1/9
Course # 804314-2 : Heat Transfer (1)
Second Semester 1435/1436 H (2014/2015 G)
Solution of Tutorial 3
Exercise 3.1:
See lecture
Exercise 3.2:
The ratio of the length of the fin and its diameter is : L D D
L 25
So, we can assume that the fin is very long fin (infinite fin), then its effectiveness is:
c
f hA
kP
where:
k = 237 W/mK,
h = 12 W/m2 K,
L = 10 cm = 0.1 m,
D = 4 mm = 0.004 m,
m 0125 . 0 m 004 .
0
P D ,
2 5 - 2
2 2
m 10 256 . 1 m 4
004 . 0 4
and D
Ac
So, the fin effectiveness is
198 . 10 140
256 . 1 12
0125 . 0 237
5
f
f
Umm Al-Qura University
College of Engineering & Islamic Architecture
Mechanical Engineering Department
2/9
Exercise 3.3
(a) The perimeter P of this fin: P D0.005 m1.57102m (b) The area of conduction heat transfer Ac :
2 5 - 2
2 2
m 10 96 . 1 m 4
005 . 0
4
D Ac
(c) The fin heat transfer rate:
The ratio of the length of the fin and its diameter is : L D D
L 20
005 . 0
1 .
0 .
So, we can assume that the fin is very long fin (infinite fin). From table (case D):
b b
f M hPkAc hPkAc
q
where
C 0 8 C ) 20 100
(
Tb T
b
W 964 . 1 W 80 10
96 . 1 200 10
57 . 1
10 2 5
qf
(d) The temperature of the end of fin:
From table (case D), the temperature distribution
mx b
b
T e T
T x
T
( )
T(x) (Tb T )e mx T
where 6.394
10 96 . 1 200
10 57 . 1 10
5
2
kAc
m hP
m in and C in where
; 20 80
)
(x e 6.394 T x
T x
The temperature in the end of fin is for x = L, thus
80 20
C 60 C)
( 6.3940.1
TL T x L e
3/9
Exercise 3.4
We can model the pot handle as an extended suface. Assume that ther is no heat transfer at the free end of the handle.
The condition matches that specified in the fins tables, case B. Use the following data:
h = 5 W/m2 °C,
L = 15 cm = 0.15m,
D = 3cm=0.03 m,
P D0.03 m0.094m,
2 -4 2
2 2
m 10 068 . 7 m 4
03 . 0
4
D Ac
b
Tb T
10020
C80Cor
kAc
m hP , M hPkAc b and the heat transfer rate through a plot handle is
mL Mqf tanh , therefore:
The heat transfer rate through a plot handle:
(a) For aluminium (k = 237 W/mK):
675 . 10 1 068 . 7 237
094 . 0 5
4
kAc
m hP and
W 447 . 22 80 10 068 . 7 237 094 . 0
5 4
b
hPkAc
M
So, the heat transfer rate through a plot handle is
W 5.524 W
) 15 . 0 675 . 1 tanh(
447 . 22 )
tanh(
M mL qf
4/9 (b) Special metal (k = 3 W/mK):
888 . 10 14
068 . 7 3
094 . 0 5
4
kAc
m hP and
W 525 . 2 80 10 068 . 7 3 094 . 0
5 4
b
hPkAc
M
So, the heat transfer rate through a plot handle is
W 2.467 W
) 15 . 0 888 . 14 tanh(
525 . 2 )
tanh(
M mL qf
The temperature distribution along the pot handle:
The condition matches that specified in the fins tables, case B. the temperature distrubtion along the pot handle is:
) cosh(
) ( cosh )
( ) (
mL x L m T
T T x T b
x
b
T T T
mL x L x m
T ( b )
) cosh(
) ( ) cosh (
m in and C in ) ( where
; 20 ) 20 100 ) ( 15 . 0 cosh(
) 15 . 0 ( ) cosh
( T x x
m
x x m
T
m in and C in ) ( where
; ) 20
15 . 0 cosh(
) 15 . 0 ( 80 cosh
)
( T x x
m
x x m
T
(a) For aluminium (m 1.675):
m in and C in ) ( where
; ) 20
15 . 0 675 . 1 cosh(
) 15 . 0 ( 675 . 1 80 cosh )
( x T x x
x
T
1.675(0.15 )
20 ; where ( )in C and in m cosh539 . 77 )
(x x T x x
T
At the end of the pot handle (x = L = 15cm = 0.15m):
0 20 97.539 Ccosh 539 . 77 ) 15 . 0
(
T x L
The temperature at the handle (x=15 cm) is very hot at all. So, We can”t touch the end of the handle.
5/9 Plot the temperature distribution along the plot handle:
(b) For the Special (m14.888):
m in and C in ) ( where
; ) 20
15 . 0 888 . 14 cosh(
) 15 . 0 ( 888 . 14 80 cosh
)
( x T x x
x
T
11.532(0.15 )
20 ; where ( )in C and in m cosh954 . 16 )
(x x T x x
T
At the end of the pot handle (x = L = 15cm = 0.15m):
0 20 36.954 C cosh954 . 16 ) 15 . 0
(
T x L
Plot the temperature distribution along the plot handle:
0 0.0375 0.075 0.1125 0.15 99
97
0 0.0375 0.075 0.1125 0.15 30
20 0.0 75 0.1 125 0.1
6/9 Comments: The temperature at the handle (x = 15 cm) is only 36.954 °C, not hot at all. So, We can touch the end of the handle.
This example illustrates the important role of the thermal conductivity of the material in the temperature disribution in a fin.
Exercise 3.5:
There are N fins in the array (for 1 m length), the total surface area is
At NAf Ab Where Af is the surface area of a single fin, and Ab is the area of the prime surface.
The total rate of heat transfer by convection is
b tube tube
f all
tube all
t
q q q N q q hA
q
fins where
fins and
In this equation , the heat transfer from a single fin is:
b f f
f
h A
q
where f is the efficiency of a single fin.
) (
2 ) (
2 R
1L t
finsR
1L N t A
tube
all
For L= 1m,
7/9
2 2
0.0565 m m
) 002 . 0 200 1
( 015 . 0 2
1m,
For L A
tube
So, the total heat transfer is
qt Nf hAf b hAbb
Note: So, the increase in heat transfer rate per meter as result of adding these fins is:
b f f f
fins
all
N q N h A
q
In these equations, the following data are N = 200/m
) (
2 )
( in Graph R
22R
12A
A
f
s
c
( 120 25 ) C 95 C
T
bT
bh = 50 W/m2K
To determine the fin efficiency (circular fin), we used graphical method (graph below) with the following data:
L = R2 – R1; where R1 = D1/2= 1.5 cm = 0.015 m, and R2 = D2/2= 3 cm = 0.03 m
Lc = L + t/2 = 1.5 cm + 1 mm = 1.51 cm;
2 -5
m 10 3.02 0.0151
002 .
0
cp
t L
A
R2c = R2 + t/2 = 3.1 cm ; So,
2 05 . 51 2 . 1
1 . 3
1
2
R R
c
So, we must use Cure 2 and We can calculate the surface area of a single fin: Af:
2 4 2
2 2 2
2 1 2
2
) 2 ( 2 . 02 1 . 51 ) cm 6 . 2 cm 6 . 2 10 m (
2 )
(
A in Graph R R
A
f s c8/9
k = 50 W/mK, and hc = h = 10W/m2K
So, we obtain
33 . 10 0
02 . 3 50 0151 50
. 0
2 / 1 5 2
/ 3 2
/ 1 2
/
3
p
c
c
kA
L h
Using Curve of
2
1
2
R R
c
and 0 . 33
2 / 1 2
/
3
p c
c
kA
L h
Then the fin efficiency (from Graph) is
f 0 . 88
So, we obtain the results below:
9/9
Heat transfer rate Expression Value (W)
From the tube (qtube) qtube hAbb 268.375
From a single fin (qf)
b f f
f
h A
q
2.595From N fins per meter (qall fins)
f fins
all
Nq
q
519The total per meter (qt)
tube fins
all
t
q q
q
787.375