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

[Schmoozer]

Homework Statement

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?

Homework Equations

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?

 

[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.

 

[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?

 

[gabbagabbahey]

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

 

[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.

 

[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}
Do you follow this?

What does that make T(x)? (remember, you don't have any restrictions on b_0 and b_1 yet)

 

[Schmoozer]

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

 

[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?

 

[Schmoozer]

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

 

[gabbagabbahey]

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

 

[Schmoozer]

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

 

[gabbagabbahey]

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

 

[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 that's my logic for it.

 

[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.

 

[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.

 

[gabbagabbahey]

are you sure about this answer...when i plug in x=0 i get T=2/0=infinity!...perhaps you should show me your steps for finding B_0 and B_1.

 

[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.)

 

[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.

 

[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}

?

 

[gabbagabbahey]

Where does the ln|r| come from?

Start with your 1st equation and multiply both sides by r:

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

Then integrate each side of the equation over r...what does that step give you?

 

[Schmoozer]

r^2*T(r)=-((qdot)*r)/k
T(r)=--(qdot)/(r*k)

 

[gabbagabbahey]

Not quite...

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

correct?

What is \int \frac{d}{dr}(r \frac{dT}{dr})dr?

What is \int \frac{-\dot{q}}{k} r dr?

 

[Schmoozer]

\int \frac{-\dot{q}}{k} r dr => \frac{-\dot{q}r^2}{2k}

\int \frac{d}{dr}(r \frac{dT}{dr})dr => Tr

The dT is throwing me off, I'm not quite sure about it.

 

[gabbagabbahey]

Your first integral is correct, but your second one is not...try writing f(r)=r\frac{dT}{dr}, then f'(r)=\frac{d}{dr}(r\frac{dT}{dr})

and so you can rewrite the integral as \int f'(r)dr...what does the fundamental theorem of calculus tell you about this integral?

 

[Schmoozer]

Doesn't that reduce to r\frac{dT}{dr}?

 

[gabbagabbahey]

yes...plus a constant of integration.

So, you now have :

r\frac{dT}{dr}=\frac{-\dot{q}r^2}{2k}+C_1

divide each side of the equation by r and integrate again...what do you get?

 

[Schmoozer]

Ahhhh, I feel silly, here I was thinking it was first order.

Ok so:

\frac{dT}{dr}=-\frac{\dot{q}r}{2k}+\frac{C_1}{r}

T=-\frac{\dot{q}r^2}{4k}+ln|r|*C_1+C_2

 

[gabbagabbahey]

Yes, looks good...you also know that T(r=R)=1, so you can eliminate one of your constants of integration.

 

[Schmoozer]

C_2=1+\frac{\dot{q}R^2}{4k}-ln|R|*C_1

T(r)=1+\frac{\dot{q}}{4k}(R^2-r^2)+C_1*ln|(\frac{r}{R})|

So now this mess again: T(x)=\sum_{n=0}^{\infty}b_n x^n ?

 

[gabbagabbahey]

You could also argue that C_1=0 on physical grounds because the ln(r) terms blows up as r approaches zero and physically, you would expect the temperature to be finite everywhere inside the cylinder.