Greetings.

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-.

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 ?

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. 

58 minutes ago, 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. 

Thanks.    "times" ?

1 minute ago, Externet said:

Thanks.    "times" ?

"times" = "multiplied by". 

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 November 11, 20223 yr by sethoflagos
small clarification

Pressure equals force divided by area (P=FA P = F A ). The equation shows that pressure is directly proportional to force, but inversely proportional to area. At a constant area, pressure increases as the magnitude of the force applied also increases.

3 hours ago, Tutoroot said:

Pressure equals force divided by area (P=FA P = F A )

Correct, so you mean P = F/A