1. The Wine Cellar Problem Geophysics’ most important contribution to the human race.

2. The Situation x z ?

3. Questions 1.What is the temperature anomaly as function of time, depth and the Fourier transform of q s (t)? 2.What constants determine the attenuation depth of the temperature anomaly? 3.What is the attenuation depth of the periodic temperature variations due to the –Diurnal cycle? –Annual cycle? –Glacial cycle?

4. Assumptions 1.The ground is a semi-infinite homogenous half- space … so we use the 1-D, time dependent heat conducting equation 2.Constant thermal properties ( , k) 3.As z — > infinity the temperature T(z,t) — > T o, where T o is the average surface temperature … which means we ’ re ignoring heat flux from the mantle, and we have no internal heat sources … which essentially means the ground in question is an isolated body

5. Deriving the Temperature Anomaly If q s (t) is a periodic forcing function we can assume it is of the form:. So the differential equation at the surface becomes: Because the heat flux is periodic and the PDE is linear we can guess the solution has the form:

6. Deriving the Temperature Anomaly Substituting T(z,t) into the diffusion equation we get: Which reduces to a 2nd order linear ODE: Which has the well known general solution:

7. Still Deriving… Because we’re interested in the exponential decay with increasing depth, we let a = 0, then select the second term and plug f(z) back into T(z,t) to get:

8. Still Deriving… And after separating out the oscillatory part:

9. But what about A? Apply the boundary condition at the surface: If we sub T(z,t) into this bad boy we get:

10. And so our super final answer is…

11. Finally, compare T(z,t) with q(s) There is a difference of  /4 between the oscillatory parts of these two functions: See the extra  /4? …meaning that the temperature anomaly at any given depth will lag behind the surface fluctuation by 1/8 of the period of the fluctuation.

12. Attenuation Depth The depth at which the temperature has negligible fluctuation w.r.t. the surface temp. In other words: where do we put our cellar?? where Equate this to the temperature function

13. Attenuation Depth… …and solve for z! z o is only dependent on  and  So re-write the temperature function…

14. So now what? We want to know how the attenuation depth will vary with time and soil conditions …so we chose three time scales to examine  = 2  f –Diurnal:  = 7.27x10-5 rad/sec –Annual:  = 1.99x10-7 rad/sec –Glacial:  = 1.99x10-12 rad/sec

15. …and we chose three soil conditions to consider: Clay Soil, Sandy Soil, & Rock Clay SoilSandy SoilRock k (W/m 2 /k) 0.250.302.90   x10 -6 m 2 /s) 0.180.241.43

16. Diurnal Cycle Tiny attenuation depths! Clay SoilSandy SoilRock z o (meters) 0.070.080.20

17. Diurnal Cycle

18. Glacial Cycle Huge attenuation depths! Clay SoilSandy SoilRock z o (meters) 4254911199

19. Annual Cycle Practical attenuation depths! Clay SoilSandy SoilRock z o (meters) 1.351.553.79

20. Annual Cycle We selected a wine-bearing region with substantial temperature fluctuations: Canandaguia, New York NEW YORK CITY?! Get a rope. Annual  T = 18 kelvin We’re assuming that the average surface temperature, T o, is the optimum temperature for storing wine: 55ºF.

21. Canandaguia: Clay Soil

22. Canandaguia: Sandy Soil

23. Canandaguia: Rock

24. Cheers!