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The giant space ship example; Can you calculate surface temperatures given a moon sized interstellar space ship with its own atmosphere?  

Tweet Topic Started: Dec 4 2011, 03:29 AM (1,492 Views)  
gbaikie  Dec 7 2011, 06:44 AM Post #16 

I have been thinking about this thing today and I think got an answer. Actually I had the idea a couple days ago, but today I got "why" it was the answer. It was when asked you how much energy it would take to heat dwarf moon to 150 K, because I wanted to know how quickly it heated up. And as I said it's a different way to check it. And it is. You said:
To repeat "the final temperature of anything is the the temperature at which it is no longer gaining thermal energy at all." Now with large thermal capacity the final temperature [if precise as within a 1/10th or say .0001 of a degree] takes a long, time. This idea is commonly known, but maybe not appreciated enough. Today I was asking myself the question: If you turn off the furnaces, how long would take to cool the atmosphere. You seem to think it would happening quickly, I don't. To quantify this, I think it might drop in temperature fairly quickly by a couple of degrees. And by quickly I mean in a minute or two of time. As example in solar ellipse I have been informed [never been to one] the temperature does drop rapidly. So something like 10 degrees in couple minutes. Enough that everyone notices it. Apparently. We also have temperatures dropping by 100 K during couple hours on the Moon during a Lunar ellipse. And we have every day a drop of about 20 C during nite every day/nite cycle on earth. On earth we have lots of records showing daytime high and nighttime low this amount change is not uniform. Ocean tropics has little variation. Deserts have have variation. I would say someone who never seen earth, if they had the high and lows, humidity, and maybe barometer reading globally, could determine the output of our sun. If you gave them earth's distance from the sun, it would make it easier. Or quickly a planet warms and cools gives you information. Of course knowing the sun's output also gives information same type of information. Or in case of our dwarf moon, the amount energy required to heat the moon by 1 degree over time tells you stuff. Or turning off the furnace is exact same data. Now I don't know the formula but it will be exponential. By which I mean, you have your 6,580,000 GigaWatts furnaces. You start will air temperature of 150 K, it will quickly heat the air 1 K per day, but as air temperature increases it will take longer and longer to warm the air. But it's like halving the distance [sort of] it will take 1000 years to get very close to the final temperature if what is meant by final temperature is very precise [such as .00001]. Or for practical final temperatures of say 1/10 of degree, maybe months or years. Now, if have furnace going for months and it close to final temperature, and you turn off the furnace for 2 seconds, it may take a day or week to regain it's the lost ground. So Earth can never get close to this kind of final temperature. One could turn off the furnace every 12 hours and turn it back on for 12 hours and one could do this for years and reach very near it's type of final temperature. That is very similar to having furnace with half the power running constantly. hm. That is what climate guys are doing they treating the sun as though it's half powered furnace. Which would work if figured out how much heat was absorbed [how much the sun increased the thermal capacity which same as it loses each nite]. So, question is how much is air temperature at surface loses per second the furnace is shut off. If you were at final temperature the loss would be the joules produced by furnace per second minus total heat capacity. Let's compare dwarf to earth. Dwarf 10 C. Earth desert 10 C. Earth has higher lapse rate and per sq meter has far less air. Night time earth desert [being 10 C]. Dwarf of course always night time. Desert might take an hour to lower 1 C. Dwarf should take longer than 1 hr to drop 1 C. If dwarf instead had 1/2 as much atmosphere as earth, than the temperature could lower at faster rate than earth. As I said above if dwarf loses some temperature from final temperature it would require a lot of time to come back to the final temperature. But if guess it takes 1 hr to lose 1 K, I can divide the heat capacity of atmosphere of 1 K and divide by 3600 and that how thermal power is needed. And in addition as it cools the atmosphere should drop a bit, this lowering of atmosphere should add heat. Oh but earth has more gravity and therefore would gain more energy {I thought dwarf would get more energy from this than earth}. Anyhow, as posted above: "So 5.4e17 KJ per K Or 5.4e20 watts/joules And have 6,580,000 GigaWatts Which is 6.58 x 10^15 watts So about 100,000 seconds and one day is 86400 seconds So 1 or two days per 1 K increase" So if cooled at 1 K per hour, it would require 30 times more energy than you said. So what you are actually saying is it will take about 30 hrs to lower 1 K if 6,580,000 GigaWatts is correct. So kinda proves you right, and here I thought you were wrong Though might need more reactor than 6,580,000 GigaWatts because 30 hour to lower 1 K seems kinda wildly optimist:) So what I am getting is ground surface temperature doesn't matter. Amount of atmosphere doesn't matter. In terms reducing power requirements. A real thermally insulated greenhouse would help :). Edited by gbaikie, Dec 7 2011, 06:54 AM.

gbaikie  Dec 7 2011, 07:58 PM Post #17 

Come think of it. It may far more than optimistic I can't believe it would only lower by 1 K in 30 hours. And 1/10 [5.4e16 instead of 5.4e17] of the atmosphere takes about 3 hours instead of 30. So instead it would have to be at least 10 times this amount of energy. So if atmosphere required 65,800,000 GigaWatts and it took 3 hrs to cool 1 K, that still be longer than I would expect, though much closer. It would be about the same as earth, and earth has other factors to keep it warmer. And if you think it would take 1/10th of 3 hours or 1/100th of 3 hours, 1/1000th 3 hours, then requires 10,100, or 1000 times more energy. So comparing to Earth. Dwarf at 65,800,000 GigaWatts [ "The Earth receives 174 petawatts (PW) of incoming solar radiation (insolation) at the upper atmosphere.[petawatts 10^15 watts] en.wikipedia.org/wiki/Solar_energy Earth gets 1.74e17 watts sunlight and much this isn't converted heat. The dwarf is on average 10 C and earth is 15 C And dwarf has far less total surface area. Something seems very wrong about this. Revising: How I made mistake by adding 5 orders, probably has something to do mornings and haste. Unfortunately it's morning and I don't have much time, but I wanted go over some numbers again. So we settled on Dwarf heat as getting 5.4e15 watts and Earth heat is disputable but receives 1.74e17 watts of sunlight. Roughly Dwarf is getting 2030 times less heat then earth. Earth surface area is 510 million sq km Dwarf surface area is 4 pi r 2 or 18 million sq km or 1/28th the area of earth. About twice land area of US, btw. Roughly Dwarf is getting same heat as earth does. And dwarf surface air temperature is on average 5 K cooler. One thing we could have done is start the dwarf at a temperature comparable to earth without greenhouse effect. One way would be the commonly stated 33 C [so a 18 C average temperature]. Or instead could start with Earth as blackbody without the "Visual geometric albedo: 0.367" included. Or as Wiki says: "If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth is, it would have a temperature of about 5.3 °C. However, since the Earth reflects about 30% (or 28%) of the incoming sunlight the planet's effective temperature...is about −18 or −19 °C". So starting temperatures could have had the dwarf start at 5.3 C or 18 to 19 C. If add greenhouse affect to a 10 C dwarf, the dwarf should be warmer than Earth. Edited by gbaikie, Dec 8 2011, 06:11 PM.

Chris HoStuart  Dec 7 2011, 08:43 PM Post #18 
Administrator

I haven't said anything at all about how long it would take. But I'll do so now, just quickly, for the first time. To answer this, we need to some assumptions.
From post #5 in the thread, mass of the atmosphere is 5.4e17 kg. Heat capacity C_{p} is 1 J/g/K, or 1000 J/kg/K. So the total atmosphere heat capacity is Energy being lost to space is 6.58e15 W, so the cooling rate is [indentblock]I've fixed up an error in my calculations, where I had the heat capacity ten times too high[/indentblock] However, this calculation has ignored the inversion layer that would form at the bottom of the atmosphere. In practice it would drop in temperature faster than this until you get the inversion in approximate equilibrium with a slow diffusion of heat from higher altitudes. Calculating that is just more distraction from the original problem posed.
Yes indeed. Nighttime inversions over the desert can be very shallow, just a few meters even; and that cools much more quickly than the entire atmosphere. A solar eclipse would get a rapid inversion forming also. (I've been in a total solar eclipse, but it was heavily overcast at the time, unfortunately. It went dark like at night, but there was not much temperature drop. Nor would I expect much drop in those conditions of heavy cloud.) Just for the sake of argument, let's assume that about the bottom 4 meters of atmospheres is the effective heat capacity as temperatures start to fall. At pressure 0.3 atm, that's about 1.2 kg of air per square meter, so the total effective heat capacity over the starship surface is roughly 2e17 J/K. Using this value in the calculation above give temperature falling initially at 1 degree every 300 seconds or so; or 5 minutes for a degree. That's a really rough ball park calculation. If you have a poor thermal contact with the surface, then ground temperature will drop much faster, but what we feel is the lowest levels of air, and that would drop more slowly.
There's no atmosphere there; the only heat sink is the top few centimeters of regolith. I think it falls much faster than that even, but I have no figures.
This is the first post in which I have said how long cooling takes.
The power requirements to maintain a given temperature depend crucially on the surface temperature, the amount of atmosphere, and the atmospheric composition. The rate of cooling when you switch off the generators depends on the same things, though the initial cooling rate will also depend on thermal contact between surface and atmosphere. Any additional surface heat capacity also matters, but it would most likely be negligible. Cheers  Chris Edited by Chris HoStuart, Dec 7 2011, 09:02 PM.

Chris HoStuart  Dec 7 2011, 08:57 PM Post #19 
Administrator

The starship is 6,580,000 Gigawatts surface emission. You've add a zero there. Watts/Joules makes no sense. I don't know what you mean by 5.4e21. The starship atmosphere is 5.4e17 kg, which is a heat capacity of about 5.4e20 J/K. [indentblock]Aha! I think you must have used my value for heat capacity. And I was the one who got the error by a factor of ten! Sorry about that; I've gone back to fix up the original, leaving my original mistake there with the
No, the wattage we've calculated for maintaining the starship temperature is 6.58e15 Watts. Of the 1.74e17 W that Earth gets from the Sun about 30% is lost by reflection. The rest goes to heat in the sense that the rest is what maintains Earth's temperature. The proper comparison is therefore as follows. Earth. Surface temperature of about 15C (but varying over the surface) maintained by 1.22e17 W of solar heating; with the benefit of an atmospheric greenhouse to give about 33K higher temperature that what this much heating would give to a bare surface. Starship. Surface temperature of about 10C (pretty constant over the surface) maintained by 6.58e15 W of fusion heating; with no atmospheric greenhouse effect to help.
You used the wrong number for the power in watts heating the starship. Edited by Chris HoStuart, Dec 7 2011, 10:09 PM.

gbaikie  Dec 8 2011, 12:55 AM Post #20 

Right. Ok so : 6.58e15 Watts. for starship maintain 10 C Takes about 30 hours to drop 1 C Or your number "8.2e4 seconds to drop one degree" Which is a long time Earth requires 1.74e17 watts sunlight [with some percentage of this added heat] And on Earth unless one is in the tropics with high humidity one doesn't normally see such a low rate of cooling. And explanation of such low rate is starship has far more atmosphere per square meter as compared to earth. 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. Water vapor adds enormous amount of heat capacity but that aspect of water vapor will not make a difference. So for the moment I will consider the above as resolved. I do have questions outstanding regarding laspe rate why should it have any effect on lowering the 6.58e15 Watts requirement. 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. Other than above I think I can move on to effects from greenhouse gases. PS: I said "We also have temperatures dropping by 100 K during couple hours on the Moon during a Lunar ellipse. " I gave ref over dragonslayer: http://diviner.ucla.edu/ http://www.diviner.ucla.edu/blog/?p=610 “The following plot shows data taken during successive orbits over a unit of lunar mare situated between 32 and 33 degrees north. The orbit path progressed from east to west (right to left), with each orbit separated by roughly two hours. The first two data swaths were taken before the eclipse, the three center swaths were taken during the eclipse, and the last two swaths were taken after the Moon had reemerged from Earth’s shadow. The data show an average decrease in surface temperature during the eclipse of around 100K, with some locations remaining warmer than others.” Obviously the speed of drop has to due high starting point of temperatures of lunar surface and low heat capacity of top most layer of regolith. I also think the sudden drop from solar eclipse on earth similarly starts from high temperature. Related to this is that as sun gets closer to dusk one see fairly rapid drop in temperature in some areas though late afternoon to near dusk can be warmest times of the day in some other places and I would guess that when you have though situations fairly warm dusk, the nite also keeps fairly warm. and areas where cools rapidly before sun goes down will see rapid cooling at nite. Another thing how high is atmosphere, say at point of 1/1000 of psi, and during this time of cooling in which furnace is off, does the atmosphere lower significantly in the 8.2e4 seconds? And is that mainly why 10 C surface air doesn't cool very quickly? My guess in which I got 100,000 2 GW reactors was considering the atmosphere as an averaged temperature. Which I think is in contrast to how you look at it you regard the 10 C of surface temperature as determinative factor. And I sort of see the atmosphere as one temperature, just different pressure. Or pressurized the one cubic meter of air at say 100,000' to pressure at surface and they will be close to same temperature. Or similar way is that if you want a warm surface temperature [10 C] one will make the atmosphere much higher as compared to a surface temperature of say 50 C. Edited by gbaikie, Dec 8 2011, 01:47 AM.

Chris HoStuart  Dec 8 2011, 01:42 AM Post #21 
Administrator

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.
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.
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.
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 nongreenhouse 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.
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 HoStuart, Dec 8 2011, 01:07 PM.

Chris HoStuart  Dec 8 2011, 03:47 PM Post #22 
Administrator

Hold on to your hat… this is a major jump in complexity with this post. The Grey Gas approximation A grey gas is one for which the interaction with radiation is the same at all frequencies. (It has no colour.) This is unlike any real gas, but it is a useful stepping stone towards solving the problem given for this topic. I’ll consider the starship with a grey atmosphere. 1. BeerLambert law The specific transparency of a gas can be described with an absorption coefficient k. In general, this will depend on temperature, pressure, and frequency. For a grey gas, we can treat it as constant value. The units of k are area/mass. For a given mass of the gas, k gives a cross section area which absorbs incident radiation. You can think of it this way. Every molecule has a certain cross section area representing how effectively it absorbs radiation. The coefficient k is this molecular cross section, times the number of molecules per unit mass. Often the units for k are per mol, rather than per kg. Consider a flux of radiation, passing through a very thin layer of gas. Let the radiation have power per unit area of Q W/m^{2}. The very thin layer of gas has a certain mass per unit area; call this x kg/m^{2}. The absorption cross section within this square meter will be k.x; so the amount of radiation absorbed is k.x.Q; and what gets through is Q.(1k.x). Now generalize to a thick layer of gas, with thickness X kg/m^{2}. Divide this into n thin layers stacked together, where each thin layer has X/n kg/m^{2}. The radiation that gets through all n layers is Q.(1k.X/n)^{n}. As n becomes large, this approaches in the limit Q.e^{k.X}; this is the amount of radiation that gets through the gas without being absorbed. This relation is known as the BeerLambert law, and it applies in many contexts other than radiation through a gas. It is often given in this equivalent form:
The above derivation assumed that the radiation was coming perpendicular to the gas layer. In general, it might be coming at an angle "a" from perpendicular. In this case, the effective thickness of the layer is divided by cos(a). I'm going to gloss over a couple of complications here, and simply consider "isotropic" radiation (that is, the same in all directions), in which case the angles average out to let a = 60 degrees be a reasonable average; so cos(a) = 0.5. In other words, we can double the absorptivity and consider radiation moving vertically only, as a reasonable proxy for isotropic radiation within the atmosphere. 2. Absorption and emission in the grey atmosphere It is possible to obtain "closed form" solutions in this case (see the book "Principles of Planetary Climate" by R. Pierrehumbert) but I'm going to go straight to a simple numeric integration. For a given altitude z, we can give p(z) (pressure), T(z) temperature, U(z) thermal radiation upwards, and D(z) thermal radiation downwards. Consider a layer of atmosphere, from altitude z to z+dz. The mass of gas per unit area in this layer is dp/g, where g is gravitational acceleration, and dp = p(z)p(z+dz) is the change in pressure. Any radiation passing through this layer has a fraction k.dp/g which is absorbed. [indentblock]Added in edit. Oops. The absorbed fraction is 1e^{k.dp/g}. This is close to k.dp/g if dp is sufficiently small.[/indentblock] We can apply Kirchoff's law, and conclude that the emissivity matches absorptivity; so this layer is radiating (k.dp/g).s.T(z)^{4}, where s is the StefanBoltzmann constant, 5.67e8 in SI units. I am using T(z) as the representative temperature for the entire layer, which is fine as long as dz is sufficiently small. This radiation is going both upwards, and downwards, equally. In thermal equilibrium, where the radiation from above and from below is characteristic of T(z), the radiation absorbed is equal to what is emitted; as we should expect. Hence, removing the absorbed radiation and adding the emitted radiation we have:
[indentblock]With the above correction, this should be:
If we know the temperature profile T(z), then we can use the boundary conditions to calculate U(z) and D(z) at every level, in increments of dz. Effectively, we are doing a numeric integration, with step size dz. The boundary conditions are:
3. Temperature and lapse rate in the atmosphere A good online reference for this subsection is Thermodynamics & Statistical Mechanics, lecture notes by Richard Fitzpatrick of the University of Texas. (Also available as a bookl.) In particular, see this page: "The adiabatic atmosphere". We've already picked on the DALR as a basis for temperature in the atmosphere. Convenient formulae for altitude and temperature in terms of pressure are as follows:
4. Convection and the stratosphere If the DALR extends all the way to the very top of the atmosphere, then these formulae give a finite height for the total atmosphere. At the very top of the atmosphere the pressure is zero. This occurs when z = z_{1}/f, which on Earth is about 29.4 km. This doesn't happen, because the DALR only applies in the troposphere, where there is convection. The stratosphere has a different lapse rate, which is actually determined by a radiative equilibrium. Consider the formulae used in section 2 for breaking up the atmosphere into layers, and calculating upwards and downwards radiation. One feature of those formulae is that they don't conserve energy! In the lower levels of the atmosphere, we will find that the radiation emitted from a layer is greater than the radiation being absorbed. That's actually okay. The temperature profile is fixed by convection, and convection works to circulate warmer air upwards, which provides additional flows of energy up into the atmosphere. All we have done is assume that convection works to maintain the temperature profile; and then calculate what radiation is absorbed and emitted at those temperatures. We will find, however, that at some level this effect reverses, and each layer will be absorbing more than it emits. This actually marks a point where convection stops; the end of the troposphere. From this point, an accurate calculation would not use the DALR any more, but would pick a temperature profile that conserves energy at each level. This is called "radiative equilibrium", and it is characteristic of a planet's stratosphere. I shall proceed with the calculation without worrying about stratospheres. It will be easy to identify the altitude at which the DALR ceases to be appropriate. As long as this is very high in the atmosphere, the end result should be pretty accurate. This was a bugger of a post to write, and it will be worse to read when you weren't expecting it. Never mind. With my next post, I'll just go ahead with the calculations. Rather than estimating a concentration level for a greenhouse gas, I'll just try different values for k, and see what impact they have. I'll also attach an Excel spreadsheet so others can experiment as well! Be patient. We still have another jump in complexity to deal with a real gas before the question in the original post is finally answered. Cheers  Chris PS. I have edited this to invert the definition of f, to be R/C_{p}. This makes everything a little bit more in accord with usual conventions; also fixed one typo in a formula. Grateful thanks in advance to anyone who catches more errors in my algebra! Edited by Chris HoStuart, Dec 10 2011, 08:39 AM.

gbaikie  Dec 9 2011, 01:01 AM Post #23 

I think the dwarf planet should not have a stratosphere. "Mars and Venus also have thermospheres. What is missing is the temperature bump of a stratosphere (and mesophere) because they do not have an ozone layer to absorb the ultraviolet light." http://www.astronomynotes.com/solarsys/s3c.htm There is a graph of Venus, Earth, and Mars. The line bends with Earth and Venus significantly, Mars is mostly straight, I think the dwarf would be straighter than Mars. Btw, I believe that article is wrong Mars does have ozone layer [as I recall it's quite low in elevation]. Oh drats, I better google it. "About the image: HIPWAC spectrum of Mars’ atmosphere over a location on Martian latitude 40°N; acquired on 11 December 2009 during an observation campaign with the IRTF 3 m telescope in Hawai’i. This unprocessed spectrum displays features of ozone and carbon dioxide from Mars, as well as ozone in the Earth’s atmosphere through which the observation was made. Processing techniques will model and remove the terrestrial contribution from the spectrum and determine the amount of ozone at this northern position on Mars." http://www.universetoday.com/56753/ozoneonmarstwowindowsbetterthanone/ "There is much less oxygen to start with, so there is not very much ozone on Mars. It doesn't form a layer there, it reaches all the way to the surface. And there is little enough ozone that far more ultraviolet light reaches the surface of Mars than the surface of Earth  you could get a bad sunburn very fast there. " http://athena.cornell.edu/kids/cs_lemmon_questions.html I didn't find what I was thinking about. But anyhow. Both Venus and Mars have fairly recently discovered to have an ozone Venus very weak and Mars is hard detect. Since less O2 it's fainter and like earth has processes which create and destroy it. Not significant point I just recalled reading some article that mars had ozone. Anyhow, I don't think you have stratosphere unless you have sunlight. And probably don't have a thermosphere and the other spheres, either. Edited by gbaikie, Dec 9 2011, 01:44 AM.

Chris HoStuart  Dec 9 2011, 02:50 AM Post #24 
Administrator

I use the term "stratosphere" to indicate where the atmosphere first becomes "stratified"; due to the lack of strong vertical mixing from convection. In the stratosphere, adiabatic lapse rates no longer apply. The starship will indeed have a stratosphere in this general sense of the term. There are other definitions, but for general usage across many different planets, I much prefer this one. It is used (for example) in "Principles of Planetary Climate" by Pierrehumbert. I'll try to make clear the precise definition I am using. The troposphere is the portion of the atmosphere mixed by vertical convection, and where convection is able to transport energy upwards and so drive the lapse rate towards the adiabatic lapse rate (dry or moist as appropriate, if we consider condensing gases in the atmosphere). Without fussing too much on different definitions (I think Strobel's usage is much too Earthcentric for a chapter on planetary science, frankly) I'll just note that I am using the term to mean where the atmosphere first becomes "stratified"; due to the lack of strong vertical mixing from convection. The starship will indeed have a stratosphere in this general sense of the term, even without sunlight or ozone. In any case, it will certainly have an altitude above which the atmosphere becomes stratified. There might be a sphere where the atmosphere also becomes sorted by molecular weights, but I don't really know. I won't worry about that; if this occurs it would be where the atmosphere was too thin to impact the surface temperature. Cheers  Chris 
gbaikie  Dec 9 2011, 10:24 PM Post #25 

Rerevision: I think our assumption refutes "greenhouse effect" as described in Wiki. In that greenhouse gases increase earth temperature by 33 C. This because roughly our dwarf planet receives about same warming that earth receives per sq meter, it doesn't have greenhouse gases and it's 10 C. But this is not saying the model of dwarf refutes AGW or CAGW but rather indicates the greenhouse effect from greenhouse gases doesn't cause planet to be 33 C on average warmer. That instead much of this warming is reducing night time losses that that is caused by heat capacity. In practical terms this means the greenhouse gas don't need to come up with "the missing" 33 C. Instead it could mean they have to account for say 5 C of warmer. This in a real sense could make more of a case for CAGW because to add a degree could take less greenhouse effect from greenhouse gases. Or it doesn't provide any difference in this regard, but does make "finding" warming affect for greenhouse gases "easier" Or suppose we put the dwarf planet at Mars distance. What Mars average would be is whatever number, but night time mars isn't going to cool it would lose less than 1 C. And most people would call that due to the "greenhouse effect" of the dwarf atmosphere [which has no "greenhouse gases"]. Edited by gbaikie, Dec 9 2011, 10:48 PM.

Chris HoStuart  Dec 9 2011, 11:17 PM Post #26 
Administrator

Except that our dwarf planet does not receive the same warming as earth per square meter; not even close! We've been using 364 W/m^{2} on the dwarf planet (calculated in post #6 of this thread). At least, that is what I have been using, and I think it is straightforward. There have been no objections to it so far. If we boosted the Dwarf planet to 15C (which we have been taking as Earth's representative temperature), then the energy requirement for Dwarf would be 391 W/m^2 The Earth is heated by 240 W/m^{2}. It needs so much less because of the greenhouse effect, and the whole point of this exercise, as I see it, is to apply physics to calculate power requirements for Dwarf, given greenhouse gases in its atmosphere. I'm getting ready to get to the next step, of using a hypothetical grey gas for the atmosphere. This will be a stepping stone to a realistic calculation with a mix of CO2, O2 and N2. The basis of the 33K calculation is simply the difference between surface temperature, and the effective temperature; where the effective temperature is the temperature equivalent of the supplied power. 240 W/m^{2} corresponds to effective temperature of 255K, which is indeed 33K less than the real surface temperature. To calculate the 33K effect using Dwarf would require giving Dwarf the same atmosphere and gravity as Earth. We'd have to consider the full complexity of moist lapse rates near the surface, and dry lapse rates higher in the troposphere, and nonadiabatic lapse rates even higher in the stratosphere. We'd have to consider variations of conditions over the whole surface. This can be done, and it IS done; but it is way too much to explain here. What we CAN do here is explain the underlying physics of the atmospheric greenhouse effect in a very elegant situation, in which atmospheric composition is very simple, surface conditions are homogenous, lapse rates are simple, and there's no need to worry about albedo or solar absorption. As such, this problem you have proposed is excellent for the purpose of calculating the greenhouse effect with real physics in a context where the details are laid bare and visible. We won't get the 33K, because Dwarf is not the same as Earth, and the magnitude of its greenhouse effect will be different. But the magnitude of its greenhouse effect will be calculated by the same methods in that simpler case. Dwarf is thus an ideal example for teaching, or for testing out alternative ideas on the thermodynamics of an atmosphere with greenhouse gases. Cheers  Chris 
gbaikie  Dec 10 2011, 04:48 AM Post #27 

"We've been using 364 W/m2 on the dwarf planet (calculated in post #6 of this thread). At least, that is what I have been using, and I think it is straightforward. There have been no objections to it so far." oh you right I was thinking in morning again. I must stop doing that dwarf has 1/28th surface area of earth, and nuclear reactor making 1/28th power earth receives from sun. But of course dwarf being smaller would receive less sunlight at earth distance as compared to larger Earth. But if dwarf was at Mars distance, it should warmer than Mars is meaning it wouldn't get as cold at nite. And Earth would get colder at night than it does, if was similar to Mars atmosphere or 1/100th it's atmosphere. So, you think the earth heat capacity adds nothing and only greenhouse gases will make up the 33 C. The original reason why I started that post [and got distracted] was the idea was to have dwarf at 10 C. So atmosphere plus greenhouse gases equals 10 C. So should subtract however much warmer greenhouse gases are going to add. So instead having it at 10 C it should be at 10 minus 33 or 23 C. Btw one reason I "liked" your number of 6,580,000 Gigawatts is "dealt with" low cooling rate better than my guess 200,000 Gigawatts. It would meant instead 24 hours to cool 1 C, with 200,000 GigaWatts it would have taken weeks to cool 1 C. Or one day to cool 1 C was more believable than weeks to do this. But if one calculates amount gigawatts to warm to 23 C, it's going to be less than 6,580,000 Gigawatts and have longer period to cool 1 C. At 23 C it would take longer to cool, but I suppose if at 10 C it might cool faster not sure how, but maybe somehow. Edit: well maybe CO2 cools planet but also warms it it robs from heat capacity. Edited by gbaikie, Dec 10 2011, 04:58 AM.

Chris HoStuart  Dec 10 2011, 05:16 AM Post #28 
Administrator

Yes, I think Earth's heat capacity makes no difference to equilibrium temperatures. Heat capacity only changes the rate at which it heats up or cools down. We have several posts in the thread so far talking about this difference between finding a temperature at which things are stable (equilibrium); and finding the rate of cooling or warming when conditions are unstable (when the temperature is not at equilibrium). There are a couple of qualifications I should add to that! Heat capacity does have an impact on the temperature variation between day and night; more heat capacity means temperatures get smoothed out. The effect on mean temperature depends on how you calculate it. It makes no difference (to the first order) for the "effective" temperature (obtained by averaging energy flows and then getting a temperature for the mean flow). But smoothing will tend to raise the cooler temperatures by more than it lowers the high temperatures; this is because of the 4th power relation between temperature and thermal energy flow. The other qualification is that heat capacity in the atmosphere, as well as working directly on the amount of heat able to be stored, also has a second effect because the lapse rate depends on heat capacity; and lapse rate has a big impact on the efficacy of the greenhouse effect. Perhaps counterintuitively... more heat capacity means LESS lapse rate, and LESS additional equilibrium temperature because the greenhouse effect is less effective! This only matters if there are greenhouse gases in the atmosphere, of course, but the effect actually depends on the heat capacity of all molecules in the atmosphere, not only the greenhouse gases. If that sounds a bit counterintuitive, at least we'll be able to run calculations and look at consequences for different values once I put up the spreadsheet. About that... the spreadsheet for a grey atmosphere is done, but not debugged. I'm doing all this from scratch, and at present the spreadsheet has errors in it somewhere. I want to get rid of as many bugs as possible before posting it; then I'll actively invite bug reports and test results for all comers. As for your reasons for putting the dwarf planet at 10C... you cannot presume a 33K greenhouse effect unless dwarf is a MUCH closer match to Earth. That would mean moist adiabiats, wide surface variation, diurnal cycles, many greenhouse gases all with different absorption spectra, and so on, and so on... way beyond what I can calculate for myself with a simple spreadsheet, or anything else for that matter! So I'll be able to calculate a greenhouse effect in the simple case of Dwarf; but it's not going to be the same number as the 33K we get for Earth. Cheers  Chris 
gbaikie  Dec 10 2011, 09:02 AM Post #29 

Ah. I didn't know that. I was thinking that sunlight would have cooling affect [strange as that seems]. It seems in comparison the sun heat the earth rapidly. Such rapid warming and cooling would result in more mixed atmosphere. And I would think a quieter atmosphere could get warmer. If you violated the lapse rate dumping large amount of heat in a localized area, would not result be radiating more heat into space. And doesn't the opposite a calm atmosphere allow it to get warmer? It also seems since that dwarf planet has more atmosphere per sq meter column of air any greenhouse gas could result in more heating. If you have 400 ppm of CO2, one has 3 times the amount per square meter than earth. With H2O you could more humidity per square meter because lower lapse rate though because of lower temperature the air can't hold as much as warmer earth, but has a higher density of air it's only pressure which is 1/2 of Earth. So comparing same temperature, one should also get about 3 times the amount water which air could hold. Edit Maybe better to switch pressure 1/10th of earth, and then dwarf as same amount of atmosphere per square meter and have same air density as earth. Less differences, less potential confusion. And gets 8 hours for cooling 1 C rather than 24 hrs. And considering difference in lapse rate having larger chunk of air at same temperature come to think about it's not vaguely unreasonable. Plus at 1/10th pressure, the sky isn't so high. If do that, remember to change numbers on chart Only reason for 1/3rd of earth atmosphere was so it's breathable without pressure suit and I had thought had heat capacity or density of atmosphere would reduce heat loss. Oh, I guess it wouldn't same density but it would more similar. Edited by gbaikie, Dec 10 2011, 11:14 AM.

Chris HoStuart  Dec 10 2011, 10:43 PM Post #30 
Administrator

In my earlier post #22, I got a little bit ahead of myself. I'm going to rewrite it entirely here, in a slightly simpler form, without any mention of the backradiation. We don't need it (yet). Backradiation exists when you have nonzero k; and the numbers allow you to infer it must be there. But as long as we assume a fixed lapse rate, we can calculate power requirements on the fusion heaters without calculating the backradiation. The Grey Gas approximation A grey gas is one for which the interaction with radiation is the same at all frequencies. (It has no colour.) This is unlike any real gas, but it is a useful stepping stone towards solving the problem given for this topic. I’ll consider the starship with a grey atmosphere and a fixed lapse rate. 1. BeerLambert law The specific transparency of a gas can be described with an absorption coefficient k. In general, this will depend on temperature, pressure, and frequency. In this post, I consider a simpler case in which it is constant. The units of k are area/mass. For a given mass of the gas, k gives a cross section area which absorbs incident radiation. The earlier post gives a bit more background derivation, ending up as follows.
2. Absorption and emission in the grey atmosphere Let us divide the atmosphere into N layers, all of equal mass per unit area. Hence, each layer has the same drop in pressure dp, and dp = p_{0}/N, where p_{0} is the pressure at the bottom of the atmosphere. The mass per unit area of the layer is dp/g. Hence radiation passing through any one layer has a fraction 1e^{k.dp/g} which is absorbed. We can apply Kirchoff's law, and conclude that the emissivity matches absorptivity; so this layer is radiating E.s.T(p+f)^{4}, where s is the StefanBoltzmann constant, 5.67e8 in SI units, and where T(p+f) is a representative temperature for the entire layer; somewhere between T(p) and T(p+dp), the temperatures at the top and bottom of the layer. When everything is all at the same temperature and at thermal equilibrium, the layer is absorbing energy from both top and bottom, and is emitting also out the top and the bottom. We're only going to worry about radiation propagating upwards (for now). Let Q(p) denote the radiation propagating upwards through the atmospheric level with pressure equal to p. Removing the absorbed radiation and adding the emitted radiation over a single layer we have: [indentblock]Q(p+dp) = (1  E).Q(p) + E.s.T(p+f)^{4}[/indentblock] We know the temperature profile T(z) already, because the lapse rate is fixed by assumption. All that is needed now is to iterate up the atmosphere from Q(p_{0}) at the surface, to Q(0) at the top of the atmosphere. That will be the energy lost to space, and which needed to be supplied by the onboard fusion heaters of the starship. 3. Temperature and lapse rate in the atmosphere The original post at post #22 describes how to calculate a lapse rate, based on a neutral dry atmosphere. A good online reference for this subsection is Thermodynamics & Statistical Mechanics, lecture notes by Richard Fitzpatrick of the University of Texas. (Also available as a bookl.) In particular, see this page: "The adiabatic atmosphere". We've already picked on the DALR as a basis for temperature in the atmosphere. Convenient formulae for altitude and temperature in terms of pressure are as follows:
4. Convection and the stratosphere As gbaikie has noted, the term "stratosphere" is ambiguous when we look at planets in general. Different authorities use different definitions of the word. Suffice to say… forget this section. Let's just proceed with the calculation of Q(0) exactly as given above, and then look at the results to see what implications follow, and where we might need to improve the model further. I'll get to that! 5. The spreadsheet. Version 1.1 Nerds and scientists and programmers will sneer… but I'm going to give you an Excel spreadsheet. Real experts would do this in R, at the very least. The spreadsheet has a huge advantage  a much wider audience will be able to grab it and experiment with various parameters. I'll use Excel 97, so old windows platforms can use it; and if you use Open Office or some such, it should work there also. Please let me know of any problems running the spreadsheet! I have taken a bit of time to clean it up. The spreadsheet is "protected"; a standard feature in Excel. It's easy to remove protection if you want to modify more than the intended user inputs. But for simply using it to calculate, enter information in the green cells. These cells have editing enabled. Your spreadsheet program should protect you from making changes to other more important cells. The spreadsheet is attached as a zip folder (to keep within the size limits on attached files). Try it out, tell me what you think and what you get. Especially tell me if you find bugs, or results that seem to be outside what you should expect. I'll probably be able to explain the result  or fix a bug which gave spurious numbers! The "k" value is entered in mm^2/kg, rather than m^2/kg. This means values are not tiny. The "k" value chosen for Earth is selected simply to give the same emission to space as we have in reality; the value is 97. With this, the emission to space for surface temperature of 15C and a dry adiabatic lapse rate is 241.2 W/m^2, which is about the same as what we have in reality. When this same value is applied to the dwarf planet starship, with temperature 10C, the emission to space is 105.54 W/m^2. That's actually a stronger greenhouse impact than for Earth, which surprised me at first. I had expected a weaker impact given the weak lapse rate. However, it seems that the more important fact is that the atmosphere ends up being much higher than on Earth, and so the effective emission layers are much colder. This gives a very strong greenhouse impact. The fusion generators now need 1.9 million GWatts, rather than the 6.6 million we had previously. Woohoo. The greenhouse effect in this case means only 30% of the original requirement is needed. I can play around with other parameters as I choose. I've entered a new example (and there's space for anyone to provide four new templates of their own choosing. My new example has:
With this, my spreadsheet gives a mere 3.4 W/m^2 trickling out the top of the atmosphere, a whopping decrease from the 272 W/m^2 at the surface. With the smaller surface area, my fusion generators now look a heck of a lot more reasonable. 6.8 thousand GWatts only. (10 times less area, 0.75 reduction in surface emission with the lower temperature, and power scaled down by 0.012 because of the massive greenhouse effect.) Play around, see what you get, and bug reports welcome! Cheers  Chris PS. Edited this post for corrected the emissivity formula for a single layer above. The spreadsheet looks to be doing it correctly, so no change there. Edited by Chris HoStuart, Dec 10 2011, 11:53 PM.

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