# Navier Stokes - Euler's Approximation

I am working on a question from my practice exam. We are asked if the following equation is a valid expression of Euler's equation - an approximation to Navier Stokes for high Reynold's number.

$$\cfrac{D\rho V}{Dt} = -\nabla P + \rho g$$

I don't believe so. The Navier stokes equation should be written as:

$$\rho \cfrac{DV}{Dt} = -\nabla P + [\nabla \cdot \tau] + \rho g$$

This equation simplifies to the following in the case of high Reynold's number.

$$\rho \cfrac{DV}{Dt} = -\nabla P + \rho g$$

However, a careful derivation of Navier Stokes indicates that the rho must exist outside of the substantive derivative. Therefore, it is my belief that the equation we are given is distinctly wrong. Is my intuition correct here?

• For incompressible flow, there is no difference. For compressible flow, you need to include the $\rho$ in the derivative, because this equation is conservation of momentum density (compare to Newton's law $d(mv)/dt=F$). But then you also need the continuity equation to complete the set! So in the given context, both expressions are equivalent (we almost always assume incompressible flow unless stated differently). – orion Nov 21 '16 at 10:19

## 1 Answer

There is a common misconception regarding if $\rho$ should be in the material derivative or outside, and does that restrict the flow to incompressible. Consider the following momentum equation, $$\frac{\partial (\rho \vec{V})}{\partial t} + \nabla \cdot (\rho \vec{V} \vec{V}) = -\nabla p + \nabla \cdot \bar{\tau} + \rho \vec{g}$$ Expanding $\nabla \cdot (\rho \vec{V} \vec{V})$, $$\nabla \cdot (\rho \vec{V} \vec{V}) = (\nabla \cdot \rho \vec{V})\vec{V} + \rho \vec{V} \cdot \nabla \vec{V}$$ Substituting, $$\frac{\partial (\rho \vec{V})}{\partial t} + (\nabla \cdot \rho \vec{V})\vec{V} + \rho \vec{V} \cdot \nabla \vec{V} = -\nabla p + \nabla \cdot \bar{\tau} + \rho \vec{g}$$ Rearranging this expression, $$\rho \frac{\partial \vec{V}}{\partial t} + \vec{V}\frac{\partial \rho}{\partial t} + (\nabla \cdot \rho \vec{V})\vec{V} + \rho \vec{V} \cdot \nabla \vec{V} = -\nabla p + \nabla \cdot \bar{\tau} + \rho \vec{g}$$ Regrouping terms, $$\rho \left( \frac{\partial \vec{V}}{\partial t} + \vec{V} \cdot \nabla \vec{V} \right) + \vec{V}\left(\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{V})\right) = -\nabla p + \nabla \cdot \bar{\tau} + \rho \vec{g}$$ Now from the general continuity equation we have, $$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{V}) = 0$$ Hence, $$\rho \left( \frac{\partial \vec{V}}{\partial t} + \vec{V} \cdot \nabla \vec{V} \right) = -\nabla p + \nabla \cdot \bar{\tau} + \rho \vec{g}$$ Notice, by manipulation of the equations and making use of the continuity equation (for a compressible flow), we are able to show that for both incompressible and compressible flows, $$\rho \frac{D\vec{V}}{Dt} = -\nabla p + \nabla \cdot \bar{\tau} + \rho \vec{g}$$ It is noted here that the top equation is the divergence form of the momentum equation, while the bottom equation is the convective form. However, they both apply for a compressible fluid.