Question

Consider a homogeneous spherical piece of radioactive material of radius \(r_{o}=0.04 \mathrm{~m}\) that is generating heat at a constant rate of \(\dot{e}_{\mathrm{gen}}=4 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\). The heat generated is dissipated to the environment steadily. The outer surface of the sphere is maintained at a uniform temperature of \(80^{\circ} \mathrm{C}\), and the thermal conductivity of the sphere is $k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(. Assuming steady one-dimensional heat transfer, \)(a)$ express the differential equation and the boundary conditions for heat conduction through the sphere, \((b)\) obtain a relation for the variation of temperature in the sphere by solving the differential equation, and \((c)\) determine the temperature at the center of the sphere.

Short answer

Answer: The temperature at the center of the spherical piece of radioactive material is approximately \(188.16^{\circ} \mathrm{C}\).

Step-by-step solution

Recall heat conduction equation in spherical coordinates

The heat conduction equation in spherical coordinates is given by: \[\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2} \frac{\partial T}{\partial r}\right) = \frac{\dot{e}_{\mathrm{gen}}}{k}\] Where \(T\) is the temperature, \(r\) is the radial coordinate, \(\dot{e}_{\mathrm{gen}}\) is the heat generation rate per unit volume, and \(k\) is the thermal conductivity.

Write down the given information and values

The given information and values are: - Sphere radius: \(r_{o} = 0.04\) m - Heat generation rate: \(\dot{e}_{\mathrm{gen}} = 4 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\) - Outer surface temperature: \(80^{\circ} \mathrm{C}\) - Thermal conductivity: \(k = 15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)

Express the differential equation and boundary conditions

Using the heat conduction equation from Step 1, we have: \[\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2} \frac{\partial T}{\partial r}\right) = \frac{4 \times 10^{7}}{15}\] Boundary conditions: 1. At the outer surface of the sphere (\(r = r_{o}\)), the temperature is maintained at \(80^{\circ} \mathrm{C}\): \(T(r_{o}) = 80^{\circ} \mathrm{C}\). 2. At the center of the sphere (\(r = 0\)), the temperature gradient is zero due to thermal symmetry: \(\frac{\partial T}{\partial r}(0) = 0\).

Solve the differential equation

Integrating the differential equation twice, we obtain: \[\frac{\partial T}{\partial r} = -\frac{4 \times 10^{7}}{15} \frac{r^{3}}{3} + C_{1}\] Applying boundary condition 2 (\(\frac{\partial T}{\partial r}(0) = 0\)), we find \(C_{1} = 0\). So, \[\frac{\partial T}{\partial r} = -\frac{4 \times 10^{7}}{15} \frac{r^{3}}{3}\] Integrating again, we obtain: \[T(r) = -\frac{4 \times 10^{7}}{15} \frac{r^{4}}{12} + C_{2}\] Now, applying boundary condition 1 (\(T(r_{o}) = 80^{\circ} \mathrm{C}\)), we find: \[C_{2} = 80 + \frac{4 \times 10^{7}}{15} \frac{(0.04)^{4}}{12}\] So, the temperature variation in the sphere is: \[T(r) = -\frac{4 \times 10^{7}}{15} \frac{r^{4}}{12} + 80 + \frac{4 \times 10^{7}}{15} \frac{(0.04)^{4}}{12}\]

Determine the temperature at the center of the sphere

To determine the temperature at the center of the sphere, set \(r = 0\) in the temperature variation equation: \[T(0) = -\frac{4 \times 10^{7}}{15} \frac{(0)^{4}}{12} + 80 + \frac{4 \times 10^{7}}{15} \frac{(0.04)^{4}}{12}\] \[T(0) = 80 + \frac{4 \times 10^{7}}{15} \frac{(0.04)^{4}}{12}\] Calculating this value, we get: \[T(0) \approx 188.16^\circ \mathrm{C}\] So, the temperature at the center of the sphere is approximately \(188.16^{\circ} \mathrm{C}\).