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April 9, 2001, 12:29 
About gaswater twophase Riemann problem

#1 
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There is a gaswater twophase compressible multifluid Riemann problem, governed by the stiffened gas equation of state:
p(\rho,e)=(\gamma1)\rho e  \gamma p_\infty For gas, \gamma=1.4, p_\infty=0; For water, \gamma=4.4, p_infty=6.E+8 And the initial conditions are For gas, \rho=1.2 Kg/m^3, p=1.E+5 Pa, velocity=0; For water,\rho=1000 Kg/m^3, p=1.E+9 Pa, velocity=0. I have used a simple mixturetype Eulerian numerical method proposed by Saurel/Abgrall(SIAM 1998) or Shyue(JCP 1999) to simulate this problem, but the result is very poor because the pressure curve is very different from the exact solution. It is very obvious in the pressure plot for using log curve. I think this problem is very stiff and the density ratio and pressure ratio is very high(1000/1.2 and 10000/1). Has someone done some work in this problem and to tell me how to solve this problem? Ma, DongJun 

April 11, 2001, 06:03 
Re: About gaswater twophase Riemann problem

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It seems to me it is a numerical problem. why not to have a look to this book:
Riemann solvers and numerical methods for fluid dynamics. By: E.F. Toro Springer. 

April 23, 2001, 18:39 
Re: About gaswater twophase Riemann problem

#3 
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Hi,
Are you using an explicit or implicit scheme for time integration. I used an implicit method to solve the 2phase problem in K. M. Shyue, JCP 1998. But in this case the density ratio is not as high as yours. but the pressure ratio is high. Do you know how to compute temperature on the water side. Buvana 

April 24, 2001, 05:11 
Re: About gaswater twophase Riemann problem

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I'm working with isentropic twofluid model and using exact Riemann solver, explicit scheme.


April 25, 2001, 06:48 
Re: About gaswater twophase Riemann problem

#5 
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Hi,
I was also doing this problem last year, just in my work, for the water \gamma is taken as 7.0. I am thinking the major problem in your calculation may be the treatment of flux through airwater interface. Please check whether you keep the interface like the wave behaves itself. Good luck! 

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