Calculate the number of screens to be placed between the two surfaces to achieve this reduction in heat exchange assuming emissivity of screens 0.05 and that of the surfaces as 0.8.

$ε_1$=$0.8ε_2$=0.$8ε_s1$=$0.05ε_s2$=0.05σ=5.67×$10$^(-8) w/$m^2$ K

Let N be the number of screens required. Then, with no radiation shield, we have the radiant heat transfer:

$Q_12$=$\frac{Aσ(T_1^4-T_2^4)}{(N+1)(\frac{1}{ε_1} +\frac{1}{ε_2} -1)}$

And, with N radiation shields, the radiant heat transfer is:

$Q_12,N shield$=$\frac{Aσ(T_1^4-T_2^4)}{(\frac{1}{ε_1} +\frac{1}{ε_2} -1+(\frac{1}{ε_s1} +\frac{1}{ε_s2} -1) )}$ Then, by data:

$\frac{Q_12N_shields}{Q_12}$ =$\frac{(\frac{1}{ε_1} +\frac{1}{ε_2} -1)}{(\frac{1}{ε_1} +\frac{1}{ε_2} -1)+N(\frac{1}{ε_s1} +\frac{1}{ε_s2} -1)}$=1/79

Solving Eq a we get N, the number of screens required.

We get:

N=$\frac{79(\frac{1}{ε_1} +\frac{1}{ε_2} -1)-\frac{1}{ε_1} +\frac{1}{ε_2} -1)}{(\frac{1}{ε_s1} +\frac{1}{ε_s2} -1)}$= $\frac{79(\frac{2}{0.8}-1)-(\frac{2}{0.8}-1)}{(\frac{2}{0.05}-1)}$=3

Therefore 3 screens to be placed between the two surfaces to achieve reduced heat transfer.