# Calculate the work for moist air adiabatic expansion and compression?

( During expansion some water vapor condenses and is removed from the system. )

More details:

Initial conditions: t 25C, p 1013 mB, humidity 100%, then expansion to ~ 450 540 mB through the double chamber like this: 1drv.ms/u/s!AjD7S4pRBfU9oD6Ysv1qUhOxe9pb I have the description of calculations for the first stage - expansion here: 1drv.ms/w/s!AjD7S4pRBfU9oDz9lYlSudWLj93y but I do not see there the latent heat. I have no description of calculations for compression part.

Most basic question is: what will happen to latent heat energy after expansion - compression - will the air have the higher temperature after that?

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Feb 17 at 16:36

For the adiabatic reversible expansion, the change in entropy per unit mass of the flowing stream is zero. Take as a basis 1 lb of air (on a dry basis), initially at 1 Bar, 25 C, and 100% relative humidity. The absolute humidity A represents the number of pounds of water vapor per pound of dry air in a stream. For the inlet stream to the expansion section, $$A_{in}=\frac{18p^*(25)}{29(p_{in}-p^*(25))}\tag{1}$$where $$p^*(25)$$ represents the absolute vapor pressure of water at 25 C and $$p_{in}$$ is the inlet pressure (1013.5 mB). For the outlet stream from the expansion section, $$A_{out}=\frac{18p^*(T)}{29(p_{out}-p^*(T))}\tag{2}$$where $$p_{out}=450\ mB$$. The amount of condensed water per pound of dry air is: $$C=A_{in}-A_{out}\tag{3}$$ The change in partial mass entropy of the air (on a dry basis) is $$\Delta s_{air}=\left[C_{v, air}\ln{\left(\frac{T}{298}\right)}-R\ln{\left(\frac{(p_{out}-p^*(T))}{(p_{in}-p^*(25))}\right)}\right]/29\tag{4}$$The change in mass partial mass entropy of the water vapor is $$\Delta s_{vapor}=\left[C_{v, vapor}\ln{\left(\frac{T}{298}\right)}-R\ln{\left(\frac{p^*(T)}{p^*(25)}\right)}\right]/18\tag{5}$$The change in partial mass entropy of the water that becomes condensate is given by: $$\Delta s_{cond}=\left[-\frac{[\Delta H_{vap}(25)]}{298}+C_{v,liq}\ln{\left(\frac{T}{298}\right)}\right]/18\tag{6}$$where $$\Delta H_{vap}(25)$$ is the molar heat of vaporization of water at 25 C. If the expansion is assumed to be adiabatic and reversible, then $$\Delta s_{air}+A_{out}\Delta s_{vapor}+C\Delta s_{cond}=0\tag{7}$$ These equations would be solved simultaneously for the outlet temperature T.