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Closed form solution heat problem

  1. Jul 1, 2012 #1
    The problem:

    Appreciate help on the following

    Hot water flows in an insulated copper pipe L long starting at temperature, T0
    Need the temperature history, T(t,x).
    T(0.x)=0
    T(t,0)=T0

    Heat transfer coefficient(conductance) water to pipe is U. Pipe heat capacity per unit length is C
    I wrote the PDE equations but can't seem to get a solution in closed form.

    Thank you
     
  2. jcsd
  3. Jul 2, 2012 #2
    What, no takers??

    Let me add the ODE equations I formulated :

    A*(dT/dx) +B*(dT/dt)+D*T=D*Tm
    C*(dTm/dt)+D*Tm=D*T

    A,B,D,C constants
    the d/dt and d/dx operators are partial derivative operators
    T= temperature of water
    Tm=temperature of pipe

    I eliminated Tm from the set and got a hypergeometric equation, but couldn't get a solution.

    Would appreciate some help here.
     
  4. Jul 2, 2012 #3

    Mute

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    Homework Helper

    Why are your equations for the temperatures only one-dimensional? The temperature may not vary with angle or along the length of the pipe, but the variation of temperature in the radial is not just d/dr. If you have a cylindrical pipe, the radial component of the laplacian is (1/r) d(r dT/dr), at least for the water, no?
     
  5. Jul 2, 2012 #4
    Thank you for responding

    In engineering it is common in these cases to assume no variation radially for the water inside the pipe since the flow is turbulent and no variation radially for the pipe owing to its thin wall and high conductivity.
     
  6. Jul 4, 2012 #5
    No takers,

    So I simplified and here is the new version. Forget the physics.

    A(d^2T/dt^2)+B(d^2T/dtdx)+C(dT/dx)+D(dT/dt)=0

    T(t,0)=T0
    T(0,x)=0
    Need
    T(t,x)
    0<x<L

    I did a numerical solution using excel, but I would prefer a closed form one.

    It has been a long time since I have studied PDE's,so maybe it is wishful thinking about
    closed form solutions.

    Your help would be much appreciated.
     
    Last edited: Jul 4, 2012
  7. Jul 5, 2012 #6

    Mute

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    Since the usual trick of letting [itex]T(t,x) = \theta(t)X(x)[/itex] won't work in this case due to the mixed partial derivative term, I would try Laplace transforming the equation in time. This should give you a first order ODE in x for the Laplace transformed function [itex]\hat{T}(s,x) = \mathcal L [T(t,x)](s)[/itex]. Solve the equation for [itex]\hat{T}(s,x)[/itex], then inverse Laplace transform.

    This might in practice be difficult. One thing you will have to figure out is the initial condition [itex]\partial T(t,x)/\partial t[/itex] at t = 0, as this will show up in the Laplace transform. You might also find it very difficult to invert the Laplace transform.
     
    Last edited: Jul 5, 2012
  8. Jul 7, 2012 #7
    Mute,

    Did what you suggested and came up with

    U(s,x)=1/s*e^[-x*(s^2+Ds)/(Bs+C)]

    No way can I invert this monster.

    Tried Wolfram-alpha for some values of x,D,B,C to no avail.

    Any further thoughts?

    Thanks again for you interest.
     
  9. Jul 7, 2012 #8

    Mute

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    Hm. If you took the length of the pipe to be infinity, [itex]L \rightarrow \infty[/itex], you could do another Laplace transform in x to get a simpler function, then inverse transform first in s, then try inverse transforming in x. However, given that U(s,x) looks rather intimidating, you might find that you won't be able to do the second inverse Laplace transform to get back to x.

    Is there a reason you suspect you should get a closed-form solution? You may not be able to write down any nice, closed form.

    You could perhaps write down T(t,x) as a series by expanding the exponential in U(s,x) and doing the inverse Laplace transform term-by-term (assuming that swapping the sum and the integral is valid), but this is likely to be pretty messy and won't be clean or simple unless you can find a formula for

    [tex]\frac{d^{n-1}}{ds^{n-1}} \left[ \left(s^2+Ds\right)^n\right],[/tex]

    which appears when calculating the contribution of the [itex]\frac{s^2+D}{Bs+C}[/itex] to the inverse laplace transform contour integral residue of the nth term of the series expansion. I suppose if D were zero the derivative is simple.

    You mention that you got a 'hypergeometric equation' when you eliminated Tm before? Do you mean a hypergeometric function? Why is this not an acceptable closed-form solution for you? Only some hypergeometric functions can be expressed in terms of standard functions.
     
  10. Jul 10, 2012 #9
    OK, took your advice and started to expand the fraction
    (s^2+D)/(Bs+C)=s/B+F/(s+G)

    The s/B term looks like the time delay of the function
    1/s*e^[-x*(s^2+Ds)/(Bs+C)]
    so I now have
    e^-sx/B*1/s*e^[-x*(F)/(s+G)]
    I should get the inverse Laplace of
    1/s*e^[-x*(F)/(s+G)]
    to get the f(t,x) the delay function.
    Is this correct??
    Physically, the delay makes some sense, since the delay term coincides with the time x/v, it takes for the fluid to reach x (where v is the fluid velocity).
    I will now expand the remaining exponent into the power series you suggested and see what happens.
    It now looks like a closed form solution is not possible, but maybe the expanded version will suffice if the terms converge rapidly.

    BTW, when I solved the 1st order homogeneous ODE in x , I took the liberty of making
    the coefficient 1/s meaning that the temperature T(t,0)=1 (actually it is T0). Is the math correct?

    Thanks again for your valuable input.
     
  11. Jul 13, 2012 #10
    "It now looks like a closed form solution is not possible, but maybe the expanded version will suffice if the terms converge rapidly."

    They did not so I'm back to square one, still stuck with getting the inverse transform of

    1/s*e^(sF/(s+G)

    F, G constants

    Any other ideas would be most appreciated.
     
  12. Jul 13, 2012 #11

    Mute

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    It might be that there's simply no analytic closed-form solution to the problem. What exactly do you want a closed solution for? Maybe there's an approximation scheme that would be useful for some parameter regime of interest?

    Again, you also mentioned that you found a solution in terms of a hypergeometric series. The hypergeometric functions [itex]_pF_q(a_1 \dots a_p; b_1 \dots b_q ; z)[/itex] are implemented on some software packages like Matlab or Mathematica, so if you have a solution in terms of those you can do some things with it. For certain p and q (especially p = 2, q = 1) some more specific things are known about the functions.

    The main problem with the inverse laplace transform is that you basically have an essential singularity in the exponential. That doesn't mean you can't necessarily do the integral, but it does make it much harder. In fact, you may want to take a look at your numerical solutions: how do they behave with time? Does the temperature increase with time or decrease? Remember that the Laplace transform is only valid if T(t,x) does not grow faster than exp(-st) in time.

    Another thing you could try is to go back to your original equations and plug in a fourier series for T(t,x):

    [tex]T(t,x) = \sum_{n=-\infty}^\infty c_n(t) \exp\left(-in\frac{x}{L}\right),[/tex]
    and similarly for [itex]T_m(t,x)[/itex], since the solution only has to be defined on x from 0 to L, right? That would give you an equation for [itex]c_n(t)[/itex] (coupled to the coefficient for the Tm series), which might be easier to solve?

    Of course, even if you solve for c_n(t), you might not be able to sum up the fourier series in closed form...

    By the way, you gave no initial data for Tm(t,x). What are its boundary conditions?
     
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