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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
,
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
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
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
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
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
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
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
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
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
"
"
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
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
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
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
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
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
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
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
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=16.76810-6 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
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
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
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
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
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
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
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
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
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
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
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
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 060, 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
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
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
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
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
262. Free Convection Boiling
• Exists when ΔTeΔTe,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,AΔTeΔTe,C where
ΔTe,C≈30℃
263
264. Transition Boiling
• Also known as Unstable film boiling or Partial film
boiling
• Occurs when ΔTe,CΔTeΔTe,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
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
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
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
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
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
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
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
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
=5.6710-8 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
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
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
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
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
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
