Solve Momentum Behavior: Get 2.72 & 2.73 Equations

[CFXMSC]

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

http://imageshack.com/a/img89/5594/kiw6.jpg

 

[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.

 

[CFXMSC]

i tried but it not looks simple to do

 

[Simon Bridge]

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

 

[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?

 

[Simon Bridge]

Which is the term you are having trouble with?
Do you mean $\vec\nabla\cdot[\mu \nabla^2 \vec V]$ ?

... you will be helped by the following list of identities:
http://en.wikipedia.org/wiki/Vector_calculus_identities#Second_derivatives_2

... especially: $\vec\nabla\cdot(\nabla^2\vec A)=\nabla^2(\vec\nabla\cdot\vec A)$

... you are going to have to learn how to type out equations.
Please see: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

 

[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

 

[CFXMSC]

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

 

[CFXMSC]

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

What now?

 

[Simon Bridge]

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