B2 — Energy equation for a heated sphere (angular symmetry): 23-Chem-B1 Transport Phenomena, May 2016, Q4

Question 4: B2 — Energy equation for a heated sphere (angular symmetry) (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A solid sphere (no bulk motion, $\mathbf{u}=0$) heated so that temperature depends only on radius and time, $T=T(r,t)$; there is no variation with the polar or azimuthal angles ($\partial T/\partial\theta=\partial T/\partial\phi=0$); constant density $\rho$ and heat capacity $c_p$; possible volumetric generation $\dot q_G$.

Find. The reduced energy equation in spherical coordinates.

Fig. B2: With no angular dependence the isotherms are concentric spheres, so conduction is purely radial. The energy equation reduces to a one-dimensional radial operator with the characteristic $2/r$ curvature term.

Approach. Start from the general energy equation (Appendix A, Table A.3 spherical form), delete the convection terms (solid at rest) and the two angular conduction terms (no $\theta,\phi$ dependence), leaving the radial operator.

Energy balance on a thin spherical element. Take the region between radii $r$ and $r+\Delta r$ (volume $4\pi r^2\Delta r$). Because $T$ has no angular dependence, heat crosses only the two spherical faces, whose areas $4\pi r^2$ and $4\pi(r+\Delta r)^2$ differ. Rate in − rate out + generation = accumulation: $$\left(4\pi r^2q_r\right)\Big|_r-\left(4\pi r^2q_r\right)\Big|_{r+\Delta r}+\dot q_G\,4\pi r^2\Delta r=\rho c_p\,4\pi r^2\Delta r\,\frac{\partial T}{\partial t}.$$ Dividing by $4\pi r^2\Delta r$ and letting $\Delta r\to0$: $$\rho c_p\frac{\partial T}{\partial t}=-\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2q_r\right)+\dot q_G,$$ and Fourier’s law $q_r=-k\,\partial T/\partial r$ turns this into the radial conduction equation below. The next two steps confirm it against the tabulated general equation.
General spherical energy equation. Table A.3 (spherical) reads, for a medium at rest with conduction and generation, $$\rho c_p\frac{\partial T}{\partial t}=\frac{1}{r^2}\frac{\partial}{\partial r}\!\left(k r^2\frac{\partial T}{\partial r}\right)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\!\left(k\sin\theta\frac{\partial T}{\partial\theta}\right)+\frac{1}{r^2\sin^2\theta}\frac{\partial}{\partial\phi}\!\left(k\frac{\partial T}{\partial\phi}\right)+\dot q_G.$$
Impose angular symmetry. With $\partial T/\partial\theta=0$ and $\partial T/\partial\phi=0$ the last two conduction terms vanish identically, leaving only the radial term: $$\boxed{\;\rho c_p\frac{\partial T}{\partial t}=\frac{1}{r^2}\frac{\partial}{\partial r}\!\left(k\,r^2\frac{\partial T}{\partial r}\right)+\dot q_G\;}$$
Constant-conductivity form. If $k$ is constant it comes outside the derivative and, expanding, $$\rho c_p\frac{\partial T}{\partial t}=k\left(\frac{\partial^2T}{\partial r^2}+\frac{2}{r}\frac{\partial T}{\partial r}\right)+\dot q_G=k\,\nabla^2_rT+\dot q_G,$$ where the $\tfrac{2}{r}\,\partial T/\partial r$ term is the spherical curvature contribution (twice that of a cylinder because the conduction area $4\pi r^2$ grows as $r^2$).
Steady, no generation (check). Setting $\partial T/\partial t=0$, $\dot q_G=0$ gives $\tfrac{d}{dr}(r^2\,dT/dr)=0\Rightarrow T=C_2-C_1/r$ — the familiar hollow-sphere conduction profile, confirming the operator.

Result	Expression
General (variable $k$)	$\rho c_p\,\partial_t T=\tfrac{1}{r^2}\partial_r(k r^2\partial_r T)+\dot q_G$
Constant $k$	$\rho c_p\,\partial_t T=k\big(T_{rr}+\tfrac{2}{r}T_r\big)+\dot q_G$
Steady, no source	$T(r)=C_2-C_1/r$

Concept & Theory

Topic. Reduction of the general energy equation to one-dimensional radial conduction in a sphere.

Key relations. Spherical Laplacian $\nabla^2_rT=\tfrac{1}{r^2}\partial_r(r^2\partial_rT)=T_{rr}+\tfrac{2}{r}T_r$; energy balance $\rho c_p\partial_tT=\nabla\!\cdot\!(k\nabla T)+\dot q_G$.

Why this works. Killing the convective terms (solid) and the angular derivatives (symmetry) collapses a three-dimensional PDE to a single radial coordinate. The factor $2/r$ — versus $1/r$ for a cylinder — encodes that a sphere’s conduction area $4\pi r^2$ grows faster with radius, so a given radial temperature gradient carries proportionally more heat outward.

Common pitfall. Writing $\tfrac{1}{r}\partial_r(r\,\partial_rT)$ (the cylindrical operator) instead of $\tfrac{1}{r^2}\partial_r(r^2\partial_rT)$, or retaining the $\sin\theta$ angular term after invoking symmetry.

Source. Bird, Stewart & Lightfoot, Transport Phenomena, Table B.9 (energy equation, spherical coordinates); Appendix A, Table A.3 of this paper.

Question 4 of 6: B2 — Energy equation for a heated sphere (angular symmetry)

Question 4: B2 — Energy equation for a heated sphere (angular symmetry) (25 marks)