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Chris Ho-Stuart
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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 non-zero 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. Beer-Lambert law

The specific transparency of a gas can be described with an absorption co-efficient 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.

  • Let X be a certain mass of gas, per unit area. (Such as, for example, the mass of atmosphere per unit surface area of a planet, or the mass in a layer of atmosphere at some altitude.)
  • Let Qi be a flux of radiation per unit area passing through X.
  • Then Qi.e-k.X is the flux of that passes directly through without being absorbed.

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 = p0/N, where p0 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 1-e-k.dp/g which is absorbed.

Let E = e-k.dp/g. Let E = 1 - e-k.dp/g. This is the "emissivity" of a single layer, hence the letter E. This value will depend, of course, on N; the number of layers being considered.

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 Stefan-Boltzmann constant, 5.67e-8 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(p0) 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:

  • Cp -- Specific heat of the atmosphere at constant pressure; close enough to 1000 for the atmosphere we are considering, in SI units.
  • R - the gas constant, which is close enough to 2/7 times Cp for our atmospheres.
  • f = R/Cp, which is hence close to 2/7 for our atmospheres.
  • g - gravitational acceleration
  • z1 = R.T0/g is the "scale height" for an atmosphere, with temperature T0 at the bottom. On Earth, this is (2/7).288/9.8 = 8.4 km.
  • T(z) = T0.(1 - f.z/z1) = T0 - z.(g/Cp). This is the DALR (dry adiabatic lapse rate) giving temperature at altitude z.
  • p(z) = p0.(1 - f.z/z1)1/f. This is the pressure in terms of altitude, for the adiabatic atmosphere.

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:

  • r = 400km radius. I wanted it smaller.
  • g = 2 m/s^2 gravity. Basically I got the highest credible gravity which kept the density of the planet reasonable. Making a starship out of tungsten would do the trick.
  • T0 = -10C surface temperature. That's cold, but survivable. A colder atmosphere has a much better change of not diffusing out into space. This temperature still gives me an atmosphere that is about 4.5 times higher than Earth's atmosphere; but at least it is less than what I got with Dwarf!
  • p0 = 1 atm surface air pressure. More pressure gives a stronger greenhouse effect, but without increasing the atmosphere height. (Say what? Yes indeed; this is a consequence of the fact we assume a constant lapse rate. That's not going to be physically plausible, and it needs to be fixed in the next version.) But for now; full Earth pressure is supplied.
  • Specific heat of the atmosphere remains as for Earth. I'm using Nitrogen and Oxygen.
  • Starship engineers have some magic grey gas to add to the atmosphere, rather than CO2. Fine; I'll add ten times as much, to get a bigger greenhouse impact; this means k = 970.

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.
Attached to this post:
Attachments: Starship_v1.1.zip (106.75 KB)
Edited by Chris Ho-Stuart, Dec 10 2011, 11:53 PM.
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The giant space ship example · Physical theory for climate