May a 2-phase fluid be treated as incompressible? 1

[tundres]

Hi!

Theory question...I have a fluid in a tube. The fluid is at the beginning water and at the end vapor because the tube gets heated.

Suppose I know the head loss because of the friction...it is ok to apply the bernoulli principle, if we take into account, that the fluid have a low velocity?

I hope somebody can help me, maybe it's a trivial question for some of you but that question it's driving me crazy the whole day...

 

[zdas04]

"Incompressible" is a relative term, not an absolute term. What you see most often is that if pressure at the outlet of a system is greater than 90% of inlet pressure then the change in density can be assumed to be trivial and ignored. This assumption results in really good results for single phase flow as long as the maximum velocity is less than about 0.6 Mach.

On the other hand, multi-phase flow pretty much invalidates any constant-density assumption. During annular, stratified, or wavy flow the gas which is several orders of magnitude less dense than the liquid and the amount of liquid at any given point is unknown (and the combined density can easily vary by far more than 10%). For mist flow, the constant density assumption is better, but it still is awful.

I had occasion to research the state of the art in multi-phase flow pretty carefully a few years back. The best multi-phase correlations were about +/- 50%. The worst were never even that close to "right". I've found the old Flanagan and Duckler equations to do a better job of representing multi-phase pressure drop than any of the new-fangled CFD models.

In short, for the gas flows, you have little risk using an incompressible assumption. For the liquid flows, nothing you do is going to be much more incorrect than anything else you do.

David

 

[tundres]

Hi David,

thanks for the answer and the time you took to write it. I really appreciate that help.

I'm aware, that 2-phase correlations are not really accurate, well I didn't know it was that inaccurate...but what I really what to know, if it's correct to apply the incompressible theory to that problem, not the get the pressure at the end for example but to get the head lose cause by the fluid.

Let's say I have a pipe with a 2-phase flow in it. I know all about it at the beginning and at the end of the pipe: pressure, velocity, temperature, etc. So I get the bernoulli and say (there is not a height difference):

(P2-P1) + ((c2)^2-(c1)^2)/2 = Friction head loss
where:
P2: pressure at outlet
P1: pressure at inlet
c2: mean velocity at outlet
c1: mean velocity at inlet

That's the point of my question, if I do anything wrong from a theoretical point, if I measure in an experiment, the pressure and velocity and what I want to get is the head loss because of friction with a simple bernoulli.

I read in a book, that the one phase theory can be applied there, if the velocity of the flow is small. Is that correct?

 

[zdas04]

Your equation is a rank bastardization of some really elegant arithmetic.

The Bernoulli Equation was a very clever derivation of a special case of the Euler Equation (which was a special case of the Navier Stokes Equation). Euler said that if the flow is incompressible (and inviscid and non rotational, etc) then you can disregard a bunch of the uglier terms in Navier Stokes.

Bottom line of this foray into Fluid Mechanics history is that Bernoulli is rarely valid for pipe flows. When people add in a "head loss" term and call it something like "Modified Bernoulli Equation" they are ignoring the derivation and talking nonsense. The underlying assumptions of both Euler and Bernoulli were that there is no parasitic losses in the flow. If I subtract the Bernoulli Equation at point 1 from the the Equation at point 2, I don't get the equation you referenced. The equation states that the sum of the flow energy at any point in the flow equals a constant. Therefore the difference between the values at two points is zero, not "Head Loss".

Bernoulli's Equation works really well to relate the dP across a known orifice to a flow rate. Using it for flows more than a few feet in a pipe (especially if there might be sloshing liquid in the middle of the pipe), doesn't work very well.

David

 

[BigInch]

2-phase flow incompressible? Only in a perfect universe. But then 2-phase flow wouldn't exist in a perfect universe. Why bother.

 

[tundres]

I suppose, my question looks a little bit simple but my idea was not from bernoulli itself.

In a pipe, assuming one dimension flow, the difference between bernoulli and an energy balance is the heat and the change in the internal energy of the flow. So it looks like I cannot neglect the change of the internal energy under any circumstances.

Thank you for the answers!

 

[stanier]

http://www.drbratland.com/PipeFlow2/PipeFlow2Multi-phaseFlowAssurance.pdf

Perhaps the answer is in this tome? Previous posting by Ione.

 

[BigInch]

You obviously must consider compressibility, if only to determine the flow regime, all other considerations ignored. Given a certain amount of air and water in a pipe volume, at low pressure you may have such a large volume of air that 2-phase flow is unavoidable, however at high pressure the air could compress to such a small volume in relation to the water that it becomes less than 2% of the pipe volume and its effects might be neglected with total impunity. One of the first criteria for beginning a 2-phase flow analysis is to determine the gas to liquid volume ratio.