Need some help with a math question

[Cbrunner]

I swear I've lost my marbles... they are somewhere around here.

A cylinder has a PO operated check on both rod and blind ends. The ratio of these checks is unknown. The cylinder is extending and lifting a load of 10,000lbs. The dimensions of the cylinder are 5 x 1 x 30. The cylinder is stopped mid stroke and both checks close. The cylinder develops a leak across the piston seal. What would the pressure gauges at the blind port and rod port read once the cylinder stops moving.

I know the gauges will equal out, and cylinder drift will stop before bottoming (if there is enough fluid to lock it). I know it takes about 510 psi to raise the load. Why is 510psi not the correct answer, and how do you find the correct answer?

Thanks;
Chris

 

[hydraulicsman]

Hello,
The forces must equalize for the cylinder to halt retraction.
444482.2N force
3.517MPa to hold
Leak in piston seal - Pressure must equalize on both sides now
Presuming 35.17bar in FA when po checks seat.
Pressure in AA unknown presuming 0
Cylinder Area ratio = 1.04:1
Presuming no leaks and fluid incompressible
AA new pressure due to leak = 3.517MPa
AA new force due to leak = 87252.3792N
FA new pressure due to leak = 6.88MPa
FA new force is surprisingly enough = 87252.3792N
At this point the cylinder is hydraulically locked. But can be displaced back and forward within the volume limits.
I think that's right. But do shoot me down in flames if it's wrong.

 

[Cbrunner]

Ok, let me describe the process and please correct me if I am wrong. First, you found what pressure would support the load on the blind end of the piston (described as FA). The pressure from FA would equalize to the AA, applying 510psi (3.517Mpa) additional force with the 10Klb load, causing the blind area (FA) pressure to rise to meet the new demand for pressure (AA+Load).

Correct?

 

[hydraulicsman]

Good morning,
Yes that's correct.

 

[hydtools]

When the piston seal leaks, the effective force area is that of the piston rod. So pressure will increase to support the load.

Ted