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( 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?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Feb 17 at 16:36
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This assumes that the expansion takes place adiabatically and reversibly and that the change in enthalpy of the stream is counterbalanced by the change in kinetic energy of the stream.

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.

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  • $\begingroup$ Intuitively, 450 mB seems like a low enough pressure to reduce the temperature below 0 C. This would result in ice rather than water. $\endgroup$ – Chet Miller Feb 18 at 0:23
  • $\begingroup$ Yes, indeed it should be 540 mB, I changed the description. Thank you for your formulas. $\endgroup$ – Aleksey Feb 18 at 4:41

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