非平衡偏导数

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摘要

在计算系数矩阵和隐式源项偏导数时，需要用到原始变量${\left[ {\rho_s ,u,v,w,T,T^v,p,} \right]^T}$对守恒变量${\bf{U}} = {\left[ {\rho_s ,\rho u,\rho v,\rho w,\rho e^v,\rho e^e} \right]^T}$的偏导数。这里进行推导。

推导

利用状态方程
$$p = \sum_{s = 1}^{ns} {\rho_s \frac{R_u}{M_s}T}$$
和热力学关系式
$$\rho {e^e} = \rho {e^v} + \frac{1}{2}\rho (u^2 + v^2+
w^2) + \sum_{s = 1}^n {\rho_s({c_{V,s}}T + e_s^0)} $$

得到用守恒量表示温度
$$\rho {e^{tr}} = \rho c_V^{tr} T= \rho {e^e} - \rho
e^v-\frac{1}{2}\frac{\sum_{s = 1}^n \rho_s}{(\sum_{s = 1}^n {\rho_s})^2}$$

其中
$c_V^{tr} = \sum_{s = 1}^{ns} Y_s c_{V,s}^{tr} = \frac{1}{\rho }\sum_{s = 1}^{ns} {\rho_s}c_{V,s}^{tr} $

$c_V^v = \sum_{s = mol} {Y_s c_{V,s}^v} = \frac{1}{\rho }\sum_{s = mol} {Y_s}c_{V,s}^v $

这样，可以将压强、温度、振动温度用守恒变量${\bf{U}} = {\left[ {\rho_s ,\rho u,\rho v,\rho w,\rho e^v,\rho e^e} \right]^T}$表示：

$$T = \frac{1}{\sum_{s = 1}^{ns} {\rho_s} c_{V,s}^{tr}}
[\rho e^e-\rho e^v -\frac{1}{2}\frac{1}{\sum_{s = 1}^n \rho_s}((\rho u)^2 + (\rho v)^2 + (\rho w)^2) - \sum_{s = 1}^n \rho_s e_s^0]$$

$$p = \sum\limits_{s = 1}^{ns}
\rho_s\frac{R_u}{M_s}\underbrace {\frac{1}{\sum\limits_{s = 1}^{ns} {\rho_s} c_{V,s}^{tr}}\left[\rho e^e - \rho e^v - \frac{1}{2}\frac{1}{\sum\limits_{s = 1}^n \rho_s}\left( (\rho u)^2 + (\rho v)^2 + (\rho w)^2 \right) - \sum\limits_{s = 1}^n {\rho_s e_s^0} \right]}_T $$

$$T^v = \frac{\rho e^v}{\sum\limits_{s = mol} \rho_s c_{V,s}^v}$$
利用这三个式子，可以得到压强、温度、振动温度对守恒变量的偏导数

$\left\{ \begin{array}{l}
\varphi = \frac{\partial p}{\partial (\rho e^e)} =\frac{R_u}{c_V^{tr}\bar M}\\
\varphi_u = \frac{\partial p}{\partial (\rho u)} = - \varphi u\\
{\varphi_v} = \frac{\partial p}{\partial (\rho v)}= - \varphi v\\
{\varphi_w} = \frac{\partial p}{\partial (\rho w)} = - \varphi w\\
{\beta_s} = \frac{\partial p}{\partial {\rho_s}} = \frac{R_u}{M_s}T - \varphi c_{V,s}^{tr}T + \varphi \left[ \frac{1}{2}\left( u^2 + v^2+ w^2 \right) - e_s^0 \right]\\
\quad= \frac{R_u}{M_s}T \varphi \left[ \frac{1}{2}\left( u^2 + v^2 + w^2 \right) - e_s^{tr} - e_s^0 \right]\\
{\varphi_{ev}} =\frac{\partial p}{\partial (\rho e^v)}= - \varphi
\end{array} \right.$

$\left\{ \begin{array}{l}
\varphi = {\left. \frac{\partial p}{\partial (\rho e^e)} \right|_Q} = \frac{R_u}{c_V^{tr}\bar M}\\
{\varphi _u} = {\left. \frac{\partial p}{\partial (\rho u)} \right|_Q} = - \varphi u\\
{\varphi _v} = {\left. \frac{\partial p}{\partial (\rho v)} \right|_Q} = - \varphi v\\
{\varphi _w} = {\left. \frac{\partial p}{\partial (\rho w)} \right|_Q} = - \varphi w\\
{\beta _s} = {\left. \frac{\partial p}{\partial \rho _s} \right|_Q} = \frac{R_u}{M_s} T - \varphi c_{V,s}^{tr} T \varphi \left[ \frac{1}{2}\left( u^2+ v^2+w^2 \right) - e_s^0 \right]\\
\quad=\frac{R_u}{M_s}T \varphi \left[ \frac{1}{2}\left( u^2+v^2+w^2 \right) - e_s^{tr} - e_s^0 \right]\\
\varphi_{ev} = {\left. \frac{\partial p}{\partial (\rho e^v)} \right|_Q} = - \varphi
\end{array} \right.$

$\left\{ \begin{array}{l}
\chi = \frac{\partial T}{\partial (\rho {e^e})} = \frac{1}{\rho c_V^{tr}}\\
{\chi _u} = \frac{\partial T}{\partial (\rho u)} = - \chi u\\
{\chi _v} = \frac{\partial T}{\partial (\rho v)} = - \chi v\\
{\chi _w} = \frac{\partial T}{\partial (\rho w)} = - \chi w\\
{\chi _s} = \frac{\partial T}{\partial \rho _s} = \chi \left[ \frac{1}{2}\left( u^2 + v^2 + w^2 \right) - e_s^{tr} - e_s^0 \right]\\
{\chi _{ev}} = {\frac{\partial T}{\partial \rho {e^v}}}= - \chi
\end{array} \right.$

$\left\{
\begin{array}
\chi \chi = \frac{\partial T^v}{\partial (\rho e^e)} = 0\\
\chi \chi _u = \frac{\partial T^v}{\partial (\rho u)} = 0\\
\chi \chi _v = \frac{\partial T^v}{\partial (\rho v)} = 0\\
\chi \chi _w = \frac{\partial T^v}{\partial (\rho w)} = 0\\
\chi \chi _s = \frac{\partial T^v}{\partial \rho _s}=- \frac{e_s^v}{\rho c_V^v}\\
\chi \chi _{ev} = \frac{\partial T^v}{\partial \rho e^v} = \frac{1}{\rho c_V^v}
\end{array} \right.$

其中$\rho c_V^v = \sum \limits_{s = 1}^{nmol} \rho_s c_{V,s}^v$,
$c_{V,s}^v = \frac{\partial e_s^v}{\partial T^v} = \frac{R_s}{\left( \exp (\theta_s^v/{T^v}) - 1 \right)^2}{\left( \frac{\theta_s^v}{T^v} \right)^2}\exp \left(\frac{\theta_s^v}{T^v} \right)$

说明

  1. 许多文章中使用总密度守恒方程加(ns-1)个组分方程的形式。理论上两种形式是等价的。但是在这些偏导数的推导中，不得不考虑最后一个没有用到的组分密度，使表达式复杂化。
  2. 许多作者求解过程中事实上是求解了总密度方程和ns个组分方程。但这样子默认在了非平衡偏导数的推导中，ns个组分密度和总的密度是独立的。我感觉这样有问题。