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Euler equations & expansion shocksHi Guys,
I'm reading through some literature on transonic flows and came across a paper that says (I'm paraphrasing) that because the Euler equations do not contain a viscous term that expansion shocks are possible - whereas they are not in the Navier Stokes equations. Why is this? I asked a colleague of mine and he said it is because if you replace 'u' with '-u' in the Euler equations you get the same answer (which apparently means they're reversible). I'm afraid I don't understand the logic of this. If someone could please shed some light on this matter I'd be very grateful. T |

I am assuming he is referring to the shock polar? You plot the shock polar and on the same graph you plot the incident flow/shock and this intersects with the shock polar in three places. The first two intersection correspond to the strong and weak shock solution and the third intersects the "tail" of the shock polar which gives an "expansion shock". The entropy for this shock decreases across the shock and hence is non-physical.
Or are you referring to rarefaction fans? Mat |

Thanks for your reply Mat,
I'm not referring to the shock polar - I guess what I'm really asking about is the reversibility of the Euler equations. I'm dealing with transonic flow over an airfoil which are terminated by weak shocks. In some formulations, solutions of the Euler (and related) equations can admit expansion shocks. I've been told that this is because the Euler equations are reversible. The Euler equations don't 'know' that these shocks are non-physical unless you use some sort of entropy boundary condition. The idea as I understand it, is that if we take the x-momentum equation and change the sign of each individual velocity term (u,v,w) then we're essentially reversing the direction of time in the equation (i.e. the flow is running backwards). I can partially understand this idea, the temporal term will change sign but the velocity terms (because they each contain two velocities - whose signs have both been +/- changed) will stay the same as the original equation. This is like making time run backwards - if this is true then the Euler equations can admit expansion shocks. If the velocity decreases across a shock then it will increase across the shock in the reversed equations. Apparently this does not apply in the NS equations because of the viscous term - which contains a lone velocity component which will change sign along with the velocity. We can't reverse time on the NS equations and get the same solution. I understand the basic concept of this but am confused by one thing: what happens to the pressure terms? the dp/dx terms would surely remain unchanged by this time-reversal and the idea breaks down? If someone could shed some light on this I'd be very grateful. T |

Hi,
I understand what you're saying now, I still think that the shock polar has some relevance to your question, that the expansion shock solution in the shock polar shows the time reversing think which you're on about. If you reverse the velocity direction that wouldn't this change the direction of force on a solid body (like your aerofoil) and therefore change the sign of the pressure? Mat |

Reversible flowThat's what I've been thinking too - it's the only way that the argument can work out.
It makes some degree of sense. For inviscid flow (going from left to right) over a symmetrical airfoil, say, where the nose is pitched upwards there's lower pressure over the upper surface than the lower surface - this generates lift. Reverse the flow direction (so it's now going from right to left) and it now sees an airfoil pitched nose downwards - so it will generate negative lift. Basically if we reverse the flow direction we get the opposite effect - I guess this is what my colleague means when he says the Euler eqns admit expansion shocks - simply run the flow backwards and any shocks that occur will be expansion shocks. The viscous term in the NS equations destroys this property. Me thinks we have cracked this one :o) |

Smooth solutions of euler equations are reversible in the sense that if you reverse the velocity of all fluid particles, they will exactly retrace their path in the reverse direction.
When the flow is not smooth, then the PDE does not apply. You have to use the weak formulation which contains the Rankine-Hugoniot jump conditions, and an irreversible increase of entropy. Then the flow is not reversible. So a shock is correct according to second law of thermodynamics if there is increase in entropy as fluid particles cross the shock. If the entropy decreases, then we say it is an expansion shock, which is not physically correct since it violates second law of thermodynamics. |

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