Esther Lee-Varisco Matt Zhang

� Want to build a wine cellar

� Surface temperature varies daily, seasonally, and geologically

� Need reasonable depth to build the cellar for lessened temperature variations

� Building a cellar in San Diego or New York

� At some point, can afford to be a sophisticated wine connoisseur.

� Aging wines need a constant 55°F (12.8°C) in order to mature perfectly

� Changes in temperature cause wines to expand and contract within the bottle, which draws air in through the cork and causes oxidation

� Attenuation depth gives us the depth at which the temperature fluctuation is lessened by a 1/e of that at the surface

�  Half-infinite space

�  Constant thermal properties: �  k = thermal conductivity (W/mk)

�  κ = thermal diffusivity (m2/s)

�  Based on type of medium the cellar will be placed

�  Periodic temperature variations �  Diurnal

�  Annual

�  Glacial

� Heat flux equation :

� Heating of the earth:

qs (t)= k dzdT

l1dtdT= dz 2d 2T

l = tck

� Assume Ts is a periodic function of time:

� Separate variables:

assume T à To as z à ∞

T s= To + DT cos~t

~ = 2rf

T (z, t) = To + Z (z) lT (t)

�  Temperature changes lag with depth so

� Need to use sinωt

and substitute into

lT (t)! cos~t

T (z, t) = To + Z1 (z) cos~t + Z2 (z) sin~t

l1dtdT= dz 2d 2T

-~ sin~tZ1 (z) + ~ cos~tZ 2(z)

= l (cos~t dz 2d 2Z1 (z)

+ sin~t dz 2d 2Z2 (z)

)

~ cos~tZ 2(z) - l cos~t dz 2d 2Z1 (z)

= ~ sin~tZ 1(z)+sin~t dz 2d 2Z2 (z)

Z1 and Z2 are independent of time

=>

( l~

Z2(z) - dz 2d 2Z1 (z)

) cos~t= ( l~

Z1 (z) + dz 2d 2Z2 (z)

) sin~t

~Z1 (z) + l dz 2d 2Z2 (z)

= 0

~Z 2(z) - l dz 2d 2Z1 (z)

= 0

dz 4d 4Z2 (z)

+ l2~ 2

Z2(z) = 0

dz 4d 4Z1 (z)

+ l2~ 2

Z1(z) = 0

� Assume solution to ODE is

then we get

�  Let α2=β to get

=>

where

Z2=ceaz

a4 + l2~ 2= 0

b 2 - (i l~) 2 = 0

b =!i l~= a2

a2 ! i l~= 0 !i = (

21 ! i

) 2

�  Thus,

and

�  The general solution for Z2 has four values that will satisfy α.

a2 - (2

1 ! i) 2 l~= 0

a =!(21 ! i

) l~

Temperature fluctuations must decay with depth so constants c1 and c2 are zero (T à To as z à ∞).

Z2 = c1e ( 21+ i

) l~z) + c2e ( 2

1- i) l~z)

+c3e ( 2

-(1+ i)) l~z) + c4 e ( 2

-(1- i)) l~z)

Z2 = e (- z 2l~) [c3e (- iz 2l

~) + c4 e (iz 2l

~)]

�  This solution can be rewritten to replace eiz and e-iz with sin z and cos z, where b1=c3+c4 and b2=c4-c3

� And Z1 can also be rewritten and where b2=b3 and b1= -b4 to satisfy our previous equations for Z1 and Z2

Z2 = e (- z 2l~) (b 1 cos 2l

~z + b2 sin 2l

~z)

Z1 = e (- z 2l~) (b 3 cos 2l

~z + b4 sin 2l

~z)

�  The surface temperature must fit our initial equation

�  This leads to b1= -b4=0 and b2=b3=ΔT

�  Thus,

T s= To + DT cos~t

T = To + DTe (- z 2l~) [cos 2l

~z cos~t+sin 2l

~z sin~t]

= To + DTe (- z 2l~) cos (~t - z 2l

~)

� Attenuation depth, also called skin depth, is given by

�  The temperature fluctuations decreases exponentially as depth increases at a value approximate to

� And the phase difference between fluctuations at Ts and T at dω is given by

d~ = ( ~2l) 1/2

e1(T s - T o)

z = z 2l~

� Given both

we can find the diurnal and annual cycle frequencies:

d~ = ( ~2l) 1/2 ~ = 2rf

d~ = ( 2rf2l) 1/2

d~ = (rfl) 1/2

d~ = ( rlx) 1/2 x = f

1

�  Diurnal: �  τ = 1 day = 86400 s �  f = 1.157x10-5 1/s �  ω = 7.272x10-5 rad/s

�  Annual: �  τ = 365 days = 3.153x107 s �  f = 3.171x10-8 1/s �  ω = 1.992x10-7 rad/s

�  Glacial (~110ka): �  ω = 1.811x10-12 rad/s

�  Transferring results from temperature fluctuations into heat flow equation at the surface

q (0, t) =-k dzdT

=-kDT [- 2l~

e (- z 2l~) cos (~t - z 2l

~)

+e (- z 2l~) sin (~t - z 2l

~) 2l~]

q (0, t) =-kDT [ 2l~(sin~t - cos~t)]

q (0, t) =-DT ~l (sin~t sin 4r- cos~t cos 4

r)

= Dq cos (~t + 4r)

q (0, t) = DT ~l cos (~t + 4r)

� We get the final T(z,t) equation to be:

T (z, t) = T o +~l

Dqe (- z 2l

~) cos (~t - 4

r- z 2l

~)

T o = mean temperature

e (- z 2l~) = attenuation depth

z = z 2l~= phase shift

~l

Dq= temperature variation

T0 = 10°C, ΔT = 7°C

dω=0.05m

Clay soil: κ = 1.0x10-7 m2/s

T0 = 10°C, ΔT = 7°C

Clay soil: κ = 1.0x10-7 m2/s

dω=1m

T0 = 10°C, ΔT = 7°C

Clay soil: κ = 1.0x10-7 m2/s

dω=317.4m

1. the variance decreases as the depth goes further. 2. there is a time shift according to the depth

= To + DTe (- z 2l~) cos (~t - z 2l

~)T

When we exert a heat flux on the surface, there is a time shift (π/4) between the heat influx and the temperature variance.

q(0,t) = Δq*cos(ωt)

Background: annual temperature change:

San Diego: ΔT= 3.45°C

New York: ΔT= 12.25°C

Take the sandstone as an example: κ: 1.15x10-6 m2/s

At the attenuation depth: dω = -1.075m

Temperature variation in these two place:

San Diego: ΔT =1.27 °C

New York: ΔT= 4.50 °C

However, if you also want a cellar in New York with a ΔT of 1.27 °C,

You must dig as deep as 2.45m.

�  This?

� Or this?

No bias here…

�  Either location will work but �  Digging depth for New York = 2.5m

�  Digging depth for San Diego = 1.1m

� Digging in San Diego is more cost efficient

Questions?