Esther Lee-Varisco Matt Zhang

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1 Esther Lee-Varisco Matt Zhang

2 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

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

5 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

6 Half-infinite space Constant thermal properties: k = thermal conductivity (W/mk) κ = thermal diffusivity (m 2 /s) Based on type of medium the cellar will be placed Periodic temperature variations Diurnal Annual Glacial

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8 Heat flux equation : qs (t)= k dt dz Heating of the earth: l 1 dt d dt = 2 T dz 2 k l = tc

9 Assume T s is a periodic function of time: T s= To + DT cos ~t ~ = 2rf Separate variables: T (z, t) = To + Z (z) T l (t) assume T à T o as z à

10 Temperature changes lag with depth so Need to use sinωt T l (t)! cos ~t T (z, t) = To + Z1 (z) cos ~t + Z2 (z) sin ~t and substitute into l 1 dt d dt = 2 T dz 2

11 -~ sin ~tz1 (z) + ~ cos ~tz 2(z) = l (cos ~t d 2 Z 1 (z) d 2 Z 2 (z) dz + sin ~t 2 dz ) 2 ~ cos ~tz 2(z) - l cos ~t d 2 Z 1 (z) dz 2 = ~ sin ~tz 1(z)+sin~t d 2 Z 2 (z) dz 2

12 ~ d 2 Z 1 (z) ~ ( l Z 2(z) - dz 2 ) cos ~t = ( l Z1 (z) + Z 1 and Z 2 are independent of time d 2 Z 2 (z) dz ) sin ~t 2 ~Z1 (z) + l d 2 Z 2 (z) dz = 0 2 => ~Z 2(z) - l d 2 Z 1 (z) dz = 0 2 d 4 Z 2 (z) ~ 2 dz + 4 l 2 Z 2(z) = 0 d 4 Z 1 (z) ~ dz l 2 Z 1(z) = 0

13 Assume solution to ODE is Z 2=ce az then we get Let α 2 =β to get ~ a l 2 = 0 b 2 - (i l ~ ) 2 = 0 => b =!i l ~ = a 2 a 2! i l ~ = 0!i = ( where 1! i ) 2 2

14 Thus, and a 2 - ( a =!( 1! i ) 2 2 1! i ~ ) 2 l ~ l = 0 The general solution for Z 2 has four values that will satisfy α.

15 Z 2 = c 1e ( 1 + i ) 2 l ~ z) + c 2e ( 1 - i ) 2 l ~ z) +c 3e ( -(1 + i) ) 2 l ~ z) + c 4 e ( -(1 - i) ) 2 l ~ z) Temperature fluctuations must decay with depth so constants c 1 and c 2 are zero (T à T o as z à ). Z 2 = e (- z ~ 2l ) [c3e (- iz ~ 2l ) + c 4 e (iz ~ 2l ) ]

16 This solution can be rewritten to replace e iz and e -iz with sin z and cos z, where b 1 =c 3 +c 4 and b 2 =c 4 -c 3 Z 2 = e (- z 2l ~ ) (b 1 cos 2l ~ z + b 2 sin 2l ~ z) And Z 1 can also be rewritten and where b 2 =b 3 and b 1 = -b 4 to satisfy our previous equations for Z 1 and Z 2 Z 1 = e (- z 2l ~ ) (b 3 cos 2l ~ z + b 4 sin 2l ~ z)

17 The surface temperature must fit our initial equation T s= To + DT cos ~t This leads to b 1 = -b 4 =0 and b 2 =b 3 =ΔT Thus, T = To + DTe (- z 2l ~ ) [cos 2l ~ z cos ~t +sin 2l ~ z sin ~t] = + DTe To (- z ~ 2l ) cos (~t - z ~ 2l )

18 Attenuation depth, also called skin depth, is given by d ~ = ( ~ 2l ) 1/2 The temperature fluctuations decreases exponentially as depth increases at a value approximate to 1 e (T s - T o) And the phase difference between fluctuations at T s and T at d ω is given by z = z 2l ~

19 Given both d ~ = ( ~ 2l ) 1/2 ~ = 2rf we can find the diurnal and annual cycle frequencies: 2l d ~ = ( 2rf ) 1/2 l d ~ = ( rf ) 1/2 d ~ = ( r lx ) 1/2 x = f 1

20 Diurnal: τ = 1 day = s f = 1.157x10-5 1/s ω = 7.272x10-5 rad/s Annual: τ = 365 days = 3.153x10 7 s f = 3.171x10-8 1/s ω = 1.992x10-7 rad/s Glacial (~110ka): ω = 1.811x10-12 rad/s

21 Transferring results from temperature fluctuations into heat flow equation at the surface q (0, t) =-k dt dz ~ ~ =-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)]

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

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24 We get the final T(z,t) equation to be: Dq T (z, t) = T o + e (- z ~l ~ 2l ) r cos (~t z 2l ~ ) T o = mean temperature Dq ~l e (- z ~ ) = attenuation depth 2l = temperature variation z = z 2l ~ = phase shift

25 d ω =0.05m T 0 = 10 C, ΔT = 7 C Clay soil: κ = 1.0x10-7 m 2 /s

26 T 0 = 10 C, ΔT = 7 C Clay soil: κ = 1.0x10-7 m 2 /s d ω =1m

27 T 0 = 10 C, ΔT = 7 C Clay soil: κ = 1.0x10-7 m 2 /s d ω =317.4m

28 1. the variance decreases as the depth goes further. 2. there is a time shift according to the depth T = To + DTe (- z 2l ~ ) cos (~t - z 2l ~ )

29 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)

30 Background: annual temperature change: San Diego: ΔT= 3.45 C New York: ΔT= C Take the sandstone as an example: κ: 1.15x10-6 m 2 /s

31 At the attenuation depth: d ω = m 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.

32 This? Or this? No bias here

33 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

34 Questions?