The Wine Cellar ProblemGeophysics’ most important contribution to

the human race.

The Situation

x

z ?

Questions1. What is the temperature anomaly as function of

time, depth and the Fourier transform of qs(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 the1. Diurnal cycle?2. Annual cycle?3. Glacial cycle?

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) —> To,

where To 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

Deriving the Temperature Anomaly

• If qs(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:

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:

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:

Still Deriving…

• And after separating out the oscillatory part:

But what about A?• Apply the boundary condition at the surface:

• If we sub T(z,t) into this bad boy we get:

And so our super final answer is…

Finally, compare T(z,t) with q(s)

There is a difference of /4 between the oscillatory parts of these two functions:

k Tz

Foe it 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.

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

Attenuation Depth…

• …and solve for z!• zo is only dependent

on and • So re-write the temperature function…

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

= 2f

– Diurnal: = 7.27x10-5 rad/sec– Annual: = 1.99x10-7 rad/sec– Glacial: = 1.99x10-12 rad/sec

…and we chose three soil conditions to consider: Clay Soil, Sandy Soil, & Rock

Clay Soil Sandy Soil Rock

k(W/m2/k)

0.25 0.30 2.90

x10-6 m2/s)

0.18 0.24 1.43

Diurnal Cycle

• Tiny attenuation depths!

Clay Soil Sandy Soil Rock

zo

(meters)0.07 0.08 0.20

Diurnal Cycle

0 5 10 15 20 250

2

4

6

8

10

12

Sur

face

Tem

pera

ture

Flu

ctua

tion

(kel

vin)

Time (hour)

Forcing Temperature at Surface for Diurnal Fluctuations

Glacial Cycle

• Huge attenuation depths!

Clay Soil Sandy Soil Rock

zo

(meters)425 491 1199

Annual Cycle

• Practical attenuation depths!

Clay Soil Sandy Soil Rock

zo

(meters)1.35 1.55 3.79

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, To, is the optimum temperature for storing wine: 55ºF.

Canandaguia: Clay Soil

-4 -3 -2 -1 0 1 2 3 4-12

-10

-8

-6

-4

-2

0

Temperature Fluctuation, dT (kelvin)

Dep

th (M

eter

s)Canandaguia Temperature as a Function of Depth and Time: Clay Soil

Canandaguia: Sandy Soil

-4 -3 -2 -1 0 1 2 3 4-12

-10

-8

-6

-4

-2

0Canandaguia Temperature as a Function of Depth and Time: Sandy Soil

Temperature Fluctuation, dT (kelvin)

Dep

th (M

eter

s)

Canandaguia: Rock

-4 -3 -2 -1 0 1 2 3 4-40

-35

-30

-25

-20

-15

-10

-5

0Canandaguia Temperature as a Function of Depth and Time: Rock

Temperature Fluctuation, dT (kelvin)

Dep

th (M

eter

s)

Cheers!