Pressure drop in annulus of two different materials
Pressure drop in annulus of two different materials
(OP)
We are in the process of performing the hydraulic calculations for a series of annuli that consist of an inner pipe of HDPE and outer pipe of carbon steel. Does anybody have any experience in taking into consideration the two different friction factors present on the outside flow?
Thanks,
Colin
Thanks,
Colin





RE: Pressure drop in annulus of two different materials
If the flow is turbulent, that implies a large pressure drop. If that's true, or if there's any restriction downstream, you may have other problems.
Check the buckling pressure on the inner pipe.
Mike Halloran
Pembroke Pines, FL, USA
RE: Pressure drop in annulus of two different materials
The Reynolds number is app. 4400, so it is considered turbulent flow. The annuli are part of a ciculation loop. The fluid is pumped through the center (HDPE) pipe and then it comes back up on the outside, so the inner pipe will actually have a greater pressure on the ID than the OD. We are trying to size the pumps for the loop. I was hoping to see if there was a way to take into consideration the different materials. It is not a necessity, but it would help in not overdesigning the pumps.
RE: Pressure drop in annulus of two different materials
I would be more concerned over the method of calculation of friction losses in the annuli?
RE: Pressure drop in annulus of two different materials
From what I have seen, zdas04 is the guy with the most experience of this application. See his comments in the following threads
thread378-111534
thread378-100815
thread378-99074
Depending on the temperature of your fluids, it may be necessary to check the differential expansion of HDPE and steel.
Katmar
RE: Pressure drop in annulus of two different materials
Thanks for the vote of confidence.
Chestilow,
As I said in the threads referenced above, the tricky part is defining the effective diameter of the annulus. The technique that I've found matches measured data very well is:
d = ((IDouter+ODinner)2(IDouter-ODinner)3)(1/5)
Then you just use the result in your flow equations. I've gotten good results with this technique in vertical flow with steel casing and spoolable composite inner pipe (I've never had much luck hanging HDPE in wellbores, it tends to creep under a tensile load). I used the friction factors for steel.
For horizontal pipe I've used steel inside of steel and gotten good results when the pipe had centralizers. Without centralizers the AGA equation with an epsilon for new steel significantly understated the pressure drop.
David Simpson, PE
MuleShoe Engineering
www.muleshoe-eng.com
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
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RE: Pressure drop in annulus of two different materials
David
RE: Pressure drop in annulus of two different materials
Thanks for the help. Could you please let me know what the origin of the "Petroleum" method is? I've looked in several references and haven't been able to locate it elsewhere. I know that if I present it to the Project Engineer, he will want to know where it came from.
Thanks,
Colin
RE: Pressure drop in annulus of two different materials
For a more mainstream source, try Petroleum Engineering Handbook Edited by Howard Bradely and published by the Society of Petroleum Engineers. In the Feb, 1992 printing it is on page 34-27 (equation 49).
Neither book does a great job of deriving it, but it isn't the only empirical eqation in Petroleum Engineering. I've tried to derive it myself, but I always get sidetracked before I make much progress (it is so easy to get sidetracked in Katz, there is so much good information that every time I open it I find myself wandering around).
In both references, they show D5 because that is the diameter they need for the flow equation they are describing. I've had really good luck taking the fifth root and using the result.
David
RE: Pressure drop in annulus of two different materials
* Darcy - Weisbach, inner pipe (after some algebra):
h=[(8fLQ^2)/(g*(pi)^2*d^5)]
* Darcy - Weisbach, outer pipe, hydraulic diameter:
h=[(fL/(do-Di))*(u^2)/(2*g)]
u=[(4Q)/(pi*(do^2-Di^2))]
hence h=[(8fLQ^2)/(g*(pi)^2*{(do+Di)^2*(do-Di)^3})]
I hope I haven't messed up on the brackets and parentheses, but I think anyone reading this can easily figure out what I meant in the event that I have.
In the above, the theory that I follow uses the uncorrected hydraulic diameter dh=(do-Di) in the Darcy - Weisbach equation, and then applies corrections to compute an effective diameter de=dh/y and an effective Reynolds Number Re=R/y, where a curve fitting algorithm for y produces an equation of the form:
y=[-0.763 + 1.593*(Di/do)^0.5 + 1.844*exp(-Di/do)].
When you compare the two expressions for dynamic head loss h as given above, you can see the similarity to the expression for effective diameter used in the "Petroleum Engineering" method. The difference is, I am inclined to not exactly equate the effective diameter as the fifth root of the expression and treat it as the equivalent diameter of a circular pipe. Rather, I am inclined to take the head loss directly from the above.
The reference I have used is a textbook:
FLUID MECHANICS - Frank M. White
Second Edition, 1986, McGraw-Hill Inc.
(see Example 6.14, page 329)
If anyone would like to see my derivations (or check them in the event I am out to lunch), including the iteration scheme applied to the Colebrook equation to determine the friction factor, let me know.
SNORGY