# Homework Help: [thermodynamics] spherical shell problem

1. Nov 30, 2006

### a_lawson_2k

1. The problem statement, all variables and given/known data

spherical shell with inner and outer radii a and b, and with temperatures of the inner and outer surfaces $$T_2$$ and $$T_1$$, the thermal conductivity of the shell is k, derive an equation for the total heat current through the shell

2. Relevant equations

$$H=\frac{dQ}{dt}=kA\frac{T_H-T_C}{L}$$

3. The attempt at a solution

I am not sure how to approach the 'A' element of the above equation, as it varies and the book didn't give a hint one way or another how to approach it. Barring that,

$$H=kA\frac{\Delta t}{b-a}$$

The book said $$H=k(4\pi*ab)\frac{\Delta t}{b-a}$$ so I can't be too far off. They have a similar problem with a cylinder I'd like to solve, so hopefully it will shed light on both of them.

2. Nov 30, 2006

### ponjavic

$$H=\frac{dQ}{dt}=kA\frac{dT}{\dr}$$

Then make A = 4*pi*r^2 and integrate from r1 to r2?

Doesn't seem to give me what I want

Last edited: Nov 30, 2006
3. Nov 30, 2006

### a_lawson_2k

Such an integration would produce a volume, not an area.

4. Nov 30, 2006

### vanesch

Staff Emeritus
This is true only if there is a prismatic section (that is, if A is constant between C and H, over length L). You have the problem that A is not constant from r1 to r2.
However, you can use spherical symmetry, and you can use the fact that the heat flux will be the same at all r.

So you should try to find the temperature dependence of r: T(r), with boundary conditions that T(r1) = T_2 and T(r2) = T_1. The above equation is then to be written in a differential form:

H = constant = - k A(r) dT/dr which gives you a (simple) differential equation for T(r), if you fill in the area A(r) = 4 pi r^2.

5. Nov 30, 2006

### a_lawson_2k

$$\int_{a}^{b} \frac{-H}{4k\pi r^2} dr= \int_{T_1}^{T_2} dT$$
$$H = \frac{4abk\pi \Delta T}{a-b}$$

That equation wasn't even hinted at in the book; should I move to a more rigorous book if I am attempting these problems?

6. Dec 1, 2006

### vanesch

Staff Emeritus
Different books have different purposes ; some are conceptually clear, others are encyclopedic, some are highly rigorous, still others give you good problem solving skills. So you should look at different books to get a good overall appreciation of a subject.

Concerning pure problem solving (and certainly NOT any conceptual depth), do you know about the excellent Schaum's outline series ? There is one in about any subject (including one on heat transport).

7. Dec 1, 2006

### a_lawson_2k

Haven't heard of them, but I'll definitely look into it. Thanks for the lead.

As for the temperature variation T(R), I tried finding it by integrating the result from the previous problem H(r) divided by $$-k*4\pi r^2$$, but it didn't reconcile with the book's result. It said $$\frac{T_1 b(r-a)+T_2 a(b-r)}{r(b-a)}$$, but I kept getting a different result when I integrated $$\frac{H}{-4\pi r^2 k}$$ from a to R (a=<R<=b)...

8. Dec 2, 2006

### vanesch

Staff Emeritus
They are really worth it: they're pretty cheap, easily available (on amazon for instance) and very efficient to learn to solve problems in an efficient way.
(and no, I have no stock in them :-)

This should come out right...(I mean, the approach is correct).

Let's check: we have
C/r^2 = - dT/dr, so after integration, T is of the form T(r) = C/r + C'
Note that $$C = \frac{ab}{a-b} \DeltaT$$ in here.

Now, you can check that the expression of your book does comply to this T(r) form (check the coefficient of 1/r, it is C all right, plus a constant).

In order to check the constant C', we only have to verify that the expression of the book has the right value at r = a (and should then also have the right expression at r = b, because that's how we fixed C - and hence H - in the first place).
Fill it in: it is all right.

So this should come out all right...

9. Dec 2, 2006

### a_lawson_2k

$$\int_{a}{R}-\frac{\frac{4abk\pi (T_1-T_2)}{b-a}}{4\pi r^2 k}$$

Maple spit out the same thing, I finally found the first part of the solutions to University Physics (set me back \$20, still grumbling about that, I'm switching over to 11th once I find a good price) so this may be resolved sooner than later thankfully.

I'm still not sure why it didn't reconcile. To shed light, I used Maple to expand the results:
Book's:
$${\frac {{\it T1}\,b}{b-a}}-{\frac {ab{\it T1}}{R \left( b-a \right) }} +{\frac {ab{\it T2}}{R \left( b-a \right) }}-{\frac {{\it T2}\,a}{b-a}}$$

mine:

$$-{\frac {ab{\it T1}}{ \left( -b+a \right) R}}+{\frac {ab{\it T2}}{ \left( -b+a \right) R}}$$

I appear to be off by the far left and right terms and then by a factor of -1.

Last edited: Dec 2, 2006
10. Dec 10, 2006

### a_lawson_2k

Still haven't figured it out; further insight would be appreciated.

Thanks

11. Dec 10, 2006

### vanesch

Staff Emeritus
I can only repeat what I wrote already...