INTRODUCTORY COURSE IN

HEAT AND MASS TRANSFER

By

PROF. O.C. ILOEJE

1
DEDICATION

Dedicated, with profound reverence, to my parents Hyacinth Anieke and Juliana Ne’chi Iloeje;

and to all my students of Heat and Mass Transfer at the University of Nigeria Nsukka.

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PREFACE
This book presents an introductory material to the subject of heat and mass transfer, which is most
appropriate for students of engineering or engineering technology in a university, polytechnique or
technological institute. It is written for those who have not had a structured exposure to the subject,
but who however have a requisite background in thermodynamics, fluid mechanics, physics and
mathematics. It covers the sub-areas of conduction, radiation and convection heat transfer processes,
as well as heat exchangers and mass transfer in stationary systems, with an applications treatment of
mass transfer in convective systems.
To comfortably handle the material under conduction and radiation, the student needs the physics and
mathematics normally done in the first two years of a 5-year engineering programme, which include
such topics as temperature, thermal properties, kinetic theory of matter, calculus including solution of
second order ordinary differential equations and partial differential equations. In thermodynamics, a
first course which is usually done in the second year of a 5-year engineering programme, and which
covers such topics as thermodynamic properties, energy, heat and work, thermodynamic states and
processes, equations of state, first and second laws of thermodynamics, etc, as well as the use of control
volumes, will be adequate. Convection, heat exchangers and mass transfer require the pre-requisites
specified above for conduction and radiation, together with a course or two in fluid mechanics which
deal with the equation of conservation of mass, forces in fluid motion, momentum equations in
cartesian and cylinderical co-ordinates, lamina and turbulent flows, and boundary layer flows. An
enhanced treatment of the first and second laws of thermodynamics and partial differential equations
will also be needed.
The entire material may be offered as a 4-Credit course in the 4th year of a 5-year engineering
programme, by which time the pre-requisites will have been covered. Alternatively, it may be split
into two 2-credit courses each, in which one course covers the introductory material, conduction and
radiation, while the other covers convection, heat exchangers and mass transfer. The course may be
done in semesters of the same year. Alternatively, the first course may be done in the 3rd year while
the second course is done in the 4th year. The material on introduction, conduction and radiation is
recommended to be done before that on convection, heat exchangers and mass transfer, irrespective of
whether the entire material is presented as one course or as two courses. Firstly, conduction and
radiation are the fundamental modes or mechanisms for heat transfer. Basic principles of heat transfer
are therefore best presented under these two topics. Secondly, convection heat transfer may be looked
upon as macroscopic transport of energy-carrying particles under the action of forces, superimposed
on microscopic or molecular transport of heat energy, which is conduction. Thus, an understanding of
conduction and radiation, especially of conduction, is a necessary background for the understanding
of convection. For heat and mass transfer, the particular laws describing the processes, in stationary
systems, are based on the same phenomenological law, which is the law of diffusion. Consequently,
the equations of heat and mass transfer in stationary systems are very similar. Understanding one set
of equations facilitates the understanding of the other. However, because of our greater familiarity with
the heat transfer potential, temperature, and heat transfer under temperature differences, than with the
mass transfer potentials – mass concentration, temperature and pressure, and mass transfer under these
potential differences, it is better to start by presenting the conduction heat transfer material. Mass
transfer material can then be more easily introduced, subsequently. For a similar reason as the above,
it is also recommended that convection heat transfer be presented before the material on mass transfer.
Chapter One introduces the fundamental mechanisms and other processes of heat transfer. Conduction
– the theory and applications, are exclusively dealt with in Chapter Two. The fundamental mechanism
of conduction and the more practical law of conduction – the heat diffusion law or Fourier’s Law, are
then presented. Solutions of simple one-dimensional conduction problems, using Fourier’s Law, are
then used to aid the student’s understanding of the law before the development of the energy equation

3
of conduction. The latter is done for the unsteady two dimensional case in cartesian, cylinderical and
spherical co-ordinates. The student is however taken through the procedure for simplifying the
equation of conduction to obtain its steady, one-dimensional, heterogenous and homogenous forms.
All further examples and problems for this introductory course are limited to the steady one-
dimensional cases. A stepwise method for the solution of the resulting ordinary second order
differential equations, is illustrated through many examples. The electrical analogy of heat transfer is
then treated and used to solve the common problems of heat transfer through composite bodies and
thermal insulation. The Chapter ends with the treatment of extended surfaces or fins. Thermal
radiation is next treated in Chapter Three. It covers the nature, generation, emission, absorption,
reflection and transmission of thermal radiation; thermal radiation properties of material surfaces in
general and of special surfaces in particular. Thermal radiation exchange between surfaces is next
treated, with particular focus on black body and gray diffusely emitting, reflecting and absorbing
surfaces. The electrical analogy of radiation heat transfer is introduced and the student is then taken
through the use of thermal radiation network diagrams to analyze problems for both black body and
gray surfaces. Application of the methods of thermal radiation exchange analysis to black body and
gray enclosures is further discussed.

Chapter Four presents Convection Heat transfer in a systematic way. The differences between
conduction and convection heat transfer are clearly explained. This is followed by the development of
the unsteady energy equation of convection in lamina flow, in two-dimensional cartesian co-ordinates,
using the law of conservation of energy. The roles of the continuity and momentum equations in this
derivation are properly explained. The procedures for simplifying the resulting equation by utilizing
particular laws of heat and momentum transfer and equations of state are presented. The use of order
of magnitude analysis and conditions specific to a particular heat transfer situation to achieve further
simplifications of the equation, are presented. Application of the energy equation of convection, the
continuity and momentum equations, to obtain the solution for the temperature distribution in a
selected one-dimensional flow problem is demonstrated, preceded by the introduction of the concepts
of thermal and velocity boundary layers. This sequence is concluded with the definition and
determination of the heat transfer coefficient and dimensionless groups in convection heat transfer.
Building on the lamina flow heat transfer background, turbulent flow heat transfer is introduced. The
procedure for converting the instantaneous lamina continuity, momentum and energy equations to their
turbulent counterparts is presented. From the final turbulent transport equations, the contributions of
turbulent fluctuations to the momentum and energy transfers are explained. Convective heat transfer
correlations and the use of dimensionless groups in the correlations is presented. Methods of
dimensional analysis and the resulting dimensionless groups relevant to convection heat transfer are
further discussed. The treatment ends with a presentation of various convection heat transfer
correlations for lamina and turbulent flows, internal and external flows and for different geometries.
Having now treated conduction, radiation and convection heat transfer, the chapter ends with the
treatment of combined modes of heat transfer.
The Heat Exchanger is a major equipment in heat transfer and for this reason, it is given a special
treatment in Chapter Five. Firstly, various concepts in heat exchanger design are presented. This is
followed by the method of analysis of a simple heat exchanger and the concept of the Log Mean
Temperature Difference. Complex heat exchanger design procedures utilizing the f-factor and Number
of Heat Transfer Units (NTU) methods are presented. Heat exchanger surface fouling is also discussed.
Mass Transfer is treated in Chapter Six. The similarities between the mechanisms and particular laws
of heat and mass transfer are presented. Diffusion properties of fluids in gaseous, liquid and solid
media are treated. The two-dimensional equation of diffusion in a stationary medium, in cartesian and
cylinderical co-ordinates are presented. Similarities between the conduction and mass diffusion
equations and solution are emphasized. The electrical analogy of mass transfer, mass transfer
resistances and the application of mass transfer analysis to the drying of solids are treated. The

4
similarities between heat and mass transfer are extended to mass transfer with convection.
Dimensionless groups in mass transfer and the corresponding groups in heat transfer are presented.
Subsequently, deduction of mass transfer correlations from correlations for corresponding situations
in heat transfer, using the dimensionless numbers, is demonstrated. Practice problems are given at the
end of each chapter (except Chapter One) to aid fuller understanding of the subject matter treated in
the chapter.
In presenting the materials on conduction and convection heat transfer and mass transfer, I have
generally adopted the approach of presenting the principles on which the subject matter is based and
then using the principles to develop the energy equations of conduction or convection or the equation
of mass transfer. This is followed by showing how the equations are simplified for particular cases and
then how to solve the simplified equations. This enables the understanding, ab initio, of the physics or
scientific principles which underlie the subject matter, and subsequently, how the simplified problems
are directly linked to these principles and are direct derivatives of the general equation. I believe that
this approach permits a more holistic understanding of the subject matter.

5
ACKNOWLEDGMENTS

I am happy to acknowledge the following persons who have had profound effects on my professional
development.

Prof. Warren M. Rohsenow, Massachusetts Inst. of Technology, Cambridge, Mass, USA

Prof. Peter Griffith, Massachusetts Inst. of Technology, Cambridge, Mass, USA
Prof. Borivoje Mikic, Massachusetts Inst. of Technology, Cambridge, Mass, USA
Dr. David Plummer, Massachusetts Inst. of Technology, Cambridge, Mass, USA
Prof. Robert Lahey, General Electric Nuclear Energy Div; San Jose California, USA
Dr. B. Shiralkar, General Electric Nuclear Energy Div, San Jose California, USA
Prof. R. Smith, University of Nottingham, England, UK
Prof. N. Hay, University of Nottingham, England, UK
Prof. R. Madu, University of Nigeria, Nsukka
Prof. F.N. Ndili, University of Nigeria, Nsukka
Prof. I. H. Umar, Energy Commission of Nigeria, Abuja
Mr. Nasir El-Rufai, Bureau of Public Enterprises, Abuja, Nigeria
Barr. Liyel Imoke, Federal Ministry of Power, Abuja, Nigeria
Dr. Mrs. Ijeoma.C. Iloeje, University of Nigeria, Nsukka
Igwe N.P. Iloeje, Igwe’s Palace, Umuodu, Uwani-Amokwe, Nigeria

I also acknowledge Miss Chidera Obetta for her painstaking work in typing the manuscript.

CONTENTS

6
Chapter

1 INTRODUCTION

1.1 Objectives of Heat Transfer Analysis

1.2 Modes of Heat Transfer

2 CONDUCTION
2.1 Kinetic Theory of Matter and Conduction
2.2 Fourier’s Law of Conduction
2.2.1 Homogenous and Heterogenous Materials
2.2.2 Isotropic and Anisotropic Materials
2.2.3 Variations of k with Temperature
2.2.4 Effect of Alloying
2.2.5 Effects of Density and Wetness
2.2.6 Effective Thermal Conductivity
2.2.7 Units of Thermal Conductivity

2.3 Differential Form of Fourier’s Law

2.4 Some Illustrative Examples
2.5 Two Dimensional Heat Transfer in Stationary Media
(The Energy Equation of Conduction)
2.6 Applications to Steady 1-D Conduction
2.7 Conduction Treated in Cylinderical and Spherical Co-ordinates
2.7.1 Two Dimensional Equation of Conduction in Cylinderical Co-ordinates
2.7.2 Two Dimensional Equation of Conduction in Spherical Co-ordinates
2.8 Electrical Analogy of Heat Transfer
2.8.1 Concept of Thermal Resistance
2.8.2 Thermal Resistances in Series
2.8.3 Thermal Resistances in Parallel

2.9 Composite Bodies

2.9.1 Mixed Series and Parallel Composites
2.9.2 Overall Heat Transfer Coefficient

2.10 Thermal Insulation

2.10.1 Optimum Insulation Thickness from Heat Transfer Considerations
2.10.2 Insulation Economics: Marginal Savings Criterion for Optimal Insulation
Thickness

2.11 Extended Surfaces (Fins)

2.11.1 Analysis of fins
2.11.2 Fin Efficiency ηfin

Problems-2
P. Fundamentals of Conduction and Energy Equation of Conduction
Q. Electrical Analogy of Heat Transfer. Composite Bodies. Thermal Insulation
R. Extended Surfaces
3 THERMAL RADIATION

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 3.1 Introduction
3.2 Generation and Emission of Thermal Radiation
3.3 Radiation Properties of Real Substances
3.3.1 Emissivity
3.3.2 Absorption, Reflection and Transmission of Thermal Radiation
3.3.3 Emissivity and Absorptivity

3.4 Laboratory Black Body

3.5 Exchange of Radiant Energy Between Surfaces
3.5.1 Types of Surfaces
3.5.2 View Factor
3.5.3 Radiant Energy Exchange Between Two Black Body Surfaces
3.5.4 Radiant Energy Interchange in an Enclosure with n Black Body Surfaces
3.5.5 Electrical analogy for Radiant Energy Exchange Between Black Body
Surfaces
3.5.6 Analysis of the Thermal Radiation Network
3.5.7 Radiant Energy Exchange Between Gray Surfaces

Problems-3

4 CONVECTION HEAT TRANSFER

4.1 The Energy Equation of Convection in 2-D Cartesian Co-ordinates

4.1.1 Simplifications of the General Energy Equation of Convection in Particular

Cases

4.2 Application to Particular Cases

4.2.1 Concept of Velocity and Thermal Boundary Layers
4.2.2 Solution for Heat Transfer with Couette Flow

4.3 Introduction to the Analysis of Turbulent Flow Heat Transfer

4.4 Convection Heat Transfer Correlations
4.4.1 Dimensional Analysis
4.4.2 Application of Bukingham pί-Theorem to Fully Developed Convection Heat
Transfer

4.5 Some Convective Heat Transfer Correlations

4.6 Combined Mechanisms of Heat Transfer
Problems-4

5 HEAT EXCHANGERS

5.1 Heat Transfer Rate and Log Mean Temperature Difference Method for Simple
Heat Exchanger Analysis and Design
5.2 Correction Factors to LMTD for Complex Heat Exchangers
5.3 NTU Method of Heat Exchanger Analysis and Design: Heat Exchanger
Effectiveness, εt and Number of Heat Transfer Units, NTU.

5.3.1  t = f (RR, NTU ) for a Simple Parallel Flow Heat Exchanger

8
 5.3.2  t = f (RR, NTU ) for Counter Flow Simple Heat Exchanger

5.4 Fouling of Heat Exchanger Surfaces

Problems-5

6 MASS TRANSFER

6.1 Mass Transfer in Stationary Systems

6.1.1 Molecular Diffusion in Gases
6.1.2 Molecular Diffusion in Liquids
6.1.3 Molecular Diffusion in Solid

6.2 Diffusion Equation in a Stationary Solid or Fluid Medium

6.2.1 Mass Transfer Coefficient at a Surface
6.2.2 Boundary Conditions
6.2.3 Diffusional Resistance or Mass Transfer Resistance
6.2.4 Drying of Solids

6.3 Mass Transfer Correlations in Convection (Overview)

Problems-6

List of Symbols

References

CHAPTER ONE
INTRODUCTION

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1.1 OBJECTIVES OF HEAT TRANSFER ANALYSIS

These fall into three categories:

(a) To determine the temperature distribution in a medium, or on the surface of the medium, for
such purposes as temperature control, systems modeling, design, etc.
(b) To Determine the rate of heat flow crossing a given surface, or the quantity of heat has crossed
the surface.
(c) To determine the quantity of heat in a mass of matter

Necessary Conditions for the Transfer of Heat

There must be a temperature differential for heat to be transferred from one location or medium to
another. No heat is transfered between locations in a medium at uniform temperature, ie when there is
thermal equilibrium, as in Fig 1.1.

Fig 1.1 Medium at Uniform Temperature

1.2 MODES OF HEAT TRANSFER: GENERAL

By this, we mean the fundamental mechanisms by which heat is transferred from one location to
another. There are essentially two modes of heat transfer, namely

• Conduction
• Radiation

Conduction requires the existence of an intervening material medium before heat can be transferred,
but it does not involve the physical transport of bulk matter from one location to another. On the other
hand, radiation does not require the existence of an intervening material medium between two locations
for the transfer of heat from one to the other. Rather, it can be transfered through a vacuum or through

10
a medium that is transparent or translucent to the radiation. In Fig 1.2, conduction is possible between
surface A and B, as well as between C and D, but is not possible between surfaces B and C. However,
radiation heat transfer is possible between surfaces B and C.

Fig 1.2 Role of Intervening Matter in Conduction and Radiation Heat Transfer
There’s a heat transfer process called CONVECTION (Note that we have not defined this as a
fundamental mode of heat transfer)
The process of heat transfer by convection involves the physical transport of bulk matter from one
point to another during the transfer of heat.

Fig 1.3 Particle Movement in Convection

During convection, bulk matter may move from location 1 to location 2 or from 3 to 4. If T1>T2 or T3
> T4, then the motions will involve the transport of hotter matter to a colder environment, making it
possible for the hotter matter to transfer some of its heat to the colder matter in the new environment.
When the hotter material has arrived at the colder environment, it finds itself surrounded by colder
fluid and then transfers heat to it. However, the transfer of heat will be by the mechanisms of
conduction and radiation. Thus the convective process has not introduced a new heat transfer
mechanism. We can think of convection heat transfer, therefore, as a process of heat transfer which
enhances heat transfer by the mechanism of conduction (and radiation) through bulk movement of
particles in the medium. We do not consider convection as a new fundamental mechanism of heat

11
transfer. However, it is another process of heat transfer. Other unique processes of heat transfer include
boiling and condensation.

There are two types of convective processes, namely

(a) Forced Convection

(b) Natural or Free Convection
Forced convection involves the motion of matter under the action of external driving surface forces
such as pressure and friction forces. Natural Convection involves the movement of matter due to
differences in the volumetric intensity of a body force. Examples of body forces are gravity, magnetic,
electrostatic and centrifugal forces. Density differences produce differences in the volumetric intensity
of gravity force (= g ) . Other mass centered forces, such as centrifugal force, do likewise. Such
differences can lead to natural or free convection.

CHAPTER TWO
CONDUCTION
2.1 Kinetic Theory of Matter And Conduction

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Conduction heat transfer is the mode of heat transfer by which heat is transferred through a material
medium through the microscopic movement of the particles (molecules, atoms, electrons) in the
medium, but without macroscopic movement of the particles from one point to another. According to
the kinetic theory of matter, molecules vibrate about their mean positions in solids, or travel at their
mean velocities throuth their mean free paths in fluids, depending on the mean kinetic energy, KE. The
mean KE is proportional to the absolute temperature. Molecules at higher temperatures have higher
KE. On collision with molecules at lower KE, the higher KE – molecules transfer some of their energy
to the lower KE – molecules. The decrease in the KE of the former implies a decrease in temperature
and hence a decrease in heat content. The rise in KE of the latter implies an increase in temperature
and hence an increase in heat content. The transfer of the molecular KE is therefore equivalent to the
transfer of heat. In this way, heat energy is transmitted through the material. The motion of free
electrons in pure crystalline metals, which produces high electrical conductivity, also produces high
thermal conductivity in the material. By this we mean that they also improve the ability of the
crystalline metals to transfer heat, through the mechanism of conduction.
The terminology thermal conductivity, refers to how highly or poorly a medium conducts
heat. The symbol of thermal conductivity is k.
For pure crystalline metals,
k  = 78 *10 −9 T 2.1
k = thermal conductivity
 = electrical conductivity
 = absolute temperature.
For insulators (electrical), there are no free electrons. Conduction is governed only by molecular
motion and the thermal conductivities are consequently low. For liquids and gases, the mean-free-
paths are long. The probabilities of collisions are lower than in solids and hence the thermal
conductivities are lower. They are lower for gases than for liquids, in general.
Conduction Process
We may summarize the conduction process thus. The kinetic theory informs us that the Molecular
Kinetic Energy is proportional to the Absolute Temperature of the particle. When a particle receives
thermal energy, its temperature increases, and by the kinetic theory, its kinetic energy thus increases.
The particle’s vibrational velocity becomes higher. The higher velocity (ie higher energy) particle
collides with an adjacent lower velocity (lower energy and lower temperature) particle and transfers
some of its kinetic energy to the latter particle. The previously lower energy particle then acquires a
higher energy and hence a higher temperature. In this manner, thermal energy is transfered through the
material. The kinetic theory is useful for understanding the mechanism of conduction and for analyzing
heat transfer processes and heat transport properties for simple systems such as mono-atomic gases. It
is inadequate for real engineering systems.
General and Particular Laws
General laws are those laws which do not depend, for their applicability, on the nature and/or state of
the system under consideration. Conversely, particular laws depend on the nature and/or state of the
system for their applicability. Examples of general laws are:
 dm 
• Law of Conservation of mass   =0
 dt  Isolated or closed system
• Newton’s laws of motion
• 1st, 2nd, 3rd laws of thermodynamics
An example of a particular law is the perfect gas law:

13
 V = mRT _____ Applies to perfect gases only 2.2

Another example of a particular law is the Fourier’s law of conduction, which we introduce below.

2.2 FOURIER’S LAW OF CONDUCTION

Fig 2.1 Illustration of Conduction Through a Slab

In Fig 2.1, CD and EF in the 2-D diagram or C΄CDD΄ & E΄EFF΄ in the 3-D diagram are isothermal
surface boundaries at temperature T1 and T2 respectively. They form boundaries of a material slab of
thickness L. The slab extends infinitely in directions parallel to CD and EF and perpendicular to the
plane of the paper.
Consider a section of the slab of uniform cross-sectional area A and extending over the thickness L.
Then according to Fourier’s law, the heat flow rate per unit cross-sectional area A is proportional to
the change in temperature over the length L, divided by the length L, [ie to the average temperature
gradient] and is in the direction of decreasing temperature.

Q   − 
ie   − 2 1 
A   L 
2.3
 L or X
The negative sign before the average temperature gradient in equation 2.3 ensures compliance with the
stated direction of heat flow, ie, that it is in the direction of decreasing temperature.

Q A = q and is called the heat flux. Hence, in the direction of L or x,

  − 1 
:. q L = −k  2  2.4
 L 

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   − 1 
and Q = −kA 2  2.25
 L 
where k is the constant of proportionality and is called the Thermal Conductivity of the material.
Fourier’s law is a phenomenological law, derived from experimental data. It also assumes that the
thermal conductivity k has the same value throughout the material – in particular, that it is the same
for all slices of the slab parallel to the bounding surfaces. Now, this is not always so. In practice, the
thermal conductivity may vary from point to point within the material. In addition, the value of k for
heat conduction in a given direction may be different for heat conduction in another direction.
The last observation implies that for a given material, k x  k y  k z in some cases. The first observation
is saying that in some cases,

(k x )1  (k x )2 2.5a
(k )  (k )
y 1 y 2 2.5b
k1  k 2 even if k x = k y = k z 2.5c

Generally, k varies from matrial to material. They are generally high for solids, intermediate for liquids
and low for gases, as shown below.

kgas ~ 0.008 – 0.08 W/m 0K for gases

kliquid ~ 0.08 – 0.8 ,, for liquids
ksolid ~ 0.8 – higher ,, for solids

2.2.1 Homogenous and Heterogeneous Materials

When k has the same value throughout a material medium, the material is said to be homogeneous,
and when k varies from point to point within the material, the material is said to be heterogeneous.
Solids may consist of a fine mixture of different materials. Such solids are therefore heterogeneous.
Also porous substances like fired clay, cork, foam, etc are heterogeneous. The thermal conductivities
of some materials are functions of temperature. Thus if such a material is at a non-uniform temperature,
it will then be heterogeneous. The variation of k in many engineering materials may, however, be so
slight that the differences are assumed negligible. We can then assume such materials to be
homogeneous. Examples of homogeneous materials are Cu, Zn, H2O, air and aluminium, at a uniform
temperature.

2.2.2 Isotropic and Anisotropic Materials

An isotropic material is one in which the thermal conductivity is the same in all directions, otherwise
it is anisotropic.
k x = k y = k z = k .......... .... if isotropic 2.6a
k x  k y  k z .......... .... if anisotropic 2.6

15
 Fig 2.2 Directional Thermal Conductivities
Applying Fourier’s law to heat transfer in the x and y directions of Fig 2.2, we obtain that:

Q x Q y
= kx; = ky 2.7
 
zy xz
x y

And similarly for k z .

Examples of anisotropic materials are wood, grass, other fibrous materials, etc. However, many
engineering materials can be assumed to be isotropic. Water, copper, aluminium and materials with
directionally invariant structures are isotropic.
2.2.3 Variations of k with Temperature

Fig 2.3 Variations of Thermal Conductivity with Temperature

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Generally speaking, k for crystalline materials and liquids decrease with temperature, while k for gases
and amorphous substances like glass and rubber, increase with temperature, as illustrated in Fig 2.3.
Some solids are mixtures of amorphous and crystalline substances, while others have gas pores. For
such mixtures, k may increase or decrease with temperature, or may have turning points. Also, since
they are heterogeneous, the thermal conductivities that we can measure for them are the effective k.

2.2.4 Effects of Alloying

For alloys, the thermal conductivity may be in–between those of the constituents or may be less than
that of any of them. The alloy nickel–silver contains

Cu ~ 64% k = 381 W mK at 200 C

Ni ~ 18% k = 90 W mK at 20 0 C
Zn ~ 17% k = 112.1 W mK at 20 0 C

But k for nickel–silver at 20 0C is 31.3 W/m 0K, which is below the values for all the constituents.

2.2.5 Effect of Density and Wetness

For any material, k increases with the density and with the wetness.

2.2.6 Effective Thermal Conductivity

Heterogeneous materials, such as solid mixtures, composites and porous materials, have thermal
conductivities that vary from point to point. In the analysis of heat transfer through such materials, it
is often convenient to asign a single value of k to the material. The value of k so asigned is called the
effective thermal conductivity of the material. It is the value of k which, when used in Fourier’s Law
in the analysis of heat flow through the material, will yield the same temperatures and heat fluxes as
would exist in the actual material. Thermal conductivities measured and listed for concrete, sand, foam,
glass wool, etc, are effective thermal conductivities.

2.2.7 Units of Thermal conductivity

From Fourier’s Law

Q x   − 1 
= −k  2 
Ax  L 
LQ x
:. k = −
Ax (2 − 1 )

( )
Units of L = (m); Q x = (W ); Ax = m 2 ; 2 − 1 = ( o
C or o K )
m *W W W
Hence Units of k = 2 o o
= or
m * C or K mC mK

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2.3 DIFFERENTIAL FORM OF FOURIER’S LAW

Fig 2.4 Two Dimensional Slab

We stated earlier that for the slab of uniform cross-sectional area, bounded by isothermal surfaces at
T1 and T2, Fig 2.4, Fourier’s law gives that:

  − 1 
q L = −k  2  2.4
 L 

for heat flow in the positive x direction, ie along the length L.

Fourier’s law assumes that the material is homogeneous. Also, we have defined the isothermal
surfaces, at T1 and T2, to be parallel to the y-axis. We have then assumed that all the heat flow caused
by this temperature change (T2 – T1) is in a direction perpendicular to the isothermal surface, that is,
in the x-direction and that heat flow in a direction parallel to the isothermal surface is zero. This is true
if the material is isotropic. Thus, Fourier’s law also assumes that the material is isotropic. Now,
consider the slab of Fig 2.5 and the slice of infinitesimal thickness x , as shown.

18
 Fig 2.5 Two Dimensional Differential Slab
If, the slab is isotropic and homogeneous, then the small slice is also isotropic and homogeneous and
we can apply Fourier’s law to the slice. If, on the other hand, the slab is isotropic but heterogeneous,
it follows that the heat flow is perpendicular to the isothermal surface but that k is not the same at
every point within the material. But the slice is of infinitely small thickness x, so that k can be
assumed constant within the infinitesimal slice. We can then apply Fourier’s law to that small slice.
We will get that,

  +  −  
q x = −k   2.8
 x 

or q x = −k 2.9
x
Taking limits of equation 2.9 as x → 0

lim q x = − k lim 2.10
x

or q x = −k 2.11a
x
Similarly,

q y = −k 2.11b
y

q z = −k 2.11c
z

Equations 2.11a, b, c constitute the Fourier’s law of Conduction in differential form.

19
2.4 SOME ILLUSTRATIVE EXAMPLES ON FOURIER’S LAW

(1) A rectangular wall of thickness 22.5cm, height 5m and length 20m has a thermal
conductivity of 0.2 W mK . If the outside surface A has a temperature of 300C and the
inside surface B is at 250C, then

(i) What is the direction of heat flow?

(ii) What is the heat flow rate through the wall?

Assume steady state and a homogeneous and isotropic wall. (Steady State means that
temperatures and heat fluxes do not vary with time)
Solution(See Fig 2.6)

(i) Temperature decreases from surface A to surface B. Hence heat flow is in the direction A
to B.
(ii) From Fourier’s law of conduction, applied to a rectangular slab,

(2 − 1 )
q x = −k where posivite x is in the direction of A to B 2.4
L

= − k (B − A ) / L
= −0.2 (25 − 30) / 0.225 = 4.44 W / m 2
 Qx = Ax q x = 5 * 20 * 4.44 = 444 W 2.12
Note that Ax is the area perpendicular to the heat flow direction, ie perpendicular to x-axis

Fig 2.6 Illustration of Problem 2.4(1)

(2) In problem #1 above, surface A is at 35 0C and the heat flow rate is 500 W. What should be the
thermal conductivity of the wall if the inside surface temperature is not to exceed 23 0C?

Solution:

20
From Fourier’s law applied to the rectangular slab,

Q x ( −  )
qx = = −k  2.13
x L

Qx /  x 500 0.225
 k= = * = 0.094 W mK 2.14
(A − B ) / L 5 * 20 12
(3) The temperature distribution across the thickness of a rectangular slab is
 = 100 + 250x o C , where x = 0 corresponds with one face of the slab. The faces of the slab
are isothermal surfaces and steady state exists.
(i) What is the direction of heat flow through the slab?
(ii) If the slab is 50 cm thick, what are the directions and magnitudes of the heat flux at x =
0 and x = 50 cm?
Solution:
(i) )x =0 = 100 0 C and )x =50cm = 225 0 C . From the given aquation, T continuosly
increases as x increases. Heat will flow in the direction of decreasing T which is in the
direction of x decreasing – ie, in the negative direction of x.
Alternatively
From Fourier’s law in differential form

q x = −k 2.11a
x
 d
But  =  (x ) only  = 2.15
x dx
d
 q x = −k 2.16
dx

d
Since  = 100 + 250x, = 250 which is + ve . Then q x = −250k which is − ve . Hence q x is in
dx
the negative x – direction
d 
(ii) q x ) x =0 = −k  = −250k
dx  x =0
d 
q x )x =0.5m = −k  = −250k
dx  x =0.5

Hence qx = - 250k and is in the negative x – direction at both x = 0 and x = 50cm. Alternatively, the
equation  = ( x ) or a graph of T vs x shows that T (x) is a straight line. Consequently the gradient
d
is constant and positive.
dx

21
 Fig 2.7 Temperature Profile for Problem 2.4(3)

Heat flux, from Fourier’s law, is given by

d
q x = −k = −k * (gradient of T (x) curve)
dx

Hence with a positive gradient, qx will be negative. Also with a constant gradient, qx will be the same
at all x.

(4) The temperature distribution across the thickness of a rectangular slab bounded by isothermal
surfaces is T = a + bx – cx2. One surface is at x = 0. The material is isotropic and homogeneous
and at steady state. The thermal conductivity = k, and slab thickness = L. Values a, b and c are
positive constants.

(i) Find the location of maximum temperature in the slab, xm

(ii) What is the heat flux at xm?
(iii) What are the magnitudes and directions of the heat flux at x = 0 and x = L, given
that L>xm
Solution:

(i) At maximum temperature, T(x) profile is at a turning point. At the turning point dT/dx
=0
d
 = b − 2cx = 0 2.9
dx
d 2
 x = x m = b 2c at the turning point. If the turning point is a maximum, then  0. But
dx 2
d 2
= −2c which is < 0, since c is positive. Hence the temperature profile is a maximum at x m = b 2c
dx 2
.

22
 d   b
(ii) q x ) x = xm = − k  = − k b − 2c  2.20
dx  x = xm  2c 
 q x ) xm = 0

Alternatively,
d
q x = −k 2.11a
dx
d
At the maximum temperature point, =0
dx
Hence q x )xm = −k * 0 = 0 2.20
(iii) q x )x =0 = −k b − 2cx x =0 = −kb , ie it is negative 2.21
q x ) x = L = −k b − 2cx  x = L = −k (b − 2cL )
= k (2cL − b ) 2.22
b
Given that L > xm, it follows that L >
2c
 2cL  b
Hence q x )x = L = k (2cL − b ) which is  0, ie it is positive

It follows that:
At x = 0, the heat flux is in the negative x–direction and so the temperature is increasing with x.
At x = L, the heat flux is in the positive x–direction and so the temperature is decreasing with x. At
dT
x = L 2c there is a turning point, and the temperature gradient = 0 . The temperature profile will
dx
be as shown in fig 2.8.

Fig 2.8 Temperature Profile for Problem 2.4(4)

23
5. Home Work

In problem #4, the temperature profile is given by T = a + bx + cx2. Repeat the solution. Determine if
the turning point in the temperature profile, x m , is a minimum or maximum temperature location.

2.5 TWO DIMENSIONAL HEAT TRANSFER IN STATIONARY MEDIA

The Energy Equation of Conduction

Heat transfer in stationary media is governed by a number of general and particular laws. The general
laws are:

• Law of conservation of energy (1st law of thermodynamics is a specific form of it).

• Law of conservation of mass

The particular laws are:

• Fourier’s law of conduction
• Relevant form of the Equation of State.

The law of conservation of energy enables us to do a book–keeping of the flow and the accumulation
(or depletion) of energy and conversion of energy from one form to another, for a system. Similarly,
the law of conservation of mass permits a book-keeping of the flow and accumulation (or depletion)
of mass for a system. Consider a material medium fixed in space and represented in the sketch below.

Fig 2.9 Energy and Work Flows and Energy Content for a Medium

The medium contains some amount of energy E in the form of:

Internal energy
Potential energy
Macroscopic kinetic energy (= 0 since medium is stationary)
Chemical Energy
Nuclear or atomic Energy
Electrical Energy
Also, energy can be transfered into or out of the medium. This energy can be in any of the forms listed
above, or in the form of heat energy. Work can be done on the medium by the surroundings, and the

24
medium can do work on the surroundings. We may study flows of energy for the medium by studying
its flows for a representative small or differential control volume in the medium. Subsequently, through
a process of integration over the entire medium, we may then obtain what happens in the whole
medium. Let us therefore consider a two-dimensional differential control volume in the medium, of
dimensions x & y, in x and y directions, and of unit dimension in the z – direction. The law of
conservation of energy for a control volume states that: the rate of change of total energy within the
control volume with respect to time, is equal to the net flow rate of energy into the control volume CV,
plus the net rate of work done on the control volume by the surroundings (ie by external forces).

Fig 2.10 Energy Flows for a Stationary Control Volume

Energy Quantities and Flows

- The total energy within the CV =  cv = e xy

where e = Total energy per unit mass of the CV
1
 e = u int + gy + V 2 + u ch + u nuc + u ele etc. Note that V = 0 for a stationary CV 2.23
2
and  = density of material within the CV.
gy = gravitational potential energy per unit mass

- Net energy flow into the control volume: This consists of heat flow by conduction q (W/m2)
and electrical energy. Since the medium is stationary, energy flow by convection (or
macroscopic energy flow) is zero.

25
- Net work done on the CV: Since the CV is fixed and the medium is stationary, the surface and
body forces acting on the CV do not do any work. Additionally, there is no shaft work. Hence
work done by or on the CV is zero.

Applying the law of conservation of energy to the control volume, we then obtain that,
   q  
 (uint + gy + u ch + u nuc + u elec )xy  = elec − elec
 + q x y −  q x + x y 
t   x  
  q y  
+ q y x −  q y + y x  2.24
  y  

Electrical energy loss due to electrical resistance heating makes elec

 less than elec by an amount
=  elec
  elec  q x q y 
( uint ) +  (gy ) +   (uch + u nuc + u elec ) =
 − +  2.25
t t t xy  x y 
u  y    elec  q x q y 
OR  int + uint + g + gy +  (u ch + u nuc + ...) = − +  2.26
t t t t t xy  x y 
uint  y   elec  q x q y 
 + (uint + gy ) + g +  (u ch + u nuc + ...) = − +  2.27
t t t t xy  x y 

Simplifications
- Since the medium or the CV is stationary
y
=0 2.27a
t
- The law of conservation of mass applies. This is represented by the continuity equation:
 
+ (V x ) +  (V y ) = 0 Continuity equation in 2-D 2.28
t x y

But for a stationary medium, V x = V y = 0


Hence from 2.28, =0 2.28a
t

Also  =  (t , x, y )
d   
 = + Vx + Vy 2.29
dt t x y

But Vx = V y = 0

d 
 =
dt t

But from 2.28a, =0
t

26
 d
Hence  =0 or  is constant 2.30
dt

Inserting 2.27a and 2.28a into 2.27, it simplifies to

 u int  elec  q 
 = −

 (u ch + u nuc + ..) −  q x + y  2.31
t xy t  x y 
 elec
  Electrical resistance heating per unit volume
= u elec
xy
For the 2nd group of terms on the right hand side of equation 2.31,  u ch = chemical internal energy

per unit volume. If ( u ch ) is negative, then chemical internal energy in the CV is decreasing with
t
time. The decrease is converted into heat and the process manifests as an exothermic chemical reaction.

There is thus a heating effect due to chemical reaction. If ( u ch ) is positive, then chemical internal
t
energy is increasing with time and heat is absorbed from the surroundings in order to effect the
increase. There is thus a cooling effect due to the chemical reation, ie an endothermic chemical
reaction. Similarly, we may have an exothermic or endothermic nuclear reaction or electrical process.


If the term  (u ch + u nuc + ..) is negative, then there’s net generation of heat due to the reactions.
t

The expression −  (u ch + u nuc + ..) becomes positive. It is represented by u oth  and is called the
t
net heat generation rate per unit volume from other sources. If the term is positive, then there’s net

heat absorption or negative heat generation. The expression -  (u ch + u nuc + ..) becomes - u oth .
t
The energy conservation equation, ie equation 2.31, becomes:

 u int  q x q y 
 = 
u elec  
u oth −  +  2.32
t  x y 
Rate of change of Elec resist. Internal Heat
sensible int. heating rate gen rate from Net heat inflow rate
energy per unit per unit vol. other sources by conduction per
volume per unit vol. unit vol.

Equation 2.32 may be written as:

u  q q y 
 = u  −  x +  2.33
t  x y 

where u = u int = sensible internal energy per unit mass

u  = u elec
 + u oth
 = net heat generation rate per unit volume

27
We may now apply the relevant particular laws.

(a) For a simple system u = u (v,  ) 2.34

 u   u 
 u =   v +    2.35
 v      v
 u 
u =   v + Cv  3.56
 v  
d
But as derived or shown earlier,  is a constant, ie = 0 for a stationary control volume in a
dt
 1
stationary medium. If  is constant, then   =  is constant.
 
 v = 0
Thus u = C v  --- an equation of state 2.37
u 
Hence = Cv 2.38
t t

Equation 2.37 or 2.38 is an equation of state, which is valid for a simple thermodynamic system of
constant density. It is therefore a particular law.

(b) From Fourier’s Law (for an isotropic medium)


qx = − k 2.11a
x

qy = − k 2.11b
y

Substituting 2.38, 2.11a and 2.11b into 2.33 gives the general unsteady energy equation of conduction
in two dimensions, for an isotropic and heterogeneous medium:

        
 Cv = u  +  k  + k  2.39
t x  x  y  y 

Constant density also implies that the medium is incompressible. For an incompressible medium

Cv = C p = C 2.40

If the medium is also homogeneous, then k is constant and equation 2.39 becomes

   2  2 
C = u  + k  2 + 2  2.41
t  x y 

Dividing 2.41 by c , we obtain

28
  u    2  2 
= +   2 + 2  2.42
t C  x y 
 u 
Or = +   2 2.43
t C

k
where  = is the thermal diffusivity of the material medium.
C

 2 2 
and  2 =  2 + 2  in 2–D Cartesian co-ordinates x and y 2.44
 x y 

2.6 APPLICATIONS FOR STEADY 1–D CONDUCTION


Steady state implies that ( ) =0 2.45
t

For 1-D conduction along the x-axis only, for instance, it follows that temperature is changing in the
x-direction only. It does not change with y or z. Hence for such a 2-D medium in x and y


=0 2.46
y

Hence, the dependent variable T is a function of one independent variable, x only.

 2 d 2
Consequently, = for  = f (x ) only.
x 2 dx 2

The general equation 2.41 reduces to:

d 2
k + u  = 0 2.47
dx 2

Boundary Conditions
We observe that the energy equation of conduction is a 2nd order differential equation, requiring two
boundary conditions for each direction in order to get a particular solution. For the 1-D problem of
d
equation 2.47, the boundary conditions will consist of the specification of  or or a linear
dx
combination of the two, at each of any two locations. It is possible to specify any of the three forms
d
(except ) at both boundaries, or a different form at each boundary. There are thus eight different
dx
pairs for the sets of boundry conditions. For example, we may specify

(i) ( x1 ) and ( x 2 )
 d 
(ii)  a + b  at x1 and x 2 , where a and b are cons tan ts.
 dx 

29
 d 
(iii) (x1 or x2 ) and  − − − − − − − − − − − (2 pairs)
dx  x1 or x 2
 d 
(iv)  a + b  and T (x1 or x2 ) − − − − − (2 pairs)
 dx  x1 or x 2
 d  d 
(v)  a + b  and  − − − − − −(2 pairs)
 dx  x1 or x 2 dx  x1 or x 2

d
Specification of at both boundaries will not permit the determination of the two constants of
dx
integration.
d
Furthermore, the temperature gradient , may be given indirectly by specifying the heat flux, say at
dx
x = x1 . If this is done, then from Fourier’s law,

d 
q x1 = − k  2.11
dx  k1
d 
 = − q x1 k 2.48
dx  x1
Also, when a surface is exposed to a fluid ambient at a different temperature convection heat transfer
takes place between the surface and the ambient.
According to Newton’s law of cooling, which is not really a ‘law’ but a definition of the term – “heat
transfer coefficient” h at a surface-fluid boundary (Fig 2.11) the heat flow rate from surface to ambient
is given by:

Fig 2.11 Definition of Heat Transfer Coefficient on a Surface

Q x = hA (x1 −  ) where A is the heat transfer surface area 2.48a

30
 Qx
or qx = = h(x1 −  ) 2.48b
A
But heat flux by conduction from within the material to its surface, in a direction perpendicular to the
surface, is giving by

d 
−k  from Fourier’s Law
dx  x1

With no accumulation of heat at the surface, heat flux by conduction from inside the material to the
surface is equal to the heat flux from the surface to the ambient (by convection).

d 
 −k  = h (x1 −  )
dx  x1
2.49
d  −h
or  = (x1 −  ) 2.50
dx  x1 k
d
or h (x1 ) + k (x1 ) = h  2.51
dx
d
If h, k and  are constants, then the left hand side of 2.51 is a linear combination of  and .
dx
Since k is known, then if h and  are specified, equation 2.51 amounts to the specification of a
d
linear combination of  and at the boundary x1 .
dx
d 
In summary therefore, instead of specifying the temperature gradient  as one of the boundary
dx  x1
conditions, we can specify the heat flux qx. We can also specify the heat transfer coefficient h at a
surface boundary, together with the ambient temperature T adjacent to that surface. Equation 2.48 in
the former case and 2.50 or 2.51 in the latter case may then be used as the boundary condition. The
2 2
units of h are W m K or W m C

ILLUSTRATIVE EXAMPLES
2.6 (1) The flat plate with no heat generation
A flat plate of thickness L extends infinitely in the y – and z – directions. Heat is supplied on one face
at the rate of qo (W/m2). The other isothermal surface is kept at a temperature T2. Temperature varies
with x only, across the slab thickness. Find the temperature profile across the plate. Assume steady
state, isotropic and homogeneous medium and no heat generation.

31
 Fig 2.12 Illustration for Example 2.6 (1).

Systematic Solution Steps

Step 1: Determine the Applicable Equation

For steady state =0
t
  2 
Since  does not vary with y , = =0
y y 2
For no heat generation, u  = 0
Using equation 2.41, which is applicable to a homogeneous and isotropic medium, and deleting terms
that are zero by virtue of the above conditions, as specified in the problem, we obtain:

 2
 k 2 =0 2.52
x
 2 d 2
 does not vary with t or y , but varies with x only. Consequently = 2
x 2 dx
d 2
 =0 This is the Applicable Equation 2.52a
dx 2

Step 2: Ascertain the correct boundary conditions which may or may not be explicitly given in the
problem statement
(i) Heat flux at x = 0 is q o , as given in the problem. Hence
d 
−k  = q0 ------------ boundary condition (i) 2.53
dx  x =0
(ii) )x =L = 2 --------------- boundary condition (ii) 2.54

Step 3: Obtain the General Solution of the Applicable differential equation

32
 d 2
=0 2.52a
dx 2

d
=a 2.55
dx

 = ax + b This is the general solution of equation 2.52a 2.56

Step 4: Get the particular Solution

This step involves the use of the boundary conditions to get the constants of integration.
Apply equation 2.53, using 2.55 or 2.56

− ka = q 0 2.57
q
a = − 0 2.58
k
Apply equation 2.54, using 2.56
2 = aL + b 2.59
− q0
Or 2 = L+b
k
q
 b = 2 + 0 L 2.60
k
Introduce the values for a and b in 2.56 to get the particular solution, thus:

q0 q
=− x + 0 L + 2 2.61
k k
q
or  = 0 (L − x ) + 2 2.62
k

Fig 2.13 Temperature Profile for Example 2.6 (1)

33
2.6 (2) Flat Plate with Heat Flux, Heat Transfer Coefficient and Ambient Temperature
Specified at the Boundaries
A rectangular slab is bounded by isothermal surfaces at x = 0 and x = L . The slab extends infinitely
in the y and z directions. Heat flux on the surface at x = 0 is  W m 2 . Heat transfer coefficient on
the other face at x = L is h W m 2 C . The ambient temperature adjacent to this face is Ta . The
thermal conductivity of the material of the slab is k W mC . There is steady state and no heat
generation. The material is isotropic and homogeneous. Determine the temperature distribution across
the slab thickness. Show that the heat flux at x = L is also  W m 2 .

Fig 2.14 Illustration for Example 2.6 (2)

Since the slab extends infinitely in the y and z directions, end conditions at y = +  and z = + 
do not affect temperature profiles at finite locations in, or within finite sizes of, the slab. We may thus
assume one-dimensional temperature variation and heat flow in the x-direction only.

From equation 2.41 and the conditions given in the problem, the applicable equation is

d 2
=0 (see solution for problem 2.6 (1)) 2.52a
dx 2

The two boundary conditions are obtained thus:

The surface has no mass and hence there is no accumulation or depletion of heat on the surface.
Therefore heat flux from the environment to the surface at x = 0 is equal to the heat flux from the
surface into the slab at x = 0. A similar situation exists at x = L. The boundary conditions are obtained
from these situations. It then follows that,

34
 d 
(i) −k  = 2.63
dx  x =0
d  
  =− 2.64
dx  x =0 k
d 
(ii) −k  = h ((L ) − a ) 2.65
dx  x = L
d 
= − (L ) − a 
h
  2.66
dx  x = L k

Integration of 2.52a gives the general solution

d 2
= 0 2.52a
dx 2
d
= a 2.67
dx
 = ax + b 2.68

The temperature profile is linear.

Substituting boundary condition (i) ie equation 2.64, in 2.67 yields


a = − 2.69
k

Next, boundary condition (ii), equation 2.66, is substituted in 2.67. In doing so (L ) in 2.66 is given
by 2.68 for x = L . We thus obtain,

a =−
h
(aL + b − a ) 2.70
k
 h  L 
− = − − + b − a  2.71
k k  k 
 L
 b = + + a 2.72
h k

The particular temperature distribution is obtained by substituting for the constants a and b (2.69 and
2.72) in 2.68. Thus,
 1 L
 = − x +   +  + a 2.73
k h k 

Heat flux at x = L is given by

d   
q L = −k  = − k a = −k  −  2.74
dx  L  k
 qL =  2.75

35
With a linear temperature profile, the temperature gradient and hence the heat flux, is the same at any
value of x in the slab.

2.6(3) Rectangular slab with Heat Generation

In the slab of Example 1, let us now assume that heat is being generated, through electrical resistance
heating, at the rate of u  W / m 3 . At steady state, the temperatures of the surfaces at x = 0 and x = L
are respectively 1 and 2 . Find the temperature distribution across the slab thickness. Determine the
directions as well as the expressions for heat flux at the boundary surfaces. Sketch the temperature
profile for T1 > T2.

Systematic Solution

Applicable Equation
d 2
0 = u  + k 2.76
dx 2
Boundary Conditions

(i) )x =0 = 1 2.77

(ii) )x =L = 2 2.78

General Solution

d 2
k 2 + u  = 0 2.79
dx
d 
2
u 
2
= − 2.79a
dx k

d u 
=− x+a 2.80
dx k
u x 2
=− + ax + b 2.81
2k
Particular Solution
Apply boundary condition (i) in 2.81 to get,

b = 1 2.82

Apply boundary condition (ii) in 2.81.

u L2
2 = − + aL + 1 2.83
2k
2 − 1 u L u L  1 − 2 
 a= + = − 2.84
L 2k 2k  L 

36
Substituting for a and b in 2.81, the temperature distribution in the slab is shown to be,

u x 2  u L  1 − 2 
 =− + −  x + 1 2.85
2k  2k  L  
d 
The heat flux at any cross-section, ie at any location x , is given by Fourier’s law, q x = −k 
dx  x
u L
+ (1 − 2 )
k
 q x = u x − 2.86
2 L
u L
 q ) x =0 = (1 − 2 ) −
k
2.87
L 2
u L k
and q )x = L = + (1 − 2 ) 2.88
2 L

At x = L and for 1  2 , q x is positive and so the direction is positive, ie heat flows in the direction
of increasing x . Thus, the temperature decreases as x increases, as x = L is approached.
At x = 0 , a number of possibilities exist for the direction of heat flux.

u L k
(a) If  (1 − 2 ) then the direction is positive and slope of ( x ) is negative
2 L

= (1 − 2 ) then q )x =0 = 0 slope of temperature profile (x ) is zero
u L k
(b) If
2 L
u L k
(c) If  (1 − 2 ) then the direction is negative and slope of ( x ) is positive
2 L
Fig 2.15 Illustrates the possible profiles for T(x).

Fig 2.15 Temperature Profiles for Example 2.6 (3)

37
2.7 CONDUCTION TREATED IN CYLINDRICAL AND SPHERICAL COORDINATES

2.7.1 Two Dimensional Equation of Conduction in Cylinderical Coordinates

The same assumptions and laws apply as in the treatment in Cartesian coordinates, namely

Stationary Law of conservation of energy

Incompressible Law of conservation of mass (continuity equation)
Isotropic and Fourier’s Law of conduction
Homogeneous Medium Relevant equation of State
Let us assume that temperature variations are in r and Ө directions only. It does not vary in the z or
axial direction.

Fig 2.16 Control Volume for 2-D Conduction in Cylinderical Co-ordinates

We assume that temperature does not vary in the circumferential direction θ. Thus, for given r and z,
temperature is the same as θ changes.Therefore  = f (r , z ) . Consequently, the control volume can be
a circular ring of inner radius r, thickness  r and axial length  , as shown in Fig 2.16. The energy
conservation law applied to the control volume CV, states that the rate of change of total energy in the
CV with respect to time = net flow rate of energy into the CV plus net rate of work done on the CV.
Hence we have that:

   q  
(.2rrz.e ) = (E elec  ) + q r .2rz −  q r + r r 2 (r + r )z 
 − E elec
t   r  
  q  
+ q z .2rr −  q z + z z 2rr  2.89
  z  

 q q q
(re ) = E elec − q r + r r + r r  2rz − r z 2rz 2.90
t  r r   z 

38
 q r 2
Neglecting the 3rd order small quantity in 2.90, ie 2  r z, which is the smallest quantity in the
r
equation, the equation becomes

 E elec  q q 
(re ) = − q r + r r + r z 
t 2rz  r z 
 E elec   q 
(re ) = −  (rq r ) + r z  2.91
t 2rz  r z 
 
 (re  ) + re  = right hand side (rhs) of 2.91
t t
e r 
r  + e + re = rhs 2.92
t t t
Continuity or mass conservation equation in 2-D cylindrical co-ordinates (r, z ) is given by
  V 
+ (Vr ) +  r + (V z ) = 0 2.93a
t r r z
Since the medium is stationary, V r = V Z = 0 . Hence

=0 2.93b
dt
Also, since  =  (t , r , z )
d   
= + Vr + Vz 2.93c
dt t r z

Since the medium is stationary V r = V z = = 0 as shown above. Hence
t
d
= 0 and  is therefore constant 2.93d
dt
r
The stationary state of the medium also implies that =0 2.93e
t
Inserting 2.93b and 2.93e in 2.92 give that:

e E elec   q 
r = −  (rq r ) + r z  2.94
t 2rz  r z 
e E elec 1  q
  = − (rq r ) + z  2.95
t 2rrz  r r z 
1 2
But e = u + V + gy + u others
2
e u   1 2  y u
  = +  V  + g + oth
t t t  2  t t
1 2 y
V = 0 and = 0 for a stationary medium
2 t
e u u
 = +  oth 2.96
t t t

39
Inserting 2.96 in 2.95 and re-arranging, we obtain

u E elec u 1  q
 = −  oth −  (rq r ) + z  2.97
t 2rrz t  r r z 
As deduced earlier while developing the conduction equation in Cartesian co-ordinates
E elec
 = electrical resistance heating per unit volume
= u elec
2rrz
u oth
−  = heat generation (or absorption) rate per unit volume from other sources
=  u oth
t
(chemical, nuclear reactions, etc).

Let u  = u elec

  u oth
 = net heat generation rate per unit volume from all source. Then 2.97 becomes
u 1  q
 = u  −  (rq r ) + z  2.98
t  r r z 
Equation 2.98 has three dependent variables u, q r and q z . It is not possible to get explicit solutions
for the three variables with one equation. Let us bring in some relevant particular laws:


q r = −k 2.99
r
Fourier’s Law for an isotropic medium

q z = −k 2.100
z
Also, u = u (v, ) 2.101
u  u 
u =  v +   2.102
v     v

But from 2.93d,  is constant. The medium is incompressible. Hence v is constant and v = 0
u 
Also,  = C v . For an incompressible medium C v = C p = C . Equation 2.102 then becomes:
  v
u = C   This is an equation of state for a simple incompressible medium 2.103
u 
Then =C
t t
Substituting 2.99, 2.100 and 2.101 in 2.98, we obtain

 1        
C = u  +  kr  + k  Heterogeneous medium 2.105
t r r  r  z  z 
Equation 2.105 is the energy equation of conduction in 2-D cylinderical co-ordinates (r, z ) for an
isotropic, stationary or incompressible and heterogeneous medium. If the medium is homogeneous,
then k is constant and 2.105 becomes,

  1      2  
C 

= u + k r + 2  Homogeneous medium 2.106
t  r r  r  z 

40
  u   1      2  
Or = +  r + 2  Homogeneous medium 2.107
t c  r r  r  z 
Where  = k C = thermal diffusivity of the medium.

For steady state, = 0 and 2.106 becomes,
t
 1      2  
k r  + 2  + u  = 0 2.108
 r r  r  z 
If there is variation of T in the radial direction only, then variation of T in the z-direction is zero. The
problem is one dimensional in r.  = f (r ) only and 2.108 becomes,

k d  d 
r  + u  = 0 2.109
r dr  dr 

Note that for  = f (r ) only, the partial derivative of T with respect to r becomes total derivative. If
instead, T does not vary with r, then recalling our earlier assumption that T does not vary with θ, it
follows that for any given axial location z, the temperature is uniform over the cross-sectional surface
perpendicular to the axis at that z. Then  = f ( z ) only. Equation 2.108 becomes,

d 2
k + u  = 0 2. 110
dz 2
Equations 2.109 and 2.110 are the steady state 1-dimensional equations of conduction in the radial and
axial directions, respectively, for a stationary, isotropic and homogeneous medium

2.7.2 Two Dimensional Equation of Conduction in Spherical Co-ordinates

By following a derivation procedure similar to that used for the cartesian and cylinderical co-ordinate
systems, but applied to a spherical control volume, it can be shown that the 2-D unsteady equation of
conduction in spherical co-ordinates (r ,  ) for a stationary isotropic and homogeneous mediums is
given by,

 u      2   1    
= + 2  r +  sin  
 
2.111
t C r  r  r  sin   

Example 2.7(1) Radial Temperature Variation in a Hollow Cylinder

Find the radial temperature distribution in a hollow circular cylinder of inside and outside radίί rί and
ro. The inside and outside surface temperatures are Tί and To, respectively. There is steady state and no
heat generation. The material is isotropic and homogeneous.

41
 Fig. 2.17 Illustration for Example 2.6 (4)

Because of the symmetry in the defined problem, temperature variations will exist in the radial
direction only. There will be no temperature variation in the circumferential ( ) direction.

Applicable equation
Introduction of the conditions of the problem to 2.108 or 2.109 yields the applicable equation:

d  d 
r =0 2.112
dr  dr 

Boundary Conditions

(i) )ri = i 2.113

(ii) )ro = o 2.114

General Solution
d  d 
r =0 2.112
dr  dr 

d
r =a
dr

d a
ie = 2.113
dr r
or  = a ln r + b 2.114
Particular Solution

i = a ln ri + b from boundary condition (i) 2.115

42
 o = a ln ro + b from boundary condition (ii) 2.116
i − o
a=
ln(ri / ro )
2.117

 − o
b = i − i ln ri 2.118
ln(ri / ro )

Substituting for a and b in 2.114 gives the particular temperature distribution:

i − o ( − o )
= ln r + i − i ln ri 2.119
ln(ri / ro ) ln(ri / ro )
 − o  − o
= i ln(r ri ) + i = i − i ln(r ri ) 2.120
ln(ri / ro ) ln(ro / ri )

If we determined b in terms of ro and o , we’ll have that

i − o
= ln( r / ro ) + o 2.121
ln(ri / ro )

The heat flux through the material at any radius r , is given, using 2.113 and 2.117, by

d k   − o 
q r = −k = −  i  2.122
dr r  ln ri / ro 

k  i − o 
ie q r =  
r  ln ro / ri
2.123

Heat flow rate Q = 2 rL q r

2kL
 Q =
ln (ro / ri )
Watts 2.124

where  = i − o

2.8 ELECTRICAL ANALOGY OF HEAT TRANSFER

2.8.1 Concept of Thermal Resistance

Just as there exists a resistance to the flow of electrical energy, there is also a resistance to the flow of
thermal energy. In electrical energy flow, the voltage H is the potential for the transfer of electrical
energy. The electrical energy flow rate, or current, is I and the resistance to that flow is R. In thermal
energy flow, the potential for the transfer of thermal energy is temperature T (or a parameter that
directly depends on temperature). The thermal energy flow rate is Q. There is correspondingly, a
resistance to the thermal energy flow Rth.
In electrical circuit analysis, the following relationship exists:

43
 H = R 2.125

where ΔH = Electrical Potential Difference or Voltage Difference, across a resistance

I = Electrical Energy flow rate or current
R = Electrical Resistance

Let us look at expressions for heat flow rate through known geometries.

(i) Heat Flow through a Slab

Consider an isotropic slab of thickness L, with constant cross-sectional area A over L, bounded by
isothermal surfaces at T1 and T2 and of uniform thermal conductivity k (Fig 2.18). The steady state
heat flow rate through the slab (in the direction 1 to 2) is given, from equation 2.5, as

Fig 2.18 Heat Flow Through a Slab

  − 2 
Q = k 1  2.5
 L 
Equation 2.5 may be written as

 L 
 = Q   2.126
 k 

where  = 1 − 2
Comparing equation 2.125 and 2.126 we deduce that

  H Temperature Difference is equivalent to Voltage Difference OR Thermal Potential

Difference is equivalent to Electrical Potential Difference

44
Q  I Heat flow rate is equivalent to Electrical Energy flow rate or Charge flow rate OR
Thermal Current is equivalent to Electrical Current
L L
R is equivalent to Electrical Resistance
k k

We may therefore define a Thermal Resistance Rth for the slab as,

L
Rth = 2.127
k
Then
 = Q Rth 2.128

Equation 2.128 constitutes the electrical analogy of heat transfer. A thermal circuit element diagram
can be drawn to represent equation 2.128 as shown in Fig 2.19. The figure is the electrical equivalent
of heat flow through the slab of Fig 2.18.

Fig 2.19 Thermal Resistance Diagram for a Slab

(ii) Radial Heat Flow Through a Hollow Cylinder

Fig 2.20 Radial Heat Flow Through a Hollow Circular Cylinder

Using obtained earlier that for steady radial heat flow through a hollow circular cylinder of uniform
thermal conductivity k and with dimensions and temperatures as shown in Fig 2.20

45
 2kL 
Q = 2.124
ln (ro / ri )

where  = i − o
 ln(ro ri ) 
  = Q   2.129
 2kL 
Using the electrical analogy of heat transfer, equation 2.128, we deduce from 2.129, a thermal
resistance for radial heat flow through a hollow cylinder as,
ln(ro ri )
Rth = 2.130
2kL
(iii) Heat Transfer at a surface
If the heat transfer coefficient at a surface is h, the surface temperature is Ts, surface area is A and
ambient temperature is Ta, then heat flow rate from the surface to the ambient is given by.
Q = h (s − a ) 2.48a
 1 
 = Q   2.131
 h 
where  = s − a
From the electrical analogy of heat transfer and equation 2.131, we define the surface thermal
resistance as:
1
Rth = 2.132
h

2.8.2 Thermal Resistances in Series

Fig 2.21 Thermal Resistances in Series

Consider a rectangular slab bounded by surfaces 1 and 2 at 1 and 2 , respectively. Corresponding
adjacent ambient temperatures are a1 and a 2 . The temperature of surface 1 on the ambient side is
equal to the temperature of surface 1 on the slab side, and similarly for surface 2. If there is no heat
46
generation or absorption at the surfaces, then heat flow rate from ambient 1 to surface 1 is equal to the
heat flow rate through the slab and is also equal to the heat flow rate from surface 2 to ambient 2.
Hence,
Q = h1 (a1 − 1 ) 2.133a
k
Q = (1 − 2 ) 2.133b
L
Q = h2 (2 − a 2 ) 2.133c

Consequently

a1 − 1 = Q h1 A 2.134a
 L 
1 − 2 = Q   2.134b
 k 
2 − a 2 = Q h2 A 2.134c

Adding 2.134a, b and c gives

 1 L 1 
a1 − a 2 = Q  + +  2.135
 h1 k h2 

If  = a1 − a 2 = the overall temperature difference from ambient (1) to ambient (2). Thus
 1 L 1 
Then  = Q  + +  2.136
 h1 k h2 

From the electrical analogy we can write equation 2.136 as  = Q Reff 2.137
Reff is the effective thermal resistance between ambient (1) and ambient (2). From 2.136 and 2.137,
 1 L 1 
we recognize  + +  as this effective thermal resistance.
 Ah1 kA Ah 2 

1 L 1
 Reff = + + 2.138
Ah1 kA Ah2

In view of equations 2.127 and 2.132, we recognize the terms on the right hand side of 2.138 as the
thermal resistance at surface (1), that of the slab and that at surface (2), correspondingly. Hence

Reff = R1 + RL + R2 2.139

The heat flow rates through the three thermal resistances are the same. The resistances are bounded by
common thermal potentials Ta1, T1, T2 and Ta2.They are thus in series. We see from (2.139) that the
effective thermal resistance of thermal resistances in series is the sum of the individual thermal
resistances. Thus series thermal resistances are additive, just as series electrical resistances are. A
thermal circuit diagram for the series thermal resistances is as sketched below, Fig 2.22.

47
 Fig 2.22 Circuit Diagram of Thermal Resistances in Series

2.8.3 Thermal Resistances in Parallel

Fig 2.23 Thermal Resistances in Parallel

The rods 1, 2 and 3 are bounded by the same isothermal surfaces, S1 an S2, at temperatures 1 and 2
. Let heat flow rate from medium C to medium D be only through the rods. Then the total heat flow
rate from medium C into the rods, Q , is equal to the sum of the heat flow rates through the rods: Q 1
, Q and Q and is equal to the heat flow rate from the rods into medium D, at steady state and no heat
2 3

generation.

Hence Q = Q 1 + Q 2 + Q3 2.140

48
 k1 A1
But Q1 = (1 − 2 ) 2.141
L
k A 
Q 2 =  2 2 (1 − 2 ) 2.142
 L 
k A 
=  3 3 (1 − 2 )
Q 3 2.143
 L 
k A k A k A 
 Q =  1 1 + 2 2 + 3 3 (1 − 2 ) 2.144
 L L L 
 
 1 
Or  = Q   2.145
 k1 A1 + k 2 A2 + k 3 A3 
 L L L 

From the electrical analogy of heat transfer, equation 2.128, we can write equation 2.145 as

 = Q Reff 2.128

Where Reff is the effective thermal resistance to heat flow from ίsothermal surface S1, through the rods,
to ίsothermal surface S2. Hence
1 1
Reff = = 2.146
k1 A1 k 2 A2 k 3 A3 L L L
+ + 1/ 1 + 1/ 2 + 1/ 3
L1 L2 L3 A1 k1 A2 k 2 A3 k 3

But Li k i Ai is the thermal resistance Ri of rod i . Hence 2.147

1
Reff = 2.148
1 1 1
+ +
R1 R2 R3
1 1 1 1 1
Or
Reff
= +
R1 R2 R3
+ = R
i
2.149
i

The heat flow rate into surface S1 is equal to the sum of the heat flow rates through the three rods’
thermal resistances. Also, the rods operate between the same temperature potentials. Hence the rods
are in parallel for thermal energy flow. Equation 2.149 indicates that the reciprocal of the effective
thermal resistance is equal to the sum of the reciprocals of the individual thermal resistances in parallel
– just as is the case for electrical resistances in parallel. The equivalent thermal circuit element is as
illustrated in Fig 2.24

49
 Fig. 2.24 Circuit Diagram of Thermal Resistances in Parallel

2.9 COMPOSITE BODIES

Consider two rectangular slabs 1 and 2 of different thermal conductivities, which are perfectly bonded
at the flat rectangular surfaces (Fig 2.25a), or two hollow cylindrical tubes 1 and 2 in which one is
perfectly force-fitted into the other (Fig 2.25b). The two slabs or tubes constitute composite bodies.
Any number of different materials may be involved in a composite body.

(a) Cross-section of Composite Slabs (b) Part of x-section of Composite Cylinders

Fig 2.25 Composite Bodies

If perfect contact is assumed to exist between the materials at the interface and there is no heat
generation or absorption at the interface, then:
(a) Heat flow rate from material 1 into the interface is equal to heat flow rate from the interface
into material 2 for both steady and unsteady cases

50
 (b) The temperature at the interface on the side of material 1 is equal to the temperature at the
interface on the side of material 2 and is equal to if . (Note this will not be so if there is a
contact resistance at the interface – say, as in the case of imperfect thermal contact, for instance,
with rough surfaces.)
(c) If there is steady state, then the heat flow rate between any two heat flow lines, say AB and
CD in the two-dimensional diagrams of Fig 2.25 a and b, is the same from material to material
and outside the materials.
(d) With the above assumptions, composite materials can be analyzed as materials whose thermal
resistances are in series or in parallels or constitute a mixture of series and parallel resistances.

For the composite body shown in Fig 2.25 a or b, the equivalent thermal circuit diagram is given as:

Fig 2.26 Thermal Circuit Diagram of a Composite Rectangular Slab with Series Thermal
Resistances

1 L1 L2 1
where Rs1 = ; R1 = ; R2 = ; Rs 2 = 2.150
Ahs1 Ak1 Ak 2 Ahs 2

for the composite slabs. From the electrical analogy of heat transfer,

a1 − 1 = Q Rs1 2.151

1 − if = Q R1 2.152
if − 2 = Q R2 2.153
2 − a 2 = Q Rs 2 2.154

Adding equations 2.151 to 2.154 yields,

a1 − a 2 = Q (Rs1 + R1 + R2 + Rs 2 ) 2.155

The effective thermal resistance is given from 2.155, and 2.128 as

1 L L 1
Reff = Rs1 + R1 + R2 + Rs 2 = + 1 + 2 + 2.156
Ahs1 Ak1 Ak 2 Ahs 2
1  1 L1 L2 1 
=  + + +  2.157
A  hs1 k1 k 2 hs 2 

 a1 − a 2
But Q = = 2.158
Reff Rs1 + R1 + R2 + Rs 2

51
  
 
a1 − a 2
:. Q = 
  2.159
 1 L L 1 
 + 1 + 2 + 
 Ah s1 Ak 1 Ak 2 Ah s 2 

For the composite cylinderical tubes, assumed to be complete cylinders, the equivalent thermal
circuit is similar to that of the composite rectangular slab above. The individual thermal resistances
are:
1 ln rif ri ln ro rif 1
Rsi = ; R1 = ; R2 = ; Rso = 2.160
Ai hsi 2k1 L 2k 2 L Ao hso
where Ai = 2ri L and Ao = 2ro L

Since the resistances are in series, equations 2.151 to 2.155 are applicable. Consequently we may
write that:
1 ln rif ri ln ro rif 1
Reff = + + + 2.161
Ai hsi 2k1 L 2k 2 L Ao hso
1 ln rif ri ln ro rif 1
Reff = + + + 2.162
2ri Lhsi 2k1 L 2k 2 L 2ro Lhso

1  1 ln rif ri ln ro rif 1 
=  + + +  2.163
2L  ri hsi k1 k2 ro hso 

But Q =  2.164
Reff
2L (a1 − a 2 )
 Q = 2.165
 1 ln rif ri ln ro rif 1 
 + + + 
 ri hsl k1 k2 ro ho 

2.9.1 Mixed Series and Parallel Composites

Fig 2.27 Mixed Series and Parallel Composites

52
There’s steady heat flow from ambient a1 to ambient a2. There’s no heat flow through the spaces
between a, b and c and no heat generation in any of the components of the composite. Using the method
of thermal resistances, the equivalent thermal circuit diagram can be represented with the sketch below.

Fig 2.28 Circuit Diagram for the Mixed Series and Parallel Composite of Fig 2.27

Rs1, R1 and R2 are in series; Ra, Rb and Rc are in parallel; R3 and Rs2 are in series. Each resistance is
given by the appropriate equation.
If R p , eff is the effective resistance of the resistances in parallel, then Rs1, R1, R2, R p , eff , R3 and Rs2 are
in series.
−1
 1 1 1 
R p , eff =  + +  2.166
 R a Rb R c 
 Rth , o = Rs1 + R1 + R2 + R p , eff + R3 + RS 2  Overall Thermal Resistance 2.167

Hence Q = (a1 − a 2 ) / Rth , o 2.168

Note: If Ta1 and Ta2 are given, we can find any intermediate temperature by using the equation of
the electrical analogy of heat transfer up to the intermediate point. For instance

(I) a1 − 2 = Q (Rs1 + R1 + R2 ) We know Q . Hence 2 can be determined. 2.169

(II) a1 − 3 = Q (R + R + R + R
s1 1 2 P , eff ) Hence 3 can be determined. 2.170

2.9.2 Overall Heat Transfer Coefficient

We express the heat transfer rate between a surface and the adjacent environment in terms of a heat
transfer coefficient h, that is, by using the Newton’s Law of Cooling, thus:

Q = Ah  where  = (s − a ) or (a − s ), depending on 2.48a

the direction of heat flow

In like manner, we want to express the heat flow rate through a composite, structure or device, in terms
of a heat transfer area A, the overall driving temperate difference  and an Overall Heat Transfer
Coefficient U . Thus we want to write that,

Q = AU 2.171

U is not an actual heat transfer coefficient in the same sense a heat transfer coefficient h at a surface-
fluid boundary. It is rather a parameter which plays the role of a heat transfer coefficient, as may be
deduced from comparing equations 2.48a and 2.171.

53
 Q 1  Q 
Hence U= =   2.172
A A   

U is not an actual heat transfer coefficient in the same sense as a heat transfer coefficient h at a surface-
fluid boundary. It is rather a parameter which plays the role of a heat transfer coefficient, as may be
deduced from comparing equations. 2.48a and 2.171.
1
 U= 2.173
ARth ,o

where Rth , o = the Overall Thermal Resistance for the composite, structure or device.
We can also write that,

1
Rth , o =
AU
2.174

For the composite rectangular slab of Fig 2.25a, the effective or overall thermal resistance is given by
equation 2.156. Thus,

 1 L L 1 
Rth , o =  + 1 + 2 +  2.175
 1
Ah Ak 1 Ak 2 Ah 2 

1  1 L1 L2 1 
=  + + +  2.176
A  h1 k1 k 2 h2 
 
 
1  1 
Hence U = =
 2.177
A Rth , o 1 L1 L2 1 
 + + + 
 h1 k1 k 2 h2 
For the composite cylinder of Fig 2.25b, it was shown that Reff or Rth ,o was given by

1 ln rif ri ln ro rif 1
Rth , o = + + + 2.178
Ai hi 2k i L 2k 2 L Ao ho

1
However, in using the relationship U = to obtain the overall heat transfer coefficient U, we
ARth ,o
note that the heat transfer area A varies with radius, from the inside surface area Ai to the outside
surface area Ao . We can choose any of the areas. However, U has to be defined with respect to the
particular chosen area. Thus we may have U i or U o defined with respect to Ai or Ao . However, the
heat flow rate through the composite cylinder remains the same. Hence,
Q = AiU i  2.179
and Q = AoU o  2.180

54
Hence AiU i = AoU o = Q  2.181
1
AiU i = AoU o = 2.182
Rth , o
1
Ui = 2.183
Ai Rth , o
1
And Uo = 2.184
Ao Rth , o

Using equation 2.178 we obtain that

 
 
 1 Ai 
Ui =   2.185
1 ln rif ri ln ro rif 1
 + + + 
 Ai hi 2k1 L 2k 2 L Ao ho 

1 Ai
= 2.186
1  1 Ai ln rif ri Ai ln ro rif A 
 + + + i 
Ai  hi 2k1 L 2k 2 L Ao ho 
 
 
 1 
Ui =   2.187
 1 + ri ln rif ri + ri ln ro rif + ri 
 hi k1 k2 ro ho 

Similarly
 
 
 1 
Uo =   2.188
 ro + ro ln rif ri + ro ln ro rif + 1 
 ri hi k1 k2 ho 

2.10 THERMAL INSULATION

2.10.1 Optimum Insulation Thickness from Heat Transfer Considerations

A thermally insulated body is analysed as a composite body which consists of the insulated material,
the insulation and the ambient. The heat flow rate through the composite system is given by

Q sys = UA (1 − 2 ) 2.189

U is the overall heat transfer coefficient, A is the heat transfer area to which U is referenced and
(1 − 2 ) is the overall temperature difference. If A is the insulation surface area, then for the
insulation of a rectangular slab, A is constant and U decreases with increasing ίnsulation thickness.

55
Hence Q decreases indefinitely with insulation thickness, implying that an infinitely thick slab is the
condition for minimum heat loss.
Suppose the system is bounded by curved surfaces – such as cylindrical surfaces, then as insulation
thickness increases, the U decreases. However, the insulation surface area increases (See Fig 2.29).

Fig 2.29 Effect of Insulation on Overall Conductance for a Cylinderical Tube

The product AU may increase and then later decrease, implying that there is a turning point. Thus,
with curved surfaces, increase in insulation thickness does not necessarily mean a reduction in heat
transfer.
Consider an insulated cylinderical tube of length L. From equations 2.161 and 2.164, the radial heat
transfer rate from fluid inside the tube at 1 to the outside ambient at 2 is given by,

1 − 2
Q = 2.190
 rif r 
 ln ln 2 
r1 rif
 1 + + +
1 
 h1 A1 2k1 L 2k 2 L h2 A2 
 
 
Where A1 = 2r1 L is the inside surface area of the tube
A2 = 2r2 L is the outside surface area of the insulation
r1 = inside radius of the tube
rif = tube/insulator interface radius
r2 = outside radius of insulation
Let us examine if there is a maximum or minimum in the heat transfer rate as insulation thickness
changes. Equation 2.190 may be written as:

 1 ln rif ri ln r2 rif 1 
Q  + + +  =  2.291
 2r1 Lh1 2k1 L 2k 2 L 2r2 Lh2 
Constant
56
The first two terms inside the square brackets of 2.291 remain constant as insulation thickness, and
hence r2 , changes.
 ln r2 ln rif 1 1 
 Q c + − +   =  2.292
 2k 2 L 2k 2 L 2Lh2  r2 

Differentiating 2.192 with respect to r2 , and recalling that

d
(xy ) = x
dy
+y
dx
we get that,
dv dv dv

 1 1   ln r2 ln rif 1 1  dQ
Q  − 2 
+  c + − +  
 dr = 0 2.193
 2k 2 r2 2Lh2 r2   2k 2 L 2k 2 L 2 Lh2  r2  2

dQ
At Optimum heat flow rate, = 0 2.194
dr2
 1 1 
 Q  − 2 
= 0 2.195
 2 k 2 Lr2 2 h2 Lr2 

h2 r22 − k 2 r2 = 0

r2 (h2 r2 − k 2 ) = 0 2.196
k2
r2 = or r2 = 0 2.196
h2
k2
r2 = 0 is not admissible. Hence the valid solution at optimum Q is r2 = .
h2
By further differentiating 2.193 and substituting for r2 = k 2 h2 at optimum Q , it can be shown that
d 2 Q 

2 
 0 . It follows that the optimum is a maximum in Q as illustrated earlier, graphically,
d r2  r = k2
2
h2

in Fig 2.29. See also Fig 2.30. Generally, therefore, as insulation is added to a cylindrical tube or pipe
whose uninsulated outside radius is less than k 2 h2 , the rate of heat loss from the tube or pipe will
initially increase, then reach a maximum at r2 = k 2 h2 , before decreasing as more insulation is added.
This behavior is also true for the insulation of materials bounded by convex surfaces.

57
 Fig 2.30 Effect of Insulation on Radial Heat Flow Rate Through a Cylinderical Tube.

In order to reduce the heat flow rate with insulation to a level below the heat flow rate with no
insulation, for the cylindrical tube or pipe, r2 should not only be greater th an k 2 h2 . It should be
greater than r2* , where r2* is the outside radius of the insulation which is required to permit the same
heat flow rate with insulation as when there was no insulation. Then, r2 − rif is the corresponding
*

insulation thickness, as in Fig 2.30. Further increases in insulation thickness will decrease the heat
flow rate indefinitely, so that the maximum insulation thickness, from thermal considerations, is again
seen to be infinitely large. The solution for the uptimum insulation thickness obtained by using a
thermal criterion is therefore not realistic or realizable.

2.10.2 Insulation Economics: Marginal Savings Criterion for Optimal Insulation Thickness
The reduction of unwanted heat loss or gain in an energy transfer system, for instance in refrigeration
fluid flow lines, boilers and reactors, always results in the saving of money. This is at the expense of
some expenditure on insulation. The objective then is to find the insulation thickness which will
maximize the net financial gain/benefits from the use of insulation.
If a pipe of a given diameter has insulation of thickness x on it, then the total cost of the insulation
varies with the length of the insulated pipe. Let the cost of the insulation, of thickness x , be N Cc ( x )
per unit length of the insulation. Let the length of insulation used be L.
Note: Cc ( x ) = material + installation labour + any other relevant cost.
Let the value of energy saved by using insulation material of thickness x be N S ( x ) per unit length of
the insulation.

Total Cost, Cc = Cc(x) L 2.198

Total Saving, S = S(x) L 2.199
Net Benefits, B = S – Cc 2.200

At optimum Net Benefits, as insulation thickness is varied,

dB
= 0 2.201
dx
58
 dS dCc
 − = 0 2.202
dx dx

dS dCc
and = , 2.203
dx dx
dS
= Marginal Saving, MS
dx
dCc
= Marginal Cost, MCc
dx

Thus, at the optimum net benefits, the marginal savings is equal to the marginal cost. If the savings S
and cost Cc are continuous algebraic expressions within the range of x of interest, then the optimum
insulation thickness can be obtained by solving equation 2.203. In many cases, the savings and cost
data are not given as algebraic expressions but as tabular values.

Fig 2.31 Savings and Cost Curves

In that case, we plot the graphs of S vs x and Cc vs x , then obtain tangents to the curves at various
dS dCc
values of x. (See Fig 2.31). The tangents are the values of and (ie the marginal savings MS
dx dx
and marginal costs MCc.). By ploting the curves of MS vs x and MCc vs x on the same graph, the
point of intersection gives the solution for x at optimum net benefits, as shown in Fig 2.32.
Alternatively MS and MCc may be obtained by tabular differentiation – using forward, central or
backward differences. See Table 2.1 The value of x at which MS = MCc is then obtained from the
tables by inspection.

59
 Fig 2.32 Marginal Savings (MS) and Marginal Cost (MCc) Curves

Table 2.1 Tabular Differentiation

xi y i Forward Central Backward
x1 y1
x2 y2 yi +1 − yi yi +1 − yi −1 yi − yi −1
x3 y3 xi +1 − xi xi +1 − xi −1 xi − xi −1
x4 y4

In applying Table 2.1 to determine marginal costs and marginal savings, y is equivalent to cost or
saving, while x is equivalent to the insulation thickness. The marginal cost and marginal saving are
then obtained by using either the forward, central or backward differential coefficients.

2.11 EXTENDED SURFACES (FINS)

Fins are extensions from the surface of a bulk material. They provide additional surface area for heat
transfer from the surface of the bulk material, for a smaller proportional increase in the bulk of the
material. The thickness of a fin is much less than the representative thickness of the bulk material.
Because of the added surface area, fins improve the heat transfer rate, heat transfer rate per unit mass
or volume and the heat transfer efficiency or effectiveness, over what would have been the case if there
were no fins. The more efficient heat exchangers are fitted with fins. The fins may have rectangular or
circular cross-section or any other convenient cross-sectional shape. Fig 2.33 shows some fin shapes.

60
 Fig 2.33 Fin Shapes
Examples

1) The engine block of a motorcycle has fins. These improve the cooling of the engine.
2) Tubes in compact heat exchangers have flat fins wound round the tubes in a spiral manner.
Alternatively, the planes of the fins may be perpendicular to the axis of the tube with each fin
being a separate circular or rectangular piece.
3) One design of the radiator of a car has plate fins attached in a continuous S – Curve between
walls of the flow passages.

61
2.11.1 Analysis of Fins
In the thermal analysis of fins, it is assumed that the fins are so thin and that the thermal conductivity
of the material of the fin is sufficiently high that the temperature is constant across the thickness of the
fin. A solution for the temperature distribution across the fin thickness shows that for this to be true,
kf
 6, where k f = thermal conductivity of the fin, l f = the characteristic dimension or
h(l f / 2)
thickness of the fin and h = the heat transfer coefficient on the surface of the fin. It follows that
h(l f / 2) 1 1 hl f
 or that Bi  where = Bi , and is called the Biot Number. For a flat plate fin,
kf 6 6 2k f
the characteristic thickness is the plate thickness. For a cylindrical fin, it is the cylinder diameter d.
Consider a rectangular fin as sketched below, Fig 2.34.

Fig 2.34 Rectangular Fin

Base temperature of fin = Surface temperature of bulk material = To

Ambient temperature = a
Fin length = L
Fin thickness = a
Fin width = b (both b and a are constant)
Fin surface heat transfer coefficient = h (on the top, bottom and side surfaces)
Fin material density = 
Fin material thermal conductivity = k
We apply the energy conservation equation to the control volume. The significant energy term flowing
into and out of the control volume is heat energy conducted through the CV. The significant energy
within the material is sensible internal energy u , where du ~ C v d (material is incompressible).

62
Density  and specific heat at constant volume C v are constant. For an incompressible material,
Cv = C p = C. There is no work done and no heat generation from any source. The first law or energy
conservation equation then gives:-


t
( ) (
(uba x ) = Q x − Q x + Q x − Q s = − Q x + Q s ) 2.204

u Q
ba x = − x x − Q s 2.205
dt x

But u = C; Q x = −ba k and Q s = h(2b + 2a )( − a )x 2.206
x
u 
=C 2.207
t t
  2
:. baC x = bak 2 x − 2(b + a )h( − a )x 2.208
t x

Let A = Cross-sectional area of fin = ba at any x

and p = Perimeter of fin = 2(b + a ) ~ 2b at any x , where a << b

  2
 AC x = kA 2 x − hp( − a )x 2.209
t x

  2
or AC = kA 2 − hp( − a ) 2.210
t x

  2 d 2
At steady state, = 0 .  is thus a f ( x ) only, so that =
t x 2 dx 2

d 2
 kA − hp( − a ) = 0 2.211
dx 2
d 2   hp 
Or =  ( − a ) 2.212
dx 2  kA 

Equation 2.212 is the governing or applicable equation for a rectangular fin of constant cross-
sectional area, at steady state. It is also applicable to a fin of any other shape if the cross-sectional
area and perimeter are constant over its length.

Solution of 2.212 by Substitution

Let  =  − a 2.213
d d
Then = 2.214
dx dx

d 2 d 2
and = 2.215
dx 2 dx 2

63
Substituting for  in terms of  in 2.212, we get:

d 2 hp hp
=  2  where  2 = or  = 2.216
dx 2 kA kA

The general solution of equation 2.216 is given by,

 = c1e x + c 2 e − x 2.217

Boundary condition

(1) At x = 0 ,  = o
and  = (o − a ) =  o
(2) At x = L, we make the following considerations:

We assume that fin end area is much smaller than fin surface area, ie, A  pL
Consequently, heat loss at the end face at x = L is much less than heat loss from the surface area pL.
d 
We can then assume that − k  is negligible. Thus
dx  x = L
d d
at x = L , = 0, ie = 0 This is the sec ond boundary condition
dx dx

Substituting the boundary conditions in the general solution, we obtain:

 o = c1 + c 2 from boundary condition (1) 2.218

0 = c1e L − c 2 e − L from boundary condition (2) 2.219

Simultaneous solution of 2.218 and 2.219 for c1 and c 2 gives that

e L
c2 =  o 2.220
e L + e − L

e − L
c1 =  o L 2.221
e + e − L

e x e − L + e − x e L
  = o 2.222
e L + e − L

 e −  (L− x ) + e  (L− x )
or = 2.223
o e  L + e − L

64
  Cosh  (L − x )
or = 2.224
o Cosh L

In terms of the temperature  , the solution becomes,

Cosh
hp
(L − x )
 − a kA
= 2.225
o − a hp
Cosh L
kA

Equation 2.224 can also be obtained by assuming that the general solution to equation 2.216 is given
by  = c1Cosh  x + c 2 Sinh  x, instead of equation 2.217.

Fig 2.35 Steady Dimensionless Temperature Profile for fin with Constant Cross-Sectional
Dimensions

The heat loss rate to the environment is given by the heat loss rate from the surface of the fin. This is
the heat loss rate from the surface of the elemental control volume, integrated over the entire surface
of the fin. The heat loss rate from the surface of the CV = h(2b + 2a )x( − a ) = hp( − a )x.

hp ( − a ) dx. It is also equal to the heat flow rate out of the bulk material
L
Hence Q fin = 
o

d 
into the fin through the base of the fin. The latter is given by Q fin = −kA 
dx  x =0
d 
 = o h p ( − a )dx
L
Hence Q fin = −kA 2.226
dx  x =0
d 
Using equation 2.225, we can obtain  and hence show that:
dt  x =0

65
  
  −  
 Q fin = − k A o a −

hp
Sinh
hp
(L − x ) 2.227
 kA kA 
L
hp
 Cosh 
 kA  x =0

kAhp (o − a ) tanh

hp
Q fin = L 2.228
kA

 ( −  )dx and hence show that the same equation as

L
We may also use equation 2.225 to obtain a
o
 .
2.228 is obtained for Q fin

2.11.2 Fin Efficiency ηfin

Heat loss through the fin
 fin = 2.229
Heat loss through the fin if  = o throughout the surface of the fin
It’s a measure of how good or effective the fin is. The fins’ function is to conduct heat from or to the
base material. It is most effective, indeed ideally effective, if the fin temperature everywhere is equal
to the temperature at the fin base, o . In this case the profile of Fig 2.35 is a straight line parallel to
the x-axis. The higher the thermal conductivity k and the lower the fin thickness a, or the lower the
value of  , then the closer the profile approaches this straight line. Hence from 2.228 and 2.229,

hp
k Ah p (o − a ) tanh L
kA
 fin = 2.230
h p L (o − a )
hp
tanh L
kA
 fin = 2.231
hp
L
kA
tanh (L )
or  fin = 2.232
L

If a finned surface has n fins of uniform or constant cross-sectional dimensions, then the total heat
transfer rate from the finned surface = Heat transfer rate from unfinned part of the surface + Heat
transfer rate from the fins. Thus,

Q tot = h ( As − nA)(o − a ) + n hpkA (o − a ) tanh  L 2.233

 hp 
= (o − a ) h ( As − nA) + n hpkA tanh L 2.234
 kA 
Where As = total base surface area, without fins
A = base area per fin = cross-sectional area of the rectangular fin
n = number of fins on the base surface.

66
h on the base surface and on the fin surface have been assumed to be the same. This need not be so.

67
 PROBLEMS-2

P. Fundamentals of Conduction and Energy Equation of Conduction

P2.1 (a) What is the similarity between conduction in solids and in liquids?
(b) What is the difference between conduction in solids and liquids?

P2.2 The two dimensional transient conduction equation for an incompressible medium maybe
written as:
  q q 
C = u  −  x + v 
t  x y 

Where u  = net heat generation rate per unit volume from all sources. Other parameters have the usual
meanings.

By using Fourier’s particular law of conduction in the differential form, formulate the equation the
way it should be if the medium is

(i) Anisotropic and directionally homogenous

(ii) Anisotropic and heterogeneous
(iii) Isotropic and homogenous

P2.3 (a) Give one example of a material which may be approximated to type 2(i).
(b) Briefly indicate the possible effects of

(i) alloying
(ii) presence of gas pores,

on the thermal conductivity of a material.

P2.4 Derive the two dimensional equation of conduction through a medium in rectangular co-
ordinates. Assume the medium to be homogeneous, isotropic, stationary, of constant density,
and in steady state. Heat is being generated within the medium at the rate of u W m 3 but no
shaft work is being done.

P2.5 The inside surfaces of a rectangular oven are at 4000 C . One of the vertical sides is of
dimensions 45cm long x 30cm high x 5cm thick. Thermal conductivity of the oven material is
1.3 W mC . For this side only, determines (a) the temperature of the material, midway through
the thickness and (b) the rate of heat loss from the oven through the side. The outside surface
temperature of that side is 500C. Assume 1 - D steady state heat flow, as well as isotropic and
homogenous conditions.

P2.6 Repeat problem # 5 but with the following conditions:

• Inside surface temperature of the oven walls is 4000 C

• Heat transfer coefficient on the outside surface of the side is 2 W m 2 C .
• Ambient air temperature around the box is 300C

68
P2.7 The material of a rectangular slab is undergoing an endothermic reaction (ie reaction with heat
absorption or negative heat generation). The uniform heat absorption rate per unit volume of
the material in 1250 W m 3 . The slab is 10cm thick and has a thermal conductivity of 0.25
W mC . The sides of the slab are isothermal surfaces. One surface, (at x = 0) is at 450C, the
other is at 300C. Find:

(a) Temperature distribution in the slab,

(b) Location and value of the minimum temperature in the slab,
(c) Direction and magnitude of heat flux at the surfaces of the slab,

P2.8 The medium in Q4 is a flat plate of thickness 5cm, and extends infinitely lengthwise and
breathwise. The flat surfaces are isothermal surfaces. The heat transfer coefficient at one face
=10W m 2 0 C and at the other = 30W m 2 0 C . If u  = 1 KW m 3 , ambient temperature =
300C around the plate and k = 20W m 0 C , calculate the position of maximum temperature
within the plate.

P2.9 Derive the one dimensional transient equation for heat conduction in a cylinder for which
temperatures in the tangential and axial directions are constant for any given radial position.
The cylinder material is isotropic, homogeneous, of constant density and not in motion. The
heat generation rate is zero and shaft work is also zero.

P2.10 The cylinder of P9 is hollow and condensing steam at 1 atmosphere pressure flows through it,
thereby keeping the inside surface temperature at approximately 1000C. Find the rate of
condensation of the steam at steady state. The outside surface temperature is 95 0C. Inside and
outside radii are 25mm and 25mm respectively, and the length is 10m. Thermal conductivity
for the material = 380W m 0 C . Heat of condensation of steam at 1 atmosphere is 2256 .7 kJ kg
.

P2.11 A long and solid electrical copper wire of resistivity S Ωm, thermal conductivity k and diameter
d carries a current of I Amps. The heat transfer coefficient on the outside surface of the wire is
h W m 2 C and the ambient temperature is Ta. Steady state exists.

(a) Where is the maximum temperature in the wire located?

(b) Draw a sketch of the radial temperature distribution in the wire. What boundary condition can
be associated with this maximum temperature location?
(c) Determine the maximum temperature in the wire in terms of the given parameters.

(Note: The assumption of no temperature variation, in the wire, in the tangential and axial directions
is valid).

P2.12 Solar radiation is incident on the outer surface of the wall of a room at an intensity of 600W m 2
of the wall surface area. The ambient temperature adjacent to that surface is 300 C , while the
heat transfer coefficient on the surface is 3 W m 2 K . The wall is a rectangular slab of thickness
150mm and effective thermal conductivity of 1.2 W mK . If the heat transfer coefficient on the
inside surface of the wall is 1.5 W/m2K and the room is maintained at 200C, find the rate of
heat flow into the room through this wall, at steady state. The outer and inner wall surfaces,
each of 32m2 in area, are assumed isothermal.

69
P2.13 Show that for spherical co-ordinates, the general equation of conduction for a transient 2-D
case (temperature variation in r and ө), with heat generation in a stationary, incompressible,
isotropic and homogeneous medium is,

T u  a    2   1    
= + 2  r +  sin  
t C r  r  r  sin     

Find the steady state radial temperature distribution for a spherical shell of inside and outside
radii ri and ro, with no circumferential temperature variation and for each of the following sets
of conditions

(i) Inside and outside surface temperature are Ti and To and there’s no heat generation.
(ii) As in (i) with heat generation in the material at the rate of u W m 3 ;
(iii) Inside surface temperature is Ti, heat transfer coefficient on the outside surface = h,
ambient temperature on the outside is Ta and there’s no heat generation.

Q Electrical Analogy of Heat Transfer. Composite Bodies. Thermal Insulation

Q2.1 If it is assumed that there is perfect thermal contact and no heat generation at the interface of a
composite body, what deductions can you make about temperatures and heat flows at the
interface, and about heat flows through the materials of the composite at steady state?

Q2.2 What is contact thermal resistance and what can cause it? Sketch the temperature profile across
two composite rectangular slab of different thermal conductivities when,

i. There is no contact thermal resistance at the interface

ii. There is contact thermal resistance at the interface

One external face of the composite is at T1, the other is at T2; T1>T2 and k1>k2.

Q2.3 What is the electrical analogy of heat transfer? Illustrate this with the case of heat flow through
(i) a rectangular slab of thermal conductivity k, cross-sectional area A and thickness L (ii) a
hollow circular cylinder of thermal conductivity k, inside and outside radii of ri or ro, and length
L.

Q2.4 The materials of two composite rectangular slabs have thermal conductivities of 0.78 and 0.4
W/mK and corresponding thicknesses of 2.5 cm and 22.5 cm. The surface area of the slabs is
1m2. If the external surface of the thinner slab is at 350C and that of the thicker slab is at 200C,
calculate the interface temperature. Assume perfect thermal contact at the interface.

Q2.5 A double glazed window consists of two rectangular sheets of 3mm thick glass, separated by
an air gap of thickness 5mm. The heat transfer coefficient on the outdoor and room side
surfaces of the glazing are 6.0 and 2.0 W m 2 K , respectively. The width and height of each
glass sheet are 0.9 m and 1.5 m, respectively. Thermal conductivities of glass and air are 0.744
and 0.025 W/mK, respectively. Neglect radiation and convection heat transfer between the
glass sheets. Then,

(a) Sketch the thermal circuit diagram from room to outdoor temperatures, indicating the
thermal resistances;

70
 (b) Calculate the share of each thermal resistance as a percentage of the total;
(c) If the room and outdoor temperature are 200C and 00C, respectively, calculate the rate of
heat loss through the window;
(d) Repeat 5c if there is only one glass sheet, while the heat transfer coefficients remain the
same. Compare the results in 5c and 5d.

Q2.6 Two solid discs, each of diameter D, thickness t and thermal conductivity kD, are connected by
five solid rods. Each rod is of diameter d, length L and thermal conductivity k r (see sketch
below). Heat transfer from one disc to the other is through the rods only. The disc surfaces are
isothermal surfaces. Calculate,

(a) Total thermal resistance between the outer surfaces 1 and 2 of the discs;
(b) The corresponding overall heat transfer coefficient with respect to the surface area of the
disc.

1 2

D D 2

L
t t
d
Figure for Question No Q6

Q2.7 A refrigerant copper tubing, with inside and outside diameters of 9.5mm and 12.5mm,
respectively, is insulated with 3cm thick glass wool. The tube inside surface temperature is
-100C, while the ambient temperature is 300C. Heat transfer coefficient on the outside surface
of the insulation is 4.0 W m 2 K . Thermal conductivities of copper and glass wool are 380 and
0.038 W/mK, respectively. Calculate the rate of heat gain by the refrigerant, per meter of tube
length. Which thermal resistance is negligible?

Q2.8 The potential increase in annual energy sales, resulting from the insulation of steam piping in
a steam power plant, and the cost of the insulation, vary with the thickness of the insulation
according to the table below. What insulation thickness will yield the maximum net financial
benefit? Use the graphical and numerical methods and compare the results.

Thickness, 0.5 1 2 3 4 5 6 7 8 9 10
cm
Sales incr, -600 -1100 -1500 -1100 0 2000 4700 6500 7500 8200 8600
MU
COST, 225 370 730 1180 1720 2350 3070 3880 4780 5770 6850
MU

MU = Monetary Units

71
R. Extended Surfaces

R2.1 (a) What are fins and what is their primary function in heat exchangers?
(b) What shapes (in the cross-section and length) can fins take? Name five engineering
components known to you which are fitted with fins for heat transfer purposes and
identify the shapes of such fins.

R2.2 Show that for a single fin of uniform cross-sectional area A over its length L, thermal
conductivity k and perimeter P at any distance x from the fin base, the energy equation for
conduction through the fin is given by,

d 2 T hp
− ( − a ) = 0
dx 2 kA

where h = heat transfer coefficient on fin surface

Ta = ambient temperature

What boundary conditions may be used to solve the above equation? Hence show that
temperature distribution along the length of the fin is given by the equation:

 − a Cosh  (L − x )
=
o − a CoshL

where = (hp / kA)

o = temperature of base of fin

R2.3 Three rods, each of diameter 25mm and length 300mm, are attached to a hot plate at a
temperature of 1000C. Ambient temperature is 300C. The rods are made of copper, brass and
steel of thermal conductivities 380, 125 and 43 W/mK, respectively. The heat transfer
coefficient on the rod surfaces are assumed equal and of value 7.5 W/m2K.

(i) Draw up a table of rod temperature T vs distance from the plate x, for each rod.
(ii) Plot the graph of T vs x for each rod on the same graph sheet. Discuss and compare the
plots.

R2.4 By utilizing the law of conservation of energy, show that for a circular plate fin of uniform
thickness, which is attached to a cylindrical tube at its base, the energy equation for conduction
through the fin is given by,

1 d  d  2h
r − ( − a ) = 0
r dr  dr  kt

where h = heat transfer coefficient on fin surface

t = thickness of the fin
Ta = ambient temperature
r = radial position

72
If the surface temperature of the cylindrical tube is Ti, while the base and outer radii of the fin are ri
and ro, respectively, what boundary conditions may be used to get a particular solution for the radial
temperature distribution for the fin?

R2.5 A rectangular channel of outer dimensions 15mm x 300mm x 900mm long is fitted with
rectangular aluminum fins on each of the larger surfaces. The plane of each fin is perpendicular
to the longitudinal axis of the channel. The fins are arranged at a pitch of 15mm, with the first
and last fins located at 7.5mm from the closest end of the channel. Each fin is of dimensions
300mm wide x 30mm long x 1mm thick and has a thermal conductivity of 204 W/mK.

Air flows over the assembly in a direction parallel to the fins. Heat transfer coefficient on the
channel and fin surfaces is 10 W/m2K. Channel surface and the ambient temperatures are
respectively 1000C and 300C. Calculate,

(i) Heat transfer rate from the channel to the air, without fins;
(ii) Heat transfer rate though the fins only, to the air;
(iii) Total heat transfer rate to the air, with fins;
(iv) Fin efficiency.

CHAPTER THREE
73
 THERMAL RADIATION
3.1 INTRODUCTION
There are many kinds of radiation. Some examples are,
gamma radiation - radiation emitted by some radioactive substances
x-rays - radiation emitted from bombardment of metals by high
energy electrons
ultraviolet rays - radiation emitted through electric discharge in gases
thermal radiation - radiation emitted through thermal excitation of materials
All the above radiations are electromagnetic waves. Consequently, they obey the law:

f = c 3.1

where  = wave length, f = frequency and c = speed of light in vacuum

Thermal Radiation is in the range 0.1    100
Visible light is in the range 0.4    0.7

Fig 3.1 Radiation Spectrum

3.2 GENERATION AND EMISSION OF THERMAL RADIATION

All materials, which are not at zero absolute temperature emit thermal radiation, but none is emitted at
zero absolute temperature. The thermal excitation important in the emission of thermal radiation is
restricted to a thin surface layer of the material. For electrical conductors, the layer is about
0.000127cm thick. For electrical non–conductors, the layer is about 0.127cm thick.

The radiant heat flux emitted from a surface is a function, not only of the nature of the surface and the
temperature of the surface, but also of the wavelength at which the radiation is emitted. There is a type
of surface, of all surfaces, which emits at the highest radiant flux. Such a surface is an ideal emitter
and is called a black body emitter. The radiant energy flux per unit wavelength, which is emitted by an
ideal emitter, at the temperature T, within the wavelength range λ → λ + δ λ (ie at the wavelength
level  ), is given by eb , where

74
 2h c 2  −5
eb = ch
3.2
e K
−1

and h = Plank’s constant 6.626*10- 34 (J – sec)

c = speed of light in vacuum (m/s)
 = wavelength m
K = Boltzman constant, 1.38026*10 −23 J 0 K
 = Absolute temperature, 0
K
eb = Black body Monochromatic emissive power, W / m 3

eb is called the monochromatic emissive power of a black body emitter. Its spectral and temperature
varations are shown in Fig 3.2.

Fig 3.2 Spectral Thermal Energy Distribution

Multiplying 3.2 by  −5 gives that

2h c 2 ()
−5

eb  = 5
3.3
e ch k − 1

The right hand side of 3.4 is a function of the product  only.

 eb  5 = f () or eb =  f ()

5
3.4

It follows that a graph of eb  5 vs  will collapse the graphs of Fig 3.2 into a single curve as
shown in Fig 3.3

75
 Fig 3.3 Modified Spectral Distribution of Thermal Radiation

The total optimum radiant energy flux (over all wavelengths) at a temperature T is given by e b where
 3.5
eb =  eb d W m 2
0

It has been shown by proper integration of equation 3.3, that

eb =  4 W m 2 3.6

where  = Stephan–Boltzman constant = 5.6697x10-8 W/m2.K4

Equation 3.6 is called the Stephan–Boltzman law of thermal radiation. It is the particular law of thermal
radiation.

e b is the maximum possible total radiant energy flux, over all wavelengths, emitted by a body at
temperature T. It is called the emissive power of a black body emitter. Only a black body emitter can
emit thermal radiation at this flux.

Definition of Black Body Emitter

A black body emitter is one that emits radiant energy at all wavelengths, emits the highest thermal
radiation at any wavelength (and over all wavelengths) and whose emissive power is a function of
temperature only. No other radiator emits radiant energy more than a black body emitter.

76
Fraction ff
The Fraction ff is defined as the ratio of thermal energy emitted by a black body radiator within the
wavelength range 0 – λ, to the total thermal radiation emitted by the black body over all wavelengths,
at a given temperature.

 e  d
b

ie ff = o
 3.7

o
eb d


e  b d
= o
3.8
eb


  f ()d () 
5

e 
=   b  d = o
 3.9
o b 
e
  f ()d () 
5

o
For a given temperature

  2h c 2 ()−5 
  
  ch (K )  d ()  d () () (e
5 ch K
−1 )
 e −1 
ff = F () = = o

3.10
 −5 
 2h c ()   d () () (e )
2 ch K
−1
  ch (K )  d ()
5

o
 e −1 

Values of ff and λT have been calculated and tabulated, as shown in Table 3.1

Table 3.1 Fraction ff vs λT (µK)

 K 1200 2000 2800 3600 4800 6000 8400
ff 0.002 0.067 0.228 0.404 0.608 0.738 0.871
 K 10,000 14,000 20,000 50,000
ff 0.914 0.963 0.986 0.999

3.3 RADIATION PROPERTIES OF REAL SUBSTANCES

3.3.1 Emissivity
Real substances do not emit energy at the same rate as a black body emitter. Whereas a black body
emitter emits the highest energy possible at a particular temperature and wavelength, and over all
wavelengths, real substances will generally emit less than the highest energy possible at a particular
wavelength and temperature and may not emit over some wavelengths. This leads us to the definition
of an emission property called the emissivity of a surface.

77
Definition of Emissivity
It is the ratio of the energy emitted by a real surface to that emitted by a black body emitter at the same
temperature and wavelength, or wavelength range.

Emissivity = f (λ, T, Surface properties of material, direction) 3.11

For a given temperature and surface:

Emissivity is a function of λ - hence monochromatic emissivity,  
Emissivity is a function of direction – hence directional emissivity,  
Emissivity in the normal direction is called normal emissivity and is denoted as  n

3.3.1.1 Monochromatic Emissivity  

This is the ratio of the radiant energy flux emitted by a surface at a given temperature T, per unit
wavelength at the wavelength level  , to that emitted by a black body emitter at the same temperature
T, per unit wavelength at the wavelength level  .

In other words,   is the ratio of the monochromatic emissive power of a surface at temperature T to
the monochromatic emissive power of a black body emitter at the same temperature. Thus
  = e eb 3.12

(a) Effect of Wavelength λ on Monochromatic Emissivity   , at a given Temperature

  = f ( , , Surface, ) 3.13
For Polished Surfaces:
For a polished surface, the monochromatic emissivity decreases as the wavelength increases.

Fig 3.4 Effect of λ on   for a Polished Surface

Within the visible region, 0.4    0.7  ,   varies within the range 0.8     0.4
For high λ, say   8,   = 0.00365 r  , where r is the resistivity in  − cm .

For Refractory, Oxidized and Black Surfaces:

78
For refractory materials,   is high and increases with wavelength. Also   is high if the surface is
oxidized or black. The value is in the range of 0.8 - 1.0 for λ in the range of 2 − 4 . If it is oxidized
or black,   remains high as  changes, but if white,   decreases rapidly at lower wavelengths.

Selective Surfaces
Surfaces can be artificially manufactured to have certain predetermined radiation properties over
specified wavelength ranges. These are called selective surfaces. They are manufactured either

a) by coating the surface with a thin layer of a material which will emit energy only at specified
wavelengths and will totally internally reflect at other wavelengths, or
b) by imparting special geometries to the surface of the material.

A surface can be made to have low emissivities at high wavelengths and high emίssivities at low
wavelengths, or vice versa. The absorptivities may also change in like manner. Collector plates in high
efficiency solar thermal collectors are coated with selective surface materials which have high
absorptivities for low wavelength radiation (solar radiation) and low emissivities for high wavelength
radiation (lower temperature radiation emitted by the collector plate). This feature significantly
increases the collector efficiency.

(b) Effect of Temperature on Monochromatic Emissivity

The effect of temperature depends, as we may expect, on the wavelength range and on the particular
surface.

For Polished Surfaces

For polished surfaces,   decreases with temperature in the visible range: 0.4    0.7 . It is
insensitive to temperature in the range 0.7    1.5 and increases with temperature according to the
law     for   8 .

Fig 3.5 Schematic of the Variation of   with Temperature, T0K

For Refractory and Oxidized Surfaces

79
For refractory materials and oxidized surfaces,   is not very sensitive to temperature. For many
engineering surfaces,   can be assumed independent of temperature.

3.3.1.2 Directional Emissivity,  

Fig 3.6 Directional Emissivity Variations for Polished Metal and Oxidized Surfaces

The direction  is measured from the normal to the surface ( = 0 ) . For polished metals,   is low
and only slowly increases with  . For oxidized surfaces it is high and fairly constant for the lower
( )
range of  say   60 but decreases at an increasing rate as  approaches 900 − ie horizontal
0

direction.

3.3.1.3 Total Hemispherical Emissivity, ε

This is the ratio of the energy flux emitted by a surface at a given temperature over all wavelengths
and in all directions, to the energy flux emitted by a black body emitter at the same temperature of the
surface over all wavelengths and in all directions. It is thus equal to the ratio of the emissive power of
the surface to the emissive power of a black body emitter at the same temperature.


e
 e d
Thus = = o
3.13
eb eb

  eb d
= 3.14
o
eb

80
 
e 
or  =     b  d at the temperature of the emitter. 3.15
o  eb 
Now, recall that:


e  e
b d 

ff = o
=   b d = F () 3.8 and 3.10
0 b 
eb e
e 
 dF =  b  d 3.16
 eb 
e 
From 3.15 d =    b  d 3.17
 eb 
Substituting 3.16 into 3.17 gives that

d =   dF 3.18
1
  =    dF
0
 for all wavelengths and at the temperature T of the emitter 3.19

Variation of ε with T
For polished metal surfaces, total hemispherical emissivity or simply the emissivity of a surface  , is
low and proportional to temperature. Thus,

   for polished metal surfaces 3.20

For oxidized surfaces, refractory materials and non-conductors,  is generally high, and mostly in the
range 0.7    1.0 . Oxidation increases the emissivity of the surface. Since  is a function of
temperature, we have to denote the emmissivity of a surface at a temperature 1 as  1 .

3.3.2 Absorption, Reflection and Transmission of Thermal radiation

When radiation falls on a surface, it can either be reflected, absorbed or transmitted. Hence we may
define, for a surface, the -
absorptivity α = fraction of incident radiant flux that is absorbed
reflectivity ρ = fraction of incident radiant flux that is reflected
transmitivity  = fraction of incident radiant flux that is transmitted
It follows that  +  +  = 1 .
Just like emission, absorption is restricted to a small surface layer for many materials. For solid
electrical conductors, the layer is 0.000127cm thick. For solid non–conductors the thickness is
0.127cm.

As a result of this restriction, many engineering solids and liquids have their transmitivities equal to
zero. Hence  = 1 −  for these materials.
Reflection of radiant energy can be specular or diffuse or general. For specular reflection, the angle of
incidence is equal to the angle of reflection. For diffuse reflection, the incident radiation is reflected

81
equally or uniformly in all directions. For general reflection however, there is a dominant direction of
reflection while some radiation is reflected in other directions but at lower intensities (See Fig 3.7 a,
b, and c).

(a) Specular Reflection (b) Diffuse Reflection (c) General Reflection with
a Dominant Direction
Fig 3.7 Reflection of Incident Radiation
3.3.2.1 Spectral or Monochromatic Absorptivity

 = f ( ) hence   is the monochromatic absorptivity. This is defined as the fraction of incident

radiant flux per unit wavelength at the wavelength level
 (ie between the wavelengt hs  and  + d ), that is absorbed by the surface.
Example – Ice has a low   for 0.4    0.7m (visible spectrum,   → 0 ) and has a high   for
  0.7 (infra-red radiation,   → 1 ).   also varies with temperature. However, for many
engineering materials,   is approximately insensitive to λ and T.

3.3.2.2 Total Hemispherical Absorptivity

This is the fraction of incident energy flux, over all wavelengths and in all directions that is absorbed
by the surface. For short, it is also refered to as the total absorptivity, or the absorptivity of the surface.
The total absorptivity α varies with both the temperature of the absorbing material and the temperature
of the emitting material. The latter is so because, since the absoptivity is a function of wavelength and
the incident radiation’s wavelength distribution is a function of the temperature of the emitting surface,
it then follows that the absorptivity of a surface must depend on the temperature of the emitting surface.
Hence  depends on the temperature of the emitting and absorbing surfaces.
We therefore write the absorptivity of a surface as α12 which is interpreted as the absorptivity of a
surface at temperature T1 to radiant energy emitted from a surface at temperature T2.
Effect of Surface.   and α12 are low for polished metal surfaces and high for oxidized surfaces,
refractory materials and non-conductors. Also,   and α12 are simultaneously approximately constant
for many surfaces.

82
3.3.2.3 Effect of Temperature on Total Absorptivity, α

Fig 3.8 Effect of Temperature on Absorptivity

α is high and decreases as T increases for non-metals. It is low and increases as T increases for polished
metal surfaces.
3.3.3 Emissivity and Absorptivity. Kirchhohff’s Law

Consider a body at temperature T1, with emissivity and absorptivity of  1 and α12 respectively. It is
placed in a medium where it is uniformly irradiated with black body radiation from an enclosing
surface at temperature T2.
For a unit surface area of the body, net energy given up by the body to surface 2 is given as

q12 )net =  1 eb1 −  12 eb 2 3.21

If the body is in thermal equilibrium with the black body emitter, then q12 )net = 0 and 1 = 2 .

  1 eb1 =  11 eb1
3.22
 1 =  11 3.23
Equation 3.22 and 3.23 are also valid if surface 2 does not enclose surface1.

Thus if a body is in thermal equilibrium with a black body emitter, its emissivity is equal to its
absorptivity to radiation from the black body emitter. This is called Kirchhoff’s Law.
Kirchhoff’s Law applies also to monochromatic emissivity and absorptivity

83
Thus   ,1 =   ,11
But from equation 3.18 3.23a

 =    df where f = f ()
1
3.18
0
1
  1 =    df 1 at temperature 1 3.24
0

From Kirchhoff’s law,

1
 11 =  1 =    df 1
3.23b
o

We can deduce an expression for  12 from equations 3.23 b and 3.24 by saying that
1
 12 =    df * 3.25
0

Where T* is some temperature between T1 and T2. Usually T* is chosen as 1

2
(1 + 2 )
3.3.3.1 Emissivity & Absorptivity of Special Surfaces

(i) Black Body Surfaces

A black body absorber absorbs all incident radiation no matter the temperature of the surface from
which the radiant energy is emitted. Recall that such a surface emits at the highest radiant flux possible
for the given temperature. It follows that   =  11 =  12 =   =  1 = 1 for a Black Body Surface.

(ii) Gray Sufaces

Gray surfaces are simply defined as those surfaces for which   =   and for which the values are
independent of λ. However, the values are less than 1. From Kirchoff’s law,  1 =  11 which also
applies monochromatically.

   =   =  1 =  11 =  12 =  =   1 for gray surfaces 3.26

In analyzing radiation to or from a general surface, we should take into account the variations of  
and   with  . Thus, in order to determine the hemispherical emissivities and absorptivities, we
should make use of equations 3.23, 3.24 and 3.25. Fortunately, many engineering surfaces are
sufficiently gray, so that we can make use of the gray surface relationship, namely,   =   =  =  ,
independent of  .
Example 3.1

A glass sheet used for flat plate solar collector design is transparent to solar radiation in the range of
0.2    2.0 and is opaque at all other  . If the sun is considered as a black body emitter at 60000K,
what is the transmittivity of the glass?

Solution: We need to find the average transmitivity over the entire wavelengths.

1 = 0 for 0    0.2 

84
  2 = 1 for 0.2    2.0 
 3 = 0 for 2.0    


K

Fig 3.9 Example 3.1

1
Now,  =    dff ( ) = Area within   vs ff  curve
0

Using graphical or numerical integration, we obtian

 =   i ( ff i  − ff i −1  ) for i = 1, 2, 3

( ) ( ) ( )
i

=  1 ff 1  − ff o  +  2 ff 2  − ff 1  +  3 ff 3  − ff 2 
( )
= 0 + 1 ff 2  − ff 1  + 0 = ff 2 − ff 1
1  = 1200m 0 K  ff  1 
= 0.002
 2  = 12,000m 0 K  ff  2 
= 0.945
 = (0.945 − 0.002 ) = 0.943
Alternatively
Fraction of black body radiation at 6000K; in the range:
 = 0 → 2.0 = ff 1200 K = 0.002
 = 0 → 2.0 = ff 12000 K = 0.945
 = 0 →  = ff K = 1.0
Fraction of black body radiation transmitted by the glass in the range:
 = 0 → 0.2 = 0 * 0.002 = 0
 = 0.2 → 2  = 1 * (0.945 − 0.002 ) = 0.943

85
  = 2 →  = 0 * (1 − 0.945 ) = 0
Total radiation transmitted as fraction of incident black body radiation = 0 + 0.943+0 = 0.943
This is the transmittivity of the glass.

3.4 LABORATORY BLACK BODY

Fig 3.10 Hohlraum

Absorption by a Hohlraum
Imagine radiation – say black body radiation e b , incident on the inside surface of the hohlraum through
a small hole. We assume that the surface is such that the radiation undergoes specular reflection. Since
the material of the hohlraum has a reasonable thickness, the transmittivity = 0
Let the absorptivity of the inside surface be  . Thus  = 1 − 

Energy leaving an inside surface, per unit area, after 1 reflection = eb − eb = eb (1 −  ) = eb 
Energy flux leaving an inside surface after two reflections = (1 −  )eb −  (1 −  )eb
= eb (1 −  ) = eb  2
2

Energy flux leaving an inside surface after n reflections = eb  n

Energy flux leaving an inside surface after  reflections = eb   = 0 (since   1)
Hence, the total energy incident on the inside surface of the hohlraum is completely absorbed.
Therefore the hohlraum behaves like a Black Body absorber.

Emission by a Hohlraum
Since the surface of the hohlraum is at a non-zero absolute temperature, it will be emitting radiant
energy. Let the emissivity of the hohlraum internal surface be  . Some of the energy emitted from any
region of the surface will suffer multiple reflections from the surface due to the fact that the surface is
concave. Energy leaving the hohlraum through the small hole will therefore consist of energy that has
suffered no reflection, 1 reflection, 2 reflections, ---------n reflections,-----------an infinte number of
reflections.

86
Therefore energy leaving through the hole per unit of inside surface area:
After zero reflection = eb
After 1 reflection =   eb
After 2 reflections =   2 eb
After n reflections =   n eb
After  reflections = 0 (  1)
:. Total energy leaving through the hole per unit of inside surface area

(
= eb 1 +  +  2 + − − − − − − +  n + − − − − − + 0 )
e b
=
1− 
eb
= = eb since from Kirchhoff’s law,  = 


NB:  and  refer to the same surface at the same uniform temperature.
Thus, the emissive power of the hohlraum is the same as for a black body emitter. Therefore, we
conclude that the hohlraum emits and absorbs thermal radiation like a black body surface. In fact, any
enclosure with a small hole will behave like a black body surface.

3.5 EXCHANGE OF RADIANT ENERGY BETWEEN SURFACES

3.5.1 Types of Surfaces:

Surfaces may be,

Black Body or Non-Black Body
Gray or non-gray
Diffused or Specular
Different techniques apply for the calculation of radiation exchange between different surfaces,
including techniques for:
Black Body Surfaces
Gray and diffusely emitting and reflecting surfaces
Quasi-gray (gray over a limited λ range) and diffusely emitting and reflecting surfaces
Partially or completely specular surfaces
No matter the nature of the surfaces exchanging thermal radiation, the concept of View Factor will be
required for determining the amount of radiant energy exchanged.

3.5.2 View Factor

Consider two elemental surfaces of areas A1 and A2 at positions 1 and 2, as in Fig 3.11. A1 and A2
are elements of larger areas A1 and A2 , respectively.

87
 Fig 3.11 Interception of Radiant Energy from A1 by A2

Not all the radiant energy emitted by A1 is intercepted by A2 and vice versa. n1 and n 2 are normals
to A1 and A2 , respectively. 1 and  2 are angles made by 1-2 with the normal n1 and 2-1 with the
normal n2, respectively.

The view factor, F12, is defined as the proportion of the total energy leaving surface 1 which is
intercepted or received by surface 2. With respect to radiant energy from surface 2 to surface 1, we
speak of the view factor F21.
Consider that energy is leaving surface 1 in all directions, travelling in straight lines. The amount of
energy leaving surface 1, towards surface 2, per unit time, E 12 , is proportional to the projection of
A1 on a plane perpendicular to the direction 1 – 2.

E12  A1Cos1 3.27

The amount of this energy which is intercepted by A2 is proportional to the projection of A2 on a
plane perpendicular to the direction 2 – 1 and also is inversely proportional to the square of the distance
between 1 and 2. Hence,

 (E12 )  
 1 
2 
(A1Cos1 )(A2 Cos2 ) 3.28
r 
 A Cos 
  (E 12 ) = I  A1Cos1 . 2 2 2  3.29
 r 

I , which is the constant of proportionality, is called the Intensity of Emitted Radiation in the direction
A2 Cos2
1 to 2 (ie in the direction 1 ) . The term is the solid angle subtended by the area A2 at
r2
location 1 and is equal to v1 .

Thus

88
  (E 12 ) = I A1Cos1 v1 3.30
  E 12 
or   = I  Cos1 3.30a
v1  A1 

In the limit as A1 → 0 .

d  dE 12 
  = I  Cos1 3.31
 dA 
dv1  1 

I , generally, is a function of  , but we can consider it to be a constant, independent of  .

The above equation, 3.31, then implies that the radiant energy flow rate from surface 1 to surface 2,
per unit surface area of 1, per unit solid angle at 1, is proportional to Cos 1 . This is the Lambert’s
Cosine Law.
We can rewrite equation 3.31 in the form,

d  dE 12 
 
 dA Cos  = I  3.32
dv1  1 1 

The intensity of emitted radiation I  is seen to physically mean the radiant energy flow rate from 1 to
2, per unit projected area of 1, per unit solid angle at 1. For many surfaces, I  is approximately
ίndependent of  .
Value of I  for a Black Body Emitter

As A1 → 0 , equation 3.29 may be written as

dA2 Cos 2
( )
d dE 12 = I  dA1 Cos1 3.33
r2
Consider a black body surface of area A1 which is totally enclosed by a hemisphere of radius r. Let
A1 lie on the diametral plane of the hemisphere, and 1 at the center of the hemisphere, Fig 3.12.
Take the hemisphere as surface A2. Consider, as A2 the elemental circular ring on the hemisphere at
the angular position  from the normal to A1 . The radius of this ring is r Sin and the width is rd .

89
 Fig 3.12 Radiant Flux from Elemental Area to Enclosing Hemisphere

Location 2 is a point on the circular elemental ring. n1 is the normal to A1 and n 2 is the normal to
A2 at location 2. The normal to the elemental ring at the point 2 is coincident with the line 2-1 which
is a radius of the hemisphere. Hence the angle between the line 2-1 and the normal to the elemental
ring is zero.
Consider radiant energy flow from point 1 on A1 to A2 . Comparing Fig 3.11 to Fig 3.12,

1 = 

 2 = 0  Cos  2 = 1
A2 = 2r Sin r

Substituting the above in 3.33 gives

2r 2 Sin d
d (dE 12 ) = I  dA1 Cos . 3.34
r2
 dE 
 d  12  = 2 I  Sin Cos d 3.35
 dA1 
or dq12 = 2 I  Sin Cos  d 3.36
 2
 dq 12
in all directions
= I 
o
Sin2d 3.37


 1  2
or q1 = I − Cos2  3.38
 2 0
q1 = I  3.39

Since surface 1 is a black body emitter, then q 1 is the emission flux from a black body surface in all
directions = eb1

90
  eb1 = I b1 3.40
eb1  4

or I b1 = = 1
3.41
 
Expression for View Factor, F12
Let surface A1 of Fig 3.11 be a black body surface. From the derivation of I  for a black body surface
and from 3.33, it follows that for such a surface,

eb1 Cos1Cos 2 dA1dA2

d (dE12 ) = 3.42
 r2
Integrating 3.42 over the areas A1 and A2 , we obtain that,

Cos1Cos 2 dA1 dA2

 d (dE12 ) =
eb1

A2 A1

 A2 A1 r2
3.43

e  Cos1 Cos 2 dA1 dA2 

 E12 = b1    3.44
  A2 A1 r2 

E12 is the total radiant energy leaving surface 1 that is intercepted by surface 2, per unit time. But total
radiant energy leaving black body surface A1 in all directions is A1eb1 . And by difinition F12 is given
by,
E 12
F12 = 3.45
A1 eb1

We substitute for E12 from 3.45 in 3.44 to obtain that,

1 Cos1 Cos 2 dA1 dA2

F12 =
A1 
A2 A1 r2
3.46

The right hand side of 3.46 is a function of geometrical parameters only. Thus even though it was
derived by considering radiation from a black body surface, it does not contain any property of the
surface and so is independent of the nature of the surface. Hence it is valid for all types of surfaces.

3.5.2(i) View Factor for a Small Area Enclosed by a Hemisphere (see Fig 3.12)

Substituting for A1 , A2 , 1 and  2 in 3.46 from Fig 3.12, we obtain that,

1 Cos1Cos 2 A1A2
F12 =
A1  
A 2 A1 r2
3.47

1  Cos .1.2r 2 Sin d  

F12 =  A1 since A1 is an elemental area 3.48
A1   r2 

91
 
1  2 A = − 1 Cos2 
2

A1  0
= Sin2 d   2  3.49
 1
 0

or F12 = 1 3.50

Conceptually, we can see that F12 should be 1, since all the radiation leaving the surface A1 will be
intercepted by the hemisphere which encloses it. The enclosing surface does, in fact, not have to be a
hemisphere. See Fig 3.13.

Fig 3.13 Enclosed Surface is Flat or Convex

We can conclude that for any flat or convex surface 1, which is completely enclosed by a larger surface
2, F12 = 1 .
For two parallel and infinitely large surfaces, the view factor is 1. Hence we can say that for two
parallel and very large surfaces, F12 ~ 1. Also, For two large flat surfaces joined at a common edge and
inclined to one another at a small angle δ, F12 ~ 1.

3.5.2(ii) View Factor Tables

The expression defining F12 as a function of 1, 2, A1, A2 and r can always be used to obtain the value
of F12 for any pair of surfaces. However, this exercise can be tedious for certain surfaces at various
relative positions to one another. Useful values of F12 have been calculated and tabulated or plotted in
texts and reference books for pairs of surfaces at common orientations. Table 3.2 and Figs 3.14 a and
b are examples.

92
Table 3.2 View Factor F12 for Two Rectangular Plates with a Common Edge, Inclined at 900

Reference: (Kothandaraman and Subramanyan, 1977)

93
Fig 3.14a View Factors F12 for Opposite Equal Rectangles

Reference: (Mills A.F., 1999)

94
 Fig 3.14b View Factors F12 for Opposite Circular Discs
Reference: (Mills A.F., 1999)

Relationship Between F12 and F21

1 Cos1 Cos 2 dA1 dA2
F12 = 
A1 A2 A1 r2
as derived earlier 3.51a

Similarly,
1 Cos1 Cos 2 dA1 dA2
F21 =
A2 
A1 A2 r2
3.51b

1 Cos1 Cos 2 dA2 dA1

 A1 F12 = A2 F21 = 
 A1 A2 r2
3.52

Hence we obtain the important conclusion that for any two radiating surfaces, 1 and 2

A1 F12 = A2 F21 3.53

95
3.5.3 Radiant Energy Exchange Between Two Black Body Surfaces

Fig 3.15 Two Black Body Surfaces Exchanging Thermal Radiation

For the two black body surfaces, rate of energy release by black body surface 1, which is intercepted
by black body surface 2, is given by

E12 = eb1 A1 F12 3.54

This energy is totally absorbed by 2 since 2 is a black body surface. Rate of energy release by black
body surface 2 which is intercepted and totally absorbed by 1 is given by,
E 21 = eb 2 A2 F21 3.55
Hence, the net radiant energy transfer rate between the two black body surfaces is given by

E12 , net = E12 − E 21 3.56

= eb1 A1 F12 − eb 2 A2 F21 3.57
= A1 F12 (eb1 − eb 2 ) 3.58
= A2 F21 (eb1 − eb 2 ) 3.59
ie E12 ; net = Q 12 = A1 F12 (14 − 24 ) 3.60

3.5.4 Radiant Energy Interchange in an Enclosure with n Black Body Surfaces

96
 Fig 3.16 Enclosure with Black Body Surfaces

We may differentiate the surfaces according to their temperatures. For an enclosure, all the radiation
n
from a surface, say surface 1, will be intercepted by the surfaces of the enclosure. Hence F
j =1
1j = 1.

From previous equations, net energy transfer between surface 1 and surface 1 is
E1−1,net = Q11 = A1 F11 (eb1 − eb1 ) or A1 F12 14 − 14 = 0 ( ) 3.61

Net energy transfer rate between surface 1 and surface 2 is

E1−2,net = Q12 = (
A1 F12 (eb1 − eb 2 ) = A1 F12 14 − 24 ) 3.62

Net energy transfer rate between surface 1 and any surface j is given by

Q1 j = A1 F1 j (eb1 − ebj ) = A1 F1 j 14 − j4 ( ) 3.63

Net energy transfer rate between surface 1 and all the surfaces of the enclosure, ie net heat flow rate
from surface 1 to all the surfaces is given by

 A1 F1 j (eb1 − ebj ) =  (14 −  j4 )

n n
E1,net = Q 1 =
j =1
AF
j =1
1 1j 3.64

 (14 −  j4 )
n
or Q 1 = A F
j =1
j j1 3.65

We recall that from equation 3.53, for the pair of surfaces 1 and j ,

A1 F1 j = A j F j1 3.66

Also, for the surface 1 in an enclosure

n

j =1
F1 j = 1 3.67

Equations 3.63 to 3.67, written with reference to surface 1, are also applicable to any other surface in
the enclosure. The relevant equations are 3.63, 3.64, 3.66 and 3.67. Thus, for surface 2, we have that

Q 2 j = (
A2 F2 j 24 − j4 ) 3.63a

97
 n
Q 2 =  Q
j =1
2j 3.64a

A2 F2 j = Aj Fj2 3.66a
n

F
j =1
2j = 1 3.67a

Using these equations, we may therefore write that for any given surface ί in the enclosure,

Q ij = (
Ai Fij  i4 − j4 ) 3.63b
n
Q i =  Q
j =1
ij 3.64b

Ai Fij = A j F ji 3.66b
n

F
j =1
ij = 1 3.67b

Equations 3.63b, 3.64b, 3.66b and 3.67b constitute the equations that need to be written for each
surface and each pair of surfaces and solved, simultaneously, to obtain the unknown quantities for the
enclosure. The number of Q ij ' s is given by the number of permutations of n members, taking two at
a time (where a number is allowed to be repeated). These are, for instance: 11, 21, 31, -----n1; 12, 22,
32, -----n2; --------;1n, 2n, 3n ----- nn. There will be nxn = n2 such permutations. The same is true of
the number of Fij ' s . There are thus n 2 of Q ij ' s , n 2 of Fij ' s and n each of Ai , i and Q i . However,
not all the Q ' s are unknown or independent variables. From equation 3.61, we deduce that
ij

Q ii = 0 3.68
There are n such Q ii ' s
Also, from 3.63b, Q ij = − Q ji 3.69

For each Q ij there is a Q ji . Consequently, the number of unknown Q ij ' s is

2 2
(n
)
n − n = (n − 1)
1 2

n n
or C 2 . There will be the same number, C 2 , of equations of type 3.63b. There will be n equations
from each of 3.64b and 3.67b and
n
(n − 1) or n C 2 equations from 3.66b. As shown in equations 3.70
2
to 3.72 below, the number of unknowns is greater than the number of equations. Hence for a solution
of equations 3.63b, 3.64b, 3.66b and 3.67b, some values of the unknown parameters will have to be
given. The number of values to be specified, Ns , is given by:

Ns = Total number of unknowns – Total number of equations

= ( n C 2 + n 2 + 3n ) − ( n C 2 + 2n + n C 2 ) 3.70
= n 2 + n − n C2 = n 2 + n −
n
(n − 1) 3.71
2
Ns =
n
(n + 3) 3.72
2

98
The specified quantities are often given as some of the values of surface area Ai , temperature i and
view factor Fij .
Example 3.5.4 (i)
The roof of a traditional hemispherical oven, of 1m radius is heated to a temperature of 227 0C. Bread
at 650C is baking on the flat floor of the oven. If the bread and internal oven surfaces are assumed to
be black body surfaces, what is the radiant heat transfer rate to the bread?

Figure for Example 3.5.4 (i)

Solution

This is a two-surface enclosure. The parameters of the problem are

  
Q1 , Q2 , Q12 , A1 , A2 , 1 , 2 , F11 , F12 , F21 and F22 (ie 11 parameters). From equations 3.63b,
3.64b, 3.66b and 3.67b the applicable equations are:

Q 12 = A1 F12 (14 − 24 ) 3.73

Q = Q
1 12 3.74
Q 2 = Q 21 (= −Q12 ) 3.75
F11 + F12 = 1 3.76
F21 + F22 = 1 3.77
A1 F12 = A2 F21 3.78

There are thus six equations. The statement of the problem has specified five parameters, thus:

A1 and A2 , from the given radius of the hemisphere

1 and 2 , from the given temperatures
F22 = 0 , from the statement that the base of the hemisphere is flat, so it does not see itself

99
From 3.77, we obtain that F21 = 1 . Note that if surface 2 were concave, then it will see itself, F22  0
and F21  1 .
A2 r 2
From 3.78, F12 = = = 0.5
A1 2r 2
From 3.76, F11 = 1 − 0.5 = 0.5
From 3.73, Q 12 = 2 *12 * 0.5 * 5.6697 *10 −8 (5004 − 3384 ) W
= 8807.6 W

Q 12 above is the radiant heat transfer rate to the bread. We may also determine the net heat flow rate
from the hemispherical surface, Q , and from the floor Q . Since it is a two-surface enclosure,
1 2

equation 3.74 gives Q 1 as Q 12 = 8807.6 W . Equation 3.75 gives Q 2 as − Q 12 = − 8807.6 W . The

negative sign means that net heat flows into the floor.

Example 3.5.4 (ii)

Electrical heating wires keep the horizontal roof of an oven at a temperature T1. The four vertical walls
are at the same temperature T2. Materials being heated constitute the horizontal oven floor at T3.
Temperatures are in 0K. Write the set of equations that need to be solved, simultaneously, in terms of
symbols, in the analysis of the thermal radiation exchange problem. Assume that the surfaces are Black
Body emitters.
Solution
The four vertical sides are at the same temperature. They are also contiguous. We may consider them
as constituting a single surface at T2. The symbols introduced earlier in this text may be used. From
equations 3.63b, 3.64b, 3.66b and 3.67b, we may write:

Fig 3.17 Sketch of Example 3.5.4 (ii)

Q 12 = A1 F12 (eb1 − eb 2 ) where ebi = i4 3.79

Q 13 = A1 F13 (eb1 − eb3 ) 3.80
Q = A F (e − e )
23 2 23 b2 b3 3.81

Q1 = Q12 + Q13 3.82

100
 Q 2 = −Q12 + Q 23 3.83
Q 3 = −Q13 − Q 23 3.84

F11 + F12 + F13 = 1 3.85

F21 + F22 + F23 = 1 3.86
F31 + F32 + F33 = 1 3.87

A1 F12 = A2 F21 3.88

A1 F13 = A3 F31 3.89
A2 F23 = A3 F32 3.90

The above equations, 3.79 to 3.90, constitute the necessary equations. Use has been made of the fact
that Qii = 0 and Q ji = −Qij . There are 21 parameters and 12 equations. Nine of the parameters will
have to be specified, directly or indirectly, in order that a specific solution may be obtained.

3.5.5 Electrical Analogy for Radiant Energy Exchanges between Black Body Surfaces
For any two black body surfaces, E = Q = A F (e − e )
12 − , net 12 1 12 b1 b2 3.91
Comparing 3.91 with the electrical energy flow equation of 2.125, given in 3.92 in the form,

H
I = 3.92
R
We find that
I is equivalent to E12,net = Q 12
H is equivalent to eb1 − eb 2 (ie V  eb )
1
R is equivalent to
A1 F12

The thermal current or heat transfer rate is Q 12 . The radiation heat transfer potential for the black body
surface is e b and the radiation heat transfer resistance for a pair of black body emitters is then deduced
to be,
1
Rth  3.93
A1 F12
 eb = E12,net Rth 3.94a
1
or ∆𝑒𝑏 = 𝑄̇12 𝑅𝑡ℎ, where Rth = 3.94b
A1 F12

Equation 3.94b is the electrical analogy of radiant heat transfer (for black body surfaces). The
equivalent thermal circuit diagram for net transfer of thermal radiation between two black body
surfaces is illustrated in Fig 3.18.

101
 Fig 3.18 Thermal Circuit Element for Two Black Body Surfaces

For a 3 – surface black body enclosure, the equivalent thermal circuit diagram is as illustrated in Fig
3.19.

Fig 3.19 Thermal Circuit Diagram for a 3-Surface Black Body Enclosure

For a 4 – surface black body enclosure, the diagram is as sketched in Fig 3.20. Note that Q 1 is the total
net radiant energy flow rate out of surface 1. Q , Q and Q are net radiant heat exchanges between
12 13 14

surface 1 and surfaces 2, 3 and 4. Logically Q 1 = Q 12 + Q 13 + Q 14 . Similar relationships exist for each
of the other surfaces 2, 3, and 4.

Note also that the direction of Q i is out of the surface, ie from the surface to the surface node. The
directions chosen for net radiant heat flow rate between surfaces i and j, ie whether the heat flow
rate is in the sense of Q ij or Q  , is not important. Any one can be chosen. However, the
ji
electrical analogy equation representing heat flow rate through the corresponding resistance
must be consistent with the direction chosen. At the end of the analysis, the correct direction
of Q ij will be indicated by its sign.

102
 Fig 3.20 Thermal Circuit Diagram for a 4-Surface Black Body Enclosure

For an n − surface enclosure, there are n C 2 resistance paths or (lines).

3.5.6 Analysis of the Thermal Radiation Network (Just as for an Electrical Network)

The rules regarding the heat flows, which are applicable in the analysis of thermal networks are
identical to those regarding electrical energy flows in electrical network analysis. Consider a three
Surface Enclosure, Fig 3.19. Drawing from the rules regarding the analysis of electrical networks, the
set of descriptive equations for the thermal network, is based on the following relationships:

(1) Sum of heat flow rate into a node = 0

Q 1 − Q 12 − Q 13 = 0 3.95
Q 3 + Q13 + Q 23 = 0 3.96
Q 2 + Q 12 − Q 23 = 0 3.97
These are the same as the equations given by 3.64b.
n
( Qi =  Q ij for n − surface enclosure, where Q ii = 0 and that Q ji = −Q ij )
j =1

(2) For heat flow rate through each thermal resistance, eb = Q Rth
This can be written as 𝑄̇ = ∆𝑒𝑏 ⁄𝑅𝑡ℎ . Thus,
Q12 = A1 F12 (eb1 − eb 2 ) 3.98
Q 13 = A1 F13 (eb1 − eb3 ) 3.99
Q = A F (e − e )
23 2 23 b2 b3 3.100
These are the same as the equations given by 3.63b
( Q ij = Ai Fij (ebi − ebj ) and ebi = i4 , for n – surface enclosure)

103
 n
(3) Geometric Relationships F j =1
ij = 1 and Ai Fij = A j F ji , ie equation 3.67b and 3.66b,

respectively.
F11 + F12 + F13 = 1 3.101
F21 + F22 + F23 = 1 3.102
F31 + F32 + F33 = 1 3.103

A1 F12 = A2 F21 3.104

A1 F13 = A3 F31 3.105
A2 F23 = A3 F32 3.106

It may be seen that equations 3.95 to 3.106 are the same as 3.79 to 3.90
3.5.7 Radiant Energy Exchange between Gray Surfaces

Fig 3.21 Thermal Radiation Exchange between Two Infinitely Large Parallel Gray Plates
Consider two gray, diffuse, parallel, infinitely large flat plates, 1 and 2, Fig 3.21, which are exchanging
thermal radiation. The view factor 𝐹12 𝑜𝑟 𝐹21 is equal to 1, radiation leaving one surface is fully
intercepted by the other surface. Assuming thick plates, radiation emitted by a surface will be partly
absorbed and partly reflected back upon interception by the other surface. There will be no radiation
transmitted. It will undergo infinite reflections and absorptions before it is fully absorbed. Let us
consider radiant energy flows on the material side of each surface. For surface 1, net radiant energy
outflow per unit area is given by:

( )
E1,net / A1 = Q1 / A1 =  1eb1 − 1 2 eb 2 + 1 1  2 2 eb 2 + 1 12  22 2 eb 2 ...

− (1  2 1eb1 + 1 1  22 1eb1 + 1 12  23 1eb1 + ....)

3.107
)
 
=  1eb1 − ( 1 2 e b 2  (1  2 ) +  2 1 1 eb1  (1  2 )n
n
3.108
n =0 n =o

=  1eb1 −  1 2 eb 2 +  2 1 1eb1  (1  2 )
n
3.108a
n =0

104
 
Since 1  2  1,  (  ) = 1 (1 − 1  2 )
n
1 2
n =0
For gray surfaces  =
For thick plates that are also gray  = 1 −  = 1 − 
Equation 3.108a becomes,
1
E1,net / A1 = Q 1 A1 =  1eb1 −  e +  (1 −  2 )eb1 
(1 − (1 −  1 )(1 −  2 )) 2 b 2 1
 e (1 − (1 −  1 )(1 −  2 )) −  1 ( 2 eb 2 +  1 (1 −  2 )eb1 )
= 1 b1
1 − (1 −  1 )(1 −  2 )
 e ( +  2 −  1 2 ) −  1 ( 2 eb 2 +  1eb1 −  1 2 eb1 )
= 1 b1 1
 1 +  2 −  1 2
  (e − e )
= 1 2 b1 b 2
 1 +  2 −  1 2
A (e − e )
or E1, net = Q 1 = 1 b1 b 2 3.108b
1 1
+ −1
1 2
This method can be used to determine E 2,net = Q 2 as well.
The method can be long and involved for a multi-surface enclosure. A more elegant technique has
been developed. This later method considers the phenomenon occurring on the space side of the surface
(as opposed to the material side of the surface)

It recognizes that energy leaving a surface is the sum of energy freshly emitted, energy reflected once,
two times, -------------n times, --------- ∞ number of times (the last of which is zero)

Similarly, energy arriving at a surface consists of energy freshly emitted, energy reflected once, two
times, ------------n times, ------------ ∞ number of times (the last of which is zero).
Let the sum of energy flux leaving a surface ί be rί W/m2 (ie both freshly emitted and multiply reflected;
rί is called the Radiocity of the surface ί).
Let sum of energy flux arriving at a surface ί = gί. [rί and gί are not necessarily equal]. See Fig 3.22

105
 Fig 3.22 Leaving and Incident Thermal Radiation on a Gray Surface

Then, we can write that,

ri =  i ebi +  i g i 3.109

Net energy rate leaving the surface ί is given by,

E i ,net = Ai ri − Ai g i 3.110

We can eliminate g i from 3.109 and 3.110 and obtain

E i ,net = Ai ri − Ai (ri −  i ebi ) /  i 3.111

 A 
E i ,net = Q i =  i i (ebi − ri ) 3.112
1− i 
where E i ,net or Q i = net energy flow rate leaving surface i

The Electrical Resistance Analogy can be used to interpret equation 3.112, thus –
The net energy leaving the material’s surface ί appears to be energy flowing through a surface
1− i 
resistance   between the potentials ebi and ri , before finally leaving that surface. It is identical
 Ai  i 
to a battery with an electromotive force, emf = ebi − ri and an internal resistance (1 −  i ) / Ai  i . This
is represented by Fig 3.23.

106
 Fig 3.23 Internal Radiation Resistance of a Gray Surface, ί

The net radiant energy exchange between two gray surface ί and j is given by

E ij ,net = Q ij = Ai Fij ri − A j F ji r j
3.113
But Ai Fij = A j F ji
 = AF r −r
E ij ,net = Qij i ij i j ( ) 3.114

From the electrical resistance analogy, we may deduce from 3.114 that in finally going from surface ί
1−  
to surface j, (after passing thro’ the surface or internal resistance   ), the net radiant energy
i


 i i 
A
 1 

appears to travel from potential ri to potential r j across the resistance 

A F 
 i ij 
This is represented in Fig 3.24

Fig 3.24 Net Radiant Energy Flow Between Gray Surfaces ί and j

Aj j
At j , E j ,net , = Q j = (ebj − r j ) 3.115
1−  j

This is represented in Fig 3.25.

107
 Fig 3.25 Internal Radiation Resistance of Gray Surface j

Thus for radiant energy interchange between two gray surfaces 1 and 2, the electrical resistance
analogy yields the circuit diagram of Fig 3.26, obtained by combining Figs 3.23, 3.24 and 3.25.

Q ij

Fig 3.26 Surface (or Internal) and External Resistances for Radiant Heat Flow Between Two
Gray Surfaces.

This is equivalent to radiant energy interchange between potentials eb1 and eb 2 , across a resistance

1 − 1 1 1− 2
Rth = + +
A1 1 A1 F12 A2  2 3.116

1  1 1  A  1 
=  +  − 1 + 1  − 1
A1  F12  1  A2   2  3.117
So that
A1 (eb1 − eb 2 )
E12,net = Q 12 = 3.118
1 1  A  1 
+  − 1 + 1  − 1
F12   1  A2   2 

Comparing this equation with the equivalent black body equation Q 12 ( )

BB
= A1 F12 (eb1 − eb 2 ) ,
we find that the equations are identical with the actual view factor F12 replaced by an equivalent view
factor f12 in the gray body equation,

1
where f12 = 3.119
1 1  A  1 
+  − 1 + 1  − 1
F12   1  A2   2 

108
For two parallel, flat and infinitely large or very large surfaces of equal area, as in Fig 3.21, 𝐹12 =
1; 𝐴1 = 𝐴2 and 𝐸1,𝑛𝑒𝑡 = 𝐸12,𝑛𝑒𝑡 . Inserting these values in 3.118, we obtain that,

A (e − e )
E1,net = E12,net = 1 b1 b 2 3.119a
1 1
+ −1
1 2

which is the same as equation 3.108b, obtained using a different method.

For an n-surface enclosure, equation 3.112, with Fig. 3.23, are valid for each surface; and equation
3.114, with Fig. 3.24, are valid for each pair of surfaces. For 3-Surface and 4-Surface enclosures, the
equivalent thermal circuit diagrams are therefore as sketched in Fig 3.27. As for the network diagram
Q i is from the surface to the surface node. The radiant
for the black body enclosure, the direction of

heat flow rate between ri and rj may be in the sense of Q ij or Q ji . The correct application
of the electrical analogy equation will yield the actual direction of the net heat flow rate
at the end of the analysis.

109
 (b) 4-Surface Gray Enclosure

Fig 3.27 Equivalent Circuit Diagrams for Net Thermal Radiation Exchange Between Gray
Surface Enclosures.

110
Analysis of the Equivalent Circuit, for an n-surface enclosure:

For each surface resistance, ί

A
Q i = i i (e bi −ri ) 3.112
1− i

For the resistance between any pair of surfaces, ί and j

Q ij = Ai Fij (ri − r j ) 3.114
At each node, ί
n
Q i =  Q ij 3.64b
j =1

(Note that Q ii = 0, Q ij = −Q ji and ebi = i4 )

Also, for each pair of surfaces, ί and j

Ai Fij = A j F ji 3.66b
n
and for each surface F
j =1
ij =1 3.67b

The unknown and independent parameters in the problem are ebi , ri , Qi , Qij , Fij , Ai and  i . As
shown in Section 3.5.4 page 83 and for an enclosure with diffusely emitting and reflecting gray
n 2
surfaces, the numbers of the unknowns are correspondingly n , n , n , C 2 , n , n and n of the listed
parameters, giving a total of n + C 2 + 5n unknowns. From equations 3.112, 3.114, 3.64b, 3.66b
2 n

n n
and 3.67b, the numbers of the equations are correspondingly n , C 2 , n, C 2 , and n, giving a total of
2 n C 2 + 3n equations. Hence, for explicit solutions to be obtained from the equations,
(n 2
) ( n
+ n C 2 + 5n − 2 n C 2 + 3n or
2
)
(n + 5) parameters will have to be specified. These are often given
as some values for  i , Ai , Fij , i and Q i .

Calculation of ri from Fundamental Equations

Recall that:

ri =  i ebi +  i g i
3.109
1 1
But gi =
Ai
j A j r j F ji = A r A F
j
j i ij =  r j Fij 3.120
i

ie g i is made up of radiation leaving other surfaces of the enclosure that is intercepted by surface ί.

i
 ri =  i ebi +
Ai
A r F
j
j j ji 3.121

111
 or Ai ri = Ai  i ebi +  i  A j r j F ji = Ai  i ebi +  i  r j Ai Fij 3.122
j j

or ri =  i ebi +  i  r j Fij 3.123

j

For transmittivity  = 0 , then  = 1 −  . For a gray surface,  =  . Hence  = 1 −  and equation

3.123 becomes:
ri =  i ebi + (1 −  i ) r j Fij 3.124
j

We can write n equations of the kind 3.124, one for each of the surfaces. If we knew all the ' s,  ' s
and all the F' s , we would be left with only n of r unknowns. We can thus solve for the r' s with the
n equations obtained from 3.124, one for each surface.
Equation 3.124 may be used to replace equation 3.112 above. Equation 3.124, together with 3.114,
3.64b, 3.66b and 3.67b will then form a set of 2 n C 2 + 3n simultaneous equations with which the
unknown parameters may be determined, thus:

ri =  i ebi + (1 −  i ) r j Fij 3.124

j

Q ij = Ai Fij (ri − r j ) 3.114

Q i =  Q ij 3.64b
j

Ai Fij = A j F ji 3.66b
F
j
ij =1 3.67b

112
 PROBLEMS-3
P3.1 (a) Define a Black Body emitter.
(b) Define the Monochromatic Emissive Power of a Black Body Emitter,ebλ.
(c) Comment, with reasons, on the correctness or otherwise, of the statement that “ebλ is a
function of the surface properties of the emitter.”
(d) Sketch the variation of ebλ with wavelength and show the effect of temperature on the
curve.

P3.2 (a) Define the emissive power of a black body emitter, eb.
(b) State the relationship between eb and ebλ.
(c) A glass sheet fully transmits solar radiation in the wavelength range 0.3 – 3.0 μ and is
opaque to it at all other wavelengths. Solar radiation is identical to black body radiation
at 60000K. What is the transmitivity of the glass to solar radiation? Use the table below.

λT 1200 2000 2800 3600 4800 6000 8400 10000 14000 20000 50000
f .002 .067 .288 .404 .608 .738 .871 .914 .963 .986 .999

P3.3 (a) Define the emissivity of any surface for thermal radiation. What parameters does it
depend on?
(b) Express total hemispherical emissivity of a surface in terms of its monochromatic
emissivity and the emissive power of a black body emitter.
(c) What is the effect of oxidation on the monochromatic and total emissivities of a surface?
Which surface has the higher total emissivity – a polished metal surface or one painted
black?
(d) Define the absorptivity α, reflectivity ρ and transmitivity τ of a material. Under what
conditions is α + ρ = 1?
(e) Why is the absorptivity of a material at temperature T1 also a function of the
temperature T2 of the source of the thermal radiation?

P3.4 (a) What is Kirchoff’s law for the emission and absorption of thermal radiation? Derive it.
(b) What is a gray emitter and absorber of thermal radiation? State the relationship between
monochromatic and total values of the absorptivity and emissivity of (i) gray body (ii)
black body radiators.
(c) What is a hohlraum? Show that a hohlraum behaves like a black body emitter and
absorber.

P3.5 (a) What is View Factor F12? State Lambert’s Cosine Law for the emission of thermal
radiation.
(b) Define the intensity of emitted radiation, I. Show that for a black body emitter,
I = eb/π, where eb is the emissive power of a black body.
(c) Determine the view factor F12 for the following cases:

• Surface -1 is the exposed surface of a loaf of bread on the floor of an oven. Surface-
2 is the inside surface of the oven above the horizontal floor.
• Surface-1 is a square flat plate of sides 4m. Surface-2 is a square flat plate of sides
4m, located 3m directly above surface-1.
• Surface-1 is a circular flat disc of diameter 4m. Surface-2 is a circular flat disc of
diameter 4m, located 3m directly above surface-1.

113
 (Refer to your text or reference book for 2nd and 3rd bullets above)

P3.6 A traditional oven for baking bread has the shape of a hemispherical dome, whose interior is
of radius 1.5m. After heating up the oven, the flat floor (surface-1) is at 1200C, while the
concave surface (surface-2) is at 2500C.

(a) Determine the view factors F12 and F21.

(b) What is the net radiant energy transfer rate to the floor if (a) all surfaces are assumed to
be black body radiators (b) all surfaces are assumed to be gray body radiators with an
emissivity of 0.8?
(c) Dough of emissivity 0.2 and temperature of 300C covers the floor of the oven. What is
the net radiant heat transfer rate to the dough from the concave surface of the oven?
Emissivity and temperature of the concave oven surface are 0.8 and 2500C,
respectively. Treat all surfaces as gray.

P3.7 A cylindrical box of internal diameter 3.2 m and height 2 m has one end surface at 8000C and
the other at 2000C. The concave surface is perfectly insulated. View factor from one end surface
to the other is 0.3. Using the electrical analogy of heat transfer, find the temperature of the
insulated surface and the net radiant heat transfer rate to the cooler un-insulated end plate, if:

(a) All surfaces are black body radiators, and

(b) All surfaces are gray body radiators, with an emissivity of 0.6 for the end surfaces.

P3.8 In problem No 7, write the equation for the total radiant flux ri leaving each of the surfaces i,
in terms of Ai, εi, ebi, ρi, Aj, rj and Fji. Then solve the resulting simultaneous equations for ri.
Using these values, obtain the results required in problem No. 7b.
(Note :  Fij = 1; Ai Fij = A j F ji ; and  i = 1 −  i for a gray surface. Symbols have their usual meanings )
.

P3.9 A furnace has a floor of 4m x 8m and a height of 2m. The roof, which has flat surfaces, is a
steel structure with longitudinal passages through which water flows. It receives thermal
radiation from the other surfaces. The temperature of the floor is 12000C. All the vertical sides,
which are perfectly insulated, are at the same temperature. All the surfaces are gray and the
emissivity of the floor, and roof are respectively 0.7and 0.9. View factor from the floor to the
roof of the furnace is 0.5.

Water of specific heat 4.2 kJ/kgK, temperature of 500C and pressure of 40 bar enters the
channels at the total rate of 30 kg/s. Calculate,

(a) The temperature of the vertical sides and the roof of the furnace, at steady state;
(b) Net heat transfer rate to the water and its exit temperature.

(Hint: Since the vertical sides are contiguous and at the same temperature, they can constitute
a single surface. Hence the six-sided oven is a three surface enclosure for radiant heat flow
calculations).
CHAPTER FOUR
CONVECTION HEAT TRANSFER
In convention heat transfer - energy, momentum and mass transfer take place simultaneously. Hence,

114
 Energy Equation
Momentum Equations are simultaneously relevant
Continuity Equation
In pure conduction we considered the diffusion of heat (molecular transport of heat) in a stationary
medium. Two new factors are introduced in the consideration of convection heat transfer,when
compared to conduction. These are the following:
(a) There is movement of matter into and out of a control volume in the flow field. This matter
carries with it its mass, energy and momentum. The laws of conservation of mass, momentum
and energy have to be satisfied by this motion or process.
(b) Since motion exists, the external forces acting at the boundaries and center of mass of the
control volume will do work on the control volume or result in work being done on the
environment by the control volume.
Consequently we have to consider:
- The equation of conservation of mass – or continuity equation
- The equations of conservation of momentum – or equations of motion
- In addition to the equation of conservation of energy – or energy equation
ίn order to be able to solve the convection heat transfer problem.

4.1 THE ENERGY EQUATION OF CONVECTION IN 2-D CARTESIAN CO-ORDINATES

Recall the law of Conservation of Energy as it applies to a control volume, CV:
Rate of change of energy within the CV = Net rate of flow of energy
ίnto the CV plus net Rate of work done on the CV by external forces.
Consider a 2–D control volume, Fig 4.1: Energy flows are represented thus:-

E cv
= exy

Fig 4.1 2-D Energy Flows with Convection, for a 2-D Control Volume
Energy flows consist of heat conduction q into and out of the CV; electrical energy, E elec inflow
and outflow; and bulk convection of energy e contained in the flow into and out of the CV. This energy
includes internal, kinetic, potential, chemical, nuclear, etc, energy types. The total value per unit mass
= e.

115
Work Flows
Firstly, the forces that do work are:
• Surface forces, consisting of normal and shear forces
• Shaft force – if this exists
• Body forces such as gravity and magnetic forces, etc. However work done by gravity force can
be represented as gravitational potential energy change in the consideration of energy flows.
The system of forces is represented thus:

Fig 4.2 2-D Force System for a Control Volume, with Convection
The actual work flow rates are then as indicated in Fig 4.3. Work is not a vector quantity. Therefore,
in developing the sketch, it is essential to know which external force does work on the control volume
ie (work in flow) and which force results in work being done by the control volume on the environment
(work outflow). The convention applied to determine this is as follows:

An external force acting on the CV does work on the CV if the direction of the force and the direction
of the displacement are the same (work inflow); but the CV does work on the source of the external
force, ie on the environment, if the direction of the external force is opposite the direction of the
displacement (work outflow).

116
 Fig 4.3 2-D Work Flows for a Control Volume, with Convection

In Fig 4.3, arrows pointing into the CV represent work inflows and those pointing out of the CV
represent work out flows.
Note: Rate of work done by gravity force is accounted for as a rate of change of gravitational potential
energy and so it is not reflected in Fig 4.3. It is however reflected in the energy flows. Finally, the
energy content of the matter in the CV is given by E cv = exy .

From the law of conservation of energy, the energy conservation equation then becomes

   q     q y  
t
(
(exy ) = E ln − E out )elec + q x y −  q x + x x y  + q y  x −  q y +
x y
y x 
       
        
+ eV x y −  eV x + (eV x )x y  + eV y  x −  eV y + (eV y )y x 

  x     y  
        
+  xV x y −   xV x + ( xV x )x y  +  y V y  x −   y V y + ( y V y )y x 

  x     y  
  
  
+ −  xy V y  y +  xy V y + ( xy V y )x y  + −  yx V x  x +  yx V x +
 
( yxV x )y x  W sh 4.1
  x     y  
 q q 

(e ) xy = E in − E out elec −  x + y xy −   (eVx ) +  (eV y ) xy
( )
t  x y   x y 
     
−  ( xV x ) + ( yV y ) xy +  ( xy V y ) + ( yx V x ) xy + W sh
 x y   x y 

117
   q q y        
(e ) = u elec
 −  x +  −  (eV x ) + (eV y ) −  ( xV x ) + ( y V y )
t  x y   x y
   x y 

  
+  ( xyV y ) + ( yxVx ) + wsh 4.2
 x y 

 =
where u elec
(E in
− E out )
elec
= electrical heat generation rate per unit volume
xy
W sh
wsh = = rate of shaft work per unit volume
xy
e = total energy of the flowing matter per unit mass

  q q    
(e ) +  (eVx ) +  (eV y ) = u elec 
 −  x + y  −  ( xV x ) + ( yV y )
t x y  x y   x y 
  
+  ( xyV y ) + ( yxVx )  wsh 4.2a
 x y 
      e e e   q q 
e  + (V x ) + (V y ) +   + V x  −  x + y 
+ V y  = u elec
 t x y   t x y   x y 
     
−  ( xVx ) + ( yV y ) +  ( xyV y ) + ( yxVx )  wsh 4.3
 x y   x y 

  
But from Continuity Equation + (V x ) + (V y ) = 0
t x y
e e e
since e = e(t , x, y )
de
And from total differentiation + Vx + Vy =
t x y dt

 q q        
 −  x + y  −  ( xV x ) + ( yV y ) +  ( xyV y ) + ( yxV x ) + wsh 4.4
de
:.  = u elec
dt  x y   x y   x y 

1 2
Equation 4.4 can be written using e = u + V + gy + u oth , where
2
𝑢 = internal energy per unit mass
𝑉 = flow velocity
gy= gravitational potential energy per unit mass
𝑢𝑜𝑡ℎ = energy per unit mass from other sources such as chemical and nuclear

d  1 2   q q y 
  −  x +
 u + V + gy + u oth  = u elec  − ---------
dt      
2  x y 

118
 d  1 2  du  q q 
  −  x + y  − ----------
 u + V + gy  +  oth = u elec
dt  2  dt  x y 
d 1 2   q q 
  u + V + gy  = u elec  −  x + y  − − − − − − − ( see comments below)
  u oth
dt  2   x y 
 q q y        
 −  ( xV x ) + ( yV y ) +  ( xyV y ) + ( yxV x )
d 1 
  u + V 2 + gy  = u  −  x +
dt  2   x y   x y   x y 
+ wsh
4.5

Comments:
du oth
− = − u oth
dt
u oth = total rate of change with time of other internal energy types. A positive value yields an
endothermic process and heat absorption. A negative value yields an exothermic process and heat

generation. Thus: − u oth =  u oth

 = heat generation rate per unit volume from other sources (negative for heat absorption);
u oth
  u oth
u  = u elec  = net heat generation rate per unit volume from all sourses.
du
The left hand side of equation 4.5 is seen to consist of a thermal energy component  and a
dt
mechanical energy component 
dt 2
(
d 1 2
)
V + gy . The right hand side also consists of thermal energy
terms, including a heating effect due to work done, as well as mechanical energy components. We may
identify and group these terms, as shown later in equation 4.11a.

From Fluid Mechanics, the 2–D forms of the continunity and momentum equations are given as:

  
Continuity: + ( V x ) + (V y ) = 0 4.6
t x y
  
x – Momentum: (Vx ) +  Vx2 +  (VxV y ) = − x + yx +  sh
( ) 4.7
t x y x y

y – Momentum:

(V y ) +  (VxV y ) +  V y2 = −  y +  xy − g + Ysh
( ) 4.8
t x y y x

Xsh and Ysh are shaft forces per unit volume in x & y directions, respectively. Let us use equations 4.6,
4.7 and 4.8 to identify the mechanical energy terms on the right hand side of equation 4.5 and hence
further simplify that equation.

Expanding the left hand side of equation 4.7 we obtain:

 Vx V V         yx
  + Vx x + V y x  + Vx  + (Vx ) + (V y ) = − x + +  sh
 t x y   t x y  x y

119
 dVx
The 2nd term on the left hand side of the above equation = 0 from contiunuity. The first term = 
dt
. The equation then reduces to,

dVx   yx
 =− x + +  sh 4.9
dt x y
Similar operation on equation 4.8 yields,
dVy  y  xy
 =− + − g + Ysh 4.10
dt y x

Equation 4.9* Vx + Equation 4.10* Vy yields

    x  y    yx  xy 
 + (V x X sh + V y Ysh )
dVx dV y
 V x + Vy + gV y  = − V x + Vy  + V x + Vy
 dt dt   dx y   dy x 
d 1    x  y 
 +  +  
1
   V x2 + V y2 + gy  = − V x + Vy
dt  2 2   x y 
  x  y    yx  xy 
d 1
dt  2

   V 2 + gy  = − V x
x
+ Vy
y  
 + V x
y
+ Vy
x 
+ V X x sh 
+ V y Ysh 4.11
 

Xsh Vx + Ysh Vy =  wsh , mech = mechanical component of shaft work per unit volume.
Expanding equation (4.5) gives

du d 1   q q y   V x V y    x  y 
 +   V 2 + gy  = u  −  x +  −   x + y  − V x + Vy 
dt dt  2   x y   x y   x y 
  yx  xy   V x V y 
+ V x +Vy  +  yx
  +    wsh , th  wsh ,
 4.11a
   
xy mech
 y x   y x 

The left hand side of (4.11) is a mechanical energy term and so, the right hand side is a mechanical
work term. Equation 4.11 states that, per unit volume of the control volume, CV, the total rate of
change of the mechanical energy of the CV is equal to the net rate of mechanical work done on the
CV. Equation 4.11a on the other hand, states that per unit volume of the CV, the total rate of change
of the energy of the CV (both thermal and mechanical) is equal to the net rate of generation and inflow
of thermal energy plus the net rate of the heating effect of work done plus the net rate of mechanical
work done, on the CV. Thus, the left and right hand sides of equation 4.11a contain both thermal and
mechanical energy terms. By using 4.11, which equates mechanical energy and mechanical work terms
only, we can separate the thermal from the mechanical energy terms in equation 4.11a. Thus, we
remove the mechanical energy and mechanical work terms in 4.11a by using equation 4.11. To do this
we subtract 4.11 from 4.11a. This leaves the thermal energy terms and heating effect due to work done,
thus:
du  q q y   V x V y   V x V y 
 = u  −  x +  −  x + y  +  yx +  xy  + wsh ,th 4.12
dt  x y   x y   y x 
rate of change
of thermal
internal 120
energy per
unit volume
 net heat net rate of
conduction rate net rate of
heating due to
per unit volume heating due to
shear stresses per
normal stress per
unit volume
unit volume
net heat generation net rate of heating
rate per unit volume due to shaft work
per unit volume

It is possible to express u, qx, qy, xy and yx in terms of T, P, Vx and Vy, using appropriate particular
laws and equations of state (as we shall do shortly). We shall then have one equation, ie, the energy
equation, and four dependent variables (instead of nine dependent variables as in 4.4 or 4.5 or 4.12).
We will then need three more equations to be able to get a solution for the four variables. These
additional equations are supplied by:

i. the Continuity Equation

ii. the Equation of Motion in x – direction
iii. the Equation of Motion in y - direction

We may now introduce the particular laws for an ίsotropic and homogeneous medium:


qx = −k
x

qy = −k
y
x   4.13
y  
V x
 yx  
y
V y
 xy  
x
Equation 4.12 (for a homogeneous and isotropic fluid) becomes

du   2   2    V x V y   V   V y  
2 2

 
 
= u + k  2 + 2  −   +  +   x  +     wsh , th 4.14
y   x y   y   
dt  x   x  

This is one form of the general 2 – D energy equation of convection heat transfer for a homogeneous
and isotropic fluid and for which the approximations for σx, σy, τyx and τxy apply.

We can further operate on equation (4.14) thus: From the continuity equation,

 
+ (Vx ) +  (V y ) = 0
t x y
      V V y 
or  + Vx + Vy  +   x + =0
 t x y   x y 

121
 d  V V y 
 +   x + =0
dt  x y 
 V V y  1 d
  x +  = - 4.15
 x y   dt

Substituting 4.15 in equation 4.14 we obtain a second form of the energy equation of convection heat
transfer,

du  d
 = u  + k 2  + +    wsh , th 4.16
dt  dt
 2  2
where  2  = +
x 2 y 2
  V y
2 2
 V x 
and  =   +  
 y   x 

We may introduce enthalpy in place of internal energy by using the equation of state:

h=u + 4.16a

dh du 1 d  d
 = + − 2
dt dt  dt  dt
du dh p d dp
 = + −
dt dt  dt dt

Introducing this in 4.16 yields:

dh d
 = u  + k  2  + +    wsh , th 4.17
dt dt
Equation 4.17 is another form of the energy equation of convection.

Furthermore

h = h (,  ) 4.18
h

dh
=( ) p d + ( h ) d
dt  dt  dt
dh d  h  d
= Cp + 
dt dt     dt

From Maxwell’s thermodynamic relationships:

h v
( ) = v − (  ) 4.18a
 

122
   1 v  
  = v (1 − ) = (1 − )
1
= v 1 − 
  v     

 1 v 
where  =   is the coefficient of volumetric expansion
 v   

d 1 d
+ (1 − )
dh
 = Cp 4.18b
dt dt  dt
or dh = C p d + (1 − )d
1
4.18c

We may note, from 4.18c, that the often quoted relationship, dh = C p dT , is only valid for a perfect
gas (𝛽 = 1⁄𝑇) or for cases in which the effect of pressure change is negligible.
dh
Substituting the above expression for into 4.17 gives,
dt

d d
C p = u  + k 2  +  +    wsh , th 4.19a
dt dt
d   2  2  d  V   V y  
2 2

or C p = u  + k  2 + 2  +  +   x  +     wsh , th

  4.19b
dt  x y  dt  y   x  

Equation 4.19a or 4.19b is yet another form of the general energy equation of convection heat transfer
for a 2 – D ίsotropic and homogeneous medium, and in which the approximate expressions for σx, σy,
yx, and xy, apply. The dependent variables in 4.19a or 4.19b are now only 4, namely , P, Vx and V y ,
instead of 9, as in equation 4.4 or 4.5 or 4.12. This was achieved by using the particular laws in 4.13
and the equations of state in 4.16a, 4.18 and 4.18a.

The full expressions for σx, σy, σz, yx, xz and zy in 3-D cartesian co-ordinates are given by:

V x 2  V V y V z 
 x =  − 2 +   x + + 
z 
4.20a
x 3  x y
V y 2  V x V y VZ 
 y =  − 2 +   + +  4.20b
y 3  x y z 
V z 2  V x V y V z 
 z =  − 2 +   + + 
z 
4.20c
z 3  x y
 V V y 
 yx =  xy =   x + 
x 
4.20d
 y
 V V y 
 yz =  zy =   z +  4.20e
 y z 
𝜕𝑉 𝜕𝑉𝑧
𝜏𝑧𝑥 = 𝜏𝑥𝑧 = 𝜇 ( 𝜕𝑧𝑥 + ) 4.20f
𝜕𝑥

123
4.1.1 Simplifications of the General Energy Equation of Convection in Particular Cases.
Case 1: Constant Density Fluid
For a constant density or incompressible fluid,

 =0
C p = Cv = C
d
 C = u  + k 2  +    wsh , th 4.21
dt

Case 2: Ideal Gas

v = RT
1 v  R 1
 = ( )P =   =
v  R    
d d
Hence C p = u  + k  2  + +    wsh , th 4.22
dt dt

The above 2 cases are the general simplifications that can be made. Other simplifications will depend
on the specific flows and geometries and on the relative magnitudes of each term in the equation. By
this latter consideration, terms of smaller orders of magnitude are neglected, while the larger order
terms are retained.
Nevertheless, constant density and ideal gas fluids do represent broad classes of flows. Within each
class, further simplicafications of the energy equation are possible, depending on flow types, flow
conditions and geometries peculiar to the case under consideration, such as:

Types of flow, e.g - boundary layer flow

- high or low Reynolds Number Flow
- developing or fully developed flow
Fluid properties, e.g - high or low viscosity fluid
- fluid with constant or varying properties
Geometry of flow, e.g - external flow over flat plate, cylinder, etc
- internal flow through conduίts
4.2 APPLICATION TO PARTICULAR CASES

In solving the energy equation of convection such as equation 4.19b or 4.21 or 4.22, we will require
the introduction of the continuity and momentum equations, so as to obtain final solutions for the
temperature, pressure, x – and y velocity distributions. From the temperature distribution, we can
obtain the heat flux in the x or y-direction by using the appropriate Fourier’s Law, for instance,


q y = −k at the required location in y. 4.23
y

If the heat transfer coefficient is needed at a given surface, s, then Newton’s law of cooling gives,

q s = h (s −  f ) where  f = appropriate reference fluid temperature

124
   
− k  
qs  y  s
 h = = 4.24
s −  f  s − f

4.2.1 Concept of Velocity and Thermal Boundary Layers

The concept of a Thermal Boundary Layer is similar to the concept of a Velocity Boundary Layer.

Recall that for the Velocity Boundary Layer,

0
Fig 4.4 Development of a Velocity Boundary Layer Over a Flat Plate

if a uniform flow of velocity V becomes incident on a flat surface for instance, the velocity of the
fluid at the surface becomes instantly equal to the velocity of the surface, V w . If V w ≠ V , then the
force exerted by the wall on the fluid at the surface to achieve the above equality condition is
transmitted to the rest of the fluid through molecular momentum transfer (in lamina flow) with
diminishing effect further away from the wall. This results in the development of a region of flow near
the wall or surface, within which the velocity of the fluid changes from V w to V . This region is called
the velocity boundary layer. A velocity boundary layer thickness,  is defined such that:

for 0  y   then
V x − Vw
0  Vx − Vw  c (V − Vw ) for V  Vw or 0  c
V − Vw
Vx
and 0  V x  c V or 0 c for Vw = 0
V

where c = 0.99 , and is chosen by convention.

125
For external flows such as flow over a flat plate,  increases with x . In the case of internal flows, an
x * exists such that for x  x * , the velocity boundary layer is said to be developing. This means that
for x  x * , the dimensionless velocity,

V x − Vw  x y
= f ,  4.25
VR − Vw L  
V R = a reference velocity. It may be the centerline or the mean velocity.
Vx  x y
If Vw = 0 (stationary wall) then = f  ,  for x  x * , ie in the developing region.
VR L  
For x  x * for internal flows, the boundary layer is fully developed and
V x − Vw  y
= f   only 4.26
VR − Vw  
Vx  y
or = f   only for Vw = 0 .
VR  
A Thermal Boundary Layer is similarly defined

0
Fig 4.5 Development of a Thermal Boundary Layer Over a Flat Plate

When a fluid stream at uniform temperature T is incident on a surface, the fluid at the wall surface
instantly assumes the temperature of the surface, Tw. If T ≠ Tw, the above equality condition imposes
a region near the surface within which the fluid temperature changes from Tw to T. This region is
called the thermal boundary layer. In lamina flow, the thermal boundary layer develops through
transfer of heat between the wall and the rest of the fluid by molecular heat transport, ie by heat
diffusion, with diminishing intensity further away from the wall. A thermal boundary layer thickness,
T, is defined such that:

For 0  y  
0  w −   c (w −  ) for w  
 − w
or 0   c in general, where c = 0.99 by convection.
 − w

126
For external flows such as flow over a flat plate, T increases with x . However, for internal flows and
for x  xT* the thermal boundary layer is said to be developing and the following condition applies:

for x  x * in internal flows,

 − w x y 
= f  ,  4.27
R − w  L  

But for x  x * in internal flows , the thermal boundary layer is said to be fully developed, and
 − w  y 
= f   only 4.28
 − w  
R = A reference temperature. It may be the centerline or the mean temperature.
Generally,  and T are not equal. However for fluids of Pr ≏ 1, then  ≏ T.
r  Prandtl Number, is a very useful dimensionless parameter in heat transfer.

 Cp   v  Molecular Diffusivity of Momentum 

r =  =    4.29
k k C p   Molecular Diffusivity of Heat. 
For air, r is approximately 0.7 under normal atmospheric temperature and pressure and  ≏ T within
those conditions.

Boundary Conditions

) y =0 = w 4.29a

q y ) y = 0 = −( k ) y =0 4.29b
y
Also q y ) y = 0 = h (w −  ) or h (w −  ) or hm (w − m ) 4.29c
The heat transfer coefficient depends on what is chosen as the reference temperature.

4.2.2 Solution for Heat Transfer in Couette Flow

Example 4.2.2 (1): Solution for Flow, with Heat Transfer, Between Two Parallel Plates in which
One Plate Moves and the other is Fixed.

Additional conditions:
Steady State
No heat generation and no shaft work in the fluid
High fluid viscosity and constant fluid properties
Small gap width between plates

127
Viscosity  ~ 10 −2 kg ms is high.
Gap width w ~ 10 −1 mm or 10−4 m is small.
Typical examples:

• Sleeve bearing with oil lubrication (D / w   1) ., where D = shaft diameter.

• Piston – Cylinder pair with oil lubrication of the surfaces (L / w   1) .

For the sleeve bearing, since D / w  1 , the flow of the lubricant is approximately like flow between
parallel plates. The gap width w  is small and approximately constant. Flow is consequently
approximately one – dimensional: ie Couette Flow.

Fig 4.6 Couette Flow

Summary of Applicable Conditions:

• Steady state
• 1 – D flow in x-direction
• Constant density fluid
• No heat generation
• No shaft work in fluid
• Viscosity is high and constant
• Gap width is small

General Energy Equation:

d dp
 Cp = u  + k  2  +  +  + wsh , th 4.19a
dt dt
       2  2      
 C p  + V x +Vy  = u  + k  2 + 2  +  + V x +Vy 
 t x y   x y   t x y 
 V  2  V y  2 
+   x  +    + wsh , th 4.19b
 y    x  
Continuity and Momentum Equations

128
   
+ ( V x ) + ( V y ) = 0 (continunity ) 4.30
t x y
dVx   2V x
 =− + (x − momentum) 4.31
dt x y 2

V x V x V x 1    V x
2
or + Vx +Vy =− + (x − momentum ) 4.32
t x y  x  y 2

V y V y V y 1    V y
2

and + Vx +Vy =− + −g ( y − momentum) 4.33

t x y  y  x 2

Simplification of the General Energy Equation makes use of the following relationships obtained from
the applicable conditions:

 
= =0 Steady state
t t
Vy = 0 1 – D flow in x-direction
u  = wsh = 0 No heat generation and no shaft work
 =0 Constant density
The energy equation then reduces to:
2
   2  2   V 
C p V x = k  2 + 2  +   x  4.34
x  x y   y 

ORDER OF MAGNITUDE ANALYSIS

We may further do an order of magnitude analysis to see which of the terms in equation 4.34 are
negligible in comparison with the others. Using the reference parameters listed below,

Let bottom plate temperature = w

,, top (moving) plate temperature = p
,, moving plate velocity = Vp
4.35a
,, plate length =L
,, gap width = w
we define dimensionless temperature, velocity, and displacements 𝑇 ∗ , 𝑉𝑥∗ , 𝑥 ∗ 𝑎𝑛𝑑 𝑦 ∗ as,

* = ( − p ) (w − p )   =  * (w − p ) + p

129
 Vx* = Vx VP Vx = Vx* V p
4.35b
x* = x L x = x*L

y* = y w  y = y * .w

We express the variables 𝑇, 𝑉𝑥 , 𝑥 𝑎𝑛𝑑 𝑦, in terms of their dimensionless forms (equation 4.35b) and
then insert these in 4.34. Then, the partially simplified energy equation becomes, with
p = w − p ,

 C pV p p  *  *  kp   2  *  kp   2  *  V p  V x* 

2 2

  V x
 =  2  *2 +  2  *2 +  2  *  4.35c
  L  x  w  y  w  y 
*
 L  x
A B C D

The starred parameters are of order 1. Hence the order of magnitude of each term in 4.35c is determined
by the order of magnitude of the quantity in square brackets, ie of quantities A, B, C and D.

C k p / w  2  L
2

Now ~ =  
B kp / L2  w 
Since L W  >> 1, it follows that
 2  2
C >> B or k 2  k 2 4.36
y x
C  k   L
.  
Also ~
A  C pV p w    w  

Since L w   1 and w  is very small, we may conclude that C/A >>1 or

 2 
k  C p Vx 4.37
y 2
x
Finally
C kp
~
D V p2

It is not indicated that this is >> 1or << 1, so C and D may be of comparable magnitudes and so both
are retained. Thus, from the order of magnitude analysis, only the 3rd and 4th terms of the partially
simplified equation are retained. Hence the energy equation reduces to

2
 2   V x 
= −  
k  y
4.38
y 2 

130
We need V x ( y ) in order to solve for T from (4.38). This we can get by solving the continuity and
momentum equations. We can simplify the continuity and momentum equations, using the specified
flow conditions, to get:

V x
= 0 (continuity) 4.39
x

 2V x 1 
= (x − momentum) 4.40
y 2  x
p
= − g ( y − momentum) 4.41
y

From equation (4.39), it follows that V x  f ( x ) . With steady state, V x  f (t ) also. Therefore
V x = f ( y ) only. Hence in equations 4.38 and 4.40,
V x dVx  2V x d 2V x
= and = 4.42
y dy y 2 dy 2
Integrating equation 4.41, we obtain

 = − gy + f p (x ) 4.43

From equation 4.43, the difference in the pressures at the top and bottom plates is given by

 =  ) y =0 −  ) y = w = f p (x ) − f p (x ) + wg. = wg 4.44

Since w  is very small  is very small and so is negligible. Hence variation of  in the
y − direction within the flow is negligible.  is thus a function of x only

 d
  = f p (x ) only and in equation 4.40, = 4.45
x dx
Equation 4.40 becomes:

d 2 Vx 1 d
= 4.46
dy 2
 dx

Equation (4.46) states that a function of y = function of x for all x and y. Each therefore must be
equal to a constant. Integrating equation (4.46) with respect to y, we obtain:

1 d 2
Vx = y + Ay + B where A and B are integration contents 4.47
2  dx

Boundary conditions:

131
 V x ) y =0 = 0 4.48a
V x ) y = w = V p 4.48b

From 4.48a, B=0

1 d 2
From 4.48b, Vp = w  + w A
2 dx
1  w 2 d 
 A =  p
V − 
w  2 dx 
1 d 2 V p y w d
Vx = y + − y
2 dx w 2 dx

1 d 2 Vp 
or V x =
2 dx
( )
y − wy +   y 4.49
 w 

The velocity profile is seen to be a parabolic term combined with a linear term, thus:

+ =

Fig 4.7 Couette Flow Profile as Sum of Parabolic and Linear Profiles

Vp
If
d V
is very small, then the profile is a straight line: x = y and equation (4.38) becomes:
dx w
2
 2  Vp 
= −  
k  w 
4.50
y 2


Now, since C pV x was found negligible, and C pV x is not negligible everywhere, we may
x

conclude that is negligible. Hence we may conclude that  ≏ f ( y ) only, so that  2  y 2 in 4.50
x
= d  dy . Equation (4.50) then becomes:
2 2

132
 d 2  Vp 2

= − 4.51
dy 2 kw 2

Integration of equation 4.51 yields

 V p2
=− y 2 + Cy + D where C and D are integration constants 4.52
2kw 2

Boundary Conditions:

) y =0 = w
) y = w = p

Substituting the boundary conditions yields,

D = w
1   V p2 
C= ( p − w ) + 
w  2k 
 V p2   V p2 
( p − w ) +
y
  =− y + 2
 + w 4.53
2kw 2
w  2k 

  p − w
2
  Vp  2
or  = w +  y −

  y − wy
 w    4.54
 w  2k  

Heat flux from the wall to the fluid (ie in the y-direction at y = 0) is given by,

d 
q y )y =0 = − k 
dy  y =0
4.55

 p − w  w  V p 
2

qy ) = −k  +  
 2k  w  
 4.56
y =0
 w    

 w −  p
2
 w   V p 
If w  p , then q w = k  −

  4.57
 w  2  w  

Conduction Friction heating

term of the wall

For w  p , conduction heat flows from the wall to the fluid, while friction heating flows from the
fluid to the wall.

133
If p  w , then heat flux to the wall is given by

d
qw = k ) y =o (ie in the - ve y-direction at y = 0, p  w )
dy

 p − w  w  V P 
2

 q w = k  +
    4.58
 w  2k  w   

Both conduction and friction heating terms flow to the wall for p  w .
To obtain the heat transfer coefficient, h

 p − w  w  V p  
2

qw = h (p − w ) = k  +

  
2k  w  
4.59
 w  
 1 w  V p  2  1 
   
:. h = k w + 2k  w    − 

 
   p w 
hDe
Nusselt Number is defined as Nu =
k
4(cross - sectional area ) 4 * w *Width
De = Equivalent Diameter = = ≏ 2 w 4.60
wetted perimeter 2(Width + w)

2hw 2kw  1 w  V p   1 

2

 Nu = =  +   
k k  w 2k  w   p − w 
 
 V p
2
  i V p2
Nu = 21 +  = 2 1 +  where i = and  = p − w 4.61
 2k (p − w )  2 k

The results are expressed in terms of dimensionless groups, Nu and ί. ί is the ratio (viscous heating
flux)/conduction heat flux). The use of dimensionless groups is very useful in heat transfer and will be
treated more fully in Section 4.4

4.3 INTRODUCTION TO THE ANALYSIS OF TURBULENT FLOW HEAT TRANSFER

Continuity, Momentum and Energy Equations
Turbulent flow processes are modelled as ones in which the flow and transport parameters have
temporal mean as well as fluctuating components. This arises from the fact that even in supposedly
‘steady state’ conditions, the parameters such as velocity, pressure and temperature at a point are
continually fluctuating. The sketch below illustrates this model.

Vx

134
 Vx = Vx + Vx
--------
V y = V y + V y
4.62
 =  + 
Time
 =  + 

Fig 4.8 Turbulent Velocity Fluctuations

Barred quantities are time averaged values over a large enough number of fluctuations or time, but
such that the time is small enough compared to the time for a significant change in the averaged values.
Energy, mass and momentum must be conserved in turbulent flow, just as they are in lamina flow.
Noting that the velocity, pressure and temperature that appear in the lamina flow conservation
equations are the instantaneous values, the same equations will apply in turbulent flow if we use also
the instantaneous values. But, since instantaneous parameters are continually fluctuating in turbulent
flow, it is difficult to solve for them. However, we take advantage of the fact that their net effects are
represented by the net effects of their time averaged or mean values.
The approach in the analysis of turbulent flow processes is to substitute the turbulent forms of the
instantaneous turbulent parameters, as in equation 4.62, in the conservation equations obtained for
lamina flow, and then obtain the time-averaged values of the resulting differential equations. It is
assumed that the averaging time is long enough to have a large number of fluctuations, but small
enough compared to the time for a significant change in the time-averaged values.

Let us apply the above procedure to the case of 2-D Convection Heat Transfer, with no heat generation,
no shaft work, constant density and negligible friction heating. The general conservation equations
are:

 
+ (Vx ) +  (V y ) = 0 (continunity ) 4.63
t x y
V x V x V x 1    2V x
+ Vx + Vy =− + ( x − Momentum) 4.64
t x y  x  y 2
V y V y V y 1    V y
2

+ Vx + Vy =− + −g ( y − Momentum) 4.65
t x y  y  x 2
     d
 Cp  + Vx + Vy  = u  + k  2  +  +   + wsh , th (energy) 4.66
 t x y  dt

For the conditions given above, ie :

u  = wsh , th =  =  = 0
4.66a
 = constant.
we can obtain and the conservation equations in the form:

135
 V x V y
+ =0 4.67
x y
V x   1    2V x
t
+
x
( )
Vx +
2

y
(V x V y ) = −
 x
+
 y 2
4.68

V y  1  μ  V y
2

+ (V yV x ) +  V y2
( ) = − + −g 4.69
t x y  y  x 2
 (V )  (V ) k   2  2 
+ + = +
x y

C p  x 2 y 2 
4.70
t x y

Note that equation (4.67), together with the conditions existing for the convection process, ie 4.66a,
were utilized in obtaining the forms in equations 4.68, 4.69 and 4.70 for the momentum and energy
equations.
Equation 4.62 is then introduced into equations 4.67 to 4.70 to obtain,


(Vx + Vx ) +  (V y + V y ) = 0 4.71
x y

(Vx + Vx ) +  Vx2 + 2VxVx + Vx 2  +  VxV y + VxV y + V yVx + VxV y 
t x y
1   2
=- ( +   ) + (Vx + Vx ) 4.72
 x  y 2

t

(V y + V y ) +  (VxV y + VxV y + V yVx + VxV y ) +  V y2 + 2 V y V y + V y 2
x y

1 
( + ) +   2 (V y + V y ) − g
2

=- 4.73
 y  x

t

( + ) +  Vx + Vx + Vx  + Vx +   V y +  V y + V y  + V y 
x y

 2 2 
=
k
 2 ( +   ) + ( +   ) 4.74
C p  x y 2 

Taking the time average of each term in the equation and noting that: Vx = 0 and similarly for
Vy, 
and  . Also V xV x = V x .Vx = 0 and likewise for similar terms, namely:
V yV x , V xV y , V yV x , V xV y , V x , V x , V y , V y .
However, V x V y = 0. Similarly, V x ' 2 , V x  and V y are  0. This inequality is illustrated
using Fig 4.9

136
 Vx

(a) V x Increases with y (b) V x Decreases with y

Fig 4.9 Effect of + ve or − ve Fluctuation of V y on Vx

If V x increases with y (Fig 4.9a), then a positive V y from level A , for instance, will induce a negative

V x at level B through x-momentum exchange with particles at level B . A negative V y from level
B will likewise induce a positive V x at level A . Thus if V x increases with y , the product V xV y will
always be negative. It’s time average value will not be zero. Thus V xV y  0 .

Similarly, it may be deduced that if V x decreases with y , (Fig 4.9b), a positive V y from level A will

induce a positive V x at level B , whereas a negative V y from level B will induce a negative V x at
level A , through x-momentum exchanges . The product V xV y will thus always be positive, so that
V xV y  0 .
The same is true of V x and V y , V x and T  , or V y and T΄. The time average of the product of the
fluctuating quantities is therefore not equal to zero. We then obtain the following equations from 4.71
to 4.74:

Vx V y
+ = 0 4.75
x y

V x
+

( ) 
V x2 + (V xV ) = − 1 + 1
x
  V x

 y  y
 1
− V yV x  − V x ' 2 ( ) 4.76
t x y  x
y

137
 V y
+
  V 
(
(V xV y ) +  V y2 = − 1  + 1   y − V yV x  − 1  V y ' 2 − g
( )  y  x  x
) 4.77
t x y   y

   1     1    
+ (V x ) + (V y ) =  k − C p V x  +  k − C p V y  4.78
t x y C p x  x  C p y  y 

From equations 4.75, 4.76, 4.77 and 4.78 and equations 4.67 to 4.70 we see that:

a) The terms associated with the mean motion in turbulent flow are identical to those in
lamina flow
b) The effect of the turbulent fluctuations is to introduce turbulent momentum transfers in
the normal directions to control volume surfaces, ie turbulent normal momentum
transfers (or normal stresses) V x and V y , as well as turbulent momentum
2 2

transfers in the shear (perpendicular to normal) directions, ie turbulent shear stresses

− V yV x and − V xV y in the x and y directions, respectively. These are in addition
Vx V y
to the molecular momentum transfers or viscous shear stresses  and 
y x
. The total or apparent shear stress is therefore the sum of a lamina and a turbulent shear
stress.
c) There are also turbulent energy transport terms  C p V x and C p V y in the x and
y directions, respectively, in addition to the molecular heat diffusion or conduction
 
terms k and k . The total apparent heat flux is thus the sum of a molecular
x y
diffusion, or conduction, heat flux and a turbulent heat flux.

With regards to the turbulent shear stresses, we may, for instance, note that the magnitude of V x
 Vx
induced by a given V y increases as increases. Hence, the magnitude V x V y increases with
y
V x
. Therefore, to a first degree of approximation and borrowing from the expression for viscous
y
V x
shear stress, we may assume that − V xV y is proportional to , with a constant of proportionality
y
μm. This will lead us to define a turbulent or eddy viscosity, μm, analogous to the molecular viscosity
μ, such that:

V x
−  V y V x =  m 4.79
y
 Vx
or V y V x = − m 4.80
 y
138
 Vx m
V x V y = −  m where m = 4.81
y 

Therefore the Total or Apparent Shear Stress app is given by,

Vx Vx
 app = (  +  m ) =  (v +  m ) 4.81
y y
εm is called the eddy or turbulent diffusivity of momentum. It is analogous to v =  /  , which is the
molecular diffusivity of momentum. However, whereas ν and μ are fluid properties, ε m and μm are
parameters of the fluid as well as of the particular turbulent flow.

In like manner, eddy or turbulent conductivity and diffusivity of heat, kh and εh respectively, can be
defined such that:


−  C p Vx = k h 4.82
x

−  C p V y = kh 4.83
y

Vx = −  h where  h = k h / C p 4.84
x

V y = −  h 4.85
y
 
q x ,app = − (k + k h ) = −  C p ( +  h ) 4.86
x x

 
q y ,app = − (k + k h ) = −  C p ( +  h ) 4.87
y y
kh and εh depend on the fluid and the flow. Isotropic turbulence is assumed and so kh and μm do not
vary with direction. The subsequent solutions for the velocity and temperature fields in turbulent flow
heat transfer depend, to a large degree, on the assumptions made regarding the variations of εm and εh
within the flow. If εm and εh are assumed constant, then the turbulent convection equations will have
similar solutions to their lamina equivalents. Appropriate assumptions need also to be made for the
normal turbulent stress components.

4.4 CONVECTIVE HEAT TRANSFER CORRELATIONS

Exact solutions of the convective heat transfer equations are difficult to obtain. Consequently, we
frequently resort to the use of correlations. These are experimentally determined relationships between
relevant variables in the convective heat transfer process, under specified conditions. For instance, the
heat transfer coefficient may be expressed as:

h = f (V, geometry, fluid properties) 4.88

139
The correlating equation is usually obtained from fits to experimental data, using numerical techniques.
If the correlation is expressed in terms of dimensional quantities, then in principle, the correlation
applies only to the conditions that existed during the particular experiments that produced the data. A
different correlation would have to be obtained for a different velocity, different geometry (size or
shape) and different fluid, even if the velocity range keeps the class of flows the same.
To avoid the above problem, the correlations are normally expressed in terms of dimensionless
quantities on which the convective process depends. This ensures that when dynamic, kinematic,
geometric and thermal similarities exist, the same correlation will apply to the same class of flows, ie
to similar flows, irrespective of actual flow velocity, channel size, or fluid properties. The relevant
dimensionless groups to be used in correlating the data for a particular class of flows can be obtained
using various methods of Dimensional Analysis.
4.4.1 Dimensional Analysis

Each physical quantity can be expressed in terms of some fundamental dimensions, thus:

Table 4.1 Some Physical Quantities and Their Dimensions

Quantity Engineering Units of Measure Dimensions
Length m L
Area m2 L2
Volume m3 L3
Mass kg M
Time sec(s) T
Temperature 0
C, 0K Ө
Density kg/m3 ML-3
Force N MLT-2
Dynamic Viscosity poise, pa.s, kg/ms ML-1T-1
Kinematic Viscosity stoke, m2/s L2T-
Thermal Conductivity W/mC or W/mK MLT-3 Ө-1
Heat Transfer Coefficient W/m2C or W/m2K MT-3 Ө-1
Specific Heat kJ/kgC or kJ/kgK L2T-2 Ө-1

A number of methods exist for dimensional analysis, as presented below.

(1) Non-dimensionalization of the governing equation or the exact solution of the problem,
e.g.
Consider the class of problems represented by the equation,

2
   2  2   V 
 C p Vx = k  2 + 2  +   x  4.34
x  x y   y 
An example is couette flow with lamina, steady state, constant density, 1 – D flow in x direction, and
with u  = wsh = 0 conditions. This was earlier shown to be so in equation 4.34.
Then if, as in 4.35b, we have that:

V x = VV x *
 =   * +p where p = w − T p
x = Lx *

140
 y = De y * ( De = equivalent diameter, used in place of gap width, w )

Then, substituting the above in equation 4.34, we obtain that:

2
   2 *  2 *  
 C p V p p  *  *  kp     kp     V p
2
  V x* 
V x =  *2  +  *2  + 2  *  4.89
L  x *  L2  x  De2  y  De  y 
       
Each starred group of terms is dimensionless. Therefore, if we divide each coefficient of the starred
terms by one of the coefficients, the resulting coefficients will each be dimensionless. These resulting
coefficients or groups then constitute the dimensionless groups that are relevant to the problem. Most
often, dimensionless groups obtained in this way are not easily recognizable in terms of the more
common or standard dimensionless numbers in convection heat transfer, such as Re, r , Nu, etc.
Nevertheless, these numbers are contained in the groups. Through a proper rearrangement of, and
operation on the dimensionless groups, they can be expressed in terms of the standard dimensionless
numbers.
Alternatively, we may note that by multiplying the coefficient of the first term on the left hand side by
LDe / C p p the Reynolds number V p De /  is obtained. Therefore multiply each term in the
LDe
equation by the same quantity to obtain:
C p p

V p De k De k L V p2 L 1
 [−−] = . [−−] + . [−−] + . . [−−] 4.90
 Cp L Cp De p De Cp

The coefficient of each term, except the last one, is now seen to contain
VP De c p
 (= Re ), or (= Pr ), or L
(= dimensionless length). For the last term, multiply by k/k.
 k De

Equation 4.90 then becomes:

 VP De   * *   k  De   2 *   k   L    2 * 
   V x * 
=     *2  +   .    *2 
    x   Cp  L   x   Cp   De   y 
  V p2   L
2
  k   Vx* 
+ .  .    * 
 kp   De    y 
4.91
     Cp
The terms in ( ) brackets in equation 4.91 are recognizable dimensionless quantities:
V p De
 = Re or Reynolds Number

 Cp
= Pr or Prandtl Number
k
L
= Dimensionless Length
De

141
  V p2 Viscous Heating / Unit Volume Viscous Heating Flux
= i  OR
kp Conduction Heating / Unit Vol Conduction Heat flux

These are the dimensionless numbers with which the convective heat transfer process represented by
equation 4.34 can be described. In a heat transfer correlation, one of the dimensionless parameters,
such as the dimensionless heat transfer coefficient, ie the Nusselt Number, Nu =
hDe  h Convective Heat flux 
  =  , can be correlated in terms of the other
k  k / D e Conduction Heat flux 
dimensionless groups, thus:

Nu = f (Re, r , L / De , i ) 4.92

(2) Use of Buckingham Pί - Theorem:

The theorem states that:
If a phenomenon is described by N independent quantities, having a total of n fundamental dimensions,
then there exists a maximum number of quantities, Nmax, of these independent quantities which, by
themselves, cannot form a dimensionless group, where Nmax ≤ n. Then by combining one each of the
remaining quantities with the Nmax quantities, a dimensionless group can be formed. The number of
dimensionless groups that can be formed is thus N – Nmax. Each dimensionless group is called a  -
term and is identified as 1, 2, 3 ----
Procedure for Application of the Theorem:
(a) Compile the list of independent quantities or variables, say,

V1 = V (V2 ,V3 − − − − − V7 )
or f (V1 ,V2 ,V3 ,− − − − V7 ) = 0 where Vi is one of the variables 4.93

Let the fundamental dimensions be M, L, T and , so that N = 7 and n = 4

(b) Select Nmax of the variables which do not form a dimensionless group. These are the
primary variables. In most practical cases Nmax = n. Each of the primary variables will
contain one or more of the fundamental dimensions.
(c) Form a dimensionless group, ie a -term, by multiplying one of the remaining variables
by a product of powers of the primary variables. With N = 7, and Nmax = n = 4 in the
example case above, there will be 7 – 4 = 3 -terms.

A correlation can subsequently be developed, using the -terms, thus:

 ( 1 ,  2 ,  3 ) = 0
or  1 =  ( 2 ,  3 ) 4.94
4.4.2 Application of Buckingham Pί – Theorem to Fully Developed Convection Heat Transfer
Consider fully developed steady flow in a circular tube, with convection heat transfer. From
experience,

142
 h = f (V ,  ,  , C p , D, k ) 4.95
or (h,V ,  ,  , C p , D, k ) = 0 4.96

There are a total of 7 variables or physical quantities. The fundamental dimensions are M, L, T and .
Hence n = 4. We make Nmax = 4. There will thus be 3 -terms. Let the primary variables be
V , , , C p .
Note: The selected primary variables cannot form a dimensionless group. Any other set of 4 variables
which do not form a dimensionless group (except a trivial one in which each variable is raised to power
zero) may be chosen.
Then 1 = V a  b  c C pd D 4.97

Dimensionally L (
−1
) (ML
a −1
 −1 ) (ML ) (L 
b −3 c 2 −2
 −1 ) L = M o Lo T o o
d
4.98

The –term is dimensionless, that is, the power of each dimension in π is zero. Consequently, for
dimensional equality, the power of each dimension on the left hand side of equation (4.98) is also zero.
Thus
For L: a − b − 3c + 2d + 1 = 0
: − a − b − 2d =0
M: b+c =0
: −d =0
Hence d = 0; a = 1; c = 1, b = −1
VD
 1 =   Re 4.99

It can be shown that:
 Cp
 2 = V a  b  c C pd k yields that 2 =  Pr 4.100
k
h
 3 = V a  b  c C pd h yields that 3 = = St 4.101
VC p
h h D  k hD k Nu
But St = . . . = = 4.102
VC p VC p D  k  VD  C p  Re r
  

   k 
St is called the Stanton Number.

From the above set of dimensionless numbers or groups, we may develop a correlation in the form:

St = f (Re, r ) 4.103
or Nu = f (Re, r ) 4.104

Once the Nusselt Number is obtained, the heat transfer coefficient, h , is then given, from the definition
of Nu , by:

143
 Nu . k
h = 4.105
D

4.5 SOME CONVECTIVE HEAT TRANSFER CORRELATIONS

1. Fully Developed Lamina Flow with Negligible Friction Heating; Forced Convection

Table 4.2 Some Forced Convection Lamina Heat Transfer Correlations

Geometry Boundary Wall Nusselt Number
________ condition Nu
(a) Circular tube qw = const. (Nu )b = 4.36
,, Tw = const. ,, = 3.66
(b) Parallel Plates qw = const. ,, = 8.23
,, Tw = const. ,, = 7.60
(c) Triangular duct qw = const. ,, = 3.00
,, Tw = const. ,, = 2.35
1 1

(d) Flat plate Tw = const. (Nu ) x = 0332 Pr Re

3 2
x
1 1
(Nu )L = 0.664 Pr Re 3 2
L

Where (Nu )b = hb De k b
(Nu )x = h x k (local) properties are
(Nu )L = h L k (average over L) calculated at
Re x = V x /  free stream
Re L = V L /  conditions
b = bulk or average fluid condition; and q w = hb (w − b )
L = plate length
 = free stream; and q w = h (w −  )

2. Fully Developed Turbulent Flow with negligible friction heating; Forced Convection
(a) Tubular Geometry:-

(i) hb D k b = 0.023(GD  )b
0.8
( C p k )b
0.4
4.106
ie Nub = 0.023 Re b
0.8 0.4
Prb Mc Adams Correlation 4.107
where G = m A = (V )b = mass flux
b = bulk condition

(ii) (h GC ) (
p b f C pb k b )
23
= 0.023 (GD  f )
0.2
4.108

144
 f
0.47

ie St b Pr
23
  = 0.023 / Re 0.2
b   4.108a
 b 

0.47
 
Nu b = 0.023 Re b0.8 Prb1 3  b  Colburn correlation 4.109
 
 f 
1
(w + b )
where f = film condition, ie at T =
2
and h GC p = h VC P = Stantan Number

(iii) (h GC p )b ( C p k )b ( w  b )0.14 = 0.027 (GD  b )0.2

23
4.110a
0.14
 w 
ie St b Pr 23
  = 0.027 Re b0.2 4.110b
 b
b

0.14
 
Nu b = 0.027 Re 0.8
Pr  b
13
 Sieder & Tate Correlation 4.110c
 w
b b

where w = wall condition, ie at wall temperature

Correlations ί to iii are applicable within the following ranges of conditions:

0.5  Pr  120
2300  Re  10 7
L D  50

(b) Flat Plate:-

1

(Nu )x = 0.0296 Pr Re 0x.83

( Local at x) 4.111
1

(Nu )L = 0.037 Pr Re 0L.8

3
(average over length L) 4.112

Pr and Re are suggested to be calculated at T = Tf

 C p 
Thus Pr f =  

 k f
Vx  f VL
(Re x ) f =f ; (Re L ) f =
f f

145
 and f = 1
2
(w +  )

The above equations are applicable for

Re x or Re L  8  105

3. Fully Developed Lamina Flow; Natural Convection

(a) For Vertical Plate and Tw = constant

(Note Tw can be greater or less than ambient Temperature)

1
Nu x 2
0.676 Pr
= ( Local at x) 4.113
(Grx 4) 14
(0.861 + Pr )
1
4
1
Nu L 0.902 Pr 2
= (average over Length L) 4.114
(GrL 4)1 4 (0.861 + Pr )
1
4

The above equations are applicable within the range

10 4  Gr . Pr  109. For Lamina natural convection, Gr . Pr  109

g x 3
(w −  )
Grx = Grasshoff Number =
v2
 = Coefficient of volumetric expansion; v = kinematic viscosity

(b) Flow Over Horizontal Cylinder and Tw = constant; Fluid = air or water
1
Nu = 0.56 (Gr. Pr) 2

103  Gr. Pr  109 4.115

and Pr in the range of 1 - 10

4. Fully Developed Turbulent Flow with Natural Convection
For Vertical Flat Plate and Tw = Constant

0.0295(Grx ) 5 Pr
2 7
15
Nu X = 2 x is measured from start of flow See Fig 5.1 a and b 4.116
1 + 0.494 Pr 2 3  5

 
0.0246(GrL ) 5 Pr
2 7
15
Nu L = 2 L = plate height 4.117
1 + 0.494 Pr 2 3  5
 
Applicable for 10  Gr. Pr  10
9 12

5. Approximate Natural Convection Correlations for air over Horizontal Square Plates,
Tw = constant.

146
 h = 0.27 ( / L )
1 0
4 Btu / hrft 2 F for 0.1  L3   20 4.118
h = 0.22 () for 20  L3   30,000
1
3 4.119
The above equations apply to heat transfer from a hot plate facing up, or cold plate facing
down.

6. Fully developed Lamina flow through a Circular Tube Annulus. Forced Convection

Nu on Inner Tube Outside Nu on Outer Tube Inner Surface. Inner tube

Surface. Outer tube is insulated is insulated
ro Ri (q.)r o
constant
w )ro = const. ro Ri q.)R = constant
i
w )Ri = constant
0.0 0.0 0.0 0.0 4. 36 3. 66
0.05 17. 81 17. 46 0.05 4. 792 4. 06
0.10 11. 91 11.56 0.10 4. 834 4. 11
0.20 8. 499 - 0.20 4. 883 -
0.25 - 7.37 0.25 - 4. 23
0.40 6. 583 - 0.40 4. 979 -
0.50 - 5.74 0.50 - 4. 43
0.60 5. 912 - 0.60 5. 099 -
0.80 5. 58 - 0.80 5. 24 -
1.00 5. 385 4. 86 1.00 5. 385 4.86

4.6 COMBINED MECHANISMS OF HEAT TRANSFER

Different mechanisms and processes of heat transfer may exist, simultaneously, in a given problem.
Consider, for instance, heat transfer through the medium sketched in Fig 4.10.

Fig 4.10 Combined Radiation, Conduction and Convection Heat Transfer Processes
The heat transfer processes may be indentified as follows:
1 – 2 Radiation only

147
 2 – 3 Conduction only
3 – 4 Convection only

(
Q = Q 12 = AF12 14 − 24 )
(
= AF12 1 + 2 )(
+1
2
1
2
2
2 )( −  ) 4.120
  
Q12 = Q23 = Q34 = Q 
Hence 1 − 2 = Q / Af 12 (1 + 2 ) 12 + 22 ( ) for radiation between 1 & 2 4.121
 −  = Q L / kA
2 3 for conduction from 2 to 3 4.122
3 − 4 = Q / hA for convection from 3 to 4 4.123
Adding 4.121, 4.122 and 4.123 yields:

Q  1 L 1
 1 − 4 =  + + 
( )
4.124
A F12 (1 + 2 ) 1 + 2
2 2
k h

A(1 − 4 )
Q = 4.125
 1 L 1
 + + 
(
F12 (1 + 2 ) 1 + 2
2 2
k h )
For any heat transfer situation, we can always define an overall heat transfer coefficient U , such
that:

Q = UA or U = Q A 2.171 or 2.172

In the above example, the overall heat transfer coefficient is given by,
1
U= 4.126
 1 L 1
 + + 
(
F12 (1 + 2 ) 1 + 2
2 2
k h )
 1  
The overall thermal resistance Rth = = UA  2.128
  Q
From equation 4.124 or 4.125 or 4.126, Rth is given by
1 1 L 1
Rth =  + + 
( )
2.127
A F12 (1 + 2 ) 1 + 2
2 2
k h

148
 PROBLMES-4

P4.1 Obtain the simplified 2-D (x, y) continuity, momentum and energy equations for lamina flow
heat transfer between two parallel plates, beginning with the non-simplified equations, if the
following conditions exist:

• Steady state
• Fully developed flow and heat transfer
• No heat generation and no shaft work
• Incompressible fluid
• Contribution of conduction in x-direction is much less than that in the y-direction

By non-dimensionalising the resulting energy equation, determine the dimensionless groups

which a heat transfer result such as Nusselt Number, Nu, can be expressed in terms of.

P4.2 (a) Give the expression for, and the physical interpretation of, each of the following
dimensionless numbers (i) Nusselt Number, Nu (ii) Prandtl Number, Pr (iii) Reynolds
Number Re and (iv) Grashoff Number, Gr.
(b) Show, by dimensional analysis, that for fully developed lamina flow over a flat plate, a
relation can be obtained of the form: Nu x = a(Re x ) (Pr ) , where a, n and m are
n m

constants, Pr is the Prandtl Number, Nu x is the Nusselt Number and Re x is the

Reynolds Number at a point x from the leading edge of the plate.

149
 (c) In lamina flow of water over a flat plate, the velocity boundary layer thickness, given
by  = 5 x Re1x 2 , is estimated to be 0.64 cm at a location x1 from the leading edge of
the plate. If Nu x = 0.332 Re1x 2 Pr1 3 , estimate the local and average (over 0-x1) heat
transfer coefficient, if water has a viscosity
 = 0.554 10 kg ms , C p = 4.18 kJ kg C,  = 1000kg m and k = 643*10-3 W mC.
* −3 3

Take free stream velocity to be 2 m s .

P4.3 In the temperature range 100 C to 300 C, the thermal conductivity of a steel is given as
k = 55.36 − 0.0346  W mC , where T is the temperature in 0C. High temperature superheated
steam flows through a 73mm od steel pipe from a steam super-heater to a turbine, at the rate of
0.1 kg/s. The pipe thickness is 5mm. Due to damage to pipe insulation, the pipe became
exposed to ambient air. Find the rate of heat loss per meter length of the exposed pipe if the
steam temperature is 260 C, and the exposed pipe surface temperature is 220 C. For the steam,
k = 0.039 W mC ,  = 0.000019 kg ms, C p = 1.967 kJ kgC.
Assume that for heat transfer inside the pipe, Nu = 0.023Re 0.8 Pr 0.4 .

P4.4 Natural convection heat transfer coefficient over a vertical flat plate in air is assumed to be a
function of the fluid properties k, Cp, β and the parameters ΔT, g, L. Show, by dimensional
analysis, that Nu = f(Gr, Pr) where:

Nu = Nusselt Number
Pr = Prandtl Number
Gr = Grashoff Number
ΔT = Tplate – Tambient
L = Plate Height
For natural convection heat transfer from a large vertical cylinder, the vertical flat plate
equation can be used, thus:

0.902 Pr1 2
Nu = (Gr 4)1 4
(0.861. + Pr )
Calculate the rate of heat loss from the vertical surface of a cylindrical water tank of diameter
0.3m and height 0.3m. Tank surface temperature is 50 C and the ambient temperature is 30 C.
For air
k = 0.028 W mC,  = 0.000019 kg ms, C p = 1.009 kJ kg C,  = 0.0033 C ,  = 1.1357 kg m 3
−1

and g = 9.81 m/s2.

P4.5 A rectangular box of outside dimensions 0.01mx1.5mx1.0m contains a refrigerant which is

evaporating at a temperature of 10 C. Air flows over the surfaces of the box parallel to the 1.5
m length, at a speed of 75 km/hr. The free stream temperature of the air is 30 C. Fully developed
average Nu for forced convection over a flat plate of length L is given by
Nu L = 0.037 Pr 1 3 Re 0L.8 (turbulent flow) and Nu L = 0.664 Pr 1 3 Re1L 2 (lamina flow). Determine
the rate of evaporation of the refrigerant if its talent heat of evaporation is 1347.5 kJ/kg. For air
k = 0.0273 W/mC, e = 1.1357 kg/m3, μ = 1.87*10-5 kg/ms and Cp = 1.008 kJ/kg-C.

150
 • Neglect the thermal resistances between the refrigerant and the outer surface of the box,
as well as the heat transfer at the small ends of the box perpendicular to the flow direction.
Turbulent flow exists when ReL>8*105. Fully developed flow is assumed to exist over the
length of the box.

P4.6 Air at 30 C and atmospheric pressure flows over a flat plate solar collector with no glazing, at
a free stream velocity of 3.5 m/s. The plate is 2m*2m and is at 90 C. If solar energy is incident
on the plate at a total intensity of 1100 W/m2, calculate the collector efficiency (ie useful energy
collected/total incident energy). Assume that all the energy not lost by convection to the
atmosphere is usefully extracted from the collector. Refer to your text or reference book for
properties of air.

P4.7 A parabolic concentrating solar collector concentrates the direct beam component of the
radiation incident on it unto the focal line of the parabola. Such a collector consists of a
parabolic trough with a mirrored inside concave surface. The trough has an aperture of 0.5m
wide * 2.0m long. A copper tube of length 2.0m, i.d = 10mm and o.d = 12.7mm, with a
blackened outside surface, is positioned along the focal line of the trough. Beam component of
solar radiation may be taken as 900 W/m2. Overall heat transfer coefficient from the water
flowing in the tube to the ambient is 10 W/m2C with respect to the tube outside surface area.
Ambient temperature is 30 C. If water enters the tube at 25 C, at a flow rate of 1.5 kg/min,
determine the bulk exit temperature of the water. Cp of water = 4.18 kJ/kg-C. Assume a linear
water temperature variation with length.

CHAPTER FIVE
HEAT EXCHANGERS
Heat exchangers are equipment designed for the orderly transfer of heat from a heat source to a heat
sink. They may be single surface or multi-surface heat exchangers. In single surface heat exchangers,
heat is transfered across a single surface, from the heat source to the heat sink. An example is a hot
solid rod. The heat source is the rod, while the heat sink is the ambient. Heat is transfered from the rod
to the ambient across the surface of the rod only. Other examples are illustrated in Fig 5.1. For multi-
surface heat exchangers, heat is transfered across two or more surfaces between the heat source and
the heat sink. Examples are double tube, shell and tube and cross-flow heat exchangers. These are
shown in Figures 5.2, 5.3 and 5.4.

w o
Vx x

a a

x
Vx
o
w
(a) Natural Convection on a
Vertical Flat Plate, Tw > Ta 151
 (b) Natural Convection on a
Vertical Flat Plate, Tw < Ta

a
w
a
w

(c) Natural Convection on a

Horizontal Plate. Hot Surface (d) Natural Convection on a
Facing Up, Tw > Ta Horizontal Plate. Cold Surface
Facing Down, Tw < Ta


w a

w
(e) Natural Convection on a
Horizontal Circular Rod, Tw > Ta
(f) Forced Convection over a
Horizontal Flat Plate, Tw < T∞
Figure 5.1 Single Surface Heat Exchangers

Figure 5.2 Double Tube Heat Exchanger (Counter Flow)

152
Fig 5.3 Shell and Tube Heat Exchangers

153
 Figure 5.3 (Continued) Shell and Tube Heat Exchangers

Tubes Tube Flow

End Plate

Cross-Flow

Figures 5.4 Cross-Flow Heat Exchanger

5.1 HEAT TRANSFER RATE AND LOG MEAN TEMPERATURE DIFFERENCE

METHOD FOR SIMPLE HEAT EXCHANGER ANALYSIS AND DESIGN

154
In simple heat exchangers, the temperature of the hot fluid or the cold fluid may be represented by a
single value at a given cross-section of the heat exchanger. Consequently, hot or cold fluid temperature
varies with the length only in the heat exchanger, ie with axial location only. Thus, whereas double
tube heat exchangers or shell and tube heat exchangers with single tube and single shell passes are
simple heat exchangers, cross-flow heat exchangers and multi-pass shell and tube heat exchangers are
not.

Counter Flow

1

(c) Neither Fluid Condensing or Evaporating

Figure 5.5 Axial Temperature Profiles for Simple Heat Exchangers

Consider Fig 5.5c. Over the elemental distance x , the surface area for heat transfer between the hot
and cold fluids = A . If U = overall heat transfer coefficient, then

155
 Q h −c = U (h − c )A 5.1

If there’s no shaft work, no heat generation in the fluid, negligible friction heating and negligible
kinetic energy and potential energy changes, then the energy conservation equation, or 1st Law Energy
Equation, yields that at steady state,

Q c = m c hc 5.2

Q h = + m h hh (negative sign for parallel flow) 5.3

If the heat exchanger is perfectly insulated then:

Heat lost by hot fluid = Heat gained by cold fluid

= Heat transfer from hot to cold fluid

 Q h = Q c = Q h −c = Q

But h = C p  +
1
(1 − ) p 4.18b

If the heat transfer fluids are perfect gases for which  = 1 ; or the fluids are such that 1 −  ≏ 0;

or the process is such that p ≏ 0 (ie effect of pressure change is negligible), or
1
(1 − )p  C p

, then

h ≏ C p  5.5
 Q =  mh C ph 5.6a
Q = m C  c pc c 5.6b
Q = U (h − c ) A 5.6c

From 5.6a and 5.6b

Q
h = 5.7a
 Ch
Q
and c = 5.7b
Cc

where Ch = m
 h C p h = Heat Capacity Rate of Hot Fluid
Cc = m c C pc = Heat Capacity Rate of Cold Fluid

But  (h − c ) = h − c

156
  1 1 
  (h − c ) = Q  −  5.8
 +C h C c 

Combining equations 5.6c and 5.8 and in the limit as x → 0 , we obtain:

d (h − c )  1 1 
= U −  dA
(h − c )
5.9
  Ch Cc 

Integrating equation 5.9 yields,

 1 1 
 ln(h − c ) inlet
exit
= UA −  5.10
 + Ch Cc 
    1 1 
 ln 2  = UA −  5.11
 1   + C h Cc 

But dQ = + C h dh = Cc dc Integratin g these equation yield :

Q = + Ch (h − h ) = Cc (c 2 − c1 )
2 1
5.12
1  − h1
 + C h = Q (h 2 − h1 ) ; or = h2 5.13
+ Ch Q
1  − c1
Cc = Q (c 2 − c1 ) ; or = c2 5.14
Cc Q

Substituting for  C h and C c from 5.13 and 5.14 in 5.11 gives,

  
ln  2  =
UA
h 2 − h1 − c 2 + c1  5.15
 1  Q

= UA Q (h 2 − c 2 ) − (h1 − c1 ) = 2 − 1 

UA
5.16
Q
 
 
 2 − 1 
Hence Q = UA 5.17
    
 ln  2  
  1  
or Q = UAlm 5.18

2 − 1
where lm =  Log Mean Temperature Difference, LMTD. It is the 5.19
ln (2 / 1 )
effective temperature difference for heat transfer between the hot and cold fluids in the heat exchanger

5.2 CORRECTION FACTORS TO LMTD FOR COMPLEX HEAT EXCHANGERS

157
 2 − 1
The expression lm =
  
ln  2 
 1 

is valid for simple heat exchangers, such as double tube and single pass shell and tube heat exchangers.
The equivalent expressions for multi-pass shell and tube, and cross-flow heat exchangers are more
difficult to obtain, analytically. However, the effective  for such complex heat exchangers can be
related to the LMTD calculated as if it were a simple heat exchanger in counter flow, through the use
of a correction factor F , thus:

( ) eff complex

= Flmc 5.20
and Q complex = UAFlmc 5.21

 
 
 (hi − co ) − (ho − ci )
where lmc =
  = the log mean temperature difference obtained by
  − co 
 ln  hi  
  ho − ci  
assuming that the inlet and exit temperatures for the complex heat exchanger are those for a sίmple
heat exchanger in counter flow. 5.22

The values of F for various common heat exchangers are tabulated or plotted in terms of two
dimensionless parameters, PP and RR, where:

Q c  − c −in c Temperatur e rise of cold fluid

 = = c − out = = 5.23
Q max h −in − c −in max Max. Temp. Rise of cold fluid
Cc h
RR = = = Heat Capacity Rate Ratio of cold to hot fluid 5.24
Ch c

Temperatur e drop of hot fluid

=
Temperatur e rise of cold fluid

Figure 5.6 shows a typical plot for F for a complex shell and tube heat exchanger.

158
 RR = C c C h
F
R R= = h c
hi − ho
=
co − ci

c  − ci
PP = = co
max hi − ci

Fig 5.6 Correction Factor F, for Heat Exchanger with 1 Shell Pass and 2, 4, 6, etc Tube Passes
Reference: (Kothandaraman and Subramanyan, 1977)

159
5.3 NTU METHOD OF HEAT EXCHANGER ANALYSIS AND DESIGN: HEAT
EXCHANGER EFFECTIVENESS, εt, AND NUMBER OF HEAT TRANSFER UNITS,
NTU.
LMTD method of heat exchanger analysis is possible when inlet and outlet temperatures are known.
If these are not known, other methods, such as the NTU method, are used. It is based on the heat
exchanger effectiveness, number of heat transfer units and heat capacity rate ratio, defined as follows:

Actual heat trans fer Q Q

(i) Heat Exchanger Effectiveness  t = = = 5.25a
Max possible Ht Transfer Q max C min max
C c c Tc
If Cc = Cmin, then  t = = 5.25b
C c max max
C h h h
If Ch = Cmin, then  t = = 5.25c
C h max max

Note: Maximum heat is transferred when the cold fluid outlet temperature equals the hot fluid inlet
temperature, or hot fluid exit temperature equals the cold fluid inlet temperature. In either case, the
temperature change is = hi − ci , which is the maximum available temperature change, ΔTmax.
Since C  = C  = Q , only the fluid with the lower heat capacity rate can conceptually attain
c c h h

the maximum temperature change hi − ci . So that Q max = C min max
Q
Hence  t = Expressions in 5.25b and 5.25c for  follow.
C min max
Also Q =  C t min max 5.25d

(ii) Number of Heat Transfer Units = NTU = AU/Cmin 5.26

(ii) Heat Capacity Rate Ratio = RR = Cmin/Cmax 5.27

A relationship can be developed, for a heat exchanger, between εt, RR and NTU.

5.3.1 εt = f ( RR, NTU) for a Simple Parallel Flow Heat Exchanger

160
 Fig 5.7 Axial Temperature Profiles for a Simple Parallel Flow Heat Exchanger

Q = U (h − c ) A 5.6c

Also Q = −C hh = Ccc
Q Q
h = − and c = 5.7a & 5.7b
Ch Cc
 1 1 
 h − c =  (h − c ) = −Q  +  5.28
 C h Cc 

Substituting for Q from 5.6c in 5.28, and in the limit as x → 0 , we get

d (h − c ) 1 1 
= −U  +  dA 5.29
h − c  Ch Cc 
 
ln(h − c )io = − AU  1 + 1  5.30
 C h Cc 

  − co   1 1 
ln ho  = − AU  +  3.31
 hi − ci   C h Cc 
AU  C h  AU  C c 
=− 1 +  or − 1 +  3.32
C h  Cc  Cc  Ch 

161
 ho − co  AU  C h   AU  C c 
 = exp − 1 +  or exp − 1 +  5.33
hi − ci  Ch  Cc   Cc  Ch 

From the definition of εt, we get

C h (hi − ho ) C ( − ci )

t = = c co 5.34
C min (hi − ci ) C min (hi − ci )

 t C min (hi − ci )

 ho = hi − 5.35
Ch

 t C min (hi − ci )

co = ci + 5.36
Cc

Substituting into 5.33 for ho and co , we get

   1 1 
 (hi − ci ) −  C min (hi − ci )
1
 + 

 hi − ci    h
C C c 

 AU  C h   AU  C c 
= exp − 1 +  or exp − 1 +  3.37
 C h  Cc   C c  C h 
 1 1   AU  C h   AU  C c 
1 −  t C min  +  = exp − 1 +  or exp − 1 +  5.38
 h
C C c   C h  C c   C c  C h 

 AU  C h   AU  C c 
1 − exp − 1 +  1 − exp − 1 + 
 C h  C c   C c  C h 
 t = OR 5.39
 1 1   1 1 
C min  +  C min  + 
Ch Cc  Ch Cc 

If Ch < Cc, then Ch = Cmin and Cc = Cmax

 AU  C min 
1 − exp − 1 + 
 C min  C max 
Then t = 5.40
1 + C min / C max

The same result, 5.40, is obtained of Cc < Ch so that Cc = Cmin and Ch = Cmax

1 − exp− NTU (1 + RR)

Hence t = 5.41
1 + RR

162
5.3.2 εt = f (NTU, RR) for Counter flow Simple Heat Exchanger

It can be shown that for a counter flow simple Heat exchanger,

1 − exp − NTU (1 − RR )
εt,counter flow = 5.42
1 − R exp − NTU (1 − RR )

Graphs of  T = f (NTU, RR) are plotted or values tabulated for different heat exchangers, as
illustrated in Fig 5.8

163
 R=

Effectiveness εt %

RR = C min C max

AU
Number of Heat Transfer Units, NTU =
C min
Fig 5.8 Heat Exchanger Effectiveness for 1 Shell Pass and 2, 4, 6, etc Tube Passes
Reference: (Kothandaraman and Subramanyan, 1977)

5.4 FOULING OF HEAT EXCHANGER SURFACES

Scale from dirt, rust, precipitates or oxidation build up on heat transfer surfaces and increase the
thermal resistance to heat flow. This is called fouling. Fouling is accounted for by assigning a scale
heat transfer coefficient h s , to the surface. The area is the clean surface area. The resistance due to h s
is an addition to the convective heat transfer resistance on that surface. The fouling or scale thermal
resistance is treated as a series resistance with the surface convective thermal resistance, as in
composite bodies.
For tubes, hsi and hso represent scale heat transfer coefficients on the inside and outside surfaces. The

reciprocal of h s is called the Fouling Factor R f . The latter are determined from experiments and
tabulated for various fluids, fluid quality and conditions of flow.
1
Rf = 5.43
hs
If Uclean and Udirty are overall heat transfer coefficients for a tube when the surfaces are clean and when
the inside (or outside) surface is dirty, respectively, then the overall thermal resistances may be written
as:

164
 1 1 1 ln ro ri 1
= + + + Inside surface is fouled 5.44a
AiU dirty Ai hi Ai hsi 2kL Ao ho
1 1 ln ro ri 1
= + + 5.44b
AiU clean Ai hi 2kL Ao ho
1 1 1
Subtracting 5.44b from 5.44a gives − = 5.44c
AiU dirty AiU clean Ai hs

1 1 1
R fi = = − 5.44d
hsi U dirty U clean

Consider a shell and tube heat exchanger tube. While in operation, fluid flows inside the tube and over
its outside surface. Fouling is therefore expected to occur on both its inside and outside surfaces. Fig
5.9 illustrates the thermal resistances encountered while moving from the fluid inside the tube to the
fluid outside the tube. These are inside surface convective thermal resistances 𝑅𝑖 , fouling thermal
resistance on the inside surface 𝑅𝑠𝑖 , thermal resistance of tube material 𝑅𝑡 , fouling thermal resistance
on the outside surface 𝑅𝑠𝑜 and outside surface convective thermal resistance 𝑅𝑜 .

Fig 5.9 Surface, Scale and Tube Thermal Resistances for Radial Heat Flow Through a Fouled
Tube

The overall thermal resistance is given by,

1 1  1 1 ln(ro ri ) 1 1 
Rth = = = + + + +  5.45
AiU i AoU o  Ai hi Ai hsi 2kL Ao hso Ao ho 
U i or U o is the overall heat transfer coefficient with respect to the tube inside or outside surface area,
Ai or Ao , respectively. 𝐿, 𝑟𝑖 𝑎𝑛𝑑 𝑟𝑜 are tube length, inside and outside radii, respectively.

165
 Table 5.1 Heat Exchanger Fouling Factors (m2 K/W or m2C/W)

Temperature of heating medium…… Up to 1150C 115-2040C

Temperature of water………………. 520C or less above 520C
Water velocity m/sec……………... 0.91 and less Over 0.91 0.91 and less Over 0.91

Distilled water 0.000088 0.000088 0.000088 0.000088

Sea water 0.000088 0.000088 0.000176 0.000176
City or well water 0.000176 0.000176 0.000352 0.000352
Treated boiler feed water 0.000176 0.000088 0.000176 0.000176
Mississippi river water 0.000528 0.000352 0.000704 0.000528

Liquid gasoline, organic vapors……..……….…………………………………….. 0.000088

Refrigerating liquids, cooling brine, oil-bearing steam………….………………… 0.000176
Refrigerating vapors, distillate bottoms above 200 API, air…………………....... 0.000352
Fuel oil, salty crude oil, residual bottoms less than 200 API………....................... 0.000880
Diesel exhaust gas, coke oven gas, cracking unit residuum..……………………… 0.001761

*Converted from Standards of Tubular Exchanger Manufacturers’ Association, 3rd Ed, New York,
1952.
o
API = Degree American Petroleum Institute
It is a measure of specific gravity used especially for petroleum crudes and distillates

141.5

AI = − 131.5
Sp Gr 60 / 60

where Sp Gr 60/60 = Specific Gravity of the liquid at 600F.

166
 PROBLEMS-5
P5.1 (a) Sketch the flow diagram and temperature distribution curves for a simple counter flow
heat exchanger under the following conditions of the heat capacity rate C of the fluids:
(i) C1  C 2 (ii ) C1  C 2 (iii) C1 = C 2 . The indices 1, 2 stand for hotter and colder fluids,
respectively.
(b) Show that for a simple heat exchanger, for which the overall heat transfer co-efficient
may be assumed constant at any section of the heat exchanger perpendicular to the axial
direction, the heat transfer rate between the two fluids is given by:

Q = UAlm
where U = overall heat transfer co-efficient based on A
A = heat transfer surface area
(2 − 1 )
lm = log mean temperature difference lm = 1,2 refer to

ln 2
1
ends of the heat exchanger.

(c) Water vapour is obtained in a distiller as saturated vapour at 100 C. Determine the heat
transfer area required to condense and sub-cool 400 kg/hr of the water vapour to 50 C if the
coolant is water with a flow rate of 20,000 kg/hr available at 20 C. Compare the areas
required for a simple heat exchanger in (i) counter flow and (ii) in parallel flow
configuration. Assume an overall heat transfer co-efficient U of 2000 W/m2C during
condensation and 1163 W/m2C during subcooling, specific heat for water of 4.2 kJ/kgC
and an enthalpy of condensation of the vapor of 2256.7 kJ/kg. What practical deductions
can you make from your results?

P5.2 Superheated refrigerant (R-12) vapour at 75 C enters a condenser. The saturation temperature
at the condenser pressure is 45 C. The R-12 leaves the condenser as sub-cooled liquid at 35 C.
The coolant is water and it enters the condenser at 20C.

(i) Sketch the temperature profiles for the two streams as they flow through the heat
exchanger, if it is a shell and tube heat exchanger with one shell pass and one tube pass
in counterflow.
(ii) Repeat (i) for a one shell-pass and two-tube pass heat exchanger, with the shell fluid
well mixed.
(iii) Discuss the relative values of the heat transfer co-efficient in different sections of the
shell flow and the tube flow.
(iv) Calculate the exit temperature of the coolant for the following values:

Fluid Property R-12 Water

h fg (kJ kg ) 125.2 -
C p − vapour (KJ kg − C ) 0.78 -
C p − Liquid (KJ kg − C ) 1.02 4.18
flow rate kg hr 600 2850

167
 (v) If the refrigerant enters the one shell-pass and one tube-pass heat exchanger as saturated
vapour and leaves as saturated liquid, how many tubes each of 19mm i.d and 2m long
will be needed? Cooling water is in turbulent flow through the tubes at a bulk velocity
of 3 m/s. Thermal resistance of the tube material and on the tube outside surface are
negligible. The tubes are assumed clean. For the water
kJ
 = 0.000732kg / ms;  = 998 kg / m 3 ; C v = 4.178 and k = 0.62 W mC.
kgC
P5.3 Super heated ammonia gas initially at 80 C is in laminar flow through a horizontal tube
immersed in a stagnant pool of water. The water temperature is constant at 22 C. The ammonia
leaves the tube as saturated liquid at 25 C. The tube is 18mm i.d * 25mm o.d. The following
mean property values may be assumed.

Water NH3-gas
k W/mC 0.605 0.028
Cp kJ/kg-C 4.18 2.54
ρ kg/m3 1000 6.23
μ kg/ms 9.18 x 104 1.09 x 105
β (coeff of volumetric expansion) C-1 0.00021 -

Enthalpy of evapouration of the ammonia is 1178 kJ/kg, gravitational acceleration g = 9.81

m s 2 and ammonia flow rate is 0.35 kg/hr.

(i) Sketch the temperature profiles of ammonia and the water along the heat exchanger
(ii) Calculate the length of the tube required.

The following assumptions may be made:

• Heat transfer co-efficient on the inside tube surface for condensation is very large
compared to that on the outside surface of the tube
• Tube material thermal resistance in negligible
• Heat transfer co-efficient from NH3 gas to tube inside surface is much less than that from
the tube outside surface to the water
• Heat transfer coefficient may be calculated at constant wall temperature

For natural convection, Nu for heat transfer over a circular cylinder is given by:

Nu = 0.53(Gr.Pr)1/4

Where Gr = g ( − x )D / ( /  ) ;
3 2

T = Tube surface temperature

Tx = Temperature of the water
D = outside diameter of tube.

168
 CHAPTER SIX
MASS TRANSFER
In mass transfer, we deal with the transport of one or more identified material component(s) through a
medium containing other matter. The medium may be

Stationary - hence Mass Transfer in a Stationary Medium

In Motion - hence mass Transfer in Super-imposed Convective System

In general the components of interest may be non-reacting, in which case there is no generation or
absorption (-ve generation) of the component(s); OR there could be reactions that lead to the generation
or absorption of the components.
A particular application of the principles of mass transfer in stationary systems is the drying of
agricultural and industrial products, eg. crops, foods, wood, wood pulp, photographic film, etc, - with
or without heat transfer. Humidification is reverse drying, arising from a change in the direction of the
mass transfer potential gradient.
6.1 MASS TRANSFER IN STATIONARY SYSTEMS: DIFFUSION
The mechanism of mass transfer in a stationary system is considered to be that of molecular transport
of the matter of interest in consequence of molecular motion and therefore, is modeled to be that of
diffusion. It is similar to heat diffusion in a stationary medium, ie to conduction. Transport of any
quantity by a diffusion process is such that the flux of the transported quantity is proportional to the
gradient of the transport potential and is in the direction of decreasing potential. For mass diffusion,
the most common transfer potential is mass concentration of the diffusing component. Mass diffusion
and heat diffusion are described by similar particular laws – called the rate equation.
The rate equation or particular law of mass diffusion is called Fick’s Law and is given as
C a
wa )x = − D 6.1
x
W a 
where wa )x =  = Mass Flux of diffusing Component, a, in the x-direction. (kg/sm2)
A x
C a = Concentration of component, a, in the medium, kg/m3
D = Mass Diffusion Coefficient, m2/s

Equation 6.1 may be expressed in terms of mass fraction C a = C a /  ; or molal concentration Na ;

*

or mole fraction N a = N a / N , where  = bulk density of the mixture and N = bulk molal density
*

of the mixture.
The similarities between heat and mass diffusion may be deduced from the above discussion and
relationships and from the corresponding discussions and relationships on conduction (See sections
2.2 and 2.3). These similarities manifest in the existence of the following correspondences:

169
 Table 6.1 Corresponding Terms in Heat and Mass Diffusion
Parameter Heat Diffusion Mass Diffusion
Quantity Diffused Heat, Q (J) Mass, W (kg)
Driving or Transfer Potential Temperature, T0C Mass Concentration, C kg/m3
C a
Diffusion Flux equation q x = −k

x
(
W / m2 ) wa )x = − D
x
(kg sm 2 )
Diffusion Coefficient Thermal Conductivity Mass Diffusion Coefficient
k (W / mK ) (
D m2 / s )
* Note: Temperature gradients can also cause mass diffusion (Soret Effect, used in isotope
separation). Also pressure gradients can cause diffusion (used in diffusion through semi-permeable
membranes). However temperature and pressure gradients have to be very high in order to cause
significant mass diffusion. Diffusion caused by mass concentration gradients governs most diffusion
of engineering interest.
For mass transfer through fibrous, porous, granular and other specially structured solids, the transport
process, even at uniform pressure and temperature, is not exactly diffusive, but is only approximately
so, or assumed to be so, using an effective diffusion coefficient, Deff.

6.1.1 Molecular Diffusion in Gases

The diffusion coefficient in two component gas mixtures in approximately given by the semi-empirical
equation,

1
3  1 1  2
  2
+ 
 M1 M 2 
D = 0.0069 ft 2 6.2
2 hr
 V1 
1 1
3
+ V 2 3
 

where T = temperature in 0R
M = molecular weight
V = atomic volume (see Table 6.2)
P = pressure in atmospheres
1,2 = components 1 and 2 in the mixture
 m2 s 
To get D in m2/s, multiply 6.2 by 2.58*10-5  2 
 ft h 

Table 6.2 Atomic Volumes (Refer also to Text or Reference Books)

170
 Air 29.9 Nitrogen (In Molecule N2) 15.6
Antimony 24.2 (In Primary Amines) 10.5
Arsenic 30.5 Oxygen (In Molecule O2) 7.4
Bismuth 48.0 (In acids) 12.0
Bromine 27 Iodine 37.0
Carbon 14.8 Fluorine 8.7
Hydrogen (in molecule H2) 14.3
(in compounds) 3.7

Kopp’s law of additive volumes applies to Atomic Volumes. Thus Atomic Volume of
CO 2 = 14.8 + 2 * 7.4 = 29.6.
1 3
Equation 6.2 gives only approximate values. For instance, it shows that D  and that D   2 .

Experimental values show that D   . Experimental values are to be used where available.
2

6.1.2 Molecular Diffusion in Liquids

Diffusion in liquids is much slower than in gases. The particular diffusion equation (equation 6.1) is
assumed applicable in liquid–liquid systems. The applicable values of the diffusion coefficient D are
obtained from experimental data–which are scanty. For diffusion of common organic and inorganic
solutes in dilute solutions in the common solvents water, alcohol and benzene, the following
approximate semi–empirical equation has been suggested by C.R Wilke

4 *10 −7 
D = ft 2 / hr 6.3
 V  3 − K l 
1

 
where T = temperature in 0R
µ = dynamic viscosity of solution in lbm ft.hr
V = atomic volume (as in Table 6.2)
K1 = 2.0, 2.46 and 2.84 for water, methyl alcohol and benzene, respectively.
 m2 / s 
Multiply D above by 2.58*10  ft 2 / hr  to change to m2/s
-5 
 

6.1.3 Molecular Diffusion in Solids

Diffusion in solids is much slower than in liquids. Diffusion of gases and liquids in solids finds much
engineering applications (drying of foods, paper pulp, cement slurry, etc). Because of the structure of
the solid medium which may be fibrous, granular or porous, experimental results show that the mass
flux is not proportional to the concentration gradient, hence the drying process is not diffusive and is
not governed by the diffusion law of 6.1. However equation 6.1- the rate equation, is sometimes used,
but with an effective diffusivity, Deff, which is determined empirically.

6.2 DIFFUSION EQUATION IN A STATIONARY SOLID OR FLUID MEDIUM

The equation is developed by considering the mass balance, or conservation of mass, of the diffusing
component. Consider a 2 – D stationary medium.

171
 (w1 ) y x + (w1 ) yx
y

Fig 6.1 Mass Transfer in a 2-D Stationary Control Volume

g1 = the mass generation rate per unit volume of the component 1 (this may be – ve, as in
absorption)
w1 = mass flux of component 1 (kg/s-m2)
C1 = mass concentration (kg/m3) of component 1

Conservation of mass of component 1 in the control volume gives that,

C1   (w1 )x 
xy = g1xy + ( )
 1 x
w y −  (w
 1 x) y + xy 
t   x 
  (w1 ) y 
+ (w1 ) y x −  (w1 ) y x + yx  6.4
  y 

C1 (w1 )x (w1 ) y

= g1− −
t x y
6.5
Introducing Fick’s Law – Equation 6.1, we obtain:

C1   C1    C1 

= g1 + D  + D  6.6
t x  x  y  y 

172
If the medium is homogenous in D then D is constant and 6.6 becomes,

C1   2 C  2 C1 
= g1+ D  21 + 
2 
6.7
t  x y 
C1
or = g1+ D  2 C1 6.8
t

For Steady State and no mass generation, then

 2 C1 = 0 6.9

It may be seen that equations 6.6 to 6.9 are similar to the equivalent conduction equations, with the
correspondences indicated in Table 6.1.
It can be shown that for mass diffusion in a stationary and isotropic medium, in cylinderical coordinates
(r , , z ) , the mass conservation equation for the diffusing component yields that,

C1 1   C1  1   C1    C1 

= g1 +  rD  + 2 D  + D  Heterogenous medium 6.9a
t r r  r  r     z  z 
C1  1   C1  1  2 C1  2 C1 
= g1 + D  r + 2 +  Homogenous medium 6.9b
t  r r  r  r  z 2 
2

6.2.1 Mass Transfer Coefficient at a Surface

If diffusing component’s concentration at a surface (on the ambient side) is C1s and in the ambient
adjacent to the surface, it is C1, then a mass transfer coefficient at the surface hD can be defined such
that

W1s
w1 )s − = = hD (C1s − C1 ) 6.10
A
or W1s = Ah D (C1s − C1 )
6.11
where A = area of the surface, m2
hD = mass transfer coefficient, m/s

6.2.2 Boundary Conditions

For 1-D mass diffusion in x-direction, typical boundary conditions are as follows:

(i) Specified concentration at a boundary

C1 )x =0 = C10 6.12a
(ii) Specified mass flux at a boundary

173
 C1 
−D  = w10 6.12b
x  x = 0
(iii) No mass flux at a boundary (impermeable boundary)
C1 
 =0 6.12c
x  x = 0
(iv) A mass transfer coefficient hD at a boundary surface is specified, together with the
concentration of the diffusing component in the ambient adjacent to the surface. If we
equate the diffusion flux from the material to the surface to the convective flux from
the surface to the ambient we obtain the boundary condition, thus:

C1 
−D  = h D (C1s − C1 ) 6.12d
x  x = 0

Note: C1s is the concentration on the ambient side, at the surface.

Example 6.2 (i): Steady State Diffusion through a Rectangular Slab or a Membrane

Equation 6.9 applies, for the 1-D case

d 2 C1
= 0 6.13
dx 2
dC1
= a 6.14
dx
C1 = ax + b 6.15
Boundary Conditions

Let C1 )x =0 = C10 6.16a

C1 )x = L = C1L 6.16b
The constants a and b in 6.15 are determined using 6.16a and 6.16b. Then 6.15 becomes,

 C − C10 
 x + C10 = C10 − (C10 − C1L )
x
C1 =  1L 6.17
 L  L
dC
Mass diffusion flux is then given by w1 = − D 1 (Fick’s Law) 6.1
dx
W
 w1 = 1 = − (C1L − C10 ) = (C10 − C1L )
D D
6.18
A L L
Mass diffusion rate becomes

W1 =
AD
(C10 − C1L ) 6.19
L

174
where A = Area of slab across which the diffusion takes place, or the cross-sectional area of the slab
perpendicular to the direction of mass diffusion.
Equations 6.17, 6.18 and 6.19 are identical to the solutions for heat conduction through a rectangular
slab, with the correspondences listed in Table 6.1.

Example 6.2 (ii): Steady State 1-D Radial Diffusion Through a Hollow Cylinder
Apply the Steady 1-D diffusion conditions, with no mass generation or absorption, to 6.9b. The
Applicable Equation becomes,

d  dC1 
r  = 0 6.20
dr  dr 

The Boundary Conditions are,

C1 (ri ) = C1i 6.21a
C1 (ro ) = C1o 6.22b

The applicable equation and the boundary conditions for the above problem are identical to those of
the radial heat conduction problem through a hollow cylinder, with C1 being equivalent to T (See
Example 2.7(1). From equations 2.120, 2.123 and 2.124, we can write the solutions for the mass
diffusion problem as:
 C − C1o 
C1 = C1i −  1i  ln(r ri )
 ln(ro ri ) 
6.22

D  C1i − C1o 
w1r =  
r  ln(ro ri ) 
6.23

2DL(C1i − C1o )
W1 =
ln(ro ri )
6.24

6.2.3 Diffusional Resistance or Mass Transfer Resistance

The same electrical analogy made in heat transfer can be made for mass transfer. If the following
correspondences are made between electric charge flow and mass transfer:

Parameter Electric Charge Transfer Mass Transfer

Driving Potential Voltage, H Mass Concentration, C
Flow Rate of transported quantity
(ie Rate of Transfer) Current, I Mass Flow Rate, W
Resistance to flow R RD
then for electrical current flow H = IR 2.125

The Electrical Analogy of Mass Transfer in equation form states that

[

C = W R D
C C 6.25
or R D = or W =
W RD

175
Comparing equations 6.19, 6.11, 6.24 and 6.25, we can define,
(i) A Diffusional Resistance for the rectangular slab as:
C
RD =

=
L
AD
(s / m3 ) 6.26
W
(ii) A Surface Mass Transfer Resistance as:
C
RD =

=
1
Ah D
(s m )3
6.27
W
(iii) A diffusional Resistance to radial diffusion of mass in a hollow cylinder
C ln(ro ri )
RD = =
2DL
(s m )3
6.28
W
We may note that equations 6.26, 6.27 and 6.28 are similar to their heat transfer equivalents.

6.2.4 Drying of Solids

Solids are often dried by exposing them to an ambient of air or inert gas or by passing a steady stream
of the air or gas over them. The drying process passes through two stages namely,
• Constant Drying Rate Process. This corresponds to the initial stage, during which the surface
of the solid is continuously wet. The mass transfer resistance at the liquid-gas interface, at the
surface, is much greater than diffusion resistance in the solid and controls the mass transfer or
drying rate. Indeed the mass transfer rate is determined by the rate of evaporation of liquid
from the wet surface into the ambient or gas stream.
• Decreasing Drying Rate Process. When the surface begins to get dry (dry spots appear) drying
rate starts to decrease with time. Internal diffusion resistance becomes significant. Drying
ceases when equilibrium moisture content, C e , is reached in the solid at the surface.
C e = f (T , P, solid, and relative humidity of the gas or air). During this process, if we assume
Fick’s Law to hold, with an effective D, then for a solid rectangular slab,

C w  2Cw
=D 6.29
t x 2

Initial condition: C w (x, 0 ) = C wi 6.30a

Boundary conditions: With the same ambient conditions and surface mass transfer coefficient, hD , on
either side of the slab, the centerline of the 2-D slab is a line of symmetry. Let 𝑥 = 0 be at the
centerline. Then

C w 
 =0 6.30b
x  x =0
C w  h
 = − D (C ws − C w ) At the surface x = L 6.30c
x  x = L D

where C ws = Concentration of water vapour on the air or gas side at the surface
C w = Concentration of water vapour in the free stream

176
 2L = Slab Thickness
The solution of equation 6.29 gives moisture concentrations within the solid. It does not give moisture
concentrations in the ambient. In particular, it does not give the moisture concentration on the ambient
side at the surface. Hence C ws is not easily known. This makes it difficult to use equation 6.30c, as
given, in terms of hD and C ws . However, since drying ceases when C w in the solid at the surface =
C e (equilibrium moisture content) we can define a new mass transfer coefficient H D such that

W C 
= −D w  = hD (C ws − C w ) = H D (C wL − Ce ) 6.30d
A x  X = L

where H D = Mass transfer coefficient defined relative to equilibrium moisture content C e

C wL = Concentration of moisture (water) in the solid at the surface.
Note: C w may be stated as Concentration on Dry Basis (kg of moisture/kg of dry solid) if solid does
not change volume or contract, upon drying – ie kg of solid per unit volume is constant.
C w may be expressed as Concentration on Wet Basis (kg of moisture/kg of dry solid plus liquid)
if volume of liquid + volume of solid remains approximately constant, ie kg of solid + liquid
per unit volume is approximately constant.
The solution to 6.29, with initial and boundary conditions 6.30a, 6.30b and 6.30d becomes:

C w − Ce Sin( n L ) Cos( n x )
( )

= 2 exp −  2n Dt 6.31
C wi − C e n =1  n L + Sin( n L )Cos( n L )

where n are the roots of the equation Cot ( n L ) =

D
6.32
HDL

Fig 6.2 Moisture Concentration in a Drying Solid Slab

During Decreasing Drying Rate Period

177
6.3 MASS TRANSFER CORRELATIONS IN CONVECTION (Overview)

Mass conservation equations in convection, applied to the component being transfered, are similar to
the energy conservation equations in convection with the following correspondences: C i   and
Ci
D  k . Furthermore, mass flux and heat flux are given by similar equations wix = − D and
x

q x = −k . Also mass and heat transfer coefficients are similarly defined hD = wis / (Cis − Cim ) and
x
h = q w / (w − m ) .

When the component being transfered is in dilute concentration in the medium, the mass transfer
correlations are identical to the heat transfer correlations, using corresponding dimensionless
parameters.

Table 6.3 Some Corresponding Dimensionless Groups in Heat and Mass Transfer
Dimensionless Groups in Heat Transfer Corresponding Dimensionless Groups in
Mass Transfer
C p v Sc =
 v
=  Schmidt Number
Pr = =  Prandtl Number D D
k 
hd hD d
Nu =
k Sh =
D
Nusselt Number or Dimensionless heat transfer
Sherwood Number or Dimesionless mass
coefficient
transfer coefficient
Vd Vd
Re =   Reynolds Number Re =   Reynolds Number
 

For fully developed lamina flow in a circular tube, similar results to those for heat transfer in Lamina
flow apply, thus

Sh = 3.66 C iw = Constant, ie Constant concentration at the wall 6.33

Sh = 4.36 wiw = Constant, ie Constant wall mass flux 6.34

For fully developed turbulent flow in a Circular tube

0.83
Sh = 0.023 Re Sc 0.44 6.35

178
This is similar to McAdams Correlation for fully developed turbulent flow heat transfer in a circular
tube. The exponents for Re and Sc of 0.83 and 0.44, respectively, are only slightly modified from 0.8
and 0. 4, the latter being the exponents for Re and Pr, respectively - the corresponding dimensionless
groups in heat transfer.

PROBLEMS-6
P6.1 (a) State Fick’s Law of diffusion in equation form. Explain the terms in the equation.
(b) Transport of moisture through wood is not strictly a diffusion process. Why is that so?
What is done to enable the use of the diffusion law in the analysis of moisture transport
in wood?
(c) A dilute solution of a substance in water (liquid – liquid system) flows through a
circular tube of i.d = 19 mm at 2 m/s. The flow is steady, fully developed, turbulent and
with constant properties. Mass transfer coefficient at the wall, referenced to the mean
concentration Cm of the substance (ie the solute) is hD.
(i) State the two boundary conditions which may be used to determine the particular radial
distribution of the concentration of the solute.
(ii) If the Sherwood Number is given by Sh = 0.023 Re 0.83 Sc 0.44 , calculate the mass transfer
coefficient for the solute, hD.

Diffusion coefficient D = 1.26x10 −9 m 2 s ;  = 6.54x10 −4 kg ms;  = 996 kg m 3

P6.2 (a) By applying the law of conservation of mass to a component ‘I’, which is being
transported by diffusion through a 2-dimensional control volume in a stationary
ci   2 ci  2 ci 
homogeneous medium, show that = g + D  2 + 2  , where Ci = mass
t  x y
i

concentration of the component; D = mass diffusion coefficient and; gi = generation
rate of ‘ί’ per unit volume.

(b) If there is steady state, one-dimensional diffusion in the x-direction as well as chemical
reaction involving the component ‘ί’, simplify the diffusion equation given in 2(a).

P6.3 In a double tube arrangement, a gas is diffusing at a constant mass flux from the inner
tube outer surface into an air stream. The air is in lamina flow through the annulus. For
the annulus, the inner tube o.d. = 20 mm and the outer tube i.d = 50 mm. See Fig. below.

179
 The Sherwood number for mass transfer from the inner tube outer surface, based on the
equivalent (or hydraulic) diameter for the annulus, is 4.979. The outer tube is
impervious to mass transfer. Calculate the mass transfer coefficient on the inner tube
outer surface. The diffusion coefficient for the gas in the air medium is 8.78*10-5 m2/s.

P6.4 (a) The annular space between two concentric cylinders of height 30cm is filled with
stagnant water. The cylinders are closed at both ends. Methanol diffuses radially
outwards through the water. Outer diameter of the inner cylinder and inner diameter of
the outer cylinder are 12.5mm and 21mm respectively. If the diffusion coefficient of
methanol in water is 1.28 *10 −9 m 2 s, what is the mass transfer resistance to the radial
diffusion of methanol through the water?
(b) A gas  is in dilute concentration in an air stream and the mixture flows through a tube
of i.d 50 mm and o.d 56 mm . Mass transfer coefficient for  on the inside surface of
−3
the tube is 6.8 * 10 m s . The effective diffusivity of  through the tube material is
1.2 *10 −5 m 2 s . If the concentration of  in the air stream is 1.11*10 −3 kg m 3 , what is
its concentration on the outside surface of the tube, if the gas leaks out of the tube
surface at the rate of 4 *10 −7 kg s per unit length of the tube?

P6.5 (a) A wooden plank of thickness 2.5cm is bounded by large flat surfaces over which the
moisture concentration is uniform. The moisture concentration in the air on either side
of the plank is the same and has a value of 6.5 *10 −4 kg m 3 . Mass transfer coefficient
on each surface is 3.0 *10 −3 m s and mass diffusion coefficient of the plank is.
2.5 *10 −9 m 2 s . State the two boundary conditions for determining the 1-D moisture
concentration from the central plane of the plank to one surface.

(b) Mass transfer coefficient for moisture transport on either side of a rectangular slab is
3.4 *10 −3 m s . Calculate the overall mass transfer coefficient from the ambient on one
side of the slab to the ambient on the other side.
P6.6 Neutrons diffuse through a rectangular slab of thickness 2L, bounded by large flat
surfaces. The edges of the slab are perfectly shielded from neutron flux so that neutron
flow direction is perpendicular to the flat surfaces only. The general 2-D equation of
neutron diffusion is given by

n   n    n 
= g + d  + D 
t x  x  y  y 
where n = neutron volumetric concentration, g = neutron generation rate per unit
volume, D = neutron diffusion coefficient, x and y are Cartesian co-ordinates. There’s
steady state. The slab is homogeneous and neutron generation is directly proportional
to neutron concentration. The ambient on both sides of the slab are identical and neutron
concentration at both slab surfaces, in the slab, is the same and equal to ns
Simplify the diffusion equation for diffusion in x-direction only and obtain an
expression for the variation of neutron concentration n, across the slab thickness.
[Hint: Assume that n = 1 Cos x +  2 Sin x ].

180
LIST OF SYMBOLS

A area m2
a fin thickness m
Al aluminum
B benefits N
Bi Biot Number
b, fin width m
C mass concentration kg m 3
Cc cost N
Cp specific heat at constant pressure kJ kg K
Cr chromium
Cu copper
CV control volume
Cv specific heat at constant volume kJ kg K
c speed of light in vacuum m s
D mass diffusion coefficient m2 s
d diameter m
E energy J
e energy per unit mass J kg
E energy flow rate W
F view factor; correction factor
f frequency cycles/s;
ff fraction of total black body radiation
G mass flux kg sm 2
Gr Grashoff Number
g gravitational acceleration m s2
g  mass generation rate per unit volume kg sm 3
H voltage volts
h heat transfer coefficient; W m2K
h fg enthalpy of evaporation kJ kg
h plank’s constant Js
 electrical current amps
 intensity of emitted radiation W m2
K boltzman constant J oK
k thermal conductivity W mK
L, l length, displacement m
LMTD log mean temperature difference deg C or K
MS marginal savings N
MCc marginal cost N
m mass kg

181
 N molal concentration 1 m3
Ni nickel
Nu Nusselt Number
n number of parameters
P pressure N m 2 , bar
pp Temperature ratio
 time averaged pressure N m 2 , bar
p perimeter m
Q heat flow rate W
q heat flux W m2
R resistance (electrical, thermal, mass transfer)
RR Heat Capacity rate ratio
Re Reynolds Number
r radius m
S savings N
Sc Schmidt Number
Sh Sherwood Number
St Stanton Number
o
T temperature C, o K
 time averaged temperature o
C, o K
t time s
U overall heat transfer coefficient W m2K
u internal energy component per unit mass J kg
u  heat generation rate per unit volume W m3
V velocity m s
V time averaged velocity m s
V atomic volume
v solid angle
W work J
W mass flow rate kg s
w com ponent mass flux
kg sm 2 w  rate of work done per unit volume
W m3
w gap width m
X, x length, x-displacement m
X , Y, Z body force per unit volume in x, y, z directions N m3
y y-displacement m
Zn zinc
z z-displacement m

Greek Symbols

 thermal diffusivity m2 s

182
  coefficient of volumetric expansion C −1 , K −1
( ),  ( ) small change in a quantity ( )
 boundary layer thickness m

 emissivity
h eddy diffusivity of heat m2 s
 m eddy diffusivity of momentum m2 s
t heat transfer effectiveness
γ electrical conductivity
 wavelength m, micron
 dynamic viscosity kg ms
v kinematic viscosity m2 s
 angle radian
 density kg m 3
 normal stress N m2
 angular displacement radian
 shear stress N m2
Symbols Re-used in Radiation
 absorptivvity
e emitted radiant energy flux or emissive power W m2
r totoal radiant flux leaving a surface W m2
 Stefan-Boltzman constant W m2K 4
 transmitivity
 reflectivity
Subscripts

0, 1, 2 location or variable indentifier

e, ele, elec electrical
L L-axis, direction, component or location
m maximum, mean
in inflow
out outflow
cv control volume
 total, constant temperature
int internal
ch chemical
nuc nuclear
oth others
v constant volume; plate
p constant pressure
a ambient
 ambient, at infinity

183
s surface
r radial direction or component
 circumferential direction or component
X X – axis, direction, component or location
x x – axis, direction, component or location
y y-axis, direction, component or location
z z-axis, direction, component or location
th thermal
D mass diffusion, mass transfer
eff effective
i inside, inside surface, identifying index
o outside, outside surface
if interface
si inside surface
so outside surface
th, o overall thermal
p, eff effective parallel
f fin, film
sys system
b black body; bulk
b monochromatic black body
 monochromatic
net net
j indentifying index
sh shaft
w wall, water
c cold
h hot
lm log mean temperature difference
min minimum
max maximum
a component “a”
e equilibrium

184
REFERENCES

Mills A.F. (1995): Basic Heat and Mass Transfer published by Irwin: Chicago, London, Boston,
Toronto, etc.+

Mills A.F. (1999) Heat Transfer, 2nd Edition, published by Prentice Hall, New Jersey USA

Simonson John R. An Introduction to Engineering Heat Transfer in SI Units: published by

MacGraw-Hill Book Co.+

Jakob and Hawkins: Elements of Heat Transfer, published by John Wiley & Sons.+

Rohsenow W.M. and Choi H. (1961). Heat, Mass and Momentum Transfer published by Prentice-
Hall.+ +

Kothandaraman C.P and Subramanyan S. (1977) Heat and Mass Transfer Data Book, 3rd Edition,
published by Wiley Eastern Ltd, New Delhi, Calcutta

Tubular Exchanger Manufacturers’ Association (TEMA) (1952) Standards of the TEMA, 3rd
Edition, New York.

Rohsenow W.M. and Hartnet J.P., Editors (1973) Handbook of Heat and Mass Transfer, published
by Mc Graw-Hill.

+ Recommended Texts
++ Reference Text

185