Conduction through hollow sphere

Expression for a hollow sphere

Steady radial conduction with no generation: $\dfrac{d}{dr}\left(r^2\dfrac{dT}{dr}\right)=0$, so $T = -C_1/r + C_2$. Using $Q = -k(4\pi r^2)\dfrac{dT}{dr}$ and the boundary temperatures $T_1$ at $r_1$ and $T_2$ at $r_2$:

$$Q = \frac{4\pi k\,(T_1 - T_2)}{\dfrac{1}{r_1}-\dfrac{1}{r_2}}$$

and the temperature at any radius:

$$T = T_1 - \frac{Q}{4\pi k}\left(\frac{1}{r_1}-\frac{1}{r}\right)$$

(a) Heat flow

Given $r_1 = 0.1$ m, $r_2 = 0.2$ m, $k = 15$ W/m K.

$$\begin{aligned} \frac{1}{r_1}-\frac{1}{r_2} &= 10 - 5 = 5\ \text{m}^{-1}\\ Q &= \frac{4\pi(15)(300-100)}{5} = 7540\ \text{W} \end{aligned}$$

(b) Temperature at 150 mm

$$\begin{aligned} T &= 300 - \frac{7540}{4\pi(15)}\left(10 - \frac{1}{0.15}\right)\\ &= 300 - 40\,(10 - 6.667) = 300 - 133.3 = 166.7\ ^\circ\text{C} \end{aligned}$$

The temperature at r = 0.15 m (166.7 °C) is lower than the arithmetic mean of the surfaces (200 °C), because the area is smaller near the inside and the gradient is steeper there.

Answer: $Q \approx 7.54\ \text{kW}$; $T(0.15\ \text{m}) \approx 166.7\ ^\circ\text{C}$.