# Radius of ice melted by heat from a point.

1. Feb 4, 2012

### B.E.M

The motivation:
Hi, Im 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 occured 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)
So you practically construct your environment without any construction, and the energy does not cost you anything because it is waste heat. If you aren't creating waste heat you are not alive.

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 lets 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 [Broken]

(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).

Last edited by a moderator: May 5, 2017
2. Feb 4, 2012

Staff Emeritus
Sorry, that's not possible. The volume of ice that can be melted is equal to the amount of heat you put in, minus the heat that leaks out, divided by the heat needed to melt a unit volume of ice. If no heat leaks out, doubling the energy doubles the volume, which is a linear function, not a cubic function. Since the heat lost depends on the radius, it will actually be sublinear.

1 mW = 1/1000 W, which is less than the power of a CD laser. You're not going to be melting cubic meters of ice with that.

3. Feb 4, 2012

### B.E.M

(fixed typo 1mw to be 1 megawatt.)

Remember I am talking about power, not energy. Im talking about the equilibrium that would be reached eventually. Im not considering how long it takes to reach that equilibrium.

For example, if we were considering the 1-d version of this problem, r would probably be infinite, never reaching equilibrium. It is quite plausible for a constant power source to melt an infinite amount of ice given infinite time, because constant power times infinite time gives you infinite energy to play with.

Last edited: Feb 4, 2012
4. Feb 4, 2012

Staff Emeritus
Yes, and as you say, if you run the system for an infinite amount of time, you inject infinite energy into the system, so you can melt an infinite amount of ice. Put another way, if you don't allow Mars itself to heat up, your heat source will always be warmer than Mars itself and will continue to push heat into it - it will never reach equilibrium.

Figuring out how much Mars itself heats up is an intractable problem, as Mars itself is not in equilibrium.

So I don't think you are setting up a solvable problem. However, even in the oversimplified case, the scaling you got is wrong: it must be sublinear in volume.

5. Feb 4, 2012

### B.E.M

Hi again.

Im really aiming this question to someone willing to check my integration. :)

6. Feb 4, 2012

### 256bits

H does not remain constant across the whole shell..

2 explanations
1. If you are using an element of area dx dy which would be hard to analize, then
as the shells become larger as r increases from the centre the number of elements of the shell of area dx dy increases also. so more elements, more H per shell
2. Using sperical coordinates the element will have its area increase. then more per element mean more dH/element

I will try to find a site for the sphere, but in the meantime the equation comes out to something like this:

T - Tw = ( $\dot{q}$ / 6k ) ( r$^{2}$$_{o}$ -r$^{2}$ ) where
q dot = rate of heat generation per unit volume per unit time W/m^3
T is temperature at radius r
Tw = temperature at the wall of the shere
r o = outer radius of the shere

q dot times (volume of sphere) = q q is heat flux , watt/ sq meter

Last edited: Feb 4, 2012
7. Feb 4, 2012

### 256bits

this site has some explanation
http://www.cdeep.iitb.ac.in/nptel/Mechanical/Heat%20and%20Mass%20Transfer/Conduction/Module%202/main/2.6.3.html [Broken]

not exactly what I gace you before but it gives the analysis

Last edited by a moderator: May 5, 2017
8. Feb 4, 2012

### B.E.M

I don't think my analysis is wrong, but I will certainly look into it.
Im not integrating over the surface of a sphere but over a set of thin-walled spherical shells with width dr nested within each other. I believe this lets me assume:

(1) That (at equilibrium) the watts passing through each shell is equal and also equals the watts emitted by the heat source.
(2) That each shell with width dr, being so thin, can be approximated by the equation I referenced, that assumes inner and outer areas are the same, being in this case a simple function of r. Of course I don't expect this to be true of the sum of shells.

Ill take a closer look at that link now to see if it reduces to the same relationship.

9. Feb 4, 2012

### B.E.M

Ah.. rats. I think that site has a misprint.
In equation (2.51) it starts talking about cylinders and an area of 2∏rL where I guess L is the length of the cylinder. I think they have pasted in the formula for a cylinder instead of a sphere?

(edit).. still very helpful. Just knowing that I have to google something called "Steady state conduction" let me find another link on this very site:

(edit 2, rather than bumping my own trumpet :) )
Ok from the link above I found:
http://rpaulsingh.com/teaching/LecturesIFE/CondSphere/condsph.htm

which provided this result:
Rsphere = (r2-r1)/(4∏kr1r2)

If r2 grows much larger than r1 then the equation approaches

Rsphere = (1)/(4∏kr1)

and using
ΔT = Q*R,
where Q is the heat flow (watts)

then unless Im making some very simple mistake of algebra we do end up with my original result!

r1 = Q*(1/ΔT4∏k) = Q*(1/63*4*∏*2) = Q*0.00063

Each additional kilowatt increases r by 63cm!

Last edited by a moderator: May 5, 2017
10. Feb 4, 2012

### B.E.M

Ok, just thought of an improvement.

The above is an approximation. What would be better is a reliable under-estimate. Consider the example where you want to build your bubble 2km under the ice on mars, and you want a diameter of one 1km (i.e. radius of 500m).

A reliable under estimate would be to consider a ball of ice with radius 2km, where outer surface is locked to mars temperature and an inner radius of 500m within which the ice is water. This is certainly less insulated than the actual situation so the waste heat required for the actual situation would be at most this result (and certainly less in reality).

Rsphere = (r2-r1)/(4∏kr1r2)
where r1 = 500m, r2=2000m, k=2.0

Substitute this into
ΔT = Q*R
where ΔT is -63

I get Q=-1.05 megawatts of waste heat to maintain a 1km sphere of liquid water 2km under the ice of the martian poles.

Its not far off my original estimate. Of course it would have to fail if r approached 2km,

(Ive ignored some sign issues here, assuming they would be clarified if I chased down the exact definitions of terms)

(edit)

just wanted to add that 1 megawatt may sound large but it isn't. Thats waste heat, not electricity. These suggest under 3 tons could deliver that, and you could start even smaller:
Safe_Affordable_Fission_Engine

Last edited: Feb 5, 2012