Question 6 of 6: C2 — Diffusion equation in spherical coordinates

Question 6: C2 — Diffusion equation in spherical coordinates (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. The species-A continuity equation in spherical coordinates (Appendix A, Table A.5), for a medium that is stationary ($\mathbf u=0$), has constant molecular diffusivity $D$, and undergoes no chemical reaction ($R_{A,G}=0$).

Find. Reduce the general species equation to the stated diffusion (Fick’s-second-law) form and expand the spherical Laplacian to match the printed target.

Approach. Start from the general species-continuity equation, delete the convective terms (stationary medium) and the generation term (no reaction), take $D$ outside the derivatives, then expand the compact spherical Laplacian operators to reproduce the printed expression.

1. General species equation. Table A.5 (spherical) states $$\frac{\partial C_A}{\partial t}+\Big(u_r\frac{\partial C_A}{\partial r}+\frac{u_\theta}{r}\frac{\partial C_A}{\partial\theta}+\frac{u_\phi}{r\sin\theta}\frac{\partial C_A}{\partial\phi}\Big)=D\,\nabla^2C_A+R_{A,G}.$$
2. Impose the stated conditions. The medium is stationary, so $u_r=u_\theta=u_\phi=0$ and the entire convective bracket vanishes; there is no reaction, so $R_{A,G}=0$; $D$ is constant. What remains is Fick’s second law $$\frac{\partial C_A}{\partial t}=D\,\nabla^2C_A.$$
3. Write the spherical Laplacian. In spherical coordinates $$\nabla^2C_A=\frac{1}{r^2}\frac{\partial}{\partial r}\!\Big(r^2\frac{\partial C_A}{\partial r}\Big)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\!\Big(\sin\theta\frac{\partial C_A}{\partial\theta}\Big)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2 C_A}{\partial\phi^2}.$$
4. Expand the radial operator. $\dfrac{1}{r^2}\dfrac{\partial}{\partial r}\!\big(r^2\partial_r C_A\big)=\dfrac{\partial^2 C_A}{\partial r^2}+\dfrac2r\dfrac{\partial C_A}{\partial r}$ (product rule: the $2r\,\partial_r C_A$ term divided by $r^2$ gives the $2/r$ factor).
5. Expand the polar ($\theta$) operator. $\dfrac{1}{r^2\sin\theta}\dfrac{\partial}{\partial\theta}\!\big(\sin\theta\,\partial_\theta C_A\big)=\dfrac{1}{r^2}\Big(\dfrac{\partial^2 C_A}{\partial\theta^2}+\cot\theta\,\dfrac{\partial C_A}{\partial\theta}\Big)$, since $\partial_\theta(\sin\theta\,\partial_\theta C_A)=\sin\theta\,\partial_{\theta\theta}C_A+\cos\theta\,\partial_\theta C_A$.
6. Assemble the result. Substituting the expanded operators, $$\boxed{\,\frac{\partial C_A}{\partial t}=D\Big[\frac{\partial^2 C_A}{\partial r^2}+\frac2r\frac{\partial C_A}{\partial r}+\frac{1}{r^2}\frac{\partial^2 C_A}{\partial\theta^2}+\frac{\cot\theta}{r^2}\frac{\partial C_A}{\partial\theta}+\frac{1}{r^2\sin^2\theta}\frac{\partial^2 C_A}{\partial\phi^2}\Big]\,}$$ Dividing through by $D$ and listing the five terms in the order printed gives $\frac1D\frac{\partial C_A}{\partial t}=\frac{\partial^2C_A}{\partial r^2}+\frac1{r^2}\frac{\partial^2C_A}{\partial\theta^2}+\frac2r\frac{\partial C_A}{\partial r}+\frac1{r^2\sin^2\theta}\frac{\partial^2C_A}{\partial\phi^2}+\frac{\cot\theta}{r^2}\frac{\partial C_A}{\partial\theta}$, which is exactly the equation to be shown (the paper’s azimuthal angle $\varphi$ is written $\phi$ here; the stray “$\beta$” in the printed list of coordinates plays no part in the equation).