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heat capacity in TEqn

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December 1, 2022, 06:11	heat capacity in TEqn	#1
newGuyAtCFD New Member Join Date: Sep 2022 Posts: 18 Rep Power: 4	Hi, Can anyone provide some insight on this: VoF solvers weigh some properties as: $\rho = \rho_1\alpha_1 + \rho_2\alpha_2$ $\mu = \mu_1\alpha_1 + \mu_2\alpha_2$ Whislt the heat capacity appears to be different. $\left(\frac{\alpha_1}{Cv_1} + \frac{\alpha_2}{Cv_2}\right) \left( \nabla \cdot \left(\vec{v} p \right) + \frac{\partial \left(\rho K \right)}{\partial t} + \nabla \cdot \left(\rho \vec{v} K \right) - K \left[ \frac{\partial \rho}{\partial t} + \nabla \cdot \left(\rho \vec{v}\right)\right] \right)$ Why is not being weighted as : $C_v = C_{v_1} \alpha_1 + C_{v_2} \alpha_2$ And then: $\frac{1}{C_v}\left( \nabla \cdot \left(\vec{v} p \right) + \frac{\partial \left(\rho K \right)}{\partial t} + \nabla \cdot \left(\rho \vec{v} K \right) - K \left[ \frac{\partial \rho}{\partial t} + \nabla \cdot \left(\rho \vec{v}\right)\right] \right)$ In the laplacian term $\alpha_{eff}$ appears to be treated as : $\alpha_{eff} = \alpha_{eff_1} \alpha_1 + \alpha_{eff_2} \alpha_2$ What is the rationale here? I need to weigh the inverse of since that is how the formulation is derived? For instance, If I removed density from inside the parenthesis and diveded every term in the equation with $\rho$ , I would need to use the same treatment as ?

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