Big differential pressure when opening sleeve

[kennni]

Big differential pressure when opening "sleeve"

Hi

If you have a "sleeve" that is covering a hole (or multiple holes) , and this sleeve is just sliding over the holes , acting as a lid. If the differential pressure is 8-9000 Psi , is there any way to calculate the force needed to open the sleeve ? The fluid can be water or oil or gas.

And are there any suggestions on how to minimize the force needed to drag the sleeve to get it open ( even out the pressure ). I would think the sleeve will experience a fluid "lock" of some sort , just before the fluid enters and equallizes.

Would it help to have a REALLY small hole in the beginning ? or grooves

The "holes" are 0,4inch diameter and there are 24 of them.

Thx

 

[tasp77]

If I understand the inquiry properly, if the holes covered by the sleeve are symmetrically arranged, and the sleeve is strong enough to not distort sufficiently to bind the device due to the pressure, the force needed is small.

You have something that is 'kinda' the reverse of the hydraulic controls for a backhoe. On those machines, a machined rod with various indentations, slides through a cast block of iron with internal passages to achieve various internal connections as the rod is moved in and out. Those backhoe controls can be directly operated by a human operator, and as I recall, they have springs added so they 'center' themselves when not pushed or pulled. Some have 'detents' to lock them in an activated position. These controls can also be activated by electric solenoids, compressed air, smaller hydraulic controls, etc.

The forces involved are quite low and these devices are commonplace. Enormous amounts of power can controlled by these devices, a large piece of construction equipment might be pumping 200 gallons (or more) of hydraulic fluid at 3000 PSI per minute. IIRC, 15 GPM at 3000 PSI takes 30 HP.

 

[AlephZero]

One important thing is the pattern of the holes.

Your 24 holes have a total area of abut 3 sq in. If you have a cylindrical pipe with a cylindrical sleeve, and all the holes are lined up on one side of the pipe, the fluid will extert a total force of about 8000 x 3 = 24000 lb (11 tons) pushing the sleeve sideways in one direction. That will be balanced by the contact force between the opposite side of the sleeve and the pipe.

For that (bad) design, you would have to overcome the friction between the pipe and the sleeve, which might be 1 ton or more.

On the other hand, the holes are arranged in a symmetrical pattern around the circumference of the pipe, and they are uncovered in a symmetrical way as the sleeve operates, the reultant force on the sleeve from the fluid pressure will always be close to zero, so the force required to move the sleeve will be small.