Pressure loss, Vena Contracta

[blckwtr]

I need some help with finding the pressure loss in an annulus system. Very simplified:

It consists of a bar. Around it two pipes, such that it forms two annuluses. Think of this system blocked in one end, and there is drilled a hole in the inner pipe radially, such that the flow from the "outer" annulus can flow into the "inner" annulus. The flow media is water or hydraulic oil...

Dimensions outer annulus: D1 and d1
Dimensions inner annulus: D2 and d2
Dimensions drilled hole: D
Flow: Q
Pressure outer annulus: p1
Pressure inner annulus: p2

What is the pressure loss through the "port"? OR:
What is the pressure difference between the two?

In advance, thanks...

 

[BigInch]

Assuming Q1 = Q2, I think basically you're looking at the pressure difference being equal to the equivalent of the change in velocity heads between d1 and d2.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[blckwtr]

BigInch...: This is a problem occuring quite often in tools designed in our company, and I've spoken with several experienced people on the field.

If I could measure parameters before the tool was produced, I would be very happy, however this is not the case...(!)

Litterature on this does not exist. If you search this problem, you most likely end up with bernoulli equation, conservation of mass etc. (modified) But in these cases, one or more parameters are missing. The orifice plate analogy is menat to measure flow (which I already know)....

One more thing: The Vena Contracta area parameter needed in these equation are missing...

More views on this matter, people??...;-)

 

[Latexman]

What is the velocity thru the hole? 0.1 ft/sec? 100 ft/sec? A reasonable reply depends on this.

Good luck,
Latexman

 

[blckwtr]

For now, the velocity on the incompressible hydraulic oil fluid is in the area between 7-14 m/s...

 

[MikeHalloran]

I have long been under the impression that a vena contracta exists only for a jet of liquid entering a half-universe of gas.

A jet of liquid entering a chamber filled with initially quiescent liquid is quite a different thing; it eventually sets up a toroidal flow, and ...

Oh. Back to the original question.

The drilled hole(s) can be modeled as orifice(s). The problem you face is that the orifice coefficient is only accurate for a sharp-edged orifice; no chamfer, no radius, no burrs. A laser-cut hole is a decent approximation of that ... unless somebody decides to break the edges, or radius them, "to make it flow better". Yes, it does make the orifice flow better, but less predictably.




Mike Halloran
Pembroke Pines, FL, USA

 

[dcasto]

Post this in the petroleum engining area, they do annulus flow all the time. The set up will be based on a equivalent flowing area (D^2 - d^2)*3.14. The friction factor will be a "wetted area function (D + d) * 3.14.

 

[Latexman]

Let's see. 7-14 m/s is about 23-46 ft/sec. Let's call it 35 ft/sec.

Let's assume C_o = 0.6 and [β] = 0.3

K [≈] (1-[β]²)/(C_o² x [β]⁴) [≈] 120

[Δ]P = 0.0001078K[ρ]v²

[Δ]P = 0.0001078(120)(62.4)(35)² = 1000 psi

Okay, at the other end of the spectrum . . .

Let's assume C_o = 0.7 and [β] = 0.7

K [≈] (1-[β]²)/(C_o² x [β]⁴) [≈] 4.33

[Δ]P = 0.0001078(4.33)(62.4)(35)² = 35 psi

The true answer depends on the relative size of the hole to the size of the annuli. I may have bracketed the answer, but you'll need to use your exact dimensions and sharpen the pencil on the calcs (I used a shortcut orifice method and the density and viscosity of water) to get the true answer. And, don't forget the oil has to change directions 90^o right before and after the hole. I'd use K = 0.3 x 2 and the average velocity in the annuli for that.

Good luck,
Latexman

 

[Latexman]

On second thought, I'd use K = 1.2 (a branch tee, not an elbow) x 2 and the average velocity in the annuli for the change in direction right before and after the hole.

Good luck,
Latexman

 

[blckwtr]

Hello, everyone...

Back to work again...

Thank's for all your good replies. Someone is referring to a wetted area, but the losses in the annulus area is very small.

Latexman: Where do you find the discharge coefficient? And what is the ? in your equation? Is this the modified Bernoulli?

Is there some litterature you could referr to?

 

[Latexman]

blckwtr,

The discharge coefficient and equations came out of Crane's Technical Paper 410:

http://www.flowoffluids.com/tp410.htm

I see no ? in my equations. Put a checkmark in the Process TGML option in Step 2 Options. That may help.

Good luck,
Latexman

 

[blckwtr]

Oh, that is ?... (Beta) I just took copy paste from character map...