Quick question please...

[Kram]

Sorry, excuse my simple question as I'm a newbie to hydraulic principles.

In my drawing what will happen to the free floating red piston, will it go left or not move at all?

The pressure source is the same but as you can see the piston has more area on the right hand end than the left.

Thanks for looking!

 

[Treemonkey1000]

I am not a hydraulic person at all but with all things being equal it would stay stationary wouldn't it? If pressure on left side equals pressure on right side. Now if you were to bump up pressure on the left it would certainly push the piston to the right especially if you had a low pressure or space for the large piston to push your fluid or even air for that matter on the right "tank"

 

[Kram]

Thanks for the quick reply Treemonkey.

I don't know thats why I'm asking!!

I want to know basically if the, say 2 square inches of area on the right pushes against the, say 1 square inch on the left therefore having double the force or as you say it just equalises out.

 

[Treemonkey1000]

That's the beauty of pressure difference. A small change in pressure in the smaller diameter can cause a big change. Same principle as a barometer. A resevoir then a constriction tube. So if your tank and all of it's components are equal pressure wise and temperature wise it would not move I think. Heat the fluid or air up and it is going to try and eqaulize by pushing on the piston.

 

[J_J]

That would be the same as raising up your bucket, and plugging the ports of the cylinder. Once the pressure equalizes, no more movement.

Are you trying to do anything in particular?

 

[rswoods]

I think the red piston/shaft would move left until either the shaft bottomed on the left, or the right pushed past it's cylinder. Pressure is in PSI. You have more sq in on the right side, therefore more force pushing it left. I could be off base, though...

 

[tommu56]

I think the red piston/shaft would move left. Pressure is in PSI. You have more sq in on the right side, therefore more pressure pushing it left. I could be off base, though...

It wouldn't move because of no place for the volume change left volume is less per inch than right and no were to make differance.
were is it fluid going to or coming from to make the diferance?

tom

 

[rswoods]

It wouldn't move because of no place for the volume change left volume is less per inch than right and no were to make differance.
were is it fluid going to or coming from to make the diferance?

tom

You are right. Volume of the cylinder/reservoir does come into play. While the pressure would try and move it left, the vacuum created behind the larger piston would pull it right.

I'm changing my answer. It would not move.

 

[dex3361]

It wouldn't move because of no place for the volume change left volume is less per inch than right and no were to make differance.
were is it fluid going to or coming from to make the diferance?

tom

Well I see a pressure stated of 100 PSI and I assume that it would be maintained and if so it would move left as stated in the first post by RSwoods. if its sealed how did the pressure get there? Don't you love how we can complicate a simple question here on TBN.

 

[rswoods]

Well I see a pressure stated of 100 PSI and I assume that it would be maintained and if so it would move left as stated in the first post by RSwoods. if its sealed how did the pressure get there? Don't you love how we can complicate a simple question here on TBN.

Naww.. I was wrong in my first post. The system as he drew it, because the 2 pistons are different sizes, and assuming it is a sealed system, is in a hydraulic lock, regardless of the pressure. Think about it (I did, too much). Imagine trying to move the red part to the left. The small end would try to push a small amount of oil down into the reservoir, but because the bigger piston is attached to the smaller one, it would try and suck up more oil than the small one can push down. Because oil does not expand or compress (much), it could not move.

Again, pressure in the system is irrelevant. Assuming a sealed system, the red part would be unable to move either direction.

I'm pretty sure I'm right this time.