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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Sod shock tube ： ウィキペディア英語版 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 consists of a one-dimensional Riemann problem with the following parameters, for left and right states of an ideal gas. $$ \left( \begin\rho_L\\P_L\\v_L\end\right) = \left( \begin1.0\\1.0\\0.0\end \right) , $$ \left( \begin\rho_R\\P_R\\v_R\end\right) = \left( \begin0.125\\0.1\\0.0\end\right) where :: *$\backslash rho$ 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 the various regions of the system. Namely the rarefaction wave, the contact discontinuity and the shock discontinuity. If this is solved numerically, one can test against the analytical solution, and get information how well a code captures and resolves shocks and contact discontinuities and reproduce the correct density profile of the rarefaction wave. ==Analytic derivation== The different states of the solution are separated by the time evolution of the three characteristics of the system, which is due to the finite speed of information propagation. Two of them are equal to the speed of sound of the both states ::$cs\_1\; =\; \backslash sqrt\}$ ::$cs\_5\; =\; \backslash sqrt\}$ The first one is the position of the beginning of the rarefaction wave while the other is the velocity of the propagation of the shock. Defining: ::$\backslash Gamma\; =\; \backslash frac$, $\backslash beta\; =\; \backslash frac$ The states after the shock are connected by the Rankine Hugoniot shock jump conditions. ::$\backslash rho\_4\; =\; \backslash rho\_5\; \backslash frac$ 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 ::$P\_4\; =\; P\_3$ Unfortunately the pressure in region 3 can only be calculated iteratively, the right solution is found when $u\_2$ equals $u\_4$ ::$u\_4\; =\; \backslash left(P\_3\text{'}\; -\; P\_5\backslash right)\backslash sqrt\}$ ::$u\_2\; =\backslash left(P\_1^\backslash beta-P\_3\text{'}^\backslash beta\backslash right)\; \backslash sqrt\}$ ::$u\_2\; -\; u\_4\; =\; 0$ This function can be evaluated to an arbitrary precision thus giving the pressure in the region 3 ::$P\_3\; =\; \backslash operatorname(P\_3,s,s,,)$ finally we can calculate ::$u\_3\; =\; u\_5\; +\; \backslash frac((\backslash gamma+1)P\_3\; +(\backslash gamma-1)P\_5)\}\}$ ::$u\_4\; =\; u\_3$ and $\backslash rho\_3$ follows from the adiabatic gas law ::$\backslash rho\_3\; =\; \backslash rho\_1\; \backslash left(\backslash frac\backslash right)^$ 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Sod shock tube」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.