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

The answer is below

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