Полный набор уравнений движения жидкости

Last modified by Alexey Popov on 2020/12/27 09:10

$\frac{D\vec v}{Dt} = \vec g - \frac{1}{\rho }\nabla P+\mu\nabla^2\vec v -\vec a$

$\frac{D P}{Dt} = -\rho v_s^2 \nabla \cdot \vec v$

$\frac{D\nabla \times \vec v}{Dt} = -\nabla \times \vec a + \mu\nabla^2\nabla \times\vec v$

$\frac{D^2\vec v}{Dt^2} =v_s^2 \nabla \times (\nabla \times \vec v)+v_s^2 \nabla^2 \vec v +\frac{D\mu\nabla^2\vec v}{Dt} -\frac{D\vec a}{Dt}$

$\frac{D \vec\omega}{Dt} = -\nabla \times \vec a+ \mu\nabla^2\vec\omega$

$\frac{D\vec a}{Dt} =v_s^2 \nabla \times \vec\omega+v_s^2 \nabla^2 \vec v- \frac{D^2\vec v}{Dt^2} -\mu\nabla^2\nabla p +\mu^2\nabla^2\nabla^2\vec v -\mu\nabla^2\vec a$

$\vec j = (\nabla \cdot \vec a) \vec v$

$v_s^2 \nabla^2 \vec v- \frac{D^2\vec v}{Dt^2} = - j$

$\frac{D\vec v}{Dt} = \vec g - \frac{1}{\rho }\nabla P+\mu\nabla^2\vec v -\vec a - \frac{1}{\rho}\nabla \times \vec m$

$\frac{D P}{Dt} = -\rho v_s^2 \nabla \cdot \vec v$

$\frac{D\vec m}{Dt} = -\rho v_s^2\vec \omega + \rho v_s^2 \nabla \times \vec v$

$\nabla \cdot \vec m = 0$

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Created by Alexey Popov on 2020/11/24 19:00

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