[X] My Assistant

[tanjent]

There is a recent posting on Emily's Planetary Society blog, which must be Doug's because she's not there herself, although her name is the only name on it. The subject is using water to cool a long-lived surface probe on Venus. It sounds far more practical than any of the other proposals for landing giant atomic-powered refrigerators, or developing a whole new family of high-temperature semiconductors, etc.

But I didn't understand the whispered criticism to the effect that the Ekonomov paper assumed that the water would absorb heat only from the one watt of power driving the instrument package itself. I simply can't believe that he went to the podium and presented his model without taking into account the fact that the surface of Venus is a pretty hot place, and that the proposed probe would be absorbing the ambient heat. This is an interesting proposal and I would like to understand both the original calculation of 50 days to bring the water to a boil, and the cited flaw in the calculation. I too find it hard to believe that it would take 50 days to bring water to a boil on the Venusian surface, but where exactly is the error, and what remains after we correct it?

Doug is busy of course, but I hope he will find the time to address this when he returns, if someone else hasn't done so by then.

[Greg Hullender]

Tasp: the idea was that the water is INSIDE a glorified thermos bottle, so it isn't exposed to 900F at all. Anyway, that's way over the critical temperature of water, so no amount of pressure could keep it liquid.

I'd say the first question to answer would be "given the best thermos we can make, with the atmosphere of Venus on the outside and holding 100 Kg of water initially at 4C and 1 atm on the inside, how will the water temperature vary as a function of time?" If the answer is, "it'll boil in hours," then there's nothing more to discuss. But if the answer is "it won't reach 100C for months, and even then it'll take two years for it to all boil away," then maybe there is something to it.

Anyone know enough about modern thermos technology to estimate this?

--Greg

[marsbug]

" how will the water temperature vary as a function of time?" If the answer is, "it'll boil in hours," then there's nothing more to discuss. But if the answer is "it won't reach 100C for months, and even then it'll take two years for it to all boil away," then maybe there is something to it.
--Greg

I usually try to avoid numbers but this intruiged me so heres a quick and dirty attempt:
Assume the thermal isolation system consists of a double walled shell with a good vacuum in between (say 10^-4 mbar) so that convection and conduction between the walls is close to zero, and that the shell is constructed of a low vapour pressure material such as tungsten, or is being actively pumped to keep it at that level. Assume also that connections between the walls are minimised (could fairly low tech heat tolerant machinery outside the shell be operated entirely by remote by an electronics package inside?) to ten square centimetres of cross sectional area, made of material with a low thermal conduction , say equal to aerogel at 0.03 w/m/k. As a random figure lets say the separation between walls is 10 cm, and the water starts at a temperature of 273 degrees Kelvin. The amount of heat transmitted through the connections would then be:


0.03= q/t * (L/(A*ΛT)
wher q/t is joules per secound, L is the length of the conducting connection, A is the total csa of the connections and ΛT is the temperature difference in kelvin.

So:
q/t= 0.03/ (0.1/(0.01*400)= 1.2 watts. Which looks ok.
BUT:
The major source of heat into the inner vessel will be radiation from the outer wall. Assuming that the outer wall reaches the same temperature as the venusian atmosphere fairly quickly, and that it can be approximated as a black body it will be radiating heat onto the inner wall at
W= σ*A*T^4
Where W is heat transfer in watts, σ is the Stefan-Boltzmann constant, A is the internal surface area of a sphere with an internal radius of 40 centimeters and T is the temperature of the outer wall in Kelvin (673 deg):
A= 4πr^2 = 4*π*0.4*0.4 = 2.01 meters square (ish)
W= σ*2.01*673^4= 23797.24 watts falling onto the inner shell? Or roughly 24 Kj per secound.
So water has specific heat capacity of 4.2 kJ.kg^-1.K^-1, so for 100 kg to reach 373 Kelvin from just above freezing requires 4.2kj*100*100= 42000 Kj. Assuming it heats linearly (which I know it doesn’t but can’t remember how to work it out properly) 42000/24 = 1750 seconds, or just under half an hour.
That doesn’t look very promising! Where did I go wrong anybody (bet its something very simple!)?. But it does assume one atmosphere of pressure, not ninety! Any chance I could look at how you worked it out for that figure doug?

Edit: Even if it works at ninety atmospheres I'd be impressed by a container that would hold a good vacuum for fifty days under those conditions without an active pump. I'd also be impressed by an active pump that would run for fifty days under venus surface conditions. Not saying it can't be done, just that I'd be impressed!

[djellison]

Any chance I could look at how you worked it out for that figure doug?

I didn't work out anything - it's all the presenters work. I'll drag up the abstract PDF later

Doug

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