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Viewing Single Post From: The giant space ship example
Chris Ho-Stuart
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Dec 8 2011, 12:55 AM

Or your number "8.2e4 seconds to drop one degree"
Which is a long time

About 23 hours.... but as I said, that was assuming we cool the entire atmosphere as a single unit, and I said that was unrealistic, and went on to consider what would happen considering the lowest layers only.

In the starship, when the fusion heaters shut down the atmosphere would cool from the bottom up, and the temperature drop in the lowest layers, where you would feel it, would occur before the rest of the atmosphere caught up. I guessed about 5 minutes to drop a degree. It's all in the rest of the post you are quoting.

And accordingly it doesn't matter how much atmosphere whether half or double
one still requires 6.58e15 Watts. for starship maintain 10 C.

So accordingly anything that adds heat capacity will not makes any difference.

Quite so. equilibrium temperature for a given power input is the same, no matter how much atmosphere, as long there are no greenhouse gases. No water, no CO2, nothing to interact with IR radiation.

This doesn't tell you the time it takes to heat up to equilibrium, or cool down from equilibrium. When you switch off the power, the rate at which it cools down depends very strongly on the heat capacity of the atmosphere.

We have to distinguish two totally different matters.

The first is -- what is the power input to maintain a certain stable temperature? Answer, for the starship with a N2 O2 atmosphere, 6.58e15 Watts. For this, heat capacity is irrelevant.

The second is -- what is the rate of cooling from a given temperature, if there's no heating supplied? The answer requires consideration of the heat capacity and thermal conductivity of the atmosphere.


I do have questions outstanding regarding laspe rate- why should it have any effect on lowering the 6.58e15 Watts requirement.

It has no effect on an IR transparent atmosphere.

With an atmospheric greenhouse effect, some emission to space comes from high in the atmosphere. The lapse rate is needed to find the temperatures at which the atmosphere emits thermal radiation.

And also when you turn off furnace where does most change in temperature occur. It seems since
bottom of atmosphere is warmest it would cool the fastest.

Yes, but that isn't actually the reason. If there's no greenhouse effect, then the only way the atmosphere can loose energy is through thermal contact with the surface. THAT is the primary reason why the bottom cools fastest in the non-greenhouse case. If you had an inversion in the atmosphere somehow (perhaps a lot of hot exhaust from failing reactors a week before they all shut down?) then the bottom of the atmosphere could end up colder than layers above, and yet it would still be the bottom cooling fastest when the fusion generators shut down, because air is such a good insulator.

The details of this really don't matter. We are now drifting a long way away from the original problem.

Other than above I think I can move on to effects from greenhouse gases.

Yes, please. I've been trying to clean up a few things on the board as well; sorry for the delay! I have actually started writing the next step in my solution to the original problem, but got distracted into all the other stuff about rates of cooling when reactors shut down... a different problem.

Cheers -- Chris

Hi John! Your post was moved to General Discussion. Hope that's okay. Thanks for the input!
Edited by Chris Ho-Stuart, Dec 8 2011, 01:07 PM.
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The giant space ship example · Physical theory for climate