Contents
Introduction to Heat
Transfer
Modes of Heat
Transfer
1. Conduction
2. Convection &
3. Radiations.
Heat Conduction
Laws of Heat
Transfer
Teaching / Examination Schemes
Teaching Scheme
: 4 Lectures/Week
: 2 Practical /Week
Examination Scheme : Practical :50 Marks
Insem:30 Marks
: Theory :70 Marks
Books Recommended:
1. Ozisik
8. M M Rathore
2. Holman
9. Nag
3. Sukhatme
10. Sachdeva
4. Rajput
11. Dutta
5. De Witt
12. Thirumaleshwar
6. Cengel
7. Domkundwar
Heat Energy & Heat Transfer
• In this lecture, we are going to study about Heat Energy & HT
• So, let us see first, What is Energy ? Energy is the capacity to do
work. Some examples of energies are mechanical energy, chemical
energy, electrical energy, nuclear energy and so on
• And,What is Heat Energy? It is a form of energy in transit, the
driving force for which, is the temp difference
•It means, whenever there is temp difference between the two
bodies, heat energy will flow. When they attain equilibrium of temp ,
heat flow will stop
• In the study of Heat Transfer:We are concerned, as to H O W and
AT WHAT RATE heat transfer takes place. HT is based on the Law of
Thermodynamics, which states that heat flows from a body at higher
temp to a body at lower temp. Conversely, heat can not flow from a
body at lower temp to a body at higher temp, unless & until some
external device, like heat pump, is employed
Why should we study Heat
Transfer ?
• You will not find a single field of engineering ,where
knowledge of Heat Transfer is not applied.
• Knowledge of HT is always required while designing an
eqpt. Let us take up some examples, branch-wise
Mech Engg: Engines, R &A/C, use of insulations,
cooling/heating of bodies for heat treatment etc
Elect Engg: Cooling of motors, transformers, current carrying
wires, etc
Chemical Engg:We have to take care of energy produced
during chemical reactions, heating or cooling of chemicals for
reactions to take place, etc
Nuclear Engg: Conversion of fission/fusion energy to
electricity, cooling of nuclear reactors etc
Importance of Heat Transfer
Electronics: Cooling of ICs, electronic devicesetc
Computer Engg: Cooling of chips, electronic cct, etc
Civil Engg: Curing of cement in buildings/dams etc
Hydraulics: Generation of electricity from hydraulic energy
from dams
Bio-technology: Ripening of fruits, processing ofbiomaterial
Space Engg: Space applications/ cooling of space vehicles
Aircraft Engg:Aircraft applications
You name an equipment and you find that the knowledge
of heat transfer is required
PRINCIPLE O N E
Heat ALWAYS flows
from hot to cold
when objects are in
contact or connected
by a good heat
conductor.
The rate of heat
transfer will increase
as the difference in
temp between the
two objects increases
pg.. 6 fig 2
PRINCIPLE T W O
Cold objects have
less internal heat
than hot objects of
the same mass
To make an object
colder, remove
heat;To make is
hotter, add heat
The mass of the
object remains the
same regardless of
the heat content
PRINCIPLE THREE
Everything is composed of matter
All matter exists in one of three states: solid, liquid or vapor.
LATENT HEAT OF VAPORIZATION:When matter changes
from liquid to vapor or vice versa, it absorbs or releases a
relatively large amount of heat without a change in
temperature.
PRINCIPLE
FOUR
CONDENSATION When a
vapor is cooled below its
dew point, it becomes a
liquid. (boiling point in
reverse)
When vapor condenses,
releases large amount of
heat
PRINCIPLE FIVE
Changing the
pressure on a liquid
or a vapor changes
the boiling point.
PRINCIPLE S I X
When a vapor is
compressed, its
temperature and
pressure will increase
even though heat has
not been added
Heat Transfer
Heat always flows from
high temperature objects
to low temperature
objects.
Heat flow stops when
temperatures equal.
Various ways by which
heat may flow.
30Cº
Heat flows from
child into air
37Cº
-2Cº
Heat flows from
child and air into
the ice cream
Modes of Heat Transfer
 Conduction: Flow of heat energy by direct contact
& through free electrons e.g. heat flow through solids
 Convection: Transfer of heat energy by fluid flowing
over a surface e.g. heat transfer from engine surface to
surrounding atmospheric air
 Radiation: Flow of heat energy without any
intervening medium e.g. energy of sun reaching the
earth.
NMIET-HT-Prof. R.R.Jadhao
CONDUCTION



Heat is transferred through
a solid and gets the solid
hot. (molecules get hotter,
then they in-turn give
energy to nearby molecules
and they get hotter too)
Different solids conduct
different amounts of heat in
a specific time. (copper vs.
glass)
Conduction is the process
whereby heat is transferred
directly through a material,
any bulk motion of the
material playing no role in
the transfer.



Those materials that
conduct heat well, are called
thermal conductors, while
those that conduct heat
poorly, are known as
thermal insulators.
Most metals are excellent
thermal conductors, while
wood, glass, and most
plastics are common
thermal insulators.
The free electrons in metals
are responsible for the
excellent thermal
conductivity of metals.
Conduction
Conduction is heat flow
by direct contact.
Some materials are go od
Tile floor feels colder than wood floor
thermal conductors,
others are insulators.
37º
30º
Wood is an
insulator
37º
30º
Tile is a
conductor
Fourier’s Law of Heat
Conduction
Rate of heat transfer by conduction (through a solid) in a
given direction is proportional to the area normal to the
direction of heat flow and the temp gradient in that direction.
Mathematically ;
T
QA
Watt
x
OR
dT
Q  kA Watt (J /s)
dx
where Q = heat flow rate,Watt(J/s)
A = area normal to heat flow direction,m2
k = conductivity of material (property),W/mK
dT/dx = temp gradient in x direction
ΔT = temp difference across Δx
Δx = thickness of material in heat flow direction
Conduction
Rate of heat transfer by conduction, Q through the length, L
across the cross-sectional area, A is given by the following
equation, where k is the thermal conductivity and ΔT is the
temperature difference between the two ends.
kAT
Q
L
SI Unit of Thermal Conductivity: J/(s · m ·C°)
Assumptions of Fourier’s Law
1. Unidirectional heat flow (only
one direction)
2. Steady state heat flow
3. Constant temp gradient
4. Constant conductivity,k
5. Both faces isothermal
Conduction
dT
Q  kA
dx
(T1  T 2 )
Q  kA
(x 1  x 2 )
(T1  T2 )
Q  kA
(x2  x1)
T
Q  kA
x
Heat Flux q=Q/A,W/m2
Temperature distribution in a solid
Inside of
house is
warm
heat flow
298 K
A
Temperature
distribution is
linear
Outside
is cold
A
273 K
Q  kA
Tout Tin
L
k = thermal conductivity
(Watts/ m K)
T = temperature(K)
Q/A = heat flux (Watts/m2)
L = distance (m)
Thermal Conductivity
Metals
Aluminum
240
Brass
110
Copper
390
Iron
79
Lead
35
Silver
420
Steel (stainless)
14
Gases
Air
0.0256
Hydrogen (H2)
0.180
Nitrogen (N2)
0.0258
High conductivity
High conductivity
High conductivity
Thermal Conductivities
•Metals have high
thermal
conductivity,most
electrical
insulators also
have low thermal
conductivity.
•Air is a great
insulator, except
that large air
spaces allow heat
flow by
convection.
Substance
Thermal
Conductivity: k
W / (m K)
Glass
0.84
Water
0.60
Wood
0.10
Air
0.023
Variation ofThermal Conductivity
1. It is the property of material; defined as ability of material to
conduct heat through it.
2. Thermal conductivity in decreasing order :
Metals » Non-metallic Solids » Liquids » Gases
3. Higher conductivity in metals due to free electrons in their
outer orbits
4. k depends on grain structure.When k is different in different
directions (kx ,ky ,kz ), material is known as anisotropic.
When k is constant in all directions, it is called Isotropic.
5. k is strongly dependent on temp; k=ko(1+αT)
Isotropic & Anisotropic Materials
•
Some materials exhibit same conductivity in all
directions.These are called ISOTROPIC materials
(kx = ky = kz = k)
•
While some materials have different conductivity in different
directions(kx ,ky ,kz ), such materials are known as
anisotropic.
•
Wood exhibits directional conductivity; different
along grains
HEAT C O N V E C T I O N
Convection is the process in which heat is carried
from place to place by the bulk movement of a fluid.
Convection currents are set up when a pan of water is
heated.
Heat Convection
When a fluid flows over a solid body or surface and temp of
the fluid and solid surface are different, heat transfer between
the solid surface and fluid takes place due to motion of fluid
relative to the surface.
If the fluid motion is artificially induced, then heat transfer is
said to be by FORCED convection.
If the fluid motion is set up by buoyancy effects resulting from
density difference caused due to temp difference in the fluid,
heat transfer is said to be by FREE or NATURAL convection
Newton’s Law of Cooling
Rate of heat transfer by convection from a surface to a fluid
or vice versa ,flowing along it is equal to the product of temp
difference between surface and the free stream of the fluid, the
area of the surface normal to the direction of heat flow and a
quantity h called convective heat transfer coefficient.
Mathematically;
Q = hA(Ts -T∞);Watt
h is not a property of fluid or
surface, but it depends on
properties of the fluid and vital
dimensions of the surface Fluid
Ts > T ∞
T∞
A
Q = h A (Ts - T∞); Watt
h
Ts
Q
Convection
y
Tair
Cool air flow
Q
Tplate
T
Hot Plate
h = convection coefficient Watts/m2 K
Q  hA(Tplate Tair )
Convection
Heat transfer in a fluid often
occurs mostly by convection.
Buoyancy causes warm air to
rise, which carries thermal
energy directly by its motion.
Convection Oven
Convection oven has a fan to enhance the circulation of
the air, increasing the transfer of heat.
Heat Radiation
•Energy carried by electromagnetic waves
•Light, microwaves, radio waves,x-rays
•Wavelength is related to vibrational
frequency
Radiation
Light has many different
wavelengths, most of which
are not visible to the eye.
All light carries energy, thus
transfers heat.
Heat Lamp
Heat Radiation
All bodies continuously emit energy if their temp is above
zero absolute (0K) and energy thus emitted is called thermal
radiation.
Thermal radiations are electromagnetic waves and do not
require any medium for propagation.
 Thermal radiation is a surface phenomenon.
 Theories of Thermal Radiation
1. Wave/Maxwell’s Classical Theory :Propagation by
electromagnetic waves
2. Quantum/ Planck’s Theory: Propagation by quanta
possessing certain amount of energy
Stefan Boltzmann’s Law of
Radiation
Thermal radiation emitted by a black body is proportional to
the Fourth Power of its absolute temp.
Mathematically;
q ∞ T4 W/m2;
Q = σAT4 W; where σ is StefanBoltzmann’s
constant (5.67 x 10-8 W/m2K4 )
Q =A1Є1σ (T14 –T 24)
The Stefan– Boltzmann’s Law
Of Radiation
The rate at which an object emits radiant energy is
proportional to the fourth power of its absolute
temperature.This is known as Stefan’s law and is expressed
as follows:

Q   AT 4
where σ is the Stefan-Boltzmann constant,
σ = 5.67  10-8 W/m2 K4.
The factor Єis called the emissivity, which is a number
between 0 and 1.
Perfect radiators have a value of 1 for Є.
A is the surface area and T is the temperature of the radiator
in Kelvin.
Emission of Radiant Energy
All objects radiate light;
higher the temperature,
75º
higher the frequency.
At room temperature the
radiated light is at
98º
frequencies too low for
our eyes to see.
Special cameras are
Attics in this house were kept warm for
sensitive to this infrared
growing marijuana.
radiation.
Reflection of Radiant Energy
White objects reflect light, black objects don’t.
Hole in a box with white interior
looks black because almost none
of the light entering the hole
reflects back out.
White tubes look black
inside.
Numerical Problem
Q1: Air at 20°C blows over a 50cm x 75cm hot plate at
250°C.The film heat transfer coefficient is 25 W/m2K. 300W
is lost from the plate surface by radiation. Calculate heat
transfer rate and other side plate temp.Thermal conductivity
of the plate material is 43 W/mK.The plate is 2cm thick.
Q=?
Q r =300W
Qc =?
Air at 20°C
Q = Q c+ Q r
Qc=hA(T1 – Ta)
Q=kA(ΔT/Δx)
=kA(T2-T1)/(Δx)
2cm
h=25
T1 =250°C
T2 = ?
Q=2456W; T 2=253°C
k=43
Q
50cm
Electrical Analogy
What flows?
Driving
Potenti
al
Flow
Resistance to
flow
Electrical
Energy
Electrons
Voltage Diff, ΔV
Current, I
ρ, A, L of
conducto
r
Heat Energy
Heat energy through
electrons
Temp Diff, ΔT
Heat Transfer Rate,Q
R,Thermal Resistance
Electrical Analogy
As per Ohm’s Law, I = ΔV/R
Similarly, Heat Flow Rate, Q = ΔT/R = C.ΔT;
where R is thermal resistance & 1/R=C conductance
Conductive Resistance:
T
T
T
Q  kA


x
x
R
kA
x
Hence, R co n d u ctive 
kA
Electrical Analogy
Convective Resistance:
T T
Q  hAT2  T  

1
R
hA
Hence, R convective
1

hA
HeatTransfer In Composite
Structures
Resistance In Series
Q=ΔT/R
=(T1 -T2)/R1
=(T2 -T3)/R2
=(T3 -T4)/R3
Q
T1
k1
T2
R1
b1
k2
R2
b2
T3
On adding up;
T1 -T 4 =Q/ (R1+R2+R3) or Q=(T1-T4)/(R1+R2+R3)
Q=ΔT/R;
hence R=R1+R2+R3
R1=b1/k1A; R2=b2/k2A; R3=b3/k3A
k3
R3
b3
T4
HeatTransfer In Composite
Structures
T
T
1
Resistance In Parallel
Q1
Q 1 = (T1 - T2)/R1
Q 2 = (T1 - T2)/R2
Q 3 = (T1 -T2)/R3
2
Q1
k1
Q
Q2
R2
Q2 Q
k2
On adding;
Q = Q 1+ Q 2+ Q 3
=(T1-T2)*(1/R1+1/R2+1/R3)
= ΔT*1/R;
Hence 1/R=1/R1+1/R2+1/R3
R1
Q3
R3
k3
b
Q3
Examples of Composite Structures
• Walls of buildings
• Walls of home refrigerators
• Insulated pipe carrying steam
• Walls of a furnace
• Walls of a cold storage
• Hot case for food
Unsteady State Heat Transfer
• Whenever a heat transfer system is switched on/ started,
it takes some time to attain steady value of heat transfer rate.
Heat transfer rate under these conditions keeps varying with
passage of time.This heat transfer system is said to be
transferring heat under unsteady state /transient conditions.
Here, temperature also keeps varying at various locations
in the system with time. Hence, temp is a function of both
location as well as time.
• Similar situation occurs when a heat transfer system is
switched off /shut off, but in reverse direction
• Examples are starting/firing of a furnace, heating of a body,
switching on a heater, starting of an engine,etc
Steady State Heat Transfer
• Whenever a heat transfer system is switched on/ started,
it takes some time to stabilize the heat transfer rate when it
becomes constant and does not change with time.This heat
transfer system is said to be transferring heat under steady
state conditions. Here, temperatures attain constant values
at various locations in the system and do not vary with time.
Hence, temp is afunction of only location and not of time.
• Heat transfer rate is directly proportional to temp difference.
Since temps attain constant values, temp difference also
become constant hence heat transfer rate attains steady value.
•
•
This implies that whatever amount of heat energy is being
received by the system, at same rate it is transferring out.
This means that under steady state, system transfers /
receives constant amount of heat energy per unit time
Q2. In a furnace, temp of hot gases is 2100°C.
Ambient temp is 40°C. Heat
flow by radiation from hot gases
Tg T1
T2 T ∞
to inner surface of the wall is
23kW/m2. Convective heat
qc2
q
cond
transfer coeff. between hot gases
qc1
qr2
and the inner surface of the wall is qr1
12W/m2K.Thermal conductance of
h1 C
h2
the wall is 58W/m K. Heat flow by
radiation from external surface of the wall to
surroundings is 10kW/m2.Temp of inside surface of
the wall is 900°C. For the external surface of the wall,
find surface temp and convective heat transfer
coefficient.
(Ans.T2=255.2°C; h2=127.3W/m2K
T2 ? h2= ?
Solution:
Q  h A(T  T )  12x1(2100  900)  14.4kW / m2
C1
1
g
1
Q  Q C 1  Q r 1  14.4  23  37.4 k W / m 2
T h i s is the h e a t c o n d u c t e d t h ro u g h slab.
T 1  T 2 
H e n c e Q  C (T  T ) 
1
2
1
C
(900  T2 )

 T2  255.2C
1
58
Q  QC
2
 Qr
2
 37400  QC
2
 10000
Hence Q C 2  27400  h2 A(T2  T )
 h  127.3W / m 2K
2
General Heat Conduction
Equation In Cartesiaz n
Coordinates
• Consider a smallrectangular
δz
g
volume of sides δx, δy & δz
dQx
dQ x+δx
T
parallel to the three axes
in medium, in which temp
is varying with loc & time.
δy x
δx
• Let T denote the temp at centrye
of this elemental volume.
•Also, let there be internal heat generation at the rate
of g Watt per unit volume (W/m3 ) due to heat source
•Let the material be anisotropic implying that thermal
conductivities have values kx ,ky & kz in x, y & z directions
respectively
General Heat Conduction
Equation
•Consider heat entering and leaving this volume through its six
faces.
•Let heat entering the elemental volume per unit time normal
to the area/face δyδz at x be dQx and heat leaving the volume
in the direction normal to the
area δyδz at x+δx be dQx+δx.
• As per Fourier’s Law, heat entering
dQ x = - kx(δyδz)∂T/∂x
• Similarly,heat leaving, dQx+δx =dQ x + ∂/∂x(dQx)δx
General Heat Conduction
Equation
So, net heat flow into the element in x-direction/time;
dQ x  dQ x  d x


dQ x x
x
 
T 
    k x yz.
x
x 
x 
  T 

kx
xyz
x 
x 
General Heat Conduction Equation
Similarly, net heat flow into the element per unit time
in y & z directions respectively are;
dQ y  dQ y  y
  T 

ky
 xyz
y 
y 
dQ z  dQ z  z
  T 

kz
 x  y  z
z 
z 
and
Thus, net heat flow in to the element from all directions
by conduction in certain time δt will be:
   T    T    T 
  kz
xyzt
 kx
  ky
 x  x  y  y  z  z 
General Heat Conduction Equation
Now, internal heat generation in time δt=g.δxδyδzδt
Heat gain by the element from above, will result in
energy storage and will increase its temp.
Let δT be the rise in temp in time δt, the net heat
storage in the element in time δt ;
(mCp ΔT) = ρVCp δT
= ρCp δT δxδyδz
General Heat Conduction Equation
Energy Balance Equation:
Net heat conducted in to the element from all
Directions +Heat generated within the element
= Energy stored in the element
   T    T    T 
  kz
xyzt
  k x x   y  k y
z 
x
y
z







 g.xyzt  CpT.xyz
Dividing the Equation by δxδyδzδt, we get;
General Heat Conduction Equation
  T    T    T 

T
kx   ky


   kz   g  Cp
z
x
x
y
y
z
t
This is three dimensional heat conduction equation
in Cartesian Coordinates for anisotropic material for
Unsteady state conditions.
For isotropic material, kx=ky=kz=k constant
 2T  2T  2T
g
1 T


 
2
2
k  t
x 2
y
z
k
m2 / s
W h e re  is thermal diffusivity 
Cp
General Heat Conduction Equation
Fourier’s Equation:
 2T 2T  2T 1 T



x 2 y 2 z 2  t
Poisson’s Equation:
2T 2T 2T g
    0
x2 y2 z2 k
Laplace Equation:
2T 2T 2T
 
 0
x2 y2 z2
Steady State, One Dimensional Equation w/o g:
d 2T
o
dx 2
Types of Problems In Heat Transfer
1.Plate/Slab/Wall
2.Tube/Pipe/Cylinder
3.Sphere
• To increase Heat Transfer Rate
• To decrease Heat Transfer Rate
General Heat Conduction Equation
In Cylindrical Coordinates
By substituting x=r.cosθ; y=r.sinθ and z=z, we get
General Heat Conduction Equation in Polar/
Cylindrical Coordinates:
 2T 1 T
1  2 T  2 T g 1 T

 2

 
r 2 r r r  2 z 2 k  t
F o r isotropic material with k  constt
Poisson’s Equation:
1 d  dT  g
 r   0
r dr  dr  k
Radial heat conduction w/o g:
1 d  dT 
r

0
r dr dr


General Heat Conduction Equation
In Spherical Coordinates
Similarly, by substituting x=r.sinθ.cosФ; y= r.sinθsinФ
and z=r.sinθ, we get heat conduction equation in
Spherical Coordinates:
1   2 T 
1 
T 
1 2T  g  1 T
 2
sin   2 2
r
2
  r sin  2 k  t
r r  r  r sin  
For isotropic material with k  constt
1 d  2 dT  g
Poisson’s Equation:
r
  0
2
r dr  dr  k
1 d  2 dT 
r

0
Radial heat conduction w/o g:


r 2 dr 
dr 
Thermal Diffusivity
•Thermal Diffusivity is the ratio of thermal conductivity to heat
storage capacity of the material.
k
Denoted by ,it is defined as :  
Cp
m2 /s
• Larger the value of α, faster shall be the heat diffusion
through the material.
Steady state heat conduction does not contain α,
hence temp distribution through material is determined by
k only, where as in unsteady state heat conduction,
temp distribution is determined by α. (Both by k & ρCp )
Example: Cooking steel utensils having copper bottom
One Dimensional Steady State
Heat Conduction through
Slab/Plane Wall
K
Consider a plane wall of thickness Δx of
material having conductivity k with its
faces maintained at temp T1 & T2
Steady state, one dimensional
Heat conduction eqn will be:
2
d T
0
2
dx
T1
T
T2
Δx
X=0
X = Δx
Integrating this equation twice;
dT
 C1..........(1) Slope of Temp Profile
dx
and T  C1x  C2 .........(2) Temp Profile
We have
Heat Conduction through
Slab/Plane Wall
K
Boundary Conditions:
T1
(T=C1.x+C2)….(2)
1) At x=0; T=T1
2) At x=Δx; T=T2
T2
Applying BC 1), we getT1=C1.0+C2
Δx
Hence C 2 =T1
X = Δx
Applying BC 2), weget
T2 T1 X=0
 C1 
T2=C1.Δx+C2
x
Or T2=C1.Δx+T1
Substituti ng C1 and C2 in Eqn..(2)
We get
T2  T1
T
.x  T 1...........Temp Distribution
x
Heat Conduction through
Slab/Plane Wall
dT
Heat Flow Rate Q  kA
dx
T2 T1
dT
From Eqn..(1);
 C1 
dx
x

T1  T 2 
HenceQ  k A
x
Hence Rcond
T1
T2
X=0
T
T


x
R
kA
x
 ..... for Slab
kA
K
Δx
X=Δx
One Dimensional (Radial)
Steady State Heat
Conduction through Hollow
Cylinder
Consider a hollow cylinder ofinner
radius r1 and outer r2 of length L
of a material having conductivity k.
r1 K
r2
L
T1
Q
T2
T1>T2
Inner surface of cylinder is at
tempT1 and outer atT2
Conduction Equation for one dimensional (radial)
Heat flow (without g) will be:
1 d  dT 
r dr r dr  0


One Dimensional Steady State Heat
Conduction through Hollow Cylinder
Integrating Equation: 1 d  dT 


r dr r dr  0


d  dT 
We have   r
0
dr  dr 
dT
C1 .......(1)
dT  
r
 C 1 or
r
dr
dr
On further Integration;
We have T  C1 ln r  C 2 .......(2)
L
r2
r1 K
T1
Q
T2
T1>T2
Heat Conduction through
Hollow Cylinder
L
Boundary Conditions:
Eqn (2) T=C1.lnr+C2
1) At r=r1;T=T1
2) At r=r2;T=T2
Substituting in Eqn ….(2);We have
T1=C1.lnr1+C2 …..(3)
T2=C1.lnr2+C2 ….(4)
r2
r1 K
T1
Q
T2
Subtracting eqn (4) from (3) and further substitution;
T2  T1
C1 
r2
ln
r1
T2  T1
.ln r1
and C 2  T1 
r2
ln
r1
T1>T2
Heat Conduction through Hollow
Cylinder
T=C1.lnr+C2 …….(2)
T2 T1
T2 T1
.lnr1
C1 
and C2T1
r2
r2
ln
ln
r1
r1
Substituting values of C 1 & C 2
in Eqn ….(2);We have
Q
L
r2
r1 K
T1
T2
T1>T2
r
r
ln
ln
r1
r1
T  T1
T 2  T1   T1 O R
T 

r2
r2
T

T
2
1
ln
ln
r1
r1
Heat Conduction through
Hollow Cylinder
L
Heat Flow Rate:
C1
dT
Q  kA
dr
dT C1
 .... from Eqn....(1)
dr r

T 2  T1
r
ln 2
r1
r2
Q
C1
Therefore, Q  k.2rL.
 2kLC1
r
Substituting C1 ;


T2  T1 
T1  T2 
Q  2kL.
 2kL.
ln
r2
r1
ln
r2
r1
r1 K
T1
T2
T1>T2
Heat Conduction through Hollow
Cylinder
L
Heat Flow Rate:
Q  2kL.
Q 
T 1
 T2
r2
ln
r1
2kL
T 2

 T1
r2
ln
r1
T

R
ln
H ence
RCond

r1 K
r2
T1
Q
r2
r1

2  kL
for Cylinder
T2
T1>T2
Heat Conduction through
Hollow Cylinder
L
In case of cylinder, in Q expression,
Q= -kA (dT/dr);
r1 K
r2
area transferring heat A=2 л r L
T1
changes with r, unlike in case of slab.
T1>T2
Therefore, it is convenient to work out Q
T2
Mean Area Am for use in analogous
formula
Logarithmic Mean Area (LMA)
for slab Q=k A (ΔT/Δx).
T
2kL.T
 k.Am .
r
r 2  r1
ln 2
r1
is m e a n a re a w h i c h c a n b e utilized
If w e w r i t e Q 
Then Am
in formula for slab
Heat Conduction through Hollow
Cylinder Logarithmic Mean Area (LMA)
Toobtain valueof LMAi, e. A m ;
We m ultiply & div ide Q exp ression by r2  r1 as;
2k L.T r2  r1  k.2L r2  r1 . T
Q
.

r2
r2

r2  r1 

r2  r1 
ln
ln
r1
r1
C o m p a r in g w ith Q  k . A
W e have
Am

m
.
2  L r 2  r1
r2
ln
r1
T
;
r 2  r1


Ao  Ai
Ao
ln
Ai
One Dimensional (Radial) Steady
State Heat Conduction
through Hollow Sphere
T >T
1
• Consider a hollow sphere of inner
radius r1 and outer r2 of a material
having conductivity k.
r2
•Inner surface of sphere is at tempT1
and outer atT2
•Conduction Equation for one dimensional
(radial)
Heat flow (without g) will be:
1 d  2 dT 
d  2 dT 
r
r

0


0




r 2 dr  dr 
dr  dr 
r1
K
T1
Q
T2
2
One Dimensional (Radial) Steady
State Heat Conduction
T >T
through Hollow Sphere
1
d  2 dT 
Integrating Eqn...  r
0
dr  dr 
dT C1
2 dT
We have r
 C1 or
 2 ...(1)
dr
dr r
On further Integration, wehave
C1
T 
 C 2 .............(2)
r
r2
r1
2
K
T1
Q
T2
One Dimensional (Radial) Steady
State Heat Conduction
T >T
through Hollow Sphere
1
Boundary Conditions:
r2
1) At r=r1 ;T=T1
2) At r=r2 ;T=T2
Substituting in Eqn
r1
K
T1
C1

T    C2 ...(2)
r
T 1
 T 2 .r1 r 2
r1  r 2
We hav e
C1 
And
C 2  T1 
.r 2
T 1
 T2
r1  r 2
2
Q
T2
One Dimensional (Radial)
Steady State Heat Conduction
through Hollow Sphere
C1
 C2
Substituti ng C1 & C2 in Eqn T  
r
r1 r2  r .T  r2 . r r1
.T2
T .
1
r r2  r1
r r2  r1
This is the Temp Profile across the thickness
of sphere
One Dimensional (Radial) Steady
State Heat Conduction through
Hollow Sphere
Heat Flow Rate Q   k A
dT
 k.4  r 2 .
dr
dT
dr
dT C1
2 C1
Substituting

 Q  k.4r . 2  4kC1
r
dr
r
S u b s t i t u t i n g
Q

4

Therefore
k . r2
C
1
;
T 1  T 2
r1 .
r2  r1
R co n d

T1  T 2
r2  r1
4  k . r 2 r1
r2  r1

and Am  4r1r2
4kr2 r1
Conductive Resistances
For Slab:
For Hollow Cylinder:
For Sphere:
 x
R 
k A
r2
ln
r1
R 
2kL
r2  r1
R
4kr2 r1
Overall Heat Transfer Coefficient
Heat Flow Rate can also be given as Q=UAΔT;
where U is called as overall heat transfer coefficient
For plane wall:
T
T
Q  UAT 

1
1
x
1


UA
h1 A kA h2 A
h en ce
Therefore,
1
1
x
1



UA
h1 A
kA
h2 A
1
1 x 1



U h1
k
h2
where U is Overall Heat Transfer Coeff
Overall Heat Transfer Coefficient
For Cylinder:
Uo
Ao
Q  U i Ai Ti  To   U o Ao Ti To 
r2
T


 
1 1
Ui Ai U oAo
T
r
r
ln 2
ln 3
r
r
1 1 
1
  2
hi Ai 2k1L 2k2 L ho Ao
ln r2 r ln r3 r
1
1
1
1 2 


Ui Ai Uo Ao hi Ai 2k1L 2k2 L
A i r1
T
U i i hi
k1
k2
r3
ho
To
1
ho Ao
where U i is Overall heat transfer coeff based on inner surface area Ai
and U o is Overall heat transfer coeff based on outer surface area Ao
Ai  2r1L and Ao  2r3L
Overall Heat Transfer Coefficient
For Sphere:
QUi Ai Ti To Uo Ao Ti To 

T
T



r r
1
1
r r
1
1
 2 1  3 2 
Ui Ai U o Ao hi Ai 4k1r2 r1 4k2 r3r2 ho Ao
r3  r2
r2  r1
1
1
1
1





Ui Ai U o Ao hi Ai 4k1r2 r1 4k2r3 r2 ho Ao
where U i is Overall heat transfer coeff based on inner surface area Ai
and U o is Overall heat transfer coeff based on outer surface area Ao
Ai  4r1
2
and
Ao  4r3
2
Q3. A steel tube with 8cm OD,
6cm ID and k=15W/mK,
is covered with an insulation
covering of thickness 2cm
and k=0.2W/mK.A hot gas
at 300°C with hg=400W/m2K
flows inside the tube.The
outer surface of insulation is
exposed to cool air at 30°C
with ha=50W/m2K. Calculate
over all heat transfer coeff.
U o based on outer surface of
insulation and heat loss from
the tube for its 25m length.
T∞
r2
r3
r1
Tg hg
ks
ha
Uo=?
6.76W/m2K
Q=?
19.19kW
ki
Solution:
Q=U o A o ΔT=U i A i
ΔT
r3
r2
ln
ln
1
1 2 r 1 r
1




Uo Ao ho Ao 2KiL 2ks L hi Ai
A  2r L  9.42m ; A  2r L
2
o
3
i
1
U A  63.67 U  6.759W / m2 K
o
o
o
Q  UoAoT
 63.67(300  30)
 17.19 k W
T∞
r2
r1
r3
Tg
ks
ho
hi
ki
Q4. An insulating powder is
Insulating Powder
densely packed in the annular
space between two concentric
r2
spheres with radii 75mm and
Heater
50mm.The inner sphere is
r1
Ti
uniformly heated with electric
To
power input of 30 W. Steady state
K=?
temp attained by the inner sphere
is 120°C and that by outer surface is
30°C.
Neglecting the thermal resistance offered
k=?=0.177W/mK
by
the spheres:
Analogous cct ?
a) Draw analogous electrical cct diagram
b) Calculate thermal conductivity of the
powder
Solution:
Insulating Powder
Analogous cct ?
Q
Ti
Q
r2
To
Heater
Rins
k=?
Q
Ti  T o
r2  r1
To
 30 W
Ti
r1
K=?
4 kins r1r2
120 30
 30  k ins  0.1768W / mK
0.075 0.05
4 kx0.075x0.05
Q5. The insulation boards for
air-conditioning purposes are
made of three layers, the middle T =45° C
1
being of packed grass 10cm thick
kp
(kg=0.02 W/mK) and the sides
are made of plywood ,each of
2cm
2cm thickness (kp=0.12 W/mK).
They are glued with each other.
Qa=?
T2=20°C
kg
10cm
kp
2cm
Qb=?
a)Determine heat flow rate if one surface is at 45°C
and the other at 20°C. Neglect resistance of the glue.
b)Instead of glue, if these three boards are bolted by
4 steel bolts (ks=40 W/mK) of 1cm dia each at the
corners per m2 area of the board, then find whether
heat flow rate is going to increase or decrease and
by what percent ?
Solution:
R1
R2
R3
T1
Q
T2
T1 T2
45 20

 4.69W/m 2
 p  g  p
0.02 0.1
0.02




k p .A k g .A k p .A 0.12x1 0.02x1 0.12x1
With Bolts:
T1
R1
R2
R3
T2
R4
Ab 

45  20
 6.9W /m2
3.61
6.9 4.69
x100
Increase 
4.69
 47.7 00
x0.012 x4  3.14x104 m2 ;Qnew 
4
2


210
2

10
R4 

k A 40x3.14x104
s
Rt 3.61
Q6: A wall 30cm thick of size 5mX3m made of redbrick
(kb=0.35W/mK). It is covered on both sides by layers
of plaster 2cm thick (kp=0.6W/mK). The wall has a
window of size 1mX2m.The door of window made up
of 12mm thick glass has a conductivity kg=1.2 W/mK.
Inner and outer
air temp are
10 & 40°C.
Take h on both
sides as 15 W/m2K.
Estimate the rate
of heat flow
5m
through the
Ti=10°C
wall.
2cm
2cm
30cm
h=15
Q
To=40°C
2m
Kb=0.35
Solution:
Q
10°C
Rci
To  Ti 40 10

R
R
Rp1
Rb
Rco
Rp2
40°C
Rw
2cm
2cm
30cm
h=15
5m
Ti=10°C
2m
Rp1
Rb
Kb=0.35
h=15
Q
Rp2
Rco
To=40°C
Solution:
1
1
Rci 

 4.44x103  Rco
hA 15x5x3
X
0.02

R p1 
 2.56x103  R p2
k p A 0.6x13
0.012
0.3
3
Rw 
 5x103
Rb 
 65.9x10
1.2x2x1
0.35x13
2cm
2cm
30cm
h=15
Rt=13.54x10-3
Ti=10°C
5m
2m
Rp1
Rb
Kb=0.35
h=15
40 10
Q
 2215.66W
3
13.54x10
Ans.
Q
Rp2
Rco
To=40°C
Insulating Materials
• Materials which are used to reduce the heat transfer rate
from /to the system, are known as INSULATOTRS
• Examples are glass wool, plastics, wood, brick,cement,
asbestos, rubber, grass, saw dust, cork, glass, clay,etc
• Insulators have low conductivity (generally k <2 W/mK)
• Insulating materials should be cheaper, able to withstand
higher temp and humidity, should remain in applied shape
and have long life, odorless, non-reactive,
• Practical applications are in refrigerators & air conditioners,
buildings, conduits carrying high temp fluids like steam/
chemicals, plastic handles of kitchen utensils, furnaces, cold
storages, offices etc
Conductivities of some Insulating
Materials
Materials
Conductivity k (W/mK)
Wood
1.2 – 0.8
Brick
0.9 – 1.3
Concrete
0.8 – 0.9
Glass
0.7 – 0.8
Asbestos
0.2 – 0.4
Glass fiber
0.04
Cork
0.03
Plastics
0.9 – 0.04
Air
0.02
Clay
1.02
Gypsum
0.3
Saw Dust
0.07
Thermal Contact Resistance
When heat flows through
two solids in contact, temp
profile experiences a sudden
drop across the interface of
the solids.This temp drop
occurs due to thermal contact
resistance.
T1
Q
T2
T1
ΔT
T2
Due to rough surfaces at contact, direct contact is
made at few points only and voids get filled with air or
surrounding fluid, whose conductivity is much lower
than solids in contact.Therefore, interface acts as
high resistance to heat flow causing sudden temp drop.
This resistance is called Thermal Contact Resistance.
Q2: A plane composite
slab with unit cross sectional
area is made up of material ‘A’ TA=300°C
(thickness=100mm,
kA=60W/mK) and material ‘B’
Q
(thickness=10mm, kB=2W/mK).
Thermal contact resistance at
their interface is 0.003m2K/W.
A
Tc1
kA=60
100mm
The temp of open side of slab ‘A’
is 300°C and that of open side
of slab ‘B’ is50°C.
Calculate:
a) The rate of heat flow through the slab (Q)
B
TB=50°C
Tc2
kB=2
10mm
Rc=0.003
(25862W)
b) Temp on both sides of the interface (Tc1 &Tc2) (257;179C)
Solution:

A TB  T
Q
R A  Rc  R B
A
TA=300°C
Q
300  50

 25862W
0.1
0.01
 0.003
60x1
2x1
Tc1
kA=60
100mm
TA  TC1  TC 2  TB
Q
RA
RB
B
TB=50°C
Tc2
kB=2
10mm
Rc=0.003
300 TC1
TC 2  50
Hence

 25862
3
4
1.667x10
50x10
TC1  256.89C &TC2  179.3C
Variable Thermal Conductivity
Thermal conductivity of materials is strongly dependent
on temperature.
Metals: k is inversely proportional to temp
In general, k = ko (1 + aT + bT2 + … … )
= ko (1 ± aT); where –ve formetals
& +ve for non-metals
Non-metallic Solids: k is directly proportional to temp
Liquids: k is inversely proportional to temp (exceptWater)
Gases: k is directly proportional to temp. k = f (T,Pr,
Humidity)
Variable Thermal Conductivity
Q3: A plane wall of isothermal faces of temps T1 atx=0
and T2 at x=b has a thermal conductivity k=k0(1+aT).
A is the area of faces. Show that heat conducted
through wall is given by Q=k0.A/b[1+a/2(T1+T2)](T1-T2)
From Fourier’s Law
dT
Q  kA
dx
BCs: 1)At x=0;T=T1
2) At x=b;T=T2
Separating Variables
We have
Q
.dx  kdT
A
T1
k
Q
A
b
X=0
T2
X=b
Variable Thermal Conductivity
Substituti ng valueof k  k 0 (1 aT ) and Integrating
b
We have
T2
Q
. dx  k 0  1  aT dT
A 0
T1
T2
o r Q .x b0   k  T  a .T 2 
0 

A
2

 T1
Or


.b  k T  T  a . T 2 T 2 
1
2
1 
0
2
A
2


Q
 a

A
 Q  k 0 . .T1  T2 1 .T1  T2  Hence proved
b 
 2

PlottingTemp Distribution Across Thickness of
Slab with Variable Conductivity
dT
Q  kA
dx
Under steady state conditions, Heat Flow Rate Q remains const.
Since k changes with temp T, for the
LHS of the eqn, i,e. Q to remain const, something onthe RHS
must change accordingly in opposite direction. As area A is
const, it is dT/dx, which should change.
1. For a=0;
k=k0 from expression k=k0(1+aT)
Therefore, for Q to remain const, since k is not
changing, dT/dx=const; hence const slope of temp
profile across thickness of the wall
PlottingTemp Distribution Across Thickness of
Slab with Variable Conductivity
2. For a>0:
We have k proportional to T
Since temp decreases in +ve x-dir; k also decreases;
So dT/dx must increase to keep RHS const
3. For a<0:
We have k inversely proportional to T.
As T decreases in +ve x-dir; k will
increase.Therefore, to keep Q const,
dT/dx must decrease
T1
a>0
a=0
a<0
T2
Q4.Variation of thermal conductivity of a wall material
is given by:

k  k0 1T T 2

If the thickness of the wall is L and its two surfaces
are maintained at temp T1 and T2, find the expression
for heat flow through the wall.
Solution:
As per Fourier's Law Q  kA
K
dT
dx
dT
Substituti ng Q  k 0 (1  T  T )
dx
T1
Q
2
Separating Variables , we have
Q
.dx  k 0 1  T   T 2 .dT
A
T2
L
Example:Variable Conductivity (Contd)
Integrating both sides;
L
T2


Q . dx  k 1 T  T 2 dT.
0

A 0
T1
T 
T

Q
 .L  k 0 .T  .
 . 
3 T
A
2

1
3
2
T2
Onsubstitution & simplifica tion, we get;
T  T  T1T2 
A 

Qk .
 .T  T 
2
2

. T1  T2 
.1
0
1
2
1
2

L  2
3



End of Unit - I
THANK YOU !