Course No Course Title Credits

For Class use only
ACHARYA N.G. RANGA AGRICULTURAL UNIVERSITY
Course No. FDEN- 221
Course HEAT AND MASS TRANSFER
Title:
Credits: 2 (1 + 1)
Prepared by
Er. B. SREENIVASULA REDDY
Assistant Professor (Food Engineering)
College of Food Science and Technology
Chinnarangapuram, Pulivendula – 516390
YSR (KADAPA) District, Andhra Pradesh

DEPARTMENT OF FOOD ENGINEERING
1 Course No : FDEN - 221
2 Title : Heat and Mass Transfer
3 Credit hours : 2 (1+1)
4 General Objective : To impart knowledge to students on different
modes of heat transfer through extended surfaces,
study of heat exchanges and principles of mass
transfer
5 Specific Objectives :
a) Theory : By the end of the course, the students will acquire
knowledge from different modes of heat transfer,
extended surfaces, boiling and condensation
process and principles of heat exchangers which
are very essential in dairy and food industries
b) Practical By the end of the course, the students will learn
efficient design of heat exchangers on the basis of
overall heat transfer coefficient and LMTD
A) Theory Lecture Outlines
1 Introduction to Heat Transfer- Basic Mechanisms of Heat Transfer -
Conduction Heat transfer - Fourier's Law of Heat Conduction- Convection
Heat Transfer- Radiation Heat Transfer
2 The basic equation that governs the transfer of heat in a solid - Thermal
conductivity
3 One-dimensional steady-state conduction of heat through some simple
geometries - Conduction Through a Flat Slab or Wall - Conduction Through
a Hollow Cylinder - Conduction Through a Hollow Sphere
4 Conduction Heat Transfer Through A Composite Plane Wall - Conduction
Heat Flow Through A Composite Cylinder
5 The Overall Heat-Transfer Coefficient - Critical Thickness of Insulation
6 Heat Source Systems: One-dimensional steady state heat conduction with
heat generation : Heat Flow through slab / Plane Wall
7 Steady state heat conduction with heat dissipation to environment-
Introduction to extended surfaces (FINS) of uniform area of cross section -
Different fin configurations - General Conduction Analysis Equation
8 Equation of temperature distribution with different boundary conditions, Fin
Performance and Overall surface efficiency of FINS
9 Principles Of Unsteady-State Heat Transfer: Derivation of Basic Equation;

Simplified Case For Systems With Negligible Internal Resistance ; Total
Amount of Heat Transferred ; Dimensional Analysis in Momentum Transfer
10 Some important empirical relations used for determination of heat transfer
coefficient: Nusselt’s number, Prandtl number, Reynold’s number, Grashoff
number
11 Radiation - heat transfer, Radiation Properties, radiation through black and
grey surfaces, determination of shape factors
12 Introduction to condensing and boiling heat transfer, Condensation Heat-
Transfer Phenomena, Film Condensation Inside Horizontal Tubes , Boiling
Heat Transfer
13 Heat exchangers- general introduction; Double-pipe heat exchanger; Shelland-tube
heat exchanger; Cross-flow exchanger; fouling factors, LMTD
14 Design problems on heat exchangers: Calculation of heat exchanger size
from known temperatures, Problem on Shell-and-tube heat exchanger ,
Design of shell-and-tube heat exchanger
15 Introduction To Mass Transfer: A Similarity of Mass, Heat, and Momentum
Transfer Processes; Fick's Law for Molecular Diffusion
16 Molecular Diffusion In Gases : Equimolar Counter diffusion in Gases
B) Practical Class Outlines
1 Determination of thermal conductivity of milk and dairy products
2 Tutorials on heat conduction through slab, cylinder and sphere
3 Tutorials on heat conduction through slab, cylinder and sphere
4 Tutorials on heat conduction through slab, cylinder and sphere
5 Tutorials on extended surfaces (FINS)
6 Tutorials on unsteady state heat conduction
7 Determination of specific heat of food materials
8 Tutorials on determination of Nusselt’s number by dimensional analysis
9 Tutorials on LMTD and NTU method of analysis of heat exchangers
10 Study of shell and tube heat exchanger
11 Study of plate heat exchanger
12 Study on temperature distribution and heat transfer in HTST pasteurizer
13 Study on temperature distribution and heat transfer in HTST pasteurizer
14 Design problems on heat exchangers - I
15 Design problems on heat exchangers - II
16 Design problems on heat exchangers - III
References
1 Geankoplis, C.J. 1978. Transport Processes and Unit Operations. Allyn and
Bacon Inc., Newton, Massachusetts.
2 Holman, J. P. 1989. Heat Transfer. McGraw Hill Book Co., New Delhi.
3 Incropera, F. P. and De Witt, D .P. 1980. Fundamentals of Heat and Mass

Transfer. John Wiley and Sons, New York.
4 Gupta, C. P. and Prakash, R. 1994. Engineering Heat Transfer. Nem Chand
and Bros., Roorkee

LECTURE NO.1
INTRODUCTION TO HEAT TRANSFER- BASIC MECHANISMS OF HEAT
TRANSFER - CONDUCTION HEAT TRANSFER - FOURIER'S LAW OF
HEAT CONDUCTION- CONVECTION HEAT TRANSFER- RADIATION
HEAT TRANSFER
Introduction
Heat transfer is the science that seeks to predict the energy transfer
that may take place between material bodies as a result of a temperature
difference. Thermodynamics deals with systems in equilibrium; it may be used
to predict the amount of energy required to change a system from one
equilibrium state to another; it may not be used to predict how fast a change
will take place since the system is not in equilibrium during the process. Heat
transfer supplements the first and second principles of thermodynamics by
providing additional experimental rules which may be used to establish
energy-transfer rates.
As an example of the different kinds of problems that are treated by
thermodynamics and heat transfer, consider the cooling of a hot steel bar that
is placed in a pail of water. Thermodynamics may be used to predict the final
equilibrium temperature of the steel bar-water combination. Thermodynamics
will not tell us how long it takes to reach this equilibrium condition or what the
temperature of the bar will be after a certain length of time before the
equilibrium condition is attained. Heat transfer may be used to predict the
temperature of both the bar and the water as a function of time.
Basic Mechanisms of Heat Transfer
Heat transfer may occur by anyone or more of the three basic
mechanisms of heat transfer: conduction, convection, and radiation.
1. Conduction Heat transfer. In conduction, heat can be conducted through
solids, liquids, and gases. The heat is conducted by the transfer of the energy
of motion between adjacent molecules. In a gas the "hotter" molecules, which
have greater energy and motions, impart energy to the adjacent molecules at
lower energy levels. This type of transfer is present to some extent in all
solids, gases, or liquids in which a temperature gradient exists. In conduction,

energy can also be transferred by "free" electrons, which is quite important in
metallic solids. Examples of heat transfer mainly by conduction are heat
transfer through walls of exchangers or a refrigerator, heat treatment of steel
forgings, freezing of the ground during the winter, and so on.
Fourier's Law of Heat Conduction
The basic rate transfer process equation for processes such as
momentum transfer, heat transfer, mass transfer and electric current is as
follows:
driving force
rate of a transfer process = ----------- (1)
resis tan ce
This equation states what we know intuitively: that in order to transfer a
property such as heat or mass, we need a driving force to overcome a
resistance.
The transfer of heat by conduction also follows this basic equation and
is written as Fourier's law for heat conduction in fluids or solids:
where
q x
A
dT = − k
--------------- (2)
dx
qx
is the heat-transfer rate in the x direction in watts (W), A is
the cross-sectional area normal to the direction of flow of heat in m 2 , T is
temperature in K, x is distance in m, and k is the thermal conductivity in
W/m·K in the SI system. The quantity
The quantity
dT / dx
q x
/ A is called the heat flux in W/m 2 .
is the temperature gradient in the x direction. The minus
sign in eq. (2) is required because if the heat flow is positive in a given
direction, the temperature decreases in this direction.
Fourier's law, eq.(2), can be integrated for the case of steady-state
heat transfer through a flat wall of constant cross-sectional area A, where the
inside temperature is T 1 at point 1 and T 2 at point 2, a distance of x2 −x
1 m
away. Rearranging eq. (2),
qx
A
x2
∫
x1
dx
= −
k
∫
T2
T1
dT
-------(3)
Integrating, assuming that k is constant and does not vary with
temperature and dropping the subscript x on
qx
for convenience,

q
A
k
= ( T ) 1
−T
2 -------- (4)
x −x
2
1
2. Convection Heat Transfer. The transfer of heat by convection implies the
transfer of heat by bulk transport and mixing of macroscopic elements of
warmer portions with cooler portions of a gas or liquid. It also often refers to
the energy exchange between a solid surface and a fluid. A distinction must
be made between forced-convection heat transfer, where a fluid is forced
to flow past a solid surface by a pump, fan, or other mechanical means, and
natural or free convection, where warmer or cooler fluid next to the solid
surface causes a circulation because of a density difference resulting from
the temperature differences in the fluid. Examples of heat transfer by
convection are loss of heat from a car radiator where the air is being
circulated by a fan, cooking of foods in a vessel being stirred, cooling of a hot
cup of coffee by blowing over the surface, and so on.
Convective Heat-Transfer Coefficient
It is well known that a hot piece of material will cool faster when air is
blown or forced past the object. When the fluid outside the solid surface is in
forced or natural convective motion, we express the rate of heat transfer from
the solid to the fluid or vice versa, by the following equation:
q= hA ( T w
−T
f
) -------------- (5)
where q is the heat-transfer rate in W, A is the area in m 2 , T w is the
temperature of the solid surface in K, T f is the average or bulk temperature of
the fluid flowing past in K, and h is the convective heat-transfer coefficient in
W/m 2 .K.
The coefficient h is a function of the system geometry, fluid properties,
flow velocity, and temperature difference. In many cases, empirical
correlations are available to predict this coefficient, since it often cannot be
predicted theoretically. Since we know that when a fluid flows past a surface
there is a thin, almost stationary layer or film of fluid adjacent to the wall which
presents most of the resistance to heat transfer, we often call the coefficient h
a film coefficient.
3. Radiation Heat Transfer:- Radiation differs from heat transfer by
conduction and convection in that no physical medium is needed for its

propagation. Radiation is the transfer of energy through space by means of
electromagnetic waves in much the same way as electromagnetic light
waves transfer light. The same laws that govern the transfer of light govern
the radiant transfer of heat. Solids and liquids tend to absorb the radiation
being transferred through them, so that radiation is important primarily in
transfer through space or gases. The most important example of radiation is
the transport of heat to the earth from the sun. Other examples are cooking of
food when passed below red-hot electric heaters, heating of fluids in coils of
tubing inside a combustion furnace, and so on.
The rate of energy emitted by a black body is proportional to the fourth
power of the absolute temperature of the body and directly proportional to its
surface area. Thus
q emitted
= σAT
4
where σ is the proportionality constant and is called the Stefan-
Boltzmann constant with the value of 5.669 x 10 -8 W/m 2 .K 4 . The equation is
called the Stefan-Boltzmann law of thermal radiation, and it applies only to
blackbodies.
Problem Heat Loss Through an Insulating Wall
Calculate the heat loss per m 2
of surface area for an insulating wall
composed of 25.4 -mm-thick fiber insulating board, where the inside
temperature is 352.7 K and the outside temperature is 297.1 K. The thermal
conductivity of fiber insulating board is 0.048 W/m.K
Solution:
The thickness x2 − x1
= 0.0254 m.
Substituting into the eq.
q
A
= k
0.048
( T ) 1
−
2
=
x −x
T 0.0254
2
1
= 105.1 W/m 2
(352 .7−297
.1)

LECTURE NO.2
THE BASIC EQUATION THAT GOVERNS THE TRANSFER OF HEAT IN A
SOLID - THERMAL CONDUCTIVITY
The basic equation that governs the transfer of heat in a solid
Fig.2.1 Elemental volume for one-dimensional heat conduction analysis
Consider (Fig. 2.1) the general case where the temperature may be
changing with time and heat sources may be present within the body. For the
element of thickness dx the following energy balance may be made:
Energy conducted in left face + heat generated within element
= change in internal energy + energy conducted out right face
These energy quantities are given as follows:
Energy in left face =
q x
∂T
=−kA
∂x
Energy generated within element =
q .
Adx
Change in internal energy =
ρcA
∂T
dx ∂τ
Energy out right face =
q
x+
dx
=− kA
∂T
∂x
x+
dx
⎡ ∂T
∂ ⎛ ∂T
⎞ ⎤
= − A⎢k
+ ⎜k
⎟dx
⎥
⎣ ∂x
∂x
⎝ ∂x
⎠ ⎦

(In the derivations, the expression for the derivative at
x + dx
has been
written in the form of a Taylor-Series expression with only the first two terms
of the series employed for the development.)
where
.
q = energy generated per unit volume, W /m 3
c = specific heat of material, J /kg .°C
ρ = density, kg/m 3
Combining the relations above gives
or
heat source.
∂
− kA T
+
∂x
q .
Adx =
∂T
ρcA
dx ∂τ
⎡ ∂T
− A⎢k
⎣ ∂x
.
∂ ⎛ ∂T
⎞ ∂T
⎜k
⎟dx
+ q=ρc
∂x
⎝ ∂x
⎠ ∂ τ
∂ ⎛ ∂T
⎞ ⎤
+ ⎜k
⎟dx
∂x
⎥
⎝ ∂x
⎠ ⎦
This is the one-dimensional steady state heat-conduction equation with
Figure 2.2 Elemental volume for three-dimensional heat-conduction analysis:
(a) cartesian coordinates; (b) cylindrical coordinates;
(c) spherical coordinates.

The general three-dimensional heat-conduction equation is
∂ ⎛ ∂T
⎜k
∂x
⎝ ∂x
⎞
⎟+
⎠
∂ ⎛ ∂T
⎜k
∂y
⎝ ∂y
⎞
.
∂ ⎛ ∂T
⎞ ∂T
⎟+ ⎜k
⎟+ q=ρc
⎠ ∂z
⎝ ∂z
⎠ ∂ τ
For constant thermal conductivity equation can be written as
where the quantity
2 2 2
∂ T ∂ T ∂ T q 1 ∂T
+ + + =
2 2 2
∂x
∂x
∂x
k α ∂τ
.
α = k / ρc
is called the thermal diffusivity of the material.
The larger the value of α , the faster heat will diffuse through the material. A
high value of α could result either from a high value of thermal conductivity,
which would indicate a rapid energy-transfer rate, or from a low value of the
thermal heat capacity ρ c . A low value of the heat capacity would mean that
less of the energy moving through the material would be absorbed and used
to raise the temperature of the material; thus more energy would be available
for further transfer. α has the units of m 2 /s.
The three-dimensional heat-conduction equation with heat generation
for cylindrical or spherical co-ordinates is
Cylindrical coordinates:
Spherical coordinates:
2
2 2
∂ T 1 ∂T
1 ∂ T ∂ T q 1 ∂T
+ + + + =
2
2 2 2
∂r
r ∂r
r ∂φ
∂z
k α ∂τ
.
1 ∂
r ∂r
2
2
( rT )
+
r
2
1 ∂ ⎛ ∂T
⎞
⎜sin
θ ⎟+
sin θ ∂θ
⎝ ∂θ
⎠ r
2
1
sin
2
2
∂ T q 1 ∂T
+ =
2
θ ∂φ
k α ∂τ
The reduced form of the general equations for several cases of practical
interest.
Steady-state one-dimensional heat flow (no heat generation):
2
d T
2
dx
= 0
Steady-state one-dimensional heat flow in cylindrical coordinates (No
heat generation):
.
2
d T
2
dr
1 dT
+
r dr
= 0

Steady-state one-dimensional heat flow with heat sources:
2
d T
2
dx
.
q
+ = 0 k
Thermal conductivity
Thermal conductivity, k, is the property of a material that indicates its
ability to conduct heat. Thermal conductivity is measured in W/m·K. The
thermal conductivity predicts the rate of energy loss (in watts, W) through a
piece of material.
Thermal conductivity, k, also defined as the quantity of heat Q that flows
per unit time through a food of unit thickness and unit area having unit
temperature difference between faces.
The reciprocal of thermal conductivity is thermal resistivity. In general,
the thermal conductivity is strongly temperature-dependent.
Thermal energy may be conducted in solids by two modes: lattice
vibration and transport by free electrons. In good electrical conductors a
rather large number of free electrons move about in the lattice structure of the
material. Just as these electrons may transport electric charge, they may also
carry thermal energy from a high-temperature region to a low-temperature
region, as in the case of gases. In fact, these electrons are frequently referred
to as the electron gas. Energy may also be transmitted as vibrational energy
in the lattice structure of the material. In general, however, this latter mode of
energy transfer is not as large as the electron transport, and for this reason
good electrical conductors are almost always good heat conductors, viz.,
copper, aluminum, and silver, and electrical insulators are usually good heat
insulators. A notable exception is diamond, which is an electrical insulator, but
which can have a thermal conductivity five times as high as silver or copper. It
is this fact that enables a jeweler to distinguish between genuine diamonds
and fake stones. A small instrument is available that measures the response
of the stones to a thermal heat pulse. A true diamond will exhibit a far more
rapid response than the non genuine stone.

LECTURE NO.3
ONE-DIMENSIONAL STEADY-STATE CONDUCTION OF HEAT THROUGH
SOME SIMPLE GEOMETRIES - CONDUCTION THROUGH A FLAT SLAB
OR WALL - CONDUCTION THROUGH A HOLLOW CYLINDER -
CONDUCTION THROUGH A HOLLOW SPHERE
CONDUCTION HEAT TRANSFER
One-dimensional steady-state conduction of heat through some simple
geometries
Fig. 3.1 Heat conduction in a flat wall: (a) geometry of wall, (b) temperature
plot.
Conduction Through a Flat Slab or Wall
Consider a flat slab or wall (Fig. 3.1) where the cross-sectional area A
and k in are constant,
q k
The eq. = ( T1
−T
2)
can be rewrite as
A x −x
2
1
q
A
=
k
∆x
where ∆ x=
x 2
−x1
.
( T 1
−T
2)
The above indicates that if T is substituted for T 2 and x for x
2 , the
temperature varies linearly with distance, as shown in Fig.3.1(b).
If the thermal conductivity is not constant but varies linearly with
temperature, then substituting
integrating,
k = a=
bT into the above equation and

q
A
T 2 .
where
T1
+ T
q 2
a + b
2
k
=
( T1
−T2
) =
m ( T1
T2
)
A ∆x
∆x
−
b T 1
+ T
k a
2
m
= +
2
This means that the mean value of k (i.e., k m ) to use in
k
= ( T ) 1
−T
2 is the value of k evaluated at the linear average of T 1 and
∆x
The rate of a transfer process equals the driving force over the
q k
resistance and the equation = ( T 1
−T
2)
can be rewritten in that form
A ∆x
as:
where
R ∆x
/ kA
T −T2
T1
−T2
q = =
∆x
/ kA R
1 =
driving force
resis tan ce
= A and is the resistance in K/W.
Fig. 3.2 Heat conduction in a cylinder
Conduction Through a Hollow Cylinder
In many instances in the process industries, heat is being transferred
through the walls of a thick-walled cylinder, such as a pipe that may or may
not be insulated. Consider the hollow cylinder in Fig.3.2 with an inside radius
of r 1 , where the temperature is T 1 , an outside radius of r 2 having a temperature
of T 2 , and a length of L m. Heat is flowing radially from the inside surface to
the outside. Rewriting Fourier's law, with distance dr instead of dx,
q x
A
= −
k
dT
dr

The cross-sectional area normal to the heat flow is
A =2πrL
Substituting A value and rearranging, and integrating,
q
2πL
r2
∫
r1
dr
r
= −
k
∫
T2
T1
dT
q=
k
2πL
ln( r
2
/ r1
)
( T −T
)
1
2
Conduction Through a Hollow Sphere
Heat conduction through a hollow sphere is another case of onedimensional
conduction. Using Fourier's law for constant thermal conductivity
with distance dr, where r is the radius of the sphere,
the radius.
q x
A
= −
k
dT
dr
The cross-sectional area normal to the heat flow is
2
A = 2πr
Substituting A value and rearranging, and integrating,
q
4π
r2
∫
r1
dr
2
r
= −
k
∫
T2
T1
( T −T
)
4πk
1 2
q =
1/ r −1/
r
1
2
=
dT
T −T
( 1/ r −1/
r ) / 4πk
It can easily be shown that the temperature varies hyperbolically with
1
1
2
2

LECTURE NO.4
CONDUCTION HEAT TRANSFER THROUGH A COMPOSITE PLANE WALL
- CONDUCTION HEAT FLOW THROUGH A COMPOSITE CYLINDER
CONDUCTION HEAT TRANSFER THROUGH A COMPOSITE PLANE
WALL
Fig. 4.1 Heat flow through multilayer wall
Consider the heat flow through composite wall made of several
materials of different thermal conductivities and thicknesses. An example is a
wall of a cold storage, constructed of different layers of materials of different
insulating properties. All materials are arranged in series in the direction of
heat transfer, as shown in the above Figure.
The thickness of the walls are x 1, x 2 , and x 3 and the thermal
conductivites of the walls are K 1 , K 2 , and K 3 , respectively. The temperatures at
the contact surfaces are T 2 , T 3 , and T 4.
From Fourier’s Law,
q
=−K A
A
This may be written as
dT
dx
∆T
= T
x1
T
1
−T
2
= . q
K A
x2
T
2
−T
3
= . q
K A
1
2
2
−T
x3
T
4
−T
3
= . q
K A
3
1
q.
∆x
=−
KA

Total temperature difference, ∆T
=∆T
1
+∆T
2
+ ∆T
3
⎡ x1
x2
x3
T1 −T4
= q.
⎢ + +
⎣K1
A K2
A K31
where, T1 − T4
= thermal potential responsible for heat flow. The
⎤
⎥
A⎦
⎡ x1
x2
x
⎢ + +
⎣K1
A K2
A K31
3
⎤
⎥
A⎦
is known as the total thermal resistance of the
composite ass. It is similar to the electrical resistance in series.
The thermal circuit for multilayer rectangular system is shown in the
following figure.
Fig 4.2 Electrical analog of one dimensional heat transfer through composite
wall
T1
−T4
q =
R + R + R
CONDUCTION HEAT FLOW THROUGH A COMPOSITE CYLINDER
1
2
3
Fig.4.3 One-dimensional heat flow through multiple cylindrical sections and
electrical analog
Consider a long cylinder of inside radius r i , outside radius r o , and length
L, such as the one shown in above Figure 4.3. We expose this cylinder to a

temperature differential T i - T o and determine what the heat flow will be. For a
cylinder with length very large compared to diameter, it may be assumed that
the heat flows only in a radial direction, so that the only space coordinate
needed to specify the system is r. Again, Fourier's law is used by inserting the
proper area relation. The area for heat flow in the cylindrical system is
A =2πrL
so that Fourier's law is written
with the boundary conditions
q
r
dT
= −k
Ar
or
dr
q r
=−2πk rL
T = T i at r = r i ,
dT
dr
T = T o at r = r o
The solution to equation
q
r
dT
= −k
Ar
is
dr
q=
k
2πL
ln( r
2
/ r1
)
( T −T
)
1
2
and the thermal resistance in this case is
R
th
ln( ro
/ ri
)
=
2πkL
The thermal-resistance concept may be used for multiple-layer
cylindrical walls just as it was used for plane walls. For the three-layer system
shown in Figure the solution is
2π
L ( T1
−T4
)
q =
ln( r2
/ r1
) ln( r3
/ r2
) ln( r4
/ r2
)
+ +
K K K
a
The thermal circuit is also shown in Figure.
B
B

LECTURE NO.5
THE OVERALL HEAT-TRANSFER COEFFICIENT - CRITICAL THICKNESS
OF INSULATION
5.1 THE OVERALL HEAT-TRANSFER COEFFICIENT
Fig. 5.1 Heat flow with convective boundaries in plane wall
Fig. 5.2 Heat flow with convective boundaries in plane wall- Electrical
Analog
In many practical situations the surface temperatures (or boundary
conditions at the surface) are not known, but there is a fluid on both sides of
the solid surfaces. Consider the plane wall in the Figure 5.1, with a hot fluid at
temperature T 1 on the inside surface and a cold fluid at T 4 on the outside
surface. The convective coefficient on the outside is h o W/m 2 .K and h i on the
inside.
The heat transfer is expressed by
q=
h A
i
K
A
A
( T −T
) = ( T −T
) = h A T − )
1 2
2 3 o
(
3
T4
∆x
A
The heat-transfer process may be represented by the resistance
network in electrical analog Figure, and the overall heat transfer is calculated

as the ratio of the overall temperature difference to the sum of the thermal
resistances:
T1
−T4
T1
−T
q=
=
1 ∆x
A
1
+ +
∑R
h A K A h A
∴
4
i
A
The overall heat transfer by combined conduction and convection is
frequently expressed in terms of an overall heat-transfer coefficient U, defined
by the relation
q=
U A∆
T overall
o
where ∆ T overall
= T 1
−T
4 and U is
U
1
∆x
1
A
= + + W/m
hi
K
A
h
2 .K
o
value as
The overall heat-transfer coefficient is also related to the R
U =
1
R value
A more important application is heat transfer from a fluid outside a
cylinder, through a metal wall, to a fluid inside the tube, as often occurs in
heat exchangers.
boundaries.
Figure 5.3 Resistance analogy for hollow cylinder with convection

Note that the area for convection is not the same for both fluids in this
case, these areas depend on the inside tube diameter and wall thickness. In
this case the overall heat transfer would be expressed by
q =
1
h A
i
i
T1
−T
ln
o
+
2πKL
4
( r / r )
i
1
+
h A
o
o
The terms A i and A o represent the inside and outside surface areas of
the inner tube. The overall heat-transfer coefficient may be based on either
the inside or the outside area of the tube. Accordingly,
Ui
=
1
h
i
1
Ai
ln
o
+
2πKL
( r / r )
i
Ai
+
A
o
1
h
o
U
o
=
Ao
A
i
1
h
i
+
1
Ao
ln
o
2πKL
( r / r )
i
1
+
h
o
The general notion, for either the plane wall or cylindrical coordinate
system, is that
U
A
1
=
∑R
th
=
R
1
th , Overall
5.2 CRITICAL THICKNESS OF INSULATION
Fig. 5.4 Critical radius for insulation of cylinder or pipe
Consider a layer of insulation is installed around the outside of a
cylinder whose radius r 1 , is fixed and with a length L. The cylinder has a high
thermal conductivity and the inner temperature T 1 at point r 1 outside the
cylinder is fixed. An example is the case where the cylinder is a metal pipe

with saturated steam inside. The outer surface of the insulation at T 2 is
exposed to an environment at T o where convective heat transfer occurs. It is
not obvious if adding more insulation with a thermal conductivity of k will
decrease the heat-transfer rate.
At steady state the heat-transfer rate q through the cylinder and the
insulation equals the rate of convection from the surface:
( T − )
q = h A
2 -------------- (1)
o
T o
T1
−T4
q =
1 ln
o i
+
h A 2πKL
i
i
( r / r )
1
+
h A
As insulation is added, the outside area, which is
o
o
--------- (2)
A
2
= 2πr
L , increases,
but T 2 decreases. However, it is not apparent whether q increases or
decreases. To determine this, an equation similar to Eq. (2) with the
resistance of the insulation represented by Eq.(3) is written using the two
resistances:
r2
−r
R =
K A
1
lm
⎛r
⎞
2
ln
⎜
r
⎟
1
=
⎝ ⎠
2πKL
A
lm = log mean area
2
q =
ln
2
K
πL
( T1
−To
)
( r / r ) 1
1
+
r h
2
o
As the outside radius, r 2 , increases, then in the denominator, the first
term increases but the second term decreases.
Thus, there must be a critical radius, r c , that will allow maximum rate of
heat transfer, q.
To determine the effect of the thickness of insulation on q, we take the
derivative of q with respect to r 2 , equate this result to zero, and obtain the
following for maximum heat flow. The maximization condition is
d
q
dr
=
−2πL
( T −T
⎡
⎢
⎣
1
( r / r )
⎛ 1
)
⎜
⎝ r2
K
2 ln
2 1
1
K
o
+
r h
2
1
−
r h
0
⎤
⎥
⎦
2
2
2
o
⎞
⎟
⎠

T 1 ,T 0 , K, L, r o , r i are constant terms.
Therefore,
⎛ 1
⎜
⎝r2
K
1
−
2
r
1 1 =
r K r
2
2
h o
2
2
h o
⎞
⎟=
0
⎠
When outside radius becomes equal to critical radius, or r 2 = r c , we get
( r 2
) =
cr
k
h
o
Where (r 2 ) cr is the value of the critical radius when the heat transfer rate
is a maximum. Hence, if the outer radius r 2 is less than the critical value,
adding more insulation will actually increase the heat- transfer rate q. Also, if
the outer radius is greater than the critical, adding more insulation will
decrease the heat transfer rate. Using values of K and h o typically
encountered, the critical radius is only a few mm. As a result, adding
insulation on small electrical wires could increase the heat loss. Adding
insulation to large pipes decreases the heat transfer rate.
radius of r 2 ,
When no insulation is provided then for a metal pipe with an outside
q
bare
= 2πr
2
L ho
2
( T −T
)
The rate of heat transfer from an insulated pipe, where the annular
insulating shell has an inside radius of r 2 and an outer radius of r 3 ,
o
q
insulated
Then,
2 r3
L ho
= π
r3
h
1+
K
( T −T
)
o
2
r
ln
r
3
2
o
q
q
insulated
bare
r
=
r
3
2
⎡
⎢ 1
⎢
⎢ r3
ho
1+
⎢
⎣ K
ln
⎤
⎥
⎥
r3
⎥
r ⎥
2 ⎦

LECTURE NO.6
HEAT SOURCE SYSTEMS: ONE-DIMENSIONAL STEADY STATE HEAT
CONDUCTION WITH HEAT GENERATION: HEAT FLOW THROUGH
SLAB / PLANE WALL
HEAT-SOURCE SYSTEMS
A number of interesting applications of the principles of heat transfer
are concerned with systems in which heat may be generated internally.
Nuclear reactors are one example; electrical conductors and chemically
reacting systems are others.
Fig.6.1 Sketch illustrating one- dimensional conduction problem with
heat generation.
One-dimensional steady state heat conduction with heat generation:
Heat Flow through slab / Plane Wall
Consider the plane wall with uniformly distributed heat sources shown
in the above Figure. The thickness of the wall in the x direction is 2L, and it is
assumed that, the dimensions in the other directions are sufficiently large that
the heat flow may be considered as one-dimensional. The heat generated per
unit volume is q • , and we assume that the thermal conductivity does not vary
with temperature. (This situation might be produced in a practical situation by
passing a current through an electrically conducting material)
The differential equation which governs the heat flow is

2
d T
2
dx
.
q
+ = 0 k
2
d T q
=−
2
dx k
Integrating twice with respect to x results in
dT
dx
.
.
q
=− x+
c
k
.
q 2
T = − x + c1x
+ c -------------- (1)
2
2k
For the boundary conditions we specify the temperatures
on either side of the wall, i.e.,
T = T w at x = ± L
1
Since the temperature must be the same on each side of the wall, c 1
must be zero.
dT
= 0,
q
• = 0⇒c
1
= 0
The temperature at the mid plane (x = 0) is denoted by T o and from
Equation (1)
At mid plane = x=0 and T= T 0
T o = c 2
The temperature distribution equation (1) becomes
2
T =− x +
.
q
2k
T
0
T
.
q
2k
x
2
−T
0
= − --------------- (2)
Assumed T=T w at x= L
.
q
2k
L
2
T w
−T
0
= − ----------------- (3)
(2)
(3)
T −T0
⇒
T −T
w
0
2
⎛ x ⎞
= ⎜ ⎟
⎝ L ⎠
------------ (4)

An expression for the midplane temperature T o may be obtained through an
energy balance. At steady-state conditions the total heat generated must
equal the heat lost at the faces. Thus
⎛
2⎜−
KA
⎝
dT
dx
⎤
⎥
⎦
x=
L
⎞ •
⎟=
q.
A.2L
⎠
where A is the cross-sectional area of the plate. The temperature gradient at
the wall is obtained by differentiating Equation (4):
T −T0
T −T
w
0
⎛
= ⎜
⎝
x
L
2
⎞
⎟
⎠
T −T
x
⎝L
⎛ ⎞
0
= ( Tw
−T0
) ⎜ ⎟ ⎠
2x
dT =
w 0
.
2
L
( T −T
) dx
dT
⎤
⎦
at x= L ( T T )
dx ⎥
=
w−
o
. L
x=
L
2
2
Since,
dT
−K.
dx
= q
•
L
dT
dx
•
q L
= − K
•
qL
T w
− 2 0
=−
L K
( T )
•
2
qL
T w
−T
0
=−
2K
•
2
qL
T
0
= + T w
2K
----- (5)

LECTURE NO.7
STEADY STATE HEAT CONDUCTION WITH HEAT DISSIPATION TO
ENVIRONMENT- INTRODUCTION TO EXTENDED SURFACES
(FINS) OF UNIFORM AREA OF CROSS SECTION - DIFFERENT FIN
CONFIGURATIONS - GENERAL CONDUCTION ANALYSIS
EQUATION
Steady State heat conduction with heat dissipation to environment:
Introduction to extended surfaces (FINS) of Uniform area of cross
section
The term extended surface is commonly used to depict an important
special case involving heat transfer by conduction within a solid and heat
transfer by convection (and/or radiation) from the boundaries of the solid.
Heat transfer from the boundaries of a solid to be in the same direction as
heat transfer by conduction in the solid. In contrast, for an extended surface,
the direction of heat transfer from the boundaries is perpendicular to the
principal direction of heat transfer in the solid.
Fig. 7.1 Combined conduction and convection in a structural element.
Consider a strut that connects two walls at different temperatures and
across which there is fluid flow (Figure 7.1). With T 1 >T 2 , temperature gradients
in the x-direction sustain heat transfer by conduction in the strut. However,
with T 1 >T 2 > T∞there is concurrent heat transfer by convection to the fluid,
causing q x , and hence the magnitude of the temperature gradient,
decrease with increasing x.
dT / dx
An extended surface is used specifically to enhance heat transfer
between a solid and adjoining fluid. Such an extended surface is termed a fin.
, to

The heat transfer rate may be increased by increasing the surface area
across which the convection occurs. This is done by employing fins that
extend from the wall into the surrounding fluid. The thermal conductivity of the
fin material can have a strong effect on the temperature distribution along the
fin and therefore influences the degree to which the heat transfer rate is
enhanced. Ideally, the fin material should have a large thermal conductivity to
minimize temperature variations from its base to its tip. In the limit of infinite
thermal conductivity, the entire fin would be at the temperature of the base
surface, thereby providing the maximum possible heat transfer enhancement.
Fig. 7.2 Use of fins to enhance heat transfer from a plane wall (a) Bare
surface (b) Finned surface.
Examples of fin applications are easy to find. Consider the
arrangement for cooling engine heads on motorcycles and lawn mowers or for
cooling electric power transformers. Consider also the tubes with attached fins
used to promote heat exchange between air and the working fluid of an air
conditioner. Two common finned tube arrangements are shown in Figure 7.3
Fig.7.3 Schematic of typical finned-tube heat exchangers.

Different fin configurations
Different fin configurations are illustrated in Figure 7.4. A straight fin is
any extended surface that is attached to a plane wall. It may be of uniform
cross-sectional area, or its cross-sectional area may vary with the distance x
from the wall. An annular fin is one that is circumferentially attached to a
cylinder, and its cross section varies with radius from the wall of the cylinder.
The foregoing fin types have rectangular cross sections, whose area may be
expressed as a product of the fin thickness t and the width w for straight fins
or the circumference
2 πr
for annular fins. In contrast a pin fin, or spine, is an
extended surface of circular cross section. Pin fins may also be of uniform or
non-uniform cross section. In any application, selection of a particular fin
configuration may depend on space, weight, manufacturing, and cost
considerations, as well as on the extent to which the fins reduce the surface
convection coefficient and increase the pressure drop associated with flow
over the fins.
FIGURE 7.4 Fin configurations. (a) Straight fin of uniform cross section. (b)
Straight fin of non-uniform cross section. (c) Annular fin. (d) Pin fin / spine.
General Conduction Analysis Equation
As engineers we are primarily interested in knowing the extent to which
particular extended surfaces or fin arrangements could improve heat transfer
from a surface to the surrounding fluid. To determine the heat transfer rate
associated with a fin, we must first obtain the temperature distribution along
the fin.

Fig.7.5 Energy balance for an extended surface
A general form of the energy equation for an extended surface is as
follows:
2
d T
2
dx
⎛ 1
⎜
⎝ Ac
dA
dx
⎞dT
⎟
⎠ dx
⎛ 1
−
⎜
⎝ Ac
h dA
k dx
⎞
⎟
⎠
( T −T
) = 0
c
s
+
∞
Its solution for appropriate boundary conditions provides the
temperature distribution, which may be used with Fourier’s equation (
dT
q x
= −kA ) to calculate the conduction rate at any x.
dx

LECTURE NO.8
EQUATION OF TEMPERATURE DISTRIBUTION WITH DIFFERENT
BOUNDARY CONDITIONS, FIN PERFORMANCE AND OVERALL
SURFACE EFFICIENCY OF FINS
Fins of Uniform Cross-Sectional Area
To solve general form of fin energy equation it is necessary to be more
specific about the geometry. We begin with the simplest case of straight
rectangular and pin fins of uniform cross section (Figure 8.1). Each fin is
attached to a base surface of temperature T(O) = T b and extends into a fluid
of temperature
T
∞.
Fig. 8.1 Straight fins of uniform cross section (a) Rectangular fin (b) Pin
fin
For the prescribed fins, A c is a constant and A s = P x , where A s is the
surface area measured from the base to x and P is the fin perimeter.
Accordingly, with dA c /dx = 0 and dA s /dx = P, general form of the energy
equation for an extended surface reduces to
2
d T
2
dx
hP
−
kA
c
( T − T ) = 0
∞
--------(1)
To simplify the form of this equation, we transform the dependent
variable by defining an excess temperature θ as
where, since
into
∞
Equation (3), we then obtain
θ ( x ) = T ( x)
−T
-------------(2)
∞
T is a constant, d θ / dx = dT / dx . . Substituting Equation (2)

where
2
d θ 2
− θ= 0
2 m
dx
hP
m 2 =
-------------(4)
kA c
--------(3)
Equation (3) is a linear, homogeneous, second-order differential equation with
constant coefficients. Its general solution is of the form
mx −mx
θ ( x)
= C1
e + C2e
---------------- (5)
By substitution it may readily be verified that Equation (5) is indeed a
solution to Equation (3).
To evaluate the constants C 1 and C 2 of Equation (5), it is necessary to
specify appropriate boundary conditions. One such condition may be specified
in terms of the temperature at the base of the fin (x = 0)
θ =
( 0)= Tb
−T
∞
θb
----------------- (6)
The second condition, specified at the fin tip (x = L), may correspond to
one of four different physical situations.
The first condition, case A, considers convection heat transfer from the
fin tip. Applying an energy balance to a control surface about this tip (Figure
8.2), we obtain
hA
c
[ T ( L)−T
]
∞
= −kA
dT
dx
x=
L
or
h
dT
( L)=−
kA
dx
θ --------------- (7)
x=
L
Figure 8.2 Conduction and convection in a fin of uniform cross section.

That is, the rate at which energy is transferred to the fluid by
convection from the tip must equal the rate at which energy reaches the tip by
conduction through the fin. Substituting Equation (5) into Equations (6) and
(7), we obtain, respectively,
And
θb = C 1
+ C 2 ----------- (8)
mL −mL
−mL
h( C1e
+ C2e
) = km(
C2e
−C1
e
Solving for C 1 , and C 2 , it may be shown, after some manipulation, that
θ cosh m(
L − x)
+ ( h / mk )sinh m(
L − x)
=
θ cosh mL + ( h / mk ) sinh mL
b
mL
)
------- (9)
The form of this temperature distribution is shown schematically in
Figure 8.2. Note that the magnitude of the temperature gradient decreases
with increasing x. This trend is a consequence of the reduction in the
conduction heat transfer q x
(x)
with increasing x due to continuous
convection losses from the fin surface.
We are particularly interested in the amount of heat transferred from
the entire fin. From Figure 8.2 it is evident that the fin heat transfer rate
may be evaluated in two alternative ways, both of which involve use of the
temperature distribution. The simpler procedure, and the one that we will use,
involves applying Fourier's law at the fin base. That is,
q
f
= q
b
=− kA
c
dT
dx
=− kA
dθ
c
x=0 dx
x=0
q
f
----- (10)
Hence, knowing the temperature distribution, θ (x)
, q
f may be
evaluated, giving
q
f
sinh mL + ( h / mk )cosh mL
= hPkA
cθ b
------- (11)
cosh mL + ( h / mk )sinh mL
However, conservation of energy dictates that the rate at which heat is
transferred by convection from the fin must equal the rate at which it is
conducted through the base of the fin. Accordingly, the alternative formulation
for
q
f is
q
q
f
f
∫
A f
[ ( x T∞] dA
s
= h T )−
∫A
f
= h )
θ ( x dA
s -------------- (12)

where
Af
Af is the total, including the tip, fin surface area. Substitution of
Equation (9) into Equation (12) would yield Equation (11):
The second tip condition, case B, corresponds to the assumption that
the convective heat loss from the fin tip is negligible, in which case the tip may
be treated as adiabatic and
dθ
dx
x =L
=0
-------- (13)
Substituting from Equation (5) and dividing by m, we then obtain
C
mL −mL
1
e −C2e
=
Using this expression with Equation (8) to solve for C 1 and C 2 and
substituting the results into Equation (5), we obtain
θ cosh m(
L −x)
=
θ cosh mL
b
0
--------- (14)
Using this temperature distribution with Equation (10), the fin heat
transfer rate is then
q
f
= hPkA
cθb
tanh mL ----- (15)
In the same manner, we can obtain the fin temperature distribution
and, heat transfer rate for case C, where the temperature is prescribed at the
fin tip. That is, the second boundary condition is θ ( L)
= θ , and the resulting
expressions are of the form
θ
=
θ
b
( θ θ )
L
/
b
sinh mx + sinh m(
L − x)
sinh mL
L
------- (16)
q
f
=
cosh mL −θ
L
/ θb
hPkA
cθb
sinh mL
The very long fin, case D, is an interesting extension of these results.
In particular, as L →∞, θ →0
and it is easily verified that
L
θ
=e −
θ
b
mx
q
f
=
hPkA
c
θ
b
Fin Performance
Fins are used to increase the heat transfer from a surface by
increasing the effective surface area. However, the fin itself represents a
conduction resistance to heat transfer from the original surface. For this

eason, there is no assurance that the heat transfer rate will be increased
through the use of fins. An assessment of this matter may be made by
evaluating the fin effectiveness ε
f . It is defined as the ratio of the fin heat
transfer rate to the heat transfer rate that would exist without the fin. Therefore
q
f
εf
=
hA θ
c,
b
b
where
A
c , b
is the fin cross-sectional area at the base. In any rational design
the value of εf
should be as large as possible, and in general, the use of fins
may rarely be justified unless ε
f
≥ 2.
Fin effectiveness is enhanced by the choice of a material of high
thermal conductivity. Aluminum alloys and copper come to mind. However,
although copper is superior from the stand point of thermal conductivity.
aluminum alloys are the more common choice because of additional benefits
related to lower cost and weight. Fin effectiveness is also enhanced by
increasing the ratio of the perimeter to the cross-sectional area. For this
reason, the use of thin, but closely spaced fins, is preferred.
Another measure of fin thermal performance is provided by the fin
efficiencyη f . The maximum driving potential for convection is the
temperature difference between the base (x = 0) and the fluid, θ b
= T b
−T
∞.
Hence the maximum rate at which a fin could dissipate energy is the rate that
would exist if the entire fin surface were at the base temperature. However,
since any fin is characterized by a finite conduction resistance, a temperature
gradient must exist along the fin and the above condition is an idealization. A
logical definition of fin efficiency is therefore
q
f
η
f
= =
qmax
q
hA
f
f
θ
b
where A f is the surface area of the fin.
Overall Surface Efficiency
In contrast to the fin efficiency
η
f , which characterizes the
performance of a single fin, the overall surface efficiency η
o characterizes an
array of fins and the base surface to which they are attached. Representative
arrays are shown in Figure 8.3, where S designates the fin pitch. In each case
the overall efficiency is defined as

where
q
f
η
f
= =
qmax
qt
hA θ
t
b
q
t is the total heat rate from the surface area A t associated with both
the fins and the exposed portion of the base (often termed the prime surface).
If there are N fins in the array, each of surface area A f , and the area of the
prime surface is designated as A b , the total surface area is
A = NA + A
t
f
b
The maximum possible heat rate would result if the entire fin surface,
as well as the exposed base, were maintained at
T
b .
Figure 8.3 Representative fin arrays. (a) Rectangular fins. (b) Annular fins.

LECTURE NO.9
PRINCIPLES OF UNSTEADY-STATE HEAT TRANSFER: DERIVATION OF
BASIC EQUATION; SIMPLIFIED CASE FOR SYSTEMS WITH
NEGLIGIBLE INTERNAL RESISTANCE; TOTAL AMOUNT OF HEAT
TRANSFERRED ; DIMENSIONAL ANALYSIS IN MOMENTUM TRANSFER
PRINCIPLES OF UNSTEADY-STATE HEAT TRANSFER: DERIVATION OF
BASIC EQUATION
Introduction
In steady state heat-transfer systems the temperature at any given
point and the heat flux were always constant over time. In unsteady state or
transient processes the temperature at any given point in the system changes
with time. Before steady-state conditions can be reached in a process, some
time must elapse after the heat-transfer process is initiated to allow the
unsteady-state conditions to disappear.
Unsteady-state heat transfer is important because of the large
number of heating and cooling problems occurring industrially. In metallurgical
processes it is necessary to predict cooling and heating rates for various
geometries of metals in order to predict the time required to reach certain
temperatures. In food processing, for example, the canning industry,
perishable canned foods are heated by immersion in steam baths or chilled by
immersion in cold water. In the paper industry wood logs are immersed in
steam baths before processing. In most of these processes the material is
suddenly immersed in a fluid of higher or lower temperature.
Fig. 9.1 Unsteady State conduction in one direction

Derivation of Unsteady-State Conduction Equation
To derive the equation for unsteady-state conduction in one direction in
a solid, refer to Fig. 9.1. Heat is being conducted in the x direction in the
∆ x, ∆y
, ∆z
in size. For conduction in the x direction, write
q x
∂T
= −kA
------- (1)
∂x
∂T
The term means the partial or derivative of T with respect to x,
∂x
with the other variables, y, z, and time t, being held constant. Next, making a
heat balance on the cube, we can write
rate of heat input + rate of generation = rate of heat output + rate of heat accumulati on
---------- (2)
The rate of heat input to the cube is
rate of heat input =
q
∂T
x / x
= −k
( ∆y
∆z
) ----------(3)
∂x
x
is
Also, rate of heat output =
The rate of accumulation of heat in the volume
q
∂T
x / x +∆ x
= −k
( ∆y
∆z
)
------- (4)
∂x
x+∆
x
, in time ∂ t
∆ x ∆y
, ∆z
∂T
rate of heat accumulation = ( ∆x , ∆y
, ∆z
) ρc
------(5)
p
∂t
The rate of heat generation in volume ∆ x, ∆y
, ∆z
is
rate of heat generation =
( ∆x
∆y
∆z
) q ------(6)
.
Substituting Eqs. (3)-(6) into (2) and dividing by
Letting
respect to x or
⎛∂T
−k
⎜
x
q
⎝ ∂
+
.
x
∂T
−
∂x
∆x
x+∆x
⎞
⎟
⎠
= ρc
p
∂T
∂t
∆ x,
∆y
, ∆z
------- (7)
∆ x approach zero, we have the second partial of T with
2
∂T / ∂x
2
on the left side. Then, rearranging,
∂T
∂t
k
=
ρc
p
2
∂ T
2
∂x
.
q
+
ρc
p
2
∂ T
= α
2
∂x
.
q
+
ρc
p
--------(8)

where α is
k / ρc
p , thermal diffusivity. This derivation assumes constant
k ,and
,ρ c p . In SI units, α = m 2 /s, T = K, t = s, k = W/m.K, ρ = kg/m 3 , q . =
W/m 3 , and
c
p = J/kg.K.
For conduction in three dimensions, a similar derivation gives
2
∂T
⎛∂
T
= α
2
t
⎜
∂ ⎝ ∂x
2
∂ T
+
2
∂y
2
∂ T ⎞ q
+
2
z
⎟
∂
+
⎠ ρ
.
c p
--------- (9)
In many cases, unsteady-state heat conduction is occurring but the
rate of heat generation is zero. Then Eqs. (8) and (9) become
∂T
∂t
2
∂ T
= α ------(10)
2
∂x
2 2 2
∂T ⎛∂
T ∂ T ∂ T ⎞
= α ⎜ + +
⎟
2 2 2 ----- (11)
∂t
⎝ ∂x
∂y
∂z
⎠
z and time t.
Equations (10) and (11) relate the temperature T with position x, y, and
SIMPLIFIED CASE FOR SYSTEMS WITH NEGLIGIBLE INTERNAL
RESISTANCE
Basic Equation
Consider a solid which has a very high thermal conductivity or very low
internal conductive resistance compared to the external surface resistance,
where convection occurs from the external fluid to the surface of the solid.
Since the internal resistance is very small, the temperature within the solid is
essentially uniform at any given time.
An example would be a small, hot cube of steel at T o K at time t = 0,
suddenly immersed in a large bath of cold water at
T ∞which is held constant
with time. Assume that the heat transfer coefficient h in W/m 2 .K is constant
with time. Making a heat balance on the solid object for a small time interval of
time dt s, the heat transfer from the bath to the object must equal the change
in internal energy of the object:
hA
( T T ) dt = c V dT
∞ −
p
ρ
where A is the surface area of the object in m 2 , T the average
temperature of the object at time t in s, ρ the density of the object in kg/m 3 ,

and V the volume in m 3 . Rearranging the equation and integrating between
the limits of T = T o when t = 0 and T = T when t = t,
T = T
∫
dT
T −T
T = T0
∞
hA
=
c ρV
p
∫t
t = t
= 0
dt
⎛ hA
t
c p V ⎟ ⎞
−⎜
⎝ ρ ⎠
T −T
∞
= e
T0
−T
∞
This equation describes the time-temperature history of the solid
object. The term
c p
ρ V is often called the lumped thermal capacitance of the
system. This type of analysis is often called the lumped capacity method or
Newtonian heating or cooling method.
Equation for Different Geometries
In using the above equation the surface / volume ratio of the object
must be known. The basic assumption of negligible internal resistance was
made in the derivation. This assumption is reasonably accurate when
N Bi
hx
= 1

calculate the temperature of the ball after 1 h (3600 s). The average physical
properties are k = 43.3 W/m.K, ρ = 7849 kg/m3 , and c p = 0.4606 kJ/kg.K.
Solution: For a sphere characteristic dimension
V 25 .4
x1 = =
r = = 8.47 × 10
A 3 1000 × 3
The Biot Number
hx 11 .36 (8.47 × 10
N Bi
= =
k 43.3
−3
m
−3
1 =
)
0.00222
Then,
This value is

q( t)
= hA ( T0
−T
∞
) e
⎛ hA ⎞
−⎜
⎟t
c p V
⎝ ρ ⎠
To determine the total amount of heat Q in W.s or J transferred from
the solid from time t = 0 to t = t, we can integrate the above equation:
Q=
Q=
c
t
∫
= t
t = 0
p
q(
t)
d(
t)
=
t
∫
= t
t = 0
hA ( T
⎛
⎜
hA
0
−T
⎞
⎟
− t
⎜c
V ⎟
⎝ pρ
⎠
ρV
( T 0
−T
∞
)[1 −e
]
∞
) e
⎛ ⎞
⎜
hA
− ⎟t
⎝
cpρV
⎠
dt
Problem 9.2 Total Amount of Heat in Cooling
For the conditions in problem 9.1, calculate the total amount of heat
removed up to time t = 3600 s.
Solution:
From Problem 9.1,
hA
cpρV
11 .36
−4
−1
= 3.71 × 10
− 3
= (0.4606 × 1000 )(7849 )(8.47 × 10 )
s
Q=
c
3
3
−5
3
Also, V = 4πr
/3=
4( π)(0.0254
) / 3=
6.864 × 10 m
Substituting the values into
p
⎛
⎜
hA
⎞
⎟
− t
⎜c
V ⎟
⎝ pρ
⎠
ρV
( T 0
−T
∞
)[1 −e
]
−4
−5
−(3.71
× 10 )( 3600 )
( 0.4606 × 1000 )( 7849 )( 6.864 × 10 )(699
.9−394 .3) × [1 −
]
Q = e
= 5.589X10 4 J
DIMENSIONAL ANA LYSIS IN MOMENTUM TRANSFER
Dimensional Analysis of Differential Equations
Dimensional homogeneity requires that every term in a given equation
have the same units. Then, the ratio of one term in the equation to another
term is dimensionless. Knowing the physical meaning of each term in the
equation, we are then able to give a physical interpretation to each of the
dimensionless parameters or numbers formed. These dimensionless
numbers, such as the Reynolds number and others, are useful in correlating
and predicting transport phenomena in laminar and turbulent flow.
Often it is not possible to integrate the differential equation describing a
flow situation. However, we can use the equation to find out which

dimensionless numbers can be used in correlating experimental data for this
physical situation.
Systems that are geometrically similar are said to be dynamically
similar if the parameters representing ratios of forces pertinent to the situation
are equal. This means that the Reynolds, Euler, or Froude numbers must be
equal between the two systems.
This dynamic similarity is an important requirement in obtaining
experimental data for a small model and extending these data to scale up to
the large prototype. Since experiments with full-scale prototypes would often
be difficult and / or expensive, it is customary to study small models. This is
done in the scale-up of chemical process equipment and in the design of
ships and airplanes.
Dimensional Analysis Using the Buckingham Method
The method of obtaining the important dimensionless numbers from
the basic differential equations is generally the preferred method. In many
cases, however, we are not able to formulate a differential equation which
clearly applies. Then a more general procedure is required, known as the
Buckingham method. In this method the listing of the important variables in
the particular physical problem is done first. Then we determine the number of
dimensionless parameters into which the variables may be combined by using
the Buckingham pi theorem.
The Buckingham theorem states that the functional relationship among
‘q’ quantities or variables whose units may be given in terms of ‘u’
fundamental units or dimensions may be written as (q - u) independent
dimensionless groups, often called
π ' s . [This quantity u is actually the
maximum number of these variables that will not form a dimensionless group.]
Let us consider the following example, to illustrate the use of this
method. An incompressible fluid is flowing inside a circular tube of inside
diameter D. The significant variables are pressure drop∆ p , velocity v,
diameter D, tube length L, viscosity µ , and density ρ . The total number of
variables is q = 6.

The fundamental units or dimensions are u = 3 and are mass M, length L, and
2
time t. The units of the variables are as follows: ∆pin M / Lt , v in L/t, D in L,
µ in M/Lt, and ρ in M/L 3 . The number of dimensionless groups or π ' s is
q− u , or 6-3 = 3. Thus,
)
π1 = f ( π2
, π3
Next, we must select a core group of u (or 3) variables which will
appear in each π group and among them contain all the fundamental
dimensions. Also, no two of the variables selected for the core can have the
same dimensions. In choosing the core, the variable whose effect one desires
to isolate is often excluded (for example,
∆ p ). This leaves us with the
variables v, D, µ and ρ to be used. (L and D have the same dimensions.)
We will select D, v, and p to be the core variables common to all three
groups. Then the three dimensionless groups are
π
a b c 1
1
= D v ρ ∆p
π =
π
d e f 1
2
D v ρ L
i 1
3
=D
g v h ρ µ
To be dimensionless, the variables must be raised to certain exponents
a, b, c, and so forth.
First we consider the π1
group:
π
a b c 1
1
= D v ρ ∆p
To evaluate these exponents, we write the above equation dimensionally by
substituting the dimensions for each variable:
M
o
o
L t
o
= 1=
L
a
b
c
⎛ L ⎞ ⎛ M ⎞
⎜ ⎟ ⎜ 3 ⎟
⎝ t ⎠ ⎝ L ⎠
Next we equate the exponents of L on both sides of this equation, of M,
and finally of t:
(L) 0 = a + b - 3c - 1
(M) 0 = c + 1
(t) 0 = -b - 2
Solving these equations, a = 0, b = -2, and c = -1.
a b c 1
Substituting these values into Eq. π = D v ρ ∆ ,
1
p
M
Lt
2
= ∆p =
v ρ
. π 1 2
N
Eu

Repeating this procedure for π
2 and π
3 ,
L
π 2
=
D
Dv ρ
π
3
= = N Re
µ
Finally, substituting π
1 , π
2 and π
3 into equation π1 = f ( π2
, π3
)
∆p
⎛
= f ⎜
2
v ρ ⎝
L
D
Dvp ⎞
, ⎟
µ ⎠
This type of analysis is useful in empirical correlations of data.
However, it does not tell us the importance of each dimensionless group,
which must be determined by experimentation, nor does it select the variables
to be used.

LECTURE NO.10
SOME IMPORTANT EMPIRICAL RELATIONS USED FOR DETERMINATION
OF HEAT TRANSFER COEFFICIENT: NUSSELT’S NUMBER, PRANDTL
NUMBER, REYNOLD’S NUMBER, GRASHOFF NUMBER
The Nusselt Number
For forced convection of a single-phase fluid with moderate
temperature differences, the heat flux per unit area
q
"
w
is nearly
proportional to the temperature difference ∆ T = Tw −T
* . From the Newton’s
law of cooling:
q
"
w
=
h
( T −T* )
w
where h is called the heat transfer coefficient, with units of W/m 2 .
But h is dimensional and thus its value depends on the units used. The
traditional dimensionless from of h is the Nusselt number, Nu, which may be
defined as the ratio of convection heat transfer to fluid conduction heat
transfer under the same conditions. Consider a layer of fluid of width L and
temperature difference ( T w
− T*
) . Assuming that the layer is moving so that
convection occurs, the heat flux would be,
q
"
w
=
h
( T −T* )
w
If, on the other hand, the layer were stagnant, the heat flux would be
entirely due to fluid conduction through the layer:

q
k ( Tw
T* =
)
L
" −
w
The Nusselt number is defined as the ratio of these two:
N
"
qw(
convection
=
"
q ( conduction
hL
k
uL
=
w
)
)
A Nusselt number of order unity would indicate a sluggish motion little
more effective than pure fluid conduction: for example, laminar flow in a long
pipe. A large Nusselt number means very efficient convection: For example,
turbulent pipe flow yields Nu of order 100 to 1000.
Prandtl Number(Pr)
The Prandtl number Pr is a dimensionless number approximating the
ratio of momentum diffusivity (kinematic viscosity) and thermal diffusivity. It is
named after the German physicist Ludwig Prandtl. It can be expressed as
Pr
υ
=
α
where
Pr = Prandtl's number
υ = momentum diffusivity (m 2 /s)
α = thermal diffusivity (m 2 /s)
The Prandtl number can alternatively be expressed as
µc
Pr =
p
k
where

μ = absolute or dynamic viscosity (kg/m. s, cP)
c p = specific heat capacity (J/kg K,)
k = thermal conductivity (W/m K)
The Prandtl Number is often used in heat transfer and free and forced
convection calculations.
Note that whereas the Reynolds number and Grashof number are
subscripted with a length scale variable, Prandtl number contains no such
length scale in its definition and is dependent only on the fluid and the fluid
state. As such, Prandtl number is often found in property tables alongside
other properties such as viscosity and thermal conductivity.
Typical values for Pr are:
• around 0.015 for mercury
• around 0.16-0.7 for mixtures of noble gases or noble gases with
hydrogen
• around 0.7-0.8 for air and many other gases,
• between 4 and 5 for R-12 refrigerant
• around 7 for water (At 20 degrees Celsius)
• between 100 and 40,000 for engine oil
• around 1 × 10 25 for Earth's mantle.
For mercury, heat conduction is very effective compared to convection:
thermal diffusivity is dominant. For engine oil, convection is very effective in
transferring energy from an area, compared to pure conduction: momentum
diffusivity is dominant.
In heat transfer problems, the Prandtl number controls the relative
thickness of the momentum and thermal boundary layers. When Pr is small, it
means that the heat diffuses very quickly compared to the velocity
(momentum). This means that for liquid metals the thickness of the thermal
boundary layer is much bigger than the velocity boundary layer.

The mass transfer analog of the Prandtl number is the Schmidt number.
Boundary Layer
In physics and fluid mechanics, a boundary layer is that layer of fluid
in the immediate vicinity of a bounding surface where effects of viscosity of
the fluid are considered in detail. In the Earth's atmosphere, the planetary
boundary layer is the air layer near the ground affected by diurnal heat,
moisture or momentum transfer to or from the surface. On an aircraft wing the
boundary layer is the part of the flow close to the wing. The boundary layer
effect occurs at the field region in which all changes occur in the flow pattern.
The boundary layer distorts surrounding non-viscous flow. It is a phenomenon
of viscous forces. This effect is related to the Reynolds number.
Fig.10.1 Boundary layer visualization, showing transition from laminar to
turbulent condition
Reynolds number
In fluid mechanics, the Reynolds number Re is a dimensionless
number that gives a measure of the ratio of inertial forces ρV 2 /L to viscous
forces μV/L 2 and consequently quantifies the relative importance of these two
types of forces for given flow conditions. The concept was introduced by
George Gabriel Stokes in 1851, but the Reynolds number is named after
Osborne Reynolds (1842–1912), who popularized its use in 1883.
Reynolds numbers frequently arise when performing dimensional
analysis of fluid dynamics problems, and as such can be used to determine

dynamic similitude between different experimental cases. They are also used
to characterize different flow regimes, such as laminar or turbulent flow:
laminar flow occurs at low Reynolds numbers, where viscous forces are
dominant, and is characterized by smooth, constant fluid motion, while
turbulent flow occurs at high Reynolds numbers and is dominated by inertial
forces, which tend to produce chaotic eddies, vortices and other flow
instabilities.
Grashof number Gr
The Grashof number Gr is a dimensionless number in fluid dynamics
and heat transfer which approximates the ratio of the buoyancy to viscous
force acting on a fluid. It frequently arises in the study of situations involving
natural convection. It is named after the German engineer Franz Grashof.
Gr
L
3
g β(
T −T∞
) L
= for vertical flat plates
s
2
υ
Gr
L
3
g β(
T −T∞
) D
= for pipes
s
2
υ
where the L and D subscripts indicates the length scale basis for the Grashof
Number.
g = acceleration due to Earth's gravity
β = volumetric thermal expansion coefficient
(equal to approximately 1/T, for ideal fluids, where T is absolute
temperature)
T s = surface temperature
T ∞ = bulk temperature
L = length
D = diameter
υ = kinematic viscosity

The transition to turbulent flow occurs in the range 10 8 < Gr L < 10 9 for
natural convection from vertical flat plates. At higher Grashof numbers, the
boundary layer is turbulent; at lower Grashof numbers, the boundary layer is
laminar.
LECTURE NO.11
RADIATION - HEAT TRANSFER, RADIATION PROPERTIES, RADIATION
THROUGH BLACK AND GREY SURFACES, DETERMINATION OF
SHAPE FACTORS
Introduction
Thermal radiation is that electromagnetic radiation emitted by a body
as a result of its temperature.
PHYSICAL MECHANISM
There are many types of electromagnetic radiation; thermal radiation is
only one. Regardless of the type of radiation, we say that it is propagated at
the speed of light, 3 x 10 8
wavelength and frequency of the radiation,
where
c = speed of light
λ= wavelength
υ = frequency
m/s. This speed is equal to the product of the
c=λυ
The unit for A may be centimeters, angstroms (1
ο
A = 10 -8 cm), or
micrometers (1 µ m = 10
-6
m). A portion of the electromagnetic spectrum is
shown in Figure 11.1. Thermal radiation lies in the range from about 0.1 to
100 µ m, while the visible-light portion of the spectrum is very narrow,
extending from about 0.35 to 0.75 µ m.

Fig.11.1 Electromagnetic spectrum
The propagation of thermal radiation takes place in the form of discrete
quanta, each quantum having an energy of
E=hυ
where h is Planck's constant and has the value
h = 6.625 X 10 -34 J.s
The total radiation energy emitted is proportional to absolute
temperature to the fourth power:
4
E b
= σT
The above equation is called the Stefan-Boltzmann law, E b is the
energy radiated per unit time and per unit area by the ideal radiator, and σ is
the Stefan-Boltzmann constant, which has the value
σ = 5.669 x 10 -8 W/m 2 . K 4
where E b , is in watts per square meter and T is in degrees Kelvin.
We are interested in radiant exchange with surfaces-hence the reason
for the expression of radiation from a surface in terms of its temperature. The
subscript b in Equation denotes that this is the radiation from a blackbody. We
call this blackbody radiation because materials which obey this law appear
black to the eye; they appear black because they do not reflect any radiation.
Thus a blackbody is also considered as one which absorbs all radiation
incident upon it. E b is called the emissive power of a blackbody.
It is important to note at this point that the "blackness" of a surface to
thermal radiation can be quite deceiving in so far as visual observations are
concerned. A surface coated with lampblack appears black to the eye and
turns out to be black for the thermal-radiation spectrum. On the other hand,
snow and ice appear quite bright to the eye but are essentially "black" for

long-wavelength thermal radiation. Many white paints are also essentially
black for long-wavelength radiation.
Fig. 11.2 Sketch showing effects of incident radiation.
RADIATION PROPERTIES
When radiant energy strikes a material surface, part of the radiation is
reflected, part is absorbed, and part is transmitted, as shown in Figure 11.2.
We define the reflectivity ρ as the fraction reflected, the absorptivity α as
the fraction absorbed, and the transmissivity τ as the fraction transmitted.
Thus
ρ+ α+
τ=1
Most solid bodies do not transmit thermal radiation, so that for many
applied problems the transmissivity may be taken as zero. Then
ρ +α=1 ( τ=0
)
Two types of reflection phenomena may be observed when radiation
strikes a surface. If the angle of incidence is equal to the angle of reflection,
the reflection is called specular. On the other hand, when an incident beam is
distributed uniformly in all directions after reflection, the reflection is called
diffuse. These two types of reflection are depicted in Figure 11.3.
Figure 11.3 (a) Specular ( φ=
1
φ2
) and (b) diffuse reflection.

Note that a specular reflection presents a mirror image of the source to
the observer. No real surface is either specular or diffuse. An ordinary mirror
is quite specular for visible light, but would not necessarily be specular over
the entire wavelength range of thermal radiation. Ordinarily, a rough surface
exhibits diffuse behavior better than a highly polished surface. Similarly, a
polished surface is more specular than a rough surface. The influence of
surface roughness on thermal-radiation properties of materials is a matter of
serious concern and remains a subject for continuing research.
The emissive power of a body E is defined as the energy emitted by
the body per unit area and per unit time.
The ratio of the emissive power of a body to the emissive power of a
blackbody at the same temperature is equal to the absorptivity of the body.
This ratio is defined as the emissivity ε of the body,
E
ε=
E b
so that
ε=α
The above equation is called Kirchhoff's identity.
The Gray Body
A gray body is defined such that the monochromatic emissivity ε λ of
the body is independent of wavelength. The monochromatic emissivity is
defined as the ratio of the monochromatic emissive power of the body to the
monochromatic emissive power of a blackbody at the same wavelength and
temperature. Thus
ε λ
E
=
λ
E b λ
The glass, which is essentially transparent for visible light, is almost
totally opaque for thermal radiation emitted at ordinary room temperatures.

Figure 11.4 Method of constructing a blackbody enclosure.
Construction of a Blackbody
The concept of a blackbody is an idealization; i.e., a perfect blackbody
does not exist. All surfaces reflect radiation to some extent, however slight. A
blackbody may be approximated very accurately, however, in the following
way. A cavity is constructed, as shown in Figure 11.4, so that it is very large
compared with the size of the opening in the side. An incident ray of energy is
reflected many times on the inside before finally escaping from the side
opening. With each reflection there is a fraction of the energy absorbed
corresponding to the absorptivity of the inside of the cavity. After the many
absorptions, practically all the incident radiation at the side opening is
absorbed. It should be noted that the cavity of Figure 11.4 behaves
approximately as a blackbody emitter as well as an absorber.
Fig.11.5 Sketch showing area elements used in deriving radiation shape
factor
RADIATION SHAPE FACTOR

Consider two black surfaces A 1 and A 2 , as shown in Figure 11.5. We
wish to obtain a general expression for the energy exchange between these
surfaces when they are maintained at different temperatures. The problem
becomes essentially one of determining the amount of energy which leaves
one surface and reaches the other. To solve this problem the radiation shape
factors are defined as
F 1-2 = fraction of energy leaving surface 1 which reaches surface 2
F 2-1 = fraction of energy leaving surface 2 which reaches surface 1
F m-n = fraction of energy leaving surface m which reaches surface n
Other names for the radiation shape factor are view factor, angle
factor, and configuration factor. The energy leaving surface 1 and arriving at
surface 2 is
E b1 A 1 F 12
and the energy leaving surface 2 and arriving at surface 1 is
E b2 A 2 F 21
Since the surfaces are black, all the incident radiation will be absorbed, and
the net energy exchange is
E b1 A 1 F 12 - E b2 A 2 F 21 = Q 1-2
If both surfaces are at the same temperature, there can be no heat
exchange, that is, Q 1-2 = 0. Also,
E b1 = E b2
so that
A 1 F 12 = A 2 F 21
The net heat exchange is therefore
Q 1-2 = A 1 F 12 (E b1 - E b2 ) = A 2 F 21 (E b1 - E b2 )
The above equation is known as a reciprocity relation, and it applies in
a general way for any two surfaces m and n:
A m F mn = A n F nm
Although the relation is derived for black surfaces, it holds for other
surfaces also as long as diffuse radiation is involved.

LECTURE NO.12
INTRODUCTION TO CONDENSING AND BOILING HEAT TRANSFER,
CONDENSATION HEAT-TRANSFER PHENOMENA, FILM
CONDENSATION INSIDE HORIZONTAL TUBES, BOILING HEAT
TRANSFER
Introduction
The convection processes are associated with a change of phase of a
fluid. The two most important examples are condensation and boiling
phenomena.
In many types of power or refrigeration cycles one is interested in
changing a vapor to a liquid, or a liquid to a vapor, depending on the particular
part of the cycle under study. These changes are accomplished by boiling or
condensation, and the engineer must understand the processes involved in
order to design the appropriate heat-transfer equipment. High heat-transfer
rates are usually involved in boiling and condensation, and this fact has also
led designers of compact heat exchangers to utilize the phenomena for
heating or cooling purposes not necessarily associated with power cycles.
CONDENSATION HEAT-TRANSFER PHENOMENA
Consider a vertical flat plate exposed to a condensable vapor. If the
temperature of the plate is below the saturation temperature of the vapor,
condensate will form on the surface and under the action of gravity will flow
down the plate. If the liquid wets the surface, a smooth film is formed, and the
process is called film condensation. If the liquid does not wet the surface,
droplets are formed which fall down the surface in some random fashion. This
process is called dropwise condensation.
In the film-condensation process the surface is blanketed by the film,
which grows in thickness as it moves down the plate. A temperature gradient
exists in the film, and the film represents a thermal resistance to heat transfer.
In dropwise condensation a large portion of the area of the plate is
directly exposed to the vapor; there is no film barrier to heat flow, and higher
heat-transfer rates are experienced. In fact, heat-transfer rates in dropwise

condensation may be as much as 10 times higher than in film
condensation.
Because of the higher heat-transfer rates, dropwise condensation
would be preferred to film condensation, but it is extremely difficult to maintain
since most surfaces become wetted after exposure to a condensing vapor
over an extended period of time. Various surface coatings and vapor additives
have been used in attempts to maintain dropwise condensation, but these
methods have not met with general success. Measurements indicate that the
drop conduction is the main resistance to heat flow for atmospheric pressure
and above. Nucleation site density on smooth surfaces can be of the order of
10 8 sites per square centimeter, and heat-transfer coefficients in the range of
170 to 290 kW/m 2 .°C.
FILM CONDENSATION INSIDE HORIZONTAL TUBES
Condensation inside tubes is of considerable practical interest because
of applications to condensers in refrigeration and air-conditioning systems, but
unfortunately these phenomena are quite complicated and not amenable to a
simple analytical treatment. The overall flow rate of vapor strongly influences
the heat transfer rate in the forced convection condensation system, and this
in turn is influenced by the rate of liquid accumulation on the walls.
BOILING HEAT TRANSFER
When a surface is exposed to a liquid and is maintained at a
temperature above the saturation temperature of the liquid, boiling may occur,
and the heat flux will depend on the difference in temperature between the
surface and the saturation temperature. When the heated surface is
submerged below a free surface of liquid, the process is referred to as pool
boiling. If the temperature of the liquid is below the saturation temperature,
the process is called sub-cooled, or local, boiling. If the liquid is maintained
at saturation temperature, the process is known as saturated, or bulk,
boiling.
The different regimes of boiling are indicated in Figure 12.1, where
heat-flux data from an electrically heated platinum wire submerged in water

are plotted against temperature excess T w – T sat . In region I free-convection
currents are responsible for motion of the fluid near the surface. In this region
the liquid near the heated surface is superheated slightly, and it subsequently
evaporates when it rises to the surface. The heat transfer in this region can be
calculated with the free-convection relations. In region II bubbles begin to form
on the surface of the wire and are dissipated in the liquid after breaking away
from the surface. This region indicates the beginning of nucleate boiling. As
the temperature excess is increased further, bubbles form more rapidly and
rise to the surface of the liquid, where they are dissipated. This is indicated in
region III. Eventually, bubbles are formed so rapidly that they blanket the
heating surface and prevent the inflow of fresh liquid from taking their place.
At this point the bubbles coalesce and form a vapor film which covers the
surface. The heat must be conducted through this film before it can reach the
liquid and effect the boiling process. The thermal resistance of this film causes
a reduction in heat flux, and this phenomenon is illustrated in region IV, the
film-boiling region. This region represents a transition from nucleate boiling to
film boiling and is unstable. Stable film boiling is eventually encountered in
region V. The surface temperatures required to maintain stable film boiling are
high, and once this condition is attained, a significant portion of the heat lost
by the surface may be the result of thermal radiation, as indicated in region VI.
Figure 12.1 Heat-flux data from an electrically heated platinum wire

in
An electrically heated wire is unstable at point a since a small increase
∆ Tx
, at this point results in a decrease in the boiling heat flux. But the wire
still must dissipate the same heat flux, or its temperature will rise, resulting in
operation farther down to the boiling curve. Eventually, equilibrium may be
reestablished only at point b in the film-boiling region. This temperature
usually exceeds the melting temperature of the wire, so that burnout results. If
the electric-energy input is quickly reduced when the system attains point a, it
may be possible to observe the partial nucleate boiling and unstable film
region.
In nucleate boiling, bubbles are created by the expansion of entrapped
gas or vapor at small cavities in the surface. The bubbles grow to a certain
size, depending on the surface tension at the liquid-vapor interface and the
temperature and pressure. Depending on the temperature excess, the
bubbles may collapse on the surface, may expand and detach from the
surface to be dissipated in the body of the liquid, or at sufficiently high
temperatures may rise to the surface of the liquid before being dissipated.
When local boiling conditions are observed, the primary mechanism of heat
transfer is thought to be the intense agitation at the heat-transfer surface,
which creates the high heat-transfer rates observed in boiling. In saturated, or
bulk, boiling the bubbles may break away from the surface because of the
buoyancy action and move into the body of the liquid. In this case the heattransfer
rate is influenced by both the agitation caused by the bubbles and the
vapor transport of energy into the body of the liquid.

LECTURE NO.13
HEAT EXCHANGERS- GENERAL INTRODUCTION; DOUBLE-PIPE HEAT
EXCHANGER; SHELL-AND-TUBE HEAT EXCHANGER; CROSS-FLOW
EXCHANGER; FOULING FACTORS, LMTD
Introduction
In the process industries the transfer of heat between two fluids is
generally done in heat exchangers. The most common type is one in which
the hot and cold fluids do not come into direct contact with each other but are
separated by a tube wall or a flat or curved surface. The transfer of heat from
the hot fluid to the wall or tube surface is accomplished by convection,
through the tube wall or plate by conduction, and then by convection to the
cold fluid.
Different Types
Double-pipe heat exchanger
The simplest exchanger is the double-pipe or concentric pipe
exchanger. This is shown in Fig. 13.1, where one fluid flows inside one pipe
and the other fluid flows in the annular space between the two pipes. The
fluids can be in cocurrent or countercurrent flow. The exchanger can be made
from a pair of single lengths of pipe with fittings at the ends or from a number
of pairs interconnected in series. This type of exchanger is useful mainly for
small flow rates.
Fig.13.1 Flow in a double pipe heat exchanger
Shell-and-tube exchanger
If larger flows are involved, a shell-and-tube exchanger is used, which
is the most important type of exchanger in use in the process industries. In

these exchangers the flows are continuous. Many tubes in parallel are used,
where one fluid flows inside these tubes. The tubes, arranged in a bundle, are
enclosed in a single shell and the other fluid flows outside the tubes in the
shell side. The simplest shell-and-tube exchanger is shown in Fig. 13.2(a) for
one shell pass and one tube pass, or a 1-1 counterflow exchanger. The cold
fluid enters and flows inside through all the tubes in parallel in one pass. The
hot fluid enters at the .other end and flows counterflow across the outside of
the tubes. Cross baffles are used so that the fluid is forced to flow
perpendicular across the tube bank rather than parallel with it. The added
turbulence generated by this cross-flow increases the shell-side heat-transfer
coefficient.
Fig.13.2. Shell-and-tube heat exchangers:
(a) 1 shell pass and 1 tube pass (1-1 exchanger);
(b) 1 shell pass and 2 tube passes (1-2 exchanger).
Fig. 13.2(b) a 1-2 parallel-counterflow exchanger is shown. The liquid
on the tube side flows in two passes as shown and the shell-side liquid flows
in one pass. In the first pass of the tube side, the cold fluid is flowing

counterflow to the hot shell-side fluid; in the second pass of the tube side, the
cold fluid flows in parallel (cocurrent) with the hot fluid. Another type of
exchanger has two shell-side passes and four tube passes. Other
combinations of number of passes are also used sometimes, with the 1-2 and
2-4 types being the most common.
Cross-flow exchanger
When a gas such as air is being heated or cooled, a common device
used is the cross-flow heat exchanger shown in Fig. 13.3 (a). One of the
fluids, which is a liquid, flows inside through the tubes, and the exterior gas
flows across the tube bundle by forced or sometimes natural convection. The
fluid inside the tubes is considered to be unmixed, since it is confined and
cannot mix with any other stream. The gas flow outside the tubes is mixed,
since it can move about freely between the tubes, and there will be a
tendency for the gas temperature to equalize in the direction normal to the
flow. For the unmixed fluid inside the tubes, there will be a temperature
gradient both parallel and normal to the direction of flow.
Fig. 13.3 Cross-flow heat exchanger
A second type of cross-flow heat exchanger shown in Fig. 13.3 (b) is
typically used in air-conditioning and space-heating applications. In this type
the gas flows across a finned-tube bundle and is unmixed, since it is confined
in separate flow channels between the fins as it passes over the tubes. The
fluid in the tubes is unmixed.
Fouling Factors and Typical Overall U Values
In actual practice, heat-transfer surfaces do not remain clean. Dirt,
soot, scale, and other deposits form on one or both sides of the tubes of an

exchanger and on other heat-transfer surfaces. These deposits offer
additional resistance to the flow of heat and reduce the overall heat-transfer
coefficient U. In petroleum processes coke and other substances can deposit.
Silting and deposits of mud and other materials can occur. Corrosion products
which could constitute a serious resistance to heat transfer may form on the
surfaces. Biological growth such as algae can occur with cooling water and in
the biological industries.
To avoid or lessen these fouling problems, chemical inhibitors are often
added to minimize corrosion, salt deposition, and algae growth. Water
velocities above 1 m/s are generally used to help reduce fouling. Large
temperature differences may cause excessive deposition of solids on surfaces
and should be avoided if possible.
The effect of such deposits and fouling is usually taken care of in
design by adding a term for the resistance of the fouling on the inside and
outside of the tube in Equation as follows:
where,
U
i
=
1 1
+
h h
i
di
1
( r0
−ri
) Ai
+
k A
A
Alm
Ai
+
A h
o
o
Ai
+
A h
hdi
is the fouling coefficient for the inside and
hdo
the fouling coefficient for the outside of the tube in W/m 2 . K.
THE LOG MEAN TEMPERATURE DIFFERENCE
Consider the double-pipe heat exchanger shown in Figure 13-1. The
fluids may flow in either parallel flow or counter-flow, and the temperature
profiles for these two cases are indicated in Figure 13-4. The heat transfer in
this double-pipe arrangement can be calculated with the following equation
q
= UA ∆T m ----- (13.1)
where
U = overall heat-transfer coefficient
A = surface area for heat transfer consistent with definition of U
∆ T m = suitable mean temperature difference across heat
exchanger
The above Fig.13.4 shows that the temperature difference between the
hot and cold fluids varies between inlet and outlet, and the average value has
o
do

to be calculated in the above equation. For the parallel- flow heat exchanger
shown in Figure 13.4 (a), the heat transferred through an element of area dA
may be written
dq =−m c
h
h
.
dT
h
.
= m
c
c
c
dT
c
------ (13.2)
Fig. 13.4 Temperature Profiles for (a) parallelflow and (b) counterflow
in double-pipe heat exchanger
where the subscripts h and c designate the hot and cold fluids, respectively.
The heat transfer could also be expressed
From Equation (13.2-6)
dT
h
dq = U ( Th −T
c)
dA ------------ (13.3)
.
= −
.
.
m
dq
h
c
h

where
Thus
dT
c
.
= −
.
.
m
dq
c
c
c
ṁ represents the mass-flow rate and c is the specific heat of the fluid.
1 1
dT
h
− dT
c
= d( Th
−Tc
) =−dq
( + )
. . ---------- (13.4)
m c m c
h
h
c
c
Solving for dq from Equation (13.3) and substituting into Equation (13.4)
gives
⎛ ⎞
d(
Th
−T
c)
u
⎜ 1 1
=−
⎟
dA
Th
T ⎜
+
.
. ⎟ ------------ (13.5)
−
c
⎝mh
ch
mc
cc
⎠
This differential equation may now be integrated between conditions 1
and 2 as indicated in Figure 13.4.
The result is
⎛ ⎞
Th
2
−T
c2
=−
⎜ 1 1
ln uA
⎟
⎜
+
. .
−
⎟ --------------- (13.5)
Th
1
Tc
1
⎝mh
ch
mc
cc
⎠
Returning to Equation (13.2), the products
ṁ cc c
and
.
mhc h
may be
expressed in terms of the total heat transfer q and the overall temperature
differences of the hot and cold fluids. Thus
.
m
h
c
h
=
T
h1
q
−T
h2
.
q
mc
cc
=
Tc
2
−T
c1
Substituting these relations into Equation (13.5) gives
( Th
2
−T
c2)
−(
Th
1−T
c1)
q= UA
------- (13.6)
ln [( T −T
) / ( T −T
)]
h2
c2
h1
c1
Comparing Equation (13.6) with Equation (13.1), the mean temperature
difference is the grouping of terms in the brackets. Thus
( Th
2
−T
c2)
−(
Th
1−T
c1)
∆ Tm
=
--------
ln[( Th
2
−T
c2) / ( Th
1−T
c1)]
(13.7)
This temperature difference is called the log mean temperature
difference (LMTD). It is the temperature difference at one end of the heat
exchanger less the temperature difference at the other end of the exchanger

divided by the natural logarithm of the ratio of these two temperature
differences.
The above derivation for LMTD involves two important assumptions:
(1) the fluid specific heats do not vary with temperature, and (2) the
convection heat-transfer coefficients are constant throughout the heat
exchanger.
If a heat exchanger other than the double-pipe type is used, the heat
transfer is calculated by using a correction factor applied to the LMTD for a
counterflow double-pipe arrangement with the same hot and cold fluid
temperatures. The heat-transfer equation then takes the form
q= UAF ∆T m -------- (13.8)
values of the correction factor F according are plotted in Figures 13.5 to 13.8
for several different types of heat exchangers. When a phase change is
involved, as in condensation or boiling (evaporation), the fluid normally
remains at essentially constant temperature and the relations are simplified.
For this condition, P or R becomes zero and we obtain
F = 1.0 for boiling or condensation
Figure 13.5 Correction-factor plot for exchanger with one shell pass and
two, four or any multiple of tube passes.

Figure 13.6 Correction-factor plot for exchanger with two shell passes and
four, eight, or any multiple of tube passes.

Figure 13.7 Correction-factor plot for single-pass cross-flow
exchanger, both fluids unmixed
Figure 13.8 Correction-factor plot for single-pass cross-flow
exchanger one fluid mixed the other unmixed.

LECTURE NO.14
DESIGN PROBLEMS ON HEAT EXCHANGERS: CALCULATION OF HEAT
EXCHANGER SIZE FROM KNOWN TEMPERATURES, PROBLEM ON
SHELL-AND-TUBE HEAT EXCHANGER, DESIGN OF SHELL-AND-TUBE
HEAT EXCHANGER
Problems on Design of Heat Exchangers
Problem 14.1 Calculation of heat exchanger size from known
temperatures
Water at the rate of 68 kg/min is heated from 35 to 75 °C by an oil
having a specific heat of 1.9 kJ /kg. °C. The fluids are used in a counterflow
double-pipe heat exchanger, and the oil enters the exchanger at 110 °C and
leaves at 75 °C. The overall heat-transfer coefficient is 320 W/m 2 .°C.
Calculate the heat exchanger area.
Solution
The total heat transfer is determined from the energy absorbed by the
water:
.
q = m c ∆T
= (68) (4180) (75 - 35) = 11.37 MJ /min (a)
w
w
w
= 189.5 kW
Since all the fluid temperatures are known, the LMTD can be calculated by
using the temperature scheme in the following Figure.
(110 −75)
−(75
−35)
∆T m
=
= 37 .44 C
ln[( 110 −75) / (75 −35)]

Then, Since q= UA ∆T
m ,
5
1.895 × 10
A = = 15 .82 m
(320 )(37 .44 )
Problem 14.2 Shell-and-tube heat exchanger
Instead of the double-pipe heat exchanger of problem 14.1, it is desired to
use a shell-and-tube exchanger with the water making one shell pass and the
oil making two tube passes. Calculate the area required for this exchanger,
assuming that the overall heat-transfer coefficient remains at 320 W/m 2 . °C.
Solution
To solve this problem, determine a correction factor from Figure 13.5 to be
used with the LMTD calculated on the basis of a counterflow exchanger. The
parameters according to the nomenclature of Figure 13.5 are
T 1 = 35 °C T 2 = 75 °C t 1 = 110 °C t 2 = 75 °C
t
P=
T
1
T
R=
t
−t
−t
1
75 −110
=
35 −110
2 1
=
2
−T
−t
1
35 −75
=
75 −110
1 2
=
0.467
1.143
2
Hence, the correction factor is
and the heat transfer is
F = 0.81
q=
UAF
∆
T m
so that
5
1.895 × 10
A =
= 19 .53 m
(320 )(0.81)(37 .44 )
2

Problem 14.3 Design of shell-and-tube heat exchanger
Water at the rate of 3.8 kg/s is heated from 38 to 55 °C in a shell-andtube
heat exchanger. On the shell side one pass is used with water as the
heating fluid, 1.9 kg/s entering the exchanger at 93 °C. The overall heattransfer
coefficient is 1419 W / m 2 . °C, and the average water velocity in the
1.9 cm diameter tubes is 0.366 m/s. Because of space limitations the tube
length must not be longer than 2.5 m. Calculate the number of tube passes,
the number of tubes per pass, and the length of the tubes, consistent with this
restriction.
Solution
First assume one tube pass and check to see if it satisfies the
conditions of this problem. The exit temperature of the hot water is calculated
from
q=
m c
h
h
.
dT
h
.
= m
c
c
c
dT
(a)
(3.8)(4.18 )(55 −38 )
∆T h
=
= 34 C
(1.9)(4.18 )
c
so
T
h . exit = 93 - 94 = 59 °C
The total required heat transfer is obtained from Equation (a) for the cold fluid:
q = (3.8)(4.18)(55 - 38) = 270 kW
For a counterflow exchanger, with the required temperature
LMTD
=∆T
m
(93 −55)
−(59
−38)
=
= 28 .66
ln[( 93 −55) / (59 −38)]
C
q=
UA
∆
T m
(b)
3
270 × 10
A = = 6.639
(1419 )(28 .66 )
m
2
Using the average water velocity in the tubes and the flow rate, calculate the
total flow area with
.
m c
= ρAu

3.8
A = = 0.0104
(1000 )(0.366 )
m
2
This area is the product of the number of tubes and the flow area per tube:
0.0104
=n πd
4
(0.0104 )(4)
n= π (0.019 )
2
2
=
36 .7
or n = 37 tubes. The surface area per tube per meter of length is
2
πd =π (0.019 ) = 0.0597 m / tube . m
Recall that the total surface area required for a one-tube-pass exchanger was
calculated in Equation (b) as 6.639 m 2 . Thus compute the length of tube for
this type of exchanger from
nπdL
=6.639
6.639
L = = 3m
(37 )(0.0597 )
This length is greater than the allowable 2.438 m, so we must use
more than one tube pass. When we increase the number of passes,
correspondingly increase the total surface area required because of the
reduction in LMTD caused by the correction factor F. Next try two tube
passes. From Figure 13.5, F = 0.88, and thus
A
q
2.70 × 10
3
total
= =
=
UF ∆T
m
(1419 )(0.88)(28 .66 )
7.54 m
The number of tubes per pass is still 37 because of the velocity
requirement. For the two-tube-pass exchanger the total surface area is now
related to the length by
so that
A total
=2 nπdL
2
7.54
L = = 1. 707 m
(2)(37 )(0.0597 )
This length is within the 2.438 m requirement, so the final design choice is
Number of tubes per pass = 37
Number of passes = 2

Length of tube per pass = 1.707 m

LECTURE NO.15
INTRODUCTION TO MASS TRANSFER: A SIMILARITY OF MASS, HEAT, AND
MOMENTUM TRANSFER PROCESSES; FICK'S LAW FOR MOLECULAR
DIFFUSION
INTRODUCTION TO MASS TRANSFER
A Similarity of Mass, Heat, and Momentum Transfer Processes
The various separation processes have certain basic principles which
can be classified into three fundamental transfer (or "transport") processes:
momentum transfer, heat transfer, and mass transfer. The fundamental
process of momentum transfer occurs in such operations as fluid flow,
mixing, sedimentation, and filtration. Heat transfer occurs in conductive and
convective transfer of heat, evaporation, distillation, and drying.
The third fundamental transfer process, mass transfer, occurs in
distillation, absorption, drying, liquid-liquid extraction, adsorption, ion
exchange, crystallization, and membrane processes. When mass is being
transferred from one distinct phase to another or through a single phase, the
basic mechanisms are the same whether the phase is a gas, liquid, or solid.
This was also shown for heat transfer, where the transfer of heat by
conduction followed Fourier's law in a gas, solid, or liquid.
General molecular transport equation. All three of the molecular transport
processes - momentum, heat, and mass-are characterized by the general
type of equation,
rate of
atransfer
process
driving force
= --------(1)
resis tan ce
Molecular diffusion equations for momentum, heat, and mass transfer
Newton's equation for momentum transfer for constant density can be
written as follows
µ d( v x
ρ)
τzx
= −
------- (2)
ρ dz
where
τzx
is momentum transferred / s.m 2 ,
µ
ρ is kinematic viscosity in m2 /s,

ρ and c p :
z is distance in m, and
v xρ is momentum/m 3 , where the momentum has units
of kg.m/s.
Fourier's law for heat conduction can be written as follows for constant
where
q d(ρc
pT
)
z
= −α
--------- (3)
A dz
q z
/ A is heat flux in W/m 2 ,
α is the thermal diffusivity in m 2 /s, and
ρc p
T is J/m 3 .
The equation for molecular diffusion of mass is Fick's law and can be
written as follows for constant total concentration in a fluid:
where,
J
*
Az
= −D
AB
dc
dz
A
------------ (4)
*
J
Az is the molar flux of component A in the z direction due to
molecular diffusion in kg mol A/s·m 2 ,
D
AB the molecular diffusivity of the molecule A in B in m 2 /s,
cA
the concentration of A in kg mol / m 3 , and
z the distance of diffusion in m.
The similarity of Eqs. (2), (3), and (4) for momentum, heat, and mass
transfer should be obvious. All the fluxes on the left-hand side of the three
equations have as units transfer of a quantity of momentum, heat, or mass
per unit time per unit area. The transport properties µ/ ρ, α and DAB
have units of m 2 /s, and the concentrations are represented as momentum/m 3 ,
J/m 3 , or kg mol/m 3 .
Examples of Mass- Transfer Processes
Mass transfer is important in many areas of science and engineering.
Mass transfer occurs when a component in a mixture migrates in the same
phase or from phase to phase because of a difference in concentration
between two points. Many familiar phenomena involve mass transfer. Liquid
in an open pail of water evaporates into still air because of the difference in
all

concentration of water vapor at the water surface and the surrounding air.
There is a "driving force" from the surface to the air. A piece of sugar added to
a cup of coffee eventually dissolves by itself and diffuses to the surrounding
solution. Many purification processes involve mass transfer. In uranium
processing, a uranium salt in solution is extracted by an organic solvent.
Fick's Law for Molecular Diffusion
Molecular diffusion or molecular transport can be defined as the
transfer or movement of individual molecules through a fluid by means of the
random, individual movements of the molecules. Imagine the molecules
traveling only in straight lines and changing direction by bouncing off other
molecules after collision. Since the molecules travel in a random path,
molecular diffusion is often called a random-walk process.
Fig 15.1 Schematic diagram of molecular diffusion process
In Fig. 15.1 the molecular diffusion process is shown schematically. A
random path that molecule A might take in diffusing through B molecules from
point (1) to (2) is shown. If there are a greater number of A molecules near
point (1) than at (2), then, since molecules diffuse randomly in both directions,
more A molecules will diffuse from (1) to (2) than from (2) to (1). The net
diffusion of A is from high- to low-concentration regions.
As another example, a drop of blue liquid dye is added to a cup of
water. The dye molecules will diffuse slowly by molecular diffusion to all parts
of the water. To increase this rate of mixing of the dye, the liquid can be
mechanically agitated by a spoon and convective mass transfer will occur.

The two modes of heat transfer, conduction and convective heat transfer are
analogous to molecular diffusion and convective mass transfer.
First, we will consider the diffusion of molecules when the whole bulk
fluid is not moving but is stationary. Diffusion of the molecules is due to a
concentration gradient. The general Fick's law equation can be written as
follows for a binary mixture of A and B:
where
J
*
Az
= −cD
AB
dx
dz
A
-------- (5)
c is total concentration of A and B in kg mol A+B/m 3 , and
xA
is the mole fraction of A in the mixture of A and B.
If c is constant, then since
c = cx ,
A
A
c dx = d( cx ) = dc -------------- (6)
A
A
A
Substituting into Eq. (5) obtain the following equation for constant total
concentration:
J
*
Az
= −D
AB
dc
dz
A
------------ (7)
This equation is the one more commonly used in many molecular diffusion
processes.
Problem Molecular Diffusion of Helium in Nitrogen
A mixture of He and N 2 gas is contained in a pipe at 298 K and 1 atm
total pressure which is constant throughout. At one end of the pipe at point 1
the partial pressure pA
1 of He is 0.60 atm and at the other end 0.2 m (20 cm)
p
A2 = 0.20 atm. Calculate the flux of He at steady state if DAB
of the He-N 2
mixture is 0.687 X 10 -4 m 2 /s (0.687 cm 2 /s). Use SI units.
Solution:
Since total pressure P is constant, then c is constant, where c is as
follows for a gas according to the perfect gas law:
where
PV = nRT ----------------------- (8)
n
V
n is kg mol A+B,
V is volume in m 3 ,
P
= = c
-------------- (9)
RT

T is temperature in K,
R is 8314.3 m 3 Pa / kg mol.K or
R is 82.057 x 10 -3 m 3 . atm / kg mol.K, and
c is kg mol A+B/m 3 .
For steady state the flux in Eq.(4) is constant. Also,
constant. Rearranging Eq. (4) and integrating,
DAB
for a gas is
J
*
Az
z2
∫
z1
dz
= −D
AB
CA 2
∫
cA 1
dc
A
J
*
Az
( cA
1
−c
A2)
= DAB
----------- (10)
( z −z
)
2
1
Also, from the perfect gas law, pA V = nA
RT , and
pA
nA
cA = 1
1
=
------------------- (11)
RT V
Substituting Eq. (11) into (10),
J
*
Az
( pA
1
−pA2)
= DAB
----------------- (12)
RT ( z −z
)
2
1
This is the final equation to use, which is in a form easily used for
gases. Partial pressures are p
A1
= 0.6 atm = 0.6 X 1.01325 X 10 5 = 6.08 X
10 4 Pa and p
A2
using SI units,
= 0.2 atm = 0.2 X 1.01325 X 10 5 = 2.027 X 10 4 Pa. Then,
*
J AZ
(0.687
=
−4
4
× 10 )(6.08 × 10 −2.027
8314 (298 )(0.20 −0)
× 10
4
)
= 5.63 X 10 -6 kg. mol A/s. m 2
If pressures in atm are used with SI units,
*
J AZ
(0.687 × 10
(82.06 × 10
=
−3
4
)(0.60 −0.20)
)(298 )(0.20 −0)
−
= 5.63 X 10 -6 kg. mol A/s. m 2
Other driving forces (besides concentration differences) for diffusion
also occur because of temperature, pressure, electrical potential, and other
gradients.

LECTURE NO.16
MOLECULAR DIFFUSION IN GASES: EQUIMOLAR COUNTER DIFFUSION
IN GASES
MOLECULAR DIFFUSION IN GASES: Equimolar Counter diffusion in
Gases
In Fig. 16.1 a diagram is given of two gases A and B at constant total
pressure P in two large chambers connected by a tube where molecular
diffusion at steady state is occurring. Stirring in each chamber keeps the
concentrations in each chamber uniform. The partial pressure
p > p and p > p . Molecules of A diffuse to the right and B to the left.
A1 A2
B2
B1
Since the total pressure P is constant throughout, the net moles of A diffusing
to the right must equal the net moles of B to the left. If this is not so, the total
pressure would not remain constant. This means that
J
= −
------- (13)
* *
AZ
J BZ
FIGURE 16.1 Equimolar counter diffusion of gases A and B.
The subscript z is often dropped when the direction is obvious. Writing
Fick's law for B for constant c,
Now since
J
*
B
B
= −
D
BA
dc
dz
P = p A
+ p = constant, then
Differentiating both sides,
B
c = c A
+ c B ------- (15)
dc
A
dc B
------(14)
= −
---------(16)

Equating Eq. J
J
*
Az
= −D
AB
dc
dz
A
to
J
*
B
= −
D
BA
dc
dz
dc
A * dc
B
= −DAB
= − J
B
=−(
− DBA
------- (17)
dz
dz
*
A
)
Substituting Eq. (16) into (17) and canceling like terms,
coefficient
D
AB
= D BA
This shows that for a binary gas mixture of A and B, the diffusivity
DAB
for A diffusing into B is the same as DBA
for B diffusing into A.
Problem Equimolar Counter diffusion
Ammonia gas (A) is diffusing through a uniform tube 0.10 m long
containing N 2 gas (B) at 1.0132 X 10 5 Pa pressure and 298 K. The diagram is
similar to Fig. 16.2. At point 1, p
A1
= 1.013 X 10 4 Pa and at point 2, p
A2
=
B
,
0.507 X 10 4 Pa. The diffusivity D
AB
= 0.230 X 10 -4 m 2 /s.
(a) Calculate the flux
*
J
A at steady state.
(b) Repeat for
*
J B
Solution:
J
*
Az
where
= D
AB
( pA
1
−pA2)
RT ( z −z
)
2
1
P = 1.0132 X 10 5 Pa,
z 2 – z 1 = 0.10 m, and
T = 298 K. Substituting into above equation
J
*
A
= D
AB
−4
4
( pA
1−pA2)
(0.23 × 10 )(1.013 × 10 −0.507
=
RT ( z −z
) 8314 (298 )(0.10 −0)
2
1
× 10
4
)
= 4.70 X 10 -7 kg mol A/s.m 2
Rewriting the above equation for component B for part (b) and noting that
p
p
= P−p
= 1.0132 X 10 5 - 1.013 X 10 4 = 9.119 X 10 4 Pa and
B1 A1
= P −p
= 1.0132 X 10 5 - 0.507 X 10 4 = 9.625 X 10 4 Pa,
B2 A2
J
*
B
= D
AB
−4
4
( pB
1−pB2)
(0.23 × 10 )(9.119 × 10 −9.625
× 10
=
RT ( z −z
) 8314 (298 )(0.10 −0)
2
1
4
)
=- 4.70 X 10 -7 kg mol B/s.m 2

The negative value for
*
J
B means the flux goes from point 2 to point 1.