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GATE 2025 ME – Question 48

Heat Transfer · Radiative heat transfer: laws, view factors, radiation network · 2 marks · Numerical answer

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$^{-4}$.

A cylinder with the top surface (2) at $T_2 = 1500$ K, the bottom surface (1) at $T_1 = 800$ K and the curved surface (3) at $T_3 = 500$ K.

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Correct answer: 307.55 to 310.65

Explanation

The bottom surface sees only the top and the curved surface, so $F_{11} = 0$, $F_{12} = 0.2$ and $F_{13} = 1 - 0.2 = 0.8$. The bottom area is $A_1 = \pi \times 2.5^2 = 19.63$ m$^2$. For black surfaces, $Q_{13} = A_1F_{13}\sigma(T_1^4 - T_3^4) = 19.63 \times 0.8 \times 5.67 \times 10^{-8} \times (800^4 - 500^4) = 309.1$ kW.