Question

Consider a spherical shell of inner radius \(r_{1}\) and outer radius \(r_{2}\) whose thermal conductivity varies linearly in a specified temperature range as \(k(T)=k_{0}(1+\beta T)\) where \(k_{0}\) and \(\beta\) are two specified constants. The inner surface of the shell is maintained at a constant temperature of \(T_{1}\), while the outer surface is maintained at \(T_{2}\). Assuming steady one-dimensional heat transfer, obtain a relation for \((a)\) the heat transfer rate through the shell and (b) the temperature distribution \(T(r)\) in the shell.

Short answer

Short Answer: The heat transfer rate through the spherical shell is given by \(\varphi = -\beta k_0^2\). The temperature distribution \(T(r)\) in the shell is given by:$$T(r) = \frac{1}{\beta}\left[e^{-\beta\frac{k_0}{\varphi}(r-r_1)} (1+\beta T_1) - 1\right]$$.

Step-by-step solution

Apply Fourier's Law of Heat Conduction

We will use Fourier's Law of Heat Conduction, which relates the heat flux \(\varphi\) to the temperature gradient. In spherical coordinates, it is given by: $$\varphi=-k(T) \frac{dT}{dr}$$.

Substitute given thermal conductivity

We are given that \(k(T) = k_0(1 + \beta T)\). Substitute this into the Fourier's Law equation: $$\varphi = -k_0(1+\beta T) \frac{dT}{dr}$$.

Apply the boundary conditions

We are given that the inner surface of the shell at \(r = r_1\) is maintained at a constant temperature \(T_1\), and the outer surface of the shell at \(r = r_2\) is maintained at a constant temperature \(T_2\). These will be our boundary conditions.

Rearrange the equation

In order to find \(T(r)\), we need to integrate. Rearrange the Fourier's Law equation, and separate variables: $$\frac{dT}{1+\beta T} = -\frac{k_0}{\varphi} dr$$.

Integrate both sides of the equation

Now, integrate both sides of the equation with respect to \(r\) and \(T\): $$\int_{T_1}^{T(r)} \frac{dT}{1+\beta T} = -\frac{k_0}{\varphi} \int_{r_1}^{r} dr$$.

Solve the integrals

The left-hand side integral can be evaluated using a substitution like \(u = 1+\beta T\), whilst the right-hand side becomes a simple linear integral to solve: $$\frac{1}{\beta}\ln(1+\beta T)\Big|_{T_1}^{T(r)} = -\frac{k_0}{\varphi}(r-r_1)$$.

Solve for temperature distribution \(T(r)\)

In order to find \(T(r)\), simplify the equation and isolate \(T(r)\): $$\ln\frac{1+\beta T(r)}{1+\beta T_1} = -\beta\frac{k_0}{\varphi}(r-r_1)$$ $$T(r) = \frac{1}{\beta}\left[e^{-\beta\frac{k_0}{\varphi}(r-r_1)} (1+\beta T_1) - 1\right]$$. Thus we have obtained the temperature distribution \(T(r)\) in the shell.

Find the heat transfer rate

Now that we have the temperature distribution \(T(r)\), let's find the heat transfer rate through the shell. Express the heat flux using \(T(r_1)\): $$\varphi = - k\left(T(r_1)\right) \frac{dT}{dr}\bigg|_{r=r_1}$$, where \(k\left(T(r_1)\right)=k_0(1+\beta T_1)\) and \(T(r_1)=T_1\). Differentiate \(T(r)\) with respect to \(r\) and evaluate it at \(r=r_1\): $$\frac{dT}{dr}\bigg|_{r=r_1} = \frac{\beta k_0 e^{-\beta\frac{k_0}{\varphi}(r-r_1)}}{1+\beta T_1}\bigg|_{r=r_1} = \frac{\beta k_0}{1+\beta T_1}$$. Substitute the values into the expression for the heat flux: $$\varphi = -k(T_1) \frac{dT}{dr}\bigg|_{r=r_1} = -(k_0)(1+\beta T_1)\left(\frac{\beta k_0}{1+\beta T_1}\right) = -\beta k_0^2$$. We now have the heat transfer rate \((a)\) through the shell, and the temperature distribution \((b)\) \(T(r)\) in the shell.