第6章 問題4 解答

よどみ点状態（添え字$\, 0\,$ ）と一様流（添え字$\, \infty \, $）との間のベルヌーイの式は \begin{equation} p_0=p_\infty+\frac{1}{2}\rho_\infty U^2 \end{equation} したがって，圧力係数$\, C_p\, $は \begin{equation} C_p=\frac{p_0-p_\infty}{\frac{1}{2}\rho_\infty U^2} =\frac{p_\infty}{\frac{1}{2}\rho_\infty U^2} \left(\frac{p_0}{p_\infty}-1\right) =\frac{2}{\gamma M_\infty^2}\left(\frac{p_0}{p_\infty}-1\right) \label{eq:2} \end{equation} 一方，式(6.64)より， \begin{equation} \frac{p_0}{p_\infty}=\left(1+\frac{\gamma-1}{2}M_\infty^2\right)^{\gamma/(\gamma-1)} \end{equation} 二項定理 \begin{equation} (1+x)^\alpha=1+\alpha x + \frac{\alpha(\alpha-1)}{2!}x^2 +\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3 +\frac{\alpha(\alpha-1)(\alpha-2)(\alpha-3)}{4!}x^4+\dots \end{equation} で展開すれば \begin{eqnarray} \frac{p_0}{p_\infty}&=&1&+&\frac{\gamma}{\gamma-1}\left(\frac{\gamma-1}{2}M_\infty^2\right) +\frac{\frac{\gamma}{\gamma-1}\left(\frac{\gamma}{\gamma-1}-1\right)}{2!} \left(\frac{\gamma-1}{2}M_\infty^2\right)^2\nonumber\\ && &+&\frac{1}{3!}\frac{\gamma}{\gamma-1}\left(\frac{\gamma}{\gamma-1}-1\right) \left(\frac{\gamma}{\gamma-1}-2\right) \left(\frac{\gamma-1}{2}M_\infty^2\right)^3\nonumber\\ && &+&\frac{1}{4!}\frac{\gamma}{\gamma-1} \left(\frac{\gamma}{\gamma-1}-1\right) \left(\frac{\gamma}{\gamma-1}-2\right) \left(\frac{\gamma}{\gamma-1}-3\right) \left(\frac{\gamma-1}{2}M_\infty^2\right)^4\dots \nonumber\\ &=&1&+&\frac{\gamma}{2}M_\infty^2+\frac{1}{2}\frac{\gamma}{\gamma-1} \frac{\gamma-(\gamma-1)}{\gamma-1} \left(\frac{\gamma-1}{2}M_\infty^2\right)^2\nonumber\\ && &+&\frac{1}{6}\frac{\gamma}{\gamma-1}\frac{\gamma-(\gamma-1)}{\gamma-1} \frac{\gamma-2(\gamma-1)}{\gamma-1} \left(\frac{\gamma-1}{2}M_\infty^2\right)^3\nonumber\\ && &+&\frac{1}{24}\frac{\gamma}{\gamma-1}\frac{\gamma-(\gamma-1)}{\gamma-1} \frac{\gamma-2(\gamma-1)}{\gamma-1} \frac{\gamma-3(\gamma-1)}{\gamma-1} \left(\frac{\gamma-1}{2}M_\infty^2\right)^4\dots\nonumber\\ &=&1&+&\frac{\gamma}{2}M_\infty^2+\frac{\gamma}{8}M_\infty^4 +\frac{1}{48}\gamma(2-\gamma)M_\infty^4 +\frac{1}{24\times16}\gamma(2-\gamma)(3-2\gamma)M_\infty^4\dots\nonumber \end{eqnarray} 式\eqref{eq:2}に代入して \begin{eqnarray} C_p&=&\frac{2}{\gamma M_\infty^2} \left(\frac{\gamma}{2}M_\infty^2+\frac{\gamma}{8}M_\infty^4 +\frac{1}{48}\gamma(2-\gamma)M_\infty^6 +\frac{1}{384}\gamma(2-\gamma)(3-2\gamma)M_\infty^8\dots\right) \nonumber\\ &=&1+\frac{1}{4}M_\infty^2+\frac{1}{24}(2-\gamma)M_\infty^4 +\frac{1}{192}(2-\gamma)(3-2\gamma)M_\infty^6+\dots \end{eqnarray}

証明終り