Topic: Heat Transfer
The radioactive heat transfer rate per unit area $(W/m^{2})$ between two plane parallel grey surfaces (emissivity $=0.9$ maintained at 400 K and 300 K is (Stefan Boltzmann constant $\sigma= 5.67\times10^{-8}W/m^{2}K^{4})$
- A. 992
- B. 812
- C. 464
- D. 567
Answer: B. 812
Explanation: Given, $\epsilon=0.9$, $T_{1}=400K$, $T_{2}=300~K$, $\sigma=5.67\times10^{-8}W/m^{2}K^{4}$. Heat transfer per unit area between parallel grey surface is given by $(\epsilon_{1}=\epsilon_{2}=\epsilon)$: $Q=\frac{\sigma(T_{1}^{4}-T_{2}^{4})}{(\frac{1}{\epsilon_{1}}+\frac{1}{\epsilon_{2}}-1)} = \frac{5.67\times10^{-8}\{(400)^{4}-(300)^{4}\}}{(\frac{1}{0.9}+\frac{1}{0.9}-1)} = 811.84~W/m^{2} \approx 812~W/m^{2}$.