Question

A very long rectangular fin is attached to a flat surface such that the temperature at the end of the fin is essentially that of the surrounding air, i.e. 20C. Its width is 5 cm; thickness is 1 mm; thermal conductivity is 200 W/m.K; and base temperature is 40C. The heat transfer coefficient is 22 W/m2 .K. Estimate the temperature at a distance of 5 cm from the base and the rate of heat loss from the entire fin.

Answer 1

Answer:

The answer is below

Explanation:

Given that:

x = 5 cm = 0.05 m, width (w) = 5 cm = 0.05 m, thickness (t) = 1 mm = 0.001 m, k = thermal conductivity = 200 W/mK, h = heat transfer coefficient = 22 W/m².K, Tb = base temperature = 40°C, T∞ = 20°C

$m=\sqrt{ \frac{hp}{kA} }=\sqrt {\frac{22*(2*0.05+2*0.001)}{200*(0.05*0.001)}} =15\ m^{-1}\\\\\frac{T-T_{\infty}}{T_b-T_{\infty}} =e^{-mx}\\\\\frac{T-20}{40-20} =e^{-15*0.05}\\\\T-20=9.45\\\\T=29.45^oC\\\\The\ rate\ of\ heat\ loss\ is:\\\\\dot{Q}=\sqrt{hpkA}(T_b-T_{\infty})\\ \\\dot{Q}=\sqrt{22*(2*0.05+2*0.001)*200*0.05*0.001}*(40-20)\\\\\dot{Q}=3\ W$