Two large parallel grey plates with a small gap exchange radiation at a rate of 1000 W/m^{2}, when their emissivities are 0.5 each. One plate is cooled such that its emissivity becomes 0.25. Find the new rate of heat transfer if the temperature difference remains the same.

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AAI JE (Technical) Official Paper 2018

Option 3 : 600 W/m2

__Explanation:__

We know that,

Radiation heat exchange between parallel plates

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

__Calculation:__

__Given:__

ϵ_{1} = 0.5, ϵ_{2} = 0.25

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

\({Q_1} = \frac{{\sigma \left( {\;T_1^4\;-\;T_2^4\;} \right)}}{{\left( {\;\frac{1}{{0.5}}\;+\;\frac{1}{{0.5}}\;-\;1\;} \right)}}=\frac{\sigma(T_1^4-T_2^4)}{3}=1000\;W\) ........(1)

\({Q_2} = \frac{{\sigma \left( {\;T_1^4\;-\;T_2^4\;} \right)}}{{\left( {\;\frac{1}{{0.5}}\;+\;\frac{1}{{0.25}}\;-\;1\;} \right)}}=\frac{\sigma(T_1^4-T_2^4)}{5}\) ......(2)

Using Equation (1) and (2)

Q_{2} = 600 W

ST 1: Logical reasoning

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