How to modify single phase fluid/solid coupled PDEs to account for a phase change?

Question

I have two PDEs that model both fluid and solid temperature change due to fluid flow through a packed bed. Schematic and equations here (where the f and s subscripts are for the fluid and solid respectively):

It seems these PDEs are limited to a single fluid phase (either gas or liquid) and do not account for any potential phase change (there is no phase change term). As far as I understand, this means that these equations can be used to model gas or liquid flow through a packed bed, but cannot be used where a phase change occurs.

My question is how could these equations be modified or used to account for a potential phase change occurring? Ideally rather than modelling a gas flow through a packed bed, I would like to model a gas flow through a cold packed bed causing liquefaction of the gas.

In addition, if there are any publications that model this I would love to see them. So far I've only found single phase models.

Link to source: https://www.sciencedirect.com/science/article/pii/S0306261921008138

Answer 1

I'm going to present the limiting solution for a much simpler case, and then you can see if you can modify it for the case you are interested in, in which there is a phase change.

At time zero, I have a bed containing a liquid, both of which are at $T_0$. After time zero, I start flowing liquid at temperature $T_1$ through the bed at pore velocity u. For this case, my basic starting equations are $$\epsilon \rho_fC_{pf}\left[\frac{\partial T_f}{\partial t}+u\frac{\partial T_f}{\partial x}\right]=ha_s(T_s-T_f)$$and$$(1-\epsilon) \rho_sC_{ps}\frac{\partial T_s}{\partial t}=ha_s(T_f-T_s)$$If we add these two equations together, we obtain:$$\epsilon \rho_fC_{pf}\frac{\partial T_f}{\partial t}+(1-\epsilon) \rho_sC_{ps}\frac{\partial T_s}{\partial t}+\epsilon \rho_fC_{pf}u\frac{\partial T_f}{\partial x}=0$$In our limiting situation, the liquid and solid bed temperatures will approach one another both behind- and ahead of the wave front. Therefore, in this limiting situation, we can write:$$\epsilon \rho_fC_{pf}\frac{\partial T}{\partial t}+(1-\epsilon) \rho_sC_{ps}\frac{\partial T}{\partial t}+\epsilon \rho_fC_{pf}u\frac{\partial T}{\partial x}=0$$or$$\frac{\partial T}{\partial t}+\frac{\epsilon \rho_fC_{pf}}{\epsilon \rho_fC_{pf}+(1-\epsilon) \rho_sC_{ps}}u\frac{\partial T}{\partial x}=0$$The solution to this is a sharp wave, for which $T=T_0$ for $x>Vt$ and $T=T_1$ for $x<Vt$, where the wave velocity V is given by $$V=\frac{\epsilon \rho_fC_{pf}}{\epsilon \rho_fC_{pf}+(1-\epsilon) \rho_sC_{ps}}u$$ Of course, as expected, according to this, because of the thermal inertia of the bed, the wave front is traveling much more slowly than the fluid.

Answer 2

Well, if you know the pressure, then when you reach the temperature at which this is the equilibrium vapor pressure, you start forming a liquid phase. After that, you have two phases flowing, and the enthalpy per unit mass of this mixture is $h_Lf+h_V(1-f)$, where f is the mass fraction liquid. Rather than temperature varying in this region (for the flowing phase), f will be varying with time and position, and T will be constant (until all the vapor has condensed).