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Question

A fin has 5 mm diameter and 100 mm length. The thermal conductivity of fin material is 400 Wm1K1. One end of the fin is maintained at 130C and its remaining surface is exposed to ambient air at 30C. If the convective heat transfer coefficient is 40 Wm2K1, the heat loss (in W) from the fin is

A

0.08
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B

5.0
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C

7.0
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D

7.8
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Solution

The correct option is B
5.0
Given data:
diameter of fin,
d=5 mm=0.005 m
cross-sectional area,
A=π4d2=3.144×(0.005)2
=1.96×105m2
Length of fin,
L=100 mm=0.1 m
Thermal conductivity
K=400 W/mK
T=130C
T=30C
h=40 W/m2K

Perimeter: P=πd=3.14×0.005=0.0157 m
m=hPkA=40×0.0157400×1.95×105
=8.97 m1
mL=8.97×0.1=0.897
For long fin, mL>2.64
So, the fin is insulated tip,

Qfin=hPkA(TT)tanh mL
=40×0.0157×400×1.95×105
(13030)tanh(8.97×0.1)
=0.0699×100×tanh 0.897
=0.0699×100×0.7148=4.99W 5W

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