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Radiative Heat Transfer Laws - I

Heat is condu...

Question

Heat is conducted at steady state through a solid wall with temperature gradient of $- 5^{\circ} / c m .$ The thermal conductivity of solid being $22.5 \frac{W}{m K} .$ If the heat is exchanged by radiation from surface to the surroundings at $30^{\circ} C .$ The surface temperature of solid wall is

A

829.4 K

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B

674.4K

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C

474.7K

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D

538.9 K

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Solution

The correct option is B 674.4K
$\begin{matrix} k & = 22.5 \frac{W}{m K} \frac{\partial T}{\partial x} & = - 5^{\circ} C / c m = - 5 \times 100^{\circ} C / m - k . \frac{\partial T}{\partial x} & = σ (T^{4} - T_{α}^{4}) - (22.5) \times (- 5 \times 100) & = 5.67 \times 10^{- 8} (T^{4} - (303)^{4}) T & = 674.4 K \end{matrix}$

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