Sod Shock Tube

Sod Shock Tube

The Sod Shock Tube problem, named after Gary A. Sod, is a common test for the accuracy of computational fluid codes, like Riemann solvers, and was heavily investigated by Sod in 1978.

The test consist of a one dimensional Riemann problem with the following parameters, for left and right states of an ideal gas.


left( egin{array}{c} ho_L\P_L\v_Lend{array} ight)=left( egin{array}{c}1.0\1.0\0.0end{array} ight),left( egin{array}{c} ho_R\P_R\v_Rend{array} ight)=left( egin{array}{c}0.125\0.125\0.0end{array} ight)
where::* ho is the density::*P is the pressure::*v is the velocity

The time evolution of this problem can be described by solving the Euler equations.Which leads to three characteristics, describing the propagation speed of thevarious regions of the system. Namely the rarefaction wave, the contact discontinuity andthe shock discontinuity.If this is solved numerically, one can test against the analytical solution,and get information how good a code captures and resolves shocks and contact discontinuitiesand reproduce the correct density profile of the rarefaction wave.

Analytic derivation

The different state of the solution are separated by the time evolution of thethree characteristics of the system, which is due to the finite speedof information propagation. Two of them are equal to the speedof sound of the both states::cs_1 = sqrt{gamma frac{P_L}{ ho_L::cs_5 = sqrt{gamma frac{P_R}{ ho_LThe first one is the position of the beginning of the rarefaction wave whilethe other the velocity of the propagation of the shock.

Defining:::Gamma = frac{gamma - 1}{gamma + 1}, eta = frac{gamma - 1}{2 gamma}The states after the shock are connected by the Rankine Hugoniotshock jump conditions.:: ho_4 = ho_5 frac{P_4 + Gamma P_5}{P_5 + Gamma P_4}But to calculate the density in Region 4 we need to know the pressure in that region.This is related by the contact discontinuity with the pressure in region 3 by::P_4 = P_3Unfortunately the pressure in region 3 can only be calculated iteratively, the rightsolution is found when u_2 equals u_4 ::u_4 = left(P_3' - P_5 ight)sqrt{frac{1-Gamma}{ ho_R(P_3'+Gamma P_5)::u_2 =left(P_1^eta-P_3'^eta ight) sqrt{frac{(1-Gamma^2)P_1^{1/gamma{Gamma^2 ho_L::u_2 - u_4This function can be evaluated to an arbitrary precision thus giving the pressure in theregion 3::P_3 = calculate(P_3,s,s,,)finally we can calculate ::u_3 = u_5 + frac{(P_3 - P_5)}{sqrt{frac{ ho_5}{2}((gamma+1)P_3 +(gamma-1)P_5)::u_4 = u_3and ho_3 follows from the adiabatic gas law:: ho_3 = ho_1 left(frac{P_3}{P_1} ight)^{1/gamma}

References

*cite journal | title=A Survey of Several Finite Difference Methods for Systems of Nonlinear Hyperbolic Conservation Laws | first=G. A. | last=Sod | year=1978 | journal=J. Comput. Phys. | volume=27| pages=1–31 | url=http://www.osti.gov/energycitations/product.biblio.jsp?osti_id=6812922 | doi=10.1016/0021-9991(78)90023-2
*cite book | first=Eleuterio F.| last=Toro| year=1999 | title=Riemann Solvers and Numerical Methods for Fluid Dynamics| publisher=Springer Verlag|location=Berlin | id=ISBN 3-540-65966-8

ee also

*Shock tube
*Computational fluid dynamics


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