# Fluid Mechanics::'Deriving' Incompressible Flow Criteria

Tags: criteria, flow, fluid, incompressible, mechanicsderiving
P: 3,007
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 $$\frac{\partial{}}{\partial{x}} (\rho u)\approx\rho\frac{\partial{u}}{\partial{x}} \qquad (1)$$ This is equivalent to the strong inequality $$|u\frac{\partial{\rho}}{\partial{x}}|\ll |\rho\frac{\partial{u}}{\partial{x}}| \qquad (2)$$ or $$|\frac{\delta\rho}{\rho}|\ll|\frac{\delta V}{V}| \qquad (3)$$
I am completly baffled as to how we went from (1) to (2) ...let alone from (2) to (3)?
Any thoughts?

Casey
 Sci Advisor HW Helper Thanks PF Gold P: 26,107 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)
 P: 3,007 Oh...that darned chain rule! Thanks tiny-tim. Also, silly question, but why did we change the ∂'s into $\delta$'s ? Is it because the (∂x)'s 'canceled' and thus it is no longer a derivative, but just a relation between 'changes?'