GATE 2021 CH – Question 55
Consider a solid slab of thickness 2L and uniform cross section A. The volumetric rate of heat generation within the slab is $\dot q$ (W/m³). The slab loses heat by convection at both ends to air with heat transfer coefficient h. Assuming steady state, one-dimensional heat transfer, the temperature profile within the slab along the thickness is $T(x)=\frac{\dot qL^2}{2k}\left[1-(x/L)^2\right]+T_s$ for $-L\le x\le L$, where k is the thermal conductivity and $T_s$ is the surface temperature. If $T_s=350$ K, ambient air temperature $T_\infty=300$ K, and Biot number (based on L) is 0.5, the maximum temperature in the slab is _____ K (round off to nearest integer).
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Correct answer: 363
Explanation
By symmetry, the maximum occurs at the midplane $x=0$, so $T_{max}-T_s=\dot qL^2/(2k)$.
Each face removes the heat generated in a half-slab: $\dot qL=h(T_s-T_\infty)$. Substitute this into the midplane rise:
$$T_{max}-T_s=\frac{hL}{2k}(T_s-T_\infty)=\frac{Bi}{2}(350-300)=12.5\ \mathrm K.$$
Thus $T_{max}=362.5$ K, which rounds to **363 K** using the conventional half-up rule.