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!

This message has been edited. Last edited by: Kram,

Have a look
Now creat the equation to prove why .

A.pdf (129 KB, 54 downloads)

Thanks Nahum for taking the time to do that, much appreciated

I should point out that I figure after more thought that if that was a sealed fluid medium that hydraulic lock would occur because the smaller end would displace more than the larger end of the piston.

But the tank will have supplementry supply ability.

Just happened upon this forum post. I think that if this is a closed system and pressure is kept constant, then no movement. Movement would imply WORK. If the system was pressurized with the piston in that position, the system is in energy equilibrium. Maybe what you are calling "hydraulic lock"?

The piston will move to the left.
Nahum asked for the equation that proves why, anybody? This is easy.
The piston will not move as far to the left as Nahum showed in his picture. Why? If one is good you can derive the equation that calculates how far the actual will move.
If you are very good you can derive equations that will calculate the positions, velocities, forces and pressures every millisecond assuming a mass for the piston.

Does the system start pressurized or not?

If you move piston to the left, left side will displace less volume than it takes to fill the right side, and system will turn pressure-less after de-compressing the 100 psi starting prssure.
So when we know dimensions on the piston and the volume of contained oil, we can calculate/estimate the maximum piston travel!!
Think about the opposit direction, if an external force try to move piston to the right, we'll displace more oil on the right side than can fit in on the left side....
edit: Moving piston to the right is how you create the 100psi pressure.....

Balance of Forces...FORCE = Pressure x Area.
3 areas... area of rod end piston , area of rod and area of piston.

Area of rod + area of rod end piston = area of piston.

In this case we have one constant pressure.. atmosphere... on the area of the rod end piston.

Piston will move to the left ...( pressure in the system will drop as the volume increases) ..until all pressures are equal..which is at Atmospheric pressure

This message has been edited. Last edited by: Woodygb,

quote:

Originally posted by Woodygb:
Piston will move to the left ...( pressure in the system will drop as the volume increases) ..until all pressures are equal..which is at Atmospheric pressure

So how far does the piston move left? Since there are no values for F1, F2, and F3 the solution must be calculated symbolically.

OK....Ummm

VOL = Original Volume.
P2 = 100 psi Pressure the volume is currently at.
P1 = 14.7 psi (Atmosphere ) Pressure wanted.

Volume increase = ((e^((P2 - P1)/β))*VOL)-VOL

or ((e^((100-14.7)/β))*VOL)-VOL

Rod end is decreasing the volume as the piston moves but the piston is increasing the volume...so it's the area of the piston ( Ap )minus the area of the rod ( Ar ).

Distance moved = volume increase /(Ap-Ar)
OR
Distance moved = (((e^((P2-P1)/β))*VOL)-VOL ) /(Ap-Ar)

I think...

Cheers Woody

This message has been edited. Last edited by: Woodygb,

quote:

Originally posted by Woodygb:
OK....Ummm

VOL = Original Volume.
P2 = 100 psi Pressure the volume is currently at.
P1 = 14.7 psi (Atmosphere ) Pressure wanted.

Volume increase = ((e^((P2 - P1)/β))*VOL)-VOL

or ((e^((100-14.7)/β))*VOL)-VOL

Rod end is decreasing the volume as the piston moves but the piston is increasing the volume...so it's the area of the piston ( Ap )minus the area of the rod ( Ar ).

Distance moved = volume increase /(Ap-Ar)
OR
Distance moved = (((e^((P2-P1)/β))*VOL)-VOL ) /(Ap-Ar)

I think...

Cheers Woody

You get an A! I see you used e. You would have got an A+ for using e but you said "I think".

quote:

Originally posted by Peter Nachtwey:
You get an A! I see you used e. You would have got an A+ for using e but you said "I think".

quote:

Originally posted by Woodygb:

quote:

Originally posted by Peter Nachtwey:
You get an A! I see you used e. You would have got an A+ for using e but you said "I think".

When Nahum posted his picture I immediately estimated the piston would move 0.0005*Vol/(Ap-Ar). Oil compresses 0.5% for every 1000 psi. The pressure drop in the system was a little less than 100 psi which is 1/10 of 1000 so the system would compress 0.05% or 0.0005. I then did the algebra to get the Vol/(Ap-Ar) so I knew approximately how far the piston would move. It could move almost twice as far depending an what is assumed for the bulk modulus of oil.

That is the gut feel, sloppy, quick and dirty, way to estimate what will happen.

Wasn't there one little piece of information missing on the input data....those 100psi, is that absolute pressure or pressure over atmospheric pressure....on such a low system pressure (100psi), that could make some relativ differens huh???
Just some thoughts.....

Btw, that "e" component in the forumla....can you guys please explain that one....(e-logaritm???)

Good job Woodygb.....

quote:

Originally posted by AKKAMAAN:
Btw, that "e" component in the forumla....can you guys please explain that one....(e-logaritm???)

http://www.ndt-ed.org/Educatio...rces/Math/Math-e.htm

The volume expansion is non linear hence the use of e.

Woody