# Homework Help: Thermal diffusion equation

1. Dec 30, 2017

### Physgeek64

1. The problem statement, all variables and given/known data
Gas with thermal conductivity κ fills the space between two coaxial cylinders
(inner cylinder radius a, outer cylinder inner radius b). A current I is passed through
the inner cylinder, which has resistivity ρ. Derive an expression for the equilibrium temperature of the inner cylinder Ta when the outer cylinder is held at a constant temperature Tb.

2. Relevant equations

3. The attempt at a solution
so for the equilibrium case we have $\frac{\partial T}{\partial t} =0$

Solving the heat diffusion equation in cylindrical coordinates with a heat source H we get

$T=-\frac{Hr^2}{4\kappa}+c_1 \ln{r} +c_2$ where $H=\frac{I^2\rho}{A^2}$ where A is the C.S.A of the inner cylinder.

I can see that we have one boundary condition, namely $T=T_b$ at r=b but i cannot see the second boundary condition.

Many thanks

2. Dec 30, 2017

### NFuller

I don't think the term $-\frac{Hr^2}{4\kappa}$ should be here since solutions of the polar Laplace equation contains powers of $r$ only when there is angular dependence in the solution. Here the problem is radially symmetric.
The other boundary condition is a neumann boundary condition at $r=a$. The condition satisfies Fourier's Law for heat flow
$$\mathbf{q}=-k\nabla_{r=a} T$$
$\mathbf{q}$ is the heat flux density which is related to the power dissipated by the resistive cylinder.

3. Dec 30, 2017

### Physgeek64

I dont understand. We don't have angular dependence here so surely we can have non linear dependence on r. If you work through the algebra I can't see how I could not include this term.

I'm not too sure how to use this without knowing how to calculate q

4. Dec 30, 2017

### NFuller

Correct, so the solution depends on the natural log of $r$, which is non-linear. Terms involving powers of $r$ would generally have a sin and/or cosine term attached to it. Since there is no angular dependence, those terms should not appear in the solution.
The power dissipated by the resistive inner cylinder can be found from ohms law and the definition of electrical power as
$$P=I^{2}R$$
and the resistance $R$ is related to resistivity $\rho$ by
$$R=\rho\frac{l}{A}$$
where $l$ is the length of the cylinder and $A$ is the cross sectional area.

Since the system is assumed to be at equilibrium, the electrical power dissipation should be equal to the rate of energy dissipation as heat. The heat flux out of the inner cylinder is then
$$P=\oint_{r=a}\mathbf{q}\cdot d\mathbf{a}$$
Since the flux is radially symmetric, $\mathbf{q}$ can be taken out of the integral giving
$$P=qA=q2\pi a l$$
Using these formulas can you construct an equation for $q$?

5. Dec 31, 2017

### Staff: Mentor

Your solution assumes that the heat is being generated within the conducting medium (i.e., the gas) rather than within the inner cylinder. This is not what the problem statement says.

In this problem, they didn't state it, but you are supposed to assume that the thermal conductivity of the inner cylinder is very high so that its temperature is uniform. So you have a heat flow out of the inner cylinder through the intervening annular region of gas. This is straight heat conduction with no heat generation within the gas. You are supposed to determine the temperature difference across the annular region of gas, knowing the rate of heat flow.