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I am trying to derivative the equations from [1] for a compressible Volume-of-fluids formulation but I am stuck in one of the last steps and would like to request some help to solve it.

The governing equations are:

Mass conservation: $$ \frac{\partial (\alpha_k \rho_k)}{\partial t} + \nabla \cdot (\alpha_k \rho_k \vec{v}) = 0 \: \:, k={1,2} \: \:\:\: (1) $$

Phase transport: $$ \frac{\partial (\alpha_1)}{\partial t} + \nabla \cdot (\alpha_1 \vec{v}) = 0 \:\:\:\: (2) $$

With

$$ \alpha_1 + \alpha_2 = 1 \:\:\:\: (3) $$

I can take equation (1) and expand it into:

$$ \alpha_k \left [\frac{\partial (\rho_k)}{\partial t} + \vec{v} \cdot \nabla(\rho_k) \right ] + \rho_k \left[ \frac{\partial (\alpha_k)}{\partial t} + \nabla \cdot (\alpha_k \vec{v}) \right ] = 0 $$

$$ \frac{\partial (\alpha_k)}{\partial t} + \nabla \cdot (\alpha_k \vec{v}) = -\frac{\alpha_k}{\rho_k} \left [\frac{\partial (\rho_k)}{\partial t} + \vec{v} \cdot \nabla(\rho_k) \right ] \:\:\:\: (4) $$

Given that $\rho$ depends on $p$, and that $\frac{\partial \rho_k}{\partial p_k} = \psi_k$, through the chain derivation rule I get:

$$ \frac{\partial (\alpha_k)}{\partial t} + \nabla \cdot (\alpha_k \vec{v}) = -\frac{\alpha_k \psi_k}{\rho_k} \left [\frac{\partial (p)}{\partial t} + \vec{v} \cdot \nabla(p) \right ] \:\:\:\: (5) $$

Summing equation (5) for both phases, and attending to equation(3), I end up with:

$$ \nabla \cdot \vec{v} = - \left( \frac{\alpha_1 \psi_1}{\rho_1} + \frac{\alpha_2 \psi_2}{\rho_2} \right )\left [\frac{\partial (p)}{\partial t} + \vec{v} \cdot \nabla(p) \right ] \:\:\:\: (6) $$

Here is the part I cannot yet grasp:

The end result should be:

$$ \frac{\partial \alpha_1}{\partial t} + \nabla \cdot (\alpha_1 \vec{v}) - \alpha_1 (\nabla \cdot \vec{v}) = \alpha_1(1-\alpha_1) \left( \frac{\psi_2}{\rho_2} - \frac{\psi_1}{\rho_1} \right ) \left [\frac{\partial (p)}{\partial t} + \vec{v} \cdot \nabla(p) \right ] $$

If I expand equation (2), I can get:

$$ \frac{\partial (\alpha_1)}{\partial t} + \alpha_1 (\nabla \cdot \vec{v}) + \vec{v} \cdot \nabla \alpha_1= 0 $$

Moving $\alpha_1 (\nabla \cdot \vec{v})$ to the RHS and replacing the velocity divergence with Equation (6), I get:

$$ \frac{\partial (\alpha_1)}{\partial t} + \vec{v} \cdot \nabla \alpha_1 = -\alpha_1 \left (- \left( \frac{\alpha_1 \psi_1}{\rho_1} + \frac{\alpha_2 \psi_2}{\rho_2} \right )\left [\frac{\partial (p)}{\partial t} + \vec{v} \cdot \nabla(p) \right ] \right) $$

Knowing that: $$ \frac{\partial (\phi)}{\partial t} + \nabla \cdot (\vec{v} \phi) = \frac{\partial (\phi)}{\partial t} + \phi (\nabla \cdot \vec{v}) + \vec{v} \cdot \nabla \phi $$

$$ \frac{\partial (\phi)}{\partial t} + \nabla \cdot (\vec{v} \phi) - \phi (\nabla \cdot \vec{v}) = \frac{\partial (\phi)}{\partial t} + \vec{v} \cdot \nabla \phi $$

I can replace the LHS with:

$$ \frac{\partial (\alpha_1)}{\partial t} + \nabla \cdot( \alpha_1 \vec{v} ) - \alpha_1(\nabla \cdot \vec{v}) = -\alpha_1 \left (- \left( \frac{\alpha_1 \psi_1}{\rho_1} + \frac{\alpha_2 \psi_2}{\rho_2} \right )\left [\frac{\partial (p)}{\partial t} + \vec{v} \cdot \nabla(p) \right ] \right) \: \: \: \: (7) $$

The question is: How can I get the RHS correctly?

Best Regards!

[1] - https://doi.org/10.1016/j.compfluid.2013.04.002

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  • $\begingroup$ Have you asked the authors of that paper? $\endgroup$ Commented Nov 3, 2021 at 4:47

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You have a problem with the equation after "the end result should be", which seems clear because you don't have any mixed terms involving $\alpha_i$ and $\rho_j$ with $i \ne j$. They get these when going from (15) to (16) in the paper. For example, taking (15) from the paper and substituting $$\frac{\alpha_1 \rho_{1,p}}{\rho_1} + \frac{\alpha_2 \rho_{2,p}}{\rho_2} \to \alpha_1\alpha_2 \left( \frac{\rho_{1,p}}{\alpha_2 \rho_1} + \frac{\rho_{2,p}}{\alpha_1 \rho_2} \right) \to \dots$$ and you see these terms. One thing I do find confusing in their formulation is the use of both $\psi_k$ and $\rho_{k,p}$...

One sees the same situation unfolding in the original Kapila model: https://aip.scitation.org/doi/pdf/10.1063/1.1398042

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