Hydraulic Press Formula

CatWoman

Hi, this really isn't a homework problem - it's posed in a text book I am using to revise having not used physics for 15 years - I'm now training as a science teacher.

Can anyone prove the hydraulic formula to raise a mass m by distance d where the force on the piston must be increased by delta F = rho g (A1 + A2) d

where A1 is the area of the piston where the force is being applied, A2 is the area of the piston on the side supporting the mass m. rho is density of hydraulic fluid and g is accn due to gravity.

Have I explained this clearly? Any help to a very rusty physicist greatly appreciated!

redargon

My guess is that you start equating both sides of the equation for a simple hydraulic system.
You know that the pressure in the system is constant, so the force generated on the platforms is dependant on their areas due to F=P*A
F - Force
P - Pressure
A - Platform area (or mass area, etc)
So F1 = P1A1 and F2=P2A2
Where the 1 and 2 represent platform 1 and 2
but, P1 = P2
a little rearranging and...
F1/A1 = F2/A2
The hydraulic fluid also plays a role itself due to it's weight, rho*g*h1 and rho*g*h2
F due to fluid is rho*g*(h1-h2)
d is h1-h2
so Force due to fluid is rho*g*d
So F1/A1 = F2/A2 + rho*g*d
ummm, then I get stuck... doh!

CatWoman

yeah, I got that far then tried equating the forces along a horizontal line level with the pushing piston when the mass had been raised a height h:

P1 = mg/A2 + rho g h2 => F1 = mg + rho g h2 A2 => m = F1/g – rho h2 A2 (1)

Also, conservation of energy:

Work done pushing piston down = Force x distance = F1 A2 h2 / A1
Gain in PE of mass and fluid on other side is m g h2 + rho A2 h2 g h2
= g h2 (m + rho A2 h2)
So F1 A2 h2 / A1 = g h2 (m + rho A2 h2)
cancelling and rearranging => F1 = g A1 (m/A2 + rho h2)
Putting in (1) => F1 = gA1 ( (F1 g – rho h2 A2)/A2 + rho h2) => F1 = g A1 F1 g /A2 which is obviously a load of rubbish!

Thanks for trying redargon - is there anyone else who can help please?

mehmoanp

Hey Dear All,

I have question regarding Hydraulic Press.
In a closed hydraulic press three piston valves (area A1, A2 und A3) are
pressed by the forces F1, F2 and F3.
a) Determine Δh1 and Δh2 for the given situation (see sketch above).
Given data:
A1=200cm2, F1=2000N, A2=500cm2,F2=4000N, A3=350cm2,F3=2000N, density=1000kg/m3,g=9.81m/s2
b) Give the relation F1 : F2 : F3 such that all pistons at the same height.

Attached Files

Pic.pdf (70.1 KB, 20 views)

sophiecentaur

If you don't include consider the weight of the hydraulic fluid, you can say that the work in is the same as the work out - Pressure X volume change for each piston should be the same (neglecting friction). This is equivalent to Force times distance but the pressure is the same both sides so it's an easier calculation.

If you want to include the work done to raise and lower the fluid in each cylinder then the fluid mass displaced on each side must be the same and the height moved up or down will be proportional to
1/Piston Area. So the net work will be equal to the difference between the mgh/2 (just express the m in terms of density and volume) in each case.
More work will be involved in changing the height in the narrower cylinder than in the fatter one. I haven't a pencil and paper handy but that's all you need to do.

mehmoanp

Can you please check the picture!!

sophiecentaur

I was commenting on the OP.
You should start another thread with a new problem, surely.

mehmoanp

Dear Sir thankx for your kind reply, I have confusion in order to generate the equations for delh1 & h2, and what it mean if you don't include. plz help me

jeannagui

-1- no force:

|------|...................|---------| A1-> area of small piston
|_A1__|...................|___A2 __| A2-> area of large piston
|_____|...................|________| no force applied
|------|...................|---------|
|------|...................|---------|
|------|____________|---------|
|-------------------------------|
|___________________________|

-2- force applied:
mass m
|deltaF|...................|___A2 __| a force(delta F is applied to raise the mass m a
|------|...................|________| distance d)
|-↑----|.....↑ ...........|---↕d----| let y be the distance moved by the small piston
|-↓Y---|.....| h.........|----------| y>d
|_A1__|......↓ ..........|----------|
|__X__|____________|---Z-----|
|-------------------------------|
|___________________________|

pressure is equal in X and Z:
Px = Py
delta F/A1 = roh g h
delta F = roh g h A1 h=d+y
delta F = roh g (A1 d + A1 y) volume of liquid is constant so V=v => A1 y=A2 d
delta F = roh g (A1 d + A2 d)
delta F = roh g d (A1+A2)

it's not that difficult - i'm still in highschool :S

sophiecentaur

You have the right stuff - good lad!
This isn't Rocket Science, is it? It's just a matter of reading it up and applying it.