Shock-tube relations

From the analysis of the incident shock, we have a relation for the induced flow behind the shock

$$u_2=u_p=\frac{a_1}{{\gamma }_1}(\frac{p_2}{p_1}-1){\left(\frac{\left(\frac{2{\gamma }_1}{{\gamma }_1+1}\right)}{\left(\frac{{\gamma }_1-1}{{\gamma }_1+1}\right)+\left(\frac{p_2}{p_1}\right)}\right)}^{1/2}$$

The velocity in region 3 can be obtained from the expansion relations

$$\frac{p_3}{p_4}={[1-\frac{{\gamma }_4-1}{2}\left(\frac{u_3}{a_4}\right)]}^{2{\gamma }_4/({\gamma }_4-1)}$$

Solving for $u_3$ gives

$$u_3=\frac{2a_4}{{\gamma }_4-1}[1-{\left(\frac{p_3}{p_4}\right)}^{({\gamma }_4-1)/(2{\gamma }_4)}]$$

There is no change in pressure or velocity over the contact surface, which means $u_2=u_3$ and $p_2=p_3$.

$$u_2=\frac{2a_4}{{\gamma }_4-1}[1-{\left(\frac{p_2}{p_4}\right)}^{({\gamma }_4-1)/(2{\gamma }_4)}]$$

Now, we have two ways of calculating $u_2$. Setting Eqn. \ref{eq:shocktube:up:a} equal to Eqn. \ref{eq:shocktube:up:d} leads to the shock tube relation

$$\frac{p_4}{p_1}=\frac{p_2}{p_1}{\{1-\frac{({\gamma }_4-1)(a_1/a_4)(p_2/p_1-1)}{\sqrt{2{\gamma }_1[2{\gamma }_1+({\gamma }_1+1)(p_2/p_1-1)]}}\}}^{-2{\gamma }_4/({\gamma }_4-1)}$$