1
$\begingroup$

I am still trying to understand Navier-Stokes equations. I understand the equations in general, BUT there is one aspect I still cannot digest: The equations assume that density is constant (okay, I understand why, because if it varies, then the derivation of the equations would be more difficult), HOWEVER, what I don't understand is : how it is possible to change pressure without changing density? meaning we use the negative gradient of P to represent forces acting on a blob of fluid, but is it even possible to change P without changing density? I understand that the ONLY way to increase pressure at a one point in space is via increasing the density of the smoke at this point?? Is there other way to change pressure without changing density? Maybe I am missing another definition of pressure that does not depend on density!

Just one last note, I understand if the fluid is liquid then changing its density is not "easy", but for gases, I "think" this is rather doable...

Thanks in advance :)

$\endgroup$
1
  • $\begingroup$ but for fluids, changes in pressure do not appreciably change the density of the fluid. Your concern should be more towards gases, and in that case you can change temperature to compensate. $\endgroup$
    – docscience
    Commented Mar 17, 2017 at 23:25

2 Answers 2

2
$\begingroup$

In the case of incompressible fluids such as liquids, pressure changes do not change the density of the fluid very much.

And in the case of gases, and if you can treat your fluid as an ideal gas it can be written as

$$\rho=\frac{P}{RT}$$

So you can see by properly adjusting $T$ to compensate for changes in $P$ you can hold density, $\rho$ constant

$\endgroup$
2
  • $\begingroup$ Thanks... Ignoring Temperature is a bad idea :) Just thinking of fluid as a group of colliding particles makes pressure easier to understand, thank you :) $\endgroup$
    – Khaled
    Commented Mar 17, 2017 at 23:51
  • 1
    $\begingroup$ Khaled : dont overlook the first part of docscience answer. In a condensed phase, you increase the pressure without changing density or température. Think of what happens if you press with your hand on a stack of métal. The pressure you apply is transmitted trough the stack, which means that you increased pressure in the material ; yet other properties are unchanged. $\endgroup$
    – Pen
    Commented Mar 18, 2017 at 0:18
1
$\begingroup$

I will try to explain what I think can be a possible explanation.I think it's kind of like this...

Consider a strong solid.What if you apply a thrust on it? It obviously applies a counter thrust... But did you see any density change? No. Well, I think no matter how strong a solid is, when you apply a trust, you do compress the bonds... Now when the solid is strong enough, it means that even the conjugation of very negligible compression of bonds which might be invisible to the naked eye, create a much larger effect... A restoring force actually builds up... The solid being extremely strong, so the restoring constant k in F=-kx is huge... I think of it like this. The molecular vibrations will have negligible effects.

Coming to gases. Here the molecular interactions are so negligible that when you apply a force it compresses to a significant extent thus increases pressure, one due to smaller volume(so more collisions per unit time), other due to increase in kinetic energy(and so velocity, so stronger collisions) due to work done on compression(though this effect will slowly fade away because of a tendency to attain thermal equilibrium). There is no or rather a very negligible effect due to bonds(interactions)

What I think is that in case of liquids the effect is somewhere in between. There is a restoring force due to compression of bonds. Also there is an effect(a pressure change) due to density change. These two effects combine to give the final results. It might be that in some cases the bond compression effect is strong and the density change effect is very weak or vice versa in some other case. Also maybe that in some other case the two effects are equally strong... etc etc.

So this is my idea...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.