Fluid Mechanics'Deriving' Incompressible Flow Criteria

  • #1
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Here we go....

My text attempts to 'derive' an expression that explains when a flow is compressible or not:

]When is a given flow approximately incompressible? We can derive a nice criterion by playing a little fast and loose with density approximations....
Great :rolleyes: ... if there's anything I like better than making density approximations, it's playing 'fast and loose' with them. :smile:

He then goes on to say:
..In essence, we wish to slip the density out of the divergence in the continuity equation and approximate a typical term as
[tex]\frac{\partial{}}{\partial{x}} (\rho u)\approx\rho\frac{\partial{u}}{\partial{x}} \qquad (1)[/tex]

This is equivalent to the strong inequality

[tex]

|u\frac{\partial{\rho}}{\partial{x}}|\ll |\rho\frac{\partial{u}}{\partial{x}}| \qquad (2)

[/tex]

or

[tex]

|\frac{\delta\rho}{\rho}|\ll|\frac{\delta V}{V}| \qquad (3)
[/tex]
I am completly baffled as to how we went from (1) to (2) ...let alone from (2) to (3)?
Any thoughts?

Casey
 

Answers and Replies

  • #2
tiny-tim
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Hi Casey! :smile:

∂(ρu)/∂x = u∂ρ/∂x + ρ∂u/∂x,

so if u∂ρ/∂x << ρ∂u/∂x, we can ignore it, and then ∂(ρu)/∂x ~ ρ∂u/∂x. :wink:

(2) to (3) is simply rearrangement (and changing u to V for some reason which escapes me)
 
  • #3
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2
Oh...that darned chain rule! :smile: Thanks tiny-tim.

Also, silly question, but why did we change the ∂'s into [itex]\delta[/itex]'s ?

Is it because the (∂x)'s 'canceled' and thus it is no longer a derivative, but just a relation between 'changes?'
 
  • #4
arildno
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(2) to (3) is based upon the fact that this argument must be repeated for the v and w-components as well, and hence, that the relative infinitesemal change in density must be much less than the relative infinitesemal change in the maximal velocity component, and hence, much less than the relative infinitesemal change in the fluid speed.
 

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