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Consider you have an adiabatic container with a piston. At the moment, the the container has liquid water and air, of known mass, in the headspace at 1 bar and 298 K. Suppose I perform some PV work by pushing the piston down and increase pressure to 10 bar. I just did some adiabatic compression, so the system will be at some elevated temperature as well. I want to get a handle on the thermodynamic properties of the system. My questions are:

Can I say the process is isentropic since no heat is being lost to the surroundings? I need one more constraint apart from knowing the pressure ratio to define the system, right?

This is something I need clarification on. Back in undergrad, we dealt with psychrometric charts. Air has a certain capacity to hold water at a given temperature. My question is, will the vaporization of water be limited by this fact? Given the temperature we are at, the air in the headspace can only hold so much water? Or do I not worry about that and do my energy balance, H1 + W = H2, where H1 is the enthalpy at state 1, W is the work performed on the system and H2 is the enthalpy at the second state?

Essentially, can I have a mixture of A kg of air and B kg of steam at a given temperature, even though the psychrometric chart says that air can hold only C kg of steam (C < B)?

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  • $\begingroup$ Air does not act like a sponge. The relevant parameters are the partial pressure of water and its saturation pressure. $\endgroup$
    – user137289
    Oct 13, 2019 at 9:57

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If you are asking whether you can determine the final thermodynamic state of the system, the answer is yes, provided you know the details of how the external pressure is applied as a function of the volume. The process is going to be isentropic if the added external pressure is applied gradually and in tiny pressure steps. If it is applied rapidly, then the process will not be irreversible, and the entropy of the system will increase. For example, if the external pressure is suddenly increased to a new constant value, the work done on the system will be $-P_{ext}\Delta V$, where $\Delta V$ is negative, and the compression will be irreversible.

Your equation for the energy balance is incorrect. It should involve the change in internal energy $\Delta U$ rather than the change in enthalpy, $\Delta H$. If you are willing to assume that up to 10 bars, the gas phase mixture can be modeled as an ideal gas and that the dissolution of air in the liquid water is negligible, this problem can be solved rather easily either for the isentropic case or for the constant external pressure increase case.

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  • $\begingroup$ Hi Chet. But regarding last paragraph of OP he should be using steam tables, not pyschrometric chart right? The chart only applies to atmospheric air, not 10 bar. That was something I intended to point out in an answer but hadn't completed it yet. $\endgroup$
    – Bob D
    Oct 13, 2019 at 13:07
  • $\begingroup$ Yes, the psychometric chart should not be used. The steam tables can be used, or, if some minor additional assumptions are made, the problem can be solved even without the use of the steam tables (if we know heat capacities, heat of vaporization, and vapor pressure vs temperature relationship). $\endgroup$ Oct 13, 2019 at 13:10
  • $\begingroup$ I was also thinking about referring to use of ideal gas law, but felt uncomfortable about doing so when talking about steam, which is one of the reasons I didn't submit an answer. $\endgroup$
    – Bob D
    Oct 13, 2019 at 13:14
  • $\begingroup$ @BobD what do you think the partial pressure of the water vapor is going to be at the final state? $\endgroup$ Oct 13, 2019 at 14:04
  • $\begingroup$ For saturated water, 10 bar- no? $\endgroup$
    – Bob D
    Oct 13, 2019 at 14:26

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