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Heat equation, periodic heating of a surface

  1. Mar 5, 2015 #1
    1. The problem statement, all variables and given/known data
    The temperature variation at the surface is described by a Fourier
    series
    [tex]\theta(t)=\sum^\infty_{n=-\infty}\theta_n e^{2\pi i n t /T}[/tex]
    find an expression for the complex Fourier
    series of the temperature at depth [itex]d[/itex] below the surface

    2. Relevant equations
    Solution of the diffusion equation
    [tex]\theta(x,t)=\cos\left(\phi+\omega t-\sqrt{\dfrac{\omega}{2D}}x\right)\exp\left(-\sqrt{\dfrac{\omega}{2D}}x\right)[/tex]

    3. The attempt at a solution
    A the surface [itex]x=0[/itex] so
    [tex]\theta(0,t)=\cos\left(\omega t + \phi\right)[/tex]

    To find the coefficients [itex]\theta_n[/itex] I'm guessing I use the Fourier formula

    [tex]\theta_n=\frac{1}{T}\int_0^T dt \, \theta(0,t) \exp\left( -2\pi i n t/T \right) [/tex]
     
  2. jcsd
  3. Mar 6, 2015 #2

    mfb

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    I don't think you can calculate the ##\theta_n##. They are some given (but unknown) constants, and you have to modify your solution to fit this constraint.

    You can start with an easy case: imagine ##\theta_1=1## and ##\theta_n=0## for all other n. That makes the temperature at x=0 a sine. Can you find the temperature at depth d?
    What happens with ##\theta_2=3## and ##\theta_n=0## for all other n? What happens in the general case?
     
  4. Mar 6, 2015 #3
    Typo, [itex]\theta(x,t)[/itex] should be
    [itex]\theta(x,t)=A\cos\left(\phi+\omega t-\sqrt{\dfrac{\omega}{2D}}x\right)\exp\left(-\sqrt{\dfrac{\omega}{2D}}x\right)[/itex] (c)
    Advice from my tutor was
    'The temperature is a superposition of solutions found in (c).
    On the surface x=0...use this to determine the coefficients in the above and complete.'
    Assuming I have got (c) correct!
    James
     
  5. Mar 7, 2015 #4
    Appologies, [itex]A=7.5^{\circ}[/itex]C, [itex]\omega=\frac{\pi}{43200} s^{-1}[/itex], [itex]\phi=\frac{4}{3}[/itex] and [itex]D=5\times10^{-7} m^2/s[/itex]
     
  6. Mar 7, 2015 #5

    mfb

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    Where do those numbers come from? You cannot fix them like that for the problem.

    A superposition of those solutions is the right approach.
     
  7. Mar 7, 2015 #6
    These were given earlier in the question and you are right have no baring on this.

    So to get the coefficients I should set x=0 and use the Fourier formula to find them?
     
  8. Mar 7, 2015 #7

    mfb

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    Yes.
     
  9. Mar 7, 2015 #8
    I have done some calculations but getting trivial answers.
    I am taking
    [tex]\theta(0,t)=\cos\left(2\pi n \ t/T + \phi\right)[/tex]
    and converting it to its complex exponential form and inserting into
    [tex]\theta_n=\frac{1}{T}\int_0^T dt \, \theta(0,t) \exp\left( -2\pi i n t/T \right)[/tex]
    Should phi be included?
     
    Last edited: Mar 7, 2015
  10. Mar 7, 2015 #9

    pasmith

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    Homework Helper

    You probably want to use the solution of the diffusion equation in the form [tex]Ce^{i\omega t}\exp\left(-\sqrt{\frac{\omega}{2D}}x - i\sqrt{\frac{\omega}{2D}}x \right)[/tex] with [itex]x[/itex] measuring distance below the surface.
     
  11. Mar 7, 2015 #10

    mfb

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    You'll need phi if θn can be complex.

    I still don't see what "inserting into [long equation]" is supposed to mean, but I agree that you don't need long calculations.
     
  12. Mar 7, 2015 #11
    By inserting into, I mean to work out the coefficients.
    I'm a bit lost at the moment.
     
  13. Mar 7, 2015 #12

    pasmith

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    You should treat the coefficients [itex]\theta_n[/itex] as known. You are asked for "an expression for the complex Fourier
    series of the temperature at depth d below the surface" so you are looking for an expression of the form [tex]
    \theta(d,t) = \sum_{n=-\infty}^\infty e^{2n\pi it/T}f_n(d)[/tex] where [itex]e^{2n\pi it/T}f_n(x)[/itex] is a solution of the diffusion equation with [itex]f_n(0) = \theta_n[/itex].
     
  14. Mar 7, 2015 #13
    So would
    [tex]f_n(0)=\cos\left( 2\pi n t /T + \phi \right)[/tex]
     
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