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.

 

[Kram]

Wow, lots of replies - thanks Guys

The reason I posted was getting a headache from thinking about it!

I think the displacement factor wins - I can see now that for the piston to move left 1" would displace 1 cubic inch (actual volume not relevant) but the piston on the right end would try to displace 2 cubic inches therefore hydraulic lock.

Ok although I agree with the above result I want to add a twist - what if the pressure was a constant and supplementry medium was added to allow for displacement increases/decreases?

Lets say it's air and the tank has a compresssor always maintaining pressure and volume.

 

[J_J]

Air can be compressed, and do you have a moving force, or is this all in a static state.

 

[rswoods]

...what if the pressure was a constant and supplementry medium was added to allow for displacement increases/decreases?

Lets say it's air and the tank has a compresssor always maintaining pressure and volume.

Then the piston assembly would move left until either the small end bottomed out or the big end escaped it's cylinder.

 

[J_J]

Exactly, what are you trying to do or prove? You have set up a static condition. It is what it is, until you change something that is going to change the balance of things.

 

[Kram]

It's in a floating state and I want it to move left/right depending on addition/subtraction of the medium. If I use a gas (air) then depending on pressure or vacuum.

I have one of 'those' ideas and everything has to start somewhere.

First time here and you have all been very kind to me giving me your time, I thank you

 

[J_J]

If that is hyd oil, and a sealed container, and the lines are full, then you are in hydraulic lock. If you cut the cyl in half, and push on the small one a lot, the big piston will only move a little.

 

[SPYDERLK]

the cyl would move left and pressure would drop to zero. If forced left beyond that point the pressure would go negative [a vacuum]... If pressure were maintained by "something" it would continue left til it hit something.
larry

 

[brin]

Exactly, what are you trying to do or prove? You have set up a static condition. It is what it is, until you change something that is going to change the balance of things.

I'm with you JJ ...what is he trying to do ? Sure hope he is not trying to build a trigger device or something nefarious..LOL

 

[Kram]

I'm with you JJ ...what is he trying to do ? Sure hope he is not trying to build a trigger device or something nefarious..LOL

Darn my plans to take over the world have been exposed!

Not quite as exotic as that sorry to say, I simply need that red piston to uncover a port at a certain pressure and just exploring ways to do it. I have never studied hydraulics but get the general idea and wanted to confirm some things with 'smarter than me' people

 

[dex3361]

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.

This is correct. My above post was my attempt at humor