# Fluid dynamics equations, number of variables and number of equations

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?

• 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. Commented Nov 3, 2015 at 16:50
• Related: physics.stackexchange.com/q/658078/226902 (energy equation). See also: en.wikipedia.org/wiki/… (First law of thermodynamics in fluid mechanics). Commented Mar 6, 2023 at 16:19
• See also: "Derivation of continuum expression of the first law of thermodynamics" physics.stackexchange.com/q/265026/226902 and "Forms of the first law of thermodynamics" physics.stackexchange.com/q/67163/226902 Commented Jun 25, 2023 at 12:51

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.

• With energy equation a new variable $T$ is added. Equation of state seems the missing link. Commented Nov 3, 2015 at 14:23
• 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. Commented Nov 3, 2015 at 14:41
• For entropy production, see Clausius–Duhem inequality, a way of expressing the second law in continuum mechanics (also useful in determining whether the constitutive relation of a material is thermodynamically allowable). Commented Jul 26, 2023 at 15:18

Assuming we want to calculate "standard" fluid dynamics - compressible single phase flow in 3D space - the variables to solve are:

• Pressure
• Density
• Temperature
• Velocity (3 components in 3 dimensions)

a total of 6 components.

The equations for conservative states are:

• continuity / mass conservation - one component)
• momentum conservaiton (3 components - in 3D, more general 1 component per dimension)
• energy conservation

a total of 5 equations, and (as pointed out before) the 6th equation will be the equation of state (could be ideal gas law..)

Depending on the task at hand, the total number of equation and states will change:

• For lower dimensions (1D or 2D), the velocity vector reduces components by the same amount as the momentum equation does.
• Incompressible flow removes one state and one equation
• For more sophisticated calculations, often extra equations and states are added, e.g. for turbulence modelling (see k-epsilon), multiphase flow, or when adding chemistry