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Heat EQ

  1. Jan 7, 2015 #1

    joshmccraney

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    hi pf!

    i'm wondering if you can help me with the heat eq for a basic cylinder wire problem. namely, we have a wire with radius ##r_i## and length ##L##and resistance is ##R## and current is ##I##. Thus heat produced $$Q = R I^2 \pi r_i^2 L$$. When using the heat eq, we assume time rate of change is negligable. flux is governed by fouriers law, and the divergence theorem gives us the following: $$\int_V k \nabla^2 T dv + \int_V \frac{Q}{\pi r^2 L}dv = 0$$.is this right though? namely, is ##Q## divided by an arbitrary ##r## or the radius ##r_i##?

    thanks so much!
     
  2. jcsd
  3. Jan 8, 2015 #2

    Bystander

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    Care to tell us how you came up with this expression?

    @Chestermiller , @Orodruin
     
    Last edited: Jan 8, 2015
  4. Jan 8, 2015 #3

    joshmccraney

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    I knew units for heat generation in 3-D need to be watts per cubic meter. So I simply tracked units. This Q multiplied by dv gives us watts, which is the unit we're after.
     
  5. Jan 8, 2015 #4
    But what are the units of [itex]k\nabla^2T[/itex] (k should be the thermal diffusivity)?

    [edit: I mean k=thermal conductivity of course]
     
    Last edited: Jan 8, 2015
  6. Jan 8, 2015 #5

    joshmccraney

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    Watts per cubic meter, right?
     
  7. Jan 8, 2015 #6
    and of the term [itex]\frac{Q}{\pi r^2 L} = I^2 R \frac{\pi r_i^2 L}{\pi r^2 L}[/itex] (with [itex]I^2R[/itex] the electric power)?
     
  8. Jan 8, 2015 #7

    joshmccraney

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    watts? am i missing something here? seems like you are eluding to something.
     
  9. Jan 8, 2015 #8

    joshmccraney

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    oh shoooot! i should have defined ##Q : Q = I^2 R / \pi r_i ^2 L## right? but is it ##r_i## or ##r##?
     
    Last edited: Jan 8, 2015
  10. Jan 8, 2015 #9
    ri. The rate of heat generation per unit volume in the wire is constant. ri should also be what appears in the equation with the integrals.

    Chet
     
  11. Jan 9, 2015 #10

    joshmccraney

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    Hi chet!

    ok, so what i should have modeled from the start is $$\int_v k \nabla ^2T dv = \int_v Q dv : Q = I^2 R / \pi r_i^2 L$$ do you all agree? if so, solving would be (using 1-D radial flow in polar coordinates) $$-k\frac{1}{r}\frac{d}{dr} ( r T') = Q \implies \\ -k d(r T') = rQdr \implies \\ \int_?^{??} -k d(r T') = \int_0^{r_i}rQdr$$
    but what are my bounds for integration? any ideas?
     
    Last edited: Jan 9, 2015
  12. Jan 9, 2015 #11
    Q is a constant, so it comes out of the integral. You integrate both sides from 0 to ri.

    Chet
     
  13. Jan 9, 2015 #12

    joshmccraney

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    Are you sure? I'm thinking if we had a differential equation over some interval of time ##[0,T]##, say, $$\frac{dy}{dt} = k \implies \\ \int_0^T \frac{dy}{dt} dt = \int_0^T k dt \implies \\ \int_{y(0)}^{y(T)} dy = \int_0^T k dt$$ but notice we do not have ##[0,T]## on both sides.
     
  14. Jan 9, 2015 #13
    OK. 0 to rT' evaluated at ri.

    Chet
     
  15. Jan 9, 2015 #14

    joshmccraney

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    But in this case we wouldn't have a function of ##r##. Would we instead just integrate from ##0,r## generally so we can still have a profile rather than a number?
     
  16. Jan 9, 2015 #15
    That's fine, but you seemed to be applying the equation over the entire volume. Integrating out to R is just fine.

    Chet
     
  17. Jan 9, 2015 #16

    joshmccraney

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    So if I'm understanding this correctly we would have $$-k\int_{0*T'(0)}^{r*T'(r)} d(r T') = Q\int_0^r r dr \implies \\ -k r T'(r) = Qr^2/2 \implies \\ -kT'(r) = Qr/2 \implies \\ -kT'(r) = \frac{R I^2}{2 \pi r_i^2 L r}$$ is this right so far?
     
    Last edited: Jan 9, 2015
  18. Jan 9, 2015 #17

    joshmccraney

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    But then if I made a flux balance we could write ##q = R I^2 / (2 \pi r L)## watts/sq. meter. Fourier's law implies also ##q = -k T'(r)## (1-D radial flow). Thus, $$-kT'(r) = \frac{R I^2} { 2 \pi r L}$$ which doesn't agree with the above. Can you help me with what I'm doing wrong?
     
  19. Jan 9, 2015 #18
    No. Check your algebra.
     
  20. Jan 9, 2015 #19

    joshmccraney

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    Sorry, we would have $$-kT'(r) = \frac{RI^2 r}{2 \pi r_i^2 L}$$ right? But this still doesn't agree with the flux balance.
     
  21. Jan 9, 2015 #20
    Who says? Multiply both sides by 2πrL and see what you get.

    Chet
     
  22. Jan 9, 2015 #21

    joshmccraney

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    I must be missing something. The flux balance states ##-kT'(r) = R I^2 / (2 \pi r L)## yet the heat eq method states ##-kT'(r) = R I^2 r / (2 \pi r_i^2 L)##. These two are different. I must have made a mistake but I'm not seeing it.

    Thanks so much for your help (and please continue)!
     
  23. Jan 9, 2015 #22

    joshmccraney

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    If the ##r_i^2## was simply ##r^2## then we would have agreeing equations.
     
  24. Jan 9, 2015 #23
    The first equation is correct only at r = ri. The second equation is correct at all radial locations.
     
  25. Jan 9, 2015 #24

    joshmccraney

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    Can you elaborate here. I'm wondering what I have done wrong in the flux balance. It really looked right to me.
     
  26. Jan 9, 2015 #25
    If the rate of heat generation within the wire is spatially uniform, what fraction of the heat is generated between r = 0 and arbitrary radial position r? What is the rate of heat generation within the wire per unit volume? What is the rate of heat generation between r = 0 and arbitrary radial position r?

    Chet
     
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