PHYSICS 312

SOLUTION TO PROBLEM SET 2 1999:


Problem 1:
In the steady state heat flows radially at the same rate as it is generated. The heat balance condition at radius $\rho$ is

$$-\kappa\frac{du}{d\rho}4\pi\rho^2=\frac{4\pi H\rho^3}{3}$$

which simplifies to the first order differential equation

$$\frac{du}{d\rho}=-\frac{H}{3\kappa}\rho$$

with solution

$$u=-\frac{H}{6\kappa}\rho^2+a$$

where a is aa constant to be determined. The boundary condition at the surface $\rho=c$ is

$$-\kappa\frac{du}{d\rho}=h(u-T)$$

We find

$$a=\frac{H}{6\kappa}c^2+T+\frac{H\rho}{3h}$$

and



Problem 2:
The condition that the heat flow is constant can be written

$$-4\pi\kappa r^2\frac{du}{dr}=const$$

We call this constant $4\pi\kappa a$ giving the differential equation

$$\frac{du}{dr}=-\frac{a}{r^2}$$

with solution

$$u(r)=\frac{a}{r}+b$$

The boundary conditions give b=T, a=-Tc and



Note that the solution is independent of $\kappa$!

For problem 3: see attached Maple worksheet.

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