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Continuity and Navier-Stokes equation for fluid are, \begin{eqnarray} \frac{\partial \rho}{\partial t} + \nabla\cdot (\rho \mathbf{u}) &=& 0 \\ \rho\left(\frac{\partial \mathbf{u}}{\partial t} + \mathbf{u}\cdot \nabla u \right) &=& -\nabla p + \nabla \cdot\sigma + \mathbf{{F}_{ext}}, \end{eqnarray}

where letters have the usual meaning. In total we have 4 equations (one continuity equation and 3 momentum balance components of Navier-Stokes) available and number of unknowns are 5 (pressure+ density, 3 velocity components). How can we then determine all the 5 variables in general?

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  • $\begingroup$ Good question! For incompressible flows this becomes even worse; then an equation of state cannot provide an additional equation. The pressure field must follow from the velocity field and vice versa simultaneously. Solving this issue is the foundation of incompressible flow solvers. $\endgroup$ – nluigi Nov 3 '15 at 16:50
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The missing equation is energy conservation $$ \frac{\partial}{\partial t} {\cal E} + \vec\nabla \cdot\vec\jmath^{\,\cal E}=0 $$ where ${\cal E}$ is the energy density and $\vec\jmath^{\,\cal E}$ is the energy current $$ \vec\jmath^{\,\cal E} = \vec{u}\left[ {\cal E}+P \right] -\eta u\cdot\sigma-\kappa\vec\nabla T\, . $$ Now the equations close if you have an equation of state, $P=P({\cal E}^0,\rho)$, where ${\cal E}^0={\cal E}-\frac{1}{2}\rho u^2$ is the internal energy density. Note that the equation of state also fixes $T({\cal E}^0,\rho)$ using thermodynamic identities (although this is tedious in practice; for a non-interacting gas things are simple, $T=mP/\rho$). The energy equation can be rewritten in various ways, for example as an equation for entropy production.

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  • $\begingroup$ With energy equation a new variable $T$ is added. Equation of state seems the missing link. $\endgroup$ – alekhine Nov 3 '15 at 14:23
  • $\begingroup$ A one component fluid has only two independent thermodynamic variables. Once the equation of state $P({\cal E},\rho)$ is given, $T$ follows from thermodynamic identities. The equation of energy conservation is crucial, because you cannot compute $P$ from $\rho$ alone. $\endgroup$ – Thomas Nov 3 '15 at 14:41

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