Heat conduction through sphere

[unscientific]

Homework Statement

In general, a sphere with conductivity $\kappa$, heat capacity per unit volume $C$ and radius $R$ obeys the differential equation at time t:

$$C\frac{\partial T}{\partial t} = \kappa \frac{\partial^2 T}{\partial r^2} + \frac{2\kappa}{r}\frac{\partial T}{\partial r}$$

Part (a): A sphere with a cavity of radius $a$ generates heat at a rate Q. Heat is lost from the outer surface of the sphere to the surroundings, with surrounding temperature $T_s$, given by Newton’s Law of Cooling with constant $\alpha$ per unit area. Find the temperature $T_a$ at the surface of the cavity at thermal equilibrium.

Part (b): A second sphere without a cavity generates heat uniformly at $q$ per unit volume. Like the first sphere, heat is lost to the surroundings at its surface. Find temperature at the center of the sphere $T_0$ at equilibrium.

Homework Equations

The Attempt at a Solution

Part (a)

Heat generated in cavity = Heat loss at surface
$$Q = \alpha (T_s - T_R)(4\pi R^2)$$
$$T_R = T_s - \frac{Q}{\alpha (4\pi R^2)}$$

We will use this to solve for the constants in the differential equation later on.

At steady state, $C\frac{\partial T}{\partial t} = \kappa \frac{\partial^2 T}{\partial r^2} + \frac{2\kappa}{r}\frac{\partial T}{\partial r} = 0$.

$$\frac{\partial^2 T}{\partial r^2} + \frac{2}{r}\frac{\partial T}{\partial r} = 0$$

Solving, we get:

$$T = A + \frac{b}{r}$$

We need one more equation with the one for $T_R$ to solve for constants $A$ and $B$.

Using $\int \vec J \cdot d\vec S = k \int \nabla \vec T \cdot d\vec S$:
$$Q = -\kappa \frac{\partial T}{\partial r}(4\pi r^2)$$
$$Q = 4\pi \kappa b$$
$$b = \frac{Q}{4\pi \kappa}$$

Solving for A:
$$A = T_s - \frac{Q}{4\pi \alpha R^2} - \frac{Q}{4\pi \kappa R}$$

Together, temperature at surface of cavity is:

$$T_a = T_s - \frac{Q}{4\pi \alpha R^2} + \frac{Q}{4\pi \kappa}\left( \frac{1}{a} - \frac{1}{R} \right)$$

Part(b)

I'm not sure how to approach this, as $T = A + \frac{b}{r}$ doesn't work at $r=0$..

I have found the temperature at surface though:
$$Q = \frac{4}{3}\pi R^3 q = \alpha (T_s - T_R)(4\pi R^2)$$
$$T_R = T_s - \frac{Rq}{3\alpha}$$

 

[maajdl]

You need to add the source term (q) in heat transfer partial differential equation.
You then again solve for the steady state solution.

 

[unscientific]

You need to add the source term (q) in heat transfer partial differential equation.
You then again solve for the steady state solution.

I don't understand what you mean. And is this for part (a) or part (b)?

 

[dauto]

I think you made a sign mistake in the Newton's law of cooling (part a). For part b you need a source term. That means you will solve the non-homogeneous heat conduction differential equation.

 

[maajdl]

I don't understand what you mean. And is this for part (a) or part (b)?

This is for part b.

$$C\frac{\partial T}{\partial t} = \kappa \frac{\partial^2 T}{\partial r^2} + \frac{2\kappa}{r}\frac{\partial T}{\partial r} + q$$

 

[unscientific]

I think you made a sign mistake in the Newton's law of cooling (part a). For part b you need a source term. That means you will solve the non-homogeneous heat conduction differential equation.

For part (a), is the sign error here:

$$Q = \alpha (T_s - T_R)(4\pi R^2)$$

It should be $T_r - T_s$, since the sphere is hotter. Other than that, is the rest of my working fine?


For part (b), I read up more on the chapter and I realized there should be a source per-unit-volume term per unit time term.

Thanks alot!

 

[unscientific]

I think you made a sign mistake in the Newton's law of cooling (part a). For part b you need a source term. That means you will solve the non-homogeneous heat conduction differential equation.

I have given this question another go. At steady state, the DE now becomes:

$$\kappa \frac{\partial^2T}{\partial r^2} + 2\frac{\kappa}{r} \frac{\partial T}{\partial r} + q = 0$$

Since the homogeneous solution is $A + \frac{b}{r}$, we now seek the inhomogeneous solution.

I try $\beta r^2$ where I found $\beta = -\frac{q}{6k}$.

So the solution for T is:
$$T = A + \frac{b}{r} - \frac{q}{6k}r^2$$

But now as $r \rightarrow \infty$, $T\rightarrow \infty$..