Sheet ch.
2 – fins
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Q1/
2
Q2/ Fin of 20 mm length and 3 mm base thickness with thermal conductivity = 45 W/m.K.
Convection coefficient = 100 W/m2.K, base temperature = 120°C surrounding fluid temperature
= 35°C.
Determine the heat flow also the fin effectiveness (Use the charts) for:
(i) rectangular fins and
(ii) triangular
Solution:
(i) rectangular fins
( ) ( )
Entering the chart at 0.585, fin efficiency is read as 0.81
q = 0.81× 100 × 2 × 0.0215 × 1 (120 – 35)
= 296.06 W
Effectiveness = 296.06/(100×0.003 × 1(120 – 35)) = 11.61
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(ii) For triangular fin:
Lc = L = 0.02, As = 2 × 0.02 × 1 , Ap = (0.003/2) × 0.02
( ) ( )
Entering the chart at 0.243, efficiency is read as 0.97
Q = 0.97 × 100 × 2 × 0.02 × 1 × (120 – 35) = 329.8 W
Effectiveness = 329.8 /(100×0.003 × 1 (120 – 35)) = 12.933
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Q3/ An aluminum fin 1.5 mm thick, is placed on a circular tube with 2.7-cm OD. The fin is 6
mm long. The tube wall is maintained at 150◦C, the environment temperature is 15◦C, and the
convection heat-transfer coefficient is 20W/m2. ◦C. Calculate the heat lost by the fin.
Solution:
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Q4/One end of a copper rod 30 cm long is firmly connected to a wall that is maintained at
200◦C. The other end is firmly connected to a wall that is maintained at 93◦C.
Air is blown across the rod so that a heat-transfer coefficient of 17 W/m2 ・ ◦C is maintained.
The diameter of the rod is 12.5 mm. The temperature of the air is 38◦C.
What is the net heat lost to the air in watts?
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Q5/ A circumferential fin on a pipe of 50 mm OD is 3 mm thick and 20 mm long. Determine the
(i) heat flow and effectiveness
(ii) If the pitch is 10 mm, determine the fin number and increase in heat flow for 1 m
length of pipe. Also determine the total
efficiency.
Lc = L + t/2 = 0.02 + 0.003/2 = 0.0215 m
r2c = r1 + Lc = 0.025 + 0.0215 = 0.0465 m
Am = t * Lc = 0.003 (0.0215) = 6.45 × 10-5 m2
As = 2π (r2c 2 – r12)
= 2π (0.04652 – 0.0252)
= 1.53725× 10-3 m2
(i) ( ) ( )
r2c /r1 = 0.0465/0.025 = 1.86
Entering the chart at 0.585 and using the curve corresponding to r2c /r1 =
1.86
(interpolation) the fin efficiency is read as 0.75
Q = 0.75 × 2π (0.04652 – 0.0252) 100 × (120 – 35) = 61.54 W/fin
Effectiveness = 61.54 /2(π × 0.025 × 0.003 × 100 × (120 – 35)) = 15.37
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(ii) Considering 1 m length
The heat flow without the fins = 2π × 0.025 × 1 × 100 × (120 – 35)
= 1335.2 W/m length.
As the pitch is 10 mm, the number of fins per m length will be 100 fins remaining
base length will be 0.7 m.
Q with fins = 100 × 61.54 + 1335.2 × 0.7 = 7088.64 W
This is about 5.8 times that of bare pipe.
Total area = (π × 0.025 × 1) + 100 × π (0.0452 – 0.0252) = 0.94247 m2
Maximum heat flow = hAΔT = 100 × 0.94247 × (120 – 35) = 8011 W.
Total efficiency = 7088.64 / 8011= 0.8848 or 88.48%.
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Q6/ Circumferential fins made of aluminum (k = 200 W/m K) are soldered the heat dissipation
from a 2.5 cm OD tube to the outer surface. The fins thick are 0.1 cm and have an outer diameter
of 5.5 cm. If the tube temperature is 100°C, the environmental temperature is 25°C, and the heat
transfer coefficient between the fin and the environment is 65 W/m2 K, calculate the rate of heat
loss from two fins.
Solution:
A parameters required to obtain the fin efficiency
curve are
( )
( ) * +
( )
( ) * +
( ) ( )( )
( )
( ) * +
( ) ( )
( )
The fin efficiency is found to be 91%.
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The rate of heat loss from single fin is
( )
[( ) ]( )
*( ) +( )
[ ] ( )
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Q7/ A 250 circumferential fins made of aluminum (k = 186 W/m K) are soldered the heat
dissipation from a (0.05m OD) tube to the outer surface. The fins thick are 0.1 cm and have an
outer diameter of 0.06 m. If the tube temperature is 180°C, the environmental temperature is
25°C, and the heat transfer coefficient between the fin and the environment is 40 W/m2 K,
calculate the rate of heat loss from two fins
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H.W
Q1) One end of a copper rod (k = 380 W/m.K), 300 mm long is connected to a wall which is
maintained at 300°C. The other end is firmly connected to other wall at 100°C. The air is
blown across the rod so that the heat transfer coefficient of 20 W/m^2.K20W/m2.K is
maintained. The diameter of the rod is 15 mm and temperature of air is 40°C. Determine :
(i) Net heat transfer rate to air,
(ii) The heat conducted to other end which is at 100°C.
Q2) Consider a copper pin fin 0.25 cm in diameter k = 396 W/m. K that protrudes from a wall a
95°C into ambient air at 25°C. The heat transfer is mainly by natural convection with a
coefficient equal to 10 W/m2 .K. Calculate the heat loss, assuming that :
(a) The fin is "infinitely long"
(b) The fin is 2.5 cm long and the coefficient at the end is the same as around the circumference.
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