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 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?

 

[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.

 

[gabbagabbahey]

There is no need for a power series here...see my last post ^^^

 

[Schmoozer]

Ok so I'm on the last one and have:
$$T(r)=1+\frac{\dot{q}}{4k}(R^2-r^2)+C_1(\frac{1}{R}-\frac{1}{r})$$

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

 

[gabbagabbahey]

I think you should have:

$T(r)=1+\frac{\dot{q}}{6k}(R^2-r^2)+C_1(\frac{1}{R}-\frac{1}{r})$

Again, you would expect the temperature to be finite everywhere inside the sphere; so you can set C_1=0 since the 1/r blows up at r=0.

PS. don't forget to answer part (b) of the question for each case ;0)

 

[Schmoozer]

b. T(r)=1 when q_dot=0 for all three cases right?

Edit: Except the first one where T(r)=1+x

 

[alpha754293]

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?

LOL. Navaz's class? MECH522? Awesome!