Contrib/CompressibleMixingPhaseChangeFoam

Solver for two fluids with phase change (for example - water <---> steam), pressure and temperature density dependence Valid versions: 1 Model Equations Derivation

VALID OPENFOAM VERSION - 2.1.0

• Equation of state

Low-compressible fluid: $\rho = \rho_0 + \frac{\partial \rho}{\partial T} \Delta T + \frac{\partial \rho}{\partial p} \Delta p$

Ideal gas: $\rho=\frac{p}{(R/M)T}$

By combining this equations, we can get general relation: $\rho = \hat \rho + \frac{\partial \rho}{\partial p} \Delta p$

where $\hat \rho$ computed with respect to previous formulations

mixture density $\rho$ calculated as $\rho = \alpha_l \rho_l + \alpha_v \rho_v$

Here, and afterthere indices:

1,l,f - for liquid (heavy media with low compressibility)

2,g,s - for gas (light media (like steam) with big compressibility)

without index - mixture variable (or all variables local to some phase)

• Liquid volume transport

Let us consider transport of liquid (heavy phase) volume fraction $\alpha_l$: $\frac{\partial \alpha_l \rho_l}{\partial t} + \nabla \cdot \left ( \alpha_l \rho_l \mathbf{U} \right ) = \dot m_l$

By converting to volume fluxes we get: $\frac {\partial \alpha_l}{\partial t} + \nabla \cdot \left ( \alpha_l \textbf{U} \right ) = -\frac {\alpha_l}{\rho_l} \frac {d \rho_l}{dt} + \frac {\dot m_l}{\rho_l}$

Using equation of state, we can reformulate substantial derivative for density in terms of pressure for any phase: $\frac {d \rho}{dt}= \frac {d \hat \rho} {dt} + \psi \frac {dp} {dt} + p \frac {d \psi}{dt}$

• General rule for converting from mass to volume fluxes in transport equation $\frac{\partial \alpha_i \rho_i \zeta}{\partial t} + \nabla \cdot \left ( \alpha_i \rho_i \textbf{U} \zeta \right ) = \rho_i \alpha_i \left ( \frac {\partial \zeta}{\partial t} + \nabla \cdot \left ( \textbf{U} \zeta \right ) \right ) + \zeta \alpha_l \frac {d \rho_i}{dt} + \zeta \rho_i \frac {d \alpha_i}{dt}$

• Momentum equation (velocity prediction) $\frac {\partial \rho \textbf{U}}{\partial t} + \nabla \cdot \rho \textbf{U} \textbf{U} = \nabla \cdot R^{Eff} - \nabla p + \rho \textbf{g}$

by substituting piezometric pressure $\hat p = p - \rho \textbf{g} \textbf{x}$ we get: $\frac {\partial \rho \textbf{U}}{\partial t} + \nabla \cdot \rho \textbf{U} \textbf{U} = \nabla \cdot R^{Eff} - \nabla \hat p - \textbf{g} \cdot \nabla \rho$

• Pressure equation obtained by summation of equation for volume phase fraction of liquid and gas phases: $\nabla \cdot \left ( \textbf{U} \right ) = - \left ( \frac {\alpha_1 \psi_1}{\rho_1} + \frac {\alpha_2 \psi_2}{\rho_2} \right ) \frac {d p}{d t} - \left ( \frac {\alpha_1}{\rho_1} \frac {d \psi_1}{d t} + \frac {\alpha_2}{\rho_2} \frac {d \psi_2}{d t} \right ) p - \left ( \frac {\alpha_1}{\rho_1} \frac {d \hat \rho_1}{d t} + \frac {\alpha_2}{\rho_2} \frac {d \hat \rho_2}{d t} \right )+ \dot m_1 \left ( \frac{1}{\rho_1} - \frac{1}{\rho_2} \right )$

• Energy equation

Energy equation for mixture temperature obtained from sum of energy equations for each phase. Consider internal energy equation for phase-1: $\frac{\partial \alpha_1 \rho_1 e_1} {\partial t} + \nabla \cdot \left ( \alpha_1 \rho_1 e_1 \textbf{U} \right ) + \nabla \cdot \textbf{q}_1 = -\alpha_1 p (\nabla \cdot \textbf{U} - \dot m_1 / \rho_1) + \dot m_1 e_1$

By combining equations of phases, we get internal energy balance for mixture in temperatures: $\frac{\partial T}{\partial t} + \nabla \cdot \left ( T \textbf{U} \right ) - T \nabla \cdot \textbf{U} -\frac{1}{\alpha_1 \rho_1 C_{v,1} + \alpha_2 \rho_2 C_{v,2}} (\nabla \cdot \kappa_1 \nabla T + \nabla \cdot \kappa_2 \nabla T) =$ $-\left ( \nabla \cdot \textbf{U} - \left(\frac{1}{\rho_1}-\frac{1}{\rho_2}\right) \dot m_1 \right ) \frac{p}{\alpha_1 \rho_1 C_{v,1} + \alpha_2 \rho_2 C_{v,2}} - \left ( \frac{\partial L}{\partial t} + \nabla \cdot (\textbf{U} L) - L \nabla \cdot \textbf{U} \right ) \frac{\alpha_2 \rho_2}{\alpha_1 \rho_1 C_{v,1} + \alpha_2 \rho_2 C_{v,2}}$

Where $L$ - latent heat of evaporation

• Linking liquid volume transport to pressure equation is done by introducing $+\alpha_1 \nabla \cdot \textbf{U}$ and $-\alpha_1 \nabla \cdot \textbf{U}$ at r.h.s of volume fraction balance equation. Then, $-\nabla \cdot \textbf{U}$ replaced by value from pressure equation $\frac {\partial \alpha_l}{\partial t} + \nabla \cdot \left ( \alpha_l \textbf{U} \right ) = \alpha_1 \nabla \cdot \textbf{U} + \dot m_1 \left (\frac{1}{\rho_1} - \alpha_1 \left (\frac{1}{\rho_1} - \frac{1}{\rho_2} \right ) \right ) +$ $\alpha_1 \alpha_2 \left ( -\frac{1}{\rho_1} \frac{d \hat \rho_1}{d t} + \frac{1}{\rho_2} \frac{d \hat \rho_2}{d t}\right ) +\alpha_1 \alpha_2 p \left ( -\frac{1}{\rho_1} \frac{d \psi_1}{d t} + \frac{1}{\rho_2} \frac{d \psi_2}{d t}\right ) +\alpha_1 \alpha_2 \frac{d p}{d t}\left ( -\frac{\psi_1}{\rho_1} + \frac{\psi_2}{\rho_2} \right )$

2 Implemented phase change models

• Phase change model Schnerr-Sauer
• Phase change model Zwart
• Phase change model Singhal
• Phase change model Merkle
• Phase change model Kunz

4 Tutorial cases

• Case #1 - cavitation on the hemispherical body

Problem description, initial and boundary conditions are shown on the figure

Volume fraction of liquid (water) is shown on the next figure Figure 1.2

• Case #2 - hot water flashing in the nozzle

Problem description, initial and boundary conditions are shown on the figure 2.1

Volume and mass fractions of vapour are shown on the figures 2.2 and 2.3 Pressure distribution is shown on figure 2.4 Velocity magnitude and Mach number for compressible case are shown on figures 2.5 and 2.6 Velocity magnitude for incompressible case is shown on the case 2.7 Temperature distribution is located of figure 2.8
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