Fluid Mechanics::'Deriving' Incompressible Flow Criteria

Saladsamurai

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

... if there's anything I like better than making density approximations, it's playing 'fast and loose' with them.

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

tiny-tim

Hi Casey!

∂(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.

(2) to (3) is simply rearrangement (and changing u to V for some reason which escapes me)

Saladsamurai

Oh...that darned chain rule!

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?'

arildno

(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.

Similar Threads for: Fluid Mechanics::'Deriving' Incompressible Flow Criteria
Thread	Forum	Replies
Fluid Mechanics: Inviscid flow v.s. Laminar flow	Engineering, Comp Sci, & Technology Homework	1
Difference between incompressible viscous and incompressible inviscid flow.	General Physics	1
Fluid Mechanics: streamlines for a flow	Advanced Physics Homework	0
incompressible fluid flow.	Mechanical Engineering	7
Fluid Mechanics - Pipe Flow	Mechanical Engineering	11

tiny-tim Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor	Hi Casey! ∂(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. (2) to (3) is simply rearrangement (and changing u to V for some reason which escapes me)

Saladsamurai Recognitions: Gold Member	Oh...that darned chain rule! 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?'