Heat transfer can occur via three modes: conduction, convection, and radiation. Conduction involves the transfer of energy between adjacent particles in solids, liquids, and gases due to temperature differences. Convection refers to the transfer of heat by the movement of fluids and involves both conduction and fluid motion. Radiation is the emission and transmission of electromagnetic waves from the surface of an object as a result of its temperature. The key modes of heat transfer are analyzed using concepts such as the thermal conductivity of materials, heat transfer coefficients, and Stefan-Boltzmann's law of thermal radiation.

1. Process Heat Transfer
Dr. V. Charles Augustin
Sr. Asst. Professor
Department of Applied Science and Technology,
A.C.Tech, Anna University Chennai

4. THERMODYNAMICS AND HEATTRANSFER
Page  4
• Heat: The form of energy that can be transferred from one
system to another as a result of temperature difference.
• Thermodynamics is concerned with the amount of heat
transfer as a system undergoes a process from one
equilibrium state to another.
• Heat Transfer deals with the determination of the rates of
such energy transfers as well as variation of temperature.
• The transfer of energy as heat is always from the higher-
temperature medium to the lower-temperature one.
• Heat transfer stops when the two mediums reach the same
temperature.
• Heat can be transferred in three different modes:
conduction, convection, radiation.

5. Heat Transfer - Application
Page  5

6. Modes of Heat Transfer
Page  6

7. 5
ENGINEERING HEAT TRANSFER
The heat transfer problems encountered in practice can be consideredin
two groups: (1) rating and (2) sizing problems.
The rating problems deal with the determination of the heat transfer ratefor
an existing system at a specified temperaturedifference.
The sizing problems deal with the determination of the size of a system in
order to transfer heat at a specified rate for a specified temperature
difference.
An engineering device or process can be studied either
The experimental approach has the advantage that we deal with the actual
physical system, and the desired quantity is determined by measurement,
within the limits of experimental error. However, this approach is expensive,
time-consuming, and often impractical.
or
The analytical approach (including the numerical approach) has the
advantage that it is fast and inexpensive, but the results obtainedare
subject to the accuracy of the assumptions, approximations,and
idP
a
eg
e
a
l5
izations made in the analysis.

8. CONDUCTION
Conduction: The transfer of energy from the
more energetic particles of a substance to the
adjacent less energetic ones as a result of
interactions between the particles.
In gases and liquids, conduction is due to the
collisions and diffusion of the molecules during
their random motion.
In solids, it is due to the combination of
vibrations of the molecules in a lattice and the
energy transport by free electrons.
The rate of heat conduction through a plane
layer is proportional to the temperature
difference across the layer and the heat transfer
area, but is inversely proportional to the
thickness of the layer.
6
Page  8

9. CONDUCTION
6
Page  9

10. Page  8 8
Fourier’s law of heat conduction
Thermal conductivity, k: Measure of the ability
of a material to conduct heat.
Temperature gradient dT/dx: The slope of the
temperature curve on a T-x diagram.
positive quantity.
conductivity.
In heat conduction
analysis, A
represents the area
normal to the
direction of heat
transfer.
The rate of heat conduction through a
solid is directly proportional to its thermal

11. Heat transfer by Conduction
Conduction heat transfer: Fourier’s law of Conduction
Heat transfer rate per unit area is proportional to temperature
gradient:-
q = heat transfer rate
= Temperature gradient
K = Thermal conductivity
(-) sign indicate heat flows downhill in temperature scale.

12. Heat transfer by Conduction

13. Heat transfer by Conduction

14. Heat transfer by Conduction

15. Heat transfer by Conduction

16. Heat transfer by Conduction

17. Thermal Conductivity

18. Conduction (contd.)
Thermal Conductivity, k
Silver = 410 Wm-1K-1
k/ksilver
METALS
k/ksilver
Silver 1 Air 0.19
Gold 0.7 Water 0.0014
Copper 0.93 Granite, Sandstone 0.011
Aluminum 0.86 Average rock 0.012
Brass (70% Cu:30% Ni) 0.33 Limestone 0.007
Platinum, Lead 0.25 Ice 0.015
Mild steel (0.1% Cu), Cast iron 0.12 Glass (crown) 0.0058
Bismuth 0.07 Concrete (1:2:4) 0.0042
Mercury 0.04 Brick 0.0038
Snow (fresh or average) 0.005
Soil (sandy, dry) 0.002
Soil (8% moist) 0.0033
Wood 0.0045
Page  9
NON-METALS

19. Thermal Conductivity
Conductivity of Gases: kinetic theory at moderately low
temperature
 At high temperature region, the molecules have the higher
velocity than the low temperature region.
 Molecules are in continuous random motion, colliding with
each others and exchanging energy and momentum.
 If molecules move from higher temperature region to low
temperature region, it transport kinetic energy to low
temperature region through collision with low temperature
molecules.

20. A material that has a high thermal
conductivity or a low heat capacity will
obviously have a large thermal diffusivity.
The larger the thermal diffusivity, the faster
the propagation of heat into the medium.
A small value of thermal diffusivity means
that heat is mostly absorbed by the
P
a
g
e
m1a1terial and a small amount of heat is
conducted further.
Thermal Diffusivity
cp Specific heat, J/kg · C: Heat capacity
per unit mass
cp Heat capacity, J/m3· C: Heat
capacity per unit volume
 Thermal diffusivity, m2/s: Represents
how fast heat diffuses through a
material

21. Thermal Conductivity – Points to remember
• It’s a physical property of a substance. k=f(temp, position, nature of the
substance, pr.[gases only])
• For isotropic material, kx=ky=kz=k (Properties of a material are identical in all directions)
• k is higher for pure metals. For alloys k is less than that of pure metals
• The k of liquids and gases is smaller than that of solids because of their larger
intermolecular spacing
• k is very low for gases and vapours; insulating materials and inorganic liquids
have k that lie in between those of metals and gases
• The k for most pure metals (except aluminium and uranium) decreases with
increasing temperature
• Air is a bad conductor of heat (0.022 W/mK)
• The k of a gas increases with increasing temp. and decreasing mol. wt.
• Super conductors are materials having high k at very low temp (eg: k for
Page  10
aluminium at 10 K is 20,000 W/mK whereas at 293 K is 175.6 W/mK only

22. Steady State Conduction Of Heat Through A Composite Solid

23. Steady State Conduction Of Heat Through A Composite Solid

24. Steady State Conduction Of Heat Through A Composite Solid

25. Conduction heat transfer
One dimensional Conduction heat
transfer:
Energy conducted in left face + Energy generated
in within element = Change in internal energy
+Energy conducted if right face

26. Conduction heat transfer
Combining above equation:

27. 3D Conduction heat transfer
For three dimension heat conduction:

28. 3D Conduction heat transfer
General 3D conduction Equation:
For constant conductivity:
= Thermal diffusivity of a material

29. 3D Conduction HT: Cylindrical Co-ordinate

30. 3D Conduction HT: Spherical Co-ordinate

31. General Equations of heat transfer for some specified
conditions

32. Steady State Conduction Of Heat Through A Composite Solid

33. Steady State Conduction Of Heat Through A Composite Solid

36. Page  12
Convection
Convection occurs in liquids and gases.
Energy is carried with fluid motion when convection occurs.
PHYSICAL
PHENOMENON
Q  hA(Tw Ta
)
MATHEMATICAL
EQUATION

37. Page  13
Convection (contd.)
• The quantity h is called the convective heat transfer coefficient (W/m2-K).
• It is dependent on the type of fluid flowing past the wall and the velocity
distribution.
• Thus, h is not a thermo physical property.
Newton’s Law of Cooling
Q  hA(Tw Ta )
Convection Process h(W/m2-K)
Free convection
Gases 2–25
Liquids
50–1000
Forced convection
Gases 25–250
Liquids
50–20,000
Convection phase change 2,500–200,000

38. CONVECTION
Convection: The mode of
energy transfer between a
solid surface and the
adjacent liquid or gas that is
in motion, and it involves
the combined effects of
conduction and fluid motion.
The faster the fluid motion,
the greater the convection
heat transfer.
In the absence of any bulk
fluid motion, heat transfer
between a solid surface and
the adjacent fluid is by pure
conduction.
Heat transfer from a hot surface to air
by convection.
Page  38

39. Forced convection: If the
fluid is forced to flow over
the surface by external
means such as a fan,
pump, or the wind.
Natural (or free)
convection: If the fluid
motion is caused by
buoyancy forces that are
induced by density
differences due to the
variation of temperature
in the fluid.
The cooling of a boiled egg by
forced and natural convection.
Heat transfer processes that involve change of phase of a fluid are also
considered to be convection because of the fluid motion induced during
the process, such as the rise of the vapor bubbles during boiling or the
fall of the liquid droplets during condensation.
Page  39

40. Convection (contd.)
Page  40
Convective Processes
 Single phase fluids (gases and liquids)
– Forced convection
– Free convection, or natural convection
– Mixed convection (forced plus free)
 Convection with phase change
– Boiling
– Condensation

41. RADIATION
Page  41
• Radiation: The energy emitted by matter in the form of electromagnetic
waves (or photons) as a result of the changes in the electronic
configurations of the atoms or molecules.
• Unlike conduction and convection, the transfer of heat by radiation does
not require the presence of an intervening medium.
• In fact, heat transfer by radiation is fastest (at the speed of light) andit
suffers no attenuation in a vacuum. This is how the energy of the sun
reaches the earth.
• In heat transfer studies we are interested in thermal radiation, which is
the form of radiation emitted by bodies because of their temperature.
• All bodies at a temperature above absolute zero emit thermal radiation.
• Radiation is a volumetric phenomenon, and all solids, liquids, and
gases emit, absorb, or transmit radiation to varying degrees.
• However, radiation is usually considered to be a surface phenomenon
for solids.

42. Page  18 18
Stefan–Boltzmann law
 = 5.670  108 W/m2 · K4 Stefan–Boltzmannconstant
Blackbody: The idealized surface that emits radiation at the maximum
rate.
Blackbody radiation represents the maximum
amount of radiation that can be emitted from
Emissivity : A measure ofhowsurfaces
closely a surface approximates a
blackbody for which  = 1 of the
surface. 0   1.
Radiation
emitted by real
a surface at a specified temperature.

43. Radiation
Energy transfer in the form of electromagnetic waves
PHYSICAL
PHENOMENON
MATHEMATICAL
EQUATION
4
s
ET
A,Ts
Page  43

44. Radiation (contd.)
Page  44
Stefan-Boltzman Law
4
s
Eb T
The emissive power of a black body over all wave
lengths is proportional to fourth power of temperature

45. Heat Conduction Eqn (contd.)
Page  45

46. Heat Conduction Eqn (contd.)
5
Page  46

47. Page  23
Heat Conduction Equation
General heat conduction equation in rectangular coordinates
or 2
T  0

48. Steady heat conduction in Plane wall
Page  48

49. Boundary Conditions
Prescribed Temperature BC (First kind) Dirichlet conditions
Prescribed Heat Flux BC (Second kind) Neumann conditions
Convection BC (Third kind) Robbins cinditions
Page  49

50. Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
0
Page  50
L x
T1 T2
T (x,t) | x=0 = T (0,t) =T1
T (x,t) | x=L = T (L,t) = T2

51. Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
0 x
Heat
Supply
Conduction
flux
Heat
Supply
Conduction
flux
x0
x
T
q0  k
 qL
L
x
T
 k
xL
W/m2
W/m2 L
xL
 q
x
k
T
 q0
Page  51
x0
x
T
 k
Plate

52. Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Hollow Cylinder or
hollow sphere
b
rb
 q
r
T
 k
ra
r
T
 k

 qa
b r
ra
qa  k
r
T b
rb
 q
r
k
T
W/m2
a
Heat
Supply
Page  52

53. Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Plate
Conduction
Convection
Fluid
Flow
T1,h1
xL
xL
x
T
 k
Convection
 h2 (T2 T )
Conduction
Fluid
Flow
T2,h2
1 1
x0
x0
x
h (T T )  k
T
Convection heat flux
from the fluid at T1 to
the surface at x = 0
Conduction heat flux
from the surface at
x= 0 into the plate
x0
Page  53
x0
1 1
h (T T
x
)  k
T

54. Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Plate
Conduction
Convection
xL
xL
x
T
 k
Convection
 h2 (T2 T )
Fluid
Flow
T1,h1
Conduction
Fluid
T
h1(T1T x0
)  k
x
Convection heat flux
from the fluid at T2 to
the surface at x = L
Conduction heat flux
from the surface at
x = L into the plate
xL
Flow
T2,h2
xL
x0
Page  54
x
)   k
T
h (T T
2 2

55. Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Hollow Cylinder or
hollow sphere
b
r
a
Heat
Supply
2 2 rb
rb
 h (T T )
r
k
T
Fluid
Flow
T1,h1
Fluid
ra
T
h1(T1T ra
)   k
r
Convection heat flux
from the fluid at T1 to
the surface at r = a
Conduction heat flux
from the surface at
r= a into the plate
ra
Flow
T2,h2
Page  55
T
h1(T1 T ra
)   k
r

56. Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Hollow Cylinder or
hollow sphere
b
r
a
Heat
Supply
2 2 rb
rb
 h (T T )
r
k
T
Fluid
Flow
T1,h1
Fluid
ra
T
h1(T1T ra
)   k
r
Convection heat flux
from the fluid at T2 to
the surface at r = b
Conduction heat flux
from the surface at
r= b into the plate
rb
Flow
T2,h2
Page  56
T
h2 (T2 T r b
) k
r

57. Page  33
Cont…
Temperature distribution in a plane wall
Steady 1D heat transfer, no heat generation
the conduction eqn. is written as
s,1
s,2 s,1
1
Ts,1  C2 &
T(x)  C1 x C2
B.C' sare
T(0)  Ts,1
T(L)  Ts,2
Thus the constants are
 C
L
T (x) (T T )
x
 T
L
Ts,2  Ts,1
 
dx  dx 
d  dT  0
 0
x2
2
T
T = T1
0 L
T = T2
x
Rectangular Coordinates

58. Page  34
Cont…
Ts,1)
dT

(Ts,2
L
s,1
s,2 s,1
wkt T (x) (T T )
x
 T
dx L
From Fourier’s law,
Q  kA
dT
 put
dT
, thus
dx dx
Q 
kA(T1 T2 )

T
L L /kA
Heat flow 
Thermal potentialdifference
Thermal resistance
K.A
L
R 

59. Page  35
Cont…
hA
R
Q conv
conv
conv , where R 
1

(Ts  T )

Thermal resistance for convection
Qconv  hA(Ts  T)
Thermal resistance for radiation
hrad As
rad
rad
surr
rad s s surr rad s s
where, R 
R
Tsurr )
) 
(Ts
A (T T
Q  A (T 4
T 4
)  h
1
Combination of
conductive and
convective
resistance

60. Combined mechanisms of heat transfer
Page  60

61. Thermal Resistance Network
Plane thickness – L
Plane area –A
Wall thermal conductivity – k
Assume T2  T1
Page  61

62. Cont…
Rtotal
Q 
T1  T
2
Summary
• The rate of steady state transfer between two surfaces is equal to the
temperature difference between those two surfaces divided by the total
thermal resistance between those two surfaces
• The ratio of the temperature drop to the thermal resistance across any
layer is constant
• The temperature drop across any layer is proportional to the thermal
resistance of the layer
(1)
Page  62

63. Cont…
Page  63

64. (2)
Compare eqns. 1 & 2, it reveals that
To determine the rate of heat transfer through the wall we do not need to
know the surface temperatures of the wall (Ref eqn.1)
How to get wall surface temperature?
Page  64

65. w.k.t.
Q 
T1  T2
Rtotal
Multilayer plane walls
Page  65

66. Problem to get intermediate temperature
A furnace wall consists of 200 mm
layer of refractory bricks, 6 mm
layer of steel plate and a 100 mm
layer of insulation bricks. The
maximum temperature of the wall is
1150oC on the furnace side and
minimum temperature is 40oC on
the outer side. Heat loss from the
wall is 400 W/m2. It is known that
there is a thin layer of air between
the layers of refractory bricks and
steel plate. Thermal conductivities
for the three different layers are
1.52, 45 and 0.138 W/moC. Find:
(i) the thickness of air layer (ii)
temperature of the outer surface of
the steel plate
P
a
Tg
e
a
k4
e2 conductivity of air is same as insulation brick

67. Page  67

68. Page  68

69. Thermal resistance network for parallel layers
Page  69

70. Series and Parallel composite wall
Page  70

71. Problem: Composite wall
Page  71

72. cont..
Page  72

73. cont..
Page  73

74. cont..
Page  74

75. Find the heat flow rate through the composite wall as shown infigure.
Assume 1D flow.
Page  75

76. cont..
Page  76

77. Thermal contact resistance
Interface offers some
resistance to h.t, and this
resistance for a unit
surface area is called the
thermal contact resistance
Page  77

78. Temperature distribution in cylinder
Heat conduction eqn.
Page  78

79. Page  55
Cont…
1D heat conduction eqn without heat generation
(3)
Eqn (3) integrated twice and put B.C’s to get temp. distribution

80. Page  56
Cont…
wh ere,A 2rL
dr
wkt, Q  -kA
dT
differentiate eqn (4) and put in Fourier’s eqn.
(4)

81. Cont…
Page  81

82. Cont…
Page  82

83. Cont…
Page  83

84. Critical radius of Insulation
The rate of heat transfer from the insulated pipe
to the surrounding air can be expressed as
q 
T1T
Page  84
r
Rins  R
The additional insulation increases the conduction resistance of the insulation
layer but decreases the convection resistance of the surface because of the
increase in the outer surface area for convection. The heat transfer from the pipe
may increase or decrease, depending on which effect dominates.

85. Page  61
Cont…
The variation of heat transfer rate with the outer radius of insulation r2 is plotted in below
figure. The value of r2 at which heat transfer rate reaches maximum is determined from
the requirement that dqr/dr=0(zero slope). Performing the differentiation and solving
for r2 yields the critical radius of insulation for a cylindrical body to be
cr,cylinder

k
h
r
The rate of heat transfer from the cylinder increases
with the addition of insulation for r2< rcr, reaches a
maximum when r2= rcr, and starts to decrease for
r2> rcr. Thus, insulating the pipe may actually
increase the rate of heat transfer from the pipe
instead of decreasing it when r2< rcr .
cr,sphere

2k
r
h

86. Page  62
Critical radius of Insulation - Problem
1cm

87. Page  63
Critical radius of Insulation – Problem cont…
1cm
4cm

88. An insulated steam pipe having outside diameter of 30 mm is to be covered with two layers of
insulation, each having the thickness of 20mm. The thermal conductivity of one material is 5
times that of the other,
Assuming that the inner and outer surface temperatures of composite insulation are fixed, how
much will heat transfer be increased when better insulation material is next to the pipe than it is
outer layer?
Case 1: better insulation is inside
Page  88

89. Cont…
Case 2: poor insulation is inside
Page  89

90. A 240 mm steam main, 210 m long is covered with 50 mm of high temperature insulation
(k=0.092 W/moC) and 40 mm of low temperature insulation (k=0.062 W/moC). The inner and
outer surface temperature as measures are 390oC and 40oC respectively. calculate:
(i) The total heat loss per hour (ii) The total heat loss/m2 of pipe surface (ii) The total heat loss/m2
of outer surface, and (iv) the temperature between two layers of insulation.
Neglect heat conduction through pipe material
Page  90

91. Cont…
Page  91

92. Cont…
Page  92

93. Page  93

94. One Dimensional Heat Conduction (contd.)
Rectangular Coordinates
Cylindrical Coordinates
A Compact Equation
t
  k
T   g  c
T(x,t)
p
 
x

x

t
1   rk
T   g  c
T(r,t)
p
 
r r

r

Spherical Coordinates
1 
t
T(r,t)
T
r k
r r
p

r
  g  c



 2
2
r k p
n
n





r

T
r r
T(r,t)
t
1 
 g  c
n = 0
n = 1
n = 2
Page  94

95. Heat Source Systems
Plane wall with heat generation
For 1D, steady state, Fourier-Biot eqn is reduced as below
HG (2)
HG (1)
Page  95

96. Heat Source Systems
cont…
Plane wall with heat generation
(a) Specified temperature on both sides
Put in HG (2)
Put in HG (2)
Page  96
in HG (2)

97. Heat Source Systems
Plane wall with heat generation cont…
(b) Insulated boundary on one side and Convective boundary on other side
Page  97

98. Heat Source Systems
Plane wall with heat generation cont…
HG (1)
HG (1) HG (2)
A plane wall insulated on
one face and exposed to
convection environment
on other face
Page  98

99. Heat Source Systems
Plane wall with heat generation cont…
HG (2), weget
Page  99

100. Heat Source Systems
Plane wall with heat generation - Problem
Page 
100

101. Page  77
Heat Source Systems
Plane wall with heat generation – Problem cont…
The temp. distribution in a plane
wall is given by
Substitute C1 and C2 in HG (2)
HG (2)
HG (1)

102. Heat Source Systems
Plane wall with heat generation – Problem cont…
Put x = 0 in eqn. A, thus the magnitude of max. temp.is
Page 
102

103. Heat Source Systems
Plane wall with heat generation - Problem
Page 
103

104. Heat Source Systems
Plane wall with heat generation – Problem cont…
w.k.t HG(1) and (2)
Page 
104

105. Heat Source Systems
Plane wall with heat generation – Problem cont…
Page 
105

106. Heat Source Systems
Plane wall with heat generation – Problem cont…
Page 
106

107. Heat Source Systems
Plane wall with heat generation – Problem cont…
Page 
107

108. Page  84
Steady State 1D Heat Conduction and generation
Cylindrical Coordinates (Solid Cylinder)
T = Ts
0 r
T = Ts
ro
 0
1 d

r
dT


g0
r dr  dr  k
Governing Equation

 dT
 0 &
at r 0
dr
o
s at rr
T T 
 c1 ln r  c2
r 2
4k
T (r)  
g0


   Ts
 ro  
 r  
2
g r 2
4k
T (r) 0 o
1 
Solving,
dT(r) g r
q(r) k 0
Temp. at centre of cylinder
Tc Ts
4k
g r 2
0 o
o
s
c
r2
r2
T  T
T (r)  T
s
1
B.C’s
Temp distribution in non-dimensionalform
g r
dr 2
C2  Ts 
C1  0
o o
4K
2

109. T = Ts
0 r
T = Ts
ro
o
g r
2
2rLrgo
 r2
Lg
2 dr 2k Heat transfer rate in cylinder

2k



dr
 o 
Q(r)  kA
Q(r)  kA
dT(r)

g0r
where
dT (r)
 
gor
Heat transfer rate at outer surface of cylinder
2h
Page 
109
 goro
r2
Lg  h(2r L) (T -T )
o o s 
Q(r)  h(2ro L) (Ts -T)
At outer surface,cond.  conv.
s
The surface temp.T  T
Steady State 1D Heat Conduction and generation

110. T = Ts
0 r
T = Ts
ro
Cylinder exposed to conv. environment
2h
Page 
110
4k
g r 2
g r
Max. temp. is seen at centre(i.e.r  0),T(r) - T
g
4k 2h
4k 2h
g r 2
g r
it gives, c2 0 o
0 o
 T
g
4k
2k 
dr
dT
dT g r c
0 o
0 o
o
dr 2k r
4k
Temp. distribution as T(r)  2
0
r  r 2

g0ro
 T
o 
2
0
1
 0  c  0
0 1
2
1

dr




 T 


 
2 
r  c 
rro

 h

 k 
g0 r 
rro
 
- kA dT   hA(T  T )
at r  ro , cond.  conv.
h, T
   ; w.k.t at r  0,
T (r)  
g0
r 2
 c ln r  c
rr
r ro

Steady State 1D Heat Conduction and generation

111. Cylindrical Coordinates (Hollow Cylinder)
Determination of Temperature Distribution
r
0
ro
T2
k
ri
T1
2
Page 
111
2
0
ln1 r  g
o i 2 1
2
2
0
put c1 and c2
2
1
2
0
1
 r ) 
(r
4k
T(r) 
1 g
o i 2 1 2
solving c 
4k
4k
g r 2
T2   0 o
 c1 ln ro  c2
T1   0 i
 c1 ln ri  c2
at r  ri , T  T1 and r  ro , T  T2
g r 2
g
c 
o i
o
o
o i
o
0 o
o i 2
i o

lnr r  4k

 (r  r 2
)  (T  T )  T



4k lnr r   4k
g r 2
ln1 r   g
 r 2
)  (T  T )  T
 0
(r 2



T )

lnr r 
4k

(r  r 2
)  (T
Steady State 1D Heat Conduction and generation

112. Page  88
Conduction-Convection Systems
Fins / Extended Surfaces
• Necessity for fins
• Biot Number
=
LONGITUDINAL
RECTANGULAR FIN
hx

(x/ k)
k 1/ h
Internal Conductive resistance
Surface Convective resistance
FIN TYPES
RADIAL FIN FIN WITH NON-
UNIFORM C.S PIN FIN (or) SPINE