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Collapse Pressure of Subsea Pipes

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maloneb

Mechanical
Aug 11, 2011
22
Hey guys,

I don't know if this is posted in the correct section or not, I'm new to the site.

I'm currently working on a design tool to calculate the nominal wall thickness of a subsea pipe for variable inputs. In order to find the wall thickness for collapse, Equation G.1 from from PD8010 is manipulated to find the value of Pc. The equations is given below.

((P/Pe)-1)*(((Pc/Py)^2)-1)=(Pc/Py)*((Fo*Do)/tnom)

Where P is external pressure, Pe is critical pressure for an elastic circular tube, Py is the yield pressure, Pc is the characteristic external pressure (collapse), Fo is the initial ovalisation of the pipe cross section, Do is the outer diameter and tnom is the nominal thickness of the wall. This equation is rearranged to a value of Pc through use of the quadratic equation.

The program has been written so that for a given value of P, tnom is increased until the Value of Pc is greater than the value of P.This gives a nominal value of the minimum wall thickness needed to resist collapse of the pipe. This section of the program works fine, giving the same wall thickness required as in a previous version of the tool using a different method.

The problem arises when a value of Pc is needed further into the tool. The value of Pc is about 5 MPa greater than in the previous version. I believe there is a fault with how I have rearranged the equation.

So my question is, how would you rearrange the above equation to give a value of Pc, and what would the reformulation be?

Thanks

Ben
 
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With Pc on both sides of the equation as it is there, you will not get a solution for Pc. It will remain a function of itself, thereby requiring an iterative solution, so why waste time doing anything more than dividing the left side by ((Fo*Do)/tnom) and leave it in terms of Pc = Py * ((P/Pe)-1)*(((Pc/Py)^2)-1)

We will design everything from now on using only S.I. units ... except for the pipe diameter. Unk. British engineer
 
Or perhaps you have and error in the equation. 5 mPa is less than 1 psi, so its a roundoff error or something, no. Why are you recalculating it later on anyway. You already have the value for it, so just make the variable global and recall it and it will be the same.

We will design everything from now on using only S.I. units ... except for the pipe diameter. Unk. British engineer
 
But if you rearrange the equation (which is LENGTHY) you can get a quadratic with Pc as the subject. There is only 1 real solution to the equation, which then means that the correct thickness can be found.

Its being programmed in VBA on Excel, so a different function has to be made to present Pc. But the value of Pc is larger than in the previous version. I'm not sure if it makes a difference, tollerences are only altered VERY marginally. If you couldnt tell, I'm sort of struggling to get my head round where to go next with it.
 
I think the problem lies not with the equation, provided it is correct, but with the reason you get a different answer later. Why is that? What variables have changed?

We will design everything from now on using only S.I. units ... except for the pipe diameter. Unk. British engineer
 
I've made a bit of progress. The job I'm doing is essentially updating an old design tool that uses old formulae to find the value of Pc and loops an increase in the nominal thickness until Pc is greater than the external pressure.

This is how my model works, in the same way yet with different equations. Now when it is used to find the nominal thickness, Pc is not displayed, yet once the thickness is found, the closest matching ASME Pipe Size ABVOVE the meaurement is used. So the value of Pc is always greater than in the previous document, but not large enough to cause an increase in ASME Pipe Size.

So the issue that I really have now is finding why the value of Pc differs between the two methods of calculating it. Which is extremely difficult to do when the original document explaing the use of the previous formulae has been misplaced!
 
Isn't that in the VBA code.

There shouldn't be any reason to use the next standard wall thickness. Being a pipeline, you more than likely have enough pipe to order that you should be buying directly from the mill and buying the exact wall thickness that you need, not what they happen to have in stock. The potential to save millions is there, if you pipeline is long enough. I would NOT program that into the solution, but let that be an option for the engineer of record to choose whether to buy the exact wall thickness, or to kick it up to the next size. In any case IMO the program should NOT be making that decision!!!!

We will design everything from now on using only S.I. units ... except for the pipe diameter. Unk. British engineer
 
These references may be helpful:

ref 1: "Tubing Limits for Burst and Collapse", Tech Note, CTES, L.C., Conroe TX
ref 2: "Effect of Initial Eccentricity on Collapse Pressure of Circular Beam Tubes", S. Yadav
Fermi National Accelerator Laboratory, Batavia IL


Mike Halloran
Pembroke Pines, FL, USA
 
Got it all sorted now guys. What with being a newbie at this company, I hadn't been informed of some other procedures. So finally (2 weeks in!!) the solution has been made available to me. We all have egg on our faces.

But cheers anyway!
 
What's this "we" kemosabe.

We will design everything from now on using only S.I. units ... except for the pipe diameter. Unk. British engineer
 
By we I meant me and the team I'm working with.

In the end, a missing document was discovered that illustrated the procedure for using the equation. The procedure was the same as in the original version, making everything I had been doing with the equation I posted meaningless.

Was a fun waste of 2 weeks though.
 
refer to API RP 1111 for Collapse calculation or DNV OS F101,Oct 2010.

Thanks.
 
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