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Ice melted by heat radiating from a point (on mars)



 
 
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Old January 1st 12, 05:59 PM
KelvinZero KelvinZero is offline
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Default Ice melted by heat radiating from a point (on mars)

(Sorry, am no longer visiting here so no point replying here. Ignore this post.)



Hi, first post on this forum (and I barely know what usenet is )
This is actually a repost of a question I asked on NasaSpaceFlight.com but I didn't manage to get any math help there. I noticed some willingness to use maths to answer newbie questions here. The effort is always appreciated!

The problem is this: given a source of heat deep under the surface of mars, what is the size of the body of water it would create?
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 1mw 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.

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 or
(5) R = H*0.63
where
R is the radius outside which ice remains frozen.
H is the heat flowing out (in kilowatts).

Last edited by KelvinZero : February 4th 12 at 01:24 PM. Reason: Added warning at top. Replies wont be received.
  #2  
Old March 12th 12, 10:38 AM
mickrio mickrio is offline
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Yep, I come from the old school and this is what I learned. I get ten points. Keep the points, I just wanted you to know how intelligent I cannot be.
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