Plane wall with internal heat generation

Derivation

Let the plate occupy $-L \le x \le L$ with $2L = 40$ mm, so $L = 0.02$ m. For steady one-dimensional conduction with generation $\dot q_g$:

$$\frac{d^2T}{dx^2} + \frac{\dot q_g}{k} = 0$$

Integrate: $T = -\dfrac{\dot q_g}{2k}x^2 + C_1x + C_2$. By symmetry $dT/dx = 0$ at $x = 0$, so $C_1 = 0$. At the surface $x = L$ the conduction equals convection:

$$-k\frac{dT}{dx}\Big|_{L} = h\,(T_s - T_\infty)\;\Rightarrow\;\dot q_g L = h\,(T_s-T_\infty)$$

so $T_s = T_\infty + \dfrac{\dot q_g L}{h}$ and

$$T(x) = T_s + \frac{\dot q_g}{2k}\left(L^2 - x^2\right)$$

The maximum is at the centre: $T_{max} = T_s + \dfrac{\dot q_g L^2}{2k}$.

(a) Surface temperature

$$T_s = 80 + \frac{(5\times10^6)(0.02)}{500} = 80 + 200 = 280\ ^\circ\text{C}$$

(b) Maximum temperature

$$T_{max} = 280 + \frac{(5\times10^6)(0.02)^2}{2(20)} = 280 + 50 = 330\ ^\circ\text{C}$$

It occurs at the mid-plane $x = 0$.

(c) Heat removed per face

By energy balance, heat generated in the half-plate equals heat leaving one face:

$$q'' = \dot q_g L = (5\times10^6)(0.02) = 1\times10^5\ \text{W/m}^2 = 100\ \text{kW/m}^2$$

Check: $h(T_s - T_\infty) = 500\times200 = 1\times10^5$ W/m$^2$.

Answer: $T_s = 280\ ^\circ$C; $T_{max} = 330\ ^\circ$C at the centre; $q'' = 100\ \text{kW/m}^2$ per face.