Question

A spherical shell with thermal conductivity \(k\) has inner and outer radii of \(r_{1}\) and \(r_{2}\), respectively. The inner surface of the shell is subjected to a uniform heat flux of \(\dot{q}_{1}\), while the outer surface of the shell is exposed to convection heat transfer with a coefficient \(h\) and an ambient temperature \(T_{\infty}\). Determine the variation of temperature in the shell wall, and show that the outer surface temperature of the shell can be expressed as $T\left(r_{2}\right)=\left(\dot{q}_{1} / h\right)\left(r_{1} / r_{2}\right)^{2}+T_{\infty}$

Short answer

Question: Determine the temperature distribution within a spherical shell with given thermal properties and boundary conditions, and show that the outer surface temperature can be expressed as \(T(r_2)=\left(\frac{q_1}{h}\right)\left(\frac{r_1}{r_2}\right)^2 + T_\infty\). Answer: The temperature distribution within a spherical shell with given thermal properties and boundary conditions is given by \(T(r)=-\frac{q_1}{2r_1}\frac{r^2}{2}+\frac{q_1}{2r_2}\ln{r}+T_\infty\), and the outer surface temperature can be expressed as \(T(r_2)=\left(\frac{q_1}{h}\right)\left(\frac{r_1}{r_2}\right)^2 + T_\infty\).

Step-by-step solution

Governing Equation for Heat Transfer in Spherical Coordinates

To determine the temperature distribution, we first need to consider the general heat conduction equation in spherical coordinates: \(\frac{1}{r^{2}}\frac{d}{d r}\left(r^{2} k \frac{d T}{d r}\right)=0\)

Simplify the Governing Equation

Since the thermal conductivity k is constant, we can simplify the governing equation further by dividing both sides by k: \(\frac{1}{r^{2}}\frac{d}{d r}\left(r^{2} \frac{d T}{d r}\right)=0\)

Integrate the Governing Equation

Integrate the simplified governing equation to obtain the first-order differential equation for temperature: \(\int{\frac{1}{r^{2}}\frac{d}{d r}\left(r^{2} \frac{d T}{d r}\right)dr}=C_{1} \Rightarrow r^{2}\frac{dT}{dr}=C_{1}\) Now, integrate once more to get the temperature distribution function: \(\int{r \frac{dT}{dr}=C_{1} dr} \Rightarrow T(r) = C_{2}\ln{r} + C_{1}\frac{r^2}{2} + C_{3}\)

Apply Boundary Conditions

We will now apply the boundary conditions to determine the constants C1, C2, and C3. Boundary condition 1: At r = r1, the heat flux q1 is given: \(q_{1}=-k\left.\frac{dT}{d r}\right|_{r=r_{1}}\). Apply this boundary condition using the given formula, and solve for C1: \(r^{2}\frac{dT}{dr} \Rightarrow 2C_{1}r_{1}= -\dot{q}_{1}\), so \(C_{1} = -\frac{q_{1}}{2r_{1}}\) Boundary condition 2: At r = r2, the convection boundary condition applies: \(-k\left.\frac{dT}{dr}\right|_{r=r_{2}} = h\left[T\left(r_{2}\right) - T_{\infty}\right]\). Apply this boundary condition and substitute C1 into the equation, then solve for C2: \(C_{2} = \frac{q_{1}}{2r_{2}} - h\left[T(r_2) - T_{\infty}\right]\) Boundary condition 3: We don't need to apply another boundary condition since the temperature is obtained through the constants C2 and C3. Thus, the equation for temperature in the shell becomes: \(T\left(r\right)=-\frac{\dot{q}_{1}}{2r_{1}}\frac{r^{2}}{2}+\frac{q_{1}}{2r_{2}}\ln{r}+T_{\infty}\)

Show the Outer Surface Temperature Expression

To determine the outer surface temperature of the shell, evaluate the temperature function at r = r2: \(T\left(r_{2}\right)=-\frac{\dot{q}_{1}}{2 \cdot r_{1}}\left(\frac{r_{2}^{2}}{2}\right)+\frac{q_{1}}{2 \cdot r_{2}}\ln{r_{2}}+T_{\infty}\) Now express the equation in the desired form: \(T\left(r_{2}\right)=\left(\frac{\dot{q}_{1}}{h}\right)\left(\frac{r_{1}}{r_{2}}\right)^{2}+T_{\infty}\)