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Depth at which thermal oscillations become negligible

  1. Dec 21, 2015 #1
    Hello everyone,

    I am working thru some of the mathematics of geo-exchange systems (semi passive heating and cooling systems for homes) and I'm starting with a very simple model: The ground is modeled as a perfectly insulated rod (perfectly insulated because of symmetry, there is no heat flux in the horizontal direction) and I want to know the depth at which annual oscillations in temperature in the ground become negligible.

    Assumptions:
    • h = 0 at the surface (only conductive cooling
    • homogeneous ground
    • surface temperatures oscillate on an annual cycle and can be modeled with a simple sine wave
    • thermal properties do not vary with temperature
    So I started with the heat equation:

    ##\frac{\partial T}{\partial t} = k\frac{\partial^2T}{\partial x^2}## where k is the thermal diffusivity of the ground ##\frac{conductivity}{specific-heat*density}##. It is about ##0.030\frac{m^2}{day}## (reference). Most solutions to this equation involve fixed boundary conditions, for example Paul's Notes does a wonderful tutorial on the differential equation. But his boundary conditions involve fixed temperatures, while mine are different:
    • T(x=0, t) = 12.5*sine(t) + 12.5 (Imaginary weather, 1 degree is approximately a day).
    • T(x -> inf , t) = 25. Here I'm making the assumption that thermal oscillations have stabilized at an infinite soil depth.
    So I don't know how to solve this differential equation with these boundary conditions and am reaching out for some guidance. Any kind of corrections or guidance would be sweet. One thing I have considered: I don't necessarily need an exact solution, I only want to know at what depth the temperatures get fairly stable, I want to know how deep I need to drill (how much $$ I need to spend) to make something like this work. So maybe I don't necessarily need a solution, just an understanding of the general behavior with respect to depth would be good.

    Also, I can use numerical methods but I don't want to because I'm sentimental and like analytical things.
     
  2. jcsd
  3. Dec 21, 2015 #2

    DrClaude

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    Staff: Mentor

  4. Dec 21, 2015 #3
    You're only interested in the time varying part of the solution, so forget about those constant values. Let T=A(x) sin ωt + B(x) cos ωt. You have enough information to solve for A(x) and B(x). Let A and B approach zero at large x, and let A and B match the boundary temperature variation at x = 0. The result should be an exponential decay with depth.

    Chet
     
  5. Jan 2, 2016 #4
    Sorry I've had the flu and have been out of commission for a bit.

    I pulled up "Soil Physics" by Hillel and he went thru his derivation of the problem, but I don't get how he obtained the "characteristic depth." He gives it as ##d =\sqrt{\frac{2D_h}{\omega}}## I mean the model looks perfect, I'm just not sure how he got that term for the exponential and I don't know why it belongs inside the sine function either.
     
  6. Jan 2, 2016 #5
    Can you please show the equation for the final result?

    Chet
     
  7. Jan 2, 2016 #6
    He starts out by approximating the temperature at the surface as a sine wave:

    ##T(t,0)=T_{ave}+A_0sin\omega t## [equation 12.26]

    He then says that the temperature at any depth can be represented by the function

    ##T(t,z)=T_{ave}+A_zsin[\omega t + \phi_z]##

    "In which A_z is the amplitude at depth z. Both A_z + phi_z are functions of z but not of t. They can be determined by substituting the solution of 12.26 in the differential equation ##\frac{\partial T}{\partial t} =k \frac{\partial^2 T}{\partial x^2}## this leads to the solution ##T(t,z)=T_{ave}+A_0sin[\omega t - z/d]/e^{-z/d}##"

    And d is the damping depth at which point the temperature amplitude equals 1/e or about 0.37

    I don't understand what he meant by "solution of 12.25", that looks like it already is a solution and I couldn't understand how he put that into the heat equation.
     
  8. Jan 2, 2016 #7
    Have you tried solving the problem yourself using the approach I recommended in post #3, with ##T(z,t)=T_{ave}+A(z) \sin ωt + B(z) \cos ωt##? This is equivalent to the form that he assumed in his analysis. Show me what you get if you substitute this into the PDE.

    Chet
     
  9. Jan 10, 2016 #8
    If I plug ##T_{ave}+A(z)sin(\omega t) + B(z)cos(\omega t)## into ##\frac{\partial T}{\partial t}=k\frac{\partial^2 T}{\partial x^2}##

    for the LHS, I get
    ##\frac{\partial T}{\partial t} = A(z)\omega cos(\omega t) - B(z)\omega sin(\omega t)##

    for the RHS, I get
    ##k\frac{\partial^2 T}{\partial x^2} = k(A(z)''sin(\omega t) +B(z)''cos(\omega t))##

    So I have:
    ##A(z)\omega cos(\omega t) - B(z)\omega sin(\omega t)=k(A(z)''sin(\omega t) +B(z)''cos(\omega t))##

    I'm not sure what kind of progress I've made here tho...
     
  10. Jan 10, 2016 #9
    You've done very well. The next step is to collect the terms multiplying cosωt and sinωt and re-express the equation in factored form.

    Chet
     
  11. Jan 11, 2016 #10
    ##cos(\omega t)(\omega A(z) - kB(z)'') - sin(\omega t)(\omega B(z) + kA(z)'') = 0##
     
  12. Jan 11, 2016 #11
    Good. Now, since cos and sin are orthogonal, their coefficients in this equation must individually be equal to zero in order for this equation to be satisfied at all values of t. This gives you two 2nd order ODEs in the two unknowns A and B. What are those two equations (you can now drop the z indication on A and B since we know that A and B are functions of z)?
     
    Last edited: Jan 11, 2016
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