Momentum behavior

1. Mar 23, 2014

CFXMSC

Can somobody help me how to get the 2.72 and 2.73 equation??

http://imageshack.com/a/img89/5594/kiw6.jpg [Broken]

Last edited by a moderator: May 6, 2017
2. Mar 23, 2014

Simon Bridge

2-72: have you tried: starting from 2-70 and take the divergence of both sides - note constant density and div(V)=0

2-73: this equation is the definition of "fluid vorticity" so it is not a derived thingy.
The equation just before it comes from 2-70 by taking the curl.

3. Mar 23, 2014

CFXMSC

i tried but it not looks simple to do

4. Mar 23, 2014

Simon Bridge

Most of the terms turn out to be zero - where do you get stuck?

5. Mar 23, 2014

CFXMSC

I really confused.... There are a lot of terms in dV/dT and gravity that become 0, but in vicosity how could i do?

6. Mar 23, 2014

Simon Bridge

7. Mar 24, 2014

CFXMSC

$\nabla p=\rho g + \mu \nabla^2 V-\rho\frac{\partial V}{\partial t}$

$\nabla^2 p=\nabla .(\rho g) + \mu\nabla . (\nabla^2 V)-\rho\nabla . \left(\frac{\partial V}{\partial t}\right)$

$\nabla^2 p=\nabla .(\rho g) + \mu\nabla^2(\nabla V)-\rho\nabla . \left(\frac{\partial V}{\partial t}\right)$

Using $\nabla . V=0$ and $\nabla .(\rho g)=0$

$\nabla^2 p=-\rho\nabla . \left(\frac{\partial V}{\partial t}\right)$

And i got stuck

Last edited: Mar 24, 2014
8. Mar 24, 2014

CFXMSC

Can i assume if $div(V)=0$ then it's a steady flow problem?

9. Mar 24, 2014

CFXMSC

The other i got $\frac{\partial \omega}{\partial t}=\frac{\mu}{\rho}\nabla \times (\nabla^2 V)$

What now?

10. Mar 25, 2014

Simon Bridge

You are told that div.V=0.
You can take the div inside the time-partial