Question

Heat is generated in a 10 -cm-diameter spherical radioactive material whose thermal conductivity is \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) uniformly at a rate of \(15 \mathrm{~W} / \mathrm{cm}^{3}\). If the surface temperature of the material is measured to be \(120^{\circ} \mathrm{C}\), the center temperature of the material during steady operation is (a) \(160^{\circ} \mathrm{C}\) (b) \(205^{\circ} \mathrm{C}\) (c) \(280^{\circ} \mathrm{C}\) (d) \(370^{\circ} \mathrm{C}\) (e) \(495^{\circ} \mathrm{C}\)

Short answer

Answer: The center temperature of the spherical radioactive material during steady operation is approximately 370°C.

Step-by-step solution

Recall the appropriate formula for heat conduction in spherical coordinates.

For steady-state heat conduction in spherical coordinates, we can use the following formula: \(q_{r} = -kA\frac{\mathrm{d} T}{\mathrm{d} r}\) Where \(q_{r}\) is the heat flux, \(k\) is the thermal conductivity, \(A\) is the area, and \(\frac{\mathrm{d}T}{\mathrm{d} r}\) is the temperature gradient. For a spherical system, the area is given by \(A = 4\pi r^{2}\).

Express the heat generation rate.

The heat generation rate, \(q'\), is given as \(15 \mathrm{W/cm^{3}}\). We can convert this to \(\mathrm{W/m^{3}}\) as follows: \(q' = 15 \mathrm{W/cm^{3}} \cdot \frac{10^6 \mathrm{cm^{3}}}{1 \mathrm{m^{3}}} = 15 \times 10^{6} \mathrm{W/m^{3}}\).

Set up a conservation of energy equation.

For steady-state conditions, the conservation of energy equation for the spherical radioactive material can be written as: \(\frac{\mathrm{d} q_{r}}{\mathrm{d} r} + q' = 0\) Using the heat conduction equation from Step 1, we have: \(\frac{\mathrm{d} (-kA\frac{\mathrm{d} T}{\mathrm{d} r})}{\mathrm{d} r} + q' = 0\) Substitute the area \(A\) for a spherical system: \(\frac{\mathrm{d} (-k(4\pi r^{2})\frac{\mathrm{d} T}{\mathrm{d} r})}{\mathrm{d} r} + q' = 0\)

Solve the differential equation for temperature distribution.

To solve the equation, first substitute \(u = \frac{\mathrm{d} T}{\mathrm{d} r}\) \(\frac{\mathrm{d} (-4\pi k r^{2}u)}{\mathrm{d} r} + q' = 0\) Integrating gives: \(-4\pi k r^{2}u = -\frac{q'}{3}r^{3} + C_{1}\) Now, \(u = \frac{\mathrm{d} T}{\mathrm{d} r}\), so: \(-4\pi k r^{2}\frac{\mathrm{d} T}{\mathrm{d} r} = -\frac{q'}{3}r^{3} + C_{1}\) Integrate again to find the temperature distribution function: \(T(r) = -\frac{q'}{12\pi k}r^{2} + \frac{C_{1}}{4\pi k}r + C_{2}\)

Apply boundary conditions.

We know the surface temperature, \(T(R)=120^{\circ}\mathrm{C}\), and the diameter of the sphere, \(D=10\,\text{cm}\), or \(R=5\,\text{cm}=0.05\,\text{m}\). Use these conditions to find the values of \(C_{1}\) and \(C_{2}\). At \(r=R\): \(120 = -\frac{15\times10^{6}}{12\pi\times25}0.05^{2} + \frac{C_{1}}{4\pi\times25}(0.05) + C_{2}\) At the center, \(r=0\), the temperature gradient \(\frac{\mathrm{d} T}{\mathrm{d} r}=0\). Since \(\frac{\mathrm{d} T}{\mathrm{d} r}=u\), we have \(-4\pi k (0)^{2}\frac{\mathrm{d} T}{\mathrm{d} r}= C_{1}\), which implies that \(C_{1}\)=0. Solving for \(C_{2}\), we get: \(C_{2} = T(R) + \frac{q'}{12\pi k}R^{2} = 120 + \frac{15\times10^{6}}{12\pi\times25}(0.05^{2})\)

Calculate the center temperature.

Using the temperature distribution function and the values of \(C_{1}\) and \(C_{2}\), we can find the center temperature \(T(0)\): \(T(0) = -\frac{q'}{12\pi k}(0)^{2} + \frac{C_{1}}{4\pi k}(0) + C_{2} = C_{2} \approx 370^{\circ} \mathrm{C}\) Therefore, the center temperature of the material during steady operation is approximately \(370^{\circ} \mathrm{C}\). The correct answer is (d).

Key concepts

Steady-State Heat Conduction

Steady-state heat conduction refers to a condition where the temperature in a material does not change with time. This situation implies that the heat entering any region within the material is equal to the heat leaving, to maintain a consistent temperature field. In the context of the textbook problem, the spherical radioactive material has reached a steady state.

To analyze heat conduction in steady-state, we apply the heat conduction equation in spherical coordinates. The equation considers the unique way in which heat spreads out from the center in a sphere, as area increases with the square of the radius. This is why the area term in our crucial equation becomes \( A = 4\pi r^2 \).

When dealing with steady-state conditions, boundary conditions become essential to solving for unknown constants, which is demonstrated in the textbook solution by applying the known surface temperature of the sphere and the condition of no temperature gradient at the sphere's center.

Thermal Conductivity

Thermal conductivity, symbolized as \( k \), is a property of materials that describes how well they conduct heat. It is a measure of the material's ability to transfer heat through its substance by conduction. The thermal conductivity for the spherical radioactive material in the exercise is given as \(25 \frac{W}{m\cdot K}\).

High thermal conductivity indicates that heat will spread through the material quickly, while low thermal conductivity implies heat will move slowly. This is integral to the heat equation in spherical coordinates for steady-state conduction, where it multiplies the area and the temperature gradient to determine the rate at which heat is transferred radially through the sphere.

Temperature Gradient

The temperature gradient in a material is the rate at which temperature changes with distance. In mathematical terms, it is the derivative of the temperature with respect to the radius in spherical coordinates, represented as \( \frac{dT}{dr} \).

For our spherical radioactive material, finding the temperature gradient is an essential step in calculating the center temperature. The gradient indicates how temperature falls off from the center of the sphere to its surface. A high gradient indicates a rapid change in temperature across a small distance, whereas a low gradient signifies a more gradual temperature transition. By integrating the temperature gradient, we arrive at the temperature distribution within the sphere.

Conservation of Energy in Thermodynamics

In thermodynamics, the conservation of energy principle states that energy can neither be created nor destroyed; it can only be transferred or converted from one form to another. In the case of steady-state heat conduction, this principle implies that the thermal energy generated within the spherical material must be equal to the thermal energy conducted away through the material's surface.

The conservation of energy is used to establish a relationship between the heat generated inside the material and the heat flux moving radially outward. We see this reflected in the differential equation set up in the exercise's solution, which incorporates the generation rate and the changes in heat flux as a result of conduction. By solving this equation, we can find the temperature distribution and hence predict the center temperature of the sphere.