2-4
Find the heat transfer per unit area through the
composite wall sketched. Assume one-dimensional
heat flow.
q
kA = 150 W/m ºC
kB = 30
T = 370ºC
kC = 50
kD = 70
2.5 cm
AB = AD
AC = 0.1 m2
B
C
A
D
7.5 cm
T = 66ºC
5.0 cm
(NEXT)
PANKAJ R. CHANDAR
1
2-4
Solution:
x
0.025
, RA 
 1.667 x10 - 4 /A
k A
150  A
 2  0.075  0.01
RB 
 30  A
A
0.05
0.001
RC 

 50  A
A
 2  0.075  0.0043
RD 
 70  A
A
1
R  RA  RC 
 2.667 x10 - 2
1
1

RB
RD
R 
q 
T
370  66

R
2.667 x10 - 2
 11400 W
PANKAJ R. CHANDAR
2
2-14 A spherical tank, 1 m in diameter, is maintained at a
emperature of 120ºC and exposed to a convection
environment. With h = 25 W/m2 ºC and T = 15 ºC, what
thickness of urethane foam should be added to ensure that
the outer temperature of the insulation does not exceed
40ºC? What percentage reduction in heat loss results from
installing this insulation?
q (no insul.)  h A (TW  T )
 25 (4) 0.5 2 (120 - 15)
 8247 W
(NEXT)
PANKAJ R. CHANDAR
3
2-14 Solution, cont’d:
mW
k FOAM  18
m C
4k (Ti  TO )
2
q
 h 4 rO (TO - T )
1
1
ri
rO
0.018 (120  40)
2

 (25) rO ( 40  15)
1
1
0.5 rO
rO  0.5023 m
thickness  rO  ri  0.0023 m
q (w/insul.)  25 (4) (0.5023) 2 ( 40  15)
 1982 W
PANKAJ R. CHANDAR
4
2-20 A 1.0 mm diameter wire is maintained at a temperature
of 400ºC and exposed to a convection environment at
40ºC with h = 120 W/m2 ºC. Calculate the thermal
conductivity which will just cause an insulation
thickness of 0.2 mm to produce a “critical radius”.
How much of this insulation
must be added to
reduce the heat transfer by 75 percent from that which
would be experienced by the bare wire.
r  0.5 mm  5x10 -4 m
i
k
 5 x10 - 4  2x10 - 4  7 x10 - 4
h
k   7x10 - 4  120  0.084 W
m C
q (bare wire)   (0.001)120(400 - 40)
 135.7 W
m
q (insulated)  135.7 (0.25)  33.93 W
rO 
q 
400 - 40
rO

ln 
-4 
1
5 x10

 
2  (0.084)
2 (120) rO
m
 33.93
PANKAJ R. CHANDAR
(L = 1 m)
5
By iteration : rO  135 mm, Thickness  134.5 mm
2-27 An insulation system is to be selected for a furnace
wall at 1000ºC using first a layer of mineral wool blocks
followed by fiberglass boards. The outside of the
insulation is exposed to an environment with h = 15
W/m2 ºC and T = 40ºC. Using the data of Table 2-1,
calculate the thickness of each insulating material
such that the interface temperature is not greater than
400ºC and the outside temperature is not greater than
55ºC. Use mean values for thermal conductivities.
What is the heat lost in this wall in watts per square
meter?
400ºC
mW
k M  90
1000ºC
m C
M
F
55ºC
mW
k F  42
m C
1
2
3
(NEXT)
PANKAJ R. CHANDAR
6
2-27 Solution:
q
 h (T3  T )  15 (55 - 40)  225 W
2
m
A

q
1000  400 
 kM
 x M  0.24 m
A
x M

q
400  55
 kF
 x F  0.0644 m
A
x F
PANKAJ R. CHANDAR
7
2-28 Derive an expression for the temperature distribution
in a plane wall having uniformly distributed heat
sources and one face maintained at a temperature T1
while the other face is maintained at a temperature T2.
The thickness of the wall may be taken as 2L.

2
d T
q

 0
2
dx
k

T1
T2
L
L
X
dT
q
 
x  C1
dx
k

q x2
T   C1 x  C 2
2k
(NEXT)
PANKAJ
R. CHANDAR
8
2-28 Solution cont’d:
Boundary Conditions:
1) at X= -L, T=T1
2) at X=L, T=T2

q x2
T   C1 x  C 2
2k
The general solution is
Substituting boundary conditions yields:

q
T  T1
T  T2
( L2  X 2 )  2
X 1
2k
2L
2
For temp. distribution on the wall.
T
PANKAJ R. CHANDAR
9
2-39 A 3.0 cm thick plate has heat generated uniformly at
the rate of 5x105 W/m3. One side of the plate is
maintained at 200ºC and the other side at 50ºC.
Calculate the temperature at the center of the plate for
k = 20 W/m ºC.
Use solution from Prob. 2-28
T = T0 at X = 0

q
T1  T2
2
T0 
L 
2k
2
5x10 5 (0.015) 2
200  50


 127.8 C
2 (20)
2
PANKAJ R. CHANDAR
10
2-45
Derive an expression for the temperature
distribution in a sphere of radius
r with uniform
.
heat generation q and constant surface temperature
TW.
T=TW @ r = R steady state, T varies only w/r
1  2 (rT)
1
 
T
 2
 sin θ

2
r r
r sin θ  θ 
θ 

1
 T
q
1 T


r 2 sin 2 θ  θ 2
k
 t
This reduces to :
2


2
2

1 d ( rT)
q
d ( rT)
qr


0



r d r2
k
d r2
k
Integrating yields :

q r2
C2
T 
 C1 
6k
r
(NEXT)
PANKAJ R. CHANDAR
11
2-45 Solution cont’d.
Boundary Conditions :
 4
3
2 d T
q
 R   k 4 R
3
d r

1)
d T
qr
 
d r
3k
dT
3)
d r

q R2
 0 , then C1  TW 
6k
C2  0

T - TW
2) T  TW @ r  R

q

R 2  r2
6k

PANKAJ R. CHANDAR
12
2-46 A stainless steel sphere [k = 16 W/m ºC] having a
diameter of 4 cm is expesed to a convection
environment at 20ºC, h = 15 W/m2 ºC. Heat is generated
uniformly in the sphere at a rate of 1.0 MW/m3.
Calculate the steady state temperature for the center of
the sphere.
From Prob. 2 - 45

q

R 2  r2 
6k
1x10 6 (0.02) 2
T0 - TW 
 4.17  C
6 (16)

 4
q  qV  q
 R 3  h 4 r 2 (TW - T )
3
1x10 6 (0.02)
TW - T 
 444.4  C
3 (15)
T - TW 
T0  444.4  4.17  448.6  C
PANKAJ R. CHANDAR
13
2-53 Calculate the overall heat transfer coefficient for Prob.
2-4.
1
1
W
U 

 32.11
-2
R
3.114 x10
m 2  C
PANKAJ R. CHANDAR
14
2-62 An aluminum rod 2.5 cm in diameter and 15 cm long
protrudes from a wall which is maintained at 260ºC.
The rod is exposed to an environment at 16ºC. The
convection heat transfer coefficient is 15 W/m2 ºC.
Calculate the heat lost by the rod.
k  204
W
d
2 .5
,
L

L


15

C
m C
4
4
 16.25 cm
1
 15  0.025 4  2
h P
m
 
 3.43
2 
k A




204

0
.
025


m L C   3.43 0.1625  0.5573
q 

h P k A θ 0 tanh  m L C 
(0.025) 
  15  0.025 ( 204) 

4


 42.41 W
2
1
2
 260  16  tanh  0.5573
PANKAJ R. CHANDAR
15
2-67 An aluminum fin 1.6 mm thick is placed on a circular
tube with 2.5 cm OD. The fin is 6.4 mm long. The tube
wall is intained at 150ºC, the environment temperature
is 15ºC, and the convection heat transfer coefficient is
23 W/m2 ºC. Calculate the heat lost by the fin.
W
k  210
m C
t
LC  L 
 6.4  0.8  7.2 mm
2
r2C  r1  L C  1.25  0.72  1.97
r2C
 1.576
r1
PANKAJ R. CHANDAR
16
2-67
Solution cont’d.
A M  t (r2C  r1 )   0.0016  0.0072 
 1.152 x10 -5 m 2
LC
3

2

h


 k AM 
1
2
  0.0072 
3

2

23
 210 1.152 x10 -5 



1

 0.0596
From Fig. 2 - 11, f  97 %

q MAX  2 h  r2C
2
 r1
2
 T  T   4.523 W
0

q  (0.97)(4.523)  4.387 W
PANKAJ R. CHANDAR
17
2
2-91 A 1.0 mm thick aluminum fin surrounds a
2.5 cm
diameter tube. The length of the fin is 1.25 cm. The fin
is exposed to a convection environment at 30ºC with h
= 75 W/m2 ºC. The tube surface is maintained at 100 ºC.
Calculate the heat lost by the fin.
r1 = 1.25 cm, r2 = 2.5 cm, r2C = 2.55 cm
LC = 1.3 cm, k = 204
LC
3

2


h

k AM 
1
2
 0.249,
 f  0.91

q   0.91 2 75 0.0255 2  0.0125 2 100  30 


 14.83 W
PANKAJ R. CHANDAR
18
2-118 Consider aluminum circumferential fins with r1 = 1.0
cm, r2 = 2.0 cm, and thicknesses of 1.0, 2.0, and 3.0
mm. The convection coefficient is 160 W/m2 ºC.
Compare the heat transfers for six 1.0 mm fins, three
2.0 mm fins, and two 3.0 mm fins. What do you
conclude? Repeat for h = 320 W/m2 ºC.
k = 204 W/m ºC
1.0 mm Fin LC = 1.05
cm
1
LC
3

2
h 


k
A
M

2


160



204
0
.
001
0
.
0105


  0.0105 2 
3
 0.294
f  0.88


q  6160   0.02052  0.012 2 T  0.88
 1.7 T
(NEXT)
PANKAJ R. CHANDAR
19
2-3. A composite wall is formed of a 2.5 cm copper
plate, a 3.2mm layer of asbestos, and a 5 cm layer
of fiberglass. The wall is subjected to an overall
temperature difference of 560 °C . Calculate the
heat-flow per area through the composite
structure.
q
560
W


419
A 0.025 3.2 *10 3 0.05
m2


386
0.16
0.038
PANKAJ R. CHANDAR
20
2-5. One side of a copper block 5 cm thick is maintained
at 260 °C. the other side is covered with a layer of
fibergalss 2.5 cm thick. The outside of fiberglass is
maintained at 38 °C, and the total flow through the
copper fiberglass combination is 44kW. What is the
area of slab?
44000
260  38

0.05 0.025
A

386 0.038
2
A  130.4m
PANKAJ R. CHANDAR
21
2-6. An outside wall for a building consists of a 10 cm
layer of common brick and a 2.5 cm layer of
fiberglass [k=0.05W/m °C]. Calculate the heat
flow through the copper fiberglass combination is 44
kW. What is the area of slab?
q
45
W

 69.78 2
0
.
10
0
.
025
A
m

0.69 0.05
PANKAJ R. CHANDAR
22
2-7. One side of a copper block 4 cm thick is
maintained at 175 °C. The other side is covered
with a layer of fiberglass 1.5cm thick. The outside
of the fiberglass is maintained at 80 °C, and the
total heat flow through the composite slab is
300W. What is the area of the slab?
q T

A R
300
175  80

0.04 0.015
A

386 0.038
A  1.247m 2
PANKAJ R. CHANDAR
23
2-10. A wall is constructed of 2.0 cm of copper, 3.0 mm
of asbestos sheet [k=0.166W/m °C], and 6.0 cm of
fiberglass. Calculate the heat flow per unit area
for an overall temperature difference of 500 °C.
0.02
RCu 
 5.35 *10 5
374
0.003
RAs 
 0.0181
0.166
0.06
RF 1 
 1.579
0.038
q 500
W

 313 2
A R
m
PANKAJ R. CHANDAR
24
2-12. A wall is constructed of a section of stainless steel
[k=16W/m °C] 4.0 mm thick with identical layer
of plastic on both side of steel. The overall heattransfer coefficient, considering convection on
both sides of the plastic, is 120 W.m2 °C. If the
overall temperature difference across the
arrangement is 60 °C, calculate the temperature
difference across the stainless steel.
Take A  1m 2
Δx 0.004
1
1
R SS 

 0.00025; R overall  
 0.00833
k
16
U 120
ΔTSS
R SS
0.00025


 0.03; TSS  (0.03)(60)  1.8C
ΔToverall R overall 0.00833
PANKAJ R. CHANDAR
25
2-13. An ice chest is constructed of Styrofoam
[k=0.033W/m °C] with inside dimensions of 25 by
40 by 100cm. The wall thickness is 5.0 cm. The
outside of the chest is exposed to air at 25 °C with
h=10W/m2 °C. If the chest is completely filled with
ice, calculate the time for the ice to completely
mely. State your assumptions. The heat of fusion
for water is 330kJ/kg.
Ice at 0C
  999.8kg/m 3
V  (0.25)(0.4)(1.0)  0.1m 3
m  100kg
g  (100)(330 *103 )  3.3 *107 J
PANKAJ R. CHANDAR
26
A i  (2)(0.25)(0.4)  (2)(0.4)(1.0)  (2)(0.25)(1.0)
 1.5m 2
A 0  (2)(0.35)(0.5)  (2)(0.5)(1.1)  (2)(0.35)(1.1)
 2.22m 2
Am  1.86m 2
x
0.05

 0.8146
kA (0.033)(1.86)
1
R0 
 0.04
hA0
Rs 
R  0.8596
Q 3.3 *107

 25 - 0
Δτ
Δτ
Δτ  0.135 *106sec  315hr  13days
PANKAJ R. CHANDAR
27
2-15. A hollow sphere is constructed of aluminum with
an inner diameter of 4 cm and an outer diameter
of 8 cm. The inside temperature is 100 °C and the
outer temperature is 50 °C. Calculate the heat
transfer.
4ππk(i  T0 )
q
1 1

ri r0
W
(k  204
)
mC
(4)π4)π(20100 - 50)

 5127W
1
1

0.02 0.04
PANKAJ R. CHANDAR
28
2-18. A steel pipe with 5-cm OD is covered with a
6.4mm asbestos insulation [k=0.096 Btu/h ft °F]
followed by a 2.5 cm layer of fiberglass
insulation [k=0.028 Btu/h ft °F]. The pipe wall
temperature is 315, and the outside insulation
temperature is 38 °C. Calculate the interface
temperature between the asbestos and fiberglass.
k A  0.166
k f  0.0485
W
mC
315  Ti
Ti  38

ln(31.4/25) ln(56.4/31.4)
0.166
0.0485
0.7283(315  Ti )  0.0828(Ti  38)
Ti  286.7C
PANKAJ R. CHANDAR
29
2-23.A cylindrical tank 80 cm in diameter and 2.0 m
high contains water at 80 °C. the tank is 90 percent
full, and insulation is to be added so that the water
temperature will not drop more than 2 °C per
hour. Using the information given in this chapter,
specify an insulating material and calculate the
thickness required for the specified cooling rate.
M W @90%full  (0.9)(970)π(0.8) 2 (2)  905kg
@2C/hr q  (3511)(4191)(2)/3600  2106W
A S  2ππ(0.42  π(0.8)(2)  6.032m 2
Fiberglass boards with k  40mW/mC
(40 *10-3 )(6.032)(80 - 20)
Δx 
 0.68cm
2106
PANKAJ R. CHANDAR
30
2-24. A hot steam pipe having an inside surface
temperature of 250 °C has an inside diameter of 8
cm and a wall thickness of 5.5 mm. It is covered
with a 9 cm layer of insulation having k=0.5W/m
°C, followed by a 4-cm layer of insulation having
k=0.25W/m °C. The outside temperature of
insulation is 20 °C. Calculate the heat lost per
meter of length. Assume k=47W/m °C for the pipe.
For 1m length :
ln(9.1/8)
ln(27.1/9.1)
R(pipe) 
 4.363 *10  4 ; R(ins1) 
 0.3474
2ππ(47
2ππ(0.5
ln(35.1/27.1)
R(ins2) 
 0.1647
2ππ(0.25
R(tot)  1.172
q  ΔT/R  (250  20)/1.172  449W/m
PANKAJ R. CHANDAR
31
2-25. A house wall may be approximated as two 1.2 cm
layers of fiber insulating board, a 8.0 cm layer of
loosely packed asbestos, and a 10 cm layer of
common brick. Assuming convection heattransfer coefficients of 15 W/m2 °C on both sides
of the wall, calculate the overall heat-transfer
coefficient for this arrangement.
Fiberglass k  0.038 x  1.2cm * 2
Asbestos
k  0.1547 x  8.0cm
W
brick
k  0.69
x  10cm h  15 2 * 2
m C
1
W
U
 0.70 2
2 (2)(0.012) 0.08
0.1
m C



15
0.038
0.154 0.69
PANKAJ R. CHANDAR
32
2-30. A plane wall 6.0 cm thick generates heat internally
at the rate of 0.3 MW/m3. One side of the wall is
insulated, and the other side is exposed to an
environment at 93°C. the convection heat-transfer
coefficient between the wall and the environment is
570W/m2 °C. The thermal conductivity of the wall is
21 W/m °C. Calculate the maximum temperature in
the wall.
q  0.30MW/m 3 Same as half of wall 15 cm thick
with convection on each side
g L2 (0.30 *106 )(0.060) 2
T0 - TW 

 25.7C
2k
(2)(21)
q LA  hA(TW - T )
TW - T  (0.30 *106 )(0.060)  31.6C
T0  Tmax  93  25.7  31.6  150.3C
PANKAJ R. CHANDAR
33
2-34. Heat is generated in a 2.5-cm-square copper rod at
the rate of 35.3MW/m3. The rod is exposed to a
convection environment at 20 °C, and the heattransfer coefficient is 4000 W/m2 °C. Calculate the
surface temperature of the rod.
q AL  hPL(Tw  T )
(35.3 *10 6 )(0.025) 2  (4000)(4)(0.025)(Tw  20)
Tw  75.16C
PANKAJ R. CHANDAR
34
2-40. Heat is generated uniformly in a stainless steel
plate having k=20W/m °C. The thickness of the
plate is 1.0 cm and the heat-generation rate is
500 MW/m3. If the two sides of the plate are
maintained at 100 and 200 °C respectively,
calculate the temperature at the center of the
plate for k=20W/m °C.
Use solution from Prob. 2  28
T  T0@x  0
q 2 T1  T2 (500 *10 6 )(0.005) 2 100  200
T0 
L 


2k
2
(2)(20)
2
 462.5C
PANKAJ R. CHANDAR
35
2-41.A plate having a thickness of 4.0mm has an
internal heat generation of 200 MW/m3 and a
thermal conductivity of 25 W/m °C. One side of
the plate is insulated and the other side is
maintained at 100 °C. calculate the maximum
temperature in the plate.
Behavs like half a plate having a thinkness
of 8mm. Max Temp is at x  0
qL2
T0 
 Tw
2k
L  0.004m
T w  100C
(200 *106 )(0.004)
T0 
 100  164C
(2)(25)
PANKAJ R. CHANDAR
36
2-51. Water flows on the inside of a steel pipe with an ID
of 2.5 cm. The wall thickness is 2mm, and the
convection coefficient on the inside is 500W/m2 °C.
The convection coefficient on the outside is 12
W/m2 °C. Calculate the overall heat-transfer
coefficient. What is the main determining factor
for U?
k  43
Ui 
W
mC
1
1.45
ln(
) (0.025)
1
0.025 1
 1.25

( )
500
(2 )(43)
0.029 12
1
2


13
.
54
W
/
m
C
3
5
3
2 *10  4.31*10  71.84 *10
PANKAJ R. CHANDAR
37
5-52.The pipe in Prob. 2-51 is covered with a layer of
asbestos[k=0.18W/m °C] while still surrounded
by a convection environment with h=12W/m2 °C.
Calculate the critical insulation radius. Will the
heat transfer be increased or decreased by
adding an insulation thickness of (a) 0.5mm, (b)
10mm?
r0  k / h  0.18 / 12  0.015m  1.5cm
a)r0  1.25  0.05  1.3cm(increased )
b)r0  1.25  1.0  2.25cm(decreased )
PANKAJ R. CHANDAR
38
2-56.An insulating glass window is constructed of two
5-mm glass plates separated by an air layer
having a thickness of 4 mm. The air layer may
be considered stagnant so that pure conduction
is involved. The convection coefficients for the
inner and outer surface are 12 and 50 W/m2 °C,
respectively. Calculate the overall heat-transfer
coefficient for this arrangement, and the R
value. Repeat the calculation for a single glass
plate 5 mm thick.
PANKAJ R. CHANDAR
39
A  1m 2
R glass  Δx/k  0.005  6.41*10 3
Δx
 0.004  0.1538
k
R conv1  1/h  1/12  0.0833
R air 
R conv2  1/50  0.02
1
U
(2)(6.41*10 3 )  0.1538  0.0833  0.02
 1/0.2699  3.705W/m 2 C
R  0.2699
singe glass plate : R  6.41*10-3  0.0833  0.02  0.01097
U  1/R  9.11W/m 2 C
PANKAJ R. CHANDAR
40
2-60. One end of a copper rod 30 cm long is firmly
connected to a wall which is maintained at 200 °C.
The other end is firmly connected to a wall which is
maintained at 93 °C. Air is blown across the rod so
that a heat transfer coefficient of 17W/m2 °C is
maintained. The diameter of the rod is 12.5mm.
The temperature of the air is 38 °C. What is the net
heta lost to the air in watts?
d 2θ hP

θ0
2
dx
kA
let m  hP , T  38 L  30cm h  17
kA
PANKAJ R. CHANDAR
41
  c1e mx  c 2 e -mx@x  0,   200 - 38  162
d 2
k  386   d A 
x  0.3   93 - 38  55
4
m  [(17) (0.0125)(4)]1  3.754
162  c1  c2 , 55  3.084c1  0.324c 2 , c1  0.91 c 2  161.09
  0.91e mx  161.09e -mx
L
1
q   hPdx  hP ( )[0.91e mx  161.09e  mx ]0L
m
0
 hPkA[0.91e mx  161.09e  mx ]00.3
1
2 2
 [(17) (0.0125)(386) (0.0125) ] * [0.91e mx  161.09e mx ]00.3
 122.7W
PANKAJ R. CHANDAR
42
2-66. A very long copper rod[k=372Wm °C] 2.5 cm in
diameter has one end maintained at 90 °C. The
rod is exposed to a fluid whose temperature is 40
°C. The heat-transfer coefficient is 3.5 W/m2 °C.
How much heat given up by the rod?
1
q  hPkA 0  [(24) (0.025)(385) (0.025) ] 2 (150  20)
2
 10.05W
PANKAJ R. CHANDAR
43
2-70. A 2.5-cm-diameter tube has circumferential fins of
rectangular profile spaced at 9.5-mm increments
along its length. The fins are constructed of
aluminum and are 0.8 mm thick and 12.5 mm long.
The tube wall temperature is maintained at 200 °C,
and the environment temperature is 93 °C. The
heat-transfer coefficient of 28 W/m2 °C. Calculated
the heat loss from the tube per meter of length.
Tw  200C T  93C L  12.5mm t  0.8mm
r1  1.25cm k  204 L c  12.9mm h  110 r2c  2.54
3/2
r2c/r1  2.03 A m  1.03 *10-5 m 2 L c (h/kA m )1/2  0.335
f  0.87 No. of fins  1.0/0.0095  105.3
PANKAJ R. CHANDAR
44
Tube Surface area  (105.3) (0.025)(9.5  0.8)(10 3 )
 0.0719m 2
Tube heat transfer  (110 )(0.0719)(200  93)  846.6W
q/fin  (0.87)(2) (110 )(0.0254 2  0.01252 )(200  93)
 31.46W
Total fin heat transfer  (31.46)(105.3)  3312W
Total heat transfer  846.6  3312  4159W
PANKAJ R. CHANDAR
45
2-72. A straight rectangular fin 2.0 cm thick and 14 am
long is constructed of steel and placed on the
outside of a wall maintained at 200 °C. The
environment temperature is 15 °C, and the heattransfer coefficient for convection is 20 W/m2 °C.
Calculate the heat lost from the fin per unit depth.
3/2
k  43C t  2cm h  20 L c  15cm L c (
h 1/2
)  0.723
kA m
ηf  0.75
q  (0.75)(20)(ππ)(0.15)200  15)  833
W
depth
m
PANKAJ R. CHANDAR
46
2-76. A long stainless-steel rod [k=16W/m °C] has a
square cross section 12.5 by 12.5 mm and has one
end maintained at 250 °C. The heat-transfer
coefficient is 40 W/m2 °C, and the environment
temperature is 90 °C. Calculate the heat lost by
the rod.
k  16 h  40 T0  250C T  90C
P  (4)(0.0125)  0.05m
A  (0.0125) 2  1.565 *10  4 m 2
q  hPkAθ 0
 [(40)(0.05)(16)(1.565 *10  4)]1/2 (250  90)
 11.31W
PANKAJ R. CHANDAR
47
2-81. A straight rectangular fin has a length of 2.0cm
and a thickness of 1.5mm. The thermal
conductivity is 55W/m °C, and it is exposed to a
convection environment at 20 °C and h=500W/m2
°C. Calculate the maximum possible heat loss for
a base temperature of 200 °C. What is the actual
heat loss?
L c  2.075cm
q max  (500)(2)(0.02075)(200 - 20)
 3735W/m
h 1/ 2
Lc (
)
kA m
3/2
f  42%
q act  (0.42)(3735)  1569W/m
PANKAJ R. CHANDAR
48
2-84. A circumferential fin of rectangular profile is
constructed of aluminum and surrounds 3-cmdiameter tube. The fin is 2 cm long and 1 mm
thick. The tube wall temperature is 200 °C, and the
fine is exposed to a fluid at 20 °C with a convection
heat-transfer coefficient of 80 W/m2 °C. Calculate
the heat from the fin.
r1  1.5cm L  2cm r2  3.5cm t  1mm h  80
k  200 L c  3.55cm Lc3/2 (
h 1/2
)  0.41
kA m
r2c
 2.37 ηf  0.81
r1
q  (80)π80)π(552  0.0152 )(2)(200  20)(0.81)  75.9W
PANKAJ R. CHANDAR
49
2-87. A straight fin having a triangular profile has a
length of 5 cm and a thickness of 4 mm and is
constructed of a material having k=23W/m °C.
the fin is exposed to surrounds with a convection
coefficient of 56 W/m2 °C and a temperature of 40
°C. The base of the fin is maintain at 200 °C.
Calculate the heat lost per unit depth of fin.
L  5cm Lc  5cm t  4mm k  23 h  20
h 1/2
Lc (
)  1.042 ηf  0.68 q  ηf hAθ 0
kAm
m2
2
2 1/2
A  (2)(0.002  0.05 )  0.10008
mdepth
q  (0.68)(20)(0.10008)(200  40)  217.8W/m
3/2
PANKAJ R. CHANDAR
50
2-89. A circumferential fin of rectangular profile is
constructed of stainless steel(18% Cr, 8%Ni).
The thickness of the fin is 2.0mm, the inside
radius is 2.0 cm, and the length is 8.0 cm. The
base temperature is maintained at 135 °C and
the fine is exposed to a convection environment
at 15 °C with h=20W/m2 °C. Calculate the heat
lost by the fin.
t  2mm, r1  2.0cm , r 2  10.0cm, L  8cm, Lc  8.1cm
Γ 2c  10.2cm, h  20, k  17, Lc3/2 (
h 1/2
)  1.96
kAm
ηf  0.33
g  (0.33)(2)(0.02555)(500)(125 - 20)  885W/m
PANKAJ R. CHANDAR
51
2-95. A straight fin of rectangular profile is constructed
of stainless steel(18% Cr, 8%Ni)and has a length
of 5 cm and a thickness of 2.5 cm. The base
temperature is maintained at 100 °C and the fin is
exposed to a convection environment at 20 °C
with
k  17, h  47, L  5cm, t  3.5cm, t  2.5cm,
h 1/ 2
Lc (
)  0.657, f  0.8
kAm
q  (0.8)(47)(2)(0.0625)(100 - 20)  376W/m
3/2
PANKAJ R. CHANDAR
52
2-97. A circular fin of rectangular profile is attached to
3.0-cm-diameter tube maintained at 100 °C. The
outside diameter of the fin thickness is 1.0mm. The
environment has a convection coefficient of the
material for a fin efficiency of 60 percent.
r1  1.5cm, r2  4.5cm, t  1.0mm, h  50, r2c  4.55cm
r2c
Lc  3.05cm, k  204, ηf  0.6,
3
r1
h 1/2
Lc (
)  0.78
kAm
1
3 50
k
[0.0305]
 76.5W/mC
2
(0.001)(0.0305)
0.78
3/2
PANKAJ R. CHANDAR
53
2-98. A circumferential fin of rectangular profile having a
thickness of 1.0mm and a length of 2.0 cm is place
on a 2.0-cm-diameter tube. The tube temperature is
150 °C, the environment temperature is 20 °C, and
h=200W/m2 °C. The fin is aluminum. Calculate the
heat lost by the fin.
t  1.0mm, L  2.0cm, 1  1.0cm, h  200, k  204,
h 1/ 2
Lc  2.05cm, 2c  3.05cm, Lc (
)  0.642
kAm
r2c  0.305 f  0.68
3/2
q  (0.68)(200)(2) (0.03052 - 0.012 )(150 - 20)  92.2W
PANKAJ R. CHANDAR
54
2-103. An aluminum fin is attached to a transistor which
generates heat at the rate of 300mW. The fin has a
total surface area of 9.0cm2 and fin is 0.9*10-4m2
°C/W, and the contact area is 0.5cm2. Estimate the
temperature of the transistor, assuming the fin is
uniform in temperature.
1/h c  0.9 *10 4 , A c  0.5cm 2 , q  300mW
Assume fin @27C
1
4
q  (300 *10 ) 
(
0
.
5
)(
10
)(Tt  27)
4
0.9 *10
Tt  27.54C
-3
PANKAJ R. CHANDAR
55
2-118 Solution cont’d.
2.0 mm Fin LC = 1.1 cm
LC
3

2
h 


k
A
M

f  0.92
1
2
 0.218


q  3160  0.0212  0.012 2 T  0.92
 0.95 T
3.0 mm Fin LC = 1.15 cm
LC
3

2
h 


k
A
M

f  0.95

1
2
 0.186

q  2160  0.02152  0.012 2 T  0.95
 0.69 T
Conclusion: Several thin fins are better than a
few thick
fins. More heat transfer for the same weight of fins.
PANKAJ R. CHANDAR
56