Two very large parallel plates are maintained at uniform temperature of T₁ = 800K and T₂ = 640 K. Their emmisivities are 0.2 and 0.5 respectively. What will be the net rate of radiation heat transfer between the two surface per unit surface area of the plates? [assuming, Stefan Boltzman constant = 5.67 × 10^-8 W/m²-K ;(800)⁴- (640)⁴= 2.4 × 10¹¹]

Concept:

The net rate of radiation heat transfer between two large parallel plates is given by:

\( q = \frac{σ (T_1^4 - T_2^4)}{\frac{1}{\varepsilon_1} + \frac{1}{\varepsilon_2} - 1} \)

Where:

σ = Stefan-Boltzmann constant = \(5.67 \times 10^{-8}~\text{W/m}^2\cdot \text{K}^4\),

\(T_1 = 800~\text{K}, T_2 = 640~\text{K}, \varepsilon_1 = 0.2, \varepsilon_2 = 0.5\)

Calculation: 

Given that \(T_1^4 - T_2^4 = 2.4 \times 10^{11}\), substitute into the formula:

\( q = \frac{5.67 \times 10^{-8} \times 2.4 \times 10^{11}}{\frac{1}{0.2} + \frac{1}{0.5} - 1} \)

Calculate the denominator:

\( \frac{1}{0.2} + \frac{1}{0.5} - 1 = 5 + 2 - 1 = 6 \)

Now calculate the numerator:

\( 5.67 \times 10^{-8} \times 2.4 \times 10^{11} = 13608 \)

Finally,

\( q = \frac{13608}{6} = 2268~\text{W/m}^2 \)