What is the next radiant interchange per square meter for two very large plates at temperatures 800 K and 500 K respectively ? (The emissivity of the hot and cold plates is 0.8 and 0.6 respectively. Stefan Boltzmann constant is 5.67 × 10^-8 w/m² K⁴)

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ESE Mechanical 2013 Official Paper - 1

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Answer 1

Concept:

Radiation heat exchange (q) between two infinitely long parallel plates is given by:

F1 M.J Madhu 25.03.20 D44 F1 M.J Madhu 25.03.20 D45 (1)

\(q = \frac{{\sigma \left( {T_1^4\;-\;T_2^4} \right)}}{{\frac{1}{{{ϵ_1}}}\;+\;\frac{1}{{{ϵ_2}}}\;-\;1}}\)

Calculation:

Given T₁ = 800 K, T₂ = 500 K; ϵ₁ = 0.8, ϵ₂ = 0.6;

Now the net heat exchange will be

\(q = \frac{{5.67 \times 10^{-8} \times \left( {(800)^4\;-\;(500)^4} \right)}}{{\frac{1}{{{0.8}}}\;+\;\frac{1}{{{0.6}}}\;-\;1}}\)

⇒ q = 10268.123 W/m²;

⇒ q = 10.26 kW/m²;