Heat Transfer Mechanisms — Question 7

Question 7

One illuminated face can radiate from both sides

Use $\sigma=5.67\times10^{-8}\,\mathrm{W/(m^2K^4)}$. A thin uniform-temperature plate absorbs a collimated beam with absorbed flux $q_{\mathrm{abs}}=800\,\mathrm{W/m^2}$ on one face. Each face has area $A$ and thermal emissivity .800. Both faces radiate to a 300 K black enclosure with unit view factor; neglect edges, convection and conduction. The stated beam flux is already absorbed, so do not multiply it by emissivity again.

1. Find the steady plate temperature.

2. If one face is perfectly insulated against radiation while the same beam remains absorbed on the other face, find the new equilibrium temperature and explain why area cancels.

Question 7 - Solution

One illuminated face can radiate from both sides

(a) Method and calculation

$$q_{\mathrm{abs}}A=2\epsilon\sigma A(T^4-300^4).$$$$T=\left[300^4+\frac{800}{2(.8)\sigma}\right]^{1/4}\approx360.653\,\mathrm K.$$

(b) Interpretation and further calculation

$$800A=\epsilon\sigma A(T_{\mathrm{one}}^4-300^4).$$$$T_{\mathrm{one}}=\left[300^4+\frac{800}{.8\sigma}\right]^{1/4}\approx400.533\,\mathrm K.$$Absorbed power and emitting area both scale with $A$; the number of emitting faces remains in the balance.

Physical check

Removing a heat-loss route raises the equilibrium temperature for fixed absorbed power.