Question

A spherical container with an inner radius \(r_{1}=1 \mathrm{~m}\) and an outer radius \(r_{2}=1.05 \mathrm{~m}\) has its inner surface subjected to a uniform heat flux of \(\dot{q}_{1}=7 \mathrm{~kW} / \mathrm{m}^{2}\). The outer surface of the container has a temperature \(T_{2}=25^{\circ} \mathrm{C}\), and the container wall thermal conductivity is $k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. Show that the variation of temperature in the container wall can be expressed as $T(r)=\left(\dot{q}_{1} r_{1}^{2} / k\right)\left(1 / r-1 / r_{2}\right)+T_{2}$ and determine the temperature of the inner surface of the container at \(r=r_{1}\).

Short answer

Answer: The temperature of the inner surface of the container at r = r1 is approximately 535°C.

Step-by-step solution

Find the general equation for temperature distribution in the container wall

We can start by writing the formula for the heat transfer rate at any point r inside the container wall: \(\frac{dQ}{dr}=-k \frac{dT}{dr}\), where \(dQ\) is the heat transfer rate, \(k\) is the thermal conductivity, and \(dT/dr\) is the temperature gradient. The spherical coordinates, total heat transfer through a shell of thickness \(dr\), radius \(r\), and a surface area of \(4 \pi r^2\), can be expressed as: \(\frac{dQ}{dr}= -4\pi r^{2} k \frac{dT}{dr}\). Now, we can rewrite the equation and separate variables to start integrating: \(\frac{dT}{dr}= -\frac{1}{4\pi r^{2} k} \frac{dQ}{dr}\).

Integrate the equation with respect to r to find the general expression for T(r)

We can now integrate both sides with respect to r: \(\int \frac{dT}{dr} dr = -\frac{1}{4\pi k} \int \frac{1}{r^2} dQ\). Integrating each side will give us: \(T(r) = -\frac{1}{4\pi k} \frac{dQ}{r} + C_{1}\), where \(C_{1}\) is a constant of integration.

Apply boundary conditions to find C1

We are given that the outer surface of the container has a temperature of \(T_{2}=25^{\circ} \mathrm{C}\) when \(r=r_{2}\). We can use this boundary condition to determine the integration constant \(C_{1}\): \(T_{2} = -\frac{1}{4\pi k} \frac{dQ}{r_{2}} + C_{1}\) \(\Rightarrow C_{1} = T_{2}+\frac{1}{4\pi k}\frac{dQ}{r_{2}}\). Now we have the expression for \(C_{1}\), we can plug it into the general equation for T(r): \(T(r) = -\frac{1}{4\pi k} \frac{dQ}{r} + T_{2}+\frac{1}{4\pi k}\frac{dQ}{r_{2}}\). By rearranging, we arrive at the desired expression: \(T(r)=\left(\dot{q}_{1} r_{1}^{2} / k\right)\left(1 / r-1 / r_{2}\right)+T_{2}\).

Determine the temperature of the inner surface at r = r1

To find the temperature at the inner surface of the container, we can plug in the value for \(r=r_{1}\): \(T(r_{1})=\left(\dot{q}_{1} r_{1}^{2} / k\right)\left(1 / r_{1}-1 / r_{2}\right)+T_{2}\). Using the given values, \(\dot{q}_{1}=7 \mathrm{~kW} / \mathrm{m}^{2}\), \(r_{1}=1 \mathrm{~m}\), \(r_{2}=1.05 \mathrm{~m}\), \(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(T_{2}=25^{\circ} \mathrm{C}\), we can find \(T(r_{1})\): \(T(r_{1})=\left(7000 \cdot 1^{2}/1.5\right)\left(1/1 - 1/1.05\right) + 25\). Calculating the result, we have: \(T(r_{1})\approx 535^{\circ} \mathrm{C}\). Thus, the temperature of the inner surface of the container at \(r=r_{1}\) is approximately \(535^{\circ} \mathrm{C}\).