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Incompressible flow 
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#1
Jun1710, 05:50 AM

P: 167

1. The problem statement, all variables and given/known data
A velocity field is given by [tex] \vec {u} = f(r)\vec{x}, r =  \vec{x} = \sqrt {x^2 + y^2 + z^2} [/tex] written in rectangular cartesian coordinates, where f(r) is a scalar function. Find the most general form of f(r) so that [tex] \vec {u} [/tex] represents an incompressible flow 2. Relevant equations Incompressible flow implies [tex] \nabla . \vec {u} = 0 [/tex]. 3. The attempt at a solution The solution is [tex] \nabla . \vec {u} = 3f + rf' so f(r) = A/r^3 [/tex] (A is an arbitrary constant) but I don't see how it is arrived at. Thanks 


#2
Jun1710, 08:04 AM

HW Helper
P: 3,220

If the norm of x is given, you can figure out its components, right?



#3
Jun1710, 08:06 AM

P: 167

Right, (x,y,z)



#4
Jun1710, 08:09 AM

HW Helper
P: 3,220

Incompressible flow
So, write u down in terms of its components and calculate the divergence.



#5
Jun1710, 08:15 AM

P: 167

Ok, so u = (fx,fy,fz)
Therefore div u = 3f !!! Now.. !? 


#6
Jun1810, 01:14 PM

P: 167

Or div u = 3f + xf_x + yf_y + zf_z ? Little help!



#7
Jun1910, 09:40 AM

P: 748

Think I can help here. Not sure where your getting stuck as you haven't properly written out your thoughts.
First thing you need to do is get the general expression for divergence in terms of your scalar function. Here are the key things for the xcomponent (and they have the same form for y and z). div u = d(u_{x})/dx + ...... u_{x} = f(r) r_{x} so write out div u making the substitution. But remember the product rule. http://en.wikipedia.org/wiki/Product_rule There are some further steps before you can get your answer, but this is a good start. 


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