Thread: Pascal's Law and Hydraulic Presses

Hello friends:

I'm wondering if anybody can clarify an issue regarding how Pascal's Law is applied to hydraulic presses. According to Pascal, a pressure applied to an enclosed fluid is distributed uniformly throughout the volume of the fluid. In other words, the pressure applied to the fluid in a vessel that contains the fluid will result in that same pressure on any part of the wall of the vessel. For instance, if a piston in a cylinder applies ten pounds per square inch to fluid in the cylinder, then any point of the wall of the cylinder will experience a pressure of ten pounds per square inch.

Now, a hydraulic press is much like the aforementioned vessel, except that it includes a second piston that usually has a larger diameter and hence a larger area than the first piston. The force applied by the smaller piston to the fluid in the cylinder moves the larger piston outward and with greater force than the force applied by the smaller cylinder. The net effect is to amplify the force applied by the smaller cylinder.

The amplification of force is made possible by the fact that since Pascal's Law dictates that the pressure is the same at all points on the wall of the cylinder, then larger areas of the wall experience greater total force. Since the larger piston has a larger area than the smaller piston under the uniform pressure in the cylinder, then the total force on the larger piston must be larger than the force on the smaller piston.

Is that correct?

Jagella

Quite so. The work exerted on each piston is equal to that done by the other, because the volume displaced is proportional to the distance traveled, and the smaller piston travels farther to displace equal volume.

Originally Posted by Harold14370

Quite so. The work exerted on each piston is equal to that done by the other, because the volume displaced is proportional to the distance traveled, and the smaller piston travels farther to displace equal volume.

Thanks a lot for that insight, Harold.I didn't think about the volumes displaced by the cylinders until you mentioned it. Here's the math:


(1) input pressure = F_i / A_i = F_o / A_o= output pressure


Dividing both sides by F_i and A_i:


(2) F_o / F_i = A_o / A_i = the ideal mechanical advantage


(3) work in = F_i X s_i = F_o X s_o = workout


Dividing both sides by Fi and so:


(4) F_o / F_i = s_i / s_o


Substituting A_o / A_i from (2) into (4):


(5) A_o / A_i = s_i / s_o


Multiplying both sides by s_o and A_i:


(6) Volume Displaced by Output Cylinder= A_o X s_o = A_i X s_i = Volume Displaced by Input Cylinder


What I'm still a little unclear about is the orientation of the pistons. The figure in the book I'm using has the pistons oriented vertically. Do the pistons need to be oriented vertically? I was thinking they can be oriented in any direction.


Jagella

They can be oriented in any direction. In your car's braking system there is a master cylinder which is moved fore and aft by the brake pedal. The hydraulic fluid actuates the brake cylinders which move side to side for disc brakes or fore and aft for drum brakes.

Originally Posted by Harold14370

They can be oriented in any direction. In your car's braking system there is a master cylinder which is moved fore and aft by the brake pedal. The hydraulic fluid actuates the brake cylinders which move side to side for disc brakes or fore and aft for drum brakes.

Thanks for the reply. I thought that the orientation presumably made no difference for the simple reason that the formulas included no variables for orientation.

Gravity might make a difference, though. If the output piston is heavy enough, its weight might counteract the force coming from the fluid in the cylinder if it needs to be pushed upward. If it is to be pushed downward, then its weight would add to the force coming from the fluid.

I believe the formulas I posted above neglect piston weight and friction.

Jagella