# Shock tube problem

(Difference between revisions)
 Revision as of 04:30, 21 September 2005 (view source)← Older edit Revision as of 13:23, 14 November 2006 (view source)Lejzy (Talk | contribs) (rewritten)Newer edit → Line 1: Line 1: - The test case involves the 1-D Euler equation describing the flow.The initial condition is given by + '''Shock tube problem''' is a special case of [[Riemann problem]] with velocities on both sides of discontinuity set to zero. It is often used as a test case for validation of numerical codes, because analytical solutions are available.The initial condition is given by - :$u\equiv u_L ,p=p_L,\rho=\rho_L,xx_o$ + :$u=u_R=0 , p=p_R , \rho=\rho_R, x>0.$ - where $p_L>p_R$ diaphragm being located at $x=x_o$ + - Two cases are considered and the flow is simulated using Roe first-order scheme and Steger-Warming  vector splitting scheme. - :Case 1 $p_R \equiv 1.2*10^4 Pa,p_L=10^5 Pa,u_L=u_R=0,\rho_R=0.125,\rho_L=1.0 kg/m^3 ,x_o=5 m ,t_f=0.0061 s$ + A well-known special case is the Sod problem (Sod, 1978) with initial conditions - :Case 2 $p_R \equiv 1.2*10^3 Pa,p_L=10^5 Pa,u_L=u_R=0,\rho_R=0.01,\rho_L=1.0 kg/m^3 ,x_o=5 m ,t_f=0.0039 s$ + :$p_L=1, \rho_L=1, p_R=0.1, \rho_R=0.125$. - + - The computational domain is $[0,2x_o]$.The boundary conditions are set equal to the intial conditions of the undisturbed gas.The computations are carried out with 600 grid points. +

## Revision as of 13:23, 14 November 2006

Shock tube problem is a special case of Riemann problem with velocities on both sides of discontinuity set to zero. It is often used as a test case for validation of numerical codes, because analytical solutions are available.The initial condition is given by

$u=u_L=0 , p=p_L , \rho=\rho_L, x<0,$
$u=u_R=0 , p=p_R , \rho=\rho_R, x>0.$

A well-known special case is the Sod problem (Sod, 1978) with initial conditions

$p_L=1, \rho_L=1, p_R=0.1, \rho_R=0.125$.