If ε is the emissivity of surfaces and shields and n is the number of shields, introduced between the two surfaces, then overall emissivity is given by

Explanation:

1 shield kept between the plates will bring in 3 additional resistances extra into the network out of which 2 are surface resistances and 1 is space resistance.

Hence if there are n number of radiation shields, then a total 2n + 2 number of surface resistance and n + 1 number of space resistances will be there in the radiation network drawn with n number of shields.

The formula for heat flux without shield when every surfaces has emissivity ϵ,

\({\left( {\frac{q}{A}} \right)_{without\;shield}}= \;\frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{{\frac{1}{\epsilon_1}+}\frac{1}{\epsilon_2} - 1}} = \;\frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\frac{2}{\epsilon} - 1}}\)

\({\left( {\frac{q}{A}} \right)_{with\;one\;shield}} = \frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\frac{4}{\epsilon} - 2}} = \frac{1}{2}\;\times\frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\frac{2}{\epsilon} - 1}}\)

If there are n number of shields kept between plates then, 

\({\left( {\frac{q}{A}} \right)_{with\;n\;shields}} = \frac{1}{{n + 1}}{\left( {\frac{q}{A}} \right)_{without\;any\;shield}}\)

\({\left( {\frac{q}{A}} \right)_{with\;n\;shields}} = \frac{1}{{n + 1}}\;\frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\frac{2}{\epsilon} - 1}} = \frac{\epsilon}{{\left( {n + 1} \right)\left( {2 - \epsilon} \right)}}\sigma \left( {T_1^4 - T_2^4} \right)\)

From the above equation, equivalent emissivity,

\({\epsilon_{eq}} = \frac{\epsilon}{{\left( {n + 1} \right)\left( {2 - \epsilon} \right)}}\)