Departing from the Navier-Stokes equations for an incompressible flow (with [itex]\rho[/itex] and [itex]\mu=\rho \nu[/itex] constant):(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \frac{Du_i}{Dt}=\frac{\partial u_i}{\partial t}+u_j \frac{\partial u_i}{\partial x_j} = -\frac{1}{\rho} \frac{\partial p}{\partial x_i} +\nu \frac{\partial^2 u_i}{\partial x_j^2}[/tex]

my book says it follows for a circular flow

[tex]\frac{\rho u_{\theta}^2}{r} = \frac{\partial p}{\partial r}[/tex]

[tex]\frac{u_{\theta}}{\partial t} = \nu [\frac{\partial^2 u_{\theta}}{\partial r^2} + \frac{1}{r} \frac{\partial u_{\theta}}{\partial r}-\frac{u_{\theta}}{r^2}] [/tex]

I can understand part of it. For a circular flow [itex]u_r=0[/tex] So the incompressibility yields that [itex] u_{\theta}[/itex] is not a function of theta. So the nonlinear term in the material derivative vanishes. But where does the last term [itex]-\frac{u_{theta}}{r^2} [/itex]

come from? And the entire first equation?!

**Physics Forums - The Fusion of Science and Community**

# Fluid Mechanics - Equations of motion circular flows

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: Fluid Mechanics - Equations of motion circular flows

Loading...

**Physics Forums - The Fusion of Science and Community**