This document provides an overview of heat transfer concepts including the three main modes of heat transfer (conduction, convection and radiation). It defines key terms like thermal conductivity, heat transfer coefficients, emissivity, absorptivity and summarizes common values. Governing equations for each heat transfer mode are presented. Example problems are provided to illustrate concepts. The document is part of a course on heat transfer and discusses the relationship between heat transfer and the first and second laws of thermodynamics.

1. HEAT TRANSFER
MEE2005
Dr. M. Sreekanth
Associate Professor Sr., SMBS
VIT Chennai
Email: manavalla.sreekanth@vit.ac.in

2. Module 1
Fundamental Concepts

3. Heat Transfer
• Heat Transfer is Thermal Energy in transit due
to spatial temperature difference
3

4. Applications of Heat Transfer
4

5. Modes of Heat Transfer
• Conduction
• Convection
• Radiation
• Conduction and Convection need a medium for
heat transfer
• Radiation does not need a medium for heat
transfer
5

6. Modes of Heat Transfer…
6
Water is analogous to heat.
People are analogous to the
medium
1: Radiation
2: Conduction
3: Convection

7. Conduction
• Conduction is the transfer of energy from more
energetic particles of a substance to the adjacent
less energetic ones
• It can take place in solids, liquids and gases
• In solids, conduction is due to a combination of
vibrations of molecules in a lattice and energy
transported by free electrons
• In liquids and gases, it is due to collision and
diffusion of molecules during their random
motion
7

8. Mechanism of Heat Conduction
8

9. Rate Equation of Heat Conduction
• The Fourier’s Law of Heat Conduction gives the
rate equation of heat conduction
• k is the thermal conductivity
• is the temperature gradient
• The equation implies that heat transfers in the
direction of decreasing temperature
9
dx
dT
kA
Qcond 


dx
dT

10. Thermal Conductivity (k)
• Thermal Conductivity is the rate of heat
transfer through a unit thickness of the
material per unit area per unit temperature
difference
• High value of k indicates good heat conductor
• Low value of k indicates a poor heat conductor
or an insulator
• Copper and silver are good conductors while
rubber, wood, styrofoam are bad conductors
10

11. Typical Values of Thermal Conductivity
Material Thermal
Conductivity
(W/m-K)
Diamond 2300
Silver 429
Copper 401
Gold 317
Aluminium 237
Glass 0.78
Human Skin 0.37
Wood 0.17
Air 0.026
Urethane 0.026
11

12. Variation of Thermal Conductivity with
Temperature
12
Thermal
Conductivity
(W/m-K)
Temperature
(K)
Copper Aluminium
200 413 237
300 401 237
400 393 240
600 379 231
800 366 218

13. Convection
• Convection is the mode of energy transfer
between a solid surface and the adjacent liquid or
gas that is in motion
• It involves the combined effects of conduction
and fluid motion
• Forced Convection takes place when the flow
caused by an external means like a fan or a pump
• Free Convection takes place due to density
differences
13

14. Examples of Convection
14
(a) Forced Convection, (b) Free Convection, (c) Boiling, (d) Condensation

15. Rate Equation for Convection
• Newton’s law of cooling gives the rate equation
for convection
• The Convective heat transfer coefficient (h) is not
a property of the fluid
15
 
K
re,
temperatu
Fluid
K
re,
temperatu
Surface
m
,
m
W
t,
coefficien
fer
heat trans
Convective
2
2









T
T
Area
A
K
h
T
T
hA
Q
s
s
conv


16. Typical Values of
Heat Transfer Coefficient
Type of Convection h, W/m2-K
Free convection of gases 2-25
Free convection of liquids 10-1000
Forced convection of gases 25-250
Forced convection of liquids 50-20,000
Boiling and condensation 2500-100,000
16

17. Radiation
• Thermal Radiation is the energy emitted by
matter that is at a nonzero temperature
• Thermal radiation differs from other radiation
like x-rays, gamma rays, microwaves that are
not related to temperature
• Thermal radiation can occur from solid
surfaces, liquids and gases
• The radiation is attributed to changes in the
electron configuration
17

18. Radiation Exchange
18
(a) Radiation exchange at a surface
(b) Radiation exchange between a surface and large surroundings

19. Rate Equation for
Radiation Heat Transfer
• The Stefan-Boltzmann law gives the upper
limit to the emissive power of a surface at a
temperature as
19
2
s
4
2
8
-
4
max
,
m
area,
Surface
A
K
re,
temperatu
surface
Absolute
m
W
10
5.67
constant
Boltzmann
Stefan








s
s
s
emit
T
K
where
T
A
Q




20. Black Body
• An idealized surface which emits the
maximum radiation is called a Black Body
• The radiation emitted is Black Body Radiation
• Real surfaces emit less than a black body
• Emissivity is a measure of how closely a
surface approximates a black body
20
Emissivity
T
A
Q s
s
emit



 4


21. Linearization of the
Radiation Rate Equation
21
 
 
   
 
 
  
 
   
 
  
2
2
2
2
2
2
2
2
4
4
4
4
4
4
,
simplicity
For
Exchange,
Heat
Radiation
surr
s
surr
s
r
surr
s
surr
s
surr
s
surr
s
surr
s
surr
s
surr
s
surr
s
surr
s
r
surr
s
surr
s
r
surr
s
r
rad
surr
s
rad
T
T
T
T
h
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
h
T
T
A
T
T
A
h
So
T
T
A
h
Q
T
T
A
Q
































22. Emissivity (ε) and Absorptivity (α)
• Emissivity of a black body is equal to 1
• For all real surfaces, emissivity lies between 0
and 0 ≤ ε ≤ 1
• Absorptivity is the fraction of the radiation
energy incident on a surface that is absorbed
by it
• 0 ≤ α ≤ 1
22
G
Gabs 


23. Typical Values of Emissivity
Material Emissivity
Aluminium foil 0.07
Anodized aluminium 0.82
Polished copper 0.03
Polished gold 0.03
Black paint 0.98
White paint 0.9
Asphalt pavement 0.85-0.93
Human skin 0.95
Soil 0.93-0.96
Water 0.96
23

24. Relationship to the 1st Law of
Thermodynamics
• The 1st law deals with conservation of energy
• 1st law deals with both closed and open
systems
• In many cases, we end up estimating the heat
to be removed or supplied
• However, the mechanism of heat removal or
supply is not dealt by thermodynamics
24

25. Relationship to the 2nd Law of
Thermodynamics
• The 2nd law statement by Kelvin-Planck
involves heat engines
• Heat engines need to exchange heat with high
and low temperature thermal reservoirs
• For heat transfer, there must exist a finite
temperature difference
• However, a finite temperature difference
results is irreversibility during heat transfer
25

26. Identification of Heat Transfer Modes
Hot coffee in a flask:
26

27. Identification of Heat Transfer Modes…
Fire place in a room:
27

28. Identification of Heat Transfer Modes…
Thermocouple junction:
28

29. Problem 1
The wall of an industrial furnace is constructed from
0.15 m thick fireclay brick having a thermal
conductivity of 1.7 W/m-K. Measurements made
during steady state operation reveal temperatures
of 1400 and 1150 K at the inner and outer surfaces
respectively. What is the rate of heat loss through a
wall that is 0.5 m × 1.2 m on a side.
Ans: 1700 W
29

30. Problem 2
A wall has inner and outer surface temperatures
of 16 and 6℃, respectively. The interior and
exterior air temperatures are 20 and 5℃,
respectively. The inner and outer convection
heat transfer coefficients are 5 and 20 W/m2-K,
respectively. Calculate the heat flux from the
interior air to the wall, from the wall to the
exterior air, and from the wall to the interior air.
Is the wall under steady state condition?
30

31. Problem 3
An overheated 30 m long, uninsulated steam
pipe of 120 mm diameter is routed through a
building whose walls and air are at 30℃.
Pressurized steam maintains a pipe surface
temperature of 150℃, and the coefficient
associated with natural convection is h=10
W/m2-K. The surface emissivity is 0.8. What is
the rate of heat loss from the steam line?
Ans: 25.7 kW
31

32. Problem 4
The blades of a wind turbine turn a large shaft at a
relatively slow speed. The rotational speed is increased by
a gearbox that has an efficiency of 0.93. In turn, the
gearbox output shaft drives an electric generator with an
efficiency of 0.95. The cylindrical nacelle, which houses
the gearbox, generator, and associated equipment, is of
length 6 m and dia 3m. If the turbine produces P=2.5 MW
of electrical power, and the air and surrounding
temperatures are 25℃ and 20℃, respectively, determine
the minimum possible operating temperature inside the
nacelle. The emissivity of the nacelle is 0.83, and the
convective heat transfer coefficient is 35 W/m2-K.
32

33. Problem 5
Consider a person with a core body temperature of
35℃, skin/fat layer of 3 mm and effective thermal
conductivity 0.3 W/m∙K. The person has a surface
area of 1.8 m2 and is dressed in bathing suit. The
emissivity of the skin is 0.95. When the person is in
still air at 297 K, what is the skin temperature and
the rate of heat loss to the environment? Take h=2
W/m2∙K. When the person is in water at 297 K,
what is the skin temperature and heat loss rate.
Take h=200 W/m2∙K.
33

34. Problem 6
A 0.3 cm thick, 12 cm high and 18 cm long circuit
board houses 80 closely spaced logic chips on one
side, each dissipating 0.06 W. The board is
impregnated with copper fillings and has an
effective thermal conductivity of 16 W/m-K. All the
heat generated in the chips is conducted across the
circuit board and is dissipated from the back side of
the board to the ambient air. Determine the
temperature difference between the two sides of
the circuit board.
Ans: 0.042℃
34

35. Problem 7
A 3 m internal diameter spherical tank made of 1 cm thick
stainless steel is used to store iced water at 0℃. The tank
is located outdoors at 25℃. Assuming the entire steel
tank to be at 0℃ and thus the thermal resistance of the
tank to be negligible, determine (a) the rate of heat
transfer to the iced water in the tank, (b) the amount of
ice at 0℃ that melts during a 24 hour period. Take the
heat of fusion to be 333.7 kJ/kg. Emissivity of the water
tank is 0.75, the heat transfer coefficient is 30 W/m2-K
and the surrounding temperature is 15℃.
Ans: 22.8 kW, 5903 kg
35

36. Module 2
Conduction I

37. Heat Conduction Equation and
Its Importance
• Conduction analysis will give the temperature
distribution in a medium
• The temperature distribution can be used to
find the heat flux, thermal stress, expansion
and deflection in an object
• It can also be used to optimize the thickness
of insulation
• Heat Conduction Equation is the key to the
above objectives
37

38. Heat Conduction Equation
Cartesian Coordinates
38
storage
energy
thermal
of
Rate
generation
energy
thermal
of
Rate


st
g
E
E



39. Heat Conduction Equation
Cartesian Coordinates…
39
     








































































x
T
k
x
dxdydz
dx
x
T
kdydz
x
dx
x
q
x
T
kdydz
q
dxdydz
T
c
dxdydz
q
dz
z
q
dy
y
q
dx
x
q
dxdydz
T
c
q
q
q
dxdydz
q
q
q
q
E
E
E
E
dxdydz
T
c
T
mc
E
dxdydz
q
E
x
x
p
z
y
x
p
dz
z
dy
y
dx
x
z
y
x
st
out
g
in
p
p
st
g
,
Since
:
gives
form
rate
in
balance
Energy
change)
phase
no
is
e
when ther
valid
is
(this
e
unit volum
per
rate
generation
heat
the
is
q
where


















40. Heat Conduction Equation
Cartesian Coordinates…
40

















































































































































































T
c
q
z
T
k
z
y
T
k
y
x
T
k
x
dxdydz
T
c
dxdydz
q
z
T
k
z
dxdydz
y
T
k
y
dxdydz
x
T
k
x
dxdydz
z
T
k
z
dxdydz
dz
z
T
kdxdy
z
dz
z
q
z
T
kdxdy
q
y
T
k
y
dxdydz
dy
y
T
kdxdz
y
dy
y
q
y
T
kdxdz
q
Similarly
x
T
k
x
dxdydz
dx
x
T
kdydz
x
dx
x
q
x
T
kdydz
q
p
p
z
z
y
y
x
x


on
substituti
On
,
,
,
Since
---Heat Conduction Equation
in Cartesian Coordinates

41. Meaning of the Heat Conduction
Equation
• At any point in the medium, the net rate of
energy transfer by conduction into a unit
volume plus the volumetric rate of heat
generation must equal the rate of change of
thermal energy stored in the volume
41

42. Simplified Forms of the Conduction
Equation
42
0
generation
heat
no
with
state
steady
,
direction)
(in x
D
-
1
Under
0
conditions
state
steady
Under
y,
diffusivit
Thermal
1
constant,
is
k
If
2
2
2
2
2
2
2






































































































x
T
k
x
q
z
T
k
z
y
T
k
y
x
T
k
x
/s
m
c
k
T
T
k
c
k
q
z
T
y
T
x
T
T
c
q
z
T
k
z
y
T
k
y
x
T
k
x
p
p
p












43. Standard Names of the Heat
Conduction Equation
43
Equation
Laplace
-
-
-
0
Equation
Diffusion
-
-
-
1
Equation
Poisson
-
-
-
0
Equation
Biot
-
Fourier
-
-
-
1
Equation
Conduction
General
-
-
-
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
















































































z
T
y
T
x
T
T
z
T
y
T
x
T
k
q
z
T
y
T
x
T
T
k
q
z
T
y
T
x
T
T
c
q
z
T
k
z
y
T
k
y
x
T
k
x
p










44. Heat Conduction Equation
Cylindrical Coordinates
44



 




































 T
c
q
z
T
k
z
T
k
r
r
T
kr
r
r
p

2
1
1

45. Heat Conduction Equation
Spherical Coordinates
45








 




































 T
c
q
T
k
r
T
k
r
r
T
kr
r
r
p

sin
sin
1
sin
1
1
2
2
2
2
2

46. All Heat Conduction Equations in one
Place
46







































 T
c
q
z
T
k
z
y
T
k
y
x
T
k
x
p




 




































 T
c
q
z
T
k
z
T
k
r
r
T
kr
r
r
p

2
1
1








 




































 T
c
q
T
k
r
T
k
r
r
T
kr
r
r
p

sin
sin
1
sin
1
1
2
2
2
2
2
Cartesian:
Cylindrical:
Spherical:

47. How to Solve the Heat Conduction
Equation
• The heat conduction equation is a partial
differential equation with temperature varying
in space (x, y, z directions) and time
• Boundary conditions and initial conditions are
needed
• Since the equation is 2nd order in space, two
boundary conditions are needed
• Since the equation is 1st order in time, one
initial condition is needed
47

48. Common Boundary Conditions
48
Also called as
Dirichlet condition
Also called as
Neumann condition

49. Combined 1-D Heat Conduction
Equation
49
2
:
Sphere
1
:
Cylinder
0
:
wall
Plane
1


















n
n
x
, r
n
T
c
q
r
T
k
r
r
r
p
n
n




50. 1-D Steady State Conduction
• Heat transfers in predominantly one direction
• Temperature is independent of time
• Simple for analysis and useful in numerous
engineering applications
• Eg. Fins, slabs, furnaces
50
0
0
0
0








































dx
dT
k
dx
d
x
T
k
x
q
dx
dT
k
dx
d
q
x
T
k
x

 With heat generation
Without heat generation

51. 1-D Steady State
Plane Wall
51
  1
,
1
,
2
,
1
,
2
,
1
1
,
1
2
,
2
1
,
2
1
2
1
2
1
1
)
(
At
C
;
0
At
conditions
boundary
applying
by
obtained
be
can
C
and
C
)
(
again,
g
integratin
On
once,
g
integratin
On
0
s
s
s
s
s
s
s
s,
s
s,
T
L
x
T
T
x
T
L
T
T
C
T
L
C
T
T
L, T
x
T
T
, T
x
C
x
C
x
T
C
k
C
dx
dT
C
dx
dT
k
dx
dT
k
dx
d





























52. 1-D Steady State
Cylindrical Shell
52
  1
1
2
1
2
1
1
2
1
2
1
1
2
1
2
1
2
1
2
1
2
1
2
1
1
1
2
2
1
1
ln
ln
ln
ln
ln
ln
C
;
ln
C
,
conditions
oundary
the
ng
substituti
On
ln
C
T
again,
g
Integratin
g,
integratin
On
are
conditions
boundary
The
0
r
:
k
constant
and
,
generation
heat
no
with
s,
coordinate
l
cylindrica
state,
steady
D
-
1
in
equation
conduction
The
r
r
r
T
T
T
r
r
r
T
T
r
T
r
r
r
T
T
T
r
r
T
T
C
r
r
C
r
T
C
r
T
r
r
at r
T
and T
r
at r
T
T
r
T
r



































































 
















1
2
2
1
1
1
ln
2
2
2
r
r
T
T
πkl
klC
r
C
rl
k
dr
dT
kA
Qcyl




53. 1-D Steady State
Spherical Shell
53
 
 
 
 
1
2
1
1
2
2
2
1
1
2
2
1
1
2
1
1
2
2
2
2
1
1
2
2
1
1
2
1
2
1
1
2
2
2
1
1
2
C
;
C
,
conditions
boundary
the
ng
substituti
On
C
T
again,
g
Integratin
g,
integratin
On
are
conditions
boundary
The
0
r
:
k
constant
and
,
generation
heat
no
with
s,
coordinate
spherical
state,
steady
D
-
1
in
equation
conduction
The
r
r
T
r
T
r
T
T
r
r
r
r
r
r
T
r
r
T
r
T
r
T
T
r
r
r
r
C
r
r
C
r
T
C
r
T
r
r
at r
T
and T
r
at r
T
T
r
T
r







































 
 
2
1
1
2
2
1
1
2
1
2
4
4
4
T
T
r
r
r
πkr
kC
r
C
r
k
dr
dT
kA
Qsphere













54. Thermal Resistance and
Electrical Analogy
54
 
hA
R
hA
; R
T
T
; V
q
I
hA
T
T
T
T
hA
kA
L
R
kA
L
; R
T
T
Q; V
I
R
V
I
kA
L
T
T
L
T
T
kA
dx
dT
kA
q
conv
t
s
conv
s
s
cond
t
s
s
s
s
s
s
x
1
1
1
q
Similarly,
,
conv
,
2
1
2
1
2
1































A
h
kA
L
A
h
Rtotal
2
1
1
1




55. Thermal Resistance and
Electrical Analogy…
55
 
parallel.
in
act
s
resistance
convective
and
radiation
Surface
small.
is
t
coefficien
fer
heat trans
convective
the
if
important
be
may
fer
heat trans
Radiation
1
1
A
h
R
A
h
T
T
T
T
A
h
q
rad
rad
rad
surr
s
surr
s
rad
rad







  
2
2
by
estimated
be
can
t
coefficien
fer
heat trans
Radiation
surr
s
surr
s
r T
T
T
T
h 

 

56. Composite Walls
• Walls involving any number of series and
parallel thermal resistances due to layers of
different materials
56

57. Composite Wall in Series
57
A
h
A
k
L
A
k
L
A
k
L
A
h
T
T
q
A
h
T
T
A
k
L
T
T
A
k
L
T
T
A
k
L
T
T
A
h
T
T
q
C
C
B
B
A
A
x
s
C
C
s
B
B
A
A
s
s
x
4
1
2
,
1
,
4
4
,
4
,
4
,
3
3
2
2
1
,
1
1
,
1
,
1
1
1
1





















58. Composite Wall in Parallel
58
 
2
1
2
1
2
1
2
1
2
1
2
1
2
2
1
1
2
1
2
1
1
1
1
1
1
R
R
R
R
R
R
R
R
R
T
T
q
R
R
T
T
R
T
T
R
T
T
q
q
q
total
total
total


























59. Composite Wall in Series and Parallel
59

60. Overall Heat Transfer Coefficient
60
4
1
4
1
2
,
1
,
1
1
1
1
t
Coefficien
Transfer
Heat
Overall
1
1
h
k
L
k
L
k
L
h
A
R
U
U
T
UA
q
A
h
A
k
L
A
k
L
A
k
L
A
h
T
T
q
C
C
B
B
A
A
total
C
C
B
B
A
A
x














 


61. Hollow Cylinder
61
   
L
r
h
kL
r
r
L
r
h
T
T
A
h
kL
r
r
A
h
T
T
Qcyl
2
2
1
2
1
1
2
,
1
,
2
2
1
2
1
1
2
,
1
,
2
1
2
ln
2
1
1
2
ln
1



























 





62. Hollow Composite Cylinder
62
   
L
r
h
L
k
r
r
L
k
r
r
L
k
r
r
L
r
h
T
T
Q
C
B
A
cyl
4
4
3
4
2
3
1
2
1
1
4
,
1
,
2
1
2
ln
2
ln
2
ln
2
1


































 



63. Critical Insulation Thickness
• In radial systems, the conduction resistance
increases with increase of insulation thickness
• The convection resistance decreases with
increase in insulation thickness and thereby
surface area
• Optimum Insulation thickness minimizes the
heat loss
63

64. Critical Insulation Thickness
64
h
k
r
h
r
kr
dr
rh
k
r
r
d
dr
R
R
rh
k
r
r
R
rh
k
r
r
T
T
q
rlh
kl
r
r
T
T
q
i
total
total
i
total
i
i
i
i
































0
2
1
2
1
0
2
1
2
ln
0
d
maximum
be
must
,
insulation
optimum
For
2
1
2
ln
2
1
2
ln
2
1
2
ln
2













65. Critical Insulation Thickness…
65
fer
heat trans
maximizes
r
radius
Insulation
Critical
k/h
r
at
minimum
is
R
0
2
1
2
1
1
1
dr
R
d
,
h
k
r
at
1
2
1
dr
R
d
resistance
the
minimizes
or
maximizes
thickness
insulation
the
if
check
To
cr
total
2
3
2
2
total
2
3
2
2
total
2
h
k
h
k
k
k
h
k
h
r
kr































66. Critical Insulation Radius
66

67. Thermal Contact Resistance
67

68. Thermal Contact Resistance…
• In composite systems, the temperature drop
across the interface may be considerable
• This is due to contact resistance
• Surface roughness is responsible for contact
resistance
• Contact spots are interspersed with gaps filled
with air
• Heat transfer is due to conduction across the
contact area and conduction and or radiation
across the gaps
68

69. Thermal Contact Resistance…
• Contact resistance is equivalent to parallel
resistances due to contact spots and gaps
• For rough surfaces, the resistance is mostly
due to gaps
69
W
m
is
area
unit
per
resistance
contact
Thermal
2
"
"
,
K
q
T
T
R
x
B
A
c
t




70. 1-D Steady State with no Heat
Generation-Summary
70

71. 1-D Steady State with Heat Generation
• Thermal energy generation is observed in
electrical conductors, due to Ohmic heating
• Heat generation also occurs in nuclear reactors
during deceleration and absorption of neutrons
• Heat generation also occurs during exothermic
reactions
71
3
2
2
W/m
V
R
I
V
E
q
R
I
E
e
g
e
g







72. 1-D Steady State with Heat Generation
Plane Wall
72
 
2
2
1
2
2
2
C
and
2
are
conditions
boundary
The
2
n,
integratio
On
0
ty
conductivi
hermal
constant t
with
generation
heat
uniform
For
2
,
1
,
1
,
2
,
2
2
2
2
,
1
,
2
2
1
,
2
,
1
2
1
2
1
2
1
2
2
s
s
s
s
s
s
s
s
s,
s,
T
T
L
x
T
T
L
x
k
L
q
x
T
T
T
L
k
q
L
T
T
C
T
(L)
and T
T
T(-L)
C
x
C
x
k
q
T
C
x
k
q
dx
dT
k
q
dx
T
d




































73. 1-D Steady State with Heat Generation
Plane Wall…
73
 
 
2
0
0
2
0
2
2
2
2
1
is
on
distributi
re
temperatu
The
2
0
1
2
When

























L
x
T
T
T
T
T
k
L
q
T
T
T
L
x
k
L
q
x
T
T
T
T
s
x
s
s
s
s,
s,


0
0








x
dx
dT

74. Implementing the Convective
Boundary Condition
74
 
 
h
L
q
T
T
T
L
x
k
L
q
x
T
dx
dT
T
T
h
dx
dT
k
s
s
s
L
x




























2
2
2
1
2
from
obtained
be
can

75. 1-D Steady State with Heat Generation
Cylinder
75
 
 
h
r
q
T
T
T
r
r
k
r
q
r
T
T
r
T
dr
dT
k
q
dr
dT
r
dr
d
r
o
s
s
o
o
s
o
r
2
1
4
and
0
are
conditions
boundary
The
0
1
2
2
2
0



































76. 1-D Steady State with Heat Generation
Sphere
76
  s
o
o
T
r
r
k
r
q
r
T 









 2
2
2
1
6


77. Problem 1
At a given instant of time, the temperature
distribution within an infinite homogeneous
body is given by the function
Assuming constant properties and no internal
heat generation, determine the regions where
the temperature changes with time.
Ans: None
77
  yz
xy
z
y
x
z
y
x
T 2
2
,
, 2
2
2






78. Problem 2
The steady state temperature distribution in a 1-
D wall of thermal conductivity 50 W/m-K and
thickness 50 mm is observed to be T=a+bx2,
where a=200 ℃ and b=-2000℃/m2 and x is in
m. What is the heat generation in the wall.
Determine the heat fluxes at the two wall
faces.
Ans: 2×105 W/m3
0 W/m2, 10,000 W/m2
78

79. Problem 3
A hollow cylinder with 30 mm inner radius and
50 mm outer radius, k=15 W/m-K, is heated on
the inner surface at a rate of 105 W/m2 and
dissipates heat by convection from the outer
surface to a fluid at 100℃ with h=400 W/m2-
K. Find the temperatures at the inside and
outside surfaces of the cylinder.
Ans: 352.2℃, 250℃
79

80. Problem 4
An industrial furnace is made of fireclay brick of
thickness 25 cm and thermal conductivity 1
W/m-K. The outside surface is insulated with
material having k=0.05 W/m-K. Determine the
thickness of the insulation layer in order to limit
the heat loss from the furnace wall to 1000
W/m2 when the inside surface of the wall is
1030℃ and the outside surface at 30℃.
Ans: 3.75 cm
80

81. Problem 5
A house has a composite wall of wood(outside), fibreglass
insulation and plaster board (inside) having 10 mm, 100
mm and 20 mm thickness and 0.17, 0.038 and 0.12
thermal conductivities respectively. The inside and
outside temperatures are 20 and -15℃ respectively. The
total area is 350 m2. The outer and inner convective heat
transfer coefficients are 60 and 30 W/m2-K respectively.
Determine the total heat loss through the wall. If the
wind were blowing violently and outer heat transfer
coefficient increases to 300 W/m2-K, find the % rise in
heat loss. What is the controlling resistance?
Ans: 4.2 kW, 0.7%, glass fibre
81

82. Problem 6
Consider a plane composite wall that is composed of two
materials having thermal conductivities ka=0.09 and
kb=0.03 W/m-K and thicknesses La=8 mm and Lb=16 mm.
The contact resistance at the interface between the two
materials is known to be 0.3 m2-K/W. Material A adjoins a
fluid at 200℃ for which h=10 W/m2-K and material B
adjoins a fluid at 40℃ for which h=20 W/m2-K. What
is the rate of heat transfer through a wall that is 2 m
high by 2.5 m wide. Sketch the temperature
distribution.
Ans: 745.9 W
82

83. Problem 7
An uninsulated, thin-walled pipe of 100 mm
diameter is used to transport water to equipment
that operates outdoors and uses the water as a
coolant. During particularly harsh winter conditions,
the pipe wall achieves a temperature of –15℃ and
a cylindrical layer of ice forms on the inner surface
of the wall. If the mean water temperature is 3℃
and a convection coefficient of 2000 W/m2-K is
maintained at the inner surface of the ice, which is
at 0℃, what is the thickness of the ice layer?
Ans: 5 mm
83

84. Problem 8
A plane wall of thickness 0.1 m and thermal
conductivity 25 W/m-K having uniform
volumetric heat generation of 0.3 MW/m3 is
insulated on one side, while the other side is
exposed to a fluid at 92℃. The convection heat
transfer coefficient between the wall and the
fluid is 500 W/m2-K. Determine the maximum
temperature in the wall.
Ans: 212℃
84

85. Module 3
Conduction II

86. Steady State Heat Conduction in 2-D
• Many real life heat transfer problems are 2-D
• Treating them as 1-D will lead to errors
• Hence multi-dimensional effects must be
included
• Solution is difficult compared to 1-D
86

87. Solution Methods of 2-D Heat Transfer
• Analytical Solutions: Mathematically involved
and limited problems only can be solved.
Yields accurate results at any point
• Graphical Solutions: Needs skill and yields
approximate solutions
• Numerical Solutions: Any problem can be
solved and accuracy depends on the fineness
of the grid
87

88. Analytical Solution
88
   
   
   
L
W
n
L
y
n
L
x
n
n
y
x
x,W
; θ
L,y
θ
x,
; θ
,y
θ
y
x
T
T
T
T
y
T
x
T
n
n








sinh
sinh
sin
1
1
2
,
method
separable
variable
Using
1
0
0
0
0
0
;
0
0
1
1
2
2
2
2
1
2
1
2
2
2
2



























89. Graphical Solution
• Graphical methods are suitable only for 2-D
problems
• Only isothermal and adiabatic boundary
conditions are possible
• This method is superseded by computers
• Presently used to obtain a first estimate of the
temperature distribution
89

90. Graphical Solution…
• The underlying principle is that the lines of
constant temperature must be perpendicular
to the lines of heat flow
• The first step is to construct a plot of
isotherms and heat flow lines
• This “Flux Plot” is used to infer the
temperature distribution and heat transfer
90

91. Graphical Solution…
91

92. Steps involved in Graphical Solution
• Identify all relevant lines of symmetry
• Lines of symmetry are heat flow lines and hence
are adiabatic
• Sketch the lines of constant temperature
(isotherms)
• Isotherms must be perpendicular to adiabats
• Heat flow lines (adiabats) must be drawn to
obtain a network of curve-linear squares
• All sides of each square must be of the same
length
92

93. Steps involved in Graphical Solution…
93
2-D Conduction in a Square Channel
2
2
bd
ac
y
cd
ab
x







94. Estimating the Heat Transfer Rate
94
 
 
 
2
1
1
2
1
1
increments
re
temperatu
of
number
total
N
lanes
of
number
M




























T
k
N
Ml
q
y
x
T
N
T
T
x
T
l
y
k
x
T
kA
q
Mq
q
q
N
j
j
j
j
j
i
i
i
M
i
i

95. Evolution of a Flux Plot
95

96. Conduction Shape Factor
• Finding analytical solution for 2-D and 3-D
conduction problems is difficult
• Rapid solutions to some problems may be
obtained by using existing solutions (obtained
from analytical as well as graphical methods)
• These solutions are reported in terms of “Shape
Factors”
• Shape Factor (S) is a dimensionless conduction
heat rate
• It is defined as
96
2
1

 T
Sk
q

97. Conduction Shape Factor…
• By definition,
• From graphical method
• Hence, from graphical method,
97
2
1

 T
Sk
q
2
1

 T
k
N
Ml
q
N
Ml
S 
 
Sk
R D
cond
t
1
2
, 


98. Some Configurations
98

99. Problem 1
A hole of diameter 0.25 m is drilled through the center of
a solid block of square cross section with 1 m on a side.
The hole is drilled along the length, l=2 m, of the block,
which has a thermal conductivity of k=150 W/m-K. A
warm fluid passing through the hole maintains an inner
surface temperature of T1=75°C, while the outer surface
of the block is kept at T2= 25°C. Using the flux plot
method, determine the Shape Factor for the system. Also
estimate the rate of heat transfer through the block.
Ans: S=8 m, Q=64 kW
99

100. Problem 1…
100
Schematic Diagram Flux Plot

101. Problem 2
A structure consists of metal walls 8 cm apart with
insulating material (k=0.12 W/m-K) between. Ribs 4
cm long protrude from one wall every 14 cm. They
can be assumed to stay at the temperature of the
wall. Find the heat flux through the wall if the first
wall is at 40℃ and the one with the ribs is at 0℃.
Find the temperature at the location ‘A’ shown in
the figure.
Ans: Q=10.54 W/m, TA=15℃
101

102. Problem 2…
102

103. Problem 3
Calculate the shape factor for a spherical shell.
Ans:
103
1
2
2
1
4
r
r
r
r
S




104. Problem 4
Radioactive wastes are temporarily stored in a
spherical container, the center of which is buried a
distance of 10 m below the earth’s surface. The
outside diameter of the container is 2 m, and 500
W of heat are released as a result of radioactive
decay. If the soil surface temperature is 20°C, what
is the outside surface temperature of the container
under steady-state conditions? Also, show the
isothermal and heat flow lines.
Ans: 92.7℃
104

105. Numerical Solution
• Analytical solutions may be obtained in only a few
cases
• In most practical cases, such solutions are not possible
• This is due to complex geometries and/or boundary
conditions
• In such cases, “Numerical Techniques” come handy
• Moreover, Numerical Methods can be readily extended
to 3-D problems
• In contrast to analytical solutions, numerical solutions
enable determination of temperature only at discrete
points
105

106. Taylor Series Expansion
106
       
     
x
x
f
x
x
f
dx
x
df
dx
x
f
d
x
dx
x
df
x
x
f
x
x
f













beyond,
and
term
third
Ignoring
...
2
2
2
 2
1
1
2
2
1
1
2
1
2
1
2
2
1
2
1
1
2
1
2
and
x
T
T
T
dx
T
d
x
x
T
T
x
T
T
x
dx
dT
dx
dT
dx
T
d
x
T
T
dx
dT
x
T
T
dx
dT
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
























































107. The Nodal network
107

108. Finite Difference Form of the Heat
Equation
• The heat equation in 2-D with no storage, no
generation and constant thermal conductivity
is
• To obtain a numerical solution, the above
differential equation must be approximated
using a finite-difference form
108
0
2
2
2
2






y
T
x
T

109. Finite Difference Form of the Heat
Equation…
109
 
 2
,
1
,
1
,
,
2
2
2
,
,
1
,
1
,
2
2
,
1
,
,
2
1
,
,
1
,
2
1
,
2
1
,
2
1
,
,
2
2
2
2
;
y
T
T
T
y
T
Similarly
x
T
T
T
x
T
x
T
T
x
T
x
T
T
x
T
x
x
T
x
T
x
T
x
x
T
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m





















































110. Finite Difference Form of the Heat
Equation…
110
0
2
2
2
2






y
T
x
T
   
4
equation
algebraic
an
by
ed
approximat
is
equation
al
differenti
exact
The
0
4
0
2
2
,
1
,
1
1
,
1
,
,
,
,
1
,
1
1
,
1
,
2
,
1
,
1
,
2
,
,
1
,
1
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
n
m
T
T
T
T
T
T
T
T
T
T
y
x
If
y
T
T
T
x
T
T
T

































111. Problem 5
Determine the temperature distribution in the
plate shown below.
111

112. Unsteady State Heat Transfer
• Many heat transfer problems are TIME
DEPENDANT
• Such ‘Unsteady’ or ‘Transient’ problems arise
when boundary conditions change
• Transient state continues until steady state is
attained
• Example: A hot metal billet exposed to cool air
112

113. Unsteady State Heat Transfer…
The problem of Transient heat conduction is to
determine the
• Temporal temperature distribution and
• Heat Transferred during a given interval
Types of Transient Conduction Problems
• Lumped Systems
• Infinite Bodies
• Semi-infinite bodies
113

114. Solutions Methods for Unsteady State
Heat Transfer
Analytical methods:
• Mostly available for Lumped systems and
• 1-D systems or simple 2-D systems
• Mathematically involved
Numerical Methods:
• Applicable for any type of problem
• Easily extendable to multi-dimensional problems
• Knowledge of programming required
114

115. Lumped Systems
• A system where temperature of the solid is
spatially uniform at any instant
• Hence no thermal gradients exist
115
Lumped System Non-Lumped System

116. Lumped Systems…
• Heat conduction in the absence of thermal
gradients implies infinite conductivity
• This ideal (and impossible) situation can be
approximated if the conduction resistance is
low compared to the convective resistance
• The heat conduction equation can’t be used
to solve such problems
116

117. Analysis of Lumped Systems
• Lumped systems can be analysed using energy
balance
• For the system shown
117
st
g
out
in E
E
E
E 


 


 



d
dT
Vc
T
T
hA
d
dE
E
E
E
s
st
out
st
out












118. Analysis of Lumped Systems…
118
 

 





























0
s
s
hA
Vc
hA
Vc
constant)
is
T
(if
d
d
d
d
d
d
d
dT
T
T
d
dT
Vc
T
T
hA
i
s













































Vc
hA
T
T
T
T
e
T
T
T
T
e
Vc
hA
s
i
i
Vc
hA
i
i
Vc
hA
i
s
i
i
s
s
exp
ln
ln
hA
Vc
s

119. Analysis of Lumped Systems…
119
  


 



 
t
s
t
s
t
s
t
dt
hA
dt
hA
dt
T
T
hA
qdt
Q
0
0
0
0
is
t
given time
a
in
ferred
Heat trans



120. Interpretation of Lumped Systems
• The difference between solid and fluid
temperature will decrease exponentially to
zero as time approaches infinity
120

121. Validity of Lumped Systems
121
   
Bi
k
hL
R
R
hA
kA
L
T
T
T
T
T
T
hA
T
T
L
kA
conv
t
cond
t
s
s
s
s
s
s











,
,
2
,
2
,
1
,
2
,
2
,
1
,
1
For low Bi, the uniform temperature
condition is satisfied. Therefore,
lumped systems are valid for Bi<0.1
1
.
0


k
hL
Bi
Object Dimension Characteri
stic length
Plane wall 2L L
Cylinder r r/2
Sphere r r/3

122. Fourier Number
122
 
Fo
Bi
T
T
T
T
Fo
Fo
Bi
L
k
hL
L
c
k
k
hL
cL
h
Vc
hA
A
V
L
Vc
hA
T
T
T
T
i
i
c
c
c
c
c
s
s
c
s
i
i






























exp
time
ess
Dimensionl
length
stic
Characteri
exp
2
2














123. Response Time of a Thermocouple
• A thermocouple should rapidly attain the
temperature of the system it is measuring
• Response time is the time taken to reach
thermal equilibrium
• For rapid response, the term must be small
• This can be achieved by small diameter, low
density, low cp and high h
• is called the Time Constant
123
s
hA
Vc

s
hA
Vc


124. Response Time of a Thermocouple…
• If 𝜏 =
𝜌𝑉𝑐
ℎ𝐴𝑠
• The temperature difference between the body
and the ambient is 36.8% of the initial
temperature difference
• In other words, the temperature difference
would be reduced by 63.2%
124
  368
.
0
1
exp
exp
exp 







 






















Vc
hA
T
T
T
T s
i
i

125. Response Time of a Thermocouple…
• Time required by the thermocouple to reach 63.2% of
the initial temperature difference is called as Sensitivity
of the thermocouple
• Lower value of Time constant results in faster response
• Most thermocouples have the same ρ and cp
• Hence response time is a function of wire diameter
• Most thermocouples have a time constant in the range
of 0.04 to 2.5 s
• Thermocouple reading must be taken after 4 times the
Time constant (i.e. 0.16-10 s)
125

126. Problem 6
An aluminium plate (k=160 W/m-K, ρ=2790
kg/m3, Cp=0.88 kJ/kg-K) of thickness 30 mm and
at a uniform temperature of 225℃ is suddenly
immersed at time t=0 in a well stirred fluid at a
constant temperature of 25℃. The heat transfer
coefficient between the plate and the fluid is 320
W/m2-K. Determine the time required for the
centre of the plate to reach 50℃.
Ans: 4 min
126

127. Problem 7
A person is found dead at 5 PM in a room whose
temperature is 20°C. The temperature of the body
is measured to be 25°C when found, and the heat
transfer coefficient is estimated to be 8 W/m2-K.
Modeling the body as a 30 cm diameter, 1.70 m
long cylinder and using the lumped system analysis
as a rough approximation, estimate the time of
death of that person.
Ans: 4 AM
127

128. Problem 8
The temperature of a gas stream is measured with a
thermocouple. The junction may be approximated
as a sphere of diameter 1 mm, k=25 W/m-K,
ρ=8400 kg/m3 and cp=0.4 kJ/kg-K. The heat
transfer coefficient between the junction and the
gas stream is 500 W/m2-K. How long will it take
for the thermocouple to record 99% of the
applied temperature difference?
Ans: 5.16 s
128

129. Problem 9
A thermocouple junction (k=20 W/m-K, cp=400
J/kg-K, ρ=8500 kg/m3) is used to measure the
temperature of a gas stream. The convection
coefficient between the junction surface and the
gas is 400 W/m2-K. Determine the junction
diameter needed for the thermocouple to have a
time constant of 1 s. If the junction is at 25℃ and is
placed in a gas stream that is at 200℃, how long
will it take for the junction to reach 199℃?
Ans: 0.7 mm, 5.2 s
129

130. Infinite Bodies
Most heat transfer problems can’t be treated as
lumped systems. Hence thermal gradients are
considered in the following geometries
• Plane Wall
• Infinite Cylinder
• Sphere
130

131. Infinite Flat Plate
• In a plane wall, if the thickness is small
compared to the width and height of the wall,
it is treated as an Infinite Plate
131

132. Temperature Distribution in an Infinite
Plate
132
 
 






















T
T
T
T
T
L
T
h
x
T
k
x
T
T
x
T
i
i
L
x
x






*
0
2
2
,
;
0
1
 
   
 
 
   
*
1
2
1
1
*
2
1
*
2
*
cos
exp
:
0.2
Fo
solution
e
Approximat
tan
;
2
sin
2
sin
4
;
cos
exp
:
0
any time,
for
valid
solution
Exact
x
Fo
C
Bi
C
L
Fo
x
Fo
C
Fo
n
n
n
n
n
n
n
n
n
n



























 





T
T
cV
Q
Q
Q
i




0
*
0
1
1
0
sin
1

133. Infinite Cylinder
• An infinite cylinder is one in which the
conduction is in radial direction
• For an infinite cylinder,
133
10

o
r
L

134. Temperature Distribution in an Infinite
Cylinder
134
 
   
 
   
 
 
kind
first
the
of
functions
Bessel
the
are
;
;
2
;
exp
0
Fo
solution
Exact
0
1
0
0
1
2
1
2
0
1
2
0
1
*
0
2
*
,J
J
k
hr
Bi
Bi
J
J
J
J
J
C
r
Fo
r
J
Fo
C
n
n
n
n
n
n
n
n
n
n
n
n






















 
   
*
1
0
2
1
1
*
exp
:
0.2
Fo
solution
e
Approximat
r
J
Fo
C 

 


 
1
1
1
*
0
0
2
1 


J
Q
Q



135. Temperature Distribution in an Infinite
Sphere
135
 
   
   
 
 
;
cot
-
1
;
2
sin
2
cos
sin
4
;
sin
1
exp
0
Fo
solution
Exact
0
2
0
1
*
*
2
*
k
hr
Bi
Bi
C
r
Fo
r
r
Fo
C
n
n
n
n
n
n
n
n
n
n
n
n
n
























   
*
1
*
1
2
1
1
*
sin
1
exp
:
0.2)
(Fo
solution
e
Approximat
r
r
Fo
C 


 


 
1
1
1
3
1
*
0
0
cos
sin
3
1 







Q
Q

136. Coefficient Values for Approximate
Solutions
136

137. Heisler’s Charts
• Heisler’s charts are the graphical
repesentation of the approximate solution
• Charts to determine the centre plane
temperature and at any other plane are
available separately
• Grober charts are used to determine the heat
transferred
137

138. Heisler’s Charts for a Plane Wall
at Midplane
138

139. Heisler’s Charts for a Plane Wall
at any Plane
139

140. Grober Charts to Estimate Heat
Transfer in a Plane Wall
140

141. Problem 10
An iron plate (k=60 W/m-K, cp=0.46 kJ/kg-K, ρ=7850
kg/m3, and α=1.6×10-5 m2/s) of 50 mm thickness is
initially at 225℃. Suddenly, both surfaces are exposed
to an ambient temperature of 25℃ with a heat transfer
coefficient of 500 W/m2-K. Calculate (a) the centre
temperature at 2 minutes after the start of cooling, (b)
the temperature at a depth of 1 cm from the surface at
2 minutes after the start of cooling and (c) the energy
removed from the plate per unit area during this time.
Ans: 143℃, 137.1℃, 15.17 MJ
141

142. Problem 11
Estimate the time required to cook a hot dog in
boiling water. Assume that the hot dog is initially
at 6℃, that the convection heat transfer
coefficient is 100 W/m2-K, and that the final
temperature is 80℃ at the centerline. Treat the
hot dog as a long cylinder of 20 mm diameter
having the properties: ρ=880 kg/m3, cp=3350
J/kg-K, and k=0.52 W/m-K.
Ans: 7.4 minutes
142

143. Problem 12
A spherical hailstone that is 5 mm in diameter is
formed in a high-altitude cloud at -30℃. If the
stone begins to fall through warmer air at 5℃ , how
long will it take before the outer surface begins to
melt? What is the temperature of the stone’s center
at this point in time, and how much energy (J) has
been transferred to the stone? A convection heat
transfer coefficient of 250 W/m2-K may be
assumed, and the properties of the hailstone may
be taken to be those of ice.
Ans: 15 s, -0.81℃, 2.6 J
143

144. Semi-Infinite Bodies
• A solid that extends to infinity in all but one
direction is a semi infinite solid
• Analytical solutions may be obtained for them
• If a sudden change of conditions is imposed
on a surface, transient 1-D conduction will
occur
• The solutions for infinite solids are valid but
several terms must be considered
• Simpler solutions are available
144

145. Semi-Infinite Bodies
145

146. Semi-Infinite Bodies
146
Constant Surface
Temperature
Constant
Surface Heat flux
Surface
Convection

147. Temperature Distribution in a
Semi-Infinite Solid
147
 
 
   
 
 
 
 















































































 























k
h
erf
k
h
k
hx
erf
T
T
T
x
T
T
T
h
x
T
k
erf
k
x
q
x
k
q
T
x
T
q
q
T
T
k
q
erf
T
T
T
x
T
T
T
i
i
i
s
i
s
s
s
i
s
s
















2
x
1
exp
2
x
1
,
,
0
:
Convection
Surface
2
x
1
4
exp
2
,
:
Flux
Heat
Surface
Constant
2
x
,
,
0
:
e
Temperatur
Surface
Constant
2
2
"
0
2
2
/
1
"
0
"
0
"
"

148. Graphical Solution for Convective
Conditions
148

149. Problem 13
A thick steel slab ( ρ=7800 kg/m3, cp=480 J/kg-K, k=
50 W/m-K) is initially at 300℃ and is cooled by
water jets impinging on one of its surfaces. The
temperature of the water is 25℃, and the jets
maintain an extremely large, approximately uniform
convection coefficient at the surface. Assuming that
the surface is maintained at the temperature of the
water throughout the cooling, how long will it take
for the temperature to reach 50℃ at a distance of
25 mm from the surface?
Ans: 30 minutes
149

150. 150

151. Problem 14
A very thick slab with thermal diffusivity 5.6 ×
10-6 m2/s and thermal conductivity 20 W/m-K is
initially at a uniform temperature of 325℃.
Suddenly, the surface is exposed to a coolant at
15℃ for which the convection heat transfer
coefficient is 100 W/m2-K. Determine
temperatures at the surface and at a depth of 45
mm after 3 min have elapsed.
Ans: 276℃, 313℃
151

152. 152

153. Problem 15
In areas where the air temperature remains below 0°C for
prolonged periods of time, the freezing of water in underground
pipes is a major concern. Fortunately, the soil remains relatively
warm during those periods, and it takes weeks for the
subfreezing temperatures to reach the water mains in the
ground. Thus, the soil effectively serves as an insulation to
protect the water from subfreezing temperatures in winter.
The ground at a particular location is covered with snow pack at
-10°C for a continuous period of three months, and the average
soil properties at that location are k = 0.4 W/m·K and α= 0.15 ×
10-6 m2/s. Assuming an initial uniform temperature of 15°C for
the ground, determine the minimum burial depth to prevent the
water pipes from freezing.
Ans: 82 cm
153

154. Module 4
Convection I

155. Convection Heat Transfer
• Convective heat transfer occurs whenever there
is a fluid moving and thermal gradients exist
• Convective heat transfer can occur in forced or
free convection mode
• A major problem in the study of convective heat
transfer is to determine the Heat Transfer
Coefficient
• A fair understanding of Fluid Mechanics is
necessary to study convective heat transfer
155

156. Conservation of Mass
156
   
0
:
Flow
ible
Incompress
D
-
1
0
:
flow
ible
Incompress
0
:
flow
le
Compressib















x
u
y
v
x
u
y
v
x
u 


157. Conservation of Momentum
157



























































































































2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
z
w
y
w
x
w
g
y
P
z
w
w
y
w
v
x
w
u
t
w
z
v
y
v
x
v
g
y
P
z
v
w
y
v
v
x
v
u
t
v
z
u
y
u
x
u
g
x
P
z
u
w
y
u
v
x
u
u
t
u
z
y
x









:
equation
Stokes
-
Navier
the
of
component
z
:
equation
Stokes
-
Navier
the
of
component
y
:
equation
Stokes
-
Navier
the
of
component
x

158. Conservation of Energy
158

159.  

















































































































2
2
2
2
Generation
Heat
n
Dissipatio
Viscous
energy
Thermal
and
Mechanical
between
Conversion
Energy
Conducted
Energy
Convected
3
2
2
y
v
x
u
y
v
x
u
x
v
y
u
q
y
v
x
u
P
y
T
k
y
x
T
k
x
y
e
v
x
e
u




 






 



 


 

 

Conservation of Energy…

160. Boundary Layer
• The Boundary Layer is a thin fluid layer in
which fluid velocity gradients and shear
stresses are large
• Outside the boundary layer, the velocity
gradients and shear stresses are negligible
• Boundary layer thickness is where u=0.99u
160

161. Velocity Boundary Layer
161
0
2
Stress
Shear
2
t
Coefficien
Friction




y
s
s
f
dy
du
u
C





162. Thermal Boundary Layer
162
99
.
0
where
kness
Layer thic
Boundary
Thermal t 




T
T
T
T
y
s
s

 







 T
T
h
y
T
k
q s
y
f
s
0
"







T
T
y
T
k
h
s
y 0

163. Some Inferences from the Thermal
Boundary Layer (TBL)
• Conditions in the TBL influence the wall
thermal gradient
As x increases
• t increases
• Thermal gradients decrease
• Heat flux decreases
• Heat transfer coefficient (h) decreases
163

164. Local and Average
Heat Transfer Coefficient
164
 
 















L
s
A
s
s
s
s
A
s
s
A
hdx
L
h
dA
h
A
h
T
T
A
h
q
dA
h
T
T
dA
q
q
s
s
s
0
1
distance.
only with
h varies
plate,
flat
a
For
1
"

165. Laminar and Turbulent Flow
165
5
, 10
5
Re
:
Re
Critical
Re
:
Re
Local










c
c
x
x
x
u
x
u

166. Some Dimensionless Numbers
• Reynolds Number (Re): It is a ratio of inertia
forces to viscous forces
• Prandtl Number (Pr): It is a ratio of
momentum diffusivity to thermal diffusivity
• Nusselt Number (Nu): It is a ratio of
convection heat transfer to conduction heat
transfer
• Grashof Number (Gr): It is a ratio of buoyancy
forces to viscous forces
166

167. Experimental Determination of
Heat Transfer Coefficient
167

168. Forced Convection
External Laminar Flow Over Isothermal
Flat Plate
168
     
3
/
1
2
/
1
2
/
1
2
,
,
3
/
1
Pr
Re
332
.
0
Re
664
.
0
2
Pr
Re
5
x
x
x
x
x
s
x
f
t
x
k
x
h
Nu
u
C
x
x
x
x














169. External Laminar Flow over
Isothermal Flat Plate…
169
 
100
Pe
Pr
0468
.
0
1
Pr
Re
3387
.
0
:
Pr
all
for
n
correlatio
single
A
100
Pe
0.05,
Pr
564
.
0
Pr
Re
564
.
0
:
metals
liquid
For
x
4
/
1
3
/
2
3
/
1
2
/
1
x
2
/
1
2
/
1





















x
x
x
x
x
Nu
Pe
Nu

170. Turbulent Flow Over
Isothermal Flat Plate
170
60
Pr
0.6
Pr
Re
0296
.
0
Re
37
.
0
10
Re
Re
Re
0592
.
0
3
/
1
5
/
4
5
/
1
8
c
x,
5
/
1
,









x
x
x
x
x
x
f
Nu
x
C


171. Other Flat Plate Conditions
171
9
/
1
10
/
9
0
3
/
1
4
/
3
0
1
:
Flow
Turbulent
1
:
Flow
Laminar
:
length
starting
unheated
With


































x
Nu
Nu
x
Nu
Nu
x
x
x
x




60
Pr
0.6
Pr
Re
0308
.
0
:
Flow
Turbulent
0.6
Pr
Pr
Re
453
.
0
:
Flow
Laminar
:
flux
heat
constant
with
plate
Flat
3
/
1
5
/
4
3
/
1
2
/
1





x
x
x
x
Nu
Nu

172. Methodology for Convection
Calculation
• Identify the flow geometry
• Find the reference temperature and evaluate
the properties needed at that temperature
• Calculate Re
• Decide whether local or average values are
needed
• Select the appropriate correlation
172

173. Problem 1
Water flows at a velocity of 1 m/s over a flat
plate of length 0.6 m. The water is at 300 K.
Experiments show that the local convection
coefficients are given by:
Determine the average convection heat transfer
coefficient over the entire plate.
Ans: 1675 W/m2-K
173
    2
.
0
5
.
0
2330
395 


 x
x
h
x
x
h turb
lam

174. 174

175. 175

176. 176

177. Problem 2
Nitrogen gas at 0℃ is flowing over a 1.2 m
long, 2 m wide plate maintained at 80℃ with a
velocity of 2.5 m/s. For nitrogen, ρ=1.142
kg/m3, cp=1.04 kJ/kg-K, ν=15.63×10-6 m2/s
and k=0.0262 W/m-K. Find the average heat
transfer coefficient and the total heat
transferred from the plate.
Ans: 5.65 W/m2-K, 1.08 kW
177

178. Problem 3
Consider a rectangular fin that is used to cool a
motorcycle engine. The fin is 0.15 m long and at
a temperature of 250℃, while the motorcycle is
moving at 80 km/h in air at 27℃. The air is in
parallel flow over both surfaces of the fin, and
turbulent flow conditions may be assumed to
exist throughout. What is the rate of heat
removal per unit width of the fin?
Ans: 5.88 kW/m width
178

179. External Cross Flow over Cylinder
179
0


 
x
u
0


 
x
u
0
0




y
y
u

180. External Flow over Cylinder…
180









2
Re
2
V
A
F
C
VD
VD
f
D
D





181. Correlations for Flow Over Cylinder
181
 
2
re.
temperatu
film
at the
evaluated
be
must
properties
the
above,
equations
For the
000
,
282
Re
1
Pr
4
.
0
1
Pr
Re
62
.
0
3
.
0
0.2
Pr
Re
all
for
Bernstein,
and
Churchill
by
proposed
is
equation
ive
comprehens
single
A
in tables.
found
be
to
m
and
C
of
Values
0.7
Pr
for
Pr
Re
5
/
4
8
/
5
4
/
1
3
/
2
3
/
1
2
/
1
D
3
/
1
w
f
D
D
D
m
D
T
T
T
Nu
C
k
hD
Nu








































182. External Flow over Sphere
182
 
3
/
1
2
/
1
s
4
/
1
4
.
0
3
/
2
2
/
1
D
Pr
Re
6
.
0
2
droplets,
liquid
falling
freely
For
T
at
found
be
to
properties
all
,
Except
Pr
Re
06
.
0
Re
4
.
0
2
5
.
0
Re
Re
24
D
D
s
D
D
D
D
D
Nu
Nu
C




















183. Problem 4
A circular pipe of 37.5 mm outside diameter is
placed in an airstream at 30℃ and 1 atm
pressure. The air moves in cross flow over the
pipe at 10 m/s, while the outer surface of the
pipe is maintained at 95℃. What is the drag
force exerted on the pipe per unit length? What
is the rate of heat transfer from the pipe per
unit length?
Ans: 2.18 N/m, 456 W/m
183

184. 184

185. 185

186. 186

187. Problem 5
A spherical, underwater instrument pod used to make
soundings and to measure conditions in the water has a
diameter of 85 mm and dissipates 300 W.
(a) Estimate the surface temperature of the pod when
suspended in a bay where the current is 1 m/s and the
water temperature is 15℃.
(b) Inadvertently, the pod is hauled out of the water and
suspended in ambient air without deactivating the power.
Estimate the surface temperature of the pod if the air
temperature is 15℃ and the wind speed is 1 m/s.
Ans: 18.7℃, 973℃
187

188. Tutorial Problem 1
Air at 20 ℃ and pressure of 1 bar is flowing over a flat plate at a velocity of 3
m/s. If the plate is 280 mm wide and at 56 ℃, calculate the following
quantities at x=280 mm, given that properties of air at the bulk mean
temperature of 38 ℃ are:
ρ=1.1314 kg/m3, k=0.02732 W/m-K, Cp=1.005 kJ/kg-K, v=16.76810-6 m2/s,
Pr=0.7.
• Boundary layer thickness
• Local friction coefficient
• Average friction coefficient
• Shear stress due to friction
• Thermal boundary layer thickness
• Local convective heat transfer coefficient
• Average convective heat transfer coefficient
• Rae of heat transfer by convection
• Total drag force on the plate
188

189. 189

190. 190

191. 191

192. Tutorial Problem 2
Approximating a human body as a cylinder of
0.3 m dia and 1.75 m long, at surface
temperature of 30°C exposed to winds at 15
kmph at 10°C, determine the rate of heat loss.
192

193. 193

194. 194

195. Tutorial Problem 3
A long 10-cm-diameter steam pipe whose
external surface temperature is 110°C passes
through some open area that is not protected
against the winds. Determine the rate of heat
loss from the pipe per unit of its length. When
the air is at 1 atm pressure and 10°C and the
wind is blowing across the pipe at a velocity of 8
m/s.
195

196. 196

197. External Flow over Bank of Tubes
• Relevant in steam generation in boiler, air cooling in
the coil of an air conditioner etc.
• One fluid flows over the tubes while another flows
inside them
197

198. Tube Bank Arrangement
198
(a) Aligned (b) Staggered

199. Flow Pattern in Tube Banks
199
Aligned Tubes
Staggered Tubes

200. Correlations for Heat Transfer in
Tube Banks
200
tables.
from
obtained
be
can
C
and
C
of
Values
rows,
10
than
less
For
7
.
0
Pr
10
000
,
40
Re
2000
V
at
D
on
based
Re
Re
Re
13
.
1
fluids,
other
For
7
.
0
Pr
10
000
,
40
Re
2000
V
at
D
on
based
Re
Re
Re
:
equation
Grimson
:
rows
more
or
10
of
composed
bundles
tube
across
flow
air
For
1
)
10
(
1
)
10
(
m
max
,
max
m
max
,
m
max
,
m
max
,
max
m
max
,
m
max
,

 








































N
D
N
D
D
D
D
D
D
D
D
D
u
N
C
u
N
N
C
u
N
N
C
u
N

201. Correlations for Heat Transfer in
Tube Banks…
201




























6
max
D,
4
/
1
36
.
0
max
,
1
10
2
Re
0
1
500
Pr
7
.
0
rows
tube
of
Number
,
20
Pr
Pr
Pr
Re
:
L
L
s
m
D
D
N
N
C
u
N
Equation
Zukauskas

 D
V
D
max
max
,
Re 
 
 
2
/
1
2
2
max
max
2
2
2
2
Else,
2
2
2
If
:
t
arrangemen
Transverse








































t
l
D
D
l
t
t
l
t
t
S
S
S
V
D
S
S
V
V
D
S
S
V
D
S
S
D
S
V
D
S
S
V
T
T


max
:
t
arrangemen
Inline

202. Heat Transferred in Bank of Tubes
• Since the fluid may experience large changes
in temperature, using ΔT in the Newton’s law
of cooling will over estimate the heat
transferred
• Instead, use ΔTlm, which is the log mean
temperature difference
202

203. Heat Transferred in Bank of Tubes…
203
   
o
s
i
s
o
s
i
s
lm
lm
T
T
T
T
T
T
T
T
T
T
hA
Q









ln
 
lm
p
T
T
i
s
o
s
πDΔT
h
N
q'
N
c
S
VN
h
DN
T
T
T
T















:
length
unit
per
ferred
Heat trans
row
each
in
tubes
of
Number
N
bank
in the
tubes
of
Number
exp
T



204. Pressure Drop
204
factor
Friction
:
factor
Correction
:
2
2
f
f
V
N
P L


 










205. Pressure Drop (using correlations)…
205
0.16
-
max
08
.
1
1
.
0
max
13
.
1
43
.
0
14
.
0
0
2
max
Re
118
.
0
25
.
0
:
t
arrangemen
staggered
For
Re
08
.
0
044
.
0
:
t
arrangemen
inline
For
09
.
2



















 























 


























d
d
S
f
d
d
S
d
S
f
g
N
fG
P
l
S
d
l
t
b
w
t



 
rature
mean tempe
bulk
at
fluid
the
of
viscosity
dynamic
re
temperatu
at wall
fluid
the
of
viscosity
dynamic
rows
e
transvers
of
number
system
SI
in
1
s
kg/m
velocity,
mass
b
w
0
max
2
max










N
F
ma
g
g
ρV
G
c

206. Problem 6
A preheater involves the use of condensing steam
at 100℃ on the inside of a bank of tubes to heat air
that enters at 1 atm and 25℃. The air moves at 5
m/s in cross flow over the tubes. Each tube is 1 m
long and has an outside diameter of 10 mm. The
bank consists of 196 tubes in a square, aligned array
for which ST = SL =15 mm. What is the heat transfer
coefficient? What is the amount of heat
transferred?
Ans: 189 W/m2-K, 54.1 kW
206

207. Problem 7
Consider a staggered arrangement for which the tube
outside diameter is 16.4 mm and the longitudinal and
transverse pitches are SL = 34.3 mm and ST = 31.3 mm.
There are seven rows of tubes in the airflow direction and
eight tubes per row. Under typical operating conditions
the cylinder surface temperature is at 70℃, while the air
upstream temperature and velocity are 15℃ and 6 m/s,
respectively. Determine the air-side convection
coefficient and the rate of heat transfer for the tube
bundle. What is the air-side pressure drop?
Ans: 157.4 W/m2-K, 22.13 kW/m, 50 Pa
207

208. 208

209. 209

210. 210

211. 211

212. 212

213. 213

214. Internal Flow Through Circular Pipes
• In internal flow, fluid is confined by the tubes
• In contrast to external flow, the boundary
layer development is limited within the
physical boundaries
• Internal flows are involved in heating/cooling
of fluids in chemical processes, environmental
control, energy conversion technologies, etc.
214

215. Velocity Boundary Layer in a
Circular Pipe
215
2300
Re
Re
, 

c
D
m
D
D
u


 

























2
2
1
4
1
o
o
r
r
r
dx
dp
r
u


216. Thermal Boundary Layer in a
Circular Pipe
216

217. Heat Transfer Correlations for
Laminar Internal Flow
217
66
.
3
Nu
:
re
temperatu
surface
constant
with
flow
developed
fully
laminar,
For
36
.
4
Nu
:
flow
developed
fully
laminar,
flux,
heat
constant
with
ube,
circular t
For
D
D




k
hD
k
hD

218. Heat Transfer Correlations for
Turbulent Internal Flow
218
 
 
10
000
,
10
Re
160
Pr
6
.
0
3
0
4
0
Pr
Re
023
.
0 5
/
4















D
L
T
T
ng
for cooli
.
n
T
T
ng
for heati
.
n
Nu
D
m
s
m
s
n
D
D



























10
000
10
Re
700
16
Pr
7
0
Pr
Re
027
.
0
14
.
0
3
/
1
5
/
4
D
L
,
,
.
Nu D
s
D
D


For large property variations:

219. Problem 8
Water is heated in an economiser (under
pressure) from 40°C to 160°C. The tube wall is at
360°C. Determine the length of 0.05 m dia tube,
if the flow velocity is 1 m/s.
Ans: 3.9 m
219

220. 220

221. 221

222. 222

223. Problem 9
Steam condensing on the outer surface of a
thin-walled circular tube of diameter D=50 mm
and length L=6 m maintains a uniform outer
surface temperature of 100℃. Water flows
through the tube at a rate of m=0.25 kg/s, and
its inlet and outlet temperatures are Tm,i=15℃
and Tm,o=57℃. What is the average convection
coefficient associated with the water flow?
Ans: 755 W/m2-K
223

224. 224

225. Problem 10
Hot air flows with a mass rate of m=0.050 kg/s
through an uninsulated sheet metal duct of
diameter D=0.15 m, which is in the crawlspace of a
house. The hot air enters at 103℃ and, after a
distance of L=5 m, cools to 85℃. The heat transfer
coefficient between the duct outer surface and the
ambient air at T=0℃ is known to be ho=6 W/m2 K.
Calculate the heat loss (W) from the duct over the
length L. Also, determine the heat flux and the duct
surface temperature at x=L.
Ans: 908 W, 337 W/m2-K, 56.2℃
225

226. Module 5
Convection II
226

227. Natural Convection
• Natural convection is the mode of convective
heat transfer in the absence of forced velocity
• Also known as Free Convection, occurs when a
body acts on a fluid in which there are density
gradients
• The net effect is buoyancy force, which induces
free currents
• In most cases, density gradients are due to
thermal gradients and body force is due to gravity
227

228. Importance of Natural Convection
• Free convection involves low velocities and
hence lower heat transfer coefficients
• Systems involving multimode heat transfer will
have the largest resistance offered by free
convection
• In systems where heat transfer and/or
operational costs must be minimized, free
convection is preferred
228

229. Applications of Natural Convection
• Heat dissipation in Electronics
• Estimating heating and ventilation in Buildings
• Dispersion of combustion products
• Environmental sciences
• Oceanic and atmospheric movements
229

230. Conditions for Free Convection
230
T1<T2
Heat Transfer Mostly
by Free Convection
T1>T2
Heat Transfer Mostly
by Conduction

231. Flow Over Vertical Plate
231
Velocity Boundary Layer Thermal Boundary Layer

232. Laminar Free Convection on Vertical
Surface
232
 
 
forces
viscous
the
to
forces
buoyamcy
the
of
ratio
the
of
measure
a
is
number,
Grashof
the
is
Gr
1
gases,
ideal
For
book.
data
HMT
of
38
page
in
Listed
t.
coefficien
expansion
thermal
c
volumetri
the
is
1
Pr
4
3
4
2
3
4
/
1
T
T
L
T
T
g
Gr
g
Gr
k
L
h
u
N
P
s
L
L
L





























9
10
Pr
flow,
laminar
For 

 Gr
Ra

233. Boussinesq Approximation
233
   
only
s
variation
re
temperatu
to
due
are
variations
density
when the
originates
ion
approximat
Boussinesq
ion
approximat
Boussinesq
-
-
-
-
-
-
1
1
1

























T
T
T
T
T
T P












234. Laminar and Turbulent Flows
234
  9
3
,
,
9
10
Pr
10
number
Rayleigh
for
turbulent
becomes
Flow







 x
T
T
g
Gr
Ra s
cr
x
cr
x

235. Correlations for Vertical Plate
235
     
     
9
9
4
16
9
4
1
2
27
8
16
9
6
1
L
13
9
3
/
1
3
/
1
9
4
4
/
1
4
/
1
10
for
Pr
492
0
1
67
0
68
0
flow
laminar
in
accuracy
better
slightly
a
For
Pr
492
0
1
387
0
825
0
,
Ra
all
For
10
10
flow
ent
for turbul
1
.
0
Pr
1
.
0
10
10
flow
laminar
for
59
.
0
Pr
59
.
0





























































L
/
/
/
L
L
/
/
/
L
L
L
L
L
L
L
L
L
L
Ra
.
Ra
.
.
Nu
.
Ra
.
.
Nu
Ra
Ra
Gr
Nu
Ra
Ra
Gr
Nu
4
/
1
35
L
D
if
cylinders
al
for vertic
applicable
ns
correlatio
Same
L
Gr


236. Flow Over Horizontal Plate
236
Cold Plate: Ts<T
Hot Plate: Ts>T

237. Correlations for Horizontal Plate
237
0.7)
Pr
,
10
(10
52
.
0
Nu
:
plate
cold
of
surface
upper
or
plate
hot
of
surface
Lower
Pr)
all
,
10
(10
15
.
0
Nu
0.7)
Pr
,
10
(10
54
.
0
Nu
:
plate
cold
of
surface
lower
or
plate
hot
of
surface
Upper
9
4
5
/
1
L
11
7
3
/
1
L
7
4
4
/
1
L











L
L
L
L
L
L
Ra
Ra
Ra
Ra
Ra
Ra

238. Flow Over Inclined Plate
238
Ts<T
Ts>T
For 060, replace Gr
with Gr cos  in
correlations for vertical
plate

239. Flow Over Cylinders
239
 
2
27
/
8
16
/
9
6
/
1
D
12
Pr
559
.
0
1
387
.
0
6
.
0
Nu
:
)
10
(Ra
Ra
of
range
wide
a
for
n
correlatio
single
A
Pr
:
cylinders
horizontal
long
For





































D
m
D
D
Ra
Gr
C
Nu
For vertical
cylinders, see
correlations for
vertical plate

240. Flow Over Spheres
240
11
9
/
4
16
/
9
4
/
1
10
7
.
0
Pr
Pr
469
.
0
1
589
.
0
2
:
spheres
For



















D
D
D
Ra
Ra
Nu

241. Combined Free and Forced Convection
241
equation
above
the
using
combined
and
separately
estimated
be
must
convection
free
and
forced
for the
Nu
3
n
flows
opposing
for
sign
flows
e
transvers
and
assisting
for
sign
1
Re
Gr
hen
relevant w
are
convection
free
and
forced
both
Hence
1
Re
Gr
when
negligible
is
convection
Forced
1
Re
Gr
when
negligible
is
convection
Natural
2
2
2







-
Nu
Nu
Nu n
N
n
F
n

242. Problem 1
A glass door fire screen, used to reduce
exfiltration of room air through a chimney, has a
height of 0.71 m and a width of 1.02 m and
reaches a temperature of 232℃. If the room
temperature is 23℃, estimate the convection
rate from the fire place to the room.
Ans: 1060 W
242

243. 243

244. 244

245. 245

246. Problem 2
A square Aluminium plate 5 mm thick and 200
mm on a side is heated while vertically
suspended in quiescent air at 40℃. Determine
the average heat transfer coefficient for the
plate when its temperature is 15℃, by using an
empirical correlation.
Ans: 4.51 W/m2-K
246

247. Problem 3
An electrical heater in the form of a horizontal
disk of 400 mm diameter is used to heat the
bottom of a tank filled with engine oil at a
temperature of 5℃. Calculate the power
required to maintain the heater surface
temperature at 70℃.
Ans: 468 W
247

248. 248

249. 249

250. 250

251. Problem 4
Beverage in cans 150 mm long and 60 mm in
diameter is initially at 27℃ and is to be cooled
by placement in a refrigerator compartment at
4℃. In the interest of maximizing the cooling
rate, should the cans be laid horizontally or
vertically in the compartment. As a first
approximation, neglect the heat transfer from
the ends.
Ans: Makes no difference
251

252. 252

253. 253

254. 254

255. Problem 5
A sphere of 25 mm diameter contains an
embedded electrical heater. Calculate the power
required to maintain the surface temperature at
94℃ when the sphere is exposed to a quiescent
medium at 20℃ for (a) air at atmospheric
pressure, (b) water, and (c) ethylene glycol.
Ans: (a) 1.55 W, (b) 187 W, (c) 57 W
255

256. Boiling and Condensation
• Boiling and condensation involve fluid motion.
Hence they are classified under convection
• They are phase change processes.
• Large heat transfer rates can be achieved with
small temperature difference
• The surface tension and density difference
between the two phases induces buoyancy
• Latent heat and buoyance driven flow results
in high heat transfer coefficients
256

257. Boiling
• Boiling is evaporation at a solid-liquid interface
• Pool Boiling: The liquid is quiescent and its motion
near the surface is due to free convection and due to
mixing induced by bubble growth and detachment
• Forced Convection Boiling: Fluid motion is induced by
external means
• Subcooled Boiling: Most of the liquid is below
saturation temperature and bubbles may condense in
the liquid
• Saturated Boiling: Liquid temperature is slightly
greater than saturated temperature
257

258. Pool Boiling
258

259. Boiling Curve
259
e
s
sat
s
e
T
h
q
T
T
T
Δ
Δ
e
Temperatur
Excess
"





260. Modes of Pool Boiling
• Free Convection Boiling
• Nucleate Boiling
• Transition Boiling
• Film Boiling
260

261. Modes of Pool Boiling…
261

262. Free Convection Boiling
• Exists when ΔTeΔTe,A, where ΔTe,A,≈5℃
• As ΔTe increases, bubble formation will
occur
• Below point A (ONB: Onset of Boiling), fluid
motion is by free convection
• For large horizontal plate, fluid flow is
turbulent. The following correlation is valid
262
Pr)
all
,
10
(10
15
.
0
Nu 11
7
3
/
1
L 

 L
L Ra
Ra

263. Nucleate Boiling
• Most preferred regime to operate devices
• Occurs when ΔTe,AΔTeΔTe,C where
ΔTe,C≈30℃
263

264. Transition Boiling
• Also known as Unstable film boiling or Partial film
boiling
• Occurs when ΔTe,CΔTeΔTe,D where
ΔTe,D≈120℃
• Bubble formation is rapid and a vapour film
begins to form on the surface
• As thermal conductivity of vapour is much less
than liquid, heat transfer coefficient and hence
heat flux decreases with increasing ΔTe
264

265. Film Boiling
• Occurs when ΔTeΔTe,D
265

266. Critical heat Flux
• The maximum heat flux is called as the Critical
Heat Flux
• It is greater than 1 MW/m2 for water
• At this point, large amount of vapour is
formed and hence the liquid does not wet the
surface
• Hence, devices may be operated close to this
point but not reaching it
• The value strongly depends on pressure
266

267. Correlations in Boiling
267
 
 
 
 
sat
s
sat
s
rad
rad
conv
conv
rad
/
rad
/
conv
/
/
v
l
v
l
v
fg
"
/
v
v
l
v
fg
"
T
T
T
T
h
h
h
h
h
h
h
h
h
h
ρ
ρ
ρ
σg
ρ
Ch
q
ρ
ρ
ρ
σg
ρ
Ch
q






















 

4
4
3
1
3
4
3
4
4
1
2
min
4
1
2
max
4
3
:
used
be
can
equation
simpler
a
If
:
Boiling
Pool
Film
:
Flux
Heat
Minimum
:
Flux
heat
Critical


geometry
on
depends
constant,
Zuber

C

268. Condensation
• Condensation occurs when vapour comes into
contact with a cool surface
• Latent heat is released and condensate
formed
• It can take place in several forms
268

269. Modes of Condensation
269
Direct Contact
Condensation
Homogenous
Condensation
Surface
Condensation

270. Film and Dropwise Condensation
• Film condensation is common and is
characteristic of clean surfaces
• Dropwise condensation occurs if surfaces are
coated with substances that inhibit wetting
• Dropwise condensation gives high heat
transfer rates and hence high condensation
• Hence commonly, coatings of silicon, teflon,
waxes, fatty acids are provided
270

271. Correlations in Condensation
271
 
 
 
25
.
0
2
3
25
.
0
2
3
25
.
0
2
3
728
.
0
tubes
N
of
Bank
)
(
728
0
Tubes
Horizontal
)
(
943
0
Surfaces
Vertical
(i)
:
on
condensati
Film






























s
v
l
fg
s
v
l
fg
s
v
l
fg
T
T
ND
μ
gh
ρ
k
h
iii
T
T
D
μ
gh
ρ
k
.
h
ii
T
T
L
μ
gh
ρ
k
.
h
 
C
T
h
C
T
C
C
T
h
o
sat
dc
o
sat
o
o
sat
dc
100
255510
100
22
2044
51104
surfaces,
copper
on
condensing
steam
For
:
on
Condensati
Dropwise







272. Problem 6
The bottom of a copper pan, 0.3 m in diameter,
is maintained at 118℃ by an electrical heater.
Estimate the power required to boil water in this
pan. What is the evaporation rate? Estimate the
critical heat flux.
Ans: 59.1 kW, 94 kg/h, 1.52 MW/m2
272

273. 273

274. 274

275. 275

276. 276

277. Problem 7
A metal clad heating element of 6 mm diameter
and emissivity ε=1 is horizontally immersed in a
water bath. The surface of the metal is 225℃
under steady state boiling conditions. Estimate
the power dissipation per unit length of the
heater.
Ans: 742 W/m
277

278. Problem 8
Saturated steam at 1 atm condenses on the
outer surface of a vertical, 100 mm dia pipe 1 m
long, having a uniform surface temperature of
94℃. The average heat transfer coefficient is
8500 W/m2-K. Estimate the total condensation
rate and the heat transfer rate to the pipe.
Ans: 7.06 g/s, 16 kW
278

279. 279

280. 280

281. Problem 9
Saturated steam at 0.1 bar condenses with a
convection coefficient of 6800 W/m2-K on the
outside of a brass tube having inner and outer
diameters of 16.5 and 19 mm, respectively. The
convection coefficient for water flowing inside
the tube is 5200 W/m2-K. Estimate the steam
condensation rate per unit length of the tube
when the mean water temperature is 30℃.
Ans: 0.516 g/s
281

282. Module 6
Radiation

283. Radiation Heat Transfer
• Unlike conduction and convection, radiation
heat transfer requires no medium
• It is an important and interesting mode of
heat transfer
• Relevant to many industrial heating, cooling
and drying processes
• Also finds application in energy conversion
methods involving fossil fuels and solar
radiation
283

284. Radiation Cooling of a Hot Solid
284

285. Spectrum of Electro-Magnetic
Radiation
285
Short wave length radiation is of interest to physicist.
Long wave length radiation is of interest to electrical
engineer.

286. Terminology and Laws
• Emissive Power (E): The rate at which radiation is
emitted from a surface per unit area, over all
wave lengths and in all directions.
• Irradiation (G): The rate at which radiation is
incident upon the surface per unit surface area
over all wave lengths and from all directions
• Radiosity (J): Rate at which radiation leaves a
surface per unit area
286

287. Terminology and Laws
287
Semi-
transparent
medium
Opaque
Medium

288. Radiation in a Semi-Transparent
Medium
• On a Semi-Transparent medium, radiation can be
reflected, absorbed and transmitted
• Reflectivity (ρ): It is the fraction of irradiation that is
reflected
• Absorptivity (α): Fraction of irradiation that is
absorbed
• Transmissivity (τ): Fraction of irradiation that is
transmitted
ρ+α+τ=1
For opaque surface: ρ+α=1
288

289. Blackbody
• A blackbody absorbs all radiation irrespective
of wavelength and direction
• For a prescribed temperature and wavelength,
no surface can emit more energy than a
blackbody
• Although blackbody radiation is a function of
wavelength and temperature, it is
independent of direction. Hence blackbody is
a diffuse emitter
289

290. Characteristics of an Isothermal
Blackbody
290
Complete Absorption Diffuse Emission Diffuse irradiation

291. Planck Distribution
291
   
K
m
k
hc
C
m
m
W
hc
C
T
C
C
T
I
T
E
s
m
c
K
J
k
S
J
.
h
T
k
hc
hc
I
b
b
b
b
-
b
b































































4
0
2
2
4
8
2
0
1
2
5
1
,
,
8
0
23
34
0
2
2
0
,
10
439
.
1
/
10
742
.
3
2
1
exp
,
,
law
s
Planck'
or
on
distributi
Planck
:
power
emissive
Spectral
/
10
2.998
in vacuum
light
of
Speed
/
10
381
.
1
constant
Boltzmann
10
626
6
constant
Planck
Universal
1
exp
2
:
intensity
spectral
Blackbody

292. Planck Distribution…
292

293. Features of the Planck’s Distribution
• Emitted radiation varies continuously with wavelength
• At any wavelength, the magnitude of emitted radiation
increases with increasing temperature
• The spectral region in which the radiation is
concentrated depends on temperature, with
comparatively more radiation appearing at shorter
wavelengths as the temperature increases
• A significant fraction of radiation emitted by the Sun,
which may be approximated as blackbody at 5800 K, is
in the visible region of the spectrum.
• For T  800 K, emission is predominantly in the infrared
region of the spectrum and is not visible to the eye
293

294. Wein’s Displacement Law
• Blackbody spectral distribution has a
maximum that corresponds to max depends
on temperature
maxT=C3; where C3=2898 m-K
• This is Wein’s displacement law
• The maximum spectral emissive power is
displaced to shorter wavelengths with
increasing temperature
294

295. Stefan-Boltzmann Law
• Emissive power of a blackbody is given by
Eb=T4
• This is known as Stefan-Boltzmann law
• The Stefan-Boltzmann constant is given as
=5.6710-8 W/m2-K4
• It enables the calculation of the amount of
radiation emitted in all directions and over all
wavelengths just using the temperature
295

296. Radiation from Real Surfaces
• Emissivity is the ratio of radiation emitted by a
surface to the radiation emitted by a
blackbody at the same temperature
296

297. View Factor (F)
• Also known as Shape factor or Configuration
factor
• View factor is defined as the “fraction of the
radiation leaving surface i that is intercepted
by surface j”
297

298. View Factor Integral
298

















j
i A
j
i
j
i
A
i
j
i
j
i
j
i
i
j
i
j
i
j
i
r,i
e
j
i
j
j
j-i
j-i
r,i
e
i
j
i
i
r,i
e
j
i
dA
dA
R
θ
J
q
dA
dA
R
θ
J
dq
dA
dA
R
θ
I
dq
R
dA
dω
dω
I
where
dω
dA
θ
I
dq
2
2
2
2
i
j
j
i
cos
cos
is
j
by
d
intercepte
is
and
i
leaves
radiation
at which
rate
total
The
cos
cos
diffusely,
reflects
and
emits
i
surface
If
cos
cos
cos
dA
from
ed
when view
dA
by
subtended
angle
solid
reflection
and
emission
by
i
surface
leaving
radiation
of
intensity
cos
is
dA
by
d
intercepte
is
and
dA
leaves
radiation
at which
rate
The







299. View Factor Integral…
299
radiosity.
uniform
have
and
reflectors
and
emitters
diffuse
are
that
surfaces
any two
with
associated
factors
view
determine
to
used
be
can
equations
above
The
cos
cos
1
,
cos
cos
1
factor,
view
of
definition
the
From
2
2
,








j
i
j
i
A
j
i
j
i
A
j
ji
A
j
i
j
i
A
i
ij
i
i
j
i
j
i
dA
dA
R
θ
A
F
Similarly
dA
dA
R
θ
A
F
J
A
q
F





300. View Factor-Special Case
300
directly
itself
strikes
that
i
surface
leaving
radiation
of
fraction
The

i
i
F
The view factor from a surface to itself is
zero for plane and convex surfaces and
non-zero for concave surfaces
1
1
1
2
2
1




F
F

301. View Factor-Some Important Points
• Radiation a surface receives from a source is
directly proportional to the angle the surface
subtends when viewed from the source
• Radiation coming from a source must be uniform
in all directions
• The medium between the surfaces must not
absorb, emit or scatter radiation
• That means, the surfaces must be isothermal and
diffuse emitters and reflectors and the medium
must be non-participating
301

302. Estimating View Factors
• Enclosures consisting of N surfaces will have
N2 View factors
• Method of Integration
• Results in Graphical form
• Results in Tabular form
• Using View factor relations
302












NN
N
N
N
N
... F
F
F
... F
F
F
... F
F
F
2
1
2
22
21
1
12
11




303. View Factor Relations
303

























3
1
3
1
2
1
1
1
1
1
i
i
j
i
i
j
i
j
1
1
surface
for
enclosure,
surface
3
a
For
1
unity.
equal
must
itself,
including
enclosure,
the
of
surfaces
all
to
enclosure
an
of
i
surface
from
factors
view
the
of
sum
The
Rule
Summation
.
2
if
if
Implies,
Relation
y
Reciprocit
.
1
j
j
N
j
j
i
j
i
j
j
i
j
j
j
i
i
F
F
F
F
F
A
A
F
F
A
A
F
F
F
A
F
A  
1
2
1
equations
of
Number 
 N
N
N
equations
of
Number 

304. View Factor Relations…
304
 
i
k
i
j
k
i
j
i
F
F
and
F
F
F
F
F











then
i,
surface
about the
symmetric
are
k
and
j
surfaces
the
If
surface.
that
from
factors
view
identical
have
will
surface
third
a
about
symmetry
possess
that
surfaces
more
or
Two
rule
symmetry
The
.
4
j.
surface
of
parts
to
i
surface
from
factors
view
the
of
sum
the
to
equal
is
j
surface
a
to
i
surface
a
from
factor
view
The
Rule
ion
Superposit
The
.
3
3
1
2
1
3
,
2
1

305. Problem 1
Consider a diffuse circular disk of diameter D
and area Aj and a plane diffuse surface of area
Ai<<Aj. The surfaces are parallel, and Ai is
located at a distance L from the centre of Aj.
Obtain an expression for the view factor Fij.
Ans:
305
2
2
2
4L
D
D
Fij



306. Problem 2
Determine the view factors associated with an
enclosure formed by two concentric spheres.
Ans:
306
2
2
1
22
2
2
1
21
12
11
1
1
0




















r
r
-
; F
r
r
F
; F
F

307. Problem 3
Determine the fraction of the radiation leaving
the base of the cylindrical enclosure that
escapes through a coaxial ring opening at its top
surface. Length is 10 cm.
Ans: 0.17
307

308. Problem 4
Determine the view factors F13 and F23 between the
rectangular surfaces shown in the figure:
Ans: F13= 0.27; F23= 0.05
308

309. Problem 5
Consider a hemispherical furnace with a flat
circular base of diameter D. Determine the view
factor from the dome of this furnace to its base.
Ans: 0.5
309

310. Blackbody Radiation
• In all real bodies, radiation may leave as
reflection and emission
• On reaching a second surface, they experience
reflection and absorption
• But in Blackbodies, there is no reflection
• Hence all radiation is emitted
310

311. Blackbody Radiation
311
 
 
4
4
is
exchange
heat
radiative
net
The
Similarly
surfaces,
black
For
j
i
ij
i
ij
bj
ji
j
bi
ij
i
ij
i
j
j
i
ij
bj
ji
j
i
j
bi
ij
i
j
i
bi
i
ij
i
i
j
i
T
T
σ
F
A
q
E
F
A
E
F
A
q
q
q
q
E
F
A
q
E
F
A
q
E
J
F
J
A
q
















312. Radiation Exchange between Opaque
Gray Diffuse Surfaces in an Enclosure
Assumptions:
• Isothermal surfaces with uniform radiosity
and uniform irradiation
• Surfaces are opaque (τ=0) and ε, α, ρ are
independent of direction (Diffuse) and
independent of wavelength (Gray)
• Hence ε=α (Kirchoff’s law)
• Medium is non-participating
312

313. Net Radiation Exchange at a Surface
313
 
 
 
 
  Resistance
Radiative
Surface
Potential
Driving
1
1
1
1
1
surface
gray
diffuse,
opaque,
an
for
and
But
1
surface,
opaque
an
For
surface,
a
leaves
that
radiation
Net
i



































i
i
i
i
bi
i
i
bi
i
i
i
i
i
i
i
bi
i
i
i
i
bi
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
A
J
E
q
E
J
J
A
q
G
E
J
E
E
G
E
A
q
G
G
E
A
q
G
E
J
G
J
A
q
q
















314. Some Insight into the Net Radiation
Exchange at a Surface
314
 
 
bi
i
i
i
i
i
i
i
i
i
i
i
bi
i
E
J
A
A
J
E
q















blackbody
a
as
behaves
surface
large
the
Hence,
1
or
0
1
0
1
resistance
Surface
A
surface
large
the
of
area
surfaces,
small
multiple
containing
room
a
In
Resistance
Radiative
Surface
Potential
Driving
1







315. Radiation Exchange between Surfaces
315
  



















N
j
j
ij
i
i
i
i
i
i
N
j
j
ij
i
i
N
j
j
j
ji
i
J
F
J
A
G
J
A
q
J
F
A
G
A
J
A
F
G
A
1
1
i
1
i
relation,
y
reciprocit
the
Using
is
i
surface
including
surfaces,
all
from
i
surface
reaches
radiation
at which
rate
Total
 
 
Resistance
Space
Potential
Driving
1
1
1
rule,
summation
the
From
1
1
1 1
1
1























 




 


N
j
ij
i
j
i
i
N
j
ij
i
j
i
i
i
i
i
bi
N
j
N
j
ij
j
i
ij
i
i
N
j
j
ij
i
N
j
ij
i
i
F
A
J
J
q
F
A
J
J
A
J
E
q
J
J
F
A
q
J
F
J
F
A
q



316. Electrical Analogy
316

317. Two Surface Enclosure
317

318. Three Surface Enclosure with One
Surface Re-radiating
318

319. Problem 6
Two very large parallel plates are maintained at
uniform temperatures T1=800 K and T2=500 K
and have emissivities ε1=0.2 and ε2=0.7,
respectively. Determine the net rate of radiation
heat transfer between the two surfaces per unit
surface area of the plates.
Ans: 3625 W/m2
319

320. Problem 7
A furnace is shaped as a long equilateral triangular duct.
The width of each side is 1 m. The base surface has an
emissivity of 0.7 and is maintained at a uniform
temperature of 600 K. The heated left side surface closely
approximates a blackbody at 1000 K. The right side
surface is well insulated. Determine the rate at which
heat must be supplied to the heated side externally per
unit length of the duct in order to maintain these
operating conditions.
Ans: 28 kW
320

321. Problem 8
Two parallel plates 2 m  1 m are spaced 1 m
apart. The plates are at temperatures 727℃ and
227℃ and their emissivities are 0.3 and 0.5
respectively. The plates are located in a large
room, the walls of which are at 27℃. Determine
the rate of radiant heat loss from each plate and
the heat gain by the walls.
Ans: 33 kW, -1.87 kW, 31.1 kW
321

322. Module 7
Practical Applications

323. Extended Surfaces (Fins)

324. Fins
• An extended surface used to
specifically enhance heat
transfer between a solid and
an adjoining fluid is called a
Fin
• Fins provide an extra surface
area for heat transfer
• Fins should be used when ‘h’
cannot be increased and T
cannot be decreased
324

325. Fin Applications
• Engine heads on motor cycles
• Lawn mowers
• Electric power transformers
• Air conditioners etc
325

326. Fin Configurations
326
Straight Fin with
Uniform Cross
Section
Straight Fin with
Varying Cross
Section
Annular Fin Pin Fin

327. General Conduction Analysis of a Fin
Assumptions:
• 1-D
• Steady state
• Constant ‘k’
• No radiation
• No heat generation
• Uniform ‘h’ over the surface
327

328. General Conduction Analysis of a Fin…
328
 
 
  0
1
1
0
on,
substituti
On
out
convected
Heat
out
conducted
Heat
in
conducted
Heat
Out
Heat
in
Heat
:
element
on the
Balance
Heat
2
2























































T
T
dx
dA
k
h
A
dx
dT
dx
dA
A
dx
T
d
T
T
dx
dA
k
h
dx
dT
A
dx
d
T
T
hdA
dq
dx
dx
dT
A
dx
d
k
dx
dT
kA
q
dx
dx
dq
q
q
dx
dT
kA
q
dq
q
q
s
c
c
c
s
c
s
conv
c
c
dx
x
x
x
dx
x
c
x
conv
dx
x
x

329. Fins of Uniform Cross Section
329
 
 
  0
dA
;
0
;
0
1
1
2
2
2
2































T
T
kA
hP
dx
T
d
P
dx
Px
d
dx
Px
A
dx
dA
const
A
T
T
dx
dA
k
h
A
dx
dT
dx
dA
A
dx
T
d
c
s
s
c
c
s
c
c
c

330. Fins of Uniform Cross Section…
330
 
 
 
  































T
T
T
x
T
T
T
T
x
T
dx
dT
T
x
T
dx
dT
k
T
T
)
T(x
L
x
b
L
L
x
b
L
x
b
L
x
b
;
0
:
conditions
Boundary
fin
Long
:
D
Case
;
0
:
conditions
Boundary
re
temperatu
fin tip
Specified
:
C
Case
0
;
0
:
conditions
Boundary
insulated)
(tip
at tip
convection
No
:
B
Case
T
h
;
0
:
Conditions
Boundary
fin tip
from
Convection
:
A
Case
L

331. Temperature Distribution and Heat
Transfer
331
   
 
 
  c
-mx
b
L
b
L
hPkA
T
e
mL
T
T
T
T
mL
M
mL
L-x
m
mx
T
T
T
T
mL
L-x
m
mL
mk
h
mL
mk
h
mL
mk
h
mL
l-x
m
mk
h
L-x
m

















b
L
T
Fin
Long
(D)
sinh
cosh
sinh
sinh
sinh
T
T
(C)
mL
tanh
M
cosh
cosh
Adiabatic
(B)
sinh
mL
cosh
cosh
mL
sinh
M
sinh
cosh
sinh
cosh
Convection
(A)
d
Transferre
Heat
on
Distributi
e
Temperatur
Condition
Tip
  c
hPkA
T
M 

 b
T

332. Fin Performance
Factors which increase fin effectiveness:
• High Thermal conductivity
• High P/Ac, thin but closely spaced fins
• Low heat transfer coefficient (natural
convection)
332
 
2
1
f
f
b
f
),
(L
D
case
For
2
Preferred
T
fin
without
rate
fer
Heat trans
rate
fer
heat trans
Fin
ess,
Effectiven
Fin
/
C
c
f
hA
kP
T
hA
q




















333. Fin Performance…
• Fin Efficiency is the ratio of heat transfer due
to the fin and maximum heat transfer if the
entire fin is at the base temperature
333
 
mL
mL
tanh
tip,
adiabatic
and
section
cross
uniform
fin with
straight
a
For
max





f
b
f
f
f
T
T
hA
q
q
q f



334. Problem 1
A brass rod 100 mm long and 5 mm in diameter
extends horizontally from a casting at 200℃.
The rod is in an air environment with T=20℃
and h=30 W/m2.K. What is the temperature of
the rod 25, 50, and 100 mm from the casting?
Ans: 155.2℃, 126.4℃, 104℃
334

335. Problem 2
A pin fin of uniform, cross-sectional area is
fabricated of an aluminum alloy (k=160 W/m K).
The fin diameter is D=4 mm, and the fin is
exposed to convective conditions characterized
by h=220 W/m2.K. It is reported that the fin
efficiency is f= 0.65. Determine the fin length L
and the fin effectiveness f. Account for tip
convection.
Ans: L=36.2 mm, f=23.74
335

336. Problem 3
Two long copper rods of diameter 10 mm are
soldered together end to end, with solder
having a melting point of 650℃. The rods are in
air at 25℃ with a convection coefficient of 10
W/m2.K. What is the minimum power input
needed to effect the soldering?
Ans: 118.6 W
336

337. Heat Exchangers

338. Heat Exchanger
• Heat Exchanger is a device used to transfer
heat between two fluids at different
temperatures separated by a solid wall
• Applications is space heating and air
conditioning, power production, waste heat
recovery, chemical processing etc
338

339. Types of Heat Exchangers
Heat Exchangers are classified based on Flow
Arrangement and Type of Construction
• Concentric tube heat exchangers
• Parallel flow
• Counter flow
• Cross flow
• Shell and tube heat exchangers
339

340. Types of Heat Exchangers…
340
Parallel and Counter Flow Heat Exchangers
Cross Flow Heat Exchangers

341. Types of Heat Exchangers…
341
Shell and Tube Heat Exchanger
One Shell Pass, Two Tube Passes
Two Shell Passes,
Four Tube Passes

342. Types of Heat Exchangers…
342
Compact Fin-Tube Heat Exchangers
Compact Plate-Fin heat Exchangers

343. Compact heat Exchangers
• Compact heat Exchangers have high Surface
Area to Volume ratio
• They are used if one of the fluids is a gas
• Flow passages are small (dh<5 mm) and hence
flow is usually laminar
343







gases
for
m
700
liquids
for
m
400
3
2
3
2
m
m
Volume
Area
Surface

344. Overall Heat Transfer Coefficient (U)
344
 
   
o
o
i
fo
o
i
i
o
i
fi
i
i
i
o
fi
i
o
i
o
o
fo
o
h
c
h
c
h
w
c
h
c
m
h
r
r
R
r
r
r
r
k
r
R
h
h
r
r
R
r
r
r
r
k
r
R
h
hA
R
hA
T
UA
q
1
ln
1
U
1
1
ln
1
U
1
fouling,
with
sections
r
For tubula
A
A
if
U
U
A
U
A
U
1
1
A
U
1
A
U
1
UA
1
streams
fluid
two
separating
wall
a
For
i
o
c
h
h
c
h
c




















fouling
f
outer
o
inner
i
hot
h
cold
c
mean
:
Subscripts






m

345. Log Mean Temperature Difference
(LMTD)
345
o
c
o
h
c
h
i
c
i
h
c
h
lm
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
,
,
2
,
2
,
2
,
,
1
,
1
,
1
1
2
1
2
2
1
2
1
ln
ln
:
Exchanger
Heat
Flow
Parallel
























346. Derivation of LMTD for Parallel Flow
Assumptions:
• Heat exchanger is adiabatic to surroundings
• Axial conduction along the tubes is negligible
• Negligible changes in KE and PE
• Fluid specific heats remain constant
• Overall heat transfer coefficient constant
346

347. Derivation of LMTD for Parallel Flow…
347
  c
h
c
h
c
c
c
c
p
h
h
h
h
p
dT
dT
ΔT
d
-T
T
ΔT
where
TdA
U
dq
Also
dT
C
dT
c
m
dq
dT
C
dT
c
m
dq











form,
al
differenti
In
,
,
element,
fluid
cold
on the
balance
energy
Applying
element,
fluid
hot
on the
balance
energy
Applying
,
c
,
h



348. Derivation of LMTD for Parallel Flow…
348
 
 
 
 

 



















































2
1
2
1
1
1
exchanger,
heat
of
length
the
along
g
integratin
On
1
1
,
variables
separating
On
1
1
1
1
dA
C
C
U
ΔT
ΔT
d
dA
C
C
U
T
T
d
C
C
TdA
U
T
d
TdA
U
dq
But
C
C
dq
T
d
c
h
c
h
c
h
c
h

349. Derivation of LMTD for Parallel Flow…
349
 
 
   
 
o
c
o
h
i
c
i
h
i
c
o
c
o
h
i
h
o
c
i
c
o
o
c
i
c
c
o
h
i
h
h
o
h
i
h
h
c
h
T
T
T
T
q
UA
ΔT
ΔT
q
T
T
q
T
T
UA
ΔT
ΔT
q
T
T
C
T
T
C
q
and
q
T
T
C
T
T
C
q
But
C
C
UA
ΔT
ΔT
,
,
,
,
1
2
,
,
,
,
1
2
,
,
,
,
,
,
,
,
1
2
ln
ln
1
,
1
1
1
ln




















 









































   




















































1
2
1
2
1
2
1
2
1
2
2
1
1
2
,
,
2
,
,
1
ln
ln
ln
and
ΔT
ΔT
T
T
T
T
UA
ΔT
ΔT
T
T
UA
q
T
T
q
UA
T
T
q
UA
ΔT
ΔT
T
T
T
T
T
T
But
m
m
o
c
o
h
i
c
i
h
1
2
1
2
T
ln
flow,
Parallel
In
T
T
T
Tlm









350. Log Mean Temperature Difference
(LMTD)…
350
i
c
o
h
c
h
o
c
i
h
c
h
lm
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
,
,
2
,
2
,
2
,
,
1
,
1
,
1
1
2
1
2
2
1
2
1
ln
ln
:
Exchanger
Heat
Flow
Counter
























351. Special Operating Conditions
351
Condensation Boiling lm
T
T
T 



 2
1
h
p
h
h
c
p
c
c
c
m
C
c
m
C
,
,





352. Problem 1
A steel tube (k=50 W/m-K) of inner and outer
diameters Di=20 mm and Do=26 mm,
respectively, is used to transfer heat from hot
gases flowing over the tube (hh=200 W/m2-K) to
cold water flowing through the tube (hc=8000
W/m2-K). What is the cold-side overall heat
transfer coefficient Uc?
Ans: 249 W/m2-K
352

353. Problem 2
A counter flow double pipe heat exchanger is to
heat water from 20℃ to 80℃ at a rate of 1.2 kg/s.
The heating is to be accomplished by geothermal
water available at 160℃ at a mass flow rate of 2
kg/s. The inner tube is thin walled and has a
diameter of 1.5 cm. If the overall heat transfer
coefficient of the heat exchanger is 640 W/m2-K,
determine the length of the heat exchanger
required to achieve the desired heating.
Ans: 109 m
353

354. Problem 3
Steam in the condenser of a power plant is to be
condensed at a temperature of 30℃ with cooling
water from a nearby lake, which enters the tubes of
the condenser at 14℃ and leaves at 22℃. The
surface area of the tubes is 45 m2, and the overall
heat transfer coefficient is 2100 W/m2-K. Determine
the mass flow rate of the cooling water needed and
the rate of condensation of the steam in the
condenser.
Ans: 32.6 kg/s, 0.45 kg/s
354

355. Problem 4
A 2-shell passes and 4-tube passes heat exchanger is used
to heat glycerin from 20℃ to 50℃ by hot water, which
enters the thin walled 2 cm diameter tubes at 80℃ and
leaves at 40℃. The total length of the tubes in the heat
exchanger is 60 m. The convection heat transfer
coefficient is 25 W/m2-K on the glycerin (shell) side and
160 W/m2-K on the water (tube) side. Determine the rate
of heat transfer in the heat exchanger (a) before any
fouling and (b) after fouling with a fouling factor of
0.0006 m2-K/W occurs on the outer surfaces of the tubes.
Ans: 1848 W, 1824 W
355

356. Effectiveness-NTU Method
• LMTD method can be used if both inlet and
outlet temperatures of both fluids is known
• If only the inlet temperatures are known,
cumbersome iterative methods will be
necessary to use LMTD method
• In such a case, Effectiveness-NTU method can
be used
356

357. Effectiveness of a Heat Exchanger
• In an ideal heat exchanger, maximum heat
transfer takes place when the hot fluid is
cooled to the entry temperature of the cold
fluid
357
 
 
 
 
 
 
 
c,i
h,i
i
c
o
c
c
c,i
h,i
o
h
i
h
h
c,i
h,i
c,i
h,i
h
c
h
c,i
h,i
c
h
c
T
T
C
T
T
C
T
T
C
T
T
C
q
q
T
T
C
q
T
T
C
q
C
C
T
T
C
q
C
C
















min
,
,
min
,
,
max
min
max
max
max
ess,
Effectiven
:
If
:
If


358. Effectiveness-NTU Relations
• Once the effectiveness is known, actual heat
transferred can be found using the preceding
equations
• It has been found that effectiveness is a
function of the heat capacity rates of both
fluids and NTU (the Number of Transfer Units)
358
min
max
min
,
C
UA
NTU
C
C
NTU
f











exchanger.
heat
larger
implies
NTU
Larger
ess.
dimensionl
is
NTU

359. Effectiveness-NTU Relations
• Effectiveness-NTU and NTU-Effectiveness
relations are available in the data book
• Graphical representation of Effectiveness-NTU
relations too are available
359
Parallel Flow Counter Flow

360. Problem 5
A counter-flow double-pipe heat exchanger is to
heat water from 20℃ to 80℃ at a rate of 1.2 kg/s.
The heating is to be accomplished by geothermal
water available at 160℃ at a mass flow rate of 2
kg/s. The inner tube is thin-walled and has a
diameter of 1.5 cm. The overall heat transfer
coefficient of the heat exchanger is 640 W/m2-K.
Using the effectiveness-NTU method determine the
length of the heat exchanger required to achieve
the desired heating.
Ans: 109.7 m
360

361. Problem 6
Hot oil is to be cooled by water in a 1-shell-pass and 8-
tube-passes heat exchanger. The tubes are thin-walled
and are made of copper with an internal diameter of 1.4
cm. The length of each tube pass in the heat exchanger is
5 m, and the overall heat transfer coefficient is 310
W/m2-K. Water flows through the tubes at a rate of 0.2
kg/s, and the oil through the shell at a rate of 0.3 kg/s.
The water and the oil enter at temperatures of 20℃ and
150℃, respectively. Determine the rate of heat transfer in
the heat exchanger and the outlet temperatures of the
water and the oil. Take cpwater=4.18 and cpoil=2.13 kJ/kg-K.
Ans: 40.4 kW, Tout,water=68.3℃, Tout,oil=86.8 ℃
361

362. Radiation Shields

363. Radiation Shields
• Radiation shields are used to reduce net
radiation heat transfer between two surfaces
• They are made of low emissivity, high
reflectivity materials
• They reduce radiation heat transfer by
introducing an additional resistance
363

364. Heat Transferred with Radiation Shield
364
 
 
2
,
3
2
,
3
1
,
3
1
,
3
2
1
4
2
4
1
12
2
1
4
2
4
1
12
1
1
1
1
Shield,
Radiation
With
1
1
1
:
Shield
radiation
Without



 










ε
ε
T
T
Aσ
q
ε
ε
T
T
Aσ
q

365. Heat Transferred with Radiation
Shield…
365
   
 
  Shield
Radiation
no
with
ferred
heat trans
Shields
Radiation
of
Number
1
1
q
,
emissivity
equal
having
Shields,
Radiation
multiple
For
1
1
1
1
1
1
:
used
be
can
equations
following
the
,
T
estimate
To
0
12
0
12
12
2
3
4
2
4
3
32
3
1
4
3
4
1
13
3
32
13
12















q
N
q
N
T
T
A
q
T
T
A
q
q
q
q







366. Problem 7
A thin aluminum sheet with an emissivity of 0.1 on
both sides is placed between two very large parallel
plates that are maintained at uniform temperatures
T1=800 K and T2=500 K and have emissivities ε1= 0.2
and ε1=0.7, respectively. Determine the net rate of
radiation heat transfer between the two plates per
unit surface area of the plates and compare the
result to that without the shield.
Ans: With shield: 811.6 W/m2
Without shield: 3625.3 W/m2
366

367. End of Course

368. References
• Principles of Heat and Mass Transfer by Frank P
Incropera, David P Dewitt, Theodore L Bergman,
Adrianne S Lavine, 7th Edition, 2014, Published by
John Wiley & Sons.
• Heat and Mass Transfer by P.K. Nag, 3rd Edition,
2013, Published by McGraw Hill Education (India)
Pvt Ltd.
• Heat and Mass Transfer by Yunus A Cengel and
Afshin J Ghajar, 4th Edition, 2014, Published by
McGraw Hill Education (India) Pvt Ltd.
368