method of combination of variables will be used to solve Eq. where a liquid A (volatile) evaporates into pure B in a tube of infinite length. (15. c and DAB. In such problems. xA = 0. . . the liquid level is maintained at z = 0. .2) Boundary condition 1: At z = 0. . 5.15 UNSTEADY-STATE EVAPORATIONOF A LIQUID In Chapter 12. B is insoluble in A. . (15. are constant. These equations have been solved with the help of boundary conditions to give the concentration profiles. .3) Boundary condition 2: At z = . . The molar densities. With the above assumptions. we can write the equation of continuity as . . we shall try to understand how to solve simple diffusion problems under unsteady-state conditions. Vapours of A and B form an ideal gas mixture. xA = 0. But we come across binary or multi-component diffusion problems under unsteady-state conditions. (15. The following assumptions are made: 1. . Let us take a case. Somehow.4) As we have already solved similar equations for momentum and heat transfer. (15. . These problems can be solved with the help of computers and analytical mathematical tools. 3. (15. the ordinary differential equations have been formed by shell mass balances. In this chapter. . .7). X = 0. . . (15. .8) . . (15. .6) Then.5) .14) Combining Eqs.. (15. . (15. (15.15) Now. . we get . . . . . (15. .16) . . we can introduce which will give the first-order differential equation as: . Eq. we get . . . (15. .11) Boundary condition 2: At z = . (15. (15. X = 1. . . . .10) Boundary condition 1: At z = 0.9) The initial and boundary conditions become Initial condition: At t = 0.1) will become . . . . .11) and (15. . . we can write the solution as: .13) On integration. . (15. (15. (15.12) In Eq. . (15. . . . . X = 0.7) where . . with the definition of error functions and their properties. (15. (15.12). . 17) It is easier to solve for xA0 as a function of z. we get . . Unsteady-state diffusion situations are more difficult to formulate. the concentration profile can be evaluated and drawn. . (15. therefore. .8). A sample case of evaporation of liquid is explained in this chapter as an unsteady-state type of diffusion problem which has been solved analytically. . . we obtain . .18) For different values of xA0 and z. .Now substituting the value of z (xA0) from Eq. (15. (15. Section D ANALOGIES AMONG MOMENTUM. HEATAND MASS TRANSFER . Analogies were developed for momentum and heat transfer. heat and mass. The concept of friction factor in momentum transfer can lead us to evaluate the heat transfer coefficient by applying the Chilton–Colburn analogy. Mathematically too. 2. Analogies help us understand the new processes in momentum. Applications of Analogies 1. 3. it can also be understood for heat transfer and mass transfer. In such cases. Then. Also. heat and mass transfer. modern instrumentation for the measurement of concentration of species was developed. by analogy. It can help us calculate the heat transfer . and to the professional as a sound means to predict the behaviour of the systems for which limited quantitative data are available.16 ANALOGIES AMONGMOMENTUM. All the three of the molecular transport processes —momentum. In laminar flow conditions. Analogies will be used to elucidate the mechanism of transfer. This helped understand mass transfer operations.e. In the later period. simple equations explained the momentum transfer under laminar and turbulent flow conditions. the mathematics involved is simple in nature. the analogy will be drawn in terms of mechanism and mathematical expression. i. The analogies are useful tools to understand the concept of transfer phenomena. If we learn some phenomena in momentum transfer. But the situation becomes complex under turbulent flow conditions. Correlations were developed for friction factor and Reynolds number. Analogies were developed among momentum. the flow behaviour of momentum transfer has been easy to understand. A typical example is the Reynolds analogy and the Prandtl analogy for the case of momentum and heat transfer. In all the cases. with the development of flow meters like orifice meter and Pitot tube. it is easier to understand the mechanism of all these transfer processes. we are considering the transfer of momentum. Hence an analogy is an inference from one particular process to another where the conclusion is general. heat and mass transfer. First. empirical correlations have been developed by scientists and technologists. HEATAND MASS TRANSFER What is an analogy? Resemblance and similarity are closely related to the term analogy. heat and mass—are characterized by the same general type of equation as already discussed. Here. temperature-measuring devices were developed to understand the mechanism of heat transfer. the mass transfer coefficient can be evaluated from the friction factor in momentum transfer. The von Kármán analogy for the turbulent flow condition is explained in Section 16.4. heat and mass transfer operations. 4. HEAT AND MASS TRANSFER The momentum.2 discusses the Reynolds analogy and its limitations. Laminar Flow (a) Mechanism of molecular transport: In laminar flow conditions. In this case. Section 16.2) Heat transfer: Fourier’s law of heat conduction is: . Analogies will be drawn among the transfer-rate correlations for laminar and turbulent flow conditions.1) For constant density: . heat and mass transport take place by the molecular transport mechanism. (b) Laws: Momentum transfer: Newton’s law of viscosity is: . e. heat and mass transfer processes take place almost simultaneously with a general similar behaviour. So there is a gradient in which the transfer takes place.1 deals with the general analogy concepts among momentum. The common features are in terms of mechanism and mathematical equations for momentum. molecules of the fluid are stationary. momentum.area in the design process of a heat exchanger. momentum. heat and mass transfer. why does the transfer of momentum. . For momentum transfer.g. heat and mass in all the directions. the mechanism and mathematical concepts are explained for laminar and turbulent conditions. . The Prandtl analogy is dealt with in Section 16. . 16. heat or mass. whereas mass transfer takes place from a higher concentration to a lower concentration. Another question which is generally asked. A close relationship exists among the three transfer phenomena for laminar and turbulent regimes.1 ANALOGY AMONG MOMENTUM. The Chilton–Colburn analogy among momentum. there is a velocity gradient. In this chapter. The flow takes place from a higher velocity to a lower velocity.3. . Heat energy flows from a higher temperature to a lower temperature. (16. heat and mass transfer is discussed in Section 16. In this part. (16. As discussed in Chapter 1. Individual molecules containing the property are transferred. heat and mass take place? There is a flux of momentum. 1. There are many common similarities between these different processes. The Chilton–Colburn analogy can be extended to the mass transfer and momentum transfer. Section 16.5. in J/s · m2 Mass transfer: Mass flux J*Ax. . . cA. in (kg · m/s)/m3 Heat transfer: Concentration of thermal energy tcpT. (16.3) For constant density and thermal conductivity: or . Heat transfer: ·Heat flux q/A.H.7) . .5) Now let us look at Eqs. .: flux Momentum transfer: Momentum flux tyx (shear stress).. DAB. in m2/s Heat transfer: Thermal diffusivity k/tcp. in J/m3 Mass transfer: Concentration of A.e. mathematically given by . in m2/s We conclude that all the three molecular transport processes—momentum. (16. .: driving force Momentum transfer: Momentum concentration tv. L. . heat or mass. . in kg mol of A/m3 (c) Constants: Momentum transfer: Momentum diffusivity n/t. in m2/s Mass transfer: Molecular diffusivity of A in B. (16.6) for molecular diffusion of the property momentum.4) and (16. . .5). heat and mass—are characterized by the same general type of equation.2).S. (16. i. can be written as . (16.4) Mass transfer: Fick’s law of diffusion is: .H. (16. . force/area. in R.6) Equation (16.S. (16. . (16. .8) Heat transfer: . However. these similarities are not well defined physically or mathematically and are more difficult to correlate with each other. these equations and notations are similar. heat and mass and transports it. = gradient in the x-direction d = constant. Each molecule carries the property of momentum. Mathematically. . The flux equations are written using the turbulent eddy momentum diffusivity ft. In all these equations. 9 and 14. Convective Transfer This mechanism of transport has been explained in Chapter 1. (16. Turbulent Flow We have already discussed turbulent flow conditions in Chapters 4.where zx = flux in the x-direction. 2. . These equations can be written as: Momentum transfer: . but the process of momentum.9) Mass transfer: . .10) where . The fluxes are given by: Momentum transfer: vt v Mass transfer: c(v – v*) where v = vector fluid velocity t = internal energy flux per unit volume c = concentration 3. heat and mass may be different. time-smoothed average temperature for heat transfer and time-smoothed average concentration for mass transfer. . we have used the time-smoothed average velocity for momentum transfer. (16. the turbulent eddy thermal diffusivity at and the turbulent eddy mass diffusivity fm. . The fluid behaviour near the solid surface is different than that in the bulk fluid.x = 0. . (16.ft = momentum eddy diffusivity. the hydrodynamic boundary-layer thickness d where vx = 0.99v Blausius also proved that the thermal boundary-layer thickness dT is equal to the hydrodynamic boundary-layer thickness. . . we can write the general equations for momentum. in m2/s = time-smoothed average velocity = time-smoothed average temperature = time-smoothed average concentration Combining the Eqs.8) to (16. (16.x)1/2 . Boundary-Layer Thickness The boundary-layer thickness for momentum.e. The Nusselt number is given as NNu. . .14) i. (16. (16. Blausius developed the theory for the momentum transfer. heat and mass transfer has been discussed in Chapters 4.13) 4. . 9 and 14 respectively. (16. and the boundary-layer thickness d is given as: .10) with the laminar flow conditions. . heat and mass transfer: Momentum transfer: .12) Mass transfer: .15) . in m2/s at = thermal eddy diffusivity.332(NRe. . (16. .11) Heat transfer: . in m2/s fm = mass eddy diffusivity. This means that the transfer of momentum and heat are directly analogous. we can say that transfer of momentum. .18) We conclude that the local mass transfer coefficient is directly proportional to the square root of distance. (16. (16. heat and mass transfer have been discussed in Chapter 4.x = Sherwood number = where kc = mass transfer coefficient at a point x. (16.17) where NSh. . These equations can be written as: Momentum transfer: .where hx = local heat transfer coefficient at point x. . (16.20) where . . 9 and 14 respectively. Prandtl Mixing Length Prandtl mixing length models for momentum. .19) where = time-smoothed average velocity in the x-direction L = Prandtl mixing length = momentum flux for turbulent flow Heat transfer: . In turbulent flow conditions it is assumed that eddies move a little distance called the Prandtl mixing length before losing their identity. . . 5. Similar notions hold good for the mass transfer case too. the Sherwood number is given as . .16) We conclude that the local heat transfer coefficient is directly proportional to the square root of distance. For mass transfer. . . . (16. As seen above. . heat and mass are directly analogous. Similar phenomena occur in heat and mass transfer. These eddies mix afterwards with the mainstream of the fluid. 6.21) where = mass transfer flux = time-smoothed concentration of A. (16. Mass transfer: . heat and mass transfer.19). respectively. many empirical correlations can be developed between these processes by the dimensionless groups. . heat and mass transfer are very much helpful in understanding these processes. Heat transfer: . Momentum transfer: Reynolds number = = (for circular pipes) Froud number = = Weber number = NWe = where v = interfacial tension. . Equations (16. are similar in nature. Also. Dimensionless Groups The dimensionless groups in momentum.20) and (16. (16.= time-smoothed average temperature cp = specific heat capacity qy = heat flux in the y-direction.21) for momentum. these processes are easy to formulate. heat transfer and mass transfer are very well understood. .2 REYNOLDS ANALOGY The processes of molecular momentum transfer.Prandtl number = = Nusselt number = = Peclet number = = (Reynolds number) (Prandtl number) Mass transfer: Schmidt number = = Lewis number = = Sherwood number = where D is the characteristic diameter. Fourier for heat. 16. The molecular diffusion equations of Newton for momentum. Mathematically too. and Fick for mass transfer are very similar. (16. For molecular heat transfer. All efforts have been made in the literature to develop the analogies between momentum. However. the similarities in turbulent flow conditions are more difficult to correlate mathematically or physically.22) and (16. . The flux equations for momentum. (16. . and 14 respectively. . Reynolds was the first to develop the similarity between heat and momentum transfer for turbulent flow conditions.These have been discussed in Chapters 1.24) Similar equations can also be written for heat transfer. the fluxes are written using the turbulent eddy momentum diffusivity ft. Momentum and Heat Transfer For the molecular momentum transfer. and the turbulent eddy mass diffusivity fm.23) where = turbulent momentum flux nt = eddy viscosity. heat and mass under turbulent flow conditions.22) where = n = molecular momentum diffusivity. Let us first discuss the Reynolds analogy between momentum and heat transfer and then that between momentum and mass transfer. There are also similarities in turbulent transport. 9. . the momentum flux for the turbulent flow is written as or . in m2/s For constant density. the Newton’s law can be written as . 1. combining Eqs. the turbulent eddy thermal diffusivity at. In turbulent flow. . (16. in m2/s t = constant Similarly. property of flow conditions and not the fluid property = time-smoothed average velocity in the x-direction ft = momentum eddy diffusivity. 6 and 11 respectively. the Fourier’s law of heat conduction can be written as: . we obtain . (16. heat and mass transfer in turbulent flow conditions have been discussed in Chapters 4.23). . 26) where a = thermal diffusivity. n/t and a are small. 2. (16. the molecular momentum and heat transfer are negligible. (16. (16.25) For constant density and specific heat. (16.27). we get . in m2/s = For turbulent flow conditions: . Now. . at = ft.28) The following assumptions are made: 1. . Let T = bulk temperature of the fluid Ti = wall temperature of the pipe . (16. we get .26) and (16.e. in m2/s = time-smoothed average temperature. . we can write ..28) and simplifying with normal notations for temperature T and velocity v. . combining Eqs. . . momentum eddy diffusivity and thermal eddy diffusivity are equal. .27) where at = turbulent thermal diffusivity. The heat flux q/A is analogous to the momentum flux t which is considered constant in the turbulent flow condition. dividing Eq. i. . (16. For constant t and cp.24) by Eq. (16. i. . (16. . In turbulent flow.e.29) Let us consider a circular pipe through which the fluid is flowing. we get . . . (16. (16. (16. .35) Substituting the value of q/A from Eq. We can write the R. (16. (16. the heat transfer is given by the Newton’s law: = h(T – Ti) .e.S.34) .29) between the limits. i.33) i. . (16. where the velocity at the wall is zero. (16. or .32) From the fluid flow. . .36) as . . (16. . we can evaluate the shear stress at the wall from the Fanning friction factor concepts.ts = shear stress at the wall vav = average fluid bulk velocity. Integrating Eq.e. .37) where . .31) From the heat transfer coefficient h. .31). (16. (16. (16.32) and that of ts from Eq.36) where G = mass velocity. . .35) in Eq.30) or . . (16. . in kg/s. . we get .H. of Eq. (16. . As we have defined the heat transfer coefficient by Eq. In literature for circular tubes. NNu = Reynolds number. . 2. . i. . NRe = Prandtl number. .40). (16. And the Fick’s law is written as: . (16.36) holds good only when the Prandtl number is equal to 1.24). cAi = concentration of A at the interface.= Stanton number. . NPr .41) . we can write the mass transfer coefficient as . Momentum and Mass Transfer For mass transfer.32).40) By combining Eqs. (16.39) and (16.9 and for other conditions. in kg mol/s · m2. (16. it varies from 0. conc. in kg mol/m3. similarly.38) The Reynolds anology assumes that Eq. integrating between the limits as done for heat transfer and assuming the Schmidt number to be 1. NSt = Nusselt number.0. NSc = Schmidt number = we obtain . kc = mass transfer coefficient. This is the limitation of Reynolds analogy. (16. (16. similar equations can be written as done above for molecular mass diffusion and turbulent mass diffusion by using the Fick’s law. the Prandtl number varies from 0.7 to 0. . (16.0.e. . in kg mol/m3. . (16. diff.39) where cA = concentration of A.5 to 1.0. In 1910. Laminar sublayer (near to the wall) 2. There is no mixing and turbulence in this sublayer. all the laws of the laminar momentum transfer are applicable to this viscous sublayer. The Prandtl analogy considered momentum and heat transfer. there is no eddy formation and no turbulence and the velocity of the fluid is very small. For example. i. (16. all the equations and laws of momentum and heat transfer for laminar flow conditions pervail.36) and (16. combining Eqs. There are two layers of movement of momentum and heat: 1. Zone I. It is postulated that there are four zones of the fluid behaviour as shown in Figure 16.3 PRANDTL ANALOGY When a flowing fluid comes in contact with the solid surface. heat and mass transfer. very close to the solid surface. So.Zone I: viscous sublayer. Figure 16. the velocity is very small and as we go away from the wall.42) 16. . the velocity of the fluid at the wall is zero and at the centre of the pipe. Let us consider the transport of momentum and heat in a circular pipe under the turbulent conditions. . (16. Zone III: inertial sublayer.41). the Newton’s law of viscosity is applicable. In the viscous sublayer. . Near the wall. we get the Reynolds analogy between momentum. if the fluid is flowing in a circular pipe. . Prandtl modified the Reynolds analogy by considering the velocity distribution in the laminary sublayer and the turbulent core region. it is maximum.1. Zone IV: turbulent core zone. the fluid characteristics of flow vary in the vicinity of the solid surface. its velocity is zero at the surface.Now.1 Four zones near the solid surface of the fluid for turbulent flow. Turbulent core region (centre of the pipe) In the laminar sublayer. Zone II: buffer sublayer. the velocity of the fluid increases. Hence.e. Eddy transport is negligible in the laminar sublayer. We can write the Newton’s law of viscosity for momentum transfer and the Fourier’s law of heat conduction for heat transfer for laminar sublayer as follows: Momentum transfer: . 5. Temperature and velocity profiles in the laminar sublayer are linear.e. 4.2 Temperature and velocity profiles in the laminar sublayer in a circular pipe. Steady-state conditions prevail. The thermal diffusivity is equal to the momentum diffusivity in the turbulent core region. i. ft = 0 and at = 0. Figure 16.2. The molecular momentum diffusivity and the thermal diffusivity are negligible in the turbulent core region. Tw is the wall temperature. momentum eddy diffusivity and thermal eddy diffusivity are zero.The following assumptions are made: 1. Let us consider the flow and heat transfer in a circular pipe. 2. Let d = laminar sublayer thickness n = viscosity of the fluid (constant) t = density of the fluid (constant) v = axial velocity of the fluid (it is a function of radial position) cp = specific heat of the fluid (constant) k = thermal conductivity (constant) r = radial distance of the pipe. The velocity and temperature profiles are shown in Figure 16. 3. . . . .48) where ft = turbulent eddy momentum diffusivity.49) and (16. (16. . . (16. (16. (16. . . (16. .at = 0 in Eqs. . Eliminating d from Eqs.e. . ft = 0. we obtain .45) .50) where qw = heat flux at the wall vd = velocity of the fluid at the edge of the laminar sublayer Td = temperature of the fluid at the edge of the laminar sublayer tw = momentum flux at the wall.46) For turbulent flow conditions.48) and integrating over the limit d for r. . in m2/s at = thermal eddy diffusivity.. i.43) .47) Heat transfer: .51) . . . in m2/s. these equations are modified to include eddy terms: Momentum transfer: .49) . . we get .50).44) Heat transfer: . (16. .47) and (16. . (16. (16. (16. . (16. (16. Substituting the values of ft and at for the laminar sublayer. tw. .55). (16.48) become .54) and the heat flux at the wall. we get or . is given by . . . . .55) where f = Fanning friction factor h = heat transfer coefficient Tb = bulk fluid temperature Also. (16. is given by the Newton’s law of cooling .60) Eliminating Td from Eqs. .57) Substituting these terms in Eqs. .58) .56) where vm = mean velocity of the fluid.51) and (16. .47) and (16. (16. It is assumed that the momentum eddy diffusivity ft is equal to the thermal eddy diffusivity at.52) and (16. . . i. qw. (16. . . . we obtain . . (16. .59) Now.e. Eq. (16.53) and introduce the following concepts. v = vm – vd .53) Let us integrate Eqs. the velocity gradient is (vm – vd) and the temperature gradient is(Td – Tb). Here. . (16.60). we get .54) and (16.52) . (16. . Let R be the radius of the pipe. Neglecting the molecular momentum and molecular heat transfer terms. (16. (16. ft = at . (16. . (16. The shear stress at the wall. (16.Now consider momentum and heat transfer for the turbulent core region beyond r = d. we define the dimensionless velocity vd+ and dimensionless distance y+ as . . . .4 VON KÁRMÁN ANALOGY . vd can be expressed in terms of tw whereas vm can be expressed by using the universal velocity in the laminar sublayer.63) in Eq.62) which is applicable for y+ = 5. Also it gives a relationship between the Stanton number and the Fanning friction factor as .63) Substituting the value of vd from Eq.65) This relationship is called the Prandtl analogy. . .. (16. . (16. .61) and using the following values (Prandtl number) we get . (16. .61) Now. In the laminar sublayer. (16. . (16. . (16. . .64) or . So. (16.66) where 16. (16. . the turbulence just starts and viscosity plays a minor role. the universal velocities are: Zone 1: Viscous sublayer . In the laminar flow. There are many empirical expressions developed in the literature for turbulent flow conditions. As discussed above. the fluid behaviour is very different than that under the laminar flow conditions. . (16. the viscous sublayer and the turbulent bulk zone have been considered.67) Zone 2: Buffer sublayer . the situation is simple and the mechanism is easily understood. (16. Zone 3: The inertial sublayer: In this zone.3 Flow regions for turbulent flow near a flat surface. Figure 16. . . Zone 4: The main turbulent stream: In this zone. In the Prandtl analogy. . . the time-smoothed velocity distribution is important and viscosity is negligible.68) Zone 3: Turbulent core region . Zone 2: The buffer sublayer: This is the transition zone which occurs between the viscous and inertial sublayers. Von Kármán modified the Prandtl analogy by considering the buffer sublayer as well.3. . It is convenient to describe the flow of fluid near the solid surface by four zones. (16. The zones are small and close to the solid surface (but for clarity these are shown bigger).Under the turbulent flow conditions. Zone 1: The viscous sublayer: This zone is very close to the wall in which the viscosity of the fluid plays an important role. . Von Kármán developed the analogy based upon the universal velocities for a smooth circular pipe. Let us analyse the flow near a flat surface as shown in Figure 16.69) where v+ is the dimensionless velocity and y+ the dimensionless distance. 72) . . (16. . . along with viscous sublayer and turbulent core zone. .4. Von Kármán included buffer zone too. Figure 16.73) where n = kinematic viscosity.Now. (16.69) for these zones are shown in Figure 16. The Nusselt number is given by the following relation. . . Equations (16. (16. n/t t0 = shear stress at the wall. in terms of friction factor: . (16. . . .4 Universal velocity profile for turbulent flow in circular pipe. .70) . (16.71) or . .67) to (16.74) . 75) whereas for turbulent conditions.g.78) for Reynolds number: 104 < NRe < 105 Prandtl number: 0. This is called von Kármán analogy. Some of the relations are as follows: For laminar flow: . .5 CHILTON–COLBURN ANALOGY In fluid flow or momentum transfer. (16. . . the heat transfer coefficient is correlated by the Nusselt number.1 103 < NRe < 105) . we learnt the close relationship between the Fanning friction factor f and the Reynolds number. . various empirical correlations are given by different scientists. . the relationship is given by (for NRe 2100) . 16. In heat transfer. .77) For turbulent flow: . (16. e. In laminar flow conditions. (16. the friction factor charts are available between the friction factor and the Reynolds number. There are several other analogies given in the literature. .6 < NPr < 100. (for 2. (16. .where NRe = Reynolds number NPr = Prandtl number. the Reynolds number and the Prandtl number. where NRe = Reynolds number NPr = Prandtl number D = diameter of the pipe L = length of the pipe nb = viscosity of the fluid at the bulk temperature n0 = viscosity of the fluid at the average wall temperature. .76) Apart from these correlations. it was extended to mass transfer as well. .5 Reynolds number vs j H factor. these charts are also drawn between the jH factor and the Reynolds number as shown in Figure 16. for highly turbulent flow. (16.Apart from correlations. This analogy is based upon experimental data for gases and liquids for laminar flow conditions and turbulent flow. this analogy was observed between momentum and heat transfer. . . Here. Figure 16. (16.81) where . Accordingly. (16.80) Mass transfer and momentum transfer: For mass transfer.000: . NRe > 10.5.79) Chilton and Colburn observed that there is a relationship between the heat transfer coefficient and the friction factor. the jD factor is defined as . . . Later on. . Initially. . Water flows at a rate of 0. the Chilton–Colburn analogy is given as: . 3. f / 2 is greater than jH or jD. different boundary conditions pervail. . where there is no form of the drag present. The tubes are 3 m long and 25 mm outside diameter. 2. t = 995 kg/m3 Viscosity.332(NRe)1/2(NSc)1/3 . 14 BWG (Birmingham Wire Gauge). Water enters at 32°C.83) This analogy is valid for turbulent flow when NRe > 10.g.057 m3/min per tube. 4. The tube wall temperature may be assumed to be constant at 80°C. . Calculate the heat transfer coefficient by the Reynolds analogy.with n = viscosity of the fluid t = density of the fluid k¢c = mass transfer coefficient DAB = diffusivity D = diameter of the pipe L = length of the pipe.6 < NPr < 100 and tube L/D > 60.623 W/m· C . SOLVED EXAMPLES EXAMPLE 16. The mass transfer coefficient cannot be obtained from the analogous to heat transfer coefficient correlations. . Data: Properties of water: Density. .1 An oil is manufactured by the vapour phase catalytic reaction. The empirical correlation for the Sherwood number is given as NSh = 0. For a situation where the drag is present (e. The condensation occurs on the shell side while the cooling water flows through the tubes. It is used for the flow over a flat plate or the flow in a pipe. k = 0. (16. (16.65 10–4 kg/m· s Thermal conductivity.000 and 0. Chilton–Colburn analogy is useful to obtain the heat transfer coefficient for an unknown system where the hydrodynamic situation is understood. For the mass transfer. n = 7. Applications: 1. The reaction gas mixture leaving the catalytic reactor in the plant is condensed in a shell-and-tube heat exchanger. packed bed) or flow past blunt objects.82) For flow past a flat plate or in a pipe. 65 10–4 kg/m· s k = 0.025 m Inside diameter. cp = 4. Q = 0. f = 0.057 m3/min = 9. the heat transfer coefficient h can be calculated as: . By the Reynolds analogy.5 10–4 m3/s Velocity.17 kJ/kg · C Area A = Volumetric flow rate.623 W/m· C cp = 4. Di = 21.17 kJ/kg · C The Fanning friction fartor f can be calculated by the equation.0212 m L (tube) = 3 m Ts = 80°C t = 995 kg/m3 n = 7. Do = 25 mm = 0. .Specific heat.0014 + Solution: Outside diameter.2 m = 0. (i) But f can be calculated as: . . . . . (ii) f = 4.85 10–3 Substituting the values in Eq. (i), we get h = 27.17 W/m2 · °C EXAMPLE 16.2 Repeat Example 16.1 by using the Prandtl analogy. Solution: By the Prandtl analogy, the heat transfer coefficient can be calculated as: . . . (i) We have calculated f as f = 4.85 10–3 NPr = 5.12 Substituting the values in Eq. (i), we get h = 13.45 W/m2 · °C EXAMPLE 16.3 Repeat Example 16.1 using the von Kármán analogy. Solution: By the von Kármán analogy, we can calculate the heat transfer coefficient as: . . . (i) f = 4.85 10–3 NPr = 5.12 NRe = 7.44 104 Substituting the values in Eq. (i), we get h = 11.4 W/m2 · °C EXAMPLE 16.4 Repeat Example 16.1 by using the Chilton–Colburn analogy. Solution: By using the Chilton–Colburn analogy, we can calculate the heat transfer coefficient as: . . . (i) But f = 4.85 10–3 NRe = 7.44 104 NPr = 5.12 Substituting the values in Eq. (i), we get h = 9.14 W/m2 · °C PROBLEMS 1. Aniline is manufactured by the vapour phase catalytic reaction of nitrobenzene and hydrogen. The reaction gas mixture leaving the catalytic reactor in an aniline plant is condensed in a shelland-tube heat exchanger. The condensation occurs on the shell side while the cooling water flows through the tubes. The tubes are 3 m long and 25 mm outside diameter, 14 BWG. Water flows at a rate of 0.057 m3/min per tube. Water enters at 32°C. The tube wall temperature may be assumed to be constant at 80°C. Calculate the rise in the temperature of water as it flows through the tube. The heat transfer coefficient may be estimated from the Dittus–Boelter equation. Compare the results from different analogies. Data: Internal diameter of the tube (14 BWG) = 21.2 mm Density of water at 32°C, t = 995 kg/m3 Viscosity of water at 32°C, n = 7.65 10–4 kg/m· s Specific heat of water at 32°C, cp = 4.17 kJ/kg · C Thermal conductivity, k = 0.623 W/m· °C Assume: one tube. The Fanning friction factor, f = 0.0014 + 0.125(Re)–0.32 2. What is analogy? Discuss the analogy among momentum transfer, heat transfer and mass transfer with respect to transport mechanism. 3. Discuss the Reynolds analogy. What are its limitations? 4. What is Prandtl analogy? State its assumptions. 5. Discuss in detail the von Kármán analogy. 6. What is Chilton–Colburn analogy? How can heat transfer coefficient be calculated by using this analogy? APPENDICES 2929 g/litre = 0.414 m3 Density of dry air at 0°C.2808 ft = 39. = 2.97 g/g mol 1 g/cm3 = 62.1 VOLUME AND DENSITY 1 g mol ideal gas at 0°C.080711 lbm/ft3 Molecular weight of air = 28.43 lbm/ft3 = 1000 kg/m3 1 g/cm3 = 8.37 in.APPENDIX A CONVERSION FACTORSAND FUNDAMENTAL UNITS A.2 LENGTH 1 in.59 g = 0.S.0185 kg/m3 A.414 cm3 1 lb mol ideal gas at 0°C.345 lbm/U.2046 lbm 1 ton (short) = 2000 lbm 1 ton (long) = 2240 lbm 1 ton (metric) = 1000 kg A.4140 litres = 22.3 MASS 1 lbm = 453. 760 mm Hg = 359. 760 mm Hg = 1. gal 1 lbm/ft3 = 16.540 cm 100 cm = 1 m (metre) 1 micron = 10–6 m = 10–4 cm = 10–3 mm = 1 m (micrometre) 1 Å (angstrom) = 10–10 m = 10–4 m 1 mile = 5280 ft 1 m = 3.97 lbm/lb mol = 28.05 ft3 1 kg mol ideal gas at 0°C. 760 mm Hg = 22.45359 kg 1 lbm = 16 oz = 7000 grains 1 kg = 1000 g = 2.4 STANDARD ACCELERATION OF GRAVITY . A. 760 mm Hg = 22. 80665 m/s2 g = 980.90 ft H2O at 4°C 1 psia = 6.2330 10–5 lbm ft/s2 (poundal) 1 kg m/s2 = 1 N (newton) 1 lbf = 4.665 cm/s2 g = 32.4 cm3 1 m3 = 35.311 ft H2O at 70°F 1 psia = 51.S.317 L (litre) 1 ft3 = 0.028317 m3 1 U.7854 L (litre) 1 ft3 = 7.g = 9.921 in.665 gm cm/gf s2 A. Hg at 0°C 1 atm = 33.S.01325 102 kPa 1 atm = 29. gal = 4 qt 1 U.5955 g/cm3) 1 atm = 14.715 mm Hg at 0°C (tHg = 13.1740 lbm ft/lbf s2 = 980.387 cm3 1 m3 = 1000 L (litre) 1 ft3 = 28.4482 N 1 g cm/s2 = 2.20094 U.5 VOLUME 1 L (litre) = 1000 cm3 1 in. gal 1 m3 = 264.3 = 16. gal 1 British gal = 1. gal = 3785.2 1 psia = 2.S. gal = 3.89476 104 g/cm s2 .2481 10–6 lbf A.7 PRESSURE 1 bar = 1 105 Pa (pascal) = 1 105 N/m2 1 psia = 1 lbf/in.0360 in. Hg at 0°C 1 psia = 2.01325 105 N/m2 = 1.313 ft3 A.17 U.01325 bar 1 atm = 760 mm Hg at 0°C = 1.S.S.696 psia = 1.2481 10–6 lbf 1 dyne = 2.6 FORCE 1 g cm/s2 (dyne) = 10–5 kg m/s2 = 10–5 N (newton) 1 g cm/s2 = 7.S.174 ft/s2 gc (gravitational conversion factor) = 32.481 U.01325 105 Pa = 1. gal 1 U. 9890 J/kg A.29307 W (watt) A.7068 btu/s 1 J/s (joule/s) = 1 W 1 btu/h = 0.6783 10–4 W/cm2 °C 1 btu/h ft2 °F = 5.89476 103 Pa 1 lbf/ft2 = 4.5 btu 1 ft lbf = 1. WORK 1 J = 1 N m = 1 kg m2/s2 1 kg m2/s2 = 1 J (joule) = 107 g cm2/s2 (erg) 1 btu = 1055.06 J = 1.12 VISCOSITY .0886 10–3 lbf/ft2 1 psia = 6.16 cal (thermochemical) 1 kcal (thermochemical) = 1000 cal = 4.17 ft lbf 1 hp h = 0.996 cal (IT) 1 btu = 778.2048 btu/h ft2 °F A.9 HEAT.35582 J 1 ft lbf/lbm = 2.7880 102 dyn/cm2 = 47.7457 kW h 1 hp h = 2544.1840 kJ 1 cal (thermochemical) = 4.6783 W/m2 K 1 kcal/h m2 °F = 0.1333224 kPa A.1365 10–3 cal/s cm °C 1 btu/h ft °F = 1.333224 102 N/m2 = 0.3571 10–4 cal/s cm2 °C 1 btu/h ft2 °F = 5.1840 J 1 cal (IT) = 4.74570 kW 1 hp = 550 ft lbf/s 1 watt (W) = 14.340 cal/min 1 hp = 0.89476 104 dyne/cm2 1 dyne/cm2 = 2.89476 103 N/m2 = 6.1 psia = 6.8 POWER 1 hp = 0.73073 W/m K A.1868 J 1 btu = 251.11 HEAT TRANSFER COEFFICIENT 1 btu/h ft2 °F = 1. ENERGY.10 THERMAL CONDUCTIVITY 1 btu/h ft °F = 4.880 N/m2 1 mm Hg (0°C) = 1.05506 kJ 1 btu = 252. 14 MASS FLUX AND MOLAR FLUX 1 g/s cm2 = 7.3734 103 lbm/h ft2 1 g mol/s cm2 = 7.1622 10–3 W A.875 104 ft2/h 1 centistoke = 10–2 cm2/s 1 m2/h = 10.67197 lbm/ft s A.16 HEAT CAPACITY AND ENTHALPY 1 btu/lbm °F = 4.875 ft2/h 1 cm2/s = 10–4 m2/s 1 m2/s = 3.9890 J/kg 1 cal (IT)/g °C = 4.764 ft2/h A.1840 103 kJ/kg mol A.13 DIFFUSIVITY 1 cm2/s = 3.7197 10–4 lbm/ft s 1 cP = 10–3 Pas = 10–3 kg/m s = 10–3 N s/m2 1 cP = 2.3562 10–3 kg mol/s m2 A.0 J/kg 1 ft lbf/lbm = 2.000 cal/g °C 1 btu/lbm = 2326.1546 W/m2 1 btu/h = 0.0886 10–5 lbf s/ft2 1 Pa s = 1 N s/m2 = 1 kg/m s = 1000 cP = 0.1868 kJ/kg K 1 btu/lbm °F = 1.4668 10–5 m/s .1868 kJ/kg K 1 kcal/g mol = 4.3734 103 lb mol/h ft2 1 g mol/s cm2 = 10 kg mol/s m2 = 1 104 g mol/s m2 1 lb mol/h ft2 = 1.15 HEAT FLUX AND HEAT FLOW 1 btu/h ft2 = 3.29307 W 1 cal/h = 1.1 cP = 10–2 g/cm s (poise) 1 cP = 2.17 MASS TRANSFER COEFFICIENT 1 kc cm/s = 10–2 m/s 1 kc ft/h = 8.4191 lbm/ft h 1 cP = 6. 3562 10–3 kg mol/s m3 mol frac 1 kx a lb mol/h ft3 mol frac = 4.8 (°C) °C = (1/1.67°F = 0 K = 0°R (absolute zero) .67 K = °C + 273.1 kx g mol/s cm2 mol frac = 10 kg mol/s m2 mol frac 1 kx g mol/s cm2 mol frac = 1 104 g mol/s m2 mol frac 1 kx lb mol/h ft2 mol frac = 1.0 K = 1.18 TEMPERATURE 0°C = 32°F (freezing point of water) 1.15 100°C = 212°F = 373.449 10–3 kg mol/s m3 mol frac 1 kG kg mol/s m2 atm = 0.15 K = 671.15 K = 491.8°R (Rankine) (as per scale) °F = 32 + 1.98692 10–5 kg mol/s m2 Pa 1 kG a kg mol/s m3 atm = 0.0°C = 1.8)(°F – 32) °R = °F + 459.67°R 0°C = 32°F = 273.8°F = 1.67°R –273.15°C = –459.98692 10–5 kg mol/s m3 Pa A. 7302 3 ft atm / lb mol °R 1545.34 J / kg mol K –3 82.9872 g cal / g mol K 1.34 3 m Pa / kg mol K .057 3 cm atm / g mol K 8314.34 2 2 kg m /s kg mol K 10.APPENDIX B GAS LAW CONSTANT R Gas Law Constant R Numerical Value Units 1.731 3 2 ft lb f / in. lb mol °R 0.9872 btu / lb mol °R 82.057 10 3 m atm / kg mol K 8314.3 ft lb f / lb mol °R 8314. APPENDIX C PROPERTIES OF WATER (LIQUID) Table C.1 Density of Liquid Water Temperature Density K °C 3 g/m 3 kg/m 273.15 0 0.99987 999.87 277.15 4 1.00000 1000.00 283.15 10 0.99973 999.73 293.15 20 0.99823 998.23 298.15 25 0.99708 997.08 303.15 30 0.99568 995.68 313.15 40 0.99225 992.25 323.15 50 0.98807 988.07 333.15 60 0.98324 983.24 343.15 70 0.97781 977.81 353.15 80 0.97183 971.83 363.15 90 0.96534 965.34 373.15 100 0.95838 958.38 Table C.2 Viscosity of Liquid Water Temperature Viscosity K °C 3 3 [(Pa s)10 , (kg/m s)10 , or cP] 273.15 0 1.7921 275.15 2 1.6728 277.15 4 1.5674 279.15 6 1.4728 281.15 8 1.3860 283.15 10 1.3077 285.15 12 1.2363 287.15 14 1.1709 289.15 16 1.1111 291.15 18 1.0559 293.15 20 1.0050 293.35 20.2 1.0000 295.15 22 0.9579 297.15 24 0.9142 298.15 25 0.8937 299.15 26 0.8737 301.15 28 0.8360 303.15 30 0.8007 305.15 32 0.7679 307.15 34 0.7371 309.15 36 0.7085 311.15 38 0.6814 313.15 40 0.6560 315.15 42 0.6321 317.15 44 0.6097 319.15 46 0.5883 321.15 48 0.5683 323.15 50 0.5494 325.15 52 0.5315 327.15 54 0.5146 329.15 56 0.4985 331.15 58 0.4832 333.15 60 0.4688 335.15 62 0.4550 337.15 64 0.4418 339.15 66 0.4293 341.15 68 0.4174 343.15 70 0.4061 345.15 72 0.3952 347.15 74 0.3849 349.15 76 0.3750 351.15 78 0.3655 353.15 80 0.3565 355.15 82 0.3478 357.15 84 0.3395 359.15 86 0.3315 361.15 88 0.3239 363.15 90 0.3165 365.15 92 0.3095 367.15 94 0.3027 369.15 96 0.2962 371.15 98 0.2899 373.15 100 0.2838 Table C.3 Heat Capacity of Liquid Water at 101.325 kPa (1 Atm) Temperature Heat Capacity, cp °C K cal / g °C kJ / kg K 0 273.15 1.0080 4.220 10 283.15 1.0019 4.195 20 293.15 0.9995 4.185 25 298.15 0.9989 4.182 30 303.15 0.9987 4.181 40 313.15 0.9987 4.181 50 323.15 0.9992 4.183 60 333.15 1.0001 4.187 70 343.15 1.0013 4.192 80 353.15 1.0029 4.199 90 363.15 1.0050 4.208 100 373.15 1.0076 4.219 Table C.4 Thermal Conductivity of Liquid Water Temperature Thermal Conductivity °C °F K btu / h ft °F W/m K 0 32 273.15 0.329 0.569 37.8 100 311.0 0.363 0.628 93.3 200 366.5 0.393 0.680 148.9 300 422.1 0.395 0.684 215.6 420 588.8 0.376 0.651 326.7 620 599.9 0.275 0.476 Table C.5 Heat Transfer Properties of Liquid Water (English Units) t cP 3 n 10 (lb m / ft s) k (btu / h ft °F) NPr 4 b 10 (1/°R) 2 2 (gbt /n ) –6 10 3 (1/°R ft ) T (°F) 3 (lb m / ft ) (btu / lb m °F) 32 62.4 1.01 1.20 0.329 13.3 –0.350 – 60 62.3 1.00 0.760 0.340 8.07 0.800 17.2 80 62.2 0.999 0.578 0.353 5.89 1.30 48.3 100 62.1 0.999 0.458 0.363 4.51 1.80 107 150 61.3 1.00 0.290 0.383 2.72 2.80 403 200 60.1 1.01 0.206 0.393 1.91 3.70 1010 250 58.9 1.02 0.160 0.395 1.49 4.70 2045 300 57.3 1.03 0.130 0.395 1.22 5.60 3510 400 53.6 1.08 0.0930 0.382 0.950 7.80 8350 500 49.0 1.19 0.0700 0.349 0.859 11.0 17350 600 42.4 1.51 0.0579 0.293 1.07 17.5 30300 APPENDIX D PROPERTIES OF LIQUIDS Table D.1 Heat Capacities of Liquids (cp = kJ/kg K) K cp Acetic acid 273 311 1.959 2.240 Acetone 273 293 2.119 2.210 Aniline 273 323 2.001 2.181 Benzene 293 333 1.700 1.859 Butane 273 2.300 i-Butyl alcohol 303 2.525 Ethyl alcohol 273 298 2.240 2.433 Formic acid 273 289 1.825 2.131 Glycerol 288 305 2.324 2.412 Hydrochloric acid (20 mol %) 273 293 2.43 2.474 Mercury 293 0.01390 Methyl alcohol 293 313 2.512 2.583 Nitrobenzene 283 303 363 1.499 1.419 1.436 Sodium chloride (9.1 mol %) 293 330 3.39 3.43 Sulfuric acid (100%) 293 1.403 Toluene 273 323 1.616 1.763 o-Xylene 303 1.721 Liquid Table D.2 Thermal Conductivities of Liquids (k = W/m K) K k 100% 293 0.171 50% 293 0.346 Ammonia 243–258 0.502 n-Amyl alcohol 303 0.163 373 0.154 303 0.159 Liquid Acetic acid Benzene 140 100% 293 0. 100% 293 0.135 293 0.175 100% 293 0.265 Glycerol.305 20% 293 0.197 n-Octane 303 0.215 60% 293 0.149 348 0.329 20% 293 0.147 333 0.182 60% 293 0.333 0.151 Ethylene glycol 273 0.589 90% 303 0.5% 303 0.284 n-Hexane 303 0.151 273 0.163 303 0.183 Carbon tetrachloride n-Decane Ethyl acetate Ethyl alcohol Kerosene Methyl alcohol NaCl brine Sulfuric acid .144 333 0.144 293 0.492 100% 323 0.185 341 0.140 25% 303 0.486 100% 323 0.433 Vaseline 332 0.364 60% 303 0.138 333 0.571 12. APPENDIX E Properties of Gases . . . . 0197 0.0208 0.0166 0.4 450 232.0196 0.0191 422.0115 0.0195 0.0156 0.0119 Gas or Vapour Table E.0218 0.00744 Ethane 239 273 373 0.0101 0.00960 0.9 350 176.2 500 260.0171 0.0099 0.0220 0.0128 0.0315 0.0234 Carbon monoxide 173 273 373 0.0256 0.0232 0. k = W/m K K k Acetone 273 319 373 457 0.325 kPa (1 Atm Abs) [Viscosity in (Pa s)103.0158 0.0203 449.0332 0.8 150 65.0138 Sulfur dioxide 273 373 0.0276 0.0236 533.0137 283.0175 0.0124 0.00862 0.6 400 204.0 100 37.0171 0.4 Viscosity of Gases at 101.0247 .0106 0.0227 0.2 50 10.0 0.0254 Ammonia 273 373 473 0.0152 0.7 0.00800 0.0264 0.0087 0.8 0.4 0 –17.0242 0.0192 0.0307 0.0141 311.0227 Ethylene 273 323 373 0.0130 0.0125 0.0183 0.0260 0.0267 0.0293 0.3 0.1 300 148.0215 477.0169 0.0154 0.0154 338.0119 0.3 Thermal Conductivities of Gases and Vapours at 101.0133 0.0183 0.00915 0.0220 0.0305 Chlorine 273 0.0273 0.0213 0.0165 0.3 250 121.0215 Ethyl ether 273 319 373 0.0303 Ethyl alcohol 293 373 0.0171 0.6 0.4 0.0 0.0484 Butane 273 373 0.5 200 93.0228 0.9 0.0128 273.00840 0.0241 0.0167 366.0149 0.0231 0.0111 0.0250 0. or cP] Temperature K °F °C H2 O2 N2 CO CO2 255.2 32 0 0.8 0.0230 0. (kg/m s)103.2 0.1 0.Table E.0225 505.0240 0.0251 0.0181 0.0179 394.0279 n-Hexane 273 293 0.0135 0.325 kPa(1 Atm Abs).0282 0.0183 0.0208 0. vulcanized 2.92 Brick.151 (0°C) 51 0.190 (93.1 Heat Capacities of Solids (cp = kJ/kg K) K cp 373 0.779 Clay 0.3°C) 1.84 1773 1.00 (200°C) 0.218 Formic acid 273 1.2 Thermal Conductivities of Building and Insulating Materials (Solids) Material 3 t (kg/m ) Asbestos 577 Asbestos sheets 889 Brick.69 1.202 Oxalic acid 323 1.64 (1000°C) .612 Tartaric acid 309 1. building Brick. yellow Porcelain 298 2.81 293–373 0.591 Caprylic acid 271 2. portland 0.243 Camphene 308 1.340 Table F.47 (600°C) 1.50 Wool 1.361 Benzoic acid 293 1.8°C) 0.202 Urea 293 1.166 20 0.775 Rubber.05 Asphalt 0.787 Oak 2.248 Cement. fireclay * t (°C) k (W/m K) 0.829 1773 1.800 Glycerol 273 1.39 Pine.382 Lactose 293 1.167 Glass 0.168 (37.629 Dextrin 273 1.63 Corkboard 303 0.938 Concrete 0.84 Magnesia 373 0.147 Solid Aluminia Asbestos 1. fireclay 373 0.980 1773 0.APPENDIX F Properties of Solids Table F.01 Steel 0. 026 Rock wool 192 128 0.2 Cotton 80. across grain 545 15 0.1 30 0.151 0.062 (93.208 Pine.0310 (–6.3°C) 0.7°C) 0.0296 (–6.0518 (93.048 0.059 (37.47 Polyurethane sprayed foam Sand soil .7°C) Rubber.9°C) 0.8°C) 0. window 0.0414 (37. across grain 825 15 0.57 0.0395 (37.068 (37.52–1.51 10% H2O 1922 4.130 16 0.0317 (–6.040 24–40 0.8°C) 0.5 2. 85% 271 208 Oak.83 Snow 559 0 0.7°C) Ice 921 0 2.762 Corkboard 160.5 1.25 Magnesia. wool 330 30 0.06 Glass fibre 64.0433 Felt.055 (0°C) Glass.1 30 0.0486 (93.4°C) 0.3°C) 0.068 (93. 4% H2O 1666 4.151 4% H2O 1826 4.0391 (37.3°C) 0. hard 1198 0 0.8°C) 0.3°C) 0.8°C) Paper Polystyrene board 0.3°C) 0.Clay soil.066 (148.052 Fibre insulation board 237 21 0.0549 (93.023–0.3°C) 0.16 Sandstone 2243 40 1.061 (37. 1:4 dry 0.8°C) 0.5 Concrete.071 (93.8°C) 0.080 (204. q. z): Spherical coordinates (r. f): . z): Cylindrical coordinates (r. y.APPENDIX G THE EQUATION OF CONTINUITY Cartesian coordinates (x. q. y. z): Cylindrical coordinates (r. z): . q.APPENDIX H EQUATION OF MOTIONFOR A NEWTONIAN FLUIDWITH CONSTANTS t AND n Cartesian coordinates (x. f): Appendix H—NEWTONIAN FLUID WITH CONSTANTS t AND n .Spherical coordinates (r. q. except in systems with large velocity gradients. y. z): Spherical coordinates (r.APPENDIX I THE EQUATION OF ENERGYFOR PURE NEWTONIAN FLUIDSWITH CONSTANTS* t AND k * This form of the energy equation is also valid under the less stringent assumptions k = constant and ( ln t/ ln T)p Dp/Dt = 0. Cartesian coordinates (x. f): . z): Cylindrical coordinates (r. The assumption t = constant is given in the table heading because it is the assumption more often made. q. q. The term nv is usually negligible. t. q. [jA = –tDAB—wA] Cartesian coordinates (x. and xA. z): Cylindrical coordinates (r. and wA by JA*. q. y. z): Spherical coordinates (r. f): . c. replace j A.APPENDIX J FICK’S (FIRST) LAWOF BINARY DIFFUSION* * To get the molar fluxes with respect to the molar average velocity. q. f): . .APPENDIX K THE EQUATION OF CONTINUITYFOR SPECIES a IN TERMS* OF ja * To obtain the corresponding equations in terms of j*a. make the following replacements: Replace . . z): Spherical coordinates (r. . . . z): Cylindrical coordinates (r. . . . . . Cartesian coordinates (x. . . . y. . . . . . . c . . t . q. . xa . ra By . . . j*a. . j a . . wa . 9 Boundary-layer thickness heat transfer.INDEX Analogy. 99 Equation of change for isothermal system. 18 Concentration distribution. 136 with a homogeneous chemical reaction. 141 through a stagnant gas film. 50 Chilton–Colburn analogy. 13 of two immiscible fluids. 13 Fick’s law. 36 of continuity. 173 Von Kármán. 165 among momentum. 63 mass. 138 Energy equation. 38 Falling film. 117 Flow of a liquid falling film. 147 through a spherical stagnant gas film. 133 with a heterogeneous chemical reaction (slow reaction). 63 Convective mass flux. 184 Prandtl. 184 Circular tube. 121 Convective momentum flux. 120 momentum. 166 Chilton–Colburn. 128 Convective energy. 129 with a heterogeneous chemical reaction. 36 of motion. 105 mass transfer. 121 Cooling fin. 30 through a circular tube. 18 through an annulus. 26 Fourier’s law of heat conduction. heat and mass transfer. 66 mass transfer. 123 momentum transfer. 177 Reynolds. 7 Convective molar flux. 61 . 182 Boundary conditions eat transfer. 79 Diffusion equation. 152 momentum transfer. 8 Convective transport heat. 8 Sherwood number. 5 Momentum transfer. 13 in a narrow slit. 156 momentum transfer. 61 Molecular fluxes. 150 momentum transfer. 85 with electrical heat source. 172 Reynolds analogy. 173 Reynolds number. 46 Navier–Stokes equations. 88 with a viscous heat source. 61 mass. 7 Molecular transfer heat. 6 Shell balances heat. 69 with nuclear heat source. 53 Prandtl number. 76 Isothermal system. 172 Schmidt number. 122 momentum. 79 with a chemical reaction heat source. 172 Mass fluxes. 72 through composite walls. 65 mass. 103 mass transfer. 3 in turbulent flow.Heat conduction from a sphere to a stagnant fluid. 46 Unsteady-state condition . 177 Prandtl mixing length heat transfer. 172 Shear stress. 172 Prandtl analogy. 68 Time-smoothed concentration. 121 Mechanism of momentum transfer. 109 mass transfer. 117 momentum. 4 Molecular energy transfer. 150 Time-smoothed equation of change for turbulent flow. 5 Nusselt number. 47 Turbulent heat transfer. 156 Temperature distribution. 38. 42 Newton’s law of viscosity. 22 Lewis number. 82 in a cooling fin. 36 Laminar flow. 111 mass transfer. 13 circular tube.heat transfer. 18 liquid falling film. 55 Velocity distribution. 182 . 159 momentum transfer. 13 Von Kármán analogy.