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I am trying formulate an equation for enthalpy involving an isenthalpic throttle valve that completely vaporizes a liquid.

Here is what I have so far:

$\int_{T_1}^{T_{sat-vap}} c_{p,liq}(T)dT + h_{vap} = \int_{T_{sat-vap}}^{T_2} c_{p,gas}(T) dT$

According to https://www.linkedin.com/pulse/adiabatic-vs-isenthalpic-process-nikhilesh-mukherjee/, an isenthalpic process means that there is no work done and no heat transfer to/from the system thus $\partial Q=0$ and $\partial W=0$ (therefore no $PV$ terms).

So it really just becomes a sort of conservation of internal energy. The first term represents the required change in energy to get to the saturation vapor temperature under pressure $P_1$, the second term is the latent enthalpy of vaporization and the third term represents the required change in energy to get to temperature $T_2$ under pressure $P_2$.

Is this right?

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    $\begingroup$ Your equation should read: $$\int_{T_1}^{T_{sat-vap}} c_{p,liq}(T)dT + h_{vap} + \int_{T_{sat-vap}}^{T_2} c_{p,gas}(T) dT=0$$This neglects non-ideal gas effects. In the initial state, the pressure must at least be equal to the saturation pressure. Get a pressure-enthalpy diagram for water, and follow a constant enthalpy contour down to a lower pressure. $\endgroup$
    – Chet Miller
    Commented Sep 26 at 11:09
  • $\begingroup$ @ChetMiller so I uploaded a P-h diagram for hydrogen. I feel like I’m missing something obvious - if I draw a vertical line anywhere for constant enthalpy, there is no way to get on the other side of vapor dome without having started at an enthalpy that makes it a gas. Can that be right? Or do I need to shift the enthalpy by the amount $h_v$ as I drop the pressure? $\endgroup$
    – Sterling Butters
    Commented Sep 26 at 13:56
  • $\begingroup$ The diagram conclusion is correct. There is no way of getting 100% vapor. $\endgroup$
    – Chet Miller
    Commented Sep 26 at 14:26

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