- #1

B.E.M

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**The motivation:**

Hi, I am a bit of a space freak and I have been enthusiastic about this question for a while because there are lots of bodies in the solar system that are a few or many kilometers deep in ice. It occurred to me that if you landed your powerplant on this ice it could sink to some depth and surround it self with a bubble of water that provides all the basics for the support of life:

- Liquid water,
- temperature between boiling and freezing,
- pressure (under 30 meters of water on Mars you would get Earth sea level pressure)
- protection from cosmic rays,
- A temperature gradient that could possibly power biological activity.
- A way of getting rid of waste heat, including cooling you nuclear reactor (the major problem with exploiting nuclear power in space)

**The physics problem.**

The problem is this: given a source of heat deep under the ice of a Martian pole, what is the size of the body of water it could create? Using Mars let's us plug in some numbers, and ice at one of the martian poles approaches 4km deep.

I tried to do the math and got the surprising result that

__every kilowatt of waste heat increases the radius of liquid water by about 60cm__! .. ie the volume goes up by the

__cube__of the wattage. A power plant creating 1 megawatt of waste heat could maintain a ball of water 600 meters in radius.

However

__my math is very rusty__. Could someone check this for me? There are also some simplifying assumptions but we can pick those apart later.

(rather than follow my ascii equations you might find it easier to just follow the wiki link and read my general approach and start from there.)

**Here goes:**

http://en.wikipedia.org/wiki/Thermal...vity#Equations

(1) H = kA(ΔT/x)

where

H is the heat flow in watts,

k is the thermal conductivity, for ice roughly 2 (W/(mK))

A is the total cross sectional area of conducting surface,

ΔT is temperature difference,

x is the thickness of conducting surface separating the two temperatures.

Picture heat in watts radiating from a single point and passing though consecutive thin spherical shells of ice. At equilibrium there is some radius R at which the temperature is that of frozen ice, 0 Celsius, the watts passing through each shell is equal, and the sum of all ΔT across all shells (from r=R to r=infinity) adds up to total change of temperature from 0 Celsius to the Mars ambient temperature which I am taking as -63 Celsius. Note my simple model ignores what is happening within that radius. It just assumes it is all liquid water swirling around in such a way as to deliver heat equally to the surface at r=R.

Rearranging the equation (1) above I get:

ΔT/x = H/k.A

A bit hazy over this step, but I think this means for a thin spherical shell

dT/dr = H/k.A(r)

where A(r) = 4.pi.r^2 is the surface area of the slice at r. so..

(3) dT/dr = H/(k.4.pi.r^2)

We need to integrate (3) from r=R to r=infinity, which is pretty easy because everything but (1/r^2) can be taken outside as a constant, which integrated = (-1/r)

T(r) = (H/k.4.pi)(-1/r) +Tc

So the total temperature drop from R (where ice freezes) to infinity (where we have Mars average temperature) is

T(infinity) - T(R) = C

then solving for R gives:

(4) R = H/(k.4.pi.C)

Substituting in some values:

C = -63 degrees, ie the drop from temperature at which ices freezes to Mars ambient temperature)

k = 2 (W/(mK)) ie the thermal conductivity of ice.

(note: I think I have a sign error, or perhaps H is negative since it is flowing out?)

Anyway finally I get R = H*0.00063 (m/watt) or

(5) R = H*0.63 (m/kilowatt)

where

R is the radius outside which ice remains frozen.

H is the heat flowing out (in kilowatts).

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