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Incompressible flow

by coverband
Tags: flow, incompressible
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coverband
#1
Jun17-10, 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
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radou
#2
Jun17-10, 08:04 AM
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P: 3,220
If the norm of x is given, you can figure out its components, right?
coverband
#3
Jun17-10, 08:06 AM
P: 167
Right, (x,y,z)

radou
#4
Jun17-10, 08:09 AM
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radou's Avatar
P: 3,220
Incompressible flow

So, write u down in terms of its components and calculate the divergence.
coverband
#5
Jun17-10, 08:15 AM
P: 167
Ok, so u = (fx,fy,fz)

Therefore div u = 3f !!! Now.. !?
coverband
#6
Jun18-10, 01:14 PM
P: 167
Or div u = 3f + xf_x + yf_y + zf_z ? Little help!
billiards
#7
Jun19-10, 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 x-component (and they have the same form for y and z).

div u = d(ux)/dx + ......

ux = f(r) rx

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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