$Q_(n-shields)$=$\frac{Aσ(T_1^4-T_2^4)}{((n+1)[\frac{2}{ϵ}-1])}$

Radiation heat transfer between two large parallel plates of emissivitiesε_1 and ε_2 maintained at uniform temperatures T1 and T2 is given by

Q_(12,no shield)=(Aσ(T_1^4-T_2^4))/(1/ε_1 +1/ε_2 -1)

Now consider a radiation shield placed between these two plates, as shown in Figure

Fig: The radiation shield placed between two parallel plates and the radiation network associated with it.

Let the emissivities of the shield facing plates 1 and 2 be $ε_3,1$and $ε_3,2$, respectively. Note that the emissivity of different surfaces of the shield may be different. The radiation network of this geometry is constructed, as usual, by drawing a surface resistance associated with each surface and connecting these surface resistances with space resistances, as shown in the figure. The resistances are connected in series, and thus the rate of radiation heat transfer i

$Q_12,one shield$=$\frac{(E_b1-E_b2)}{(\frac{(1-ε_1)}{(A_1 ε_1 )}+\frac{1}{(A_1 F_12 )}+\frac{(1-ε_3,1)}{(A_3 ε_31 )}+\frac{(1-ε_3,2)}{(ε_3,2 A_3 )}+\frac{1}{(A_3 F_32 )}+\frac{(1-ε_2)}{(A_2 ε_2 )}+\frac{1}{(A_3 F_32 )}+\frac{(1-ε_2)}{(A_2 ε_2 )}))}$

Noting that F13 =F23 = 1 and A1 = A2 = A3 =A for infinite parallel plates,

$Q_12,one shield$ =$\frac{(Aσ(T_1^4-T_2^4))}{((\frac{1}{ε_1} +\frac{1}{ε_2} -1)+(\frac{1}{ε_3,1} +\frac{1}{ε_3,2} -1) )}$

where the terms in the second set of parentheses in the denominator represent the additional resistance to radiation introduced by the shield. The appearance of the equation above suggests that parallel plates involving multiple radiation shields can be handled by adding a group of terms like those in the second set of parentheses to the denominator for each radiation shield. Then the radiation heat transfer through large parallel plates separated by N radiation shields becomes,

$Q_12,N shields = \frac{A\sigma T_1^4-T_2^4}{\frac{1}{ε_1}+\frac{1}{ε_2}-1+\frac{1}{ε_3,1}+\frac{1}{ε_3,2}-1+....\frac{1}{ε_N,1}+\frac{1}{ε_N,2}-1}$

If the emissivities of all surfaces are equal,

$Q_12,N shields$=$\frac{Aσ(T_1^4-T_2^4)}{(N+1)(\frac{1}{ε}+\frac{1}{ε-1})}$=$\frac{(Aσ(T_1^4-T_2^4))}{((N+1)(\frac{2}{ε}-1) )}$

Hence proved.