Answer:
The rate of transfer of heat is 0.119 W
Solution:
As per the question:
Diameter of the fin, D = 0.5 cm = 0.005 m
Length of the fin, l =30 cm = 0.3 m
Base temperature, [tex]T_{b} = 75^{\circ}C[/tex]
Air temperature, [tex]T_{infty} = 20^{\circ}[/tex]
k = 388 W/mK
h = [tex]20\ W/m^{2}K[/tex]
Now,
Perimeter of the fin, p = [tex]\pi D = 0.005\pi \ m[/tex]
Cross-sectional area of the fin, A = [tex]\frac{\pi}{4}D^{2}[/tex]
A = [tex]\frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}[/tex]
To calculate the heat transfer rate:
[tex]Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})[/tex]
where
[tex]m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237[/tex]
Now,
[tex]Q_{f} = \sqrt{20\times 388\times 0.005\pi\times 6.25\times 10^{- 6}\pi}tanh(41.237\times 0.3)(75 - 20) = 0.119\ W[/tex]
