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#2357 - 12/28/04 01:24 AM Reaction Loads Due to Disc Rupture
Rajesh Malhotra Offline
Member

Registered: 10/16/03
Posts: 20
Loc: Malaysia
Hi,
I would like to have your opinion for the reaction loads to be considered or indusrty practise followed to calculate reaction forces for inline disc rupture. We are considering the total axial force as dP * Pipe Area * 2 ( Dynamic Amplification Factor ) but it's leading to very high reaction forces for which we have to design our stop.
Appreciate a quick response on this.

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#2358 - 01/04/05 09:27 AM Re: Reaction Loads Due to Disc Rupture
SUPERPIPER Offline
Member

Registered: 08/13/03
Posts: 405
Loc: Europe
Hello Rajesh.
I myself have only just asked the same question

Ie. if F=2pa from manufacturer, (a)
then,
is f(dyn)=2(2pa) ie:- (4pa)?

the replies i recieved would seem to conflict each other, although after re-reading my original question, i was a little unclear.

My particular manufacturer bases f=2pa on 'bernoulli' and states the assumption that all the potential energy is to be transformed into kinetic energy. he does not mention a dynamic load factor.

So according to Mr Luf, F(dyn) = 4pa.
and according to mr Wilcox F(dyn) = 2pa.

the '2' term in eq (a) just happens to be a part of bernoulli and is not a DLF

I would also appreciate clarity on this.

Best regards.
_________________________
Best Regards


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#2359 - 01/07/05 01:09 PM Re: Reaction Loads Due to Disc Rupture
Jim Wilcox Offline
Member

Registered: 12/20/99
Posts: 46
Loc: Calgary, AB, Canada
SP; I'm not sure that Mr. Luf argued in support of 4PA, and I'm not sure that Bernoulli helps us out here, but I agree that the potential energy is converted to kinetic. This is what makes the system move/accellerate. If you place a weight on a spring, the static displacement is F/K. Applying the same weight load 'instantly' and equating the 'work done' by the displacement of the weight with the 'energy stored' in the spring gives us:
WX = KX^2/2
2W = KX (twice the static weight load)
resulting in twice the static displacement maganitude when the load is applied dynamically.
Put in 100 N, get a maximum dynamic response equivalent to 200 N applied statically.
Our force is PA, and we can get as much as 2PA equivalent out.
I believe that this is what the manufacturer is addressing; their 2 is the DLF.

Rajesh; the PA loads can be very significant indeed. Here's one example: a CAESAR II user reported that an in-line 10" relief valve on a pipeline bypass opened with 530 psi upstream. The PA load computes to about 41,750 lbf before consideration of the DLF of 2, resulting in the equivalent of about 83,500 lbf applied statically. A number like that does sound hard to believe, but CAESAR's time-history analysis predicted minus 8 inches of movement in one direction, followed by +3.5 inches in the opposite direction and the damage resulting from this in-line relief valve opening supported the analysis completely.
Note that even with large loads, your DLF will not necessarily be 2; it may even be much less than 1. It depends on the speed of application of the load, the duration of the load and the piping system characteristics. If the load duration is very short, you may have a low system response. If the load duration is long and is quickly applied, you're likely to see a DLF of 2 for most relevant frequencies.
To investigate your case, you could use CAESAR's DLF generator to determine the DLF curve for your force-time profile. This will tell you what the corresponding DLF is likely to be for each natural frequency of your system. Run the Modal analysis & use the mode shapes to help determine the natural frequencies that are likely to be excited by the load and match them up on the DLF curve. This will tell you what type of DLF(s) your system is likely to see.
I wouldn't typically go to this much trouble; if I have a force-time profile and a CAESAR model, it is quite simple to enter the force-time data, perform a time-history analysis & see how the system is likely to respond.

Hope this helps.
_________________________
J.

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#2360 - 01/08/05 12:33 AM Re: Reaction Loads Due to Disc Rupture
Rajesh Malhotra Offline
Member

Registered: 10/16/03
Posts: 20
Loc: Malaysia
Dear Jim and SP,
Thanks for your valuable response.
However, I have a doubt. Since the dynamic loads are occuring due to instant rupture of the in-line disc, is it OK to treat it like the response we would typically get for a relief valve? I mean the time history analysis and calculation of DLF.

Regards.

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#2361 - 01/10/05 12:36 AM Re: Reaction Loads Due to Disc Rupture
Jim Wilcox Offline
Member

Registered: 12/20/99
Posts: 46
Loc: Calgary, AB, Canada
RM; to your question: "OK to treat it like the response we would typically get for a relief valve?"
I say yes; the differences are that:
1) a relief valve will close after some time, while the rupture disc does not.
2) The ramp-up time for the load will be faster in a rupture disc, as the disc will rupture faster than the valve will open.
Other than this, I see no difference from CAESAR's perspective.
_________________________
J.

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