[物理学与PDEs]第3章第2节 磁流体力学方程组 2.2 考虑到电磁场的存在对流体力学方程组的修正

时间:2022-09-22 16:22:32

1.  连续性方程 $$\bex \cfrac{\p \rho}{\p t}+\Div(\rho{\bf u})=0.  \eex$$

2.  动量守恒方程 $$\bex \cfrac{\p }{\p t}(\rho{\bf u}) +\Div(\rho {\bf u}\otimes{\bf u}-{\bf P}) -\mu\rot{\bf H}\times{\bf H}=\rho {\bf F}, \eex$$ 或 $$\bex \rho \cfrac{\rd {\bf u}}{\rd t} -\Div{\bf P} -\mu_0\rot{\bf H}\times{\bf H}=\rho {\bf F}. \eex$$

3.  能量守恒方程 $$\beex \bea \cfrac{\p}{\p t}&\sex{\rho e+\cfrac{1}{2}\rho u^2+\cfrac{1}{2}\mu_0 H^2} +\Div\sez{\sex{\rho e+\cfrac{1}{2}\rho u^2}{\bf u}-{\bf P} {\bf u}}\\ +\Div&\sez{\cfrac{1}{\sigma}\rot{\bf H}\times{\bf H}-\mu_0({\bf u}\times{\bf H})\times{\bf H}} =\Div(\kappa \n T)+\rho {\bf F}\cdot{\bf u}, \eea \eeex$$ 或 $$\bex \rho T\cfrac{\rd S}{\rd t} -\bar \mu \cdot \tr \sex{{\bf S}\cdot\n {\bf u}} -\sex{\bar \mu'-\cfrac{2}{3}\bar \mu}|\Div{\bf u}|^2 -\cfrac{1}{\sigma}|\rot{\bf H}|^2=\Div(\kappa\n T). \eex$$