4
$\begingroup$

Conservation of momentum?

I am considering the Euler equations in conservative form and solving the Sod shock tube problem I have written a Godunov finite volume type solver. It solves for density ρ, momentum ρu, and total energy E; therefore, I would expect all of these quantities to be conserved wrt time. Density and total energy are indeed conserved, however, momentum is not. The Euler equations are given by $$ \frac{\partial}{\partial t}\begin{bmatrix} \rho \\ \rho u \\ E \end{bmatrix} + \frac{\partial}{\partial x}\begin{bmatrix} \rho u \\ \rho u^2 + p \\ u(E +p) \end{bmatrix} = 0 $$

where pressure is related to the conserved quantities by $ p = (\gamma - 1)(E - 0.5 \rho u^2) $

The Sod shock problem splits the domain into two regions separated by a density and pressure discontinuity with initial velocity zero. That is, $\rho_L = 1, \rho_R = .125; p_L=1, p_R =.1; u_L=u_R =0.$ These initial conditions imply that ρu=0, therefore, momentum should be zero throughout the simulation.

The solution profiles are well known and can be found here. We see that ρ >0 and u > 0, therefore, there is no way for momentum ρu = 0 (which it should be from the intial conditions). As a result, I do not even see why it is reasonable to expect that momentum would be conserved. A paper by Sod which surveys some methods for solution, on page 20, list a table which shows momentum to be increasing linearly. I generally do not work in this area, so maybe I am missing something basic. Can anyone shed some light on this? Thanks!

$\endgroup$
  • $\begingroup$ The imbalance of pressure in the initial condition is being converted into momentum. $\endgroup$ – Joce Jul 1 '14 at 7:15
  • 1
    $\begingroup$ Note that the code used to display the animation is not very stable and there a lot of numerical instabilities in these simulations which are non-physical. Just something to be aware of... $\endgroup$ – MoonKnight Jul 9 '14 at 9:26
6
$\begingroup$

In hydrodynamics, conservation means that what flows into the control volume is equivalent to the flow out of the control volume. With respect to momentum, we mean precisely that

any change in momentum of the fluid within a control volume is due to the net flow of fluid into the volume and the action of external forces on the fluid within the volume (source)

Formally, this is mathematically described using the integral formulation, $$ \frac{\partial}{\partial t}\iiint_V\rho \mathbf u\,dV=-\oint_S\left(\rho \mathbf u\,d\mathbf S\right)\mathbf u-\oint_S p\,d\mathbf S+\iiint_V\rho\mathbf f_{body}\,dV+\mathbf F_{surf} $$ where $\mathbf f_{body}$ are body forces and $\mathbf F_{surf}$ are surface forces, $d\mathbf S$ is the cell surface and $dV$ its volume; all other variables take their normal meaning.

In the case of Eulerian hydrodynamics, $\mathbf f_{body}=\mathbf F_{surf}=0$. Then we can use the divergence theorem to obtain (for a 1D flow), $$ \frac{\partial\rho u}{\partial t}+\frac{\partial}{\partial x}\left(\rho u^2+p\right)=0 $$ So this equation does contain the conservation law, just not quite how you were expecting it. You may also want to read over the Wikipedia article on the Rankine-Hugoniot jump conditions, as this might explain a bit more clearly the quote I give at the top.

$\endgroup$
  • $\begingroup$ The integral formulation allows to see how the source of momentum growth is the action of different pressures at the solid walls at the ends of the tube. The initial pressure imbalance is converted into momentum. $\endgroup$ – Joce Jul 10 '14 at 8:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.