Calculating Max. Distance of Flying Lead

[Chris718]

Hi Guys,

New to the forum, so a big hello to everyone on here! I was wondering if someone could point me in the right direction of how to calculate tha maximum distance (Distance 'X' in attachment) of a flying lead (oil and gas subsea equipment), suspended between two stabplates (another O&G term), with the length of flying lead given (50.4m), and requiring a minimum bend radius of 1400mm from the stabplate. The distances from the centre of the two stabplates to the seabed are also given. See attachment.

Many Thanks.

 

[Dougt115]

Questions

Are the connection points to the stabplates 90 degrees? Meaning straight down. Or do they pivot to allow a constant radius from end to end of 1400mm?


If this is homework, you cannot post it here and it will soon be deleted.

 

[racookpe1978]

What is the shape of the curve EFL: two 90 deg bends, two 90 bends and a flat pipe, and catenary suspended from
(1) a vertical but fixed "pipe" that then begins bending like a flex hose
or (2) a flexible hose NOT vertically fixed to the original horizontal pipe

 

[chicopee]

According to your sketch, it appear that the EFL is freely suspended between two anchors in water and consequently would assume the shape of a catenary if the EFL has uniform unit weight, uniform cross section, and no cross current flow to alter the shape. Assuming a catenary see the attachment. I have calculated your "X" value to be approximately equal to 2800mm. From that data the length of the EFL is estimated at 4175mm which is about 1/10th of your length of 50.4m. Somewhere the units either in in the length of the EFL or depth to the sea floor are off.

 

[IRstuff]

Your scenario is underspecified, specifically with respect to allowable tension in the EFL.

Attached is a mathematically correct solution in Mathcad, but I suspect that the droop that you've specified requires tension well beyond the breaking strength of the EFL. The end result to meet the droop constraint gives a spacing of 50.2 m, which seems dubious, but I don't see any math errors; given an overall length of 50.4 m, this is a completely unrealistic solution. Nevertheless, the solution is in the general ballpark, given the rather tiny droop required. I suspect that the physically plausible result is that your EFL will actually be lying on the sea bottom, with the stab plates about 48 meters apart

Catenary.pdf
catenary.xmcd

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

Of course I can. I can do anything. I can do absolutely anything. I'm an expert!

 

[racookpe1978]

It's a totally unrealistic problem - which is a clue to it being a homewrok/textbook issue only - but ...

If it is a catenary droop of a theoretical constant-weight, constant cross-section pipe, doesn't the path solved by theory (by the program itself and the math endpoints and their slopes) depend completely on the required end point limits?

Thus: The ends (the two stations or mounts) have to fixed rigidly in positions, the ends of the catenary (pipes) have to be fixed with position and direction (their final slope of the joint), and the middle have to be freee to fall? So you's solve for the catenary curve, set the end angles, re-solve the catenary, and then check again that the pipes and ends align. (No inflection points)

If the ends are on anchors, then both ends would be pulled towards one another.
The ends need to be freely "pinned" so they can rotate to the calculated angle.
If the water is that shallow (limited to his 50 meter droop) then the pipe comes out, goes down to the bottom, trails across the bottom, then goes up to the other end.
If the water is deep enough to require or allow the pipe not to be touching the bottom, it is has to be suspended in a catenary (or "U" or droop) then it cannot be a down, turn (bend), run horizontal, turn up (bend), run vertical as he has drawn the sketch.