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

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 1.2 Convective Momentum Transfer 7 1.3 Shell Momentum Balances and Boundary Conditions 8 1.3.1 Boundary Conditions 9 Solved Examples 10 Problems 11 2 Shell Momentum Balance and Velocity Distributionin Laminar Flow (Typical Cases) 13–35 2.1 Flow of a Liquid Falling Film 13 2.2 Flow through a Circular Tube (Gravity Flow) 18 2.3 Laminar Flow in a Narrow Slit 22 2.4 Flow through an Annulus 26 2.5 Flow of Two Immiscible Fluids 30 Problems 35 3 Equations of Change for Isothermal System 36–45 3.1 Equation of Continuity 36 3.2 Equation of Motion (Navier–Stokes Equation) 38 3.2.1 Navier–Stokes Equations 42 Solved Examples 43 Problems 45 4 Momentum Transfer in Turbulent Flow 46–54 4.1 Time-Smoothed Equation of Change for Turbulent Flow (Incompressible Fluid) 47 4.1.1 Time-Smoothed Equation of Change for Incompressible Fluids for Turbulent Flow 48 4.2 Boundary-Layer Thickness for Flow Near a Solid Surface 50 4.3 Prandtl Mixing Length Model 53 5 Unsteady-State Flow of Newtonian Fluids 55–58 5.1 Flow Near a Wall Suddenly Set in Motion 55

Section B: Heat Transfer

6 Heat Transfer 61–67 6.1 Fourier’s Law of Heat Conduction (Molecular Energy Transport) 61 6.2 Convective Energy Transport 63 6.3 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 7.1 Heat Conduction with an Electrical of Heat Source 69 7.2 Heat Conduction with a Nuclear Heat Source 72 7.3 Heat Conduction through Composit Walls 76 7.4 Heat Conduction in a Cooling Fin 79 7.5 Heat Conduction from a Sphere to a Stagnant Fluid 82 7.6 Heat Conduction with a Viscous Heat Source 85 7.7 Heat Conduction with a Chemical Reaction Heat Source 88 Solved Examples 94 Problems 97 8 The General Energy Equation 99–102 8.1 Special Cases for Energy Equation 102 9 Temperature Distribution in Turbulent Flow 103–110 9.1 Time-Smoothed Equations for Energy for Incompressible Fluids in Turbulent Flow 103 9.2 Boundary-Layer Thickness for Heat Transfer Near the Solid Surface 105 9.3 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 Fick’s Law of Binary Diffusion (Molecular Mass Transport) 117 11.1.1 Some Features of Fick’s Law of Diffusion 119 11.2 Convective Mass Transport 120 11.2.1 Mass and Molar Fluxes 121 11.3 Shell Mass Balances and Boundary Conditions 122 Solved Examples 124 Problems 127 12 Shell Mass Balances and Concentration Distribution forLaminar Flow 128–146 12.1 Diffusion through a Stagnant Gas Film 129 12.2 Diffusion with a Heterogeneous Chemical Reaction (Instantaneous Reaction) 133 12.3 Diffusion with a Heterogeneous Chemical Reaction (Slow Reaction) 136

12.4 Diffusion with a Homogeneous Chemical Reaction 138 12.5 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 14.1 Time-Smoothed Concentration in Turbulent Flow 150 14.2 Boundary-Layer Thickness for Mass Transfer 152 14.3 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 16 Analogies among Momentum, Heat andMass Transfer 165–190 16.1 Analogy among Momentum, Heat and Mass Transfer 166 16.2 Reynolds Analogy 173 16.3 Prandtl Analogy 177 16.4 Von Kármán Analogy 182 16.5 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

Preface “Transport Phenomena” is a subject of interest to many scientists and engineers in their respective fields of study. This subject, apart from being of prime interest to chemical engineers, is also increasingly gaining popularity and application in the fields of agriculture, biology, biotechnology, nanotechnology and micro-electronics. The subject of transport phenomena covers mainly three aspects: momentum transfer (also known as fluid dynamics), heat transfer and mass transfer. In the field of chemical engineering, the fluid dynamics or momentum transfer occurs in industrial operations such as mixing, sedimentation, fluid flow and filtration. Heat transfer occurs in conduction and in convection transfer of heat during evaporation and drying, whereas mass transfer takes place in operations such as distillation, absorption, chemical reactions, liquid–liquid extraction, crystallization and adsorption. Generally a question is asked, why do we need to study these three phenomena together? There are reasons for it. Let us take a simple process where the raw materials are transported from the storage vessel to the reactor. So, the fluid flow phenomenon comes into the picture. Then, if the reaction in the reactor takes place at high temperatures (generally that happens), then the raw materials have to be heated. After the reaction, the separation processes take place with a view to isolating the product in pure form. The mass transfer operations such as distillation, absorption or crystallization may also take place. In other words, all these three transport phenomena take place. So, the subject of transport phenomena should be studied together on account of the following reasons: 1. The mechanisms of these three transport phenomena are closely related. Also, the governing mathematical equations are similar in nature. 2. By analogies, we can understand one transport phenomenon from another. For example, we can understand mass transfer or heat transfer from momentum transfer. There are many books and literature available which cover these three topics separately. But there are only a few books on the market which cover all the topics together. The subject has grown rapidly in the recent past. It is not possible to cover all the aspects of the subject in one book of this size. So, this book covers only the introductory part to the subject. Moreover, in many universities and engineering colleges including Indian Institutes of Technology, this subject is taught at two levels, i.e. introductory level and advanced level. This book caters to the needs of the introductory level. An attempt has been made in this book to explain the phenomena of momentum transfer, heat transfer and mass transfer by flux expressions. Another approach adopted is to write shell balances for these transfer processes. The partial differential equations so obtained are solved using the boundary conditions. The one-dimensional equations evolved are taken up in all the examples. The expressions for the conservation of momentum, heat and mass are also developed. Since it is an introductory book, care has been taken to use simple mathematical expressions and also to provide complete solutions of the problems. Students are expected to have a good knowledge of

differential equations and methods of integration. Since there are many similarities in molecular transport of momentum, heat or mass, analogies among these three transport processes are also demonstrated to understand them better. This book is an outgrowth of my interactions with the students during class work and tutorials. Simple illustrations and mathematical tools have been adopted to explain the transport phenomena in order to fulfil the needs of the students. This book is divided into four sections: Section A: Momentum Transfer Section B: Heat Transfer Section C: Mass Transfer Section D: Analogies among Momentum, Head and Mass Transfer. Section A: Momentum Transfer (Chapters 1 to 5) In Chapter 1, the basic law of momentum transfer, namely the Newton’s law of viscosity, is covered. The mechanism of momentum transfer is explained. The shell momentum balance for momentum transfer is also covered. 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. The general equation of continuity and the equation of motion are explained in Chapter 3. Momentum transfer in turbulent flow conditions is explained in Chapter 4. The various models like the Prandtl mixing length are also covered in this chapter. The unsteady-state flow of Newtonian fluids is explained in Chapter 5. Section B: Heat Transfer (Chapters 6 to 10) The mechanism of molecular heat conduction is discussed in Chapter 6. It explains the Fourier’s law of heat conduction. In this chapter, convective energy transport and shell energy balances are also discussed. Specific problems of heat conduction in solids are highlighted in Chapter 7. These problems are solved by taking shell energy balances. The general energy equation is developed in Chapter 8. The temperature distribution in turbulent flow conditions is dealt with in Chapter 9. The unsteady-state case of heat transfer in conduction is discussed in Chapter 10. Section C: Mass Transfer (Chapters 11 to 15) The governing law of molecular mass transfer is explained in Chapter 11, which covers the Fick’s law of diffusion. In this chapter, convective mass transfer and shell mass balances are also covered. Simple problems for mass transfer are discussed in Chapter 12 by taking shell mass balances for laminar flow conditions. The general equation of diffusion is explained in Chapter 13. Mass transfer in turbulent flow condition is discussed in Chapter 14. A specific case of diffusion for the unsteadystate condition is explained in Chapter 15. Section D: Analogies among Momentum, Heat and Mass Transfer (Chapters 16) The mechanism and governing equations for momentum, heat and mass transfer are discussed in Chapter 16. Different analogies like Reynolds, Prandtl, von Kármán and Chilton–Colburn are explained in this chapter. This introductory book on transport phenomena will go a long way to cater to the needs of the students. Constructive suggestions from readers, teachers, students and well wishers to improve this first edition of the book will be gratefully acknowledged.

BODH RAJ E-mail: [email protected]

Section A MOMENTUM TRANSFER

1 INTRODUCTION TOMOMENTUM TRANSFER The matter consists of a multitude of extremely small particles called molecules, and matter may be divided into three different classes of substances or states, i.e. solids, liquids and gases. In gases, the molecules have more free space between them compared to the molecules in liquids. In solids, the molecules are arranged in a packed fashion. The behaviour or the movement of the matter can be studied at the molecular level, i.e. microscopically or it can be studied macroscopically at the bulk level. The scientist studies the properties of matter at the molecular level, whereas in engineering applications we are concerned with the study of the bulk or macroscopic behaviour of solids, liquids and gases rather than their molecular behaviour. While studying engineering applications, we generally ask two questions. One, how does a particular phenomenon take place? Second, why does it take place? The same is the case when we deal with momentum transfer. When the molecule of a fluid moves, it carries some momentum with it as the mass of the molecule moves with a certain velocity in a particular direction. Why does the molecule move? It moves because of velocity gradient in the system. We can, therefore, write a general statement such as rate of transfer process =

. . . (1.1)

In momentum transfer, the driving force is the difference in velocities. Thus, molecules move because of velocity gradient. Now, 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 transfer of heat energy takes place because of temperature gradient and mass transfer takes place because of concentration gradient. The heat and mass transfers will be discussed in detail in subsequent chapters. Momentum, heat and mass transfer phenomena take place by two means:

1. Molecular transfer 2. Convective transfer Now, let us discuss these two mechanisms in detail. As shown in Figure 1.1, suppose there is a fire to be extinguished at a particular place. Let us also assume that there is a water pond in the vicinity. There are three different modes available to us to extinguish the fire.

Figure 1.1 Mechanism of momentum transfer.

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, and so on. In this way, water will be transported to the fire place and used to extinguish the fire. Here all men remain stationary but the bucket of water keeps moving. Mode II: In this case, each man is allowed to fill up the bucket with water from the pond. Each man carries the water bucket to the fire place. Here both men and buckets of water remain moving. Men are moving carrying the buckets of water. Mode III: We install a pump to spray water from the pond onto the fire. Let us analyse the above mechanisms in a scientific way. In Mode I, men are stationary but the bucket of water is moving from one man to another. A similar type of mechanism is simulated in a molecular transfer process. In molecular transfer, the molecules remain stationary but the energy is transferred from one molecule to another. This mechanism is called molecular transfer. It is applicable to phenomena of momentum, heat and mass transfer. In Mode II, each man is moving with a bucket of water. It is similar to each molecule moving with the energy, i.e. here the molecules and the energy move together. Such a mechanism is called convective transfer. Here again, it is applicable to momentum, heat and mass transfer. Both the mechanisms of Modes I and II will be discussed further in subsequent sections. In Mode III, where water is sprayed from the pond to the fire place, a medium is required to transport

energy from one place to another. Such a mechanism is called radiation. It is one of the modes used for heat transfer but not much important for momentum and mass transfer. In summary, the mechanism of transport is accomplished by two means: 1. Molecular transfer 2. Convective transfer We will now discuss the phenomenon of molecular momentum transfer and subsequently that of the convective momentum transfer.

1.1 NEWTON’S LAW OF VISCOSITY (MOLECULAR MOMENTUM TRANSFER) Let us discuss the basic molecular momentum transfer process. This will introduce “Newton’s law of viscosity”. The viscosity is a physical property of a fluid. We know that oils are more viscous than water. The Newton’s law of viscosity will tell us the definition of viscosity. The viscosity of fluids, gases and liquids varies with temperature and pressure. Under laminar flow conditions, molecules of a fluid move in layers that slide past one another in a orderly fashion. Let us simulate this situation, when the fluid is flowing between two parallel plates. Consider the lower plate to be moving with a velocity v and suppose the area of the plates is A. Initially, both the plates are at rest. We select the x–y coordinates as shown in Figure 1.2. The lower plate is set in motion suddenly with a constant velocity v in the x-direction. As the time passes, the velocity of the fluid settles down. At steady-state condition, the velocity distribution becomes linear as shown in Figure 1.2.

Figure 1.2 Laminar velocity distribution: flow between two plates.

Let F be the force r equired to maintain the motion of the lower plate and y be the distance between the plates. It has been experimentally found that the force F can be expressed as F A F v F Combining these factors, we get

or

. . . (1.2)

where n is a proportionality constant called the viscosity of the fluid. It is a fluid property. Now, we write F/A = xyx, which is defined as the force per unit area required perpendicular to the ydirection. It is called shear stress. Further, we can designate v/y in the differential form as –dvx/dy. The negative sign indicates that velocity decreases in the positive direction of y. We can now write Eq. (1.2) in symbol form as . . . (1.3) where xyx = shear stress and

= velocity gradient

Equation (1.3) states that the shear stress is proportional to the negative of the velocity gradient. Since the constant of proportionality is called viscosity, this law is called “Newton’s law of viscosity”. Newton suggested that those fluids which obey the law of viscosity be called Newtonian fluids. And those fluids, especially polymers, etc. which do not obey this law be called non-Newtonian fluids. Let us interpret the behaviour of this concept. At the neighbourhood of a solid surface (i.e. y = 0), the fluid adjacent to the solid acquires x-momentum in the positive direction of y. So xyx is also called xmomentum in the positive direction of y. SI units of viscosity n can be arrived at as follows. We have xyx = [N/m2 = Pa] vx = [m/s] y = [m]

We can write another form of Newton’s law of viscosity as

. . . (1.4) where

= o = kinematic viscosity.

Let us summarize the molecular fluxes in Cartesian coordinates as shown in Table 1.1. Table 1.1 Molecular Fluxes in Cartesian Coordinates

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. It can be estimated by the following methods: 1. Reduced temperature–pressure and reduced viscosity, obtained from critical values 2. Leonnard–Jones potentials 3. Molecular theory of liquids In this book, we will not be discussing the estimation of viscosity. This is well discussed in other chemical engineering books and handbooks. Viscosity of fluids, liquids or gases, is affected by temperature and pressure. For liquids, as the temperature increases, the viscosity decreases, whereas the viscosity of a gas at low pressure increases with increasing temperature.

1.2 CONVECTIVE MOMENTUM TRANSFER In Section 1.1, we discussed momentum transfer by the molecular way. In addition, momentum can also be transported by the bulk flow of the fluid. This process is called convective transfer. When a fluid enters the space of coordinates x, y, z with vector velocity v, we have to consider three mutually perpendicular planes through x, y, z. Each of these planes is taken as unit area. Let the fluid enter with a velocity vx, then the momentum flux per unit volume is vx t v. Similarly for the velocity in y- and z-directions, we get momentum fluxes vy t v and vz t v. Once we know these fluxes, then multiplying by the area perpendicular, we can calculate the forces. The summary of the convective momentum flux components is given in Table 1.2. Table 1.2 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

1.3 SHELL MOMENTUM BALANCES AND BOUNDARY CONDITIONS Once we know the molecular and convective momentum fluxes, we can easily apply the momentum balances. Here, we develop the velocity distribution in a particular case. Once the velocity profiles are known, the other engineering problems such as shear stress, average velocities, etc. can be easily worked out. In this book, we will study only the steady flow conditions in subsequent chapters. Under steady flow conditions, pressure, density and velocity do not change with time. However, the unsteady flow problems can be solved separately. Let us, therefore, write the general conservation of momentum for steady flow. Since momentum is

being transferred, the term momentum transport is also usually used. 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.g. gravity) = 0 (1.5) The following procedure is adopted for the shell momentum balance: 1. Identify the coordinate system, for example, Cartesian, cylindrical or spherical depending upon the system of the problem, for example, whether (x, y, z) or (r, z, i) or (r, i, z). 2. Identify the non-vanishing velocity component. For example, if the flow is in the x-direction, then vy = 0, vz = 0 but vx is considered. 3. Select the thin shell over which the momentum balance is to be applied, for example, as shown below.

4. Apply a momentum balance over the shell and to the area perpendicular to the variable velocity selected. 5. Formulate the differential equation for the momentum flux by considering the shell thickness approaching zero. Thus, we will get the first differential equation for the momentum flux as

6. Integrate the differential equation formed to get the momentum flux distribution. 7. There will be a constant of integration appearing now. There are two methods to evaluate it. Either solve the constant here itself from physical concepts or carry it forward to solve it later. 8. Apply the Newton’s law of viscosity and obtain a differential equation for the velocity distribution. 9. Integrate the differential equation to get the velocity distribution. It will appear with a constant of integration. 10. Apply the boundary conditions to evaluate the constant values. Get the velocity distribution equation. 11. From the velocity distribution, the other parameters such as average velocity, shear stress, maximum velocity, etc. can be easily evaluated. 1.3.1 Boundary Conditions In order to solve the differential equations formed for velocity and momentum flux, we have to use the boundary conditions from physical concepts. The following concepts are commonly used in momentum transport:

1. Solid–liquid interface: If the solid surface over which the liquid flows is stationary, then at the solid–liquid interface, the liquid velocity will be zero. In other cases, 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. 2. Liquid–liquid interface: In such cases, there is no slip between the two liquids at the interface. The velocities of both the liquids will be the same at the interface. Also the momentum flux will be the same for both the liquids at the interface. 3. Gas–liquid interface: Since the viscosity of the gases is low compared to that of the liquids, so the shear stress at the gas–liquid interface will be zero.

SOLVED EXAMPLES EXAMPLE 1.1 There are two parallel plates some distance apart. Between the plates, water is used at 24°C. The lower plate is being pulled at a constant velocity 0.4 m/s faster relative to the top plate. How far apart should the two plates be placed so that the shear stress x is 0.3 N/m2? Also calculate the shear rate. Solution: The relative velocity of two plates, vx = 0.4 m/s Viscosity of water at 24°C, n = 0.9142 cP = 0.9142 10–3 kg/m· s Let y be the distance between the two plates. Shear stress, xyx = 0.3 N/m2 Applying the Newton’s law of viscosity,

or or

or = 0.00122 m = 0.122 cm Thus, shear rate,

EXAMPLE 1.2 The distance between two parallel plates is 0.00914 m and the lower plate is being pulled at a constant velocity 0.366 m/s faster relative to the top plate. The fluid filled between the plates is glycerol at 293 K having a viscosity 1.069 kg/m · s. Calculate the shear stress and the shear rate. Solution: The relative distance between the plates, y = 0.00914 m The relative velocity, vx = 0.366 m/s The viscosity of glycerol at 293 K = 1.069 kg/m· s Applying the Newton’s law of viscosity,

or

shear rate =

PROBLEMS 1. Predict the viscosity of CO2 at 200 K and 1 atm. Given: f/k = 190 K v = 3.996 n = 1.548 where f and n are Leonnard–Jones parameters and K is temperature in kelvin. 2. Two square plates with each side 60 cm are spaced 12.5 mm apart. The lower plate is stationary and the upper plate requires a force of 100 N to keep it moving with a velocity of 2.5 m/s. The oil film between the plates has the same viscosity as that of the oil at the surface of contact. Assuming a linear velocity distribution, determine the dynamic viscosity of oil in poise. 3. Determine the ratio of eddy viscosity to molecular viscosity at a distance 0.6R from the wall for water flowing at a steady rate in a long smooth round tube under the following conditions: Tube radius, R = 7 cm Wall shear stress = 16.2 10–5 kPa

Density = 1000 kg/m3 Kinematic viscosity = 10–6 m2/s 4. A Newtonian fluid flows between two parallel plates at rest initially. 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 0.001 m. Fluid viscosity is 0.7 cP. 5. Two parallel plates are 0.5 cm apart. The lower plate moves at a velocity 10 cm/s and the upper plate is stationary. Assume linear velocity distribution. The fluid between the plates is ethyl alcohol at 273 K having a viscosity of 1.77 cP. Calculate the shear stress and the velocity gradient.

2 SHELL MOMENTUM BALANCEAND VELOCITY DISTRIBUTIONIN LAMINAR FLOW (TYPICAL CASES In this chapter, we will learn how to make shell momentum balances for laminar flow conditions. These are basically force balances over the shell considered. Next, we will apply the Newton’s law of viscosity to molecular momentum flux. Then, we will obtain the velocity profile. Later, the other parameters like maximum velocity, average velocity, will be evaluated. The examples in this chapter are considered “idealized problems”. The tools to solve these problems should be understood thoroughly. Once the simple problems are understood, then the practical problems can be solved. 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. In Section 2.1, the problem of liquid falling film has been solved. Section 2.2 deals with the flow through a circular tube. Here, only the gravity flow is considered. Flow in a narrow slit is discussed in Section 2.3. The flow in an annulus (Section 2.4) and the flow of two immiscible fluids (Section 2.5) give an idea of solving the problems with different boundary conditions. In all the cases, steadystate conditions are assumed. The unsteady-state flow equations are solved separately in Chapter 5.

2.1 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 2.1. 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

Figure 2.1 Flat plate.

Let us consider that a liquid is forming a film over the flat plate. Let the density t and the viscosity of the fluid n be constant. The thickness of the film of the liquid is d and is very small compared to the length and width of the plate. The assumptions made in the analysis of flow of a liquid over a vertical flat plate are as follows: 1. No end effects. The disturbances of the flow of liquid at the edges (i.e. z = 0, z = L) are neglected. 2. Liquid properties like density t, viscosity n, etc. remain constant. 3. Unidirectional flow, i.e. liquid flow is in the z-direction only, i.e. vx = 0, vy = 0. 4. Steady-state conditions. 5. Laminar flow of the liquid. 6. d L and d W. 7. The flow of liquid is under gravity only. 8. No convective momentum is considered. 9. The fluid is Newtonian. Let us consider the liquid film of thickness d as shown in Figure 2.2. Consider that the flow of the liquid is in the z-direction only, so the velocity vz is considered. The z-momentum considered will be as follows: Molecular momentum flux = xxz i.e. xzz = 0 and xyz = 0. Velocity = vz Let us consider a shell of the liquid at a distance x and of thickness x. Applying a momentum balance at the shell, 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 . . . (2.1)

Figure 2.2 Liquid falling film.

Rate of molecular “momentum in” across the surface at x = (W L) xxz|x . . . (2.2) Rate of molecular “momentum out” across the surface at x + x = (W L) xxz|x+x . . . (2.3) Gravity force acting on the fluid in the z-direction = WL tx g . . . (2.4) Substituting the values of Eqs. (2.2), (2.3) and (2.4) in Eq. (2.1), we get . . . (2.5) Dividing Eq. (2.5) by WL x and taking the limit x 0, we get . . . (2.6) We know from calculus that

Hence we get from Eq. (2.6),

. . . (2.7) Integrating both sides of Eq. (2.7), we have xxz = t gx + C1 . . . (2.8) where C1 is a constant of integration. Physically we know that at x = 0, the fluid velocity is maximum and at x = d, i.e. at the solid surface of the plate, the fluid velocity is zero. Therefore, at the gas–liquid interface, x = 0 and the shear stress is zero. Boundary condition 1: At x = 0, xxz = 0. C1 = 0 Equation (2.8), therefore, becomes xxz = t g x . . . (2.9) Applying the Newton’s law of viscosity to the left side of Eq. (2.9), we obtain . . . (2.10) where n is the viscosity of the liquid. . . . (2.11) This equation is the differential equation for the velocity distribution. Integrating it, we will get the velocity, . . . (2.12) where C2 is a constant, which is evaluated by the boundary condition. We know that at the solid surface of the flat plate, there is no slip between the liquid and the solid, i.e. since the plate is stationary, hence the liquid is also stationary at the surface. Boundary condition 2: At x = d, vz = 0 . . . (2.13) Substituting this value in Eq. (2.12), we have

. . . (2.14) Substituting the value of C2 in Eq. (2.12), we get velocity distribution as

. . . (2.15)

This is the parabolic velocity distribution solution. It is clear that this velocity distribution is a parabola as shown in Figure 2.2. The shear stress is given by Eq. (2.9) and is a straight line as shown in Figure 2.2. From the velocity distribution, the other quantities can be easily calculated as follows: (a) Maximum velocity At x = 0, the velocity becomes maximum. Hence . . . (2.16) (b) Average velocity vz,av

vz,av

Thus, vz,av = (c) Mass flow rate

. . . (2.17)

. . . (2.18) (d) Film thickness From Eqs. (2.17) and (2.18), the film thickness can be calculated as . . . (2.19) where

is the mass flow rate.

(e) Force of the liquid on the solid surface Fz

= LWdtg . . . (2.20) = weight of the entire liquid film For falling film, the Reynolds number is defined as . . . (2.21) For laminar flow, NRe < 20. For turbulent flow, NRe > 1500.

2.2 FLOW THROUGH A CIRCULAR TUBE (GRAVITY FLOW) In many chemical engineering processes, one comes across a fluid flowing in a circular tube. The flow pattern can be understood by applying momentum balance. Here, cylindrical coordinates are considered. Let a fluid of constant density t and constant viscosity n be flowing in a circular tube of radius R and length L. Let us consider that the fluid is flowing in the z-direction. Hence only the velocity vz is considered here. The assumptions made in the analysis of flow of a liquid in a circular tube are as follows: 1. 2. 3. 4. 5. 6. 7.

Steady-state conditions. Laminar flow, i.e. NRe 1800. Incompressible fluid, i.e. density t is constant and viscosity n is constant. Unidirectional flow, i.e. liquid flow is only in the z-direction. Hence, v(i) = 0 and v(r) = 0. Newtonian fluid. No slip between the liquid and the solid surface of the wall. No end effects. The disturbances of the flow of liquid at the edges (i.e. at z = 0, z = L) are neglected. 8. The flow of liquid is under gravity only. Under these assumptions, the z-momentum considered will be as follows: Molecular momentum flux = xrz. i.e. xiz = 0; xzz = 0 Let us consider a shell at a radius r and of thickness r as shown inFigure 2.3. Applying a momentum balance at the cylindrical shell in the z-direction, we get

Figure 2.3 Axial flow in a circular tube.

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 . . . (2.22)

Rate of molecular “momentum in” across the cylindrical surface at r = (2rrL)xrz|r . . . (2.23) Rate of molecular “momentum out” across the cylindrical surface at r + r = (2rrL)xrz|r+r . . . (2.24) Gravity force acting in the z-direction of the liquid on the cylindrical shell = (2rr rL)tg . . . (2.25) Substituting the values of Eqs. (2.23), (2.24) and (2.25) in Eq. (2.22), we get . . . (2.26) Dividing Eq. (2.26) by 2pLr and taking the limit r 0, we get the differential equation, . . . (2.27) Integrating both sides of Eq. (2.27), we obtain

. . . (2.28) where C1 is a constant of integration. At r = 0, centre of the tube, the liquid velocity will be maximum and at the wall of the tube, the velocity of liquid will be zero. So physically, the molecular momentum flux, xrz, will be finite at the centre of the tube. Boundary condition 1: At r = 0, xrz = finite . . . (2.29) So, C1 must be zero, otherwise the momentum flux would become infinite at the centre of the tube. Therefore, the momentum flux would become . . . (2.30) Applying the Newton’s law of viscosity, we get . . . (2.31) Substituting Eq. (2.31) into Eq. (2.30), the differential equation for the velocity is given by . . . (2.32) Integrating this separable differential equation, we have

. . . (2.33) The constant C2 can be evaluated from the boundary condition. Boundary condition 2: At r = R, vz = 0

or

. . . (2.34)

Substituting the value of C2 in Eq. (2.33), we obtain velocity distribution as

. . . (2.35)

This velocity profile is for laminar, incompressible flow of a Newtonian fluid. The velocity profile is parabolic as shown in Figure 2.4. The shear stress xrz is also shown in Figure 2.4 which is linear.

Figure 2.4 Momentum flux distribution and velocity distribution in a circular tube.

The various other quantities can be evaluated as follows: (a) Maximum velocity At r = 0, the velocity becomes maximum. Hence

. . . (2.36) (b) Average velocity vz,av

Substituting vz from Eq. (2.35) and integrating, we get . . . (2.37) Thus, vz,av = (c) Mass flow rate = (average velocity)(area)(density) =

. . . (2.38)

(d) Force exerted at the wall of the tube

or

. . . (2.39)

The flow must be checked before using these equations for the velocity profile, i.e. NRe 2100.

2.3 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, length L and width W as shown inFigure 2.5. The flow is laminar and the fluid is incompressible of density t and viscosity n. Here, the width of the slit in very small compared to the length and width of the plate, i.e. B << W << L. Also, we consider that the flow is in the z-direction only. There are no end effects.

Let us consider the coordinates to be (x, z). Select a shell at a distance x and of thickness x as shown in Figure 2.6. The velocity here is vz. The flow of the fluid is under pressure and gravity. Here, vy = 0, vx = 0. p0 is the pressure at z = 0. pL is the pressure at z = L.

Figure 2.5 Flow through a slit.

Figure 2.6 Velocity profile and momentum flux distribution.

We make the following simplifications. . . . (2.40) ( P = p + tgz) The molecular momentum flux will be xxz. Let us select the (x, z) coordinates with x = 0 at the centre of the slit. Applying a momentum balance at the shell, 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 . . . (2.41) Rate of molecular “momemtum in” across the surface at x = xxz|x (W L) . . . (2.42) Rate of molecular “momentum out” across the surface at x + x = xxz|x+x (W L) . . . (2.43) External forces = (P0 – PL)W x . . . (2.44) Substituting the values of Eqs. (2.42), (2.43) and (2.44) in Eq. (2.41), we get (W L)xxz |x – (W L)xxz |x+x + (P0 – PL)Wx = 0 . . . (2.45) Dividing Eq. (2.45) by WL x and taking the limit x 0, we have . . . (2.46) Integrating both sides of Eq. (2.46), we get

. . . (2.47) where C1 is a constant of integration. Physically we know that the velocity at the centre, i.e. x = 0, will be maximum. The velocity of the fluid at both the plates will be zero, i.e. at x = B or at x = –B, vz = 0 Boundary condition 1: At x = 0, xxz = 0 C1 = 0 Equation (2.47) now becomes . . . (2.48) Applying the Newton’s law of viscosity, we get . . . (2.49)

. . . (2.50) Integrating this equation by the variable separable method, we have . . . (2.51) Boundary condition 2: At x = B, vz = 0

Substituting the value of C2 in Eq. (2.51), we obtain the velocity distribution as

. . . (2.52)

This is the velocity profile solution in a slit. It is clear that this velocity distribution is a parabola as shown in Figure 2.6. The shear stress is given by Eq. (2.48) and is a straight line as shown in Figure 2.6. From the velocity distribution, the other quantities can be easily calculated as follows: (a) Maximum velocity

At x = 0, the velocity beccomes maximum. Hence, . . . (2.53) (b) Average velocity vz,av

. . . (2.54) Comparing Eqs. (2.53) and (2.54), we get

(c) Mass flow rate

. . . (2.55) A very important relation can be obtained from this analysis, i.e. slit analog of the Hagen–Poiseuille equation is

2.4 FLOW THROUGH AN ANNULUS Many a time in chemical engineering, we come across flow through an annulus. A typical example is shell and tube heat exchanger where we want to know the fluid flow behaviour through the annulus. 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, where k becomes the ratio of the radius of the inner cylinder to that of the outer cylinder. Here, we consider the fluid to be flowing upwards in the annular space, i.e. opposite to gravity. The length of the cylinder is L as shown in Figure 2.7.

Figure 2.7 Cylindrical annulus.

The assumptions made in the analysis of flow of a liquid through an annulus are as follows: 1. 2. 3. 4. 5.

Steady-state conditions. Incompressible fluid. Unidirectional flow, i.e. liquid is flowing in the z-direction only. Hence vi = 0, vr = 0. Laminar flow of the fluid. The fluid is Newtonian.

p0 is the pressure at z = 0.

pL is the pressure at z = L. We make the following simplification:

( P = p + tg) For the cylindrical coordinate, the momentum flux becomes xrz. Let us consider a cylindrical shell of radius r and thickness r as shown in Figure 2.8. Applying a momentum balance at the cylindrical shell, 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 . . . (2.56) Rate of molecular “momentum in” across the cylindrical surface at r = (2p rL) xrz | r . . . (2.57)

Figure 2.8 Velocity distribution and shear stress distribution.

Rate of molecular “momentum out” across the cylindrical surface at r + r = (2rrL)xrz | r + r . . . (2.58) External forces acting in the z-direction of the liquid on the cylindrical shell = 2rrr (P0 – PL) . . . (2.59) Substituting the values of Eqs. (2.57), (2.58) and (2.59) in Eq. (2.56), we get

(2rLr)xrz | r – (2rLr)xr+r + 2rrr(P0 – PL) = 0 . . . (2.60) Dividing Eq. (2.60) by 2rLr and taking the limit r 0, we get . . . (2.61) Integrating both sides of Eq. (2.61) by the variable separable method, we obtain

. . . (2.62) where C1 is a constant. At this stage, C1 cannot be determined, since we have no knowledge about the momentum flux xrz at the solid surfaces r = kR and r = R. But we know from the concept that wherever the velocity is maximum, the shear stress will be zero at that plane. In the space between the annulus, there will definitely be a plane where the velocity will be maximum. Let that plane (unknown) be at r = mR at which the shear stress will be zero. Hence, at r = mR, xrz = 0. Now Eq. (2.62) becomes . . . (2.63) . . . (2.64) Substituting the value of C1 in Eq. (2.62), we get . . . (2.65) Now applying the Newton’s law of viscosity, we have . . . (2.66) Substituting the value of xrz in Eq. (2.65), we obtain a differential equation for vz as . . . (2.67) Integrating this first order differential equation by the separable method, we get . . . (2.68) where C2 is a constant.

In Eq. (2.68), there are two unknowns C2 and m. Hence, we need two boundary conditions to obtain their values. Boundary condition 1: At r = kR, vz = 0 . . . (2.69) Boundary condition 2: At r = R, vz = 0 . . . (2.70) 0 = k2 – 2m2 ln k + C2 . . . (2.71) and 0 = 1 + C2 . . . (2.72) C2 = –1 Substituting the value of C2 in Eq. (2.71), we get

Substituting the values of C2 and 2m2 in Eq. (2.68), we get the velocity profile distribution as

. . . (2.73)

Substituting the value of m2 in Eq. (2.65), we obtain the momentum flux distribution as . . . (2.74) The velocity profile distribution and momentum flux distribution are shown in Figure 2.8. The other quantities here, can easily be calculated as follows: (a) Maximum velocity . . . (2.75) (b) Average velocity vz,av

or vz,av

. . . (2.76)

(c) Mass flow rate = (average velocity)(area)(density) = vz,av rR2(1 – k2)t or

. . . (2.77)

(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.

or Fz = pR2(1 – k2)(P0 – PL) . . . (2.78) We must check for the Reynolds number since we have assumed laminar flow condition.

. . . (2.79) For laminar flow, NRe 2100.

2.5 FLOW OF TWO IMMISCIBLE FLUIDS Let us consider that two immiscible, incompressible liquids are flowing in the z-direction in a horizontal narrow slit of length L and width W. The fluids are flowing under pressure gradient only. At z = 0 the pressure is p0 and at z = L, the pressure is pL. The fluid I is more dense and more viscous than the fluid II as shown in Figure 2.9. 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. We can encounter such a situation with the transportation of petroleum products and water in a chemical industry. 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. The fluids are flowing sufficiently slowly so that the interface is not disturbed.

Figure 2.9 Two immiscible fluids flowing between two horizontal plates.

The assumptions made in the analysis of flow of two immiscible fluids flowing between two horizontal plates are as follows: 1. 2. 3. 4. 5.

Laminar flow Incompressible liquids Steady-state conditions Only the pressure gradient occurs Liquids are Newtonian

The coordinates (x, z) are selected in such a manner that the interface in Figure 2.9, becomes the zaxis. The distance between the plates is 2b. 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 2.10. Applying a momentum balance at the shell, 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 . . . (2.80) Rate of molecular “momentum in” across the surface at x = (W L)xxz | x . . . (2.81) Rate of molecular “momentum out” across the surface at x + x = (W L)xxz | x+x . . . (2.82) External force acting on the fluid = (p0 – pL)Wx . . . (2.83)

Figure 2.10 Velocity distribution and shear stress distribution for two immiscible fluids flowing between two horizontal plates.

Substituting the values of Eqs. (2.81), (2.82) and (2.83) in Eq. (2.80), we obtain . . . (2.84) Dividing Eq. (2.84) by WLx and taking the limit x 0, we get . . . (2.85) Integrating this equation for two phases I and II, we have Fluid I:

. . . (2.86)

Fluid II:

. . . (2.87)

We know that at the interface (i.e. at x = 0), the momentum flux is continous. Let us now apply the boundary conditions. Boundary condition 1: At x = 0, and the constants are Now applying the Newton’s law of viscosity for both the fluids, we have . . . (2.88)

. . . (2.89)

Fluid I:

. . . (2.90)

Fluid II:

. . . (2.91)

Integrating Eqs. (2.90) and (2.91) by the variable separable method, we get Fluid I:

. . . (2.92)

Fluid II:

. . . (2.93)

The three constants in Eqs. (2.92) and (2.93) can be determined from the boundary conditions. Boundary condition 2: At x = 0,

. . . (2.94)

(there is no slip of fluid I and fluid II at the interface) Boundary condition 3: At x = b,

. . . (2.95)

Boundary condition 4: At x = –b,

. . . (2.96)

From boundary condition 2:

. . . (2.97)

From boundary condition 4: . . . (2.98) From boundary condition 3: . . . (2.99) Subtracting Eq. (2.99) from Eq. (2.98), we have

or

or

. . . (2.100)

Substituting the value of C1 in Eq. (2.98), we get

or

or

. . . (2.101)

Substituting these values of C1 and C2 in Eq. (2.92) and (2.93), we get the velocity profile distribution for fluid I and fluid II as follows:

Fluid I:

. . . (2.102)

Fluid II:

. . . (2.103)

Substituting the value of C1 in Eq. (2.86), we get the value of shear stress as follows:

. . . (2.104)

These velocity profiles and the shear stress profile are shown in Figure 2.10. Now, let us calculate the plane of zero shear stress. Equating Eq. (2.104) to zero, we get the plane of zero shear stress as

Once we know the velocity profiles of both the fluids, we can easily calculate the average velocity and maximum velocity.

PROBLEMS 1. Show that the flow of a liquid along an inclined plate is

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. 2. 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. (a) Derive an expression for the velocity distribution. (b) What is the ratio of the average velocity to the maximum velocity? [Ans. (a)

, (b)

]

3. An oil has a kimematic viscosity of 2 10–4 m2/s and density of0.8 103 kg/m3. What should be the mass flow rate of the liquid for a falling film thickness of 2.5 mm on a vertical wall per metre of vertical plate? [Ans. 0.2 kg/s]

3 EQUATIONS OF CHANGEFOR ISOTHERMAL SYSTEM In Chapter 2, we learnt about the velocity distribution in simple situations of common interest. 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. We then integrated this equation, solved it with the help of boundary conditions and derived the velocity profile. Once the velocity profile is known, other quantities like average velocity, maximum velocity, etc. can easily be found out. To derive all these equations we used the mass and force balance notation. But it is not always possible to apply the mass balance to any problem or situation. We often encounter complicated problems in chemical engineering. In such cases, the motions of fluid are nonlinear and it is not always easy to solve these problems by shell momentum balances. Such problems can easily be solved by two general equations: 1. Equation of continuity (mass balance) 2. Equation of motion (momentum balance) These equations are derived in this chapter. These equations are used as the starting point for studying any problem involving the isothermal flow of a pure liquid. Further, these general equations are simplified for constant density and viscosity and are called Navier–Stokes equations which are more famous for tackling engineering problems.

3.1 EQUATION OF CONTINUITY Let us consider a fluid element of volume x y z through which a fluid is flowing. Let t be the density of the fluid, and vx, vy, vz be the velocity components in x-, y-, z-direction, respectively. 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 3.1. 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) . . . (3.1) Rate of “mass in” at t`he surface x is . . . (3.2) Rate of “mass out” at the surface x + x is Rate of increase in mass = Volume Rate of change of density

. . . (3.3)

. . . (3.4)

Figure 3.1 Fluid element of volume x y z through which a fluid is flowing.

Similarly, we can write the expressions for the mass entering and leaving iny- and z-directions. Hence the mass balance from Eq. (3.1) becomes

. . . (3.5) Dividing both sides of Eq. (3.5) by xyz and taking the limit as x 0, y 0 and z 0, we obtain . . . (3.6) The rate of change in fluid density described by Eq. (3.6) is known as the equation of continuity. In vector notation, we can write . . . (3.7) where is the divergence of mass velocity vector tv, and t changes with respect to space and time. We can expand the term ∂(tvx) and rearrange Eq. (3.6) as follows: . . . (3.8)

or

. . . (3.9)

where D/Dt on the left-hand side of Eq. (3.9) is called the substantial derivative. Case 1: For a compressible fluid (where t is constant), Eq. (3.7) becomes (·tv) = 0

3.2 EQUATION OF MOTION (NAVIER–STOKES EQUATION) Let us consider a fluid element of volume xyz through which a fluid of density t is flowing, and vx, vy, vz be the velocity components in x-, y-, z-direction, respectively. The mass of fluid entering at x surface and the mass of fluid leaving at x + x surface are shown in Figure 3.2 as the shaded faces. Similarly, the fluid is entering and leaving at y and y + y and z and z + z, respectively.

Figure 3.2 Fluid element of volume x y z, through which a fluid is flowing.

Here, we consider the momentum transfer by two mechanisms, i.e. convective and molecular. Also, the external forces acting on the system are pressure and gravity. First, we write the momentum balance for the x-component of momentum and external forces. Similarly, we can write such expressions for the y-component and z-component. 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”)

+ (Sum of external forces acting on the system) = (Rate of momentum accumulation) . . . (3.10) Before writing the individual terms of Eq. (3.10), we must be clear that the y-direction momentum and the z-direction momentum contribute to the x-direction momentum as well. Let us consider the volume xyz of fluid as shown in Figure 3.2. 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.

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

. . . (3.12) Now, let us consider the external forces, pressure and gravity. Let p |x be the pressure exerted at the face at x and p |x+x be the pressure exerted at x + x, in the x-direction. Net fluid pressure acting in the x-direction . . . (3.13) Gravitational force acting in the x-direction = tgxxyz . . . (3.14) There is no meaning of writing the gravitational force by the symbol gx, but to make it more clear that we are considering the x-direction forces, we write in this fashion. Here we assign the symbol gx to the x-component of the gravitation vector g, as a matter of convenience. Rate of accumulation of x-momentum . . . (3.15) Substituting Eqs. (3.11) to (3.15) in Eq. (3.10) and dividing by xyz and taking the limits x 0, y 0, and z 0, we obtain the x-component of the differential equation of motion as follows:

. . . (3.16) Similarly, the y-component and z-component of the differential equation of motion can be written as follows: y-component:

. . . (3.17) z-component:

. . . (3.18) Simplifying these equations by using the equation of continuity (3.6), i.e.

we obtain the equation of motion as: x-component:

. . . (3.19) y-component:

. . . (3.20) z-component:

. . . (3.21)

The general form of the equation of motion for pure liquids can be written as . . . (3.22) where D/Dt = substantial derivative = divergence

3.2.1 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, i.e.

we obtain the following equations of motion for the Newtonian fluids. x-component:

. . . (3.23) y-component:

. . . (3.24) z-component:

. . . (3.25) By combining these three components, we arrive at the general form of the equation as . . . (3.26) where . These equations are called Navier–Stokes equations and are most commonly used for solving engineering problems.

SOLVED EXAMPLES EXAMPLE 3.1 A fluid enters a flow system as shown in Figure 3.3.

Figure 3.3 Mass balance over a flow system.

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. Find an expresion for the velocity v2 of the fluid leaving the system. Solution: Applying the mass balance over the flow system, we have Rate of mass entering – Rate of mass leaving = Rate of mass accumulation At the steady-state conditions, the rate of mass accumulation is zero. Rate of mass entering = v1t1A1 Rate of mass leaving = v2t2A2 Applying the mass balance, we obtain v1t1A1 = v2t2A2

Let the pipes be circular of D1 and D2 diameters.

or EXAMPLE 3.2 A fluid enters a nozzle 1 of 40 cm diameter at a velocity 3 m/s. Then the fluid splits into two nozzles 2 and 3 connected in a Y shape. The diameter of nozzle 2 is 30 cm and the flow rate is 2 m/s. The diameter of the nozzle 3 is 20 cm. Find the velocity in the nozzle 3. Solution: Let us consider the nozzles’ configuration as shown in Figure 3.4.

Figure 3.4 Nozzle configuration—Example 3.2.

Let the fluid enter nozzle 1 at a velocity, v1 = 3 m/s Diameter of nozzle 1 = 40 cm = 0.4 m Fluid velocity in nozzle 2, v2 = 2 m/s Diameter of nozzle 2 = 30 cm = 0.3 m Let v3 be the velocity in nozzle 3 of diameter, D3 = 20 cm = 0.2 m Applying the mass balance, we get Rate of mass of fluid entering = Rate of mass of fluid leaving or where t is the density of the fluid. 3/(0.4)2 = 2(0.3)2 + v3(0.2)2

or v3 = 7.5 m/s

PROBLEMS 1. A 20% (wt) sucrose solution having a density of 1074 kg/m3 flows through a system as shown in Figure 3.3. The fluid enters a pipe of diameter 52.5 mm at a flow rate of 1.892 m3/h. (a) Calculate the velocity at which the fluid will leave the system in a pipe of diameter of 40.9 mm. (b) Also calculate the mass velocity of the fluid. [Ans. (a) 0.4 m/s, (b) 2032 kg/h] 2. An oil of density 892 kg/m3 flows through a pipe of diameter 52.5 mm as shown in Figure 3.3. The total flow rate is 1.388 10–3 m3/s entering the pipe. If the diameter of the pipe in which the oil leaves the system is 77.9 mm, then at what velocity will the oil leave the system? [Ans. 0.0321 m/s]

4 MOMENTUM TRANSFERIN TURBULENT FLOW Laminar flow or streamline flow is easier to understand. In laminar flow, the fluid moves in a particular fashion. Molecules of the fluid behave orderly and move in straight lines and stream-wise. The mathematical formulation of the differential equations is easy and straightforward. In Chapter 2, we discussed typical cases by shell momentum balance for laminar flow conditions. Then we solved the equations formed with the help of boundary conditions. Since the mechanism is easy, mathematical formulation becomes an easier way to understand the laminar flow phenomenon. But as the number of variables becomes large, the mathematics becomes difficult. Nowadays with the help of the computer, the numerical methods of solving such problems are available. On the other hand, the situation for turbulent flow is different. The molecules move randomly with different velocities in all directions. The mechanism is difficult to understand. Eddy formation takes place. Boundary-layer formation and boundary-layer separation also take place. The velocities of the fluid changes in direction and magnitude when the fluid faces some solid obstruction. The mathematics becomes more complex and solutions are not feasible. Scientists have taken different approaches to understand the fluid behaviour under turbulent flow conditions. These methods are empirical, experimental or models. An attempt has been made to understand the flow behaviour under turbulent flow by time–smoothed velocities, boundary-layer thickness near the solid surface and Prandtl mixing length models.

4.1 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. In order to improve the efficiency of the process, turbulence is created. The turbulence is created when two streams of fluids meet with different velocities or the flowing stream comes in contact with a solid, e.g. walls or baffles. Turbulent flow is important in many areas of engineering; it has been investigated extensively. There are no exact solutions of flow problems similar to those which exist in the laminar flow conditions. Some empirical or experimental approaches have been developed to understand the turbulent flow problems. It has been postulated that fluctuations of velocity in turbulent flow form eddies. High-speed photography has evidence of measuring these eddies to as small a size as 0.1–1 mm. Let us discuss fluctuations of velocities in the turbulent flow. These fluctuations take place in all directions.

In order to understand the process, let us first consider these fluctuations in the x-direction. We can measure the instantaneous velocity at a given point w.r.t. time with the help of a Pitot tube. These plots of variations of the instantaneous velocity vx in the x-direction are shown in Figure 4.1.

Figure 4.1 Velocity fluctuations in turbulent flow.

Let vx = instantaneous velocity of main stream in the x-direction. vx = deviation of velocity from the mean velocity in the x-direction. = mean velocity (time averaged) in the x-direction of flow stream. vx =

+ vx . . . (4.1)

Similarly for the y- and z-directions, we can write The time-averaged velocity can be calculated as . . . (4.2) Some of the properties of these time-averaged velocities and fluctuations of the velocities are now considered. The time-averaged values of these fluctuations of velocities in the x-, y-, and z-directions are zero, i.e.

However, the squares of these fluctuation velocities are not zero, i.e. (v¢x)2 0, (v¢y)2 0, (v¢z) 0 but have positive values. Since the fluctuations are random, the data have been analysed by statistical techniques. The timeaveraged fluctuations of velocity vanish over a short period of time. The level of intensity of the

turbulence may be related to the square root of the mean squares of the fluctuating velocity components. The intensity of the turbulence, I, can be defined mathematically as in Eq. (4.3). . . . (4.3)

The intensity of the tubulence, I, is an important parameter and is helpful in understanding heat and mass transfer processes under turbulent conditions. Experimentally, this parameter I can be measured by a hot-wire anemometer. 4.1.1 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, we can now discuss the equation of motion and the continuity equation. For simplicity, let us consider only the case of incompressible fluids with constant density and viscosity. Similar equations hold good for pressure as well. That is, . . . (4.4) Let us reproduce here the equation of continuity and the equation of motion, i.e. Eq. (3.6) and Eq. (3.23). Equation of continuity with constant density: . . . (3.6) Equation of motion: x-component:

. . . (3.23) Substituting,

in the equation of continuity and the equation of motion, we get the equation of continuity as . . . (4.5) and the x-component of equation of motion as

. . . (4.6) Now applying the simplification procedure noting that the time-averaged product

as discussed above and

is not zero, we obtain

. . . (4.7)

. . . (4.8) Similar relations can be obtained for the y-component and the z-component. These equations are timesmoothed equation of continuity and equation of motion for constant density and viscosity. Comparing these equations with Eqs. (3.6) and (3.23), we observe: (a) The equation of continuity (4.7) is the same as the previous one except that vx is replaced by (time-smoothed average velocity). Similar is the case for vy and vz. (b) The equation of motion (4.8) also has the same changes in terms of time-smoothed velocity and pressure. The only change is that a third expression on the right-hand side is added. The three terms of this expression are described by the momentum transport associated with the turbulent fluctuations. Let these components be defined as: . . . (4.9) These components are generally known as Reynolds stresses.

4.2 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. Here, we consider the boundary-layer region near the solid surface. For study of velocity in a boundary layer near the wall, viscous flow is considered. The solutions are approximate. Let us consider a flat plate over which the fluid is entering at a uniform velocity v. The thickness of the boundary layer d is arbitrarily selected at some distance away from the surface of the flat plate

where the velocity reaches 99% of the main stream velocity. OL is the boundary-layer thickness line as shown in Figure 4.2. Some simplifications of the Navier–Stokes equations are done to the concept of a thin boundary layer:

Figure 4.2 Boundary layer for flow over a flat plate.

The assumptions made are: 1. 2. 3. 4.

The flow is in the x and y-directions only. Incompressible fluid; hence t is constant. No gravity force is acting. Flow is steady state.

With the above assumptions, the general Navier–Stokes equation

can be simplified. x-component: . . . (4.10) y-component: . . . (4.11) The continuity equation may be used to simplify the above equations, which is . . . (4.12) In Eq. (4.10), the term ∂2vx/∂x2 is negligible compared to the other terms in the equation. The terms containing vy and its derivative are small. Hence,Eq. (4.10) becomes

. . . (4.13) Further simplification can be done. Since the velocity at the entrance v is constant, dp/dx is zero, hence Eq. (4.13) becomes . . . (4.14) Equation (4.14) and the equation of continuity (4.12) are solved with the following boundary conditions. Boundary condition 1: vx = 0 at y = 0 vy = 0 at y = 0 Boundary condition 2: At some distance away from the flat plate. vx = v at y Equations (4.12) and (4.14) are nonlinear equations and hence the solutions become complex. Blasius reduced these two equations to a single ordinary differential equation which is nonlinear and obtained a series solution. The results of Blasius are outlined below. The boundary-layer thickness d, where vx 0.99v , is given approximately by . . . (4.15)

where NRe,x = It is clear that the boundary-layer thickness d is proportional to

.

Before proceeding further, one should be clear that for calculating the drag force for flow over a flat plate, only the skin friction plays an important role. Let us calculate the shear stress at the surface at y = 0 for any value of x. . . . (4.16) From the relation of vx as a function of x and y obtained from the series solution, Eq. (4.16) becomes . . . (4.17) Let L be the length of the plate and b be the width. The total drag force can be calculated as

. . . (4.18) Substituting Eq. (4.17) in Eq. (4.18) and integrating, we get . . . (4.19) The drag coefficient is defined from the drag force as . . . (4.20) But A = b L . . . (4.21) Substituting the value of FD from Eq. (4.19) in Eq. (4.21), we obtain . . . (4.22)

. . . (4.23)

or

where The Fanning friction factor f for laminar flow in circular pipes, we know, is given by . . . (4.24)

where

.

Now, let us compare Eq. (4.23) with Eq. (4.24). These are similar equations for different situations. The only constraint is that CD applies to the laminar boundary layer only for NRe, L < 5 105. Also, Eq. (4.23) has been experimentally verified.

4.3 PRANDTL MIXING LENGTH MODEL In order to solve for velocity profile in turbulent flow, the Reynolds stresses have to be evaluated. More simplifications and semi-empirical correlations are available in the literature. Boussinesq used the eddy-diffusivity model to solve these Reynolds stresses.

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

Equation (5.6) becomes . . . (5.7) In order to solve Eq. (5.7), the boundary conditions become Boundary condition 1: h = 0, z = 1 . . . (5.8) Boundary condition 2: h = , z = 0 . . . (5.9) By introducing another variable for ∂z / ∂h, one can get the first-order separable equations from Eq. (5.7). With the help of the above boundary conditions, one can easily get the solution by the definition of error functions. These functions are well known in mathematics.

= 1 – erf(h) = erfc(h) . . . (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

. . . (5.10) The complementary error function is a decreasing function which varies from 1 to 0. One can plot the dimensionless velocity vs. dimensionless time as shown in Figure 5.2.

Figure 5.2 Velocity distribution (dimensionless velocity vs. dimensionless time) for a flow near the wall when the wall is suddenly set in motion.

In this chapter, we have explained a simple situation of unsteady-state flow near a wall solving it analytically. However, in practice, numerical methods too, are extensively used to solve such problems with the help of computers.

Section B

HEAT TRANSFER

6 HEAT TRANSFER Some solid materials conduct heat easily, whereas others do not. Metals conduct heat, whereas wood does not conduct heat. Wood insulates heat. Heat conduction in fluids is slightly different than that in solids. In fluids (such as liquids and gases), energy is being transferred by two mechanisms: 1. Molecular energy transfer 2. Convective energy transfer These two modes of energy transfer have been explained well in Chapter 1 with the help of an example. In heat transfer, there is a third mechanism, namely radiation. In radiation, energy is interdiffusing and does not require any media. Radiation energy mechanism is different and will not be discussed in this book. The physical property that describes the rate at which heat is conducted is called the thermal conductivity k. Now, we shall define and explain the thermal conductivity by Fourier’s law of heat conduction.

6.1 FOURIER’S LAW OF HEAT CONDUCTION (MOLECULAR ENERGY TRANSFER) Let a solid slab of area A be placed between two parallel plates, Y distance apart. When the slab is placed, its temperature initially is T0. Suddenly, the lower plate is raised to a higher temperature T1. As time elapses, the temperature of the solid slab starts increasing. At steady state, the temperature attains a linear profile as shown in Figure 6.1.

Figure 6.1 Steady-state temperature profile for a solid slab.

As time passes, the temperature of the slab changes. After a small period of time, the temperature profile is shown in Figure 6.1. At steady-state conditions, the temperature profile becomes linear. Let

Q be the rate of heat flow in the slab. Then, temperature gradient, T = T1 – T0 It is obvious from the above that Q A Q T and Q Combining all these quantities, we get

or

. . . (6.1)

where k is a constant called thermal conductivity, a thermal property of the solid. Equation (6.1) is also valid if we replace the solid slab by a fluid (liquid or gas). The only precaution to be taken is that there is no convection or radiation of heat taking place. Let us write Eq. (6.1) in the usual differential form when the thickness of the slab approaches zero. The rate of heat flow per unit area is written as qy which is called heat flux in the positive direction of y. Equation (6.1) now becomes . . . (6.2) This equation is called “Fourier’s law” of heat conduction in one dimension. In other words, we can say that the heat flux by conduction is proportional to the temperature gradient. The negative sign indicates that heat flow decreases in the positive direction of y. Similarly, we can write Eq. (6.2) in x- and z-directions as well. . . . (6.3) . . . (6.4) In the vector notation, all these equations can be combined and we can write . . . (6.5) There is another way of defining thermal conductivity, known as thermal diffusivity a. It is defined as . . . (6.6) where cp = specific heat capacity at constant pressure, J/kg · K a = thermal diffusivity.

Units are: k = W/m· K a = m2/s qy = W/m2 Here, a, the thermal diffusivity, is similar to the momentum diffusivity as discussed in Chapter 1. Momentum and thermal diffusvity are correlated by a dimensionless number called Prendtl number. It is dfined as . . . (6.7) The thermal conductivity has to be measured expeimentally because it depends upon so many parameter. In solids, the thermal conductivity and electrical conductivity are closely related. It is discussed in detail by Jacob1.

1 M. Jacob, Heat Transfer, Vol. 1, Chapter 6, Wiley, New York. The thermal conductivities for liquids and gases can be estimated by the (a) corresponding state chart,using reduced temperature, reduced pressure and reduced thermal conductivity. (b) Chapman–Enskog kinetic theory of gases. In this book, we are not discussing the estimation of thermal conductivity. This is discussed in other heat transfer and chemical engineering handbooks. The thermal conductivities of gases at low density increase with increasing temperature, whereas the thermal conductivities of most liquids decrease with increasing temperature.

6.2 CONVECTIVE ENERGY TRANSFER So far we have discussed that energy can e transported by three modes: 1. Molecular energy transfer (conductive heat transfer). 2. Convective energy transfer. 3. Radiation: This mode will not be discussed in this book. The mechanism of transport of thermal energy is done in two steps. In the first step, molecules migrate by random motion in a given volume. In the second step, if a molecule of higher thermal energy migrates to a region of lower energy level, it must distribute its excess energy among the low energy molecules. We assume the concentration of thermal energy to be constant in the given volume. The energy of molecules consists of kinetic energy and internal energy. The iternal energy is associated with the random translational motion of molecules and internal motion of molecules. Let v be the velocity vector with which the fluid s moving. It has three components, vx, vy,vz n x-, y-, z-directions. onsderthat dS is the area perpendicular to the x-axis as shown in Figure 6.2.

Figure 6.2 Energy transfr by convective mode with velocity vx across the surfae d perpendicular to the x-axis.

Volumetric flow rat across the area dS ^ to the x-axis = vx × dS . . . (6.8) Energy asociated with kinetic enery per unt volume . . . (6.9) Energy associated with internal energy per unit volume = tU . . . (6.10) where U is the internal energy per unit mass. Total energy associated by convective transfer . . . (6.11) Similar equations can be written for the y-direction and z-direction. Total convective energy =

. . . (6.12)

By using Eq. (6.12), we can have the convective energy flux.

6.3 SHELL ENERGY BALANCES AND BOUNDARY CONDITIONS After knowing the convective and conductive heat fluxes, we can easily apply the energy balances. Here, we develop the temperature distribution in a particular case. The other parameters like average temperature and surface heat fluxes can easily be calculated. In the subsequent chapters, only heat transfer under steady-state conditions is discussed. Under steady-flow conditions, temperature, pressure, density, etc. do not change with time. However, the unsteady-flow problems can be solved separately. 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)

+ (Rate of work done on the system) + (Rate of work done on the system by external forces) + (Rate of energy production) = 0 (6.13) The molecular and convective fluxes are discussed in Sections 1.1 and 1.2. The following procedure is followed for the shell heat transfer: 1. Identify the coordinate system depending upon the system of the problem. 2. Identify the non-vanishing temperature component. 3. Select the thin shell over which the heat/energy balance is to be applied. 4. Apply the energy balance over the shell and to the area perpendicular to the variable heat transfer being transferred. Substitute in Eq. (6.13). 5. Formulate the differential equation for the heat flux by considering the fact that the shell thickness approaches zero. Thus, we will get the first differential equation for the heat flux as . . . (6.14) 6. Integrate the differential equation formed to get the heat flux distribution. 7. A constant of integration will appear in the equation. This constant can be evaluated from the boundary conditions. 8. Apply the Fourier’s law of heat conduction and obtain the differential equation for the temperature distribution. 9. Integrate this differential equation to get the temperature distribution. It will appear with the constant of integration. 10. Apply the boundary conditions to evaluate the constants. Get the temperature distribution equation. 11. From the temperature distribution, other quantities can be calculated such as average temperature, maximum temperature, etc. Boundary conditions: 1. Ambient temperature at the outer surface may be specified, i.e. air temperature is given. 2. Sometimes heat flux normal to the surface is given. 3. At the interface of two solids or fluids, the heat flux at the interface for both the fluids is the same. At the same time, the temperature at the interface is also the same for these two phases. 4. At the solid–fluid interface, the Newton’s law of cooling may be applied, i.e. q = h(T0 – Tb) . . . (6.15) where h is the heat transfer coefficient, T0 the ambient temperature and Tb the bulk fluid temperature.

SOLVED EXAMPLES EXAMPLE 6.1 The thermal conductivity of an insulating material was measured. The temperature of a flat slab of 25 mm of the material was measured to be 318.4 K and 303.2 K. The heat flux was measured to be 35.1 W/m2. Calculate the thermal conductivity of the material. Solution:

Using Fourier’s law of heat conduction

= heat flux = 35.1 W/m2 T = 318.4 – 303.2 = 15.2 K x = 25 mm = 0.025 m

EXAMPLE 6.2 Calculate the heat loss per m2 of surface area for an insulating wall of a cold storage room where the outside temperature is 299.9 K and the inside room temperature is 276.5 K. The wall is made of 25.4 mm of corkboard having thermal conductivity k of 0.0433 W/m· K. Solution: Applying Fourier’s law of heat conduction,

where k = 0.0433 W/m· K T = 299.9 – 276.5 = 23.4 K x = 25.4 mm = 0.0254 m Heat loss per m2 = 0.0433

= 39.9 W/m2

PROBLEMS 1. A plastic panel of area 0.5 m2 and thickness 0.1 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. What is its thermal conductivity? . . . [Ans. 0.3 W/m· K] 2. Calculate the heat loss per m2 of the surface area of an insulating wall composed of 25.4 mm thick fibre insulating board, where the inside temperature is 352.7 K and the outside temperature is 297.1 K. The thermal conductivity of the board k is 0.048 W/m · K. . . . [Ans. 105.1 W/m2]

7 SHELL ENERGY BALANCESAND TEMPERATURE DISTRIBUTION IN HEAT CONDUCTION IN SOLIDS (Typical Cases) In Chapter 2, we applied momentum balance over a shell and solved for momentum flux. In this way, we obtained the differential equation for the momentum flux. Then the Newton’s law of viscosity was applied. Thus, the differential equation was obtained. After integration, the constants evolved were evaluated from the boundary conditions. As discussed earlier in Chapter 2, these differential equations for momentum flux and velocity were obtained after evaluating the constants of integration from physical concepts. Similarly, in this chapter too, we select the shell over which heat balance is applied. The differential equation for heat flux is obtained. One thing should be kept in mind that the area should be taken perpendicular to the flow of heat flux. Afterwards, the Fourier’s law of heat conduction is applied. After integration of the differential equation, we obtain the temperature distribution profile. But the constants of integrations are evaluated from the boundary conditions. In both these chapters, the wordings used are similar and the methods to solve momentum and heat problems are also similar. But one should be very careful while solving energy problems. We start with a very simple heat conduction problem in Section 7.1 with an electrical heat source. Here, we consider heat conduction in one dimension only to understand the solution very clearly. Heat conduction in a nuclear heat source is discussed in Section 7.2 to understand the continuity of heat flux and temperature at the interfaces. In Sections 7.3 and 7.4, heat conduction through composite walls and cooling fins are dealt with. These problems are of practical nature. Heat conduction from a spherical ball to a stagnant fluid is solved in Section 7.5. Sections 7.6 and 7.7 illustrate the heat conduction problems with a viscous heat source and chemical reactions. In these problems, we learn how to apply the boundary conditions. In all these problems, steady-state conditions are assumed.

7.1 HEAT CONDUCTION WITH AN ELECTRICALSOURCE OF HEAT When an electrical current passes through a wire, this electrical energy is converted into heat energy. In order to design the material of construction of the insulation, we need to know the temperature distribution produced by electrical current. 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). This is different than the thermal conductivity of the material of the wire. Let us consider a cylindrical wire of radius R and length L through which electrical current is passed

as shown in Figure 7.1.

Figure 7.1 Cylindrical electric wire through which current is flowing.

The heat produced per unit volume by the electrical wire is given by (7.1) where Se = energy produced per unit volume, A-/cm3 I = current density, A/cm2 ke = electrical conductivity, –1 cm–1 The following assumptions are made in the analysis of heat conduction with an electrical source: 1. 2. 3. 4. 5.

Heat energy flows only in one direction, i.e. in the r-direction. Thermal conductivity k of the wire, is constant. Electrical conductivity ke is constant. Heat energy flows by conduction only. Steady-state conditions prevail.

Let qr be the heat flux in the r-direction. Let us consider a cylindrical shell of radius r and thickness r as shown in Figure 7.1(b). Applying energy balance at the cylindrical shell, we obtain Rate of “heat in” – Rate of “heat out” + Rate of heat produced by electrical energy = 0 . . . (7.2)

Rate of “heat in” at the cylindrical surface at r = (2rrL)qr |r . . . (7.3) Rate of “heat out” at the cylindrical surface at r + r = (2rrL)qr |r+r . . . (7.4) Rate of heat produced by electrical energy = (2rrL r)Se . . . (7.5) Substituting these values in Eq. (7.2), we get (2rrL)qr |r – (2rrL)qr |r+r + (2rrL r)Se = 0 . . . (7.6) Dividing both sides of Eq. (7.6) by 2rL r, we obtain . . . (7.7) Taking the limit r 0, we get . . . (7.8) Integrating both sides of Eq. (7.8), we get . . . (7.9) where C1 is the constant of integration. . . . (7.10) Boundary condition 1: At r = 0 (centre of wire), qr is finite. C1 = 0 Therefore, the heat flux distribution is . . . (7.11) It is clear that the heat flux is a linear relationship which is shown in Figure 7.1(b). Now, we apply the Fourier’s law of heat conduction, i.e. . . . (7.12) Substituting this value in Eq. (7.11) and integrating the first-order differential equation,

we get

. . . (7.13)

where C2 is the constant of integration, which can be solved by using the boundary condition. Boundary condition 2: At r = R, T = T0 (surface temperature). C2 = T0 +

. . . (7.14)

Substituting the value of C2 in Eq. (7.13), we get

. . . (7.15)

The temperature distribution given by Eq. (7.15) is parabolic as shown in Figure 7.1(b). Other temperatures can now easily be calculated from this temperature distribution as follows: (a) Maximum temperature at r = 0 (Tmax – T0) =

. . . (7.16)

(b) Average temperature rise

. . . (7.17) Comparing Eq. (7.16) with Eq. (7.17), we get Tav =

Tmax . . . (7.18)

(c) Heat flow at the surface It can be calculated for the whole length of the wire. Q |r = R = 2rRL qr |r = R . . . (7.19) Substituting for qr |r = R from Eq. (7.11), we obtain Q |r = R = 2rRL = rR2LSe . . . (7.20) All the heat produced by the electrical wire at the surface must be allowed to dissipate.

7.2 HEAT CONDUCTION WITH A NUCLEAR HEAT SOURCE Let us consider that heat is produced by a nuclear fissionable material. The kinetic energy is produced by this nuclear fuel in the reactor. Let the fissionable material be a spherical ball of radius R1. It is surrounded by aluminium cladding of radius R2 as shown in Figure 7.2. Aluminium cladding is done to prevent the leakage of nuclear radiation from the reactor. Let us assume that the heat energy resulting from the nuclear heat source is Sn (cal/cm3· s). We know that this energy is minimum at the centre. As the fissionable material is consumed, the temperature may not be uniform. Let us assume a simple parabolic model, i.e. . . . (7.21) where Sn0 = volume rate of heat produced at the centre of the sphere b = constant

Figure 7.2 A spherical nuclear fuel heat source.

Let us consider that heat is being conducted away only by conduction. There is no convection energy. 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. Applying the heat balance at the shell at the steady-state condition, we get Rate of “heat in” at r – Rate of “heat out” at r + r + Rate of heat produced by the nuclear fission = 0 . . . (7.22) Rate of “heat in” at r by conduction = (4rr2)qrF |r . . . (7.23) Rate of “heat out” at r + r by conduction = (4rr2qrF) |r+r . . . (7.24) Rate of heat produced by the nuclear fission = Sn(4rr2r) . . . (7.25) Substituting the values of Eqs. (7.23), (7.24) and (7.25) in Eq. (7.22), we obtain (4rr2qrF)|r – (4rr2qrF)|r+r + Sn(4rr2r) = 0 . . . (7.26) Dividing both sides of Eq. (7.26) by 4rr and taking the limit as r 0, we have . . . (7.27) Substituting the value of Sn from Eq. (7.21) in Eq. (7.27), we get . . . (7.28) Similarly, the heat flux qrC in the aluminium cladding can be evaluated by heat conduction balance but there is no heat source term here. Hence

. . . (7.29) Integrating Eqs. (7.28) and (7.29), we get . . . (7.30)

and

. . . (7.31)

where C1 and C2 are the constants. These are calculated by boundary conditions. We know that at the interface of the fissionable material and cladding, the heat flux is the same. Boundary condition 1: At r = 0, qrF = finite . . . (7.32) Boundary condition 2: At r = R1, qrF = qrC . . . (7.33) By boundary condition 1, C1 = 0. Hence . . . (7.34) By boundary condition 2, putting r = R1, we have

. . . (7.35) and qrC becomes . . . (7.36) By equating Eqs. (7.35) and (7.36), we get . . . (7.37) Substituting this value of C2 in Eq. (7.31), we obtain . . . (7.38)

Now applying the Fourier’s law of heat conduction for both these materials of different thermal conductivity, we have . . . (7.39)

and

. . . (7.40)

Substituting these values of qrF and qrC in Eqs. (7.34) and (7.38) and integrating, we get . . . (7.41)

. . . (7.42) where C3 and C4 are constants of integration, which can be evaluated by boundary conditions. Boundary condition 3: At r = R1, TF = TC (temperature at the interface) . . . (7.43) Boundary condition 4: At r = R2, TC = T0 (ambient temperature) . . . (7.44) By boundary condition 4, C4 can be evaluated as . . . (7.45) Substituting the value of C4 in Eq. (7.44), we obtain

. . . (7.46)

By boundary condition 3, at r = R1 and TF = TC, . . . (7.47)

. . . (7.48) Substituting the value of C4 from Eq. (7.45) in Eq. (7.48), we get

. . . (7.49)

or

. . . (7.50) At the interface TC = TF, we get

or

. . . (7.51)

Substituting the value of C3 in Eq. (7.41), we get

. . . (7.52)

We have learnt two very important notions in this problem. 1. At the interface of the two solid materials, the heat flux is the same. 2. Also, at the interface, the temperatures of the two materials are the same.

7.3 HEAT CONDUCTION THROUGH COMPOSITE WALLS Many times in the industry, we come across the problem of heat conduction through various materials. A typical example is that of furnace walls which are made up of different refractory materials. Another example is of refrigration chambers. These materials in multilayer walls have different thermal conductivities. If we were to construct a furnace, we need to know the thicknesses of different layers of materials and the temperature distribution across each of them. Let us consider that the furnace consists of three layers of slabs of width W and height H as shown in Figure 7.3.

Figure 7.3 Heat conduction through composite walls.

In Figure 7.3, the three slabs are of different thicknesses and of different materials of thermal conductivities, k1, k2, k3. The surface of the first slab comes in contact with fluid A which is at a temperature Ta. The third slab comes in contact with fluid B which is at a temperature Tb. 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. The slab surfaces are at places x0, x1, x2 and x3. The thicknesses of these slabs are (x1 – x0), (x2 – x1) and (x3 – x2). We are interested in finding the temperature profiles in the slabs and also the heat flux. The following assumptions are made in the analysis of heat conduction through composite walls: 1. 2. 3. 4.

Steady-state conditions prevail. Thermal conductivities of the slabs k1, k2, k3 are constant and not changing with temperature. Heat flows in one direction only, i.e. in the x-direction. Heat transfer (at boundaries, i.e. at x = x0, and x = x3) is given by the “Newton’s law of cooling” with heat transfer coefficients h0 and h3, respectively.

With the above assumptions, let qx be the heat flux flowing in the x-direction. Let us consider a shell at x and of thickness x for applying heat balance in the region 1. There is no heat generation. Rate of “heat in” at the surface x – Rate of “heat out” at the surface x + x = 0 . . . (7.53) Region 1: qx |x WH – qx |x+x WH = 0 . . . (7.54) Dividing both sides of Eq. (7.54) by WH x and taking the limit x 0, we have

. . . (7.55) or

. . . (7.56)

Integrating Eq. (7.56), we get qx = constant = q0 . . . (7.57) = heat flux at surface x = x0 This equation is valid for all the three slabs. qx = q0 . . . (7.58) Now applying the Fourier’s law of heat conduction for all the three slabs, we obtain Region 1:

. . . (7.59)

Region 2:

. . . (7.60)

Region 3:

. . . (7.61)

where k1, k2 and k3 are the respective thermal conductivities of these three slabs and are constant. Integrating these equations over the thickness of each slab,

we get for: Region 1:

. . . (7.62)

Region 2:

. . . (7.63)

Region 3:

. . . (7.64)

Applying the Newton’s law of cooling at the surfaces x = x0 and x = x3, we will get: At surface x = x0: q0 = h0(Ta – T0) or

. . . (7.65)

At surface x = x3: . . . (7.66) where h0 and h3 are the heat transfer coefficients of the fluids A and B, respectively. Adding these five Eqs. (7.62) to (7.66), we get . . . (7.67)

. . . (7.68)

Equation (7.68) can be written either in heat flux (J/m2 · s) form or in heat flow Q(J/s) form. Let us now define the overall heat transfer coefficient U as q0 = U(Ta – Tb) . . . (7.69) or Q = U(W H)(Ta – Tb) . . . (7.70) Comparing Eqs. (7.69) and (7.70), we get . . . (7.71) We can now generalize Eq. (7.71) for n number of slabs as . . . (7.72) This equation is very useful for calculating heat transfer through composite walls separating two fluid streams. If we know the heat transfer coefficients, then the thickness of the slabs can be found out. It is a very useful relation for designing furnaces.

7.4 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. A typical example is heat transfer in a motorbike. These fins may be rectangular or simple circular plates. Generally, the fins are of small thickness in comparison to their length and width. Let us calculate the efficiency of a cooling fin. Let us consider a rectangular fin of width W, length L and thickness 2B as shown in Figure 7.4.

Figure 7.4 A rectangular cooling fin (B << L and B << W).

The rectangular fin is attached to a hot wall at a temperature Tw which is known. We make certain assumptions and consider Cartesian coordinates. Assumptions: 1. Steady-state conditions prevail. 2. B << L and B << W, i.e. the thickness of the fin is very small in comparison to its length and width. 3. Heat flows only in the z-direction, i.e. qx and qy are zero. 4. Temperature is a function of the z-direction. 5. Heat is lost by conduction in the fin. 6. No heat is lost by the edges of the fin. 7. Heat flux at the surface of the fin is given by the Newton’s law of cooling, i.e. qz = h(T – Ta) . . . (7.73) where h = heat transfer coefficient (constant) Ta = ambient air temperature at the surface of the fin. 8. Thermal conductivity k is constant. With the above assumptions, we select the shell as shown in Figure 7.4. Here, the heat flux will be qz only. Now applying, the heat balance at the shell, 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 . . . (7.74) i.e. . . . (7.75) Dividing both sides of Eq. (7.75) by 2BWz and taking the limit z 0, we get . . . (7.76) Now we apply the Fourier’s law of heat conduction and get

. . . (7.77) where k is the thermal conductivity of the material of fin which is assumed constant. We now get by substituting Eq. (7.77) in Eq. (7.76), . . . (7.78) This differential equation can be solved with the help of boundary conditions. Boundary condition 1: At z = 0, T = Tw (wall temperature) . . . (7.79) Boundary condition 2: At z = L,

(no heat loss from the edges) . . . (7.80)

In order to solve the above differential equations, we introduce the following dimensionless quantities: , dimensionless temperature . . . (7.81) , dimensionless distance . . . (7.82) , dimensionless heat transfer coefficient* . . . (7.83)

**

, where NBi is called “Biot” number.

With the introduction of the above dimensionless quantities, Eq. (7.78) and the boundary conditions become: . . . (7.84) Boundary condition 1: i |Z=0 = 1 . . . (7.85) Boundary condition 2: . . . (7.86) The solution of the differential Eq. (7.84) is well known in mathematics.

or (D2 – N2)i = 0

. . . (7.87)

The roots are D = N, thus the solution becomes i = C1e–NZ + C2eNZ . . . (7.88) where C1 and C2 are constants. Now applying boundary conditions C1 + C2 = 1 . . . (7.89) . . . (7.90) Applying the second boundary condition, 0 = –NC1e–N + NC2eN . . . (7.91) Substituting C1 = 1 – C2 from Eq. (7.89), we get . . . (7.92)

. . . (7.93) Substituting the values of C1 and C2 in Eq. (7.88) and simplifying, we obtain i = cosh NZ – tanh N sinh NZ . . . (7.94) . . . (7.95)

Substituting in Eq. (7.95) the values of i and Z from Eq. (7.81) and Eq. (7.82) we get . . . (7.96) The effectiveness of the fin can now be calculated as . . . (7.97)

or

. . . (7.98)

(7.99)

or

. . . (7.100)

. . . (7.101)

7.5 HEAT CONDUCTION FROM A SPHERETO A STAGNANT FLUID Let us consider that a heated sphere of radius R is suspended in a large, motionless fluid. This hot sphere conducts heat to the surrounding fluid in the absence of convection. The following assumptions are made: 1. 2. 3. 4. 5.

Steady-state conditions prevail. No convection heat flows in the fluid. The thermal conductivity k of the fluid is constant. Heat flows only in the radial r-direction. Heat flux at the surface can be evaluated by the Newton’s law of cooling.

Let us consider that the heated sphere is suspended in a pool of fluid as shown in Figure 7.5. Let us also consider a shell of radius r and thickness r. The surface temperature of the sphere is TR and the temperature of the fluid far away from the sphere is TC. With the above assumptions, let us apply the heat balance to the shell. Heat flux = qr

Figure 7.5 Heated sphere in a stagnant fluid.

Rate of “heat in” at the spherical surface at r – Rate of “heat out” at the spherical surface at r + r = 0 . . . (7.102) i.e. (4rr2qr)|r – (4rr2qr)|r+r = 0 . . . (7.103) Dividing both sides of Eq. (7.103) by 4rr and taking the limit r 0, we have . . . (7.104) Integrating Eq. (7.104), we get r2qr = C1 . . . (7.105) where C1 is the integration constant. Now applying the Fourier’s law of heat conduction, i.e. . . . (7.106)

we get

. . . (7.107)

Integrating Eq. (7.107), we obtain . . . (7.108) where C2 is constant. Now C1 and C2 are evaluated from the boundary conditions. Boundary condition 1: At r = R, T = TR . . . (7.109) Boundary condition 2: At r = , T = TC . . . (7.110) Now,

. . . (7.111)

and TC = C2 . . . (7.112) Substituting the value of C2 in Eq. (7.111), we get C1 = (TR – TC)kR . . . (7.113) Substituting the values of C1 and C2 in Eq. (7.108), we obtain . . . (7.114)

Now let us calculate the heat flux at the surface of the sphere. Once the temperature profile is known, the heat flux at the surface of the sphere can be calculated. . . . (7.115) By differentiating Eq. (7.114) and substituting r = R, we get . . . (7.116) . . . (7.117) We can write the Newton’s law of cooling at the surface of the sphere as qr |r = R = h(TR – TC) . . . (7.118) where h is the heat transfer coefficient. Equating Eqs. (7.117) and (7.118), we get . . . (7.119) or

. . . (7.120)

where D is the diameter of the sphere . Nusselt number = 2 . . . (7.121) This is a very important relationship.

7.6 HEAT CONDUCTION WITH A VISCOUS HEAT SOURCE High speed viscometers are used to find out the viscosity of highly viscous fluids. In these viscometers, the inner cylinder is stationary and the outer cylinder is moving, and the fluid is filled in the gap of the cylinders. Here, when the fluid is moving with a high speed, its adjacent layers rub against each other. The mechanical energy is converted into heat energy. Let the volume heat source from this “viscous dissipation” be called SV. 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. The width of the gap between these cylinders is b. A highly viscous fluid is filled between these cylinders. The system is shown in Figure 7.6. In order to solve this complicated problem, it is simplified as shown in Figure 7.6(b). Here, we ignore the curvature of the cylinders and select the Cartesian coordinates. It is also assumed that the velocity distribution is linear, i.e. . . . (7.122) where vb = ~R

b = width of the gap between the cylinders.

Figure 7.6 Heat conduction with a viscous heat source.

The other assumptions made are: 1. 2. 3. 4. 5. 6.

Incompressible, Newtonian fluid between the cylinders. The curvature of the cylinder is neglected. The temperature of the fluid is a function of R only. The gap b between the cylinders is small in comparison to the length L and width W. Thermal conductivity is constant. Steady state conditions prevail.

With the above assumptions, we apply the heat balance at the shell. Here, we use the concept of combined flux e. WLex |x – WLex |x+x = 0 . . . (7.123) Dividing both sides of Eq. (7.123) by WLx and taking the limit x 0, we get . . . (7.124) Integrating Eq. (7.124), we have ex = C1 . . . (7.125) The combined flux e has three components, i.e. 1. Convective energy flux 2. Rate of work done by molecular transfer mechanism

3. Rate of heat transfer by molecular mechanism . . . (7.126) where

is enthalpy per unit volume.

The first term, convective energty flux, can be neglected since vx = 0, vy = 0. The x-component of the second term can be written as [x · V] = xxxvx + xxyvy + xxzvz . . . (7.127) Since the velocities vx = 0, vy = 0 and hence applying the Newton’s law of viscosity, we have . . . (7.128) . . . (7.129) The third term q can be written after applying the Fourier’s law of heat conduction. The x-component of q can be written as . . . (7.130) Substituting these values of qx and x · v in Eq. (7.125), . . . (7.131) We have assumed the linear relationship of the velocity, i.e.

or

. . . (7.132)

Substituting the value of dvz/dx in Eq. (7.131), we have . . . (7.133) Integrating this equation, we obtain . . . (7.134) where C1 and C2 are constants. These constants can be evaluated by the boundary conditions. Boundary condition 1:

At x = 0,T = T0 . . . (7.135) Boundary condition 2: At x = b, T = Tb . . . (7.136) Here we assume T0 Tb. By boundary condition 1, C2 = T0 . . . (7.137) Applying boundary condition 2,

. . . (7.138) Substituting the values of C1 and C2 in Eq. (7.134), we have . . . (7.139) On simplification, 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”, SV. Rate at which the work is done by molecular momentum flux

.

= xxz area velocity . . . (7.141) We know from the above discussions that

vz = vb Rate at which the work is done by molecular momentum

Rate of viscous energy dissipation, Sv

. . . (7.142)

7.7 HEAT CONDUCTION WITH A CHEMICAL REACTIONHEAT SOURCE When a chemical reaction of the reactants takes place and the product is produced, the heat energy is either produced or consumed depending upon whether the reaction is exothermic or endothermic. Thus, . . . (7.143) The reaction generally takes place in a reactor. The reactors are of various types depending upon the flow conditions. Let us consider a tabular reactor of radius R which extends from z = – to z = + as shown in Figure 7.7.

Figure 7.7 Axial flow fixed bed reactor.

The reactants enter the reactor with a uniform velocity. The reactor is divided into three zones: Zone I: Entrance zone, z = – to z = 0. Zone II: Reaction zone, z = 0 to z = L. Zone III: Exit zone, z = L to z = + . Zone I is packed with inert spherical particles which are non-catalytic. Zone II is packed with catalyst particles where the reaction takes place. Zone III is packed with inert spherical particles which are again non-catalytic. The reactor is insulated. Here, the heat of reaction depends upon the operating conditions of the reactor such as pressure, temperature, composition of the reactants and catalyst. The chemical energy is converted into heat energy. Let Sc be the chemical energy per unit volume. It is a very complicated parameter to be evaluated. Let us formulate the simple form in which the chemical energy is a function of temperature only. Sc = SC1 F(i) . . . (7.144) where SC1 is a constant and i, the dimensionless temperature, is given by . . . (7.145) where T1 is the reactant temperature at which it enters. The following assumptions are made in the analysis of heat conduction with a chemical reaction heat source: 1. The fluid enters with uniform axial velocity in the “plug flow reactor”. 2. The reactor wall is insulated, i.e. the temperature is independent of the r-direction but is a function of the z-direction only. 3. Steady-state conditions prevail. 4. Density, superficial velocity and mass flow rate of the fluid are independent of r- and z-

directions. The superficial velocity v0 is based upon the empty reactor, i.e. . . . (7.146) where w = mass flow rate, kg/s t = density of fluid, kg/m3. Let us consider a disk of radius R at a distance of z and of thickness z. Using the concept of combined flux e, we can now apply heat balance in Zone II. rR2ez |z – rR2ez |z+z + (rR2z)Sc = 0 . . . (7.147) Dividing both sides of Eq. (7.147) by rR2z and taking the limit z 0, we get . . . (7.148) This combined flux e has three components: 1. Convective energy flux. 2. Rate of work done by molecular momentum transfer. 3. Rate of heat flux due to conduction. . . . (7.149) The first term can be re-written by neglecting the pressure term which is constant in the z-direction: i.e.

. . . (7.150)

where cp = specific heat, in kJ/kg · K. The second term for z-component: [x ·v] = [vxxxz + vyxyz + vzxzz]vz . . . (7.151) Here vx = 0, vy = 0, and applying the Newton’s law of viscosity, we have . . . (7.152) where vz = constant.

and [x · v] = 0 . . . (7.153)

The third term is of the z-component and applying the Fourier’s law of heat conduction, we obtain . . . (7.154) Here, we consider thermal conductivity keff (effective), since the heat is being transferred from the catalyst particle to the other particles, liquid to liquid, liquid to solid catalyst particle, so we assume the thermal conductivity in all the three zones to be constant. Substituting these various values in Eq. (7.148), we get . . . (7.155)

or

. . . (7.156)

Here vz is constant and we have assumed vz = v0, so we get . . . (7.157) Similar equations hold good for all the three zones, except that the Sc term will not appear in Zones I and III. Zone I: z < 0 . . . (7.158) Zone II: 0 < z < L . . . (7.159) Zone III: z > L . . . (7.160) These differential equations can be solved by using boundary conditions. Boundary condition 1: At z = –, TI = T1 . . . (7.161) Boundary condition 2: At z = 0, TI = TII (temperature at the interface) . . . (7.162) Boundary condition 3:

At z = 0,

(heat flux at the interface) . . . (7.163)

Boundary condition 4: At z = L, TII = TIII . . . (7.164) Boundary condition 5: At z = L,

. . . (7.165)

Boundary condition 6: At z = , TIII = finite . . . (7.166) The equations for all the three zones can be solved with the help of these boundary conditions. But let us take a practical case of interest to engineers of limiting solution. In this case, the convective heat transfer is more significant than the conductive heat transfer. So the conductive heat transfer terms can be neglected. Let us introduce the dimensionless terms: . . . (7.167) . . . (7.168)

. . . (7.169) where N is the dimensionless thermal heat source. . . . (7.170) where

is a constant.

With the introduction of the above dimensionless terms, we can transfer the differential Eqs. (7.158) to (7.160), and neglecting the conductive heat transfer terms, we get Zone I: z < 0, . . . (7.171) Zone II: 0 < z < 1, . . . (7.172) Zone III: z > L,

. . . (7.173) The above equations are the first-order differential equations which are solved with the separable variable methods. The boundary conditions become: Boundary condition 1: At Z = – , iI = 1 . . . (7.174) Boundary condition 2: At Z = 0, iI = iII . . . (7.175) Boundary condition 3: At Z = 1, iII = iIII . . . (7.176) With these boundary conditions, a limiting solution for the case will be discussed: Let F(i) = i . . . (7.177) The solutions are: Zone I: z < 0, iI = 1 . . . (7.178) Zone II: 0 < z < 1, iII = eNZ . . . (7.179) Zone III: z > L, iIII = eN . . . (7.180) Now, let us plot these dimensionless temperature vs. dimensionless distance as shown in Figure 7.8.

Figure 7.8 Dimensionless temperature vs. Dimensionless distance.

SOLVED EXAMPLES EXAMPLE 7.1 A cold storage room is constructed of an inner layer of 12.7 mm of pinewood, a middle layer of 101.6 mm of cork board and an outer layer of 76.2 mm of concrete. The wall surface temperature is 255.4 K inside the cold room and 297.1 K at the outside surface of the concrete. The thermal conductivities are given below. Calculate the heat loss in watts for 1 m2 and the temperature at the interface between the pinewood and cork board. Data: The thermal conductives of the materials used are as follows: S.No.

Material

Thermal Conductivity, W/m · K

1. 2. 3.

Pinewood Cork board Concrete

0.151 0.0433 0.762

Solution: Let us consider the cold storage room to be made up of three materials A, B and C as pinewood, cork board and concrete, respectively.

Figure 7.9 Example 7.1.

The cold storage room temperature, T1 = 255.4 K. The outside temperature, T4 = 297.1 K. Area = 1 m2. Pinewood, A = thermal conductivity, kA = 0.151 W/m· K Cork board, B = thermal conductivity, kB = 0.0433 W/m· K Concrete, C = thermal conductivity, kC = 0.762 W/m· K Thickness: xA = 0.0127 m xB = 0.1016 m xC = 0.0762 m Let us calculate the resistances:

The heat flow q0 is given by

The heat flows from outside to inside the cold storage room. Let T2 be the temperature at the interface between the pinewood and the cork board. Applying the heat balance, we have

Substituting the values, we get

T2 = 256.8 K EXAMPLE 7.2 A layer of pulverized cork of 6 in. thick is used as a layer of thermal insulation in a flat wall. The temperature on the cold side of the cork is 40°F and that on the warm side is 180°F. The area of the wall is 25 ft2. The thermal conductivity of the cork is 0.021 Btu/ft · h · F at 32°F and is 0.032 at 200°F. 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 = 0.021 Btu/ft · h · F at 32°F. At 200°F, the thermal conductivity k2 of the pulverised cork = 0.032 Btu/Ft·h·F.

Figure 7.10 Example 7.2.

The thickness of pulverised cork, x = 6 in. Area, A = 25 ft2 The temperature of the cold side, T1 = 40°F. The temperature of the hot side, T2 = 180°F.

Let the thermal conductivity vary linearly, then k1 = a + bT1 . . . (i) k2 = a + bT2 . . . (ii) where a and b are constants. Substituting the values of k1, k2, T1 and T2 in (i) and (ii), we get 0.021 = a + b(32 + 460) . . . (iii) 0.032 = a + b(200 + 460) . . . (iv) By solving (iii) and (iv), we get the values of a and b as a = –0.0112 b = 6.55 10–5 We can calculate heat flow as . . . (v) Substituting the values in (v), we obtain q = 27.96 Btu/h (= 29.5 kJ) EXAMPLE 7.3 An aluminium fin 3 mm thick and 7.5 cm long is extending from a wall. The base is maintained at 300°C and the ambient temperature is 50°C. Calculate the heat loss from the fin per unit depth. The data given is: (a) The thermal conductivity of aluminium, k = 300 W/m· C (b) Heat transfer coefficient, h = 10 W/m2 · h · C

Figure 7.11 Example 7.3.

Solution: We consider the case when the end of the fin is insulated. Let the width of the fin be 1 cm. Here, Tw = 300°C Ta = 50°C

The heat transfer from the fin: Q = kAc m(Tw – Ta) tanh mL . . . (i) k = thermal conductivity = 300 W/m· C P = perimeter of the fin L = 7.5 cm = 0.075 m h = 10 W/m2 · h · C

or m = 3.8 Substituting the values in Eq. (i), we get Q = 300 2(0.01 0.003) 3.8(300 – 50) tanh (3.8 0.075) Q = 4.746 W or

PROBLEMS 1. Consider a heat conduction with electrical heat source from an electrical wire of radius R and length L. At the wall, T0 is unknown but the heat flux at the wall is given by the Newton’s law of cooling, i.e. at r = R qr |r=R = h(T – Tair) where h is the heat transfer coefficient and Tair is the ambient temperature. Develop the temperature profile for the above system. Assume the heat transfer coefficient and thermal conductivity to be both constant. . . . 2. Develop the overall heat transfer coefficient for the composite cylindrical pipes as

where the terms used have their usual meaning. 3. The wall of an oven consists of three layers of bricks. The inside of the oven is made up of 0.2 m fire brick, surrounded by 0.1 m thickness of insulating brick and the outer side layer is made

up of 0.15 m of building material. The oven operates at 870°C and the outside temperature is 50°C. (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.No.

Material

Thermal Conductivity, k (kcal/h · m · C)

1. 2. 3.

Fire brick Insulating brick Building brick

0.5 0.225 0.60

[Ans. (a) (= 179 kJ/m2)] (b) Interface temperatures: 570.45°C, 237.6°C] 4. Let the thermal conductivity of a certain solid be expressed as a linear function of temperature k = a + bT. 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. 5. A flat furnace wall is constructed of a 4.5 in. layer of sil-o-cel brick with a thermal conductivity of 0.08 Btu/ft · h · F backed by a 9 in. layer of common brick of thermal conductivity of 0.8 Btu/ft · h · F. The temperature of the inner face of the wall is 1400°F and that of the outer face is 170°F. (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. (a) 689.5 W/ m2, (b) 190.5°C or 463.5 K] 6. Heat is flowing through an annular wall of inside radius r0 and outside radius r1. The thermal conductivity varies linearly with the temperature from k0 at T0 and k1 at T1. Develop an expression for the heat flow through the wall at r = r0. 7. A circular stainless steel fin (R = 17.9 W/m · K) is attached to a copper tube having an outside radius of 0.04 m. The length of the fin is 0.04 m and the thickness is 2 mm. The outside wall or tube base is at 523.2K, and the external surrounding air at 343.2K has a convective coefficient of 30 W/m2 · K. Calculate the fin efficiency and the rate of heat loss from the fin. [Ans. (i) 89% (ii) 150 W]

8 THE GENERAL ENERGY EQUATION In Chapter 7, we solved the simple, steady-state heat transfer problems by shell energy balance. It is not always easy to solve the complex problems in such a way. So, it is necessary to develop general energy equations and simplify them to solve particular problems. Let us consider a fluid element of volume xyz as shown in Figure 8.1. The fluid carries different forms of energies. It enters at the left-hand shaded portion as shown in the figure and leaves at the right-hand shaded portion. Let us identify the different energies associated with the fluid. The transfer of energy takes place by two modes, convection and molecular transport or conduction. Applying the energy balance to the volume of the fluid xyz, 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 . . . (8.1)

Figure 8.1 Fluid element of volume xyz.

In order to understand the general energy equation, let us take the x-component; similar equations can be written for the y- and z-components. (a) Energy associated by convection This constitutes the internal energy and the kinetic energy. This is the energy associated with the random movement of molecules and molecular interactions. These aspects are discussed in physics.

But we will discuss only the engineering aspects. 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:

Substituting the preceding values in Eq. (8.1) and dividing by xyz and taking the limits x 0, y 0, z 0, we get

. . . (8.2) We can write the general equation of energy in the following form:

. . . (8.3)

This general equation of energy is not easy to handle in this complicated way. So it is simplified by using the equation of continuity and the equation of motion derived earlier in Chapter 3. We can write the equation of energy for Newtonian fluids and constant thermal conductivity as . . . (8.4)

where This is called Fourier’s second law of heat conduction.

8.1 SPECIAL CASES OF ENERGY EQUATION Let us now consider the special cases of energy equations which are commonly used in the field of

engineering. (a) Fluid at constant pressure In Eq. (8.4),

. . . (8.5) (b) For solids In this case t is constant and v = 0. . . . (8.6) (c) Heat generation If there is a heat produced in the fluid by chemical reaction or electrical means, then SV is added to the energy equation. Thus, . . . (8.7) where SV = rate of heat generation, W/m3. In order to solve the heat transfer problems, 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. The simplifications are done with a view to discarding the terms which are not required. General energy equations are useful tools for solving energy problems. Simplifications are done for particular situations in the field of engineering applications.

9 TEMPERATURE DISTRIBUTIONIN TURBULENT FLOW In Chapter 7, we solved some of the heat conduction problems in laminar flow conditions. The differential equations formed by shell balance have been solved by using the appropriate boundary conditions and initial conditions. To find the temperature profiles in turbulent flow, conditions are different. Sometimes, we create turbulence in the system to enhance the heat transfer rates. Mixing takes place in the fluid. In order to solve the heat transfer problems, many empirical methods are available in the literature. The thermal problems can be understood in a similar pattern as that for the flow problems in a system. Some of the concepts discussed are: time-smoothed temperature, thermal boundary-layer thickness, and Prandtl mixing length models.

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. A similar concept holds good for the temperature profiles. Let us introduce the temperature terms: = time-smoothed temperature T = temperature fluctuation Similar to Eq. (4.1), for temperature we can write as . . . (9.1) The instantaneous temperature and fluctuation in temperature for turbulent flow are shown in Figure 9.1. The time-averaged temperature is defined as

Figure 9.1 Temperature fluctuations in turbulent flow.

. . . (9.2) All the properties of the velocity fluctuation hold good for the temperature fluctuation as well, for example will not be zero. The time-smoothed equation of continuity and the equation of motion for a fluid with constant density and viscosity were given in Eqs. (4.7) and (4.8). The energy Eq. (8.4) can be written for a fluid with constant n, t, cp, and k as follows:

. . . (9.3) Now for turbulent flow, let us substitute the terms the time-smoothed equation becomes:

and

, and so on. Then

. . . (9.4) We can observe that the time-smoothed equation has the same original form of the terms, except those indicated with dashed terms. The turbulent heat flux components can be defined as follows: . . . (9.5) . . . (9.6) . . . (9.7) To summarize, all the three equations of motion, 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 .

9.2 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. A similar situation also arises for the heat transfer problems. Let us consider a flat plate over which the fluid is entering at a uniform temperature T. 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 9.2. The surface temperature of the plate is Ts.

Figure 9.2 Thermal boundary layer for a flow past a flat plate.

We write the energy equation as . . . (9.8)

. . . (9.9) where cp = heat capacity in kJ/kg mol · K. The following assumptions are made: 1. Steady-state conditions prevail, i.e. T/t = 0. 2. The fluid properties such as density, viscosity, specific heat and thermal conductivity are constant. 3. No heat is conducted in the z-direction. 4. The heat conduction in the x-direction is negligible in comparison to the heat conducted in the y-direction. With the above assumptions, the energy equation becomes . . . (9.10) The differential Eq. (9.10) can be solved with the help of boundary conditions. Boundary condition 1: At y = 0, T = Ts . . . (9.11) Boundary condition 2: At y = , T = T . . . (9.12) Boundary condition 3: At x = 0, T = T . . . (9.13) Let us see the similarity between the above energy equation and the equation of motion earlier derived in Chapter 4. These equations are . . . (4.14)

and

. . . (4.12)

These equations were solved by using the following boundary conditions: Boundary condition 1: At y = 0, vx = 0 Boundary condition 2: At y = 0, vy = 0 Boundary condition 3: At y = 0, vx = v

The similarity of Eqs. (9.10) and (4.14) is clear. Blasius assumed the Prandtl number cp n/k = 1. In such a case, the coefficients on the R.H.S. of Eqs. (4.14) and Eq. (9.10) are the same. 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. The boundary conditions become: Boundary condition 1: At y = 0,

;

Boundary condition 2: At x = 0,

;

Boundary condition 3: At y = ,

;

. . . (9.14)

. . . (9.15)

. . . (9.16)

We conclude that these equations and boundary conditions are identical for velocity profile and temperature profile for the Prandtl number being equal to 1. Also for any x and y point, the dimensionless velocity vx /v and the dimensionless temperature (T – Ts)/(T – Ts) are equal. Hence, the solution of the velocity profile is applicable to the temperature profile solution, i.e. replacement of v by T is feasible. Therefore, the momentum transfer and heat transfer are analogous in the boundary-layer thickness. 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. The above heat transfer equations are nonlinear and hence the solution becomes complex. Blausius discussed the solution in a similar way as done for momentum transfer. By combining Eqs. (4.16) and (4.17), we get . . . (4.16)

. . . (4.17) The velocity gradient at the surface is given by . . . (9.17)

where As discussed above, . . . (9.18)

By combining Eqs. (9.17) and (9.18), we get the temperature gradient at the surface as . . . (9.19) The convective heat transfer qy (J/s) can be written from the Fourier’s law as

. . . (9.20)

where hx is the heat transfer coefficient at a point x and k is the thermal conductivity. Combining Eqs. (9.19) and (9.13), we get . . . (9.21) where

= Nusselt number.

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 0.6. Thus, . . . (9.22) Then, the local heat transfer coefficient becomes . . . (9.23) Let us calculate the mean heat transfer coefficient between x = 0 andx = L for the whole length of the plate, b = width of the plate,

A = area = bL Substituting the value of hx from Eq. (9.23), we get . . . (9.24) Integrating Eq. (9.24), we get . . . (9.25) where

9.3 PRANDTL MIXING LENGTH MODEL IN HEAT TRANSFER In Chapter 4, we discussed the Prandtl mixing length model for the turbulent flow conditions. Similarly, Prandtl proposed a model for heat transfer in the turbulent flow conditions. In fluid transport, it is assumed that the eddies formed in the turbulent flow conditions move a little distance L before they lose their identity. And then they get absorbed in the bulk of the fluid. In a similar way, energy is transported to a distance L with a velocity v¢y in the y-direction. 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. After travelling this distance L, the lump of the fluid will have a deviation of temperature from the mean temperature as shown in Figure 9.3. This situation is exactly similar to the case of velocity.

Figure 9.3 Eddy temperature in the y-direction.

Suppose T = deviation of temperature from the mean = mean temperature T = instantaneous temperature. We can write the fluctuation of temperature as: T

. . . (9.26)

Dividing both sides of Eq. (9.26) by L and taking the limit L 0, where L is small, we get T

. . . (9.27)

Let the rate of heat transferred per unit area be qy /A. Now, = (Mass)(Heat capacity)(Temperature difference) = –(tvy)cpT . . . (9.28) where cp = specific heat capacity in kJ/kg · K. Substituting T from Eq. (9.27), we have

. . . (9.29) We assumed

in Section 4.3. Also, . . . (9.30)

Substituting the value of vy in Eq. (9.29), we get . . . (9.31) As per the earlier discussion in Section 4.3, we arrived at eddy diffusivity ft as . . . (4.33) When we consider heat transfer, it is called thermal eddy diffusivity at. Then, Eq. (9.31) becomes . . . (9.32) The Fourier’s law of heat conduction for the molecular heat transfer can be written as . . . (9.33) where a is known as the molecular thermal diffusivity. Combining the Fourier’s equation and Eq. (9.32), we get . . . (9.34) If can be seen from the above that the energy problems are different for turbulent flow conditions. Three models are applied in this chapter to understand temperature distribution in turbulent flow: time-smoothed equation of energy, boundary-layer thickness and Prandtl mixing length.

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. We used the shell energy balance to formulate the differential equations. Then, these differential equations were solved with the help of boundary conditions. In Chapter 8, we developed the general energy equations. These equations are useful to tackle complex problems. Sometimes we have to handle more than one variable. Generally, one variable is space and the other variable becomes time. Such a situation is called unsteady state. A typical problem is solved in this section. Let us consider a semi-infinite plate. Initially, its surface is at temperature T0 at time t = 0. Suddenly, the surface temperature is raised to T1. Let us consider the x-y coordinate system as shown in Figure 10.1. We are interested to know the temperature profile, T(y, t).

Figure 10.1 Flat plate whose surface temperature is suddenly raised.

The following assumptions are made: 1. 2. 3. 4.

No heat flows by convection. There is no external heat source. The fluid properties such as t, n, k, cp, etc. are constant. Heat does not flow in the x-direction.

With the above assumptions, the energy Eq. (8.5) can be used to solve this problem.

. . . (8.5) Dividing both sides of Eq. (8.5) by tcp, we get 100(10.1) where a is the thermal diffusivity. This differential equation can be solved with the help of the initial and boundary conditions. Initial condition: At t 0, T = T0 for all y . . . (10.2) Boundary condition 1: At y = 0, T = T1 for t > 0 . . . (10.3) Boundary condition 2: At y = , T = T0 for t > 0 . . . (10.4) We now introduce the dimensionless temperature i, defined as . . . (10.5) The above differential Eq. (10.1) becomes . . . (10.6) The boundary and initial conditions become Initial condition: At t 0, i = 0 for all y . . . (10.7) Boundary condition 1: At y = 0, i = 1 for t > 0 . . . (10.8) Boundary condition 2: At y = , i = 0 for t > 0 . . . (10.9) Just as we solved the velocity profile in Chapter 5, we also introduce the dimensionless distance y here as: . . . (10.10) The differential equation (10.6) then becomes: . . . (10.11) The solution obtained is: i=

. . . (10.12)

or i = 1 – erf h . . . (10.13) or

=

. . . (10.14)

The erf is called the error function. These functions can be solved using any mathematics handbook. In real situations, we generally come across unsteady-state heat transfer problems. The number of variables becomes more and the mathematical tools to solve the unsteady-state heat transfer problems, become complicated. A simple case of heat conduction in a semi-infinite slab is solved in this chapter by analytical methods.

Section C MASS TRANSFER

11 MASS TRANSFER In this Section C of the book, we will consider the third fundamental transport process, that is, mass transfer. Just as we have seen Newton’s equation for momentum transfer and Fourier’s law of heat conduction, we will now study the Fick’s law for molecular diffusion of mass.

11.1 FICK’S LAW OF BINARY DIFFUSION (MOLECULAR MASS TRANSFER) Let us consider a binary system in which “species” A is diffusing into “species” B. In this case, we consider helium as A and fused silica as B as shown in Figure 11.1.

Figure 11.1 Helium (A) diffusing into silica plate (B).

First, the silica plate is exposed to air and then to helium gas. Helium gas penetrates into the silica plate. The molecular transport of one substance to another is called diffusion, concen-tration diffusion or mass diffusion. The molecular transfer of mass takes place because of the concentration difference. Here, the mass fraction wA is defined for species A as:

Similarly, the mass fraction of B is given by

At time t = 0 or t < 0, the silica plate is exposed to air and there is no helium gas, i.e. wA = 0. Suddenly, the silica plate is exposed to helium gas. As the time passes, the concentration of helium is built up and later the steady-state concentration of helium is achieved. This process is shown in Figure 11.2.

Figure 11.2 Steady-state concentration profile for diffusion of helium A into fused silica B.

The concentration of A at the top of the silica slab is zero and as time passes, the concentration profile becomes straight. The x-y coordinates are selected. Let A be the area of the silica slab and Y be the thickness. The mass flow of helium in the y-direction, wAy, is given as wAy A wAy wAy where

is the initial concentration of A. Combining these, we get . . . (11.1)

or

. . . (11.2)

where t = density of the silica–helium system DAB = diffusivity of silica–helium system = proportionality constant called diffusivity, 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. The differential form of Eq. (11.2) becomes . . . (11.3) Equation (11.3) is called the Fick’s law of diffusion. Here, jAy is the molecular mass flux of helium in the positive direction of y. Helium is moving slowly and its concentration is very small. Equation (11.3) is similar to Eq. (1.2) for molecular momentum transfer and similar to Eq. (6.5) for heat conduction. Similarly in other directions too, we can write the Fick’s law as . . . (11.4)

and

. . . (11.5)

Combining all these equations, we can write . . . (11.6)

where = divergence =

.

Similarly, when B is diffusing into A, we can write the equation as jB = –tDBAwB . . . (11.7) 11.1.1 Some Features of Fick’s Law of Diffusion Let us discuss some of the features of the Fick’s law of diffusion. 1. The Newtons’s law of viscosity has been discussed in momentum transfer in Section 1.1. The Fourier’s law of heat conduction has been discussed in heat transfer in Section 6.1. Let us look into Eqs. (1.2), (6.5) and (11.6). All these equations are basically similar. In brief, 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, i.e. m2/s. 2. In a system when species A and B are moving, the velocities of A and B are taken as weighted velocities according to the mass fractions and not instantaneous velocities. Then the mass fluxes can be defined as jAy = twA(vAy – vy) . . . (11.8) and jBy = twB(vBy – vy) . . . (11.9) where vAy = weighted velocity of A according to mass fraction of A in the y-direction. vBy = weighted velocity of B according to mass fraction of B in the y-direction. These fluxes are measured with respect to the motion of centre of mass of A and B. 3. jAy + jBy = 0. 4. Normally when A and B are diffusing in a system, then the diffusivity of A in B and the diffusivity B in A are not equal, i.e.

DAB DBA The diffusivity of a system can be estimated by a number of methods. There are many empirical equations available in the literature for the estimation of diffusivity for gases and liquids. Some of them are: (a) Reduced temperature and pressure and the corresponding reduced diffusivity, obtained from critical values. (b) Chapman–Enskog theory is applied for the estimation of diffusivity for gases at low density. These are similar equations used for the estimation of viscosity as discussed earlier. (c) Hydrodynamic theory as stated by the Nernst–Einstein equation is used for diffusion in binary liquids. In this book, we will not describe the estimation of diffusivity. This is well discussed in other chemical engineering books. For binary gas mixtures at low pressures, DAB is inversely proportional to pressure, increases with increasing temperature and is almost independent of the composition for a given pair of gases.

11.2 CONVECTIVE MASS TRANSFER In addition to molecular mass transfer, mass may also be transferred by the motion of molecules of the fluid. Each molecule is having some velocity v of the fluid in the bulk. This velocity will have three components in the x-, y-, z-directions as vx, vy, vz, respectively. If t is the density of the fluid and dS is the area of cross section perpendicular to the direction of flow, 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, the convective mass flux for component A can be written as tAv = tAvxdx + tAvydy + tAvzdz . . . (11.10) where dx, dy and dz are the unit vectors in the x-, y-, z-directions. In mass transfer process, one can express fluxes in either mass flux or molar flux. So, convective mass flux = kg/m2 · s convective molar flux = kmol/m2 · s 11.2.1 Mass and Molar Fluxes The diffusion equations are written along with the equation of motion when no chemical reaction takes place. We express mass concentration, mass average velocity and mass flux for a particular species. In such cases, mass units are expressed. When a chemical reaction takes place, then molar concentration and molar fluxes are expressed. Let us define the terms: Mass concentration of species A

wA = mass of A per unit volume of the solution Molar concentration of species A, 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 . . . (11.11) Molar units:

= cA(vA – v*) = cDABxA . . . (11.12)

where vA – v = diffusion velocity of species A with respect to the mass average velocity v. vA – v* = diffusion velocity of species A with respect to the molar average velocity v*. Let us say that the combined flux is the sum of the fluxes due to molecular diffusion and convective diffusion. We can summarize the gradient and Fick’s law of diffusion for mass and molar terms as shown in Table 11.1A and Table 11.1B respectively. Table 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 11.1B 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

11.3 SHELL MASS BALANCES AND BOUNDARY CONDITIONS

Steady-state momentum balances have been made in shell momentum balances in Chapter 1. Also, we have made shell energy balances for heat conduction in Chapter 6. After formulation of the steadystate differential equations, these equations have been solved by setting boundary conditions. Similarly, we shall see that the steady-state diffusion problems may be formulated by shell mass balances. The steps involved are as follows: 1. Select the coordinate system of the problem. 2. Select the shell, where mass balance has to be applied. 3. Under steady-state conditions, 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 (11.13) 4. Express the terms on the basis of mass, if mass balance is to be applied. Otherwise, express the terms in molar form for molar balance. 5. We have selected the fluxes as molecular, convective or combined as the case may be. Generally, we select the combined flux and simplify the other terms. 6. Formulate the differential equation in terms of fluxes. 7. Integrate the differential equation formed and either keep the constant of integration or solve its value from the physical concepts. 8. Apply the Fick’s law of diffusion and formulate the differential equation for concentration. 9. Integrate the differential equation formed to get the concentration equation. 10. Apply the boundary conditions to solve for the constants of integration. 11. Get the concentration profile and solve other important items. Before discussing the boundary conditions, let us discuss the molar flux and mass flux produced by chemical reactions. Let NA be the number of moles of A diffusing per unit area per unit time in a binary system. The molar flux of A in the z-direction can be written as: . . . (11.14) We may come across the problem of solving for NBz. This can be solved by physical or chemical reasoning. One may say that in a system, A is diffusing but B is not diffusing. In such a case, NBz = 0. In other cases, NAz/NBz may be known from the physical or chemical concepts. The reactions may take place by two mechanisms: 1. Homogeneous reactions 2. Heterogeneous reactions In homogeneous reactions, the source term may appear in the differential equations. This is similar to the heat source problem for energy shell balances. For heterogeneous reactions, there is no source term in the differential equations in the shell mass balance. These differential equations can be solved with the help of boundary conditions. Let the reaction be taking place as: A products The reaction rate equations can be written for homogeneous and heterogeneous reactions.

Homogeneous reaction: RA = kncAn . . . (11.15) Heterogeneous reaction: NAz |surface = kn ≤ cAn |surface . . . (11.16) where RA = rate of reaction, in mol/m3 · s kn = reaction rate constant cA = concentration of A, in mole of A/cm3 n = order of reaction, for first order n = 1 NAz = combined molar flux, in mol/cm2 · s kn ≤ = reaction rate constant based on the surface area Boundary conditions: 1. The concentration at the surface may be specified, e.g. xA =

.

2. The mass flux at the surface may be specified, e.g. NAz = or

= given

or NBz = 0 for B is not diffusing. 3. If the solid substance A is lost to the surroundings, then we can write: molar flux concentration gradient or molar flux = kc(concentration gradient) or

. . . (11.17)

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, similar to heat transfer coefficient 4. The rate of chemical reaction may be specified, e.g. . . . (11.18) where k1 ≤ = rate constant for the first-order reaction based on the surface area.

SOLVED EXAMPLES

EXAMPLE 11.1 CO2 gas is diffusing through N2 in one direction at atmospheric pressure and temperature 15°C. The mole fraction of CO2 at point A is 0.2, at point B, 3 m away in the direction of diffusion, it is 0.0195. The concentration gradient is constant over this distance and diffusivity is 1.5 10–5 m2/s. The gas phase as a whole is stationary, i.e. N2 is diffusing at the same rate as the CO2, but in the opposite direction. (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. DAB = 1.5 10–5 m2/s

Figure 11.3 Example 11.1.

Mole fraction of A (CO2) at z2 = 0.0195 Mole fraction of A (CO2) at z1 = 0.2 Total pressure = 1 atm = 1.0132 105 Pa Temperature, T = 15°C = 273.1 + 15 = 288.1 K z2 – z1 = 3 m Let us write the Fick’s law: . . . (i) where c = concentration of the CO2–N2 system. or Now the molar flux of A becomes . . . (ii) where xA = mole fraction of A.

Integrating Eq. (ii), we have . . . (iii) or

or

. . . (iv)

R = 8314 Substituting the above values in Eq. (iv), we get

= 0.38 10–7 kg mol of A/m2 · s For the component B, i.e. N2, mole fraction of B,

= 1 – 0.2 = 0.8

mole fraction of B,

= 1 – 0.0195 = 0.9805

The mole flux for B,

, can be written as

. . . (v) Substituting the values in Eq. (v), we get

= –0.38 10–7 kmol of B/m2 · s EXAMPLE 11.2 Ammonia gas (A) and nitrogen gas (B) are diffusing in counter diffusion through a straight glass tube 0.61 m long with an inside diameter 24.4 mm at 298 K and 101.32 kPa. Both ends of the tube are connected to a large mixed chamber at 101.32 kPa. The partial pressure of NH3 is constant at 20.0 kPa in one chamber and 6.666 kPa in another. The diffusivity at 298 K and 101.32 kPa is 2.3 10–5 m2/s. Calculate the diffusion of NH3 in kg mol/s and also the diffusion of N2. Solution: Let A(NH3) be diffusing in the chamber as shown in Figure 11.4 and B(N2) be diffusing in the opposite direction.

Figure 11.4 Example 11.2.

DAB = 2.3 10–5 m2/s = 20 103 Pa = 6.666 103 Pa z2 – z1 = 0.61 m Total pressure, P = 101.32 103 Pa Temperature, T = 298 K Diameter of the pipe, D = 24.4 mm = 0.0244 m Let us write the molar flux of A: . . . (i) Substituting the values in Eq. (i), we get

= 2.03 10–5 kmol of A/m2 · s Diffusion of A (NH3):

= 9.49 10–11 kmol of A/s Diffusion of B (N2): = (101.32 103 – 20 103) = 81.32 103 Pa

= (101.32 103 – 6.666 103) = 94.654 103 Pa . . . (ii) Substituting the values in Eq. (ii), we have

= –2.03 10–5 kmol of B/m2 · s Diffusion of B = –2.03 10–5 (0.0244)2 = –9.49 10–11 kmol of B/s

PROBLEMS 1. Prove that DAB is equal to DBA for an ideal gas. Is this relation true for a diffusion in a binary liquid system? 2. A mixture of He and N2 gases is contained in a pipe at 298 K and 1 atm total pressure which is constant. At one end of the pipe at point A (1), the partial pressure the other end at 0.2 m,

of He is 0.6 atm and at

= 0.2 atm. Calculate the flux of He at steady state if DAB of theHe–

N2 mixture is 0.687 10–4 m2/s. [Ans. 5.63 10–6 kg mol A/m2 · s] 3. Calculate the steady-state mass flux jAy of helium for the fusedsilica–helium system at 500 K. The partial pressure of helium is 1 atm at y = 0 and zero at the upper surface of the plate. The data given is: Thickness of the silica plate = 10–2 mm Density of silica = 2.6 g/cm3 Solubility of helium in silica = 0.0084 m3/m3 Diffusivity of helium–silica, DAB = 0.2 10–7 cm2/s Make suitable assumptions.

12 SHELL MASS BALANCES AND CONCENTRATION DISTRIBUTION FOR LAMINAR FLOW In Chapter 2, we developed the shell momentum balance for viscous flow. In Chapter 7, we formulated the shell energy balance for heat conduction. In the same manner, in this chapter, we will develop shell mass balances for laminar flow. All these problems are solved for the steady-state conditions. The steps involved will be the same as those previously defined for momentum and heat balances. Mass balance is done over the shell to formulate the first-order differential equations. This will lead to the solution for mass flux distribution. Next, we will introduce the relation between the mass flux and the concentration gradient. This will result in the differential equation for concentration distribution. On integration, we get the constants. These constants are evaluated with the help of boundary conditions. These boundary conditions are taken from the knowledge of physical concepts of the problem. Several mass transfer problems are solved by considering the mass flux. Let NA, the mass flux, be defined as the number of moles of A diffusing per unit area per unit time. Normally, we can express the combined flux of A(NAz) in the z-direction as . . . (11.14) where if the B component is non-diffusing, then NB = 0, or where mostly the NBz/NAz ratio is specified. Section 12.1 deals with the simple diffusion problem. It will enable us to understand the mass balance easily. Sections 12.2 and 12.3 discuss the diffusion problems with heterogeneous reaction. The reaction may be instantaneous reaction or slow reaction. Section 12.4 deals with diffusion with homogeneous reaction. Diffusion from a spherical droplet through a stagnant gas film is discussed in Section 12.5. All these problems are solved for steady-state conditions.

12.1 DIFFUSION THROUGH A STAGNANT GAS FILM Let us consider that a liquid A (e.g. benzene) is being evaporated into a gas B (e.g. air). This diffusion system is shown in Figure 12.1.

Figure 12.1 Steady-state diffusion of liquid A into stagnant gas B.

The following assumptions are made: The level of liquid A is maintained throughout during evaporation. Vapours of A and B form an ideal gas mixture. The solubility of B in A is negligible, i.e. air is not soluble in benzene. After the evaporation of A, it flows only in the z-direction (i.e. unidirectional flow). Steady-state conditions prevail. As soon as A evaporates, it is carried away by the B gas stream. As the pressure and temperature are assumed constant, so diffusivity DAB is independent of composition. 8. No radial flow of A, i.e. the flows at the centre and near the wall, affect the diffusion process. 1. 2. 3. 4. 5. 6. 7.

Let xA1 be the concentration of A at the liquid–gas interface. Then, . . . (12.1) where pA = vapour pressure of A p = total pressure Also, p = cRT where c = concentration, assumed constant T = temperature, in K R = gas constant. Let us consider a shell at a distance z and of thickness z as shown in Figure 12.1. As assumed above, diffusion of A takes place in the z-direction only. Let the combined flux be NAz. Applying a mass balance at the shell, we have

Rate of “mass in” of A at z – Rate of “mass out” of B at z + z = 0 . . . (12.2) Let S be the area of cross section of the vessel in which evaporation takes place. Substituting the rate of mass in Eq. (12.2), we get SNAz |z – SNAz |z+z = 0 . . . (12.3) Dividing both sides of Eq. (12.3) by S and z and taking the limit z 0, we get . . . (12.4) We can write the equation for the combined flux NAz from the previously developed Eq. (11.14), which is . . . (11.14) When B is not diffusing into A, NBz = 0. . . . (12.5)

or

. . . (12.6)

Substituting this value of NAz in Eq. (12.4), we get . . . (12.7) For ideal gas We know that p = cRT where c is the concentration (assumed constant, also we have assumed DAB to be constant). Therefore, Eq. (12.7) becomes . . . (12.8) Integrating Eq. (12.8), we get . . . (12.9) where C1 is a constant of integration. Integrating this equation again, we get –ln (1 – xA) = C1z + C2 . . . (12.10) where C2 is a constant. Looking into the pattern of Eq. (12.10) and in order to evaluate C1 and C2, we assume: C1 = –ln K1 . . . (12.11)

C2 = –ln K2 . . . (12.12) where K1 and K2 are constants. We obtain Eq. (12.10) as: ln (1 – xA) = ln (K1z · K2) or 1 – xA = K1z · K2 . . . (12.13) Boundary condition 1: At z = z1, xA = xA1 . . . (12.14) Boundary condition 2: At z = z2, xA = xA2 . . . (12.15) . . . (12.16) . . . (12.17) Dividing Eq. (12.17) by Eq. (12.16), we obtain

or

. . . (12.18)

Substituting this value of K1 in Eq. (12.16), we get

or

. . . (12.19)

Substituting the values of K1 and K2 in Eq. (12.13), we get

(12.20)

or

. . . (12.21)

For component B xB = 1 – xA . . . (12.22) The concentration profile is shown in Figure 12.1. The slope dxA/dz is not constant. The flux can be obtained from Eq. (12.6). Once the concentration profiles are known, then the average concentration and mass flux can be easily calculated. Average concentration Let . . . (12.23) where Z is the dimensionless distance. Let us write the average concentration for B: . . . (12.24)

Boundary condition 1: At z = z1, Z = 0 . . . (12.25) Boundary condition 2: At z = z2, Z = 1 . . . (12.26) Now Eq. (12.24) becomes . . . (12.27)

or

. . . (12.28)

. . . (12.29)

Thus, the average value of xB is the logarithmic mean, (xB)ln, of the terminal concentrations.

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 12.2. There are spherical balls coated with a catalytic material. Also, we assume that 2 moles of A (reactant) produce 1 mole of B (product).

We again assume an “idealized model” for the system. Reactant A diffuses through the gas film surrounded by the catalyst. At the catalyst surface, the reaction takes place instantaneously. After the reaction, the product B diffuses back in the mainstream.

Figure 12.2 Catalytic reactor and a catalyst.

As the reaction takes place at the surface of the catalyst, the gas film and the surface of the catalyst are as shown in Figure 12.3. Two moles of A are

Figure 12.3 Idealized model for a catalytic reaction, concentration profile and a shell for mass balance.

diffusing through the gas film and the reaction takes place at the surface of the catalyst. Then, the product B formed diffuses back to the main stream. Reactant A is diffusing in the positive direction of z, while the product B is diffusing in the negative direction of z. Although some heat is produced in the catalytic reaction, we assume it to be very small and negligible. The following assumptions are made:

1. Gas film surrounding the catalyst is isothermal, i.e. the temperature remains constant. Therefore, DAB is constant. 2. Steady-state conditions prevail. 3. Diffusion of A and B takes place only in the z-direction. Let us consider a shell at a distance z and of thickness z. NAz is the combined flux operating in the system. Let us take, W as the width and L as the length of the shell for consideration. Now applying a mass balance on the shell, we have W LNAz |z – W LNAz |z+z = 0 . . . (12.30) Dividing both sides of Eq. (12.30) by WLz and taking the limit as z 0, we get . . . (12.31) The general Fick’s law can be written as . . . (12.32) As discussed above, 2 moles of A penetrate the gas film and produce 1 mole of B, and B is moving in the negative direction of z. So, the flux for B, NBz, can be written as . . . (12.33) Substituting this value of NBz in Eq. (12.32), we get

or

. . . (12.34)

Substituting this value of NAz in Eq. (12.31), we obtain

or

. . . (12.35)

Integrating this equation, we get . . . (12.36) where C1 is a constant of integration. Integrating again Eq. (12.36), we get . . . (12.37) where C2 is a constant. For mathematical similarity, we may assume: C1 = –2 ln K1 . . . (12.38) C2 = –2 ln K2 . . . (12.39) where K1, K2 are constants. Now,

or

. . . (12.40)

Let d be the gas film thickness as shown in Figure 12.3. Boundary condition 1: At z = 0, xA = . . . (12.41) Boundary condition 2: At z = d, xA = 0 . . . (12.42) Substituting these values and solving for the concentration of A, we obtain . . . (12.43) or

. . . (12.44)

or

or

. . . (12.45)

Substituting the values of K1 and K2 in Eq. (12.40), we get

or

. . . (12.46)

This is the concentration of A which is shown in Figure 12.3. The molar flux at the catalyst surface can be calculated as . . . (12.47) Differentiating Eq. (12.46) w.r.t. z and taking the value at z = d, we obtain . . . (12.48) From the above, we can conclude: 1. Although chemical reaction occurs instantaneously at the surface of the catalyst, it appears as if no reaction term appears and only diffusion takes place. 2. It is a series process. First, diffusion takes place and then the chemical reaction takes place. 3. A to B formation, we can say, is a “diffusion” controlled one for the case of an instantaneous reaction.

12.3 DIFFUSION WITH A HETEROGENEOUS CHEMICAL REACTION (SLOW REACTION) Many a time, one comes across catalytic chemical reactions which are not instantaneous. Mostly, reactions take place at the catalyst surface slowly. Let us assume that after diffusion of A through the gas film, the reaction of A takes place at the catalyst surface, i.e.

The whole process is explained as illustrated in Figure 12.2. At the catalyst surface at z = d, the reaction takes place. It is assumed that the rate at which A disappears at the catalyst surface is proportional to the concentration of A at the interface. We can write the flux as NAz = k1 ≤ cA = ck1 ≤ xA . . . (12.49) where k1 ≤ is the rate constant for the first-order surface reaction. After applying mass balance at the shell, we will get the same relation as that of Eq. (12.31), i.e.

Integrating this equation and putting proper constants of integration, we get the same relation as that of Eq. (12.40), i.e.

Boundary condition 1: At z = 0, . . . (12.50) Boundary condition 2: At z = d, NAz = K1cxA . . . (12.51) . . . (12.52) At steady state, the flux NAz becomes constant. Substituting the value of boundary condition 2 in Eq. (12.40), we get

or

. . . (12.53)

Substituting the values of K1 and K2 in Eq. (12.40), we obtain

or

. . . (12.54)

From this equation, one can find out the flux at the catalyst surface. Differentiating Eq. (12.54) and evaluating at z = d and substituting in Eq. (12.34), we get

. . . (12.55)

Let us analyse the result. We have obtained the combined effect of diffusion and chemical reaction. The surface reaction kinetics is described by the term DAB/k1 ≤ d in Eq. (12.55). It is related to the Damköhler number which describes the reaction kinetics as . . . (12.56) Two limiting cases may arise: 1. If k1 ≤ is large, the term

can be expanded, neglecting the higher-order terms.

We obtain the diffusion controlling mechanism. 2. The other limiting case is the reaction kinetics controlling mechanism, where diffusion is negligible.

12.4 DIFFUSION WITH A HOMOGENEOUS CHEMICAL REACTION It is very common in chemical engineering that homogeneous reactions take place to get the desired product. A typical example of such reactions is absorption of a gas in a liquid. If CO2 gas is absorbed in a concentrated aqueous solution of NaOH, then the reaction can be written as CO2 + 2NaOH --- Na2CO3 + H2O In general, we say that when A is absorbed in B, a chemical reaction takes place to produce AB.

Let us take liquid B in a vessel and allow gas A to be absorbed in it. Here, a chemical reaction takes place. The system is shown in Figure 12.4.

Figure 12.4 Absorption of gas A in a liquid B with a homogeneous reaction.

The following assumptions are made: 1. The diffusion of gas A in liquid B takes place isothermally. Therefore, the diffusivity DAB is constant. 2. An irreversible first-order reaction takes place, i.e. RA = k1cA . . . (12.57) where RA = rate of reaction k1 = rate constant based on volume, in s–1 3. Steady-state conditions prevail. 4. The concentration of AB formed is very small compared to the concentration of A or B. In other words, we assume that A and B form a binary mixture. Let us consider a shell at a distance z and of thickness z as shown in Figure 12.4. Let NAz be the combined molar flux. Applying a mass balance at the shell, we have Rate of molar “mass in” of A – Rate of molar “mass out” of A – Depletion of A per unit volume = 0 . . . (12.58) Let S be the area of cross section of the vessel. Then, SNAz |z – SNAz |z+z – k1cASz = 0 . . . (12.59) Dividing both sides of Eq. (12.59) by Sz and taking the limit z 0, we get . . . (12.60) Applying the Fick’s law of diffusion, i.e. . . . (12.61) From Eq. (12.60), we obtain

. . . (12.62) This differential equation can be solved with the help of boundary conditions. We know the concentration of A at the surface of the liquid, i.e. . From the physical concept, it is understood that no diffusion of A takes place at the bottom of the vessel, i.e. NAz = 0. Let us write the boundary conditions: Boundary condition 1: At z = 0, . . . (12.63) Boundary condition 2: At z = L, NAz = 0 or

. . . (12.64)

This second-order differential Eq. (12.62) can be solved with the help of definitions of the following dimensionless numbers. . . . (12.65)

. . . (12.66)

. . . (12.67) where z is called Tiele modulus, which is a dimensionless number. It is the ratio of the chemical reaction rate to the diffusion rate, i.e. . . . (12.68) By using the dimensionless variables, the differential Eq. (12.62) becomes . . . (12.69) This equation can easily be solved mathematically as (D2 – z2)b = 0 . . . (12.70) where D2 = z2 or D = z. b = C1e–za + C2eza . . . (12.71) where C1 and C2 are constants and can be solved with the help of boundary conditions.

Boundary conditon 1: At a = 0, b = 1 . . . (12.72) Boundary conditon 2: At a = 1,

. . . (12.73)

The solution of Eq. (12.71) becomes b = C1 cosh za + C2 sinh za . . . (12.74) Solving for C1 and C2 values from the boundary conditions, we get

or

. . . (12.75)

Substituting the values of a, b and z in Eq. (12.75) from the above, we get

. . . (12.76)

The concentration profile is shown in Figure 12.4.

12.5 DIFFUSION THROUGH A SPHERICAL STAGNANT GAS FILM SURROUNDING A DROPLET OF LIQUID Let us consider a spherical droplet of liquid A. Surrounding the liquid is a stagnant gas film B as shown in Figure 12.5. Liquid A is being evaporated.

Figure 12.5 Diffusion through a spherical stagnant gas film surrounding a droplet of liquid A.

The radius of the liquid droplet is r1 and the radius of the stagnant gas film is r2. Let us consider a shell of radius r and thickness r as shown in Figure 12.5. It is assumed that the diffusion takes place only in the radial direction. Let us take the spherical coordinate system for mass balance. It is also assumed that during diffusion, the temperature remains constant, i.e. DAB is constant. We consider that NAr is the mass flux operating in the system. Applying a mass balance over the shell, we get Rate of “mass in” of A at r – Rate of “mass out” of A at r + r = 0 . . . (12.77) 4p(r2NAr)|r – 4p(r2NAr)|r+r = 0 . . . (12.78) Dividing both sides of Eq. (12.78) by 4pr and taking the limit r 0, we have . . . (12.79) Integrating Eq. (12.79), we get r2NAr = C1 . . . (12.80) where C1 is a constant. 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. These will be used as boundary conditions for evaluating the constants of integration. Boundary condition 1: At r = r1, NAr = NAr1 . . . (12.81) Substituting this value in Eq. (12.80), we have

. . . (12.82) But by the Fick’s law, NAr is given as . . . (12.83) Substituting this value in Eq. (12.79), we get . . . (12.84) At constant temperature, cDAB is constant. Integrating this equation, we get . . . (12.85) This equation gives the concentration profile. If we integrate Eq. (12.83) between the limits at r1, xA =

and at r2, xA =

, we get

. . . (12.86) where

SOLVED EXAMPLES EXAMPLE 12.1 Methane gas is diffusing in a straight tube 0.1 m long containing helium at 298 K and the total pressure is 1 atm (=1.01325 105 Pa). The partial pressure of methane is 1.4 104 Pa at one end and 1.333 103 Pa at the other end. Helium gas is insoluble in methane and it is non-diffusing. The diffusivity of methane–helium is 0.675 10–4 m2/s. Calculate the flux of methane at steady-state conditions. Solution: Let methane = A and helium = B z2 – z1 = 0.1 m Temperature, T = 298 K Total pressure, p = 1.01325 105 Pa Partial pressure of methane at point (1),

pA1 = 1.4 104 Pa Partial pressure of methane at point (2), pA2 = 1.333 103 Pa DAB = 0.675 10–4 m2/s The flux of A is given by the following equation where B is non-diffusing:

Substituting the values, we get

= 1.028 10–5 kg mol A/s · m2 EXAMPLE 12.2 Liquid chloropicrin (CCl3NO2) evaporates into “pure air” from a long tube of surface area 2.29 cm2 exposed for evaporation. The temperature is 25°C. Calculate the evaporation rate in g/h of the liquid chloropicrin. The data given is: Total pressure = 770 mm Hg Diffusivity (CCl3NO2–Air) = 0.088 cm2/s Vapour pressure of CCl3NO2 = 23.81 mm Hg Distance from the liquid level to top of the tube = 11.14 cm Density of CCl3NO2 = 1.65 g/cm3 Surface area of the liquid exposed for evaporation = 2.29 cm2 Solution: Let us consider CCl3NO2 = A Air = B

Figure 12.6 Example 12.2.

We assume steady-state conditions. CCl3NO2 (liquid) is taken in a tube as shown in Figure 12.6. The evaporation rate is given by . . . (i) Total pressure = 770 mm Hg

Temperature = 25°C = 298 K = 770 mm Hg = 770 – 23.81 = 746.19 mm Hg z2 – z1 = 11.14 cm DAB = 0.088 cm2/s R = 82.06 Molecular weight of CCl3NO2 = 164.5 Substituting the values in Eq. (i), we get

= 1.027 10–8 g mol/cm2 · s Evaporation rate in g/h

= 1.027 10–8 (164.5)(2.29) 3600 = 0.0139 g/h

PROBLEMS 1. Derive an equation for the molar flux for a situation of diffusion of A in a stagnant gas B at steady-state conditions. Specises A is assumed to be a spherical particle of radius r1.*. . . .

* 2. A sphere of naphthalene of radius 2 mm is suspended in a large volume of still air at 318 K and 1 atm pressure. The surface temperature of the naphthalene can be assumed to be at 318 K. Its vapour pressure at 318 K is 0.555 mm Hg. The diffusivity of the naphthalene–air system at 318 K is 6.92 10–6 m2/s. Calculate the rate of evaporation of the naphthalene from the surface. [Ans. 9.68 10–8 kg mol/s · m2] 3. 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. At point 1 the concentration of ethanol is 16.8% (wt) and the solution density is 972.8 kg/m3. At point 2, the concentration of ethanol is 6.8% and density is 988.1 kg/m3. The diffusivity of ethanol is 0.74 10–9 m2/s. Calculate the steady-state flux NA. . . . [Ans. NA = 9 10–7 kmol/m2·s] 4. Deduce an expression to find the rate of diffusion of gas A through a non-diffusing gas B. CO2 is diffusing through N2 in one direction at atmospheric pressure at 15°C. The mole fraction of CO2 at point A is 0.2, and at point B, 3 m away in the direction of diffusion, it is 0.0195. The concentration gradient is constant over this distance and diffusivity is 1.5 10–5 m2/s. The gas phase as a whole is stationary, i.e. N2 is diffusing at the same rate as the CO2, but in the opposite direction. (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? . . . [Ans. (a) 3.8 10–8 kg mol/m2 · s, (b) 1.5 10–6 kg/m2 · s, (c) 1.45 10–5 m/s] 5. Derive a concentration profile for a reaction 2A B. The reaction is not instantaneous at the catalytic surface at z = d. Assume that the rate at which A disappears at the catalyst-coated

surface is proportional to the concentration of A in the fluid at the interface. That is, NAz = k1CA where k1 is the rate constant for the pseudo-first-order surface reaction. 6. Oxygen A is diffusing through carbon monoxide B under steady-state conditions, with CO nondiffusing. The total pressure is 1 105 N/m2 and the temperature is 0°C. The partial pressures of oxygen at two planes 2 mm apart are 13,000 and 6500 N/m2, respectively. The diffusivity of the mixture is 1.87 10–5 m2/s. Calculate the rate of diffusion of oxygen. . . . [Ans. 2.67 10–5 kg mol/m2 · s] 7. 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. The length of the tube containing CCl4 is 17.1 cm above the gas–liquid interface. The total pressure of the system is 755 mm Hg and the temperature is 0°C. The vapour pressure of CCl4 at that temperature is 33 mm Hg. The cross-sectional area of the tube is 0.82 cm2. It is found that 0.0208 cm3 CCl4 evaporates in a 10-hour period after steady state has been attained. What is the diffusivity of the system CCl4–O2? . . . [Ans. 0.0636 cm2/s]

13 THE GENERAL EQUATIONOF DIFFUSION In Chapter 12, we applied the mass balance on the shell and then formulated the differential equation. Sometimes, it is not possible to formulate the problems by the shell balance. For such complicated problems, we have to start from the general mass balance equations and simplify these general equations with suitable assumptions. The mass flux equations are set up from these general mass balances. Then, these diffusion equations can be used to solve any problem. In Chapter 3, we developed the general equations of motion. In Chapter 8, the general energy equations have been developed. Similarly, we shall develop the general diffusion equations in this chapter. Let us consider an element of x y z through which a fluid is flowing as shown in Figure 13.1.

Figure 13.1 A fluid element of xyz through which a fluid is flowing.

The following assumptions are made: 1. There is no generation of mass of the fluid. 2. No chemical reaction takes place. 3. As the temperature of the fluid remains constant, the diffusivity DAB remains constant. Let NAx be the mass flux operation in the system, in kg mol/m2 · s. The fluid A is entering the shaded area at x as shown in Figure 13.1 and leaving at x + x. Let cA be the concentration of the component A of the fluid. We consider a binary mixture of the fluid. Now applying a mass balance over the element, we have Rate of “mass in” of A at the surface x – Rate of “mass out” of A at the surface x + x

= Rate of mass of A accumulated . . . (13.1) Rate of “mass in” of A at the surface x = (yz)NAx |x . . . (13.2) Rate of “mass out” of A at the surface x + x = (yz)NAx|x+x . . . (13.3) Rate of mass of A accumulated = (xyz)

. . . (13.4)

Substituting these values in Eq. (13.1), we obtain . . . (13.5) Dividing both sides of Eq. (13.5) by xyz and taking the limit as x 0, we get . . . (13.6) Now, let us substitute the NAx value from the Fick’s law of diffusion, which is . . . (13.7) Then, we get . . . (13.8) Similarly, we can also obtain the diffusion equations for the y- and z-directions as . . . (13.9)

and

. . . (13.10)

Hence we can write the general equation as . . . (13.11)

or

. . . (13.12)

where When we compare Eq. (13.12) with the heat transfer equation,

where a is the thermal diffusivity, we can note a similarity between these two equations. Equation (13.12) is, therefore, also called the second Fick’s law of diffusion. If we include the convective mass transfer and the rate of reaction terms, then the general diffusion Eq. (13.12) becomes . . . (13.13) where the second term v · cA contributes towards the convective mass transfer term, and RA is the rate of reaction. The molar diffusion equation can be written as . . . (13.14) The simplification of Eq. (13.14) for the binary system results in: (a) When tDAB is constant: . . . (13.15) (b) When the velocity is zero: . . . (13.16) In this chapter, the general equations of diffusion have been developed. The simplifications can be done for particular situations of mass transfer.

14 CONCENTRATION DISTRIBUTION IN TURBULENT FLOW In Chapter 4, we discussed momentum transfer in turbulent flow. Similarly, in Chapter 9, we discussed heat transfer for turbulent flow conditions. It is not required to repeat the physical mass transfer phenomenon taking place in turbulent flow conditions. Instead, by analogy, we can explain the concentration distribution in turbulent flow. In Chapter 12, the equations of mass transfer or diffusion have been derived by considering the shell mass balances. Then, the Fick’s law of diffusion has been incorporated. By this method, one can obtain the concentration distribution. In arriving at the concentration equation, we assumed the laminar flow conditions, i.e. no turbulence. Also, we assumed the steady-state conditions. In this chapter, we will understand the concentration distribution in turbulent flow conditions by timesmoothed concentration, 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.

14.1 TIME-SMOOTHED CONCENTRATION IN TURBULENT FLOW In Chapters 4 and 9, we discussed the time-smoothed velocity and time-smoothed temperature in turbulent flow. Let us consider here the fluctuations of concentration in turbulent flow. These fluctuations take place in all directions and in all species. Let us consider the concentration of only one component and in only one direction for understanding the problem. The instantaneous concentration of A and time are shown in Figure 14.1.

Figure 14.1 Concentration fluctuations in turbulent flow condition.

Let cA = molar concentration of A in the turbulent stream cA = concentration fluctuation of A = time-smoothed average concentration of A Then cA =

+ cA . . . (14.1)

The time-smoothed concentration of A can be measured experimentally by taking fluid samples at various points and at various times. The will vary slightly with position in the turbulent core region, whereas it varies to a large extent near the solid surface. By analogy, as done for the velocity fluctuations, the properties of the concentration fluctuations will be similar in nature to those of velocity fluctuations, i.e. . . . (14.2) . . . (14.3) . . . (14.4) . . . (14.5) 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. 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):

. . . (14.6) where k1 is the rate constant for the first-order reaction. Now replacing cA with

, vx with

, etc., we obtain

. . . (14.7) When we compare this equation with the general equation of concentration, it appears to be the same, the only difference being that cA is replaced by which is the time-smoothed concentration. It may also be noted that the reaction terms may differ for the higher-order reactions. 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. 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. Similar is the situation for the temperature distribution equations. So, we can write the fluxes for the turbulent flow conditions as follows: Turbulent diffusion molar flux: Turbulent momentum flux: Turbulent heat flux:

14.2 BOUNDARY-LAYER THICKNESS FOR MASS TRANSFER Earlier in Chapters 4 and 9, we discussed the boundary-layer thickness for flow and heat transfer conditions. Similar concepts are applicable for the mass transfer too. We used the Blasius solution in the case of hydrodynamic boundary layer and thermal boundary layer. In an analogous manner, we can use the Blasius solution for convective mass transfer for the laminar flow over a flat plate. Let us consider a flat plate over which the fluid of constant concentration cA is entering as shown in Figure 14.2. Here, again we consider thex–y coordinate only. The following assumptions are made: 1. Laminar flow conditions exist near the surface of the flat plate, i.e. the Fick’s law of diffusion is applicable. 2. Steady-state conditions prevail. 3. No chemical reaction takes place, i.e. RA = 0. 4. No diffusion takes place in the x- and z-directions. 5. No flow occurs in the z-direction i.e. vz = 0.

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.

Figure 14.3 Eddy concentration in the y-direction.

Let cA be the deviation of concentration of A from the mean concentration and let concentration of A. The fluctuation of concentration can be written as: . . . (14.34)

be the mean

Dividing both sides of Eq. (14.34) by L and taking the limit when L 0, we get . . . (14.35) The rate of mass transferred per unit area (mass flux) is velocity

. The rate of mass transfer of A at the

for a distance L in the y-direction is given as: . . . (14.36)

Combining Eqs. (14.35) and (14.36), we get . . . (14.37) Earlier in momentum transfer in Chapter 4, we arrived at Eq. (14.38): . . . (14.38) Substituting the value of

from Eq. (14.38) in Eq. (14.37), we get . . . (14.39)

The term is given as:

is called eddy mass diffusivity, fm. For the molar mass diffusion, the Fick’s law

. . . (14.40) Combining Eqs. (14.39) and (14.40), we obtain the total mass flux as: . . . (14.41) Similar equations have been derived earlier for the momentum and heat transfer. The fluxes are given as: Momentum transfer: . . . (14.42) Heat transfer: . . . (14.43) where ft = eddy momentum diffusivity at = eddy thermal diffusivity. In this chapter we have explained the mechanism of mass diffusion by three models: time-smoothed concentration, boundary-layer thickness, and Prandtl mixing length.

15 UNSTEADY-STATE EVAPORATIONOF A LIQUID In Chapter 12, the steady-state diffusion problems have been solved. In such problems, the ordinary differential equations have been formed by shell mass balances. These equations have been solved with the help of boundary conditions to give the concentration profiles. Many diffusion problems can be solved by looking into the solutions to the analogous of heat and momentum transfer. But we come across binary or multi-component diffusion problems under unsteady-state conditions. In this chapter, we shall try to understand how to solve simple diffusion problems under unsteady-state conditions. As the chemical reactions take place along with diffusion, these problems become more and more difficult. These problems can be solved with the help of computers and analytical mathematical tools. Let us take a case, where a liquid A (volatile) evaporates into pure B in a tube of infinite length. Somehow, the liquid level is maintained at z = 0. The following assumptions are made: 1. 2. 3. 4. 5.

Temperature and pressure are constant. Vapours of A and B form an ideal gas mixture. The molar densities, c and DAB, are constant. B is insoluble in A. The molar average velocity in the gas phase does not depend on the radial coordinates.

With the above assumptions, we can write the equation of continuity as . . . . (15.1) where xA0 = interfacial gas-phase concentration. This equation can be solved with the help of boundary conditions: Initial condition: At t = 0, xA = 0. . . . (15.2) Boundary condition 1: At z = 0, xA = xA0. . . . (15.3) Boundary condition 2: At z = , xA = 0. . . . (15.4) As we have already solved similar equations for momentum and heat transfer, the same method of combination of variables will be used to solve Eq. (15.1). Let us define the dimensionless terms:

. . . . (15.5) . . . . (15.6) Then, Eq. (15.1) will become . . . . (15.7)

where

. . . . (15.8)

. . . . (15.9) The initial and boundary conditions become Initial condition: At t = 0, X = 0. . . . (15.10) Boundary condition 1: At z = 0, X = 1. . . . (15.11) Boundary condition 2: At z = , X = 0. . . . (15.12) In Eq. (15.7), we can introduce which will give the first-order differential equation as: . . . . (15.13) On integration, we get . . . . (15.14) Combining Eqs. (15.11) and (15.12), we get . . . . (15.15) Now, with the definition of error functions and their properties, we can write the solution as:

. . . . (15.16)

Now substituting the value of z (xA0) from Eq. (15.8), we obtain . . . . (15.17) It is easier to solve for xA0 as a function of z, therefore, we get . . . . (15.18) For different values of xA0 and z, the concentration profile can be evaluated and drawn. Unsteady-state diffusion situations are more difficult to formulate. 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.

Section D

ANALOGIES AMONG MOMENTUM, HEATAND MASS TRANSFER

16 ANALOGIES AMONGMOMENTUM, HEATAND MASS TRANSFER What is an analogy? Resemblance and similarity are closely related to the term analogy. Hence an analogy is an inference from one particular process to another where the conclusion is general. The analogies are useful tools to understand the concept of transfer phenomena, and to the professional as a sound means to predict the behaviour of the systems for which limited quantitative data are available. Analogies will be used to elucidate the mechanism of transfer. Here, we are considering the transfer of momentum, heat and mass. In all the cases, the analogy will be drawn in terms of mechanism and mathematical expression. All the three of the molecular transport processes —momentum, heat and mass—are characterized by the same general type of equation as already discussed, i.e.

In laminar flow conditions, it is easier to understand the mechanism of all these transfer processes. Also, the mathematics involved is simple in nature. But the situation becomes complex under turbulent flow conditions. In such cases, empirical correlations have been developed by scientists and technologists. First, with the development of flow meters like orifice meter and Pitot tube, the flow behaviour of momentum transfer has been easy to understand. Mathematically too, simple equations explained the momentum transfer under laminar and turbulent flow conditions. Correlations were developed for friction factor and Reynolds number. Then, temperature-measuring devices were developed to understand the mechanism of heat transfer. Analogies were developed for momentum and heat transfer. A typical example is the Reynolds analogy and the Prandtl analogy for the case of momentum and heat transfer. In the later period, modern instrumentation for the measurement of concentration of species was developed. This helped understand mass transfer operations. Analogies were developed among momentum, heat and mass transfer. Applications of Analogies 1. Analogies help us understand the new processes in momentum, heat and mass transfer. 2. If we learn some phenomena in momentum transfer, by analogy, it can also be understood for heat transfer and mass transfer. 3. The concept of friction factor in momentum transfer can lead us to evaluate the heat transfer coefficient by applying the Chilton–Colburn analogy. It can help us calculate the heat transfer

area in the design process of a heat exchanger. 4. The Chilton–Colburn analogy can be extended to the mass transfer and momentum transfer. In this case, the mass transfer coefficient can be evaluated from the friction factor in momentum transfer. In this chapter, Section 16.1 deals with the general analogy concepts among momentum, heat and mass transfer. In this part, the mechanism and mathematical concepts are explained for laminar and turbulent conditions. Section 16.2 discusses the Reynolds analogy and its limitations. The Prandtl analogy is dealt with in Section 16.3. The von Kármán analogy for the turbulent flow condition is explained in Section 16.4. The Chilton–Colburn analogy among momentum, heat and mass transfer is discussed in Section 16.5.

16.1 ANALOGY AMONG MOMENTUM, HEAT AND MASS TRANSFER The momentum, heat and mass transfer processes take place almost simultaneously with a general similar behaviour. There are many common similarities between these different processes. A close relationship exists among the three transfer phenomena for laminar and turbulent regimes. The common features are in terms of mechanism and mathematical equations for momentum, heat and mass transfer operations. Analogies will be drawn among the transfer-rate correlations for laminar and turbulent flow conditions. 1. Laminar Flow (a) Mechanism of molecular transport: In laminar flow conditions, momentum, heat and mass transport take place by the molecular transport mechanism. As discussed in Chapter 1, molecules of the fluid are stationary. Individual molecules containing the property are transferred, e.g. momentum, heat or mass. Another question which is generally asked, why does the transfer of momentum, heat and mass take place? There is a flux of momentum, heat and mass in all the directions. So there is a gradient in which the transfer takes place. For momentum transfer, there is a velocity gradient. The flow takes place from a higher velocity to a lower velocity. Heat energy flows from a higher temperature to a lower temperature, whereas mass transfer takes place from a higher concentration to a lower concentration. (b) Laws: Momentum transfer: Newton’s law of viscosity is: . . . (16.1) For constant density: . . . (16.2) Heat transfer: Fourier’s law of heat conduction is:

. . . (16.3) For constant density and thermal conductivity:

or

. . . (16.4)

Mass transfer: Fick’s law of diffusion is: . . . (16.5) Now let us look at Eqs. (16.2), (16.4) and (16.5). L.H.S.: flux Momentum transfer: Momentum flux tyx (shear stress), force/area. Heat transfer: ·Heat flux q/A, in J/s · m2 Mass transfer: Mass flux J*Ax, in R.H.S.: driving force Momentum transfer: Momentum concentration tv, in (kg · m/s)/m3 Heat transfer: Concentration of thermal energy tcpT, in J/m3 Mass transfer: Concentration of A, i.e. cA, in kg mol of A/m3 (c) Constants: Momentum transfer: Momentum diffusivity n/t, in m2/s Heat transfer: Thermal diffusivity k/tcp, in m2/s Mass transfer: Molecular diffusivity of A in B, DAB, in m2/s We conclude that all the three molecular transport processes—momentum, heat and mass—are characterized by the same general type of equation, mathematically given by . . . (16.6) Equation (16.6) for molecular diffusion of the property momentum, heat or mass, can be written as . . . (16.7)

where zx = flux in the x-direction. = gradient in the x-direction d = constant. Mathematically, these equations and notations are similar, but the process of momentum, heat and mass may be different. 2. Convective Transfer This mechanism of transport has been explained in Chapter 1. Each molecule carries the property of momentum, heat and mass and transports it. 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. Turbulent Flow We have already discussed turbulent flow conditions in Chapters 4, 9 and 14. The flux equations are written using the turbulent eddy momentum diffusivity ft, the turbulent eddy thermal diffusivity at and the turbulent eddy mass diffusivity fm. In all these equations, we have used the time-smoothed average velocity for momentum transfer, time-smoothed average temperature for heat transfer and time-smoothed average concentration for mass transfer. However, these similarities are not well defined physically or mathematically and are more difficult to correlate with each other. These equations can be written as: Momentum transfer: . . . (16.8) Heat transfer: . . . (16.9) Mass transfer: . . . (16.10) where

ft = momentum eddy diffusivity, in m2/s at = thermal eddy diffusivity, in m2/s fm = mass eddy diffusivity, in m2/s = time-smoothed average velocity = time-smoothed average temperature = time-smoothed average concentration Combining the Eqs. (16.8) to (16.10) with the laminar flow conditions, we can write the general equations for momentum, heat and mass transfer: Momentum transfer: . . . (16.11) Heat transfer: . . . (16.12) Mass transfer: . . . (16.13) 4. Boundary-Layer Thickness The boundary-layer thickness for momentum, heat and mass transfer has been discussed in Chapters 4, 9 and 14 respectively. The fluid behaviour near the solid surface is different than that in the bulk fluid. Blausius developed the theory for the momentum transfer, and the boundary-layer thickness d is given as: . . . (16.14) i.e. 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. This means that the transfer of momentum and heat are directly analogous. The Nusselt number is given as NNu,x = 0.332(NRe,x)1/2 . . . (16.15)

where

hx = local heat transfer coefficient at point x. . . . (16.16) We conclude that the local heat transfer coefficient is directly proportional to the square root of distance. Similar notions hold good for the mass transfer case too. For mass transfer, the Sherwood number is given as . . . (16.17) where NSh,x = Sherwood number = where kc = mass transfer coefficient at a point x. . . . (16.18) We conclude that the local mass transfer coefficient is directly proportional to the square root of distance. As seen above, we can say that transfer of momentum, heat and mass are directly analogous. 5. Prandtl Mixing Length Prandtl mixing length models for momentum, heat and mass transfer have been discussed in Chapter 4, 9 and 14 respectively. In turbulent flow conditions it is assumed that eddies move a little distance called the Prandtl mixing length before losing their identity. These eddies mix afterwards with the mainstream of the fluid. Similar phenomena occur in heat and mass transfer. These equations can be written as: Momentum transfer: . . . (16.19) where = time-smoothed average velocity in the x-direction L = Prandtl mixing length = momentum flux for turbulent flow Heat transfer: . . . (16.20) where

= time-smoothed average temperature cp = specific heat capacity qy = heat flux in the y-direction. Mass transfer: . . . (16.21) where = mass transfer flux = time-smoothed concentration of A. Equations (16.19), (16.20) and (16.21) for momentum, heat and mass transfer, respectively, are similar in nature. 6. Dimensionless Groups The dimensionless groups in momentum, heat and mass transfer are very much helpful in understanding these processes. Also, many empirical correlations can be developed between these processes by the dimensionless groups. Momentum transfer: Reynolds number =

=

(for circular pipes)

Froud number =

=

Weber number = NWe = where v = interfacial tension. Heat 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.

16.2 REYNOLDS ANALOGY The processes of molecular momentum transfer, heat transfer and mass transfer are very well understood. Mathematically too, these processes are easy to formulate. The molecular diffusion equations of Newton for momentum, Fourier for heat, and Fick for mass transfer are very similar.

These have been discussed in Chapters 1, 6 and 11 respectively. There are also similarities in turbulent transport. The flux equations for momentum, heat and mass transfer in turbulent flow conditions have been discussed in Chapters 4, 9, and 14 respectively. In turbulent flow, the fluxes are written using the turbulent eddy momentum diffusivity ft, the turbulent eddy thermal diffusivity at, and the turbulent eddy mass diffusivity fm. However, the similarities in turbulent flow conditions are more difficult to correlate mathematically or physically. All efforts have been made in the literature to develop the analogies between momentum, heat and mass under turbulent flow conditions. Reynolds was the first to develop the similarity between heat and momentum transfer for turbulent flow conditions. Let us first discuss the Reynolds analogy between momentum and heat transfer and then that between momentum and mass transfer. 1. Momentum and Heat Transfer For the molecular momentum transfer, the Newton’s law can be written as

. . . (16.22)

where = n = molecular momentum diffusivity, in m2/s t = constant Similarly, the momentum flux for the turbulent flow is written as

or

. . . (16.23)

where = turbulent momentum flux nt = eddy viscosity, property of flow conditions and not the fluid property = time-smoothed average velocity in the x-direction ft = momentum eddy diffusivity, in m2/s For constant density, combining Eqs. (16.22) and (16.23), we obtain . . . (16.24) Similar equations can also be written for heat transfer. For molecular heat transfer, the Fourier’s law of heat conduction can be written as:

. . . (16.25) For constant density and specific heat, we can write

. . . (16.26)

where a = thermal diffusivity, in m2/s = For turbulent flow conditions: . . . (16.27) where at = turbulent thermal diffusivity, in m2/s = time-smoothed average temperature. For constant t and cp, combining Eqs. (16.26) and (16.27), we get . . . (16.28) The following assumptions are made: 1. In turbulent flow, the molecular momentum and heat transfer are negligible, i.e. n/t and a are small. 2. at = ft, i.e. momentum eddy diffusivity and thermal eddy diffusivity are equal. Now, dividing Eq. (16.24) by Eq. (16.28) and simplifying with normal notations for temperature T and velocity v, we get . . . (16.29) Let us consider a circular pipe through which the fluid is flowing. The heat flux q/A is analogous to the momentum flux t which is considered constant in the turbulent flow condition. Let T = bulk temperature of the fluid Ti = wall temperature of the pipe

ts = shear stress at the wall vav = average fluid bulk velocity. Integrating Eq. (16.29) between the limits, where the velocity at the wall is zero, we get . . . (16.30)

or

. . . (16.31)

From the heat transfer coefficient h, the heat transfer is given by the Newton’s law: = h(T – Ti) . . . (16.32) From the fluid flow, we can evaluate the shear stress at the wall from the Fanning friction factor concepts, i.e. . . . (16.33)

i.e.

or

. . . (16.34)

. . . (16.35)

Substituting the value of q/A from Eq. (16.32) and that of ts from Eq. (16.35) in Eq. (16.31), we get . . . (16.36)

where G = mass velocity, in kg/s. We can write the R.H.S. of Eq. (16.36) as

. . . (16.37)

where

= Stanton number, NSt = Nusselt number, NNu = Reynolds number, NRe = Prandtl number, NPr

. . . (16.38) The Reynolds anology assumes that Eq. (16.36) holds good only when the Prandtl number is equal to 1.0. This is the limitation of Reynolds analogy. In literature for circular tubes, the Prandtl number varies from 0.7 to 0.9 and for other conditions, it varies from 0.5 to 1.0. 2. Momentum and Mass Transfer For mass transfer, similar equations can be written as done above for molecular mass diffusion and turbulent mass diffusion by using the Fick’s law. As we have defined the heat transfer coefficient by Eq. (16.32), similarly, we can write the mass transfer coefficient as . . . (16.39) where cA = concentration of A, in kg mol/m3. cAi = concentration of A at the interface, in kg mol/m3. kc = mass transfer coefficient, in kg mol/s · m2. conc. diff. And the Fick’s law is written as: . . . (16.40) By combining Eqs. (16.24), (16.39) and (16.40), integrating between the limits as done for heat transfer and assuming the Schmidt number to be 1.0, i.e. NSc = Schmidt number = we obtain . . . (16.41)

Now, combining Eqs. (16.36) and (16.41), we get the Reynolds analogy between momentum, heat and mass transfer. . . . (16.42)

16.3 PRANDTL ANALOGY When a flowing fluid comes in contact with the solid surface, its velocity is zero at the surface. For example, if the fluid is flowing in a circular pipe, the velocity of the fluid at the wall is zero and at the centre of the pipe, it is maximum. Near the wall, the velocity is very small and as we go away from the wall, the velocity of the fluid increases. So, the fluid characteristics of flow vary in the vicinity of the solid surface. It is postulated that there are four zones of the fluid behaviour as shown in Figure 16.1. In the viscous sublayer, Zone I, very close to the solid surface, there is no eddy formation and no turbulence and the velocity of the fluid is very small. Hence, all the laws of the laminar momentum transfer are applicable to this viscous sublayer, i.e. the Newton’s law of viscosity is applicable. In 1910, Prandtl modified the Reynolds analogy by considering the velocity distribution in the laminary sublayer and the turbulent core region. The Prandtl analogy considered momentum and heat transfer.

Figure 16.1 Four zones near the solid surface of the fluid for turbulent flow.Zone I: viscous sublayer; Zone II: buffer sublayer; Zone III: inertial sublayer; Zone IV: turbulent core zone.

Let us consider the transport of momentum and heat in a circular pipe under the turbulent conditions. There are two layers of movement of momentum and heat: 1. Laminar sublayer (near to the wall) 2. Turbulent core region (centre of the pipe) In the laminar sublayer, all the equations and laws of momentum and heat transfer for laminar flow conditions pervail. There is no mixing and turbulence in this sublayer.

The following assumptions are made: 1. Eddy transport is negligible in the laminar sublayer; ft = 0 and at = 0, i.e. momentum eddy diffusivity and thermal eddy diffusivity are zero. 2. Steady-state conditions prevail. 3. Temperature and velocity profiles in the laminar sublayer are linear. 4. The thermal diffusivity is equal to the momentum diffusivity in the turbulent core region. 5. The molecular momentum diffusivity and the thermal diffusivity are negligible in the turbulent core region. Let us consider the flow and heat transfer in a circular pipe. Tw is the wall temperature. The velocity and temperature profiles are shown in Figure 16.2. 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.

Figure 16.2 Temperature and velocity profiles in the laminar sublayer in a circular pipe.

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:

. . . (16.43) . . . (16.44) Heat transfer: . . . (16.45) . . . (16.46) For turbulent flow conditions, these equations are modified to include eddy terms: Momentum transfer: . . . (16.47) Heat transfer: . . . (16.48) where ft = turbulent eddy momentum diffusivity, in m2/s at = thermal eddy diffusivity, in m2/s. Substituting the values of ft and at for the laminar sublayer, i.e. ft = 0,at = 0 in Eqs. (16.47) and (16.48) and integrating over the limit d for r, we obtain . . . (16.49) . . . (16.50) 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. Eliminating d from Eqs. (16.49) and (16.50), we get . . . (16.51)

Now consider momentum and heat transfer for the turbulent core region beyond r = d. Let R be the radius of the pipe. Neglecting the molecular momentum and molecular heat transfer terms, Eq. (16.47) and (16.48) become . . . (16.52) . . . (16.53) Let us integrate Eqs. (16.52) and (16.53) and introduce the following concepts. The shear stress at the wall, tw, is given by . . . (16.54) and the heat flux at the wall, qw, is given by the Newton’s law of cooling . . . (16.55) where f = Fanning friction factor h = heat transfer coefficient Tb = bulk fluid temperature Also, v = vm – vd . . . (16.56) where vm = mean velocity of the fluid. Here, the velocity gradient is (vm – vd) and the temperature gradient is(Td – Tb). It is assumed that the momentum eddy diffusivity ft is equal to the thermal eddy diffusivity at, i.e. ft = at . . . (16.57) Substituting these terms in Eqs. (16.54) and (16.55), we get

or

. . . (16.58)

. . . (16.59) Now, we get . . . (16.60) Eliminating Td from Eqs. (16.51) and (16.60), we obtain

. . . (16.61) Now, vd can be expressed in terms of tw whereas vm can be expressed by using the universal velocity in the laminar sublayer. In the laminar sublayer, we define the dimensionless velocity vd+ and dimensionless distance y+ as . . . (16.62) which is applicable for y+ = 5. So,

. . . (16.63)

Substituting the value of vd from Eq. (16.63) in Eq. (16.61) and using the following values

(Prandtl number) we get . . . (16.64)

or

. . . (16.65)

This relationship is called the Prandtl analogy. Also it gives a relationship between the Stanton number and the Fanning friction factor as . . . (16.66) where

16.4 VON KÁRMÁN ANALOGY

Under the turbulent flow conditions, the fluid behaviour is very different than that under the laminar flow conditions. In the laminar flow, the situation is simple and the mechanism is easily understood. There are many empirical expressions developed in the literature for turbulent flow conditions. Let us analyse the flow near a flat surface as shown in Figure 16.3.

Figure 16.3 Flow regions for turbulent flow near a flat surface.

It is convenient to describe the flow of fluid near the solid surface by four zones. The zones are small and close to the solid surface (but for clarity these are shown bigger). Zone 1: The viscous sublayer: This zone is very close to the wall in which the viscosity of the fluid plays an important role. Zone 2: The buffer sublayer: This is the transition zone which occurs between the viscous and inertial sublayers. Zone 3: The inertial sublayer: In this zone, the turbulence just starts and viscosity plays a minor role. Zone 4: The main turbulent stream: In this zone, the time-smoothed velocity distribution is important and viscosity is negligible. In the Prandtl analogy, the viscous sublayer and the turbulent bulk zone have been considered. Von Kármán modified the Prandtl analogy by considering the buffer sublayer as well. Von Kármán developed the analogy based upon the universal velocities for a smooth circular pipe. As discussed above, the universal velocities are: Zone 1: Viscous sublayer . . . (16.67) Zone 2: Buffer sublayer . . . (16.68) Zone 3: Turbulent core region . . . (16.69) where v+ is the dimensionless velocity and y+ the dimensionless distance.

Now,

. . . (16.70)

. . . (16.71)

or

. . . (16.72)

. . . (16.73) where n = kinematic viscosity, n/t t0 = shear stress at the wall. Equations (16.67) to (16.69) for these zones are shown in Figure 16.4.

Figure 16.4 Universal velocity profile for turbulent flow in circular pipe.

Von Kármán included buffer zone too, along with viscous sublayer and turbulent core zone. The Nusselt number is given by the following relation, in terms of friction factor:

. . . (16.74)

where NRe = Reynolds number NPr = Prandtl number. This is called von Kármán analogy. There are several other analogies given in the literature.

16.5 CHILTON–COLBURN ANALOGY In fluid flow or momentum transfer, we learnt the close relationship between the Fanning friction factor f and the Reynolds number. In laminar flow conditions, the relationship is given by (for NRe 2100) . . . (16.75) whereas for turbulent conditions, various empirical correlations are given by different scientists, e.g. (for 2.1 103 < NRe < 105) . . . (16.76) Apart from these correlations, the friction factor charts are available between the friction factor and the Reynolds number. In heat transfer, the heat transfer coefficient is correlated by the Nusselt number, the Reynolds number and the Prandtl number. Some of the relations are as follows: For laminar flow: . . . (16.77)

For turbulent flow: . . . (16.78) for Reynolds number: 104 < NRe < 105 Prandtl number: 0.6 < NPr < 100. 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.

Apart from correlations, these charts are also drawn between the jH factor and the Reynolds number as shown in Figure 16.5.

Figure 16.5 Reynolds number vs j H factor.

Here,

. . . (16.79)

Chilton and Colburn observed that there is a relationship between the heat transfer coefficient and the friction factor. This analogy is based upon experimental data for gases and liquids for laminar flow conditions and turbulent flow. Initially, this analogy was observed between momentum and heat transfer. Later on, it was extended to mass transfer as well. Accordingly, for highly turbulent flow, NRe > 10,000: . . . (16.80) Mass transfer and momentum transfer: For mass transfer, the jD factor is defined as . . . (16.81) where

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. The empirical correlation for the Sherwood number is given as NSh = 0.332(NRe)1/2(NSc)1/3 . . . (16.82) For flow past a flat plate or in a pipe, where there is no form of the drag present, the Chilton–Colburn analogy is given as: . . . (16.83) This analogy is valid for turbulent flow when NRe > 10,000 and 0.6 < NPr < 100 and tube L/D > 60. Applications: 1. Chilton–Colburn analogy is useful to obtain the heat transfer coefficient for an unknown system where the hydrodynamic situation is understood. 2. It is used for the flow over a flat plate or the flow in a pipe. 3. For a situation where the drag is present (e.g. packed bed) or flow past blunt objects, f / 2 is greater than jH or jD. 4. The mass transfer coefficient cannot be obtained from the analogous to heat transfer coefficient correlations. For the mass transfer, different boundary conditions pervail.

SOLVED EXAMPLES EXAMPLE 16.1 An oil is manufactured by the vapour phase catalytic reaction. The reaction gas mixture leaving the catalytic reactor in the plant is condensed in a shell-and-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 (Birmingham Wire Gauge). 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 heat transfer coefficient by the Reynolds analogy. Data: Properties of water: Density, t = 995 kg/m3 Viscosity, n = 7.65 10–4 kg/m· s Thermal conductivity, k = 0.623 W/m· C

Specific heat, cp = 4.17 kJ/kg · C The Fanning friction fartor f can be calculated by the equation, f = 0.0014 + Solution: Outside diameter, Do = 25 mm = 0.025 m Inside diameter, Di = 21.2 m = 0.0212 m L (tube) = 3 m Ts = 80°C t = 995 kg/m3 n = 7.65 10–4 kg/m· s k = 0.623 W/m· C cp = 4.17 kJ/kg · C Area A = Volumetric flow rate, Q = 0.057 m3/min = 9.5 10–4 m3/s

Velocity,

By the Reynolds analogy, the heat transfer coefficient h can be calculated as: . . . (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

APPENDIX A

CONVERSION FACTORSAND FUNDAMENTAL UNITS A.1 VOLUME AND DENSITY 1 g mol ideal gas at 0°C, 760 mm Hg = 22.4140 litres = 22.414 cm3 1 lb mol ideal gas at 0°C, 760 mm Hg = 359.05 ft3 1 kg mol ideal gas at 0°C, 760 mm Hg = 22.414 m3 Density of dry air at 0°C, 760 mm Hg = 1.2929 g/litre = 0.080711 lbm/ft3 Molecular weight of air = 28.97 lbm/lb mol = 28.97 g/g mol 1 g/cm3 = 62.43 lbm/ft3 = 1000 kg/m3 1 g/cm3 = 8.345 lbm/U.S. gal 1 lbm/ft3 = 16.0185 kg/m3

A.2 LENGTH 1 in. = 2.540 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 = 3.2808 ft = 39.37 in.

A.3 MASS 1 lbm = 453.59 g = 0.45359 kg 1 lbm = 16 oz = 7000 grains 1 kg = 1000 g = 2.2046 lbm 1 ton (short) = 2000 lbm 1 ton (long) = 2240 lbm 1 ton (metric) = 1000 kg

A.4 STANDARD ACCELERATION OF GRAVITY

g = 9.80665 m/s2 g = 980.665 cm/s2 g = 32.174 ft/s2 gc (gravitational conversion factor) = 32.1740 lbm ft/lbf s2 = 980.665 gm cm/gf s2

A.5 VOLUME 1 L (litre) = 1000 cm3 1 in.3 = 16.387 cm3

1 m3 = 1000 L (litre)

1 ft3 = 28.317 L (litre) 1 ft3 = 0.028317 m3

1 U.S. gal = 3.7854 L (litre)

1 ft3 = 7.481 U.S. gal 1 m3 = 264.17 U.S. gal

1 British gal = 1.20094 U.S. gal

1 U.S. gal = 4 qt

1 U.S. gal = 3785.4 cm3 1 m3 = 35.313 ft3

A.6 FORCE 1 g cm/s2 (dyne) = 10–5 kg m/s2 = 10–5 N (newton) 1 g cm/s2 = 7.2330 10–5 lbm ft/s2 (poundal) 1 kg m/s2 = 1 N (newton) 1 lbf = 4.4482 N 1 g cm/s2 = 2.2481 10–6 lbf 1 dyne = 2.2481 10–6 lbf

A.7 PRESSURE 1 bar = 1 105 Pa (pascal) = 1 105 N/m2 1 psia = 1 lbf/in.2 1 psia = 2.0360 in. Hg at 0°C 1 psia = 2.311 ft H2O at 70°F 1 psia = 51.715 mm Hg at 0°C (tHg = 13.5955 g/cm3) 1 atm = 14.696 psia = 1.01325 105 N/m2 = 1.01325 bar 1 atm = 760 mm Hg at 0°C = 1.01325 105 Pa = 1.01325 102 kPa 1 atm = 29.921 in. Hg at 0°C 1 atm = 33.90 ft H2O at 4°C 1 psia = 6.89476 104 g/cm s2

1 psia = 6.89476 104 dyne/cm2 1 dyne/cm2 = 2.0886 10–3 lbf/ft2 1 psia = 6.89476 103 N/m2 = 6.89476 103 Pa 1 lbf/ft2 = 4.7880 102 dyn/cm2 = 47.880 N/m2 1 mm Hg (0°C) = 1.333224 102 N/m2 = 0.1333224 kPa

A.8 POWER 1 hp = 0.74570 kW 1 hp = 550 ft lbf/s

1 watt (W) = 14.340 cal/min

1 hp = 0.7068 btu/s

1 J/s (joule/s) = 1 W

1 btu/h = 0.29307 W (watt)

A.9 HEAT, ENERGY, 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 = 1.05506 kJ 1 btu = 252.16 cal (thermochemical) 1 kcal (thermochemical) = 1000 cal = 4.1840 kJ 1 cal (thermochemical) = 4.1840 J 1 cal (IT) = 4.1868 J 1 btu = 251.996 cal (IT) 1 btu = 778.17 ft lbf 1 hp h = 0.7457 kW h 1 hp h = 2544.5 btu 1 ft lbf = 1.35582 J 1 ft lbf/lbm = 2.9890 J/kg

A.10 THERMAL CONDUCTIVITY 1 btu/h ft °F = 4.1365 10–3 cal/s cm °C 1 btu/h ft °F = 1.73073 W/m K

A.11 HEAT TRANSFER COEFFICIENT 1 btu/h ft2 °F = 1.3571 10–4 cal/s cm2 °C 1 btu/h ft2 °F = 5.6783 10–4 W/cm2 °C 1 btu/h ft2 °F = 5.6783 W/m2 K 1 kcal/h m2 °F = 0.2048 btu/h ft2 °F

A.12 VISCOSITY

1 cP = 10–2 g/cm s (poise) 1 cP = 2.4191 lbm/ft h 1 cP = 6.7197 10–4 lbm/ft s 1 cP = 10–3 Pas = 10–3 kg/m s = 10–3 N s/m2 1 cP = 2.0886 10–5 lbf s/ft2 1 Pa s = 1 N s/m2 = 1 kg/m s = 1000 cP = 0.67197 lbm/ft s

A.13 DIFFUSIVITY 1 cm2/s = 3.875 ft2/h 1 cm2/s = 10–4 m2/s

1 m2/s = 3.875 104 ft2/h 1 centistoke = 10–2 cm2/s

1 m2/h = 10.764 ft2/h

A.14 MASS FLUX AND MOLAR FLUX 1 g/s cm2 = 7.3734 103 lbm/h ft2 1 g mol/s cm2 = 7.3734 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 = 1.3562 10–3 kg mol/s m2

A.15 HEAT FLUX AND HEAT FLOW 1 btu/h ft2 = 3.1546 W/m2 1 btu/h = 0.29307 W 1 cal/h = 1.1622 10–3 W

A.16 HEAT CAPACITY AND ENTHALPY 1 btu/lbm °F = 4.1868 kJ/kg K 1 btu/lbm °F = 1.000 cal/g °C 1 btu/lbm = 2326.0 J/kg 1 ft lbf/lbm = 2.9890 J/kg 1 cal (IT)/g °C = 4.1868 kJ/kg K 1 kcal/g mol = 4.1840 103 kJ/kg mol

A.17 MASS TRANSFER COEFFICIENT 1 kc cm/s = 10–2 m/s 1 kc ft/h = 8.4668 10–5 m/s

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 = 1.3562 10–3 kg mol/s m3 mol frac 1 kx a lb mol/h ft3 mol frac = 4.449 10–3 kg mol/s m3 mol frac 1 kG kg mol/s m2 atm = 0.98692 10–5 kg mol/s m2 Pa 1 kG a kg mol/s m3 atm = 0.98692 10–5 kg mol/s m3 Pa

A.18 TEMPERATURE 0°C = 32°F (freezing point of water) 1.0 K = 1.0°C = 1.8°F = 1.8°R (Rankine) (as per scale) °F = 32 + 1.8 (°C) °C = (1/1.8)(°F – 32) °R = °F + 459.67 K = °C + 273.15 100°C = 212°F = 373.15 K = 671.67°R 0°C = 32°F = 273.15 K = 491.67°R –273.15°C = –459.67°F = 0 K = 0°R (absolute zero)

APPENDIX B

GAS LAW CONSTANT R Gas Law Constant R Numerical Value

Units

1.9872

g cal / g mol K

1.9872

btu / lb mol °R

82.057

3 cm atm / g mol K

8314.34

J / kg mol K

–3 82.057 10

3 m atm / kg mol K

8314.34

2 2 kg m /s kg mol K

10.731

3 2 ft lb f / in. lb mol °R

0.7302

3 ft atm / lb mol °R

1545.3

ft lb f / lb mol °R

8314.34

3 m Pa / kg mol K

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

333

0.151

273

0.185

341

0.163

303

0.147

333

0.144

293

0.175

100%

293

0.182

60%

293

0.305

20%

293

0.486

100%

323

0.151

Ethylene glycol

273

0.265

Glycerol, 100%

293

0.284

n-Hexane

303

0.138

333

0.135

293

0.149

348

0.140

100%

293

0.215

60%

293

0.329

20%

293

0.492

100%

323

0.197

n-Octane

303

0.144

333

0.140

25%

303

0.571

12.5%

303

0.589

90%

303

0.364

60%

303

0.433

Vaseline

332

0.183

Carbon tetrachloride n-Decane Ethyl acetate Ethyl alcohol

Kerosene Methyl alcohol

NaCl brine

Sulfuric acid

APPENDIX E

Properties of Gases

Table E.3 Thermal Conductivities of Gases and Vapours at 101.325 kPa(1 Atm Abs); k = W/m K K

k

Acetone

273 319 373 457

0.0099 0.0130 0.0171 0.0254

Ammonia

273 373 473

0.0218 0.0332 0.0484

Butane

273 373

0.0135 0.0234

Carbon monoxide

173 273 373

0.0152 0.0232 0.0305

Chlorine

273

0.00744

Ethane

239 273 373

0.0149 0.0183 0.0303

Ethyl alcohol

293 373

0.0154 0.0215

Ethyl ether

273 319 373

0.0133 0.0171 0.0227

Ethylene

273 323 373

0.0175 0.0227 0.0279

n-Hexane

273 293

0.0125 0.0138

Sulfur dioxide

273 373

0.0087 0.0119

Gas or Vapour

Table E.4 Viscosity of Gases at 101.325 kPa (1 Atm Abs) [Viscosity in (Pa s)103, (kg/m s)103, or cP] Temperature K

°F

°C

H2

O2

N2

CO

CO2

255.4

0

–17.8

0.00800

0.0181

0.0158

0.0156

0.0128

273.2

32

0

0.00840

0.0192

0.0166

0.0165

0.0137

283.2

50

10.0

0.00862

0.0197

0.0171

0.0169

0.0141

311.0

100

37.8

0.00915

0.0213

0.0183

0.0183

0.0154

338.8

150

65.6

0.00960

0.0228

0.0196

0.0195

0.0167

366.5

200

93.3

0.0101

0.0241

0.0208

0.0208

0.0179

394.3

250

121.1

0.0106

0.0256

0.0220

0.0220

0.0191

422.1

300

148.9

0.0111

0.0267

0.0230

0.0231

0.0203

449.9

350

176.7

0.0115

0.0282

0.0240

0.0242

0.0215

477.6

400

204.4

0.0119

0.0293

0.0250

0.0251

0.0225

505.4

450

232.2

0.0124

0.0307

0.0260

0.0264

0.0236

533.2

500

260.0

0.0128

0.0315

0.0273

0.0276

0.0247

APPENDIX F

Properties of Solids Table F.1 Heat Capacities of Solids (cp = kJ/kg K) K

cp

373

0.84

1773

1.147

Solid Aluminia Asbestos

1.05

Asphalt

0.92

Brick, fireclay

373

0.829

1773

1.248

Cement, portland

0.779

Clay

0.938

Concrete

0.63

Corkboard

303

0.167

Glass

0.84

Magnesia

373

0.980

1773

0.787

Oak

2.39

Pine, yellow Porcelain

298

2.81

293–373

0.775

Rubber, vulcanized

2.01

Steel

0.50

Wool

1.361

Benzoic acid

293

1.243

Camphene

308

1.591

Caprylic acid

271

2.629

Dextrin

273

1.218

Formic acid

273

1.800

Glycerol

273

1.382

Lactose

293

1.202

Oxalic acid

323

1.612

Tartaric acid

309

1.202

Urea

293

1.340

Table F.2 Thermal Conductivities of Building and Insulating Materials (Solids) Material

3 t (kg/m )

Asbestos

577

Asbestos sheets

889

Brick, building Brick, fireclay

* t (°C)

k (W/m K) 0.151 (0°C)

51

0.166

20

0.69 1.00 (200°C)

0.168 (37.8°C)

0.190 (93.3°C)

1.47 (600°C)

1.64 (1000°C)

Clay soil, 4% H2O

1666

4.5

Concrete, 1:4 dry

0.57 0.762

Corkboard

160.2

Cotton

80.1

30

0.0433

Felt, wool

330

30

0.052

Fibre insulation board

237

21

0.048

0.055 (0°C)

Glass, window

0.068 (93.3°C)

0.0414 (37.8°C)

0.0549 (93.3°C)

0.071 (93.3°C) 0.062 (93.3°C)

0.080 (204.4°C) 0.066 (148.9°C)

0.0391 (37.8°C) 0.0395 (37.8°C)

0.0486 (93.3°C) 0.0518 (93.3°C)

0.52–1.06

Glass fibre

64.1

30

0.0310 (–6.7°C)

Ice

921

0

2.25

Magnesia, 85%

271 208

Oak, across grain

825

15

0.208

Pine, across grain

545

15

0.151

0.068 (37.8°C) 0.059 (37.8°C)

Paper Polystyrene board

0.061 (37.8°C)

0.130 16

0.040

24–40

0.023–0.026

Rock wool

192 128

0.0317 (–6.7°C) 0.0296 (–6.7°C)

Rubber, hard

1198

0

0.151

4% H2O

1826

4.5

1.51

10% H2O

1922

4.5

2.16

Sandstone

2243

40

1.83

Snow

559

0

0.47

Polyurethane sprayed foam

Sand soil

APPENDIX G

THE EQUATION OF CONTINUITY

Cartesian coordinates (x, y, z):

Cylindrical coordinates (r, q, z):

Spherical coordinates (r, q, f):

APPENDIX H

EQUATION OF MOTIONFOR A NEWTONIAN FLUIDWITH CONSTANTS t AND n

Cartesian coordinates (x, y, z):

Cylindrical coordinates (r, q, z):

Spherical coordinates (r, q, f):

Appendix H—NEWTONIAN FLUID WITH CONSTANTS t AND n

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. The assumption t = constant is given in the table heading because it is the assumption more often made. The term nv is usually negligible, except in systems with large velocity gradients.

Cartesian coordinates (x, y, z):

Cylindrical coordinates (r, q, z):

Spherical coordinates (r, q, f):

APPENDIX J

FICK’S (FIRST) LAWOF BINARY DIFFUSION* * To get the molar fluxes with respect to the molar average velocity, replace j A, t, and wA by JA*, c, and xA. [jA = –tDAB—wA] Cartesian coordinates (x, y, z):

Cylindrical coordinates (r, q, z):

Spherical coordinates (r, 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 . . . . t . . . . wa . . . . j a . . . . ra By . . . . c . . . . xa . . . . j*a, . . . .

Cartesian coordinates (x, y, z):

Cylindrical coordinates (r, q, z):

Spherical coordinates (r, q, f):

INDEX Analogy, 165 among momentum, heat and mass transfer, 166 Chilton–Colburn, 184 Prandtl, 177 Reynolds, 173 Von Kármán, 182 Boundary conditions eat transfer, 66 mass transfer, 123 momentum transfer, 9 Boundary-layer thickness heat transfer, 105 mass transfer, 152 momentum transfer, 50 Chilton–Colburn analogy, 184 Circular tube, 18 Concentration distribution, 128 Convective energy, 63 Convective mass flux, 121 Convective momentum flux, 8 Convective transport heat, 63 mass, 120 momentum, 7 Convective molar flux, 121 Cooling fin, 79 Diffusion equation, 147 through a spherical stagnant gas film, 141 through a stagnant gas film, 129 with a heterogeneous chemical reaction, 133 with a heterogeneous chemical reaction (slow reaction), 136 with a homogeneous chemical reaction, 138 Energy equation, 99 Equation of change for isothermal system, 36 of continuity, 36 of motion, 38 Falling film, 13 Fick’s law, 117 Flow of a liquid falling film, 13 of two immiscible fluids, 30 through a circular tube, 18 through an annulus, 26 Fourier’s law of heat conduction, 61

Heat conduction from a sphere to a stagnant fluid, 82 in a cooling fin, 79 with a chemical reaction heat source, 88 with a viscous heat source, 85 with electrical heat source, 69 with nuclear heat source, 72 through composite walls, 76 Isothermal system, 36 Laminar flow, 13 in a narrow slit, 22 Lewis number, 172 Mass fluxes, 121 Mechanism of momentum transfer, 4 Molecular energy transfer, 61 Molecular fluxes, 7 Molecular transfer heat, 61 mass, 117 momentum, 5 Momentum transfer, 3 in turbulent flow, 46 Navier–Stokes equations, 38, 42 Newton’s law of viscosity, 5 Nusselt number, 172 Prandtl analogy, 177 Prandtl mixing length heat transfer, 109 mass transfer, 156 momentum transfer, 53 Prandtl number, 172 Reynolds analogy, 173 Reynolds number, 172 Schmidt number, 172 Shear stress, 6 Shell balances heat, 65 mass, 122 momentum, 8 Sherwood number, 156 Temperature distribution, 68 Time-smoothed concentration, 150 Time-smoothed equation of change for turbulent flow, 47 Turbulent heat transfer, 103 mass transfer, 150 momentum transfer, 46 Unsteady-state condition

heat transfer, 111 mass transfer, 159 momentum transfer, 55 Velocity distribution, 13 circular tube, 18 liquid falling film, 13 Von Kármán analogy, 182