Solar radiation of 1000 W/m² is incident on a grey opaque surface with an emissivity of 0.4 and emissive power (black body) of 400 W/m². The radiosity of the surface will be:

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ESE Mechanical 2016 Paper 1: Official Paper

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Answer 1

Concept:

Irradiation (G): Total radiation incident upon a surface per unit time per unit area.

Radiosity (J): Total radiation leaving a surface per unit time per unit area.

Radiosity comprises the original emittance from the surface plus the reflected portion of any radiation incident upon it.

J = E + ρG

J = εE_b + ρG

E_b = Emissive power of a perfect black body

α + ρ + τ = 1

For opaque body: τ = 0 ⇒ α + ρ = 1 ⇒ ρ = 1 - α = 1 - ε

J = εE_b + (1 - ε)G = E + (1 - ε)G

Calculation:

Given:

G = 1000 W/m², E_b = 400 W/m², ϵ = 0.4

J = 400 × 0.4 + (1 - 0.4)1000 = 760 W/m²