All the talk of adiabatic cooling in the other thread Why are the top of mountains cold? got me thinking about adiabatic processes in general. When a rising parcel of air does work on its surroundings due to pressure expansion, is it fair to say then that as the parcel drops it's temperature, the surroundings gain in temeprature (although to a much diluted extent)? Similarly, when a parcel is compressed on descending, is it fair to say that as the parcel heats up, the surrounding air is doing the work and thus cools down? I am just trying to apply conservation of energy to my understanding.
By definition, no heat is transferred to or from the working fluid in an adiabatic process. The reason that parcels of air lose/gain temperature is not because they have lost/gained energy, but rather because that energy is distributed over larger/smaller volume. I.e., temperature of a gas is related to its energy density, so a quantity of gas with a fixed energy will change temperature as its volume (i.e., pressure) is varied, without any transfer of energy to/from the fluid.
I'm kind of confused now because I heard before that the individual gas molecules themselves lose kinetic energy due to work on the surrounding air during expansion and that this is the reason for the loss in temperature during an adiabatic expansion. to me it's quite intuitive - like the bicycle pump example - you are putting energy into the gas from your own body which is converted to heat energy and raises the temperature. Then when you let go of the pump, the air does work on the pump piston and a lot of the heat energy is lost from the air as it cools adiabatically. However, I can see your point above. Do you think it's a mixture of these two things? Or is it wrong to look at the problem this way?
A parcel does loses energy as it rises. It rises adiabatically until its temperature is equal to its dew point, for that period the only change to the parcel is temperature. After that it rises pseudo-adiabatically, at this point the surrounding air has less total air. For its pseudo-adiabatic rise a few things happen; water is expended and rained out, entrainment penetrates the parcels surface and causes it to cool (though not at the rate of the general atmosphere). For the period that rainout is occuring, stored work is being expended. When the parcel begins decent its in stasis with the atmosphere. As it descends it heats up and contracts, however I don't think it's fair to say the surrounding atmosphere is doing work. No work is being done. The change in pressure directly corresponds to a higher temperature. So at this point...if the parcel was lifted back up to its previous position no work would be done, since the parcel had no moisture.
Can you really say that no work is being done on a parcel of air that is contracting? The contraction has to be caused by something doing work doesn't it?
Work is typically equivalent to head in atmospheric dynamics. Adiabatically if it goes from point 1-->2-->3 if 3 is the same location as 1. Then there was no net work done. However from 1-->3 work was done, and from 2-->3 work was done. I misunderstood your question, at first I thought you were referring to a parcel that moves down in the atmosphere unforced. I reread your question, to be referring to a parcel that's being forced down into the atmosphere. If it's unforced, the parcel is simply moving vertically in order to remain in stasis with the surrounding atmosphere. In this case work is being done, here's the equation. \(\Delta W = \int_{\alpha_0}^{\alpha_1} p d\alpha\) Where \(\alpha\) is the specific density. This can't be directly integrated. Poisson's relationship states: \(T\alpha^{\mu -1} = K\) where K is a constant for that particular parcel and \(\mu = \frac{C_p}{C_v} = 1.4\). I'm not going to derivate this equation...but I will tell you it comes from these two statements of energy: \(C_v dT + p d\alpha = 0 = C_p dT +\alpha dP\) Cv and Cp are specific volume and specific pressures. So then you take: \(p = K\alpha^{-\mu}\) Then integrate this equation: \(\Delta W = \int_{\alpha_0}^{\alpha_1} K\alpha^{-\mu} d\alpha\) The two alphas you want to find might will be determined from atmospheric readings of different pressure levels...it's generally conserved if moving downwards in the atmosphere. \(\alpha_n = \frac{R_d T_n}{p_n}\) This is in (small) part the reason that rotating storms with significant down-welling lose energy over time.
