第6章 問題2 解答

\begin{eqnarray} M_2^2&=&\frac{u_2^2}{a_2^2}=M_1^2\Bigg/\frac{u_1^2}{a_1^2}\cdot \frac{u_2^2}{a_2^2}=M_1^2\cdot\frac{u_2^2}{u_1^2}\cdot\frac{a_1^2}{a_2^2} =M_1^2\cdot\frac{\rho_2 u_2}{\rho_1 u_1}\cdot\frac{\gamma p_1/\rho_1}{\gamma p_2/\rho_2} =M_1^2\cdot\left(\frac{\rho_2 u_2}{\rho_1 u_1}\right)^2\cdot\frac{p_1}{p_2}\cdot\frac{\rho_1}{\rho_2} \end{eqnarray} ここで，$\, \rho_2 u_2=\rho_1 u_1\, $，式(6.91)，式(6.92)を用いれば \begin{eqnarray} M_2^2&=&M_1^2\cdot\left\{1-\frac{2}{\gamma+1}\frac{M_1^2-1}{M_1^2}\right\} \Bigg/\left\{1+\frac{2\gamma}{\gamma+1}(M_1^2-1)\right\}\nonumber\\ &=&\left\{M_1^2-\frac{2}{\gamma+1}(M_1^2-1)\right\} \Bigg/\left\{1+\frac{2\gamma}{\gamma+1}(M_1^2-1)\right\} =\frac{(\gamma+1)M_1^2-2(M_1^2-1)} {(\gamma+1)+2\gamma(M_1^2-1)}\nonumber\\ &=& \left[1+\frac{1}{2}(\gamma-1)M_1\right]\Bigg/ \left[\gamma M_1^2-\frac{1}{2}(\gamma-1)\right] \end{eqnarray}