Considerations for phase change through throttle valve

Question

I am looking at characterizing, as well as I can, the exit velocity and density downstream of an isenthalpic throttle valve as a function of the downstream pressure. What is throwing me off is the fact that I am seeking complete vaporization of the working fluid (hydrogen for now). The upstream conditions consist of a known temperature and pressure such that the fluid is either saturated or compressed liquid.

My questions:

1. Do I know that I don't need to worry about compressibility effects (e.g. choked flow) inside the valve because the working fluid starts as a liquid?
2. Does phase-change violate any of the prerequisite assumptions for Bernoulli's equation?
3. Can I just use ideal gas law downstream of the throttle valve (what justification do I have for doing so) or do I need a more robust equation of state (e.g. Van der Waals)?
4. How do I account for the Joule-Thompson effect?
5. How can I be sure that I vaporize the entirety of the working fluid passing through the valve? From Wikipedia:

I can just set the quality to unity which to me indicates that $T_d >= T_u + H_vc_p^{-1}$

(I should point out that I'm not looking explicitly for $\rho(P_d)$ - a transcendental, under-constrained relation is fine... I have another equation for $P_d(\rho)$ that is based on a design constraint and numerical methods are fair game). My thought was to do the following:

$P_d = \frac{T_d - T_u}{\mu_{JT}(T_u)} + P_u$ (based on definition of $\mu_{JT}$)

Then since $T_d=\left(P_d + \frac{a}{\frac{M}{\rho}}\right)\left(\frac{\frac{M}{\rho}-b}{R}\right) >= T_u + H_vc_p^{-1}$ (based on Van der Waals EOS), I can make the substitution into the first equation to get $P_d(\rho)$. Is this reasonable?

As for obtaining the fluid velocity, I am at a complete loss if Bernoulli's equation is off the table. However, I noticed that I could alter Bernoulli's (acceptable only from a unit analysis perspective) like:

$\frac{P_1}{\rho_1} + \frac{1}{2}u_1^2 + h_v = \frac{P_2}{\rho_2} + \frac{1}{2}u_2^2$

where $h_v$ is the enthalpy of vaporization.

Edit: I definitely can't use the definition of the Joule-Thompson coefficient because the coefficient is different for a gas than it is for a liquid.

Answer 1

Declaring a valve isenthalpic involves the assumption that the change in enthalpy of the fluid is much larger than its change in kinetic energy. This assumption can be checked a'priori.

The changes of enthalpy of a single phase of material is described by $$dh=C_pdT+\left[v-T\left(\frac{\partial v}{\partial T}\right)_P\right]dP$$ where h is the specific enthalpy and v is the specific volume. If you specify a reference state in the ideal gas region, say $h(T_0,0)$ you can first integrate the first term in above equation with respect to T along a path of constant pressure ($P\approx 0$) in the ideal gas region, and then subsequently integrate the second term with respect to pressure at constant temperature to obtain h at any specified temperature and pressure. If there is a phase change, that can be included by adding or subtracting the heat of vaporization at the equilibrium vapor pressure and equilibrium temperature.

One way or another, you are going to have to be able to express the enthalpy as a function of temperature and pressure that is consistent mathematically for compressed liquid, saturated liquid, saturated vapor, and superheated vapor. How you do this depends on the accuracy you desire and the thermodynamic data available.