# 1-D steady state heat conduction equation (Cartisian, Cylindrical and Sperical)

1. Oct 26, 2008

### Schmoozer

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

The one dimensional steady-state heat conduction equation in a medium with constant conductivity (k) with a constant volumetric heat generation in three different coordinate systems (fuel rods in a nuclear power plant) is given as:

$$\frac{d^2 T}{dx^2}=-\frac{\dot{q}}{k}$$ T(x=0)=1 T(x=1)=2 Cartisian

$$\frac{1}{r}\frac{d}{dr}(r\frac{dT}{dr})=-\frac{\dot{q}}{k}$$ T(r=R)=1 Cylindrical

$$\frac{1}{r^2}\frac{d}{dr}(r^2\frac{dT}{dr})=-\frac{\dot{q}}{k}$$ T(r=R)=1 Sperical

a. Find an expression for the temperature distribution in a solid for each case
b. What is a temperature distribution if the heat generation is zero?

2. Relevant equations

3. The attempt at a solution

$$T(x)=\sum_{n=0}^{\infty}b_n x^n$$

$$T'(x)=\sum_{n=0}^{\infty}nb_n x^n^-^1$$

$$T''(x)=\sum_{n=0}^{\infty}n(n-1)b_n x^n^-^2$$ so n=0, n=1

$$b_{0}=0$$

$$b_{1}=0$$

$$b_{2}=2x$$

$$b_{3}=3x^2$$

Am I at least on the right track?

2. Oct 26, 2008

### gabbagabbahey

What is the derivative of $b_0x^0$?

....Doesn't that mean that the first term in the series drops out and hence:

$T'(x)=\sum_{n=1}^{\infty}nb_n x^{n-1}$

and

$T''(x)=\sum_{n=2}^{\infty}n(n-1)b_n x^{n-2}=-\frac{\dot{q}}{k}$

Which means that if $\dot{q}$ is not a function of x, the only non-zero coefficient would be $b_2$.

3. Oct 26, 2008

### Schmoozer

Ok, so:

$$b_{0}=0$$

$$b_{1}=0$$

$$b_{2}=2x$$

$$b_{3}=3x^2$$

I am unable find a pattern in the even and odd b subscripts as the powers will simply continue to increase.

So I guess my next step is to find a solution to $$b_{n}$$.

$$b_{n}=nx^n^-^1$$

Is the next step just to sum the first few even and odd $$b_{n}$$'s or am I way off base?

4. Oct 26, 2008

### gabbagabbahey

You are way off base. Is $\dot{q}$ a function of x?

5. Oct 26, 2008

### Schmoozer

q, heat generation, is a constant. $$\dot{q}$$ will change with distance, x. So no $$\dot{q}$$ is not constant.

Edit: Scratch that, $$\dot{q}$$ is the rate of heat transfer so it is a constant in this condition.

6. Oct 26, 2008

### gabbagabbahey

If q is constant in x, then $\dot{q}=\frac{dq}{dt}$ will be constant in x as well. This means that $\frac{\dot{q}}{k}$ is just some constant.

Meanwhile, your solution for T gives

$T''=\sum_{n=2}^{\infty} B_n n(n-1)x^{n-2}=2B_2 + 6B_3 x+ 12B_4 x^2+....$

but this must equal a constant, so shouldn't $B_3=B_4=\ldots B_{\infty}=0$ so that you are left with $2B_2= \frac{-\dot{q}}{k}$

What does that make T(x)? (remember, you dont have any restrictions on b_0 and b_1 yet)

Last edited: Oct 26, 2008
7. Oct 26, 2008

### Schmoozer

Not quite. In $2B_2 + 6B_3 x+ 12B_4 x^2+....$ where are the 2, 6, 12, etc comming from?

And why is this $B_3=B_4=\ldots B_{\infty}=0$ true especially when $B_2$ is equal to some real number?

Isn't it true that as n increases $B_n$ decreases? Are we just assuming $B_3$ and on to be too small to be worty of counting?

8. Oct 26, 2008

### gabbagabbahey

the 2,6,12 are coming from n(n-1)....2(2-1)=2, 3(3-1)=6, 4(4-1)=12...etc.

Suppose that B_3,B_4,...etc weren't zero...wouldn't that mean that T'' was a function of x?

For example, if B_3 were equal to 1 and B_4,B_5,....=0 then T'' would equal 2B_2+6x....how could this possibly equal a constant?

9. Oct 26, 2008

### Schmoozer

Ah ha! Ok I get that part. Now what do I with the $2B_2= \frac{-\dot{q}}{k}$?

10. Oct 26, 2008

### gabbagabbahey

Well, that means that $B_2= \frac{-\dot{q}}{2k}$ right?...So what does that make your series for T(x)?

Last edited: Oct 26, 2008
11. Oct 26, 2008

### Schmoozer

T(x)=$$\frac{\dot{q}x^2}{2k}$$

12. Oct 26, 2008

### gabbagabbahey

1st, your missing a negative sign....2nd why have you set b0 and b1 equal to zero?

13. Oct 26, 2008

### Schmoozer

Ok: $$T(x)=-\frac{\dot{q}x^2}{2k}$$

but

bn => n(n-1)*bn*x^(n-2)
b0 => 0(0-1)*b0*x^(-2)=0
b1 => 1(1-1)*b1*x^(-1)=0

I'm assuming this is wrong, but thats my logic for it.

14. Oct 26, 2008

### gabbagabbahey

The sum for T'' starts at n=2....it says nothing about the n=0 and n=1 terms, so just leave them as unknowns:

$T(x)=B_0+B_1x-\frac{\dot{q}x^2}{2k}$

(Remember $\frac{d^2}{dx^2}(B_0+B_1 x)=0$ for any B_0 and B_1 not just for B_0=B_1=0)

But you are also told that T(x=0)=1 andT(x=1)=2, so you can use these two conditions to determine B_0 and B_1.

15. Oct 26, 2008

### Schmoozer

$T(x)=\frac{2}{x}+x+\frac{\dot{q}x^2}{2k}(x-1)$

Thanks so much!

I should be able to figure out the other two on my own now.

16. Oct 26, 2008

### gabbagabbahey

17. Oct 26, 2008

### Schmoozer

$1=b_0+b_1(0)-\frac{\dot{q}0^2}{2k}$
$$b_0=1$$
$$2=1+b_1(1)-\frac{\dot{q}}{2k}$$
$$1+\frac{\dot{q}}{2k}=b_1$$
$$T(x)=1+x+\frac{\dot{q}x}{2k}-\frac{\dot{q}x^2}{2k}$$

(I found an algebra mistake as I was recopying it.)

18. Oct 26, 2008

### gabbagabbahey

That looks much better! :0)

If you post your solutions for the other two cases, I'll be happy to check them for you.

19. Oct 26, 2008

### Schmoozer

Ok now were going back to some of my more fundamental problems in this class.

$$\frac{1}{r}\frac{d}{dr}(r\frac{dT}{dr})=-\frac{\dot{q}}{k}$$

$$(ln|r|)rT=-\frac{\dot{q}}{k}$$

???

20. Oct 26, 2008

### gabbagabbahey

Where does the ln|r| come from????

$\frac{d}{dr}(r \frac{dT}{dr})=\frac{-\dot{q}}{k} r$