Liquid pressure in a closed system

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SUMMARY

The discussion centers on calculating the pressure increase in a sealed steel cylinder filled with water when a solid piston enters. The initial pressure is 3000 psi, and the calculations involve isothermal compressibility of water (β = 3.0621e-6 psi-1). The first user calculated a new pressure of 4429.9 psi after the piston enters, while another user provided a revised calculation yielding approximately 3142 psi. The discrepancy highlights the importance of accurately accounting for displaced volume and the correct application of the compressibility equation.

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klk
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Homework Statement


A steel cylinder filled with water contains 3000psi of pressure, completely sealed, walls ≈ infinite thick. No gas is present inside of the cylinder, and no heat exchange. To the problem; if a solid rod/piston enters on top of the cylinder with no possibilities to bleeding off the pressure, what will the pressure build up be if the rod enters the cylinder with 1 ft?

Height of cylinder 500ft, ID of cylinder 8,535in, OD of piston 4in, Temperature of water 333,15K→ Isothermal compressibility of water β= 3,0621e-6 [psi-1]


Homework Equations



β=(1/V)*(∂V/∂P)
F=P*A


The Attempt at a Solution


Volume before entering the the piston V1= 35,4bbl, after V2 = 35,384 bbl
Solved the above equation for P2 = P1 - ((V2-V1)/(V1*β)). This gave a new pressure inside the cylinder of 4429,9 psi.
I think this pressure is too high compared to the displaced volume that the piston occupies. I am a bit uncertain if the equation above is correct to use for this situation.

Please help, thank you.
 
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klk said:

Homework Statement


A steel cylinder filled with water contains 3000psi of pressure, completely sealed, walls ≈ infinite thick. No gas is present inside of the cylinder, and no heat exchange. To the problem; if a solid rod/piston enters on top of the cylinder with no possibilities to bleeding off the pressure, what will the pressure build up be if the rod enters the cylinder with 1 ft?

Height of cylinder 500ft, ID of cylinder 8,535in, OD of piston 4in, Temperature of water 333,15K→ Isothermal compressibility of water β= 3,0621e-6 [psi-1]


Homework Equations



β=(1/V)*(∂V/∂P)
F=P*A


The Attempt at a Solution


Volume before entering the the piston V1= 35,4bbl, after V2 = 35,384 bbl
Solved the above equation for P2 = P1 - ((V2-V1)/(V1*β)). This gave a new pressure inside the cylinder of 4429,9 psi.
I think this pressure is too high compared to the displaced volume that the piston occupies. I am a bit uncertain if the equation above is correct to use for this situation.

Please help, thank you.
Hi klk. Welcome to Physics Forums!

I calculate a pressure increase that's a factor of 10 lower than yours.
 
Thank you for the input, that actually helped me!

I simply added up the volume it takes to pressure up to 3000psi and the displaced volume from the piston -> dV.

P2 = dV/(V1*3,2e-6), which gave me a new pressure of 3142 approx..
 

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