# Shock tube problem

(Difference between revisions)
 Revision as of 13:23, 14 November 2006 (view source)Lejzy (Talk | contribs) (rewritten)← Older edit Latest revision as of 14:18, 19 December 2008 (view source)Peter (Talk | contribs) m (Reverted edits by C4tc4Tsitc (Talk) to last version by Jola) (3 intermediate revisions not shown) Line 1: Line 1: '''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 '''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 - :$u=u_L=0 , p=p_L , \rho=\rho_L, x<0,$ + :$u=u_L=0 , p=p_L , \rho=\rho_L, x<0,$ - :$u=u_R=0 , p=p_R , \rho=\rho_R, x>0.$ + :$u=u_R=0 , p=p_R , \rho=\rho_R, x>0.$ A well-known special case is the Sod problem (Sod, 1978) with initial conditions A well-known special case is the Sod problem (Sod, 1978) with initial conditions - :$p_L=1, \rho_L=1, p_R=0.1, \rho_R=0.125$. + :$p_L=1, \rho_L=1, p_R=0.1, \rho_R=0.125.$

## Latest revision as of 14:18, 19 December 2008

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

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