Question

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 m\(^{-3}\)). The slab loses heat by convection at both the 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 given by: \[ T(x) = \frac{\dot{q}L^2}{2k} \left[1 - \left(\frac{x}{L}\right)^2 \right] + T_s \text{for} -L \leq x \leq L \] where \(k\) is the thermal conductivity of the slab 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 as the characteristic length) is 0.5, the maximum temperature in the slab is \(\underline{\hspace{1cm}}\) K (round off to nearest integer).

Hint:

Biot number helps determine the effect of thermal resistance in the heat transfer process.

Answer 1

Correct Answer: 362

Using the Biot number and thermal properties, the maximum temperature can be calculated as: \[ \text{Bi} = \frac{hL}{k} \] Given \(\text{Bi} = 0.5\), \(L = 2\), and other properties, the maximum temperature will be:
\[ T_{\text{max}} = 362 \text{ to } 363 \text{ K}. \] Final answer: 362–363 K.