# Derivation of compressible volume-of-fluids formulation

I am trying to derivative the equations from  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!

• Have you asked the authors of that paper? Nov 3, 2021 at 4:47
• Hi. As a matter of fact, I did. But have not received any reply. Nov 3, 2021 at 13:50

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 $$\rho_{k,p}$$...