Two long parallel plates of the same emissivity 0.5 are maintained at different temperatures and have radiation heat exchange between them. The radiation shield of emissivity 0.25 placed in the middle will reduce radiation heat exchange to:

Explanation:

We know that,

Radiation heat exchange between parallel plates

\(Q = \frac{{\sigma \left( {\;T_1^4\;-\;T_2^4\;} \right)}}{{\left( {\;\frac{1}{{{\varepsilon _1}}}\;+\;\frac{1}{{{\varepsilon _2}}}\;-\;1\;} \right)}}\)

Radiation heat exchange when radiation shield is inserted between plates 

\(Q = \frac{{\sigma \left( {\;T_1^4\;-\;T_2^4\;} \right)}}{{\left( {\;\frac{1}{{{\varepsilon _1}}}\;+\;\frac{1}{{{\varepsilon _2}}}\;+\;\frac{2}{{{\varepsilon _3}}}\;-\;2\;} \right)}}\)

Calculation:

Given:

\({\varepsilon _1} = \;{\varepsilon _2} = \varepsilon = 0.5\;\left( {{\rm{\;emissivity\;of\;plates\;}}} \right)\)

\({\varepsilon _3} = 0.25\;\left( {{\rm{emissivity\;of\;radiation\;shield}}} \right)\;\)

\({Q_1} = \frac{{\sigma \left( {\;T_1^4\;-\;T_2^4\;} \right)}}{{\left( {\frac{1}{{{\varepsilon _1}}}\;+\;\frac{1}{{{\varepsilon _2}}}\;-\;1} \right)}}\)

\({Q_1} = \frac{{\sigma \left( {\;T_1^4\;-\;T_2^4\;} \right)}}{{\left( {\;\frac{1}{{0.5}}\;+\;\frac{1}{{0.5}}\;-\;1\;} \right)}}\)

\({Q_1} = \frac{{\sigma \left( {\;T_1^4\;-\;T_2^4\;} \right)}}{3}\)

Now,

If radiation shield inserted between plates

\({Q_2} = \frac{{\sigma \left( {\;T_1^4\;-\;T_2^4\;} \right)}}{{\left( {\;\frac{1}{{{\varepsilon _1}}}\;+\;\frac{1}{{{\varepsilon _2}}}\;+\;\frac{2}{{{\varepsilon _3}}}\;-\;2\;} \right)}}\)

\({Q_2} = \;\frac{{\sigma\left( {\;T_1^4\;-\;T_2^4\;} \right)}}{{\left( {\;\frac{1}{{0.5}}\;+\;\frac{1}{{0.5}}\;+\;\frac{2}{{0.25}}\;-\;2\;} \right)}}\)

\({Q_2} = \frac{{\sigma \left( {\;T_1^4\;-\;T_2^4\;} \right)}}{{10}}\)

\(\therefore\;\frac{{{Q_2}}}{{{Q_1}}} = \frac{3}{{10}}\)