Question

Consider steady one-dimensional heat conduction in a slab of thickness $2L$ (with $L=1$ m). Thermal conductivity varies as $k = C T$, with $C = 2$ W·m$^{-1}$·K$^{-2}$, and the slab generates heat at $1280$ kW/m$^3$. Both faces are held at 600 K. The temperature at $x=0$ is __________ K (in integer).

Hint:

For variable-conductivity conduction problems, symmetry at the center makes $dT/dx = 0$, simplifying integration.

Answer 1

Correct Answer: 1000

Given: \[ k = C T,\quad C = 2\ \text{W·m}^{-1}\text{·K}^{-2} \] Heat generation: \[ \dot{q} = 1280\ \text{kW/m}^3 = 1.28\times10^6\ \text{W/m}^3 \] Steady 1-D conduction with temperature-dependent $k$: \[ \frac{d}{dx}\left( k \frac{dT}{dx} \right) + \dot{q} = 0 \] Substitute $k = C T$: \[ \frac{d}{dx}\left( C T \frac{dT}{dx} \right) + \dot{q} = 0 \] Integrate once using symmetry at $x=0$ (\( dT/dx = 0 \)): \[ C\, T\, \frac{dT}{dx} = -\dot{q} x \] Separate and integrate: \[ T\, dT = -\frac{\dot{q}}{C} x\, dx \] \[ \frac{T^2}{2} = -\frac{\dot{q}}{2C} x^2 + \text{constant} \] At $x=0$, let the maximum temperature be $T_0$: \[ \frac{T^2}{2} = \frac{T_0^2}{2} - \frac{\dot{q}}{2C} x^2 \] At $x=L=1$ m, boundary condition $T=600$ K: \[ 600^2 = T_0^2 - \frac{\dot{q}}{C}L^2 \] Substitute values: \[ \frac{\dot{q}}{C} = \frac{1.28\times10^6}{2} = 640000 \] \[ 600^2 = T_0^2 - 640000 \] \[ 360000 = T_0^2 - 640000 \] \[ T_0^2 = 1000000 \] \[ T_0 = 1000\ \text{K} \] Thus, \[ \boxed{1000\ \text{K}} \]