# Depth at which thermal oscillations become negligible

1. Dec 21, 2015

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. Dec 21, 2015

### Staff: Mentor

3. Dec 21, 2015

### Staff: Mentor

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

4. Jan 2, 2016

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.

5. Jan 2, 2016

### Staff: Mentor

Can you please show the equation for the final result?

Chet

6. Jan 2, 2016

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.

7. Jan 2, 2016

### Staff: Mentor

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

8. Jan 10, 2016

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...

9. Jan 10, 2016

### Staff: Mentor

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

10. Jan 11, 2016

$cos(\omega t)(\omega A(z) - kB(z)'') - sin(\omega t)(\omega B(z) + kA(z)'') = 0$