Question

A horizontal metal plate which is completely insulated except for one exposed surface of 100 sq. ft. is heated electrically to maintain a temperature of 200 °F on the exposed surface. Heat is transferred from this surface by both radiation and conduction through air (assumed to be without motion) to a second plate of similar size maintained at a surface temperature of 100 °F. The distance between the parallel plates is 0.25-in., and each plate has an imissivity of 0.20. What is the watts input to the warmer plate?

Answer 1

Solution;

Given;

$A_s=100sq.ft=9.29m^2$

$T_h=200°F=93.33°c=366.33K$

$T_c=100°F=37.78°c=310.78K$

$L=0.25in=6.35×10^{-3}m$

$\xi=0.20$

We know;

$Q_T=Q_{rad}+Q_{conv}$

$Q_{rad}=\xi A_s\sigma(T_h^4-T_c^4)$

Where;

$\sigma=5.67×10^-8Wm^{-2}/K^4$

By substitution;

$Q_{rad}=0.20×9.29×5.67×10^{-8}(366.3^4-310.78^4)$

$Q_{rad}=914.48W$

Now;

$Q_{conv}=hA_s(T_h-T_c)$

Where h is the convectional heat transfer coefficient,value is;

$h=3.5W/m^2°c$

Hence;

$Q_{conv}=3.5×9.29(93.33-37.78)$

$Q_{conv}=1806.21W$

Therefore;

$Q_T=914.48+1806.21$

$Q_T=2720.69W$