We are connecting a straight ~ 5' run of pipe to a vendor-supplied skid, containing some pumps, bypass lines, and instruments, etc. They have given us an allowable force on their 3" flange as 50 lb (axial). This is steel pipe, design temp of 180 degrees. My question is this: do I model their flange as an anchor or as a nozzle? If I model as an anchor, the resulting load result was around 50,000 lb in operating. First off, I am not sure how Caesar came up with this number, since the axial stress is around 19,000 psi, and we're dealing with a 3" pipe. It should result in the force in the pipe wall. On the other hand, how would this be modeled as a nozzle, since there is no vessel wall thickness, nor any distance to stiffeners, etc.

As another question, aside from the modeling method: this same vendor had specified the piping components to us to use in our portion of pipe, (they will supply said compoents, all we need to buy is the pipe itself). But, they made no mention of an expansion joint. with the stresses as 27.9e6 psi*(8.75e-6 in/in/F)*110 degrees / 12.2 sq. in. of flange face area, the resulting load is still way above allowable. Am I losing my mind, or do we need to go back to our vendor and tell them that since they are giving us such low allowables, they need to supply some expansion joints?

Thanks,

Flexible's as my colleagues in the UK call EJs' are not without problems... they create pressure thrust issues which in your arrangement may be considerable....


Your best bet is to model back into the vendor piping to some point of 3 way fixity....

as for what CAESAR II is telling you its telling you it takes a lot of load to compress a cylinder of metal with no bending in it whats so difficult here?

Thanks for the quick reply John,

In reply:

- the particular portion that I am dealing with right now is a 5' run from the supply tank to the inlet to the skid (i.e. the pump suction). So, the only pressure thrust would be any pressure resulting from the 20' of head in the tank (at it's fullest). I haven't checked with vendors of EJ's yet, but am pretty sure there are restraint rods capable.

- as i mentioned, it is a vendor supplied skid. we do not know the true layout of the internals, much less the support locations. All we have is the coordinates of the flange, as well as the allowable load. But, if it is anything like any other skids I've seen, I am thinking the pipng internal is short and striahgt into the pump. Now, it is possible that there is indeed an EJ in their skid, but the vendor is very difficult to deal with and get any answers from )i.e. how we don't have the skid info), so I will not make assumptions.

- I'm still not sure of how an axial stress of 19 ksi results in an axial load of 50 ksi in a 3" line. S still equal F/A, right?

Connecting a straight pipe from Tank nozzle to pump nozzle without expansion joint is not correct.

The best method is to either provide an expansion joint or change the pump nozzle orientation to 90 deg with tank nozzle.

You might have modeled the nozzles of pump and tank as anchors, this results in high loading of nozzles.
A 3" sch 80 straight pipe subjected to a temperature change of 80 deg F between 2 anchors will produce around 50 ksi of force irrespective of length.
This is the force required to compress the pipe by an amount equal to linear thermal expansion of the pipe.

stress = Mod E * Delta

Mod E is a huge number, so the stress calculated is high and obviously the resulting force is high I do not understand why you do not understand the problem.

The codes are based upon the ASSumption that the thermal displacements act primarily in bending you have a straight cylinder of metal with no bending.

I suggest that you are swimming in water over your head and your work should be carefully reviewed by a competent practitioner in this field. To not understand this simple bit of physics indicates that you lack any credible insight.

Also watch out if your supply tank is subject to differential settlement relative to the skid. If a small "tank' then probably not; if ring wall supported storage tank then probably yes.

Delta= P*L/(A*E), solve for P.

If you cannot reorient (Touseef's good recomendation), model into the skid (Luf) and include some local flex values at the tank.

In reality as mentioned in other posts, things give/deflect but quantifying this 'give' can prove difficult at times. The CII default stifness values are quite high. In the old days one could just say the expansion of 0.0x" is insignificant and be done with it. If we run everything in CII, the numbers will be large unless the 'give' can be quantified.

Doing all this extra work for a 3" line at 180 deg F (design?) seems like a waste of time but unfortunately people (clients) seems more comfortable with computer runs.