Kt dz Tzo Sieiut bT722

If pcp is constant, this boundary condition can be satisfied by a solution of the diffusion equation of the form

where A is a constant and D is the diffusivity kt/(pcp). The complex vertical wavenumber k has been determined by substitution of the exponential form of T' into the diffusion equation. A will be determined by substitution of the solution into the boundary condition, but before doing so it is worth pausing to make some remarks on the solution Eq. 7.23. This solution was first obtained by Fourier, in his study of diurnal and seasonal variations of temperatures in the interior of the Earth. Eq. 7.23 shows that the characteristic depth to which temperature fluctuations penetrate is •y/(D/w). Low frequency fluctuations penetrate to a greater depth than high frequency fluctuations, because heat has a longer time to diffuse before the surface temperature reverses. Note also that the phase lag of the time of maximum temperature with depth also reflects the time required for

pcp(JJm3)

Conductivity (Wm iK i)

Diffusivity (m2js)

Water Ice

1.93 •lO6

2.24

1.16 •1O-6

Fresh Snow

.21 •lO6

.O8

.38 •1O-6

Old Snow

1.0 •lO6

.42

.O5 •1O-6

Sandy Soil

1.28 •lO6

.3

.24 •1O-6

Clay Soil

1.42 •lO6

.25

.18 •1O-6

Peat Soil

.575 •lO6

.O6

.1 •lO-6

Rock

2.O2 •lO6

2.9

1.43 •1O-6

Lunar Regolith

1 •lO6

.O1

.O1 •1O-6

Table 7.1: Thermal properties of some common surface materials

Table 7.1: Thermal properties of some common surface materials the surface conditions to penetrate to the interior. For the diffusivity of water ice (Table 7.1) the characteristic depth is 12 cm for the diurnal period, 2.4m for the annual period, 24m for a century and 76m for a millennium. Solid rock yields similar numbers. Hence, the temperature profile within ice or rock still contains information about temperatures centuries or even millennia in the past, albeit in a rather smoothed and degraded form. This fact has been exploited in reconstructions of past temperatures.

Exercise 7.4.3 You are designing a lunar colony to be placed at a Lunar latitude where the sun is directly overhead at noon. The moon has an albedo close to zero, and the response time of the surface is rapid, so that the noontime surface temperature is close to the instantaneous equilibrium temperature of 394K (re-derive this temperature yourself). At night, the equilibrium temperature would be absolute zero, but there is not enough time to reach equilibrium; still the night-time temperature plummets to 100K. Since the Moon is tide-locked to the Earth, the Lunar day is 28 Earth days. The diffusivity of the Lunar regolith ("soil") is about 10-8m2/s.

Approximate the day-night temperature variation by a sinusoidal curve. What would be the constant temperature far below the surface (neglecting internal heat sources)? How deeply would the colony habitat have to be buried in order for the ambient diurnal temperature fluctuations to be less than 1K?

NB: Given the low diffusivity of the regolith, your main difficulty is likely to be getting rid of the heat generated by energy use (biological and otherwise) within the colony.

Now we substitute Eq. 7.23 into the boundary condition (7.22). The result is

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