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?