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 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