变比热容计算

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理想气体比热容是温度的函数，
那么有个问题：
已知总温总压马赫数怎么求得静温静压？
望解答

李东岳

比如压力这个，你需要知道速度，然后从方程倒推出来。链接文本

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李老师@李东岳
根据我的推导，对于可压缩理想气体应该满足如下的关系，但是从公式看出这是要迭代求解的，推导是否正确
\par $∙$ 求解静温(已知总温和马赫数)
$\begin{matrix} (1) & U_{s}^{2} = 2 (h_{t} (T_{t o t}) - h_{t} (T_{s t a})) = 2 (C_{p} (T_{t o t}) T_{t o t} - C_{p} (T_{s t a}) T_{s t a}) \end{matrix}$
$\begin{matrix} (2) & M a c h^{2} = \frac{U^{2}}{γ (T) R_{g} T} = \frac{U^{2}}{\frac{C_{p} (T)}{C_{p} (T) - R_{g}} R_{g} T} \end{matrix}$
\par由( $2$ )和( $1$ )得
$\begin{matrix} (3) & T_{s t a} = \frac{C_{p} (T_{t o t}) T_{t o t}}{C_{p} (T_{s t a}) (1 + \frac{1}{2} M a c h^{2} \frac{R_{g}}{C_{p} (T_{s t a}) - R_{g}})} \end{matrix}$
\par $∙$ 求解静压(已知总温、总压和静温)
\par由( $???$ )得,等熵过程
$\begin{matrix} (4) & d s = C_{p} (T) \frac{d T}{T} - R_{g} \frac{d p}{p} = 0 \end{matrix}$
\par两边同时积分有
$\begin{matrix} (5) & \int_{T_{t o t}}^{T_{s t a}} C_{p} (T) \frac{d T}{T} = \int_{p_{t o t}}^{p_{s t a}} R_{g} \frac{d p}{p} \end{matrix}$
\par记
$\begin{matrix} (6) & S (T_{x}) = \int_{T_{x}}^{T_{0}} C_{p} (T) \frac{d T}{T} \end{matrix}$
\par则
$\begin{matrix} (7) & S (T_{s t a}) - S (T_{t o t}) = R_{g} \ln \frac{p_{s t a}}{p_{t o t}} \end{matrix}$
\par那么
$\begin{matrix} (8) & p_{s t a} = p_{t o t} e^{(\frac{S (T_{s t a}) - S (T_{t o t})}{R_{g}})} \end{matrix}$

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之前有点问题，最终如下
\par $C_{p}$ 采用4次多项式分段拟合
$\begin{matrix} (9) & C_{p} (T) = a_{1} + a_{2} T + a_{3} T^{2} + a_{4} T^{3} + a_{5} T^{4} \end{matrix}$
\par静焓
$\begin{matrix} (10) & H (T) = \int_{T_{0}}^{T_{x}} C_{p} (T) d T = a_{1} T + \frac{a_{2}}{2} T^{2} + \frac{a_{3}}{3} T^{3} + \frac{a_{4}}{4} T^{4} + \frac{a_{5}}{5} T^{5} + a_{6} \end{matrix}$
\par熵
$\begin{matrix} (11) & S (T) = \int_{T_{0}}^{T_{x}} C_{p} (T) \frac{d T}{T} = a_{1} \ln T + a_{2} T + \frac{a_{3}}{2} T^{2} + \frac{a_{4}}{3} T^{3} + \frac{a_{5}}{4} T^{4} + a_{7} \end{matrix}$
\par $∙$ 求解静温(已知总温和马赫数)
$\begin{matrix} (12) & U_{s}^{2} = 2 (H (T_{t o t}) - H (T_{s t a})) \end{matrix}$
$\begin{matrix} (13) & M a c h^{2} = \frac{U^{2}}{γ (T) R_{g} T} = \frac{U^{2}}{\frac{C_{p} (T)}{C_{p} (T) - R_{g}} R_{g} T} \end{matrix}$
\par由( $2$ )和( $1$ )得
$\begin{matrix} (14) & T_{s t a} = \frac{2 (H (T_{t o t}) - H (T_{s t a}))}{M a c h^{2} \frac{C_{p} (T_{s t a}) R_{g}}{C_{p} (T_{s t a}) - R_{g}}} \end{matrix}$
$∙$ 求解静压(已知总温、总压和静温)
\par由p等熵过程
$\begin{matrix} (15) & d s = C_{p} (T) \frac{d T}{T} - R_{g} \frac{d p}{p} = 0 \end{matrix}$
\par两边同时积分有
$\begin{matrix} (16) & \int_{T_{t o t}}^{T_{s t a}} C_{p} (T) \frac{d T}{T} = \int_{p_{t o t}}^{p_{s t a}} R_{g} \frac{d p}{p} \end{matrix}$
\par记
$\begin{matrix} (17) & S (T_{x}) = \int_{T_{0}}^{T_{x}} C_{p} (T) \frac{d T}{T} \end{matrix}$
\par则
$\begin{matrix} (18) & S (T_{s t a}) - S (T_{t o t}) = R_{g} \ln \frac{p_{s t a}}{p_{t o t}} \end{matrix}$
\par那么
$\begin{matrix} (19) & p_{s t a} = p_{t o t} e^{(\frac{S (T_{s t a}) - S (T_{t o t})}{R_{g}})} \end{matrix}$

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变比热容计算