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Force exerted by pressure...


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A hose/pipe/tube is friction push-inserted into a vessel tight hole.   Transport yourself to the interior of the pressurized vessel.  You can see the inserted pipe end that supplies the pressurized -say air-.

3" x 21 ft Schedule 40 Black Steel Plain End Pipe | Fastenal

There is in-diameter and out-diameter, corresponding to inside area; outside area, and the smaller torus area from the pipe wall.

What is the force the air pressure in the vessel pushes the pipe out towards unplugging it ? ---> force = pressure / area

Which is the area to consider ?  inside, outside, torus ?

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On 11/10/2022 at 7:16 PM, sethoflagos said:

The force acting to propel the pipe back out of the hole in the vessel is equal to absolute fluid pressure times the area of the hole in the vessel. 

To do justice to your question, @Externet I should add that strictly speaking my response corresponds to the 'design load' on the nozzle when there is either no flow (due to eg a closed valve), or when the fluid is sufficiently viscous that its shearing force on the pipe wall far exceeds its gain in momentum. 

At the other end of the spectrum, we could in theory propose the unrestricted flow of a zero viscosity fluid where none of the fluid inertial acceleration is lost to shear at the pipe wall. In this case, the only axial force acting on the pipe in opposition to the restraining force of the interference fit, would be the vessel pressure acting on the pipe thickness (your 'torus' case). 

Real flowing cases should be expected to fall somewhere between these limits. 

In short, good question. Deserved a more considered answer.

Edited by sethoflagos
small clarification
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