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Old   March 5, 2008, 06:48
Default Entropy jump across shocks
  #1
Dave
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Hi Guys,

I'm writing a code to solve the transonic small disturbance code. The approximation I've made until now is that the shock jumps are small and entropy can be considered constant. I now want to extend the code to include the entropy jump across the shock. The entropy jump is included in the pressure terms, the standard pressure coefficient is corrected downstream of the shock to include the entropy effects. According to the paper I'm reading the pressure correction is

Cps=2*(s-s0)/( Cv*gam*(gam-1)*M0^2 ) {s0=upstream entropy, Cv=heat capacity, gam=Cp/Cv and M0 is upstream mach number}

I can easily incorporate this equation into my code by simply adapting the pressure coefficient downstream of the shock. My problem is that I have no idea where the expression actually comes from!

I have tried to derive the expression from several different angles and have had little success. I prefer not to include something like this within my own work unless I know its mathematical justification.

If someone could help me work out where this equation comes from I'd be hugely grateful.

Many thanks, Dave
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Old   March 5, 2008, 12:40
Default Re: Entropy jump across shocks
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Patrick
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It looks as if this is obtained from the Rankine-Hugoniot conditions (for the energy) throught the surface of discontinuity, using some thermodynamic relations. The R-H conditions are written for the mass, force, and energy. It is too lenghty to write this down here, but look into books such as Landau Lifshitz "Fluid Mechanics" , search for the term Rankine-Hugoniot - it is of course in the chapter on shocks. The Rankine conditions are just the conservation laws across the surface of discontinuity when neglecting viscosity. It seems here that some of the terms (e.g. downstream velocity) have been neglected (downstream velocity => 0).

That's from the top of my head, I might be wrong, this needs to be checked.
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Old   March 5, 2008, 21:22
Default Re: Entropy jump across shocks
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Patrick
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ok,.. got back after peeking into what I have... It definitely is derived from the 3 equations of the Rankine-Hugoniot conditions. It does not seem that one has to put the downstream velocity to zero, but just express it as a function of the other variables.

THe RH conditions are:

rho v is conserved (say =J)

p + rho*v*v is conserved (say =I)

v*v/2 + (5/2)*p/rho is conserved (say =E)

where 5/2 = 3/2 (for the internal energy) + 1 (for pressure) and gamma=5/3 for a monoatomic gas in 3 dimensions.

One has then to write

J1 = J2 = J (a constant)

I1 = I2 = I (a constant)

E1 = E2 = E (a constant)

where 1 is on one side of the shock and 2 is on the other side and you replace J1 by rho1*v1, etc...

This system is then solved to obtain the "boundary" conditions at the shock on both sides of it. Somewhere you have to use the definition of the entropy S from thermodynamics.

Landau Lifshitz (p.327) gives eventually:

s2-s1 = (1/12)/T1 * (gamma+1)*V1/gamma^2/p1^2 * (p2-p1)^3

where s is the entropy, V=1/rho and the shock is weak, the quantities dont change too much across the shock.

Here you migth have to use the ideal gas equation to get rid of T and use the definition of the sound speed

c^2 = (dp/d rho)

It might still be a long way to get to what you found in that paper, but it seems to be in the correct direction.

Good luck.

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Old   March 6, 2008, 09:30
Default Re: Entropy jump across shocks
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Dave
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Thanks for your replies Patrick - I'll have another bash at deriving the equation and if I work it out I'll post the solution on the forum.

I had tried the Rh conditions but may have made a mistake with my algebra. I know how to determine the size of the entropy jump from the RH conditions but trying to work out the pressure caused by the jump in entropy is proving to be a pain . I could take it for granted that the author is correct but I'd reluctant to do so without knowing WHY he's correct )

Thanks again for your help, Dave (the pedantic mathematician)
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Old   March 6, 2008, 14:20
Default Re: Entropy jump across shocks
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Ananda Himansu
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The normalized entropy jump should go as the cube of the normalized pressure jump. If not, something is wrong with the expression. The cubic dependence is why, for weak shocks with small pressure jump, we can neglect the entropy jump and treat the flow across the shock as isentropic.
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Old   March 7, 2008, 15:28
Default Re: Entropy jump across shocks
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Dave
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Hi Ananda, I think you're right about the cubic dependence - I should have mentioned that the term is a second order term. I'm looking at the fluid properties as asymptotic expansions of a small parameter 'e' and the pressure coefficient term is of order 'e^2'.

I'm aware that the entropy jump is cubic in terms of shock strength (p1-p2)/p1 so I guess it's appropriate that the pressure coefficient is a second order term in e.

Many thanks for raising this point - I should have included it in my original post. Dave

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Old   March 7, 2008, 16:44
Default Re: Entropy jump across shocks
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Ananda Himansu
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Well, it seems that you already have the standard R-H jump conditions across the shock expressed in terms of the upstream Mach number, and you also already have the entropy jump across the shock expressed as a cubic function of the pressure jump, as cited from Landau-Lifshitz by Patrick.

The bit about the asymptotic expansion in a small parameter 'e' was not evident in your original post, and I do not believe that people unaware of the specific asymptotic representation you seek and of the definition of your Cps, would be able to derive the equation you wish to verify. My guess is that to derive the expression for Cps, you should use (a) the expression for Cpi that you have presumably already verified, where Cpi is the downstream pressure coefficient assuming isentropic flow across the shock, and (b) the expression for Cp, the downstream pressure coefficient obtained from normalizing the Rankine-Hugoniot conditions. Cps should be either Cp-Cpi, or the leading term of the epansion thereof in powers of 'e'.
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Old   March 11, 2008, 12:01
Default Re: Entropy jump across shocks
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sebi
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Are the R-H equations valid for the TSD formulation? Check J. Batina reports from NASA. The ASP code has an alternative correction.

Good luck
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Old   March 12, 2008, 11:21
Default Re: Entropy jump across shocks
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Dave
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Thanks Sebi, The Batina report gives another form of the pressure correction that results from the entropy change across the shock. The report gives some great additions to the TSD theory that make numerical solutions match more closely with the Euler solns.

My problem is that I don't want to include all of Batina's correction terms. They provide a good extension to the theory (and produce realistic results) but aren't entirely mathematically rigorous. My own work stems from an asymptotic expansion analysis of transonic flow and so the use of, for example, the composite TSD containing both p_xt and p_tt terms should ideally be avoided (i.e. I'm just dealing with the low frequency TSD equation). I'm merely writing my code as an academic exercise, in order to avoid multiple solutions it's neccessary to include the effects of the entropy jump across the shock.

Batina's report is great and my next step will probably be to build all of his correction terms into my code, for now though I simply want to include the shock generated entropy jump and avoid the effects of viscosity, rotationality.

The R-H jump conditions are valid for the exact case. Some TSD solvers don't satisfy them well and that's the main cause of the difference between Euler and TSD solvers. Batina's correction terms allow more accurate treatment of the R-H conditions while maintaining the efficiency of the TSD formulations.

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