# Newtons law of cooling applied to spherical region

I'm having trouble solving the following problem:

Formulate a mathmatical model for a stationary (steady) temperature distribution inside the spherical volume $$R^2\leq x^2+y^2+z^2\leq (2R)^2,$$ where $$R$$ is a given constant. The region is homogenous and the boundary $$x^2+y^2+z^2=R^2$$ has constant temperature $$T=T_0$$. Newton's law of cooling describes the temperature at the other boundary $$x^2+y^2+z^2=(2R)^2$$ (the normal component of the heat diffusion is proportional to the difference of the boundary temperature and the temperature of the region outside, $$T_1$$.

I found this formula on the heat equation for a spherical region:

$$\frac{\partial T}{\partial t}= \frac{\alpha}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial T}{\partial r}\right)\quad 0

But this is for a spehere with no inner boundary, so I am lost at how I apply this formula for my case. But the model seems correct, if I am correct this is derived from Laplace equation using spherical coordinates, correct me if I am wrong. Or is it Newton's law in spherical coordinates?

Best regards

• "But this is for a spehere with no inner boundary" No this holds in general (boundary enters as boundary conditions to the PDE). The domain in your case is simply $R<r<2R$. Jan 19, 2019 at 11:19
• Since it is steady state, the left hand side of the equation you found should be $0$. Then you also need to add the boundary conditions, $T=T_0$ at $r=R$ and $\frac{\partial T}{\partial r}=k(T-T_1)$ at $r=2R$. Jan 19, 2019 at 12:50
• If you assume that the temperature depends on $r$ only, then your equation is valid in a region $r_1 < r < r_2$. Jan 20, 2019 at 3:20

This is something that might be useful. You are looking at a spherical shell with inner radius R and outer radius 2R. The heat equation and solution are given in the following image with control volume to derive it also in the image. If you need help on derivtion here it goes:

$$\frac{d}{dr}\left(r^2\frac{dT}{dr}\right)=0$$

$$\left(r^2\frac{dT}{dr}\right) = C_1$$

$$\frac{dT}{dr} = \frac{C_1}{r^2}$$

$$T = -\frac{C_1}{r} + C_2$$

Applying boundary conditions

$$T_1 = -\frac{C_1}{R} + C_2$$

$$T_2 = -\frac{C_1}{2R} + C_2$$

Solve the algrebraic equations C_1 = , you get

$$C_1 = 2R(T_2-T_1)$$

$$C_2 = T_1+ 2(T_2-T_1)$$

$$T = -2\frac{R}{r}(T_2-T_1) + T_1+2(T_2-T_1)$$

$$T = T_1 + 2(T_2-T_1)\left(1-\frac{R}{r}\right)$$ • Very nice! Thanks for the edit, I was at a lost how to actually solve it from the figures, but your solution made it clearer what was happening. Jan 30, 2019 at 10:36