how does hydraulic press work?
nw i know how it works bt even when v apply less input pressure, v get more output pressure fr eg. if we apply force of 1N v get output of 10N etc......so isnt this a contradiction to law of conservation of energy?
nw i know how it works bt even when v apply less input pressure, v get more output pressure fr eg. if we apply force of 1N v get output of 10N etc......so isnt this a contradiction to law of conservation of energy?
Comments
Here's how a hydraulic press works.
A hydraulic system works using two pistons. One piston has a small area and a long stroke, the other has a large area and a short stroke.
The equation for pressure in a piston is:
P = F / A
P = pressure
F = piston force
A = piston area
The piston with the small diameter, long stroke is the input piston (piston 1) , the one you apply force to. The piston with the large diameter and short stroke is the output piston (piston 2), that develops the large force.
The two pistons are connected hydraulically, meaning that the pressure inside both is the same.
So we can write,
P1 = P2
or,
F1 / A1 = F2 / A2
Rearranging,
F2 = (A2 / A1) * F1
The output force of piston 2 is multiplied by the ratio of the areas of the faces of the two pistons. If the output piston has a diameter 10 times that of the input piston, its area is 100 times larger. So, for an input of 10 N, the output would be 1000 N.
Now, nothing comes for free. The distance that the output piston moves will be a lot less than that of the input piston. Also, since the change in volume of the input piston has to match the change in volume of the output piston, and since the areas of the two pistons is fixed, the only parameter that changes is the distance that the piston moves.
Using the following definitions,
A1, A2 = areas of the piston faces
dV1, dV2 = change in volume of piston 1, piston 2
dx1, dx 2 = change in displacement of piston 1, piston 2
We can write that the change in volume of the input piston has to match the change volume of the output piston:
dV1 = dV2
A1 * dx1 = A2 * dx2
Rearranging,
dx2 = (A1 / A2) * dx1
The stroke of the output piston is that of the input piston but scaled by the ratio of the input piston area to output piston area. From the example above with the output piston having a diameter 10X that of the input piston, the stroke would be 1/100th that of the input piston.
So, while the force of the output piston goes up by a factor of 100, the distance it travels decreases by the same amount.
From a conservation of energy perspective,
Work = Force * Distance
W1 = W2 <== work done to move the input piston equals work done by the output piston
F1 * D1 = F2 * D2
Substituting for F2, D2
F1 * D1 = [(A2/A1)*F1] * [(A1/A2)*D1]
F1 * D1 = F1 * D1
Energy is conserved.
Hope this helps,
-Guru
Pressure and energy are not the same.
If you have a large and small piston connected by a pipe you have to move the small piston further to make the large piston move.
see
http://www.youtube.com/watch?v=TjzKpke0nSU