Option 1 : 11 W/m^{2}

__Concept:__

Total incident radiation = Direct + Diffuse

If absorptivity is given then total incident radiation = α (Direct + Diffuse)

The emissive power of a body is given as \(\text{ }\!\!σ\!\!\text{ }\!\!~\!\!\text{ }\times \epsilon \times \left( T_{2}^{4}-T_{1}^{4} \right)\)

Where σ = Stefan Boltzmann’s constant = 6 × 10^{-8} W/m^{2}.K^{4}, ϵ = emissivity, T_{2} & T_{1} are temperatures in Kelvin

__Calculation:__

Given, Direct radiation = 400 w/m^{2}^{2}, α = 0.1

Also direct radiation makes an angle of 60° with the normal

So from Lambert’s cosine law

Direct radiation = 400 × cos 60 = 200

Total incident radiation = α (Direct + Diffuse)

Total incident radiation = 0.1 (200 + 300) = 50 W/m2

Total emissive power = \(\varepsilon \times σ \times \left({T_2^1 - T_1^4} \right)\)

E = 0.1 × σ × 10^{-8} (300^{4} - 200^{4})

E = 39 W/m^{2}

So the net radiation heat transfer = Incident – emitted

net radiation heat transfer = 50 – 39 = 11 W/m2

Radiative heat transfer is intended between the inner surfaces of two very large isothermal parallel metal plates. While the upper plate (designated as plate 1) is a black surface and is the warmer one being maintained at 727ºC, the lower plate (plate 2) is a diffuse and gray surface with an emissivity of 0.7 and is kept at 227ºC. Assume that the surfaces are sufficiently large to form a two-surface enclosure and steady state conditions to exist. Stefan Boltzmann constant is given as 5.67×10^{-8}W/m^{2}K^{4}

Option 4 : 31.7

__Concept:__

The radiation heat transfer between two infinitely large grey parallel planes are given by,

**\({\left( {\frac{Q}{A}} \right)_{1 - 2}} = \frac{{\sigma \times \left( {T_1^4 - T_2^4} \right)}}{{\frac{{1}}{{{\varepsilon _1}}} + \frac{1}{{{\varepsilon _2}}} - 1}}\)**

where T_{1} = Temperature of plate_{1}

T_{2} = Temperature of plate_{2}

ϵ _{1} = Emissivity of plate_{1}

ϵ_{2} = Emissivity of plate_{2}

__Calculation:__

__Given: __

Emissivity of plate_{1}, ϵ _{1} = 0.8

Emissivity of plate_{2} , ϵ _{2} = 0.7

Temperature of plate_{1} , T_{1} = 727 ° C = 1000 K

Temperature of plate_{2} , T_{2} = 227 ° C = 500 K

The radiation heat exchange between plate_{1} and plate_{2}

_{\({\left( {\frac{Q}{A}} \right)_{1 - 2}} = \frac{{\sigma \times \left( {T_1^4 - T_2^4} \right)}}{{\frac{{1}}{{{_1}}} + \frac{1}{{{_2}}} - 1}}\)}

\(\frac{{5.67 \times {{10}^{ - 8}} \times \left( {{{1000}^4} - {{500}^4}} \right)}}{{\frac{1}{{0.8}} + \frac{1}{{0.7}} - 1}}\)

31.68 kW/m^{2}

Radiative heat transfer is intended between the inner surfaces of two very large isothermal parallel metal plates. While the upper plate (designated as plate 1) is a black surface and is the warmer one being maintained at 727ºC, the lower plate (plate 2) is a diffuse and gray surface with an emissivity of 0.7 and is kept at 227ºC. Assume that the surfaces are sufficiently large to form a two-surface enclosure and steady state conditions to exist. Stefan Boltzmann constant is given as 5.67×10^{-8 }W/m^{2}K^{4}

Option 4 : 19.5

__Concept__

**Irradiation** – It is denoted by **G.**

Irradiation is defined as the total thermal radiation incident upon a surface **per unit time per unit area.**

**Radiosity** – it is denoted by **J**.

Radiosity is defined as the total thermal radiation leaving a surface **per unit time per unit area**.

J = Emitted thermal energy + Reflected part of incident thermal energy.

J = E + ρ G

J = ϵ E_{b} + ρ G

where E = Energy emitted by Nonblack body.

E_{b} = Energy emitted by the Black body.

ρ = Reflectivity of the surface.

For any surface,

α +ρ +τ = 1

where α = Absorptivity of the surface.

ρ = Reflectivity of the surface.

τ = Transmitivity of the surface.

For opaque surface transmitivity τ = 0

∴ For opaque surface, α +ρ = 1 or ρ = 1-α

Kirchhoff’s law states that when a body is in thermal equilibrium with its surroundings, its emissivity is equal to its absorptivity.

α =ϵ therefore ρ = 1-ϵ

Thus, Radiosity is given by

J = ϵ E_{b} + (1 - ϵ ) × G

__Calculation:__

__Given:__

The temperature of the upper plate,plate_{1} T_{1} = 727 ° C = 1000 K

The temperature of the lower plate,plate_{2}, T_{2} = 227 ° C = 500 K

The emissivity of the plate_{1}, ϵ _{1} = 0.7