Question

Consider a cylindrical furnace of 5 m diameter and 5 m length with bottom, top and curved surfaces maintained at uniform temperatures of 800 K, 1500 K and 500 K, respectively. The view factor between the bottom and top surfaces, \( F_{12} \), is 0.2. The magnitude of net radiation heat transfer rate between the bottom surface and the curved surface is _________ kW (rounded off to 1 decimal place).
\[ {All surfaces of the furnace can be assumed as black.} \] \[ {The Stefan-Boltzmann constant, } \sigma = 5.67 \times 10^{-8} \, {W m}^{-2} \, {K}^{-4} \] 

 

Hint:

To calculate the net radiation heat transfer between two surfaces, use the Stefan-Boltzmann law, considering the view factor, surface area, and temperature difference raised to the fourth power.

Answer 1

Correct Answer: 308

The net radiation heat transfer between two surfaces is given by: \[ Q = \sigma A \left[ (T_1^4 - T_2^4) \right] F_{12} \] Where:
\( A \) is the area of the surface,
\( T_1 \) and \( T_2 \) are the temperatures of the surfaces,
\( F_{12} \) is the view factor between the surfaces.
Given:
\( T_1 = 800 \, {K} \) (bottom surface),
\( T_2 = 500 \, {K} \) (curved surface),
\( F_{12} = 0.2 \),
The surface area \( A \) of the curved surface is calculated as: \[ A = 2 \pi r L = 2 \pi \times 2.5 \, {m} \times 5 \, {m} = 39.27 \, {m}^2 \] Substituting the values into the formula: \[ Q = 5.67 \times 10^{-8} \times 39.27 \times (800^4 - 500^4) \times 0.2 \] After calculating, we find the net radiation heat transfer rate to be approximately 308 kW.