changes.mady.by.user Arthur Aslanyan (Nafta College)
Saved on Nov 10, 2021
\frac{\partial (\rho_A \, \phi) }{\partial t} + \nabla (\rho_A \, {\bf u}_A) = 0, \quad A = \{ O, G, W \}
\int_V \, \frac{\partial (\rho_A \, \phi) }{\partial t} \, dV = - \int_V \, \nabla (\rho_A \, {\bf u}_A) \, dV = - \int_{\partial V} \, \rho \, {\bf u}_A \, d {\bf A}
V \cdot \delta (\rho_A \, \phi) = \delta \, m_A
V \cdot \delta \left( \phi \, \sum_\alpha \rho_{A,\alpha} \, s_\alpha \right) = \mathring{\rho}_A \cdot \delta \, q_A, \quad \alpha = \{ o, g, w \}
\delta \left( \phi \, \sum_\alpha \frac{\rho_{A,\alpha}}{\mathring{\rho}_A} \, s_\alpha \right) = V^{-1} \cdot \delta \, q_A
\delta \left( \phi \, \sum_\alpha \frac{\mathring{V}_{A,\alpha}}{V_\alpha} \, s_\alpha \right) = V^{-1} \cdot \delta \, q_A
\delta \left( \phi \, \sum_\alpha \frac{\mathring{V}xi_{A,\alpha}}{V_\alpha} \, s_\alpha \right) = V^{-1} \cdot \delta \, q_A
\xi_{A,\alpha} = \frac{\mathring{V}_{A,\alpha}}{V_\alpha}
For the MBO fluid:
\xi_{A, \alpha} = \frac{\mathring{V}_{A,\alpha}}{V_\alpha}
\xi_{O, o} = \frac{\mathring{V}_{O, o}}{V_o} = \frac{1}{B_o}
\xi_{O, g} = \frac{\mathring{V}_{O, g}}{V_g} = \frac{\mathring{V}_{O, g}}{\mathring{V}_{G, g}} \cdot \frac{\mathring{V}_{G, g}}{V_g} = \frac{R_s }{B_g}
\xi_{G, o} = \frac{\mathring{V}_{G, o}}{V_o} = \frac{\mathring{V}_{G, o}}{\mathring{V}_{O, o}} \cdot \frac{\mathring{V}_{O, o}}{V_o} =\frac{R_s }{B_o}
\xi_{G, g} = \frac{\mathring{V}_{G, g}}{V_g} = \frac{1}{B_g}
\xi_{W, w} = \frac{\mathring{V}_{W, w}}{V_w} = \frac{1}{B_w}
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