Why the 'moist lapse rate' is what it is.

As I pored over my master Greenhouse Effect spreadsheet after work today, another thing struck me (and which I have glossed over so far, it being a tricky topic).

We calculate the ‘dry lapse rate’ as 9.75 K/km, that’s easy enough to understand. But the measured real-world ‘moist lapse rate’ is only 6.5 K/km. We know that water is a moderating influence, so we correctly put it down to water, water vapour and clouds with a hand wave, but how do we know it must be one-third lower than the ‘dry lapse rate’ for everything else to makes sense? Let’s do the numbers…
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If you need a refresher…

Rule 1, energy cannot be created or destroyed, it just changes from one form to another.
Rule 2, energy tries to spread out as evenly as it can, taking into account the medium and the forms it can take within the restraints of that medium.

I don’t like the ‘parcel of air’ analogy, but it will do for now. Air cools as it rises, so converts kinetic energy (aka ‘heat’ in layman’s terms) into some other form of energy. What kind of energy does it convert into? Easy – it changes into potential energy (think about it for a minute or two).

1 kg of air higher up has the same total amount of energy (of all forms) as 1 kg at sea level. Kinetic energy (mass x specific heat capacity, referred to as Cp, dunno why) has changed into the same amount of potential energy (mass x height x acceleration due to gravity). The fall in temperature is called the ‘lapse rate’.

Maths of the ‘dry lapse rate’:
Joules converted = mass x change in height x gravity = mass x change in K (temp) x Cp.
Cancel ‘mass’ on both sides.
g x change in h = change in K x Cp.
Reshuffle that, you get
Change in K ÷ change in height = g ÷ Cp.
9.807 m/s2 ÷ 1,006 J/kg = 9.75 K/km
(see Wiki for the hopelessly confusing explanation with the same end result).
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So far so good. The easy bit is PE gained = KE lost, but there’s some energy unaccounted for.

PE gained is easy:
1 kg of air has no PE for these purposes. It can’t fall any further.
1 kg of air 1,000 metres above sea level has 1 kg x 9.807 m/s2 x 1,000 metres Joules = 9,807 Joules of PE.

So 9,807 Joules of energy, mainly of KE (‘heat’), have changed into PE.
We know it has cooled by 6.5 K and that the specific heat capacity of air is 1,006 J/K/kg. So only up 6,539 Joules of KE have been converted into PE.
That leaves 3,268 Joules unaccounted for, doesn’t it?

The missing figure is latent heat of evaporation of evaporation/condensation (‘LHEC’). The ground gets radiation (energy) from the Sun and doesn’t convert it all into KE, it converts it into 6,539 Joules of KE and 3,268 Joules of LHEC per kg of air.

1,000 metres higher up,  some water vapour condenses, freeing up 3,268 of LHEC in addition to the 6,539 Joules of KE that 1 kg or air already had, and all 9,807 Joules are converted to PE.

Which is why the ‘moist lapse rate’ observed in the real world is 6.5 K/km, not 9.75 K/km.

(6,539 Joules KE converted ÷ 9,807 total Joules converted) x 9.75 K/km dry lapse rate = 6.5 Joules KE converted per km = moist lapse rate.
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I know I’m reinventing the wheel here, and I’d hope that this is all old hat to people who’ve got a physics A-level or degree, but why don’t they give this sort of fun stuff the same coverage as all the other Greenhouse Effect-related stuff?

And why does the IPPC’s energy budget show that the hard surface only converts one-sixth of incoming radiation into LHEC and not one-third?

(Rhetorical questions, we know the answers, which I will cover soon).
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While checking my numbers, I stumbled across this fine article (ignore the ghastly typo in the title, I’ve made that same mistake many a time), well worth a read if you are into maths and are a bit dubious about the Consensus.