Saturday, 1 May 2021

How to estimate pressure and temperature using density as a starting point

As a fun maths challenge, I decided to apply the principles outlined in Acceleration ≈ gravity and see whether I could get sensible results by applying basic maths, basic physics and common sense.

Let's assume that you are only given the following measurements (taken from the US standard atmosphere):
1. Pressure at sea level = 101.325 Pa
2. Temperature at sea level = 288 K
3. Density at sea level = 1,200 kg/m3 (a bit on the low side?)

T x D ∝ T. We would do it the long way round by using Barry to find pressure at each altitude; then inserting the temperautre based on the lapse rate to find density at each altitude.

But what if you are also told that D at the top of the troposphere (10 km altitude for simplicity) = 400g/m3? It's actually much easier. You can interpolate everything else, including the likely lapse rate.

1. Set up your Excel sheet, type in the given numbers (pale yellow)
2. You can assume that density changes in straight line, but it is more realistic to assume it changes geometrically, so it goes up by a factor of 1.011 (3^0.1) for each km lower (doesn't make much difference)
3. Then work out the 'mass of the air/m3' for each 1 km 'slice' of altitude. That's just density x 1,000. I hid this column to simplify it a bit.
4. We know that the total mass of air at sea level must be enough to create a pressure of 101,325 Pa at sea level, so it must be 101,325/9.801 ≈ 10,338 kg total, so you put in 2,649 kg (pale blue) as a balancing figure (to get it to add up to 10,338 kg).
5. Pressure = force ÷ area, so it is simply 'mass of air above that altitude' x gravity (9.801 m/s2) per m2, so e.g. at 9 km, it's 2,649 kg + (446 g/m3 x 1,000 m) = 3,095 kg x 9.801 = 26 kPa
6. Temperature ∝ pressure/density i.e. temperature = pressure/density x constant 'k'
7. Work out 'k' = (288 x 1.2)/101 = 3.411 in this example (it's not a universal constant, you work it out individually each time)
8. Temperature at each altitude = pressure/density at that altitude x 3.411.

The results match the Standard Atmosphere very well. The T, and P results for top of troposphere are pretty bang-on. It does show that the lapse rate is lower than the accepted mid-figure 6.5 K/km at low altitudes and higher at higher altitudes. This is not unrealistic, as there is more absolute humidity at lower altitudes, but that's probably a happy accident and it averages out to ~6.5 K/km overall:
All this neither proves nor disproves the Physics Denier's contention that sea level temperature would be ~33 degrees cooler without 'greenhouse gases' (disproving that is far simpler and requires little or no maths, just facts and logic), but I love a maths challenge.
These workings also give a good understanding of the real world, against which you can test various Physics Denier theories - like the 'Top Of Atmosphere' theory, which says the temperature at the Effective Emitting Altitude (about 5 to 6 km up) is fixed at 255K, but sea level temperatures are 33 degrees warmer than that solely because of 'greenhouse gases'. This is an alternative explanation for the observed lapse rate of ~6.5 K/km, but it has to be that much anyway because of gravity, in other words... the 'Top of Atmosphere' theory assumes that gravity doesn't exist.

Or... do they mean that without greenhouse gases, the Effective Emitting Altitude would be sea-level (so sea level temperature would be 255K) and the gravity-induced lapse rate would stay much the same? On Mars, there is about thirty times as much CO2 as there is on Earth, but the Greenhouse Effect is negligible on Mars (5 degrees at most) and the Effective Emitting Altitude is no higher than it is on Earth. So that fails on the facts. Hmm.


MrMC said...

You really need a TV to occupy yourself, I would suggest a week watching the BBC and them you will have ample opportunity to test your own blood pressure and temperature compared to the previous week....

Mark Wadsworth said...

MC, I read a couple of Guardian articles every morning to keep myself immunised.

Dinero said...

"Or do they mean that without greenhouse gases, the Effective Emitting Altitude would be sea-level (so sea level temperature would be 255K) and the gravity-induced lapse rate would stay much the same?"

Yes thats it.

Mark Wadsworth said...

Din, yes, that is the interpretation which is leads to a more plausible end result.

But fails on the facts. What about the counter-example Mars? What about mountain tops? They already have a negative Greenhouse Effect, even though there is plenty of CO2 above them. Surely the Greenhouse Effect is always positive?

Dinero said...

What search engine do you use. Do you use Bing. I suggest Google. On Google I often get the answers to these type of science questions quickly on the first page.
Mars is cool because it has a thin atmosphere without the water vapour that Earth has. Water vapour being the greater GHG. and the CO2 GH effect is measurable for Mars.
Not sure what you mean about mountains.

Mark Wadsworth said...

Din, I have googled and binged and read lots of different theories. A lot of the hits are pure propaganda, so you have to filter it all down to things that accord with basic maths, basic physics and common sense. If you get sensible answers, then that's good enough for me.

Re Mars, grow up. Mars has about twice as much CO2 as Earth has in CO2 and H20 combined. Yet the 'official' greenhouse effect is no more than 5 degrees (against alleged 33 degrees on Earth). If you want to scale that down to CO2-caused greenhouse effect on Earth, the answer is "too small to calculate".

Re mountain tops, this is a glaring Achilles Heel of AGW theory. They get as much sunshine as at sea level, and while there is less CO2 above them than above sea level, there is no 'greenhouse effect', there is a *negative* greenhouse effect, they are cooler than Earth's effective temperature.

MrMC said...

RE mountain tops, have you factored in the methane from the yetis ?

Mark Wadsworth said...

MC, does yeti methane contribute to the negative greenhouse effect at higher altitudes? Interesting theory. Maybe we should take all the snow in the Antarctic and use it to build Abominable Snowmen?

Dinero said...

Air temp at Mountain peaks stands to reason. The surface is not flat, the ground level used for the laps rate calculation is the average of the valleys and peaks. The mountain tops are at the maximum of those heights and so it stands to reason that the air at that altitude will be towards the low temp end of the lapse rate.

Mark Wadsworth said...

Din, in real life, that's correct.

But how warm would they be if there were no atmosphere and hence no greenhouse effect?

Answer = warmer

Ergo, there is a negative greenhouse effect higher up to balance out the positive one lower down.

Dinero said...

If there was no atmosphere there would be be no lapse rate and they would be warmer, if there was no greenhouse effect they would be cooler, so no negative GH effect.

Mark Wadsworth said...

Din, you are bad at doing comparisons.

"if there was no atmosphere there would be be no lapse rate and they would be warmer"


"if there was no greenhouse effect they would be cooler"

Cooler than what?

Cooler than with "no atmosphere"? Or cooler than "with an atmosphere but a magical kind of atmosphere that does not have greenhouse effect"? or "cooler than they are now" (which is a meaningless comparison and presumes that there is a greenhouse effect, and you can't use an assumption as evidence to support that assumption).

I can't even guess what you mean. You seem to be saying that without a greenhouse effect they would be a) warmer and b) cooler.

Dinero said...

Without an effect that causes the surface level to be higher the air temp at the lower temp of the lapse rate would be lower still. That is the case and so it is not a negative effect. It comes out in accordance with the positive greenhouse effect.

Mark Wadsworth said...

Din, so you are a Flat Earther?