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8/11/2019 (Only Theory) Heat Mass Transfer
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8/11/2019 (Only Theory) Heat Mass Transfer
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Indias No.1
IES
Academy
25 - 1st Floor, Jia Sarai , Near IIT, New Delhi-110016 Ph: 011-26537570 , 9810958290 2
HEATM
ASSTRANSFER
MODES AND BASIC LAWS OF HEAT TRANSFER
The literature on heat transfer generally recognises
three distinct modes of heat transmission :
conduction , convection and radiation. These three
modes are similar in that a temperature differen-
tial must exist and the heat exchange is in be di-
rection of decreasing temperature. Each method
has, however, different physical picture and dif-
ferent controlling laws.
Conduction :Thermal conduction is a mechanism
of heat propagation from a region of higher tem-
perature to a region of low temperature with in a
medium (solid, liquid, or gaseours) or between dif-
ferent mediums in direct physical contact. Conduc-
tion does not involve any movement of macro-
scopic portions of matter relative to one another.The thermal energy may be transferred by means
of electrons which are free to move through the
lattice structure of the material. In addition, or al-
ternatively. It may be transferred as vibrational
energy in the lattice structure. Irrespective of the
exact mechanism, the observable effect of conduc-
tion is an equalization of temperature.
Consider the flow of heat along a metal rod,
one end of which is placed adjacent to a flame.
The elementary particles (molecules, atoms, elec-
trons) composing the rod, and which are in imme-diate vicinity of the flame, get heated. Because of
the resulting temperature growth, their kinetic en-
ergy increases and this puts them in a violent state
of agitation, and they start vibrating about their
mean positions. Consequently, these more active
particles collide with less active molecules lying
next to them. During collision, the less active par-
ticles also get excited, i.e.., thermal energy is im-
parted to them. The process is repeated for layer
after layer of molecules until the other end of therod is reached. Each layer of molecules is at a
slightly higher temperature than the preceding one,
i.e., a temperature gradient exists along the length
of the rod. The rate of heat flow between the two
ends depends upon the lengths of the rod, tempera-
ture difference the two ends, and the physical and
chemical composition of the bar material.
Fig 1
Fig : Conduction heat flow along a rod.
Since conduction is essentially due to random
molecular motion, the concept is termed as micro-
form of heat transfer and is usually referred to as
diffusion of energy. The rate equation for one-di-
mensional steady flow of heat by conduction is
prescribed by the Fourier Law :
......................1.2dt
Q kAdx
=
where, Q is the heat transfer rate. A is the area of
heat transfer surface. , dt is the temperature differ-
ence for a short perpendicular distance dx, and the
thermal conductivity k is a characteristic of the
surface material. Since the temperature gradient is
negative in the positive x-direction , the minus signin the equation gives positive heat flow.
If is the path length in the direction of heat flow
and
1 2( )t t
is the temperature difference, then
1 2( )kA t t
Q
=
............................1.3
The heat flux q is heat conducted per unit time per
unit area and is given by
1 2( )Q k t t
q A
= = ...........................1.4
Heat transfer in metal rods, in heat treatment of
steel forgings and through the walls of heat ex-
change equipment are some practical examples of
heat conduction.
Example:A 75 cm thick side will of an oven is
primarily made of insulation with a thermal con-
ductivity of 0.0345 kcal/m hr K (0.04 W/m K.).
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HEATM
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Conditions on the inside of wall fix the tempera-
ture on that side at 420 K. The electric coil with in
the oven dissipate 31.4 kcal/hr (36.5 watts) of elec-
trical energy to made up for the heat loss through
the wall. Calculate the wall surface area, perpen-
dicular to heat flow, so that temperature on the other
side of the wall does not exceed 310 K.
Solution :Under the stipulations of one-dimen-
sional steady state heat conduction , the electrical
energy dissipation rate with in the oven must equal
the conduction heat conduction heat transfer rate
across the wall. That is :
1 2( )kA t t
Q
=
0.0345 (420 310)31.4 50.6
0.75
AA
= =
Hence , the required wall surface area,
231.4 0.6250.6
A m= =
Convection :Thermal convection is a process of
energy transport affected by the circulation or mix-
ing of a fluid medium (gas, liquid or a powdery
substance). Convection is possible only in a fluid
medium and is directly linked with the transport
of medium itself. Macroscopic particles of a fluid
moving in space cause the heat exchange, and thus
convection constitutes the microform of the heattransfer. The effectiveness of heat transfer by con-
vection depends largely upon the mixing motion
of the fluid.
With respect to origin, two types of con-
vection are distinguished; forced, and natural or
free convection.
In natural or free convention, the circula-
tion of the fluid medium is caused by buoyancy
effects, i.e., by the difference in the densities of
the cold and heated particles. Consider heat flow
from a hot plate to atmosphere. The stagnant layer
of air in the immediate vicinity of the plate gets
thermal energy by conduction. The energy thus
transferred serves to increase the temperature and
internal energy of the air particles. Because of tem-
perature rise these particles become less dense (and
therefore lighter) than the surrounding air. The
lighter air particles move upwards to a region of
low temperature where they mix with transfer a
part of their energy to the cold particles. Simulta-
neously the cold air particles descend downwards
to fill the space vacated by the hot air particles.
The circulation pattern, upward movement of the
warm air and the downward movement of the cool
air, is called the convection currents.
A similar effect can also be demonstrated
by a hot-water heating system. Fig. where water
serves as the medium for carrying heat to all parts
of the building. Water is heated in the boiler in-
stalled at the base of the building. The hot water
becomes lighter, rises up in the left hand vertical
pipes and passes through the radiators fitted in dif-
ferent rooms of the building. The radiators get
heated and dissipate heat to the rooms. After los-
ing heat to the radiators, the water gets cooled and
returns back to the boiler through the pipe on theright. Convection currents are setup and the build-
ing
Fig 2
is kept warm continuously at a constant tempera-
ture. In this way, a constant circulation of water
through the pipes and through the radiator is main-
tained. Some other examples of free convectionare chilling effect of a cold wind on a warm body;
heat flow from a hot pavement to surrounding at-
mosphere and heating of air in a room by a stove;
cooling of billets in the atmosphere; heat exchange
on the outside of cold and warm pipes.
In forced convection, the flow of fluid is
caused by a pump, fan or by atmospheric winds.
These mechanical devices provide a definite cir-
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HEATM
ASSTRANSFER
cuit for the circulating currents and that speeds up
the heat transfer rate. Example of forced convec-
tion are; flow of water in condenser tubes, fluid
passing through the tubes of a heat exchanger; cool-
ing of internal combustion engine; air condition-
ing installation and nuclear reactors.
Regardless of the particular nature. The ap-
propriate rate equation for the convective heat trans-
fer between a surface and an adjacent fluid is pre-
scribed by Newtons Law of cooling :
1 2( )Q hA t t = ..........................1.5
where, Q is the convective heat flow rate, A is area
exposed to heat transfer, ts and tf are the surface
and fluid temperature respectively. The heat trans-
fer coefficient h depends upon the thermodynamic
and transport properties (e.g. density, viscosity,
specific heat and thermal conductivity of the fluid.)
the geometry of the surface the nature of fluid flow,
and the prevailing thermal conditions.
Convection mechanisms involving phase
changes leads to the important fields of boiling
(evaporation) and condensation.
Example :An oil cooler in a high performance
engine has an outside surface area 0.12 m2and a
surface temperature of 650C. The air rushes over
the surface of the cooler at a temperature of 300C
and gives rise to a surface coefficient of heat trans-
fer equal to 45.4 W/m2
K. Calculate the heat trans-fer rate from the cooler.
Solution :The conditions described imply a con-
vective process, that is , heat transfer from a solid
surface (the oil cooler) to an adjacent moving fluid
(the air passing over the cooler.) The rate of heat
transfer by convection from oil cooler to the air is
then
( )
45.4 0.12(65 30) 190.68
S FQ hA t t
W
=
= =
Radiation :Thermal radiation is the transmissionof heat in the form of radiant energy or wave mo-
tion from one body to another across an interven-
ing space. Unlike heat transfer by conduction and
convection, transport of thermal radiation does not
necessarily affect the material medium between the
heat source and the receiver. An intervening me-
dium is not even necessary and the radiation can
be affected through vacuum or a space devoid of
any matter. Radiation exchange, in fact, occurs most
effectively in vaccum. A material present between
the heat source and the receiver would either re-
duce or eliminate entirely the propagation of ra-
diation energy.
The mechanism of the heat flow by radiation con-
sists of three distinct phases :
(i) Conversion of thermal energy of the hot source
into electromagnetic waves :
All bodies above abosolute zero temperature are
capable of emitting radiant energy. Energy released
by a radiating surface is not continuous but is in
the form of successive and separate (discrete) pack-
ets or quanta of energy called photons. The pho-
tons are propagated through the space as rays; the
movement of swarm of photons is described as the
electromagnetic waves.
(ii) Passage of wave motion through-interveningspace :
The photons , as carries of energy, travel with un-
changed frequency in straight paths with speed
equal to that of light.
(iii) Transformation of waves into heat :When
the photons approach the cold receiving surface
there occurs reconversion of wave motion into ther-
mal energy which is partly absorbed, reflected or
transmitted through the receiving surface.
Thermal radiation is limited to range of wavelength
between 0.1 and 100 of the electromagnetic spec-
trum. Thermal radiations thus include the entire
visible and infrared, and a part of ultra violet spec-
trum. It is to be recognized that thermal radiation
is the transfer of energy by disorganized photon
propagation. In contrast, an organized photon en-
ergy such as radio transmission can be macroscopi-
cally identified and is not considered heat. Further,
emission of thermal radiations is associated with
thermally excited conditions which depend upon
the temperature and nature of the surface.The most vivid evidence of radiation heat
transfer is that represented by solar energy which
passes through inter stellar (conditions close to that
perfect vaccum) on its way to the earth space. So-
lar radiation plays an important part in the design
of heating and ventillating systems. Heat transfer
by radiation is encountered in boiler furnaces, bil-
let reheating furnaces and other types of heat ex-
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HEATM
ASSTRANSFER
change apparatus. The design and construction of
engines, gas turbines, nuclear reactors and solar
collectors is also significantly influenced by the
radiation heat transfer.
The basic rate equations for heat transfer
are based on Stefan-Boltzman Law :
4
b bE AT=
.....................1.6where, E
bis the energy radiated per unit time, T is
the absolute temperature of the surface, and b is
the Stefan-Boltzman constant.
8 2 2
8 2 4
5.67 10 /
4.86 10 /
b W m K
kcal m hr K
=
=
Equation 1.6 is essentially valid for an ideal radia-
tor or a black body - suffix b designates a black
surface. The radiant energy emitted by a real sur-
face is less than for an ideal emitter and is given byA
bE AT=
where is a radiative property of the surface andis called emissivity ; its value depends upon sur-
face characteristics and temperature. It indicates
how effectively the surface emits radiations com-
pared to an ideal or black body radiator. Normally
a body radiating heat is simultaneously receiving
heat from other bodies as radiation. Consider that
surface 1 at temperature T1is completely enclosed
by another black surface 2 at temperature T2. Thenet radiant heat transfer is
4 4
1 1 2( )bQ A T T =
....................1.8
Likewise, the net rate of heat transfer between the
real surface ( called gray surface) at temperature
T1to a surrouding black surface at temperature T
2
is
4 4
1 1 1 2( )bQ A T T = ...................1.9
The net exchange of heat between the two radiat-
ing surfaces is due to the fact that one at the higher
temperature radiates more and receives less energy
for its absorption. An isolated body which remains
at constant temperature emits just as much energy
by radiation as it receives.
Example :A radiator in a domestic heating sys-
tem operates at a surface temperature of 600C. Cal-
culate the heat flux at the surface of the radiator if
it behaves as a black body.
Solution: The heat flux at the surfaces is the rate
at which radiant energy leaves the surfaces per unit
area.
4
8 2 25.67 10 (273 60) 697.2 /
b
Qq T
A
W m
= =
= + =
STEADY AND UNSTEADY HEAT TRANS-
FER
Any physical phenomenon generally involves a
changes of its physical properties. Likewise heat
exchange is also accompanied by space-time varia-
tion of temperature, and the analytical computa-
tions for the amount of heat exchange lie in deter-
mining a mathematical relation for the tempera-
ture field prescribed as
( , , , )t f x y z = .................1.10
Equation 1.10 refers to the entries set of tempera-
ture at all points of space studied at any instant of
time
.
Heat exchange between two systems may take place
under steady (stable) thermal conditions or under
unsteady (unstable) thermal condition, Steady state
implies that temperature at each point of the sys-
tem remains constant in the course of time, and it
is a function only of space co-ordinates.
( , , ); 0dt
t f x y zd
= =
....................1.11
Steady state results in a constant rate of heat ex-
change (heat influx equals heat effux), and there is
no change in the internal energy of the system dur-
ing such a process. Typical examples of steady state
heat transfer are : Cooling of an electric bulb by
surrounding atmosphere; heat flow from the prod-
ucts of combustion to water in the tubes of a boiler,
from the hot to cold fluid in a heat exchanger, and
from a refrigerated space to cooling surface of the
evaporator.Under unsteady thermal conditions, temperature of
the system changes continuous with time. Tempera-
ture is obviously a function of space and time co-
ordinates.
( , , , ); 0dt
i f x y zd
=
.........................1.12
Unsteady state results in heat transfer rate which
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HEATM
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changes with time. Further, a change in tempera-
ture, indicates a change of internal energy of the
system. Energy storage is thus a part and parcel of
unsteady heat flow. Typical examples of unsteady
heat transfer are : Warm-up periods of furnaces,
boilers and turbines; cooling of castings in a
foundry; heat treatment and stress relieving of metal
castings.
A special kind of unsteady process is the
transient static wherein the system is subjected to
cyclic variations in the tempeature of its environ-
ment. The temperature at a particular point of the
system returns periodically to the same value; the
rate of heat flow and energy storage also undergo
periodic variations. Examples are : Heating or cool-
ing of the water of an I.C engine; heating or cool-
ing of the walls of a building during the 24 -hour
cycle of the day.Further , the heat transfer in a system may be in
one, two or more directions. In a one dimensional
heat flow, there is a single predominant direction
in which tempeature differential exists and obvi-
ously the heat flow takes place; heat flow in the
other two directions can be safety neglected. When
the temperature is a function of two co-ordinates ,
heat flow is two-dimensional . A three-dimensional
heat flow stipulates that temperature is a function
of three coordinates. A three-dimensional heat flow
stipulates that temperature is a function of three
co-ordinates, and consequently heat flow occurs
in all three directions.
Dimensionality of temperature field for
steady/unsteady heat flow can be mathematically
expressed as:
_____________________________________________
Type of heat flow Steady Unsteady
_______________________________________________
one-Dimensional t = f(x) t = f(x,
)
heat flowTwo-Deimensional t - f(x,y) t = f(x,y,
)
heat flow
Three-Dimensional t = f(x,y,z) t =
f(x,y,z,
)
het flow
________________________________________________
For simplicity, solutions to majority of heat trans-
fer problems are obtained by the one-dimensional
analysis.
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FOURIER EQUATION
Conduction is primarily a molecular phenomenon
requiring temperature gradient as the driving force
. Experimental evidence does indicate that steady-
state one dimensional flow of heat by conduction
through a homogeneous material is given by the
Fourier Law.
...................2.1
dtQ kA
dx
Q dtq k
A dx
=
= =
The heat flux q (heat conducted per unit time per
unit area) flows along normal to area A in the di-
rection of decreasing temperature. The units oneach term are :
Q : rate of heat flow, kcal/hr or kJ/hr
A : area perpendicular to the direction of heat flow,
m2
dx : thickness of material along the path of heat
flow, m
dt : temperature difference between the two sur-
faces which heat is passing, degree Kelvin K or
degree centigrade C.
The ratio dt/dx represents the change in tempera-ture per unit thickness, i.e., the temperature gradi-
ent. The negative sign indicates that the heat flow
is in the direction of negative temperature gradi-
ent, and that serves to make the heat flow positive.
The proportionality factor k is called the heat con-
ductivity or thermal conductivity of the material.
Thermal conductivity of a material is one of its
transport properties. Others are the viscosity asso-
ciated with the transport of moment , and the dif-
fusion coefficient associated with the transport of
mass. Thermal conductivity provides an indicationof the rate at which heat energy is transferred
through a medium by the diffusion (conduction)
process. For a prescribed temperature gradient and
geometric parameters, the heat flow rate increase
with increasing thermal conductivity.
The Fourier law is essentially based on the follow-
ing assumptions :
FOURIER EQUATION AND THERMAL CONDUCTIVITY
Steady state conduction which implies that the
time rate of het flow between any two selected
points is constant with time. This also means that
the temperature of the fixed points within a heat
conducing body does not change with time :
( )t f
one-directional heat flow; only one space co-
ordinate is required to describe the temperature dis-
tribution within the heat conducting body; t = f(x).
The surfaces in the y-and z-direction are perfectly
insulated.
bounding surfaces are isothermal in character,
i.e, constant and uniform temperature are main-tained at the two faces.
isotropic and homogenous material , i.e., ther-
mal conductivity has a constant value in all the di-
rections.
constant temperature gradient and a linear tem-
perature profile.
no internal heat generation.
Some essential features of the Fourier relation are
enumerated below :
Fourier law is valid for all matter regardless of
its state; solid, liquid or gas
Fourier law is a vector expression indicating
that heat flow rate is normal to an isotherm and is
in the direction of decreasing tempeature.
Fourier law cannot be derived from first prin-ciple ; it is a generalization based on experimental
evidence.
Fourier law helps to define the transport prop-
erty k,i.e., the thermal conductivity of the heat con-
ducting medium.
Assuming dx = 1m; A = 1m2and dt = 10; , we ob-
tain
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Q = k
Hence thermal conductivity may be defined as the
amount of heat conducted per unit time across unit
area and through unit thickness, when a tempera-
ture difference of unit degree is maintained across
the bounding surfaces.
The units of thermal conductivity are worked out
form the Fourier law written in the form.
Q dxk
A dt=
Thus
2
2
1[ ] / deg
deg
1 m[ ] / deg
deg
kcal mk kcal m hr
hr m
kJk kJ m hr
hr m
= =
= =
The unit kJ/m- hr-deg could also be specified as J/m-s-deg or W/m-deg and this is actually done while
quoting the numerical values of thermal conduc-
tivity. Following conversion factors help to con-
vert the thermal conductivity from MKS system
of units into SI units.
1.0 / deg 4.186 / deg
1.163 / deg
kcal m hr kJ m hr
W m
=
=
THERMAL RESISTANCE
Observations indicate that in systems involving
flow of fluid, heat and electricity, the flow quan-
tity is directly proportional to the driving and in-
versely proportional to the flow resistance. In a
hydraulic circuit, the pressure along the path is the
driving potential and roughness of the pipe is re-
sistance of the flow material . Likewise tempera-
ture difference constitutes the driving force for heat
conduction through a medium.
From Ohms law
potential (dV)( )resistance (Re)
voltagecurrent ielectrical
=
and from Fourier;s law
temperature potential (dt)flow rate (Q)=
resistance (dx/kA)heat
thermal
Obviously there is a one-one correspondence be-
tween the flow of electric current and heat i.e.,
- electrical current (amperes) is analogous to ther-
mal heat flow rate ( kJ/hr.)
- electric voltage (volts) corresponds to thermal
temperature difference (degree Kelvin).
- electric resistance (ohms) is analogous to quan-
tity dx/kA. This quantity is called thermal resis-
tance.
Thermal resistance, Rt = (dx/kA), is expressed in
the unit hr-deg/kcal ir s-deg/J or deg/W. The recip-
rocal of thermal resistance is called thermal con-
ductance and it represents the amount of heat con-
ducted through a solid wall of area A and thick-
ness dx when a temperature difference of unit de-
gree is maintained across the bouding surfaces.
Fig 3
Fig: Concept of thermal resistance
Sometimes the heat conducting capacity of a given
physical system is expressed in terms of unit ther-
mal resistance rtand unit thermal conductance c
t
1c =
rt
dx kr and
k dx= =
1 2
:.
( )
dtQ kA
dx
t tkA
dx
=
=
1 2
1 2
( )
( ).....................2.1
t
A t t
r
cA t t a
=
=
GENERAL HEAT CONDUCTION EQUA-TION
The objective of conduction analysis is two fold :
(i) to determine the temperature distribution , i.e.,
variation of temperature with time and position,
and
(ii) to make computations for heat transfer etc. :
Fourier law of heat conduction is essentially valid
for heat flow under uni-directional and steady state
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conditions. However in many practical cases the
temperature may be function of space co-ordinate
as well as time. Recourse is then made to three-
dimensional heat flow equations which consider
both non-uniformly of temperature and any irregu-
larity in the boundary of the surface. To accom-
plish this task an elemental volume is taken , the
relevant energy transfer processes are identified and
the appropriate rate equations are introduced. So-
lution of the resulting differential equations yield
the temperature distribution. Fourier rate equation
is then invoked to workout the heat transfer rate
through the conducting medium.
Cartesian Co-ordinates:Consider the flow of heat
through an infinitesimal volume element oriented
into a three-dimensional co-ordinate system fig.
The side dx, dy and dz have been taken parallel to
the x, y and z axis respectively.Let t represent the temperature at the left face of
the differential control volume. Since area of this
face can be made arbitrarily small, the temperature
t may be assumed uniform over the entire surface.
The temperature changes along the x-direction and
the rate of change is given by /t x . Then changeFig 4
Fig: Conduction analysis in Cartesian co-
ordinates
of temperature through distance dx will be
( )/ .t x dx
This temperature change has been
graphically illustrated in Fig. Therefore the tem-
perature on the right face, which lies at a distance
dx from the left face will be ( )/ .t t x dx + . For
non-strophic materials there will also be a change
in thermal conductivity as heat flows through the
control volume.
Fig 5
Fig : Change in temrpertaure as a function of
distance.
The general conduction equation can be set up by
applying Fourier equation in each cartesian direc-
tion, and then applying the energy conservation
requirement. If kx represents the thermal conduc-
tivity at the left face, then quantity of heat flowing
into the control volume through this face during
time interval
d
is given by :
Heat influx
( ) ......................2.6x x
tQ k dy dz d
x
=
During the same time interval the heat flow out of
the right face of the control volume will be,
Heat efflux
( ) ..................2.7x dx x x
Q Q Q dxd
+
= +
Equation 2.7 simply states that the x-component
of heat transfer rate at (x + dx) is equal to value ofthis component at x plus the amount by which it
changes with respect to x times dx.
Accumulation of heat in the elemental volume due
to heat flow in the x-direction is given by the dif-
ference between heat influx and heat effiux. Thus
the heat accumulation due to heat flow in x-direc-
tion is
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( )
( )
x x x x
x
dQ Q Q Q dxd
Q dxx
= +
=
2 (
....................2.8x
tk dydz d dx
x x
tk dx dy dz d
x x
=
Likewise the heat accumulation in the control vol-
ume due to heat flow along the y-and z-directions
will be:
dy dz d ............2.9y yt
dQ k dxy y
=
dy dz d ............2.10x zt
dQ k dxz z
=
Sume of heat accumulations as prescribed by equa-
tions 2.8 , 2.9 and 2.10 gives the total heat stored
in the elemental volume due to heat flow along all
the coordinate axes.
Total or net accumulation of heat is equal to
dy dz dx y zt t t
k k k dx
x x y y z z
+ +
......... . 2.11
There may be heat sources inside the control vol-
ume due to nuclear fission, flow of electric current
in the coils of electric motors and generators, and
ohmic heating of the material. If qgis the heat gen-
erated per unit volume and per unit time, then the
total heat generated in the control volume equal to
dy dz d .........................2.12g
q dx
The total heat accumulated in the lattice due to heat
flow along all the co-ordinate axes Eq. 2.11 andthe heat generated within the lattice (Eq. 2.12) to-
gether serve to increase the thermal energy of the
lattice. This increase in thermal energy is reflected
by the time rate of change in the heat capcity of the
control volume and is given by :
t( dy dz) c ..............2.13dx d
where is the density and c is the specific heat of
the material. Thus from energy balance consider-
ations :
dy dz d
dy dz d dy dz c ..........2.14
x y z
g
t t tk k k dx
x x y y z z
tq dx dx d
+ + +
=
Dividing both sides by dx dy dz d
.
x y z g c
t t t t k k k q
x x y y z z
+ + + =
............2.14
or , using the vector operator ,
.( ) ...................2.14g
tk t q c a
+ =
Equation 2.14 represents a volumetric heat balance
which must be satisfied at each point for self-gen-
erating, unsteady state three-dimensional heat flow
through a non-dimensional heat flow through non-
isotropic material. This expression known as the
general heat conduction equation, equation, estab-
lishes in differential form the relationship between
the time and space variation of temperature at any
point of the solid through which conduction takes
place. It should be noted that the heat generation
term qgmay be a function of position or time, or
both.
Homogenous and isotropic material :A homoge-
neous material implies that the properties, i.e., den-
sity , specific heat and thermal conductivity of the
material are same everywhere in the material sys-
tem. Isotropic means that these properties are not
directional characteristic of the material i.e, they
are independent of the orientation of the surface.
Therefore for an isotropic and homogenours mate-
rial, thermal conductivity is same at every pointand in all directions. In that case k
x= k
y= k
z= k
and the diffusion equation 2.14 takes the form.
2 2 2
2 2 2
1........2.15
gqt t t c t t
x y z k k
+ + + = =
The quantity /k pc== is called the thermal dif-
fusively, and it represent a physical property of the
material of which the solid is composed. Thermal
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diffusivity is an important characteristic quantity
for unsteady conduction situations. By using the
Laplacian operator
2
, the above equation 2.15
may be written as :
2 1 .................2.15g
t
q ta
k
+
Equation 2.15 governs the temperature distribution
under unsteady heat flow through a homogeneous
and isotropic material.
Different cases of particular interest are :
(i) In many situations there is no dependence of
temperature on time. Conduction then occurs in
the steady state, and the heat flow equation reduces
to :
2 2 2
2 2 20
gqt t t
x y z k
+ + + =
or2 0
gqt
k + = (Poissons equation ) ......2.16
In the absence of internal heat generation or re-
lease of energy with in the body, equation 2.16 fur-
ther reduces to :2 2 2
2 2 2
2
0
0 (Laplace equation)........2.17
t t t
x y z
or t
+ + =
=
(ii) Unsteady state heat flow with no internal heat
generation gives
2 2 2
2 2 2
2
1
1 (Fourier equation)........2.18
t t t t
x y z
tor t
+ + =
=
(iii) For one-dimensional and steady state heat flow
with no internal het generation, the general con-
duction equation takes the form :2
20; 0.............2.19
d dt d t k
dx dx dx
= =
Solution of these equations for any specific bound-
ary conditions will yield the temperature distribu-
tion in the conducting material.
Thermal diffusivity.The following reflections can
be made with regard to this physical property of
the conducting material :
(i) Thermal diffusivity of a material is the ratio
of its thermal conductivity k to the thermal storage
capacity pc. The heat storage capacity essentially
represents thermal capacitance or thermal inertia
of the material, i.e., its sluggishness to conduct heat.
A high value of thermal diffusivity could result ei-
ther from a high value of thermal conductivity or
form low value of thermal conductivity or form
low value of thermal capacity. Liquids have a low
thermal conductivity, high thermal inertia and
hence a small thermal diffusivity. Metals posses
high thermal conductivity, low thermal inertia and
hence a large thermal diffusivity.
(ii) Thermal diffusivity indicates the rate at which
heat is distributed in a material , and this rate de-
pends not only on the conductivity but also on the
rate at which heat energy can be stored. An insightinto equation 2.15 would reveal that thermal
diffusivity, higher would be the rate of change of
temperature at any point of the material.
Equalisation of temperature would then proceed
at a higher rate in materials having large thermal
diffusivity.
(iii) Temperature distribution indicates state is be-
ing governed both by thermal conductivity as well
as by thermal storage capacity. In contrast, during
steady state heat conduction (Eq. 2.16) thermal
conductivity is the only property of the medium
which influences the temperatures distribution.
(iv) The relative magnitude of thermal diffusively
is a measure of the rapidity with which the mate-
rial responds to sudden temperature changes in the
surrounding. Metals and gases have relatively large
value of
and their response to temperature
changes is quite rapid. The non-metallic solids and
liquids respond slowly to temperature changes be-
cause of their relatively small value of thermal
diffusivity.
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COMPOSITE SLAB AND EQUIVALENT RESISTANCE
A composite plane wall is shown in figure (a) be-
low.
Fig 1
In the Figure there are three layers of different
materials with different thermal conductivity and
widths. The temperature at various interfaces are
noted in the figure. In a steady flow, we assume
that the heat condition through all the materials is
same. This can be used to write the following equa-
tion
2 1
3 2
4 3
( )
( )
( )
A
A
B
B
C
C
T Tq k A
x
T Tk A
x
T Tk A
x
=
=
=
From the above equations we can obtain
4 1( )
CA B
A B C
T Tq
xx x
k A k A k A
=
+ +
also( )
A B C
Tq
R R R
=
+ +
Again electrical analogy circuit can be obtained as
shown in figure below:
Fig 2
In general , the heat flow and electrical current flow
can be represented as
thermal potential differenceflow=
resistanceHeat
thermal
Combined heat transfer process
In most of the engineering applications, heat is
transferred in successive steps by similar or differ-ent mechanisms. In a particular location, the trans-
fer may occur by one mechanism only or by more
than one mechanism working in parallel. For in-
stance in the case of the heating of water in a tube
laid in a combustion chamber of a boiler, see fig-
ure. the water will receive heat directly from the
products of combustion which may contain hot
gases which emit and absorb radiation. The heat
will flow by combination of different modes
through successive steps as indicated below.
Fig 3
Step I:Transfer of heat from hot gas to the outer
surfaces of area A of the tube by convection and
radiation working in parallel.
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( )
1 2 1 1 2
1 2
1 2
1
( ) ( )
( )( )
c r
c
c r
q q q
h A t t h A t t
A h h t t
t t
R
= +
= +
= +
=
where1
R is the resistance to heat flow through step
(1) =
( )1
c rh h A
+
and t1and t
2are temperature of
hot gas and that of outside surface of the tube re-
spectively.
Step II:Conduction of heat through the metallic
wall of the tube.
( )2 3kA
q t tl
= ( )2 3K t t= ( )2 32
t t
R
=
where R2is the resistance to heat flow in step II
and t3is the inner wall temperature.
Step III.Finally , the heat will pass to the water
from the inner wall by convection.
( )
( )
3 4
3 4
3
' 'c
q q h A t t
t t
R
= =
=
where R3is the resistances to heat flow through
step III and t4is the water temperature within the
tube.
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Heat Transfer with Fins
This is a case of heat in combined modes of con-
duction and convection.
Consider the one dimensional fin exposed to a sur-
rounding fluid at a temperature ofmT . The tem-
perature of the base of the fin is To. See figure
Fig 4
The problem of heat transfer can be approached by
making energy balance on an element of the fin of
length dx at a distance x as shown in figure.
Thus
IN OUT lost
left face right face by convection
Energy Energy Energy = +
Convective heat transfer is calculated as
( )conq hA T T =
Let the cross-sectional area of the fin be A and pe-
rimeter be P. Then the energy quantities are
Energy IN left face x fdT
q kdx
=
Energy OUT right face xx dx
dTq dx kA
dx +
+ =
2
2
dT d T kA dx
dx dx
= +
Energy lost by convection
( )conq hPdx T T =
It should be noted that the differential area for con-
vection is the product of the perimeter of the fin
and the differential length dx. When we combine
the quantities, the energy balance yields.
( )2
2
10
d hPT T
dx kA = =
Let
T T =
then the above equation becomes,
2
20
d hP
dx kA
=
To solve the above equation, we need two differ-
ent boundary conditions. One common condition
is
x = 0o o
T T at = = =
(i.e) at the base, tempera-
ture is
oT
or
o =
and we can assume three differ-
ent physical situations.
Case (i)
The fin is assumed to be very long so that the tem-
perature at the tip of the fin is essentially that of
the surrounding fluid. The boundary conditions for
the case is( ) 0
( ) 0
oa at x
b at x L or
= =
= =
Let
2
'
hPm
kA=
the above differential equation yields
a general solution as
mx
1 2
mxC e C e = +
where C1and C
2are constants and they are calcu-
lated by applying the above conditions
o
T-T( )
T
mx
o
mx
e
or e
T
=
=
and the rate of heat through the fin is given by
0
fin
x
dTq kA
dx =
=
{ }0
( ) mx
o xkA m T T e
=
=
= kAm ( )oT T
HEAT DISSIPATION FROM EXTENDED SURFACES
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( )oPhkA T T =
Case (ii)
For this case, the fin is of finite length and loses
heat by convection from its end to the ambient air,
equal to the heat conducted by the fin - and the
solution for the heat flow equation will be (same
as before)
1 2
mx mxC e C e = +
with the new boundary conditions
0oT T at x = =
and
LT T at x L = =
( )Lx L
dTk h T T
dx
=
=
[ Heat conducted at x= L
is the same convected by the
end to the ambient air.]
Constants C1and C
2are calculated with these con-
ditions and, substituting the constants, the tempera-
ture distribution along the length of the fin is given
by
( )
( )
/ 2( / )
21
mL x Lm L x L
mLo
T T e e
T T e
whereh mk
h mk
+=
The above equation can also expressed in terms of
hyperbolic function as:
hcosh m(L-x)+ sinh ( )
mkh
cosh mL+ sinhmk
o
m L xT T
T TmL
=
The rate of heat transfer is given by,
0
2
21.1
tanh
( )
1 tanh mL
fin
x
mL
mL
o
dTq kA
dx
ePhAke
hmL
mkPhAk T T
h
mk
=
=
+=
+
= +
Case (iii)
In this case, the end of the fin is assumed to be
insulated (i.e) 0dT
dx= at x = L the boundary condi-
tions are
( ) 0oi at x = =
and
( ) od
ii at x Ldx
= =
thus
( )1 2
1 20
o
mL mL
C C
m C e C e
= +
= +
Solving these equations, C1and C
2and be calcu-
lated - the solution for temperature distribution is
2 2
cosh ( )
1 1 cosh
mx mx
mL mL
o
e e m L x
e e mL
= + =
+ +
the heat transfer in this case is given by
( )0 tanhfinq h p kA mL=
In many applications straight fins are used all along
the length of the wall, instead of a number of spines.
Such fins are generally fins are generally of rect-
angular profile and are called straight rectangular
fins as shown in figure.
Fig 5
The width of the fin in Z, thickness of fin is t, then
A = Zt2( )
2
P Z t
Z if Z t
= +
= >>
then
( )2
rect
hP hm
kA kt = =
For a circular fin
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HEATM
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( )2
2
2
cir
h rm
k r
h
kr
=
=............................12
Fin effectiveness
This is also called fin efficiency. The objective offin is only to increase the area of the heat transfer
surface. The temperature at the base of the fin will
be maximum and will be decreasing gradually to-
wards the tip. The fin would have dissipated maxi-
mum amount of heat if its temperature all along
the length was the same as that at the base. Since
this cannot be achieved, the fin efficiency cannot
be at maximum efficiency.
In order to study the thermal performance of a fin,
we need a definition as fin effectiveness or fin ef-
ficiency.
Fin efficiency = fin
transferred across fin surface
heat transfer across equal maintained
at a constant temperature equal to that of base
Heat
Theoretical=
Design of in system
Before the actual design of in system, it is impor-
tant to recognise the various conditions for which
the finned surface has advantage over the unfinned
surface. The probable conditions are
1. space consideration, limiting certain dimensions
of fins,
2. weight considerations and
3. cost considerations.
The quantity of heat dissipated by a fin to its sur-
roundings goes on increasing as the length, but
there is a limit on the length of the fin, imposed by
the manufacturing difficulties, stability and so on.
The mathematical limit for the design is 0.dqdL
= It
may also be be reminded that for short fins, heat
flow becomes tow-dimensional and all our assump-
tions and expressions given earlier become use-
less.
The consideration of the weight of the fin is par-
ticularly important in the case of automobile and
aircraft practice. In all such cases the requirements
is the maximum heat exchange with minimum
weight of the heat exchange system. Triangular or
parabolic shaped fins may be used depending upon
the specific conditions imposed. Any design of a
fin system, should always be aimed at maximum
fin efficiency and minimum material requirement.
It is generally a compromise of several aspects in
the design.
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Heat exchangers are generally classified accord-
ing to the direction of flow of the hot and cold flu-
ids with respect to each other, or according to the
temperature distribution of the two fluids along the
exchanger length. Thus, we may have the follow-
ing types of heat exchangers:
(a) Parallel-flow exchanger
(b) Counter-flow exchanger
(c) Cross-flow exchanger
(d) Condenser or evaporator.
(a) Parallel-flow Heat Exchanger
In this type of heat exchanger the hot and cold flu-
ids flow in the same directions, hence the name
parallel-flow. Such an exchanger is shown sche-
matically in figure (a) and (b). In a parallel flowexchanger, the temperature difference between the
hot and cold fluids keeps as decreasing from inlet
to exit. Many devices, such as water heaters and
oil coolers, etc. belong to this class.
Fig 6
(b) Counter-Flow Heat Exchanger
In this case, as schematically shown in figure (a)
and (b), the two fluids flow through the exchange
in opposite directions hence the name, counter-
flow. A common type of counter flow exchanger is
shown in figure (d).The temperature distributionin a counter-flow exchanger is shown in figure (c).
Fig 7
(c) Cross-Flow Exchanger
This exchanger is shown in figure. Here the two
fluids flow at right angles to each other. Two dif-
ferent arrangements of this exchanger are com-
monly used. In one case, each of the fluids is un-
mixed as it flows through the exchanger. As a re-
sult, the temperature of the fluids leaving the ex-
changer are not uniform. An automobile radiator
is an example of this type of exchanger.
In the other case, one fluid is perfectly mixed while
the other is unmixed as it flows through the ex-
changer. This type is shown in figure The air flow-
ing over the tube bank is mixed while the hot gases
flowing inside the separate tubes are not mixed.
Fig 8
(d) Condenser of Evaporator
In a condenser, fluid remains at constant tempera-
ture throughout the exchanger while the tempera-
ture of the colder fluid gradually increases from
inlet to exit, Similarly, in an evaporator, the boil-
ing fluid (cold fluid) remains at constant tempera-
ture while the hot fluid temperature gradually de-
Heat Exchangers
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creases. The temperature distribution in these ex-
changes are shown in figure (a) and (b). Since the
temperature of one of the fluid remains constants,
it is immaterial whether the two fluids flow in the
same direction or in opposite directions.
Fig 9
Temperature distribution in Condensing ofEvaporating Heat Exchanges
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An important parameter in the design of heat ex-
changers is the overall heat transfer coefficient U,
between the two fluids. A general expression for U
can be easily obtained as follows: Consider a
double pipe heat exchanger in which one fluidflows through the inner pipe and the other fluid
through the annular space between the two pipes.
Let L = Length of heat exchanger, m
ri= Inside radius of inner pipe, m
or
= Outside radius of inner pipe, m
iA
= Inside surface area of inner pipe
2(2 ),ir L m
oA = Outside surface area of inner pipe
2(2 ),or L m
ih = Film coefficient of heat transfer at inside sur-
face of inner pipe, W/m2
k.
0h
= Film coefficient of heat transfer at outside
surface of inner pipe,
2/W m k
.
wk
= Thermal conductivity of inner pipe wall, W/
mK
it
= Temperature of fluid flowing through the in-
ner pipe, 0C. ot
= Temperature of fluid flowing through the an-
nular space between the two pipes
o C
.
iR
= Thermal resistance of fluid film at the inside
surface of inner pipe, K/W.
oR
= Thermal resistance of fluid film at the out-
side surface of inner pipe , K/W
wR
= Thermal resistance of inner pipe wall, K/W.
The rate of heat transfer between the two fluids is
given by
i ot tqR
=
where
i o wR R R R= + +
since
1i
i i
RA h
=
( / )
2
n o iw
w
l r rR
Lk=
1o
o o
RA h
=
Hence1 1
2
i o
o
i i
i i w o o
t tqr
nr A
A h Lk A h
=
+ +
.....................3
If Uiand U
odenotes respectively the overall heat
transfer coefficient basedon unit area ofthe inside
and outside surfaces of the inner pipe, then
( ) ( )i i i o o o i oq A U t t A U t t = =
From Equations (1) and (2)
1
ln
12
i
oi
i
ii w
o o
Ur
Ar
Ah LkA h
=
= +
and 1
1
1
( / )1 1
2
oo o n o
i w o
UA A l r r
A h Lk h
=+ + ................4
Since 2i iA r L= and
2o o
A r L=
Equations (3) and(r) can also be written as :
1............5
1 1i
i o in
i w i o o
Ur r r
lh k r r h
=
+ + +
and
1............6
1 1i
o o o
i i w i o
Ur r r
lnr h k r h
=
+ +
In most heat exchangers some scale formation will
take place on one or on both sides of the heat trans-
fer surface after the heat exchanger has been in use
for some time. This introduces two additional re-
sistance in the heat flow path. Thus the total ther-
mal resistance becomes.
i si w so oR R R R R R= + + + +
OVERALL HEAT TRANSFER COEFFICIENT
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where
siR
= Thermal resistance due to scale for-
mation on inside surface of inner pipe,
2 /M K Watt
.
soR
= Thermal resistance due to scale formation
on outside surface of inner pipe
2 /m K watt
.
Since it is difficult to ascertain accurtain accurately
the thickness and thermal conductivity of the scaleformed, the effect of the scale deposit on heat flow
is generally taken into account by specifying an
equivalent scale heat transfer coefficient hs.
The reciprocal of the scale heat transfer coefficient
is called the fouling factor. If hsiand n
sodenote the
heat transfer coefficients for the scale formed on
the inside and outside surface, respectively, of the
inner pipe, then,
1....................7
1...........8
si
i si
so
o so
RA h
RA h
=
=
and
( )
1 1 1 1
2
i o
on
i
i i i si w o so o o
t tq
rl
r
Ah Ah Lk A h A h
=
+ + + +
( )
11 1 1
2
i o
oi n
i i i
i si w o so o o
t tq
rAl
r A A
h h Lk A h A h
=
+ + + +
or
1..............11
1 1 1 1ln
i
i i i i
i si o o o so o o
U
r r r r h h r r r h r h
=
+ + +
1
ln1 1 1 1
2
o
o
o
io o
i i i si w so o
Ur
ArA A
A h A h Lk h h
=
+ + + +
or
1
1 1 1 1o
o o o o
i i i si w i so o
Ur r r r
inr h r h k r h h
=
+ + + +
The fouling factors
1
sh
for some repensentative
applications are listed in table 1 given below:
Table 1
Fouling Factors
________________________________________
Fluid Fouling Factor
21 /s
m K Wh
_______________________________________
Distilled water 0.000086
Sea water 0.000172Well water 0.000344
Treated boiler feed water 0.000172
Liquid gasoline 0.000086
Refrigerant liquids, brine 0.000172
or oil-bearing steam
Refrigerant vapours 0.000344
Fuel oil and crude oil 0.00086
Diesel exhaust gas 0.000172
_________________________________________
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(a) Forced convection over flat plates and through
tubesCase (1): Pipe and tube flow. (Dittus-Boelter rela-
tion)
Nu = 0.023 Re0.8Prn
where n = 0.4 for heating
= 0.3 for cooling
Case (2) : Flow across cylinders
0.52 0.3(0.35 0.56Re ) Pr Nu +
in the range 1 510 Re 10 <
(iii)
0.8274.82 0.0185Nu Pe= +
for the range3 5 2 4
3.6 10 Re 9.05 10 10 10and Pe < < < 109
0.330.1( Pr)Nu Gr=
Case (3): Horizontal cylinders2 0.250.53( Pr )Nu Gr=
Case (4): Inclined surfaces at
0.250.56( Pr cos )Nu Gr =
where 0 5 1188 10 Prcos 10and Gr <