Cauchy Momentum Equation - Derivation - Cylindrical Coordinates

Cylindrical Coordinates

$r:\;\;\rho \left(\frac{\partial u_r}{\partial t} + u_r \frac{\partial u_r}{\partial r} + \frac{u_{\phi}}{r} \frac{\partial u_r}{\partial \phi} + u_z \frac{\partial u_r}{\partial z} - \frac{u_{\phi}^2}{r}\right) = -\frac{\partial P}{\partial r} - \frac{1}{r}\frac{\partial {(r{\tau_{rr})}}}{\partial r} - \frac{1}{r}\frac{\partial {\tau_{\phi r}}}{\partial \phi} - \frac{\partial {\tau_{z r}}}{\partial z} + \frac {\tau_{\phi \phi}}{r} + \rho g_r$

$\phi:\;\;\rho \left(\frac{\partial u_{\phi}}{\partial t} + u_r \frac{\partial u_{\phi}}{\partial r} + \frac{u_{\phi}}{r} \frac{\partial u_{\phi}}{\partial \phi} + u_z \frac{\partial u_{\phi}}{\partial z} + \frac{u_r u_{\phi}}{r}\right) = -\frac{1}{r}\frac{\partial P}{\partial \phi} -\frac{1}{r}\frac{\partial {\tau_{\phi \phi}}}{\partial \phi} - \frac{1}{r^2}\frac{\partial {(r^2{\tau_{r \phi})}}}{\partial r} - \frac{\partial {\tau_{z r}}}{\partial z} + \rho g_{\phi}$

$z:\;\;\rho \left(\frac{\partial u_z}{\partial t} + u_r \frac{\partial u_z}{\partial r} + \frac{u_{\phi}}{r} \frac{\partial u_z}{\partial \phi} + u_z \frac{\partial u_z}{\partial z}\right) = -\frac{\partial P}{\partial z} - \frac{\partial {\tau_{z z}}}{\partial z} - \frac{1}{r}\frac{\partial {\tau_{\phi z}}}{\partial \phi} - \frac{1}{r}\frac{\partial {(r{\tau_{rz})}}}{\partial r} + \rho g_z.$

By expressing the shear stress in terms of viscosity and fluid velocity, and assuming constant density and viscosity, the Cauchy momentum equation will lead to the Navier–Stokes equations. By assuming inviscid flow, the Navier–Stokes equations can further simpify to the Euler equations.

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