Density constant along streamline of incompressible fluid

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  • #1
jmz34
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Question: Show that for a steady flow with div.u=0, the density is constant along streamlines.


I just don't see how to approach this question without Bernoulli's equation, I can see that the Lagrangian derivative of the density is zero but that doesn't specifically show that the density is constant along streamlines.

Thanks.
 

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  • #2
tiny-tim
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hi jmz34! :smile:

try proving it the same way you would for Bernoulli's equation …

consider a section of a tube bounded by streamlines :wink:
 
  • #3
jmz34
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Hi, thanks for your help- but I just need a bit more guidance.

By considering a tubular volume bounded by streamlines, I've considered the rate of change of mass inside this volume and equated it to the momentum flux through the surface.

This has led to the integral expression for the continuity equation- expanding out div.(pu) (where p=density and u=velocity), I've been able to use the fact that div.u=0 and I'm left with the remainder of the integral expression:

INT(dp/dt)dV=INT(u.grad(p))dV

I don't really see why u must be orthogonal to the gradient of p.

Thanks.
 
  • #4
tiny-tim
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hi jmz34! :smile:

(have a rho: ρ and a grad: ∇ :wink:)

i'm not sure what you've done :confused:

since no fluid can go through the sides of the tube, all you need do is consider the rate of gain or loss of fluid at the two ends of the tube :wink:
 
  • #5
jmz34
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hi jmz34! :smile:

(have a rho: ρ and a grad: ∇ :wink:)

i'm not sure what you've done :confused:

since no fluid can go through the sides of the tube, all you need do is consider the rate of gain or loss of fluid at the two ends of the tube :wink:

Surely this is not a simple matter of saying:

(d/dt)INTρdV = INT(over end A) ρu(r).ds - INT(over end B) ρu(r+dr).ds

Where A and B are the ends of the tube.

I'm under the impression that it cannot be assumed that the densities at A, within the volume V and at B are the same.

This is probably a simple problem so sorry for not seeing it already.
 
  • #6
tiny-tim
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ok, assume ρ depends on distance, and write it …

(d/dt) ∫ ρdV = ∫ (over end A) ρAu(r).ds - ∫ (over end B) ρBu(r+dr).ds …

(and then replace each ∫ by the area :wink:)
 
  • #7
jmz34
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ok, assume ρ depends on distance, and write it …

(d/dt) ∫ ρdV = ∫ (over end A) ρAu(r).ds - ∫ (over end B) ρBu(r+dr).ds …

(and then replace each ∫ by the area :wink:)

I'm still very confused.

Do we have to assume that the areas A and B are the same? Surely we can imagine a scenario where the streamlines diverge and so A and B are different.

Thanks very much for your help.
 
  • #8
tiny-tim
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no, we assume A and B are different …

you should get AρAuA = BρBuB

(oh, i've just noticed, it needn't have been u(r) and u(r+dr) … the tube can be as long as we like :wink:)
 
  • #9
jmz34
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no, we assume A and B are different …

you should get AρAuA = BρBuB

(oh, i've just noticed, it needn't have been u(r) and u(r+dr) … the tube can be as long as we like :wink:)

So you're assuming that the mass of fluid within the tube is constant? I don't see how it's ok to say this, I mean if the density can vary- the mass within the tube can also vary?

But if I can understand that then the expression you've given makes sense since the velocity times area along the tube (the flux) must be constant.
 
  • #10
tiny-tim
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So you're assuming that the mass of fluid within the tube is constant?

no, I'm saying that mass out minus mass in = that integral, and if you put du/dx = 0, you should get that integral to be zero :smile:

(for example, using the divergence theorem)
 
  • #11
jmz34
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Ok thanks alot for all your help :)
 

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