Question

Consider a spherical container of inner radius \(r_{1}\), outer radius \(r_{2}\), and thermal conductivity \(k\). Express the boundary condition on the inner surface of the container for steady onedimensional conduction for the following cases: (a) specified temperature of \(50^{\circ} \mathrm{C}\), (b) specified heat flux of \(30 \mathrm{~W} / \mathrm{m}^{2}\) toward the center, (c) convection to a medium at \(T_{\infty}\) with a heat transfer coefficient of \(h\).

Short answer

Case (a) is specified by a temperature of \(50^{\circ} \mathrm{C}\). Case (b) is specified by a heat flux of \(30 \mathrm{~W} / \mathrm{m}^{2}\) toward the center. Case (c) includes convection to a medium at \(T_{\infty}\) with a heat transfer coefficient of \(h\). Answer: The boundary conditions for each case are: Case (a): \(T\left(r_{1}\right) = 50^{\circ} \mathrm{C}\) Case (b): \(-k\frac{\partial T}{\partial r}\biggr|_{r=r_1} = -30 \mathrm{~W} / \mathrm{m}^{2}\) Case (c): \(-k\frac{\partial T}{\partial r}\biggr|_{r=r_1} = h\left(T - T_{\infty}\right)\biggr|_{r=r_1}\)

Step-by-step solution

General Heat Conduction Equation

The generic heat conduction equation in spherical coordinates is given by: $$ \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2k\frac{\partial T}{\partial r}\right) = 0 $$ Since it is steady and one-dimensional, there is no time or angle variation, so we are left with only the left side of the equation to solve.

Boundary Condition for Case (a)

In Case (a), the specified inner surface temperature is \(50^{\circ} \mathrm{C}\). This means that the boundary condition for this case is simply a Dirichlet condition, where the temperature at the inner surface is given. Therefore, $$ T\left(r_{1}\right) = 50^{\circ} \mathrm{C} $$

Boundary Condition for Case (b)

In Case (b), the specified heat flux toward the center at the inner surface is \(30 \mathrm{~W} / \mathrm{m}^{2}\). In this case, we have a Neumann boundary condition, with the heat flux specified. The heat flux can be expressed as $$ q_r = -k\frac{\partial T}{\partial r}\biggr|_{r=r_1} $$ Therefore, for this case, the boundary condition will be: $$ -k\frac{\partial T}{\partial r}\biggr|_{r=r_1} = -30 \mathrm{~W} / \mathrm{m}^{2} $$

Boundary Condition for Case (c)

In Case (c), we have convection to a medium at \(T_{\infty}\) with a heat transfer coefficient of \(h\). This introduces a convective boundary condition at the inner surface, and we can use Newton's law of cooling to relate the heat flux and the temperature difference: $$ q_r = h\left(T - T_{\infty}\right)\biggr|_{r=r_1} $$ We also have the heat flux expression in terms of temperature gradient as before: $$ q_r = -k\frac{\partial T}{\partial r}\biggr|_{r=r_1} $$ Since the heat flux is continuous across the boundary, we equate the two expressions to get the boundary condition for Case (c): $$ -k\frac{\partial T}{\partial r}\biggr|_{r=r_1} = h\left(T - T_{\infty}\right)\biggr|_{r=r_1} $$