Does thrust force appear at the flange of a pump? 4

[KernOily]

OK dumb question. When installing/specifying/modeling expansion joints in a piping system, one has to carefully handle and account for the pressure thrust force developed in the line due to the presence of the bellows.

So here is the question. For a line without a bellows, say a pump suction or discharge line, doesn't this force also appear at the flange, and therefore doesn't one have to check this force against the manufacturer's allowable for the pump flange?

Thanks!
Pete

 

[prex]

The difference with respect to the bellow case is that with bellows the longitudinal pressure thrust goes to the supports that have to be designed to carry it.
In the absence of bellows the longitudinal thrust is taken by the pipe: of course it is present at pump flanges, but is also present during a hydrotest of the pump with covers at the flanges. Hence the manufacturer will take care of it in certifying the pressure rating and that load should not be added again to check the allowable flange loads.

prex

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[KernOily]

prex thanks for the response. So it sounds like, yes, the longitudinal P*A force is there all right, but that the manufacturer has already accounted for it in his design? So, that means if the manufacturer provides an allowable in the direction of the flange axis, then that allowable is over and above the longitudinal thrust force that is present?

And when designing the foundation, the resultant force in that direction resisted by the foundation will be (P*A) + (net longitudinal force applied by piping due to dead, thermal, seismic, etc.) - yes? Thanks! Pete

Thanks!
Pete

 

[prex]

Assuming there are no bellows in the circuit:
Yes, my opinion is that the allowable loads at the flanges are only coming from external loads (weight, wind, EQ, ...), pressure thrust need not be included, and
No, the force resisted by the foundation will not include P*A, as the longitudinal pressure thrust is resisted by pipe and equipment walls in a circuit with no open discharge and no bellows.

prex

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[KernOily]

OK great that's what I thought too. Thanks - Pete

Thanks!
Pete

 

[sitidor]

When there is no bellows in the piping,then the only thrust that could act on the pump flange is when there is a reducer in the line of the pump nozzle. In these cases, the thrust load is equivalent to the difference in cross sectional area's of the two pipe sizes multiplied by the pressure. The line of action of the thrust load will be in the direction of small pipe to large pipe.

 

[tmukhtar]

When there is no bellow in the system, there are two opposit forces acting on the nozzle. One is the thrust force exerted on the elbow or tee equal to P * pipe cross sectional area, and other is the net thrust force acting on the pump equal to P * nozzle cross sectional area. Normally the two cross sectional area are equal, so net force on nozzle is zero. If there is reducer then the two areas will not be equal and there will be net force on the nozzle as explained by sitidor.
When there is bellow in the system then the thrust force acting on the elbow or tee is not transmitted to the nozzle due to flexible bellow element. Therefore there will be net force on the nozzle equal to the p * effective bellow area

 

[prex]

Well no, sitidor, the situation doesn't change in the presence of a reducer, as this item too, like the pump body, an elbow or any other pressure retaining device, is capable of transmitting the longitudinal stress generated by pressure from the inlet to the outlet, so that this load won't go to the supports and the foundation.
The only exception is with pressure retaining devices that cannot transmit the longitudinal load: bellows (longitudinal, not angular ones) are the primary example, sliding joints may be another one, no more come to my mind at the moment.

prex

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[DLANDISSR]

74Elsinore:
Without a bellows there is no thrust forces on the pump flange. The thrust forces from the internal pressure are restrained by the pipe wall and flange bolts. The pipe wall and the bolts take the forces in tension and cancel each other.

When using a bellows with no bolts to restrain expansion of the bellows due to the internal forces a force of P x the internal area of the pipe cross section are put on the pipe and the pump flange.

Say a bellows with no restraining bolts is subjected to an internal pressure of 150PSIG. Assume the internal diameter of the bellows is 6". The force exerted by the internal pressure of 150PSIG on the pipe area (28.27sqin) is 150 x 28.27 = 4241.16pounds. The pipe must be restrained to resist these forces. The pump flange will be pushed on with 4241.16pounds. The pump probably is not designed to resist these forces. Bolts can be used across the bellows to contain these forces. Bolts across the bellows would resist this force like flange bolts do.

A better way is the put the bellows in a vertical run near the pump. Put a support at the bottom and at the top of the vertical run to resist the pressure forces or use bolts across the bellows. The expansion of the pipe will be taken up latterally (offseting the centerlins on each side of the bellows).

Another thing about expansion joints. If an expansion joint is used to absorb expansion movement axially the forces due to internal pressure and the spring constant of the expansion joint have to be overcome to move the joint. This results in very large forces. It is better to locate the expansion joint so the expansion forces will be absorbed laterally (offseting the pipe centerlines). Doing it this way means that only the latteral spring constant of the expansion joint must be overcome to move the joint. This force is much smaller than the pressure thrust forces.

Example: Using the pipe mentioned above if the expansion joint is used to absorb expansion axially and assuming the spring constant of the joint is 100lb/inch. then the force required to move the joint 1" is 4241.16pounds pressure force plus 100lb/in x 1" = 4341.16 punds. If the same joint is used to absorb the expansion latterally and assuming the latteral spring constant is 200lb/inch, then the force to move the joint 1" would be 200lb/inch x 1" = 200 pounds.

Obviously the forces being much smaller using the expansion joint latterally would result in much smaller forces being exerted on any supports consequently the supports would be smaller. Remember if a joint is used to absorb expansion latterally it should be restained with bolts across it to restain the axial pressure thrust or the pipe must be restrained to resist this force.

Hope this helps.