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

  1. Oct 26, 2008 #1
    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:

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

    [tex]\frac{1}{r}\frac{d}{dr}(r\frac{dT}{dr})=-\frac{\dot{q}}{k}[/tex] T(r=R)=1 Cylindrical

    [tex]\frac{1}{r^2}\frac{d}{dr}(r^2\frac{dT}{dr})=-\frac{\dot{q}}{k}[/tex] 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

    [tex]T(x)=\sum_{n=0}^{\infty}b_n x^n[/tex]

    [tex]T'(x)=\sum_{n=0}^{\infty}nb_n x^n^-^1[/tex]

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

    [tex]b_{0}=0[/tex]

    [tex]b_{1}=0[/tex]

    [tex]b_{2}=2x[/tex]

    [tex]b_{3}=3x^2[/tex]

    Am I at least on the right track?
     
  2. jcsd
  3. Oct 26, 2008 #2

    gabbagabbahey

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    What is the derivative of [itex]b_0x^0[/itex]?

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

    [itex]T'(x)=\sum_{n=1}^{\infty}nb_n x^{n-1}[/itex]

    and

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

    Which means that if [itex]\dot{q}[/itex] is not a function of x, the only non-zero coefficient would be [itex]b_2[/itex].
     
  4. Oct 26, 2008 #3
    Ok, so:

    [tex]b_{0}=0[/tex]

    [tex]b_{1}=0[/tex]

    [tex]b_{2}=2x[/tex]

    [tex]b_{3}=3x^2[/tex]

    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 [tex]b_{n}[/tex].

    [tex]b_{n}=nx^n^-^1[/tex]

    Is the next step just to sum the first few even and odd [tex]b_{n}[/tex]'s or am I way off base?
     
  5. Oct 26, 2008 #4

    gabbagabbahey

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    You are way off base. Is [itex]\dot{q}[/itex] a function of x?
     
  6. Oct 26, 2008 #5
    q, heat generation, is a constant. [tex]\dot{q}[/tex] will change with distance, x. So no [tex]\dot{q}[/tex] is not constant.

    Edit: Scratch that, [tex]\dot{q}[/tex] is the rate of heat transfer so it is a constant in this condition.
     
  7. Oct 26, 2008 #6

    gabbagabbahey

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

    Meanwhile, your solution for T gives

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

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

    What does that make T(x)? (remember, you dont have any restrictions on b_0 and b_1 yet)
     
    Last edited: Oct 26, 2008
  8. Oct 26, 2008 #7
    Not quite. In [itex]2B_2 + 6B_3 x+ 12B_4 x^2+....[/itex] where are the 2, 6, 12, etc comming from?

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

    Isn't it true that as n increases [itex]B_n[/itex] decreases? Are we just assuming [itex]B_3[/itex] and on to be too small to be worty of counting?
     
  9. Oct 26, 2008 #8

    gabbagabbahey

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    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?
     
  10. Oct 26, 2008 #9
    Ah ha! Ok I get that part. Now what do I with the [itex]2B_2= \frac{-\dot{q}}{k}[/itex]?
     
  11. Oct 26, 2008 #10

    gabbagabbahey

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    Well, that means that [itex]B_2= \frac{-\dot{q}}{2k}[/itex] right?...So what does that make your series for T(x)?
     
    Last edited: Oct 26, 2008
  12. Oct 26, 2008 #11
    T(x)=[tex]\frac{\dot{q}x^2}{2k}[/tex]
     
  13. Oct 26, 2008 #12

    gabbagabbahey

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    1st, your missing a negative sign....2nd why have you set b0 and b1 equal to zero?
     
  14. Oct 26, 2008 #13
    Ok: [tex]T(x)=-\frac{\dot{q}x^2}{2k}[/tex]

    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.
     
  15. Oct 26, 2008 #14

    gabbagabbahey

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    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:

    [itex]T(x)=B_0+B_1x-\frac{\dot{q}x^2}{2k}[/itex]

    (Remember [itex] \frac{d^2}{dx^2}(B_0+B_1 x)=0[/itex] 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.
     
  16. Oct 26, 2008 #15
    [itex]T(x)=\frac{2}{x}+x+\frac{\dot{q}x^2}{2k}(x-1)[/itex]

    Thanks so much!

    I should be able to figure out the other two on my own now.
     
  17. Oct 26, 2008 #16

    gabbagabbahey

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    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.
     
  18. Oct 26, 2008 #17
    [itex]1=b_0+b_1(0)-\frac{\dot{q}0^2}{2k}[/itex]
    [tex]b_0=1[/tex]
    [tex]2=1+b_1(1)-\frac{\dot{q}}{2k}[/tex]
    [tex]1+\frac{\dot{q}}{2k}=b_1[/tex]
    [tex]T(x)=1+x+\frac{\dot{q}x}{2k}-\frac{\dot{q}x^2}{2k}[/tex]

    (I found an algebra mistake as I was recopying it.)
     
  19. Oct 26, 2008 #18

    gabbagabbahey

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    That looks much better! :0)

    If you post your solutions for the other two cases, I'll be happy to check them for you.
     
  20. Oct 26, 2008 #19
    Ok now were going back to some of my more fundamental problems in this class.

    [tex]\frac{1}{r}\frac{d}{dr}(r\frac{dT}{dr})=-\frac{\dot{q}}{k}[/tex]

    [tex](ln|r|)rT=-\frac{\dot{q}}{k}[/tex]

    ???
     
  21. Oct 26, 2008 #20

    gabbagabbahey

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

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

    [itex]\frac{d}{dr}(r \frac{dT}{dr})=\frac{-\dot{q}}{k} r[/itex]

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