Question

A straight fin is made from copper (k = 388 W/m-K) and is 0.5 cm in diameter and 30 cm long. The temperature at the base of the fin is 75oC and it is exposed to flowing air at 20oC with h = 20 W/m2-K. That is the rate of heat transfer from this pin?

Answer 1

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, $T_{b} = 75^{\circ}C$

Air temperature, $T_{infty} = 20^{\circ}$

k = 388 W/mK

h = $20\ W/m^{2}K$

Now,

Perimeter of the fin, p = $\pi D = 0.005\pi \ m$

Cross-sectional area of the fin, A = $\frac{\pi}{4}D^{2}$

A = $\frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}$

To calculate the heat transfer rate:

$Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})$

where

$m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237$

Now,

$Q_{f} = \sqrt{20\times 388\times 0.005\pi\times 6.25\times 10^{- 6}\pi}tanh(41.237\times 0.3)(75 - 20) = 0.119\ W$