Hello all, some colleagues and I have run across a problem that we cannot seem to explain, I was hoping to find some guidance here. The problem came about from messing with an air over hydraulic system. I have attached a very rudimentary mspaint picture with our scenario. Using 3 in place of pi for simplicity, the large cylinder is cyl A and the small B in both examples. Area of circle for cylinder A = 75in^2 Area of circle for cylinder B = 3in^2 P=f/a Force on Cyl A= 7500lbs Force on Cyl B= 2500lbs Basically the things I don't understand are: a. Is pressurizing the top of Cyl. A equivalent to placing (75) 1"x1", 100lb weights on top of cyl a. b. if I put a pressure gauge on the other side of cylinder b, what reading would I get from the gauge? and c. Are the first and second configurations functionally identical? (I assume your just replacing the fluid as the medium of transfer with a rod) I hope I have adequately explained the scenario with our questions, I'm sure its very simple, but short of mocking something up with PVC we don't quite understand the mechanics of what is going on here. I appreciate any guidance. Edit: Added Picture, also I would like to ignore the fact that the second configuration would likely need to be vented in between the top and bottom to prevent air from compressing due to a change of volume.
The two configurations are not the same. In the first one, the pressure on both side of the piston in A must be the the same, otherwise the piston would move. The same is true for piston B. So the pressure above B will be 100 psi. In the second configuration, there can be a force in the connecting rod. If the volume below piston A is large, the change in volume between the pistons will be small as A moves down, and the pressure between A and B will stay close to atmospheric. The rod will transmit almost all the downwards force from A onto B, and the pressure below B will be therefore be higher, in proportion to the areas of A and B.
I think there may have been a disconnect here. I do want Cyl. A to move and to transfer force to Cyl B. I am assuming Cyl. A in the first configuration is filled with an incompressable fluid and want to know what the PSI of both Cyl B's will be. If the pressures are different, I would like to understand why an incompressible fluid isnt equivalent to a mechanical connecting rod. Basically, i have taken the stance that the first configuration, the 100psi on the top of Cyl. Pressurizes the bottom side to 100psi, then repeats with B. I want to know why the 100psi to 100psi doesn't hold true on the second configuration when the only change made was a medium of transfer (unless there is more to it than that.) Basically, why do we go from pressure to pressure to pressure in the first, and pressure to force to pressure in the second? I appreciate your feedback.
It looks like we are just going to have to mock something up with pvc to see what happens. I just want to see for myself that i can get such a large amount of pressure in cyl b from the second configuration. Will report back with results.
First, the areas you calculated are off unless you are giving the radius of the cylinders which would be confusing thing to do unless you explicitly stated this. Secondly, you would have to assume that the length of the cylinders and their relative positions are small compared to the actual pressures used and that the weight of the incompressible fluid can be neglected. If those simplifying assumptions are valid then the answer is simply that a solid transmits forces directly to it's supports while a fluid transmits the same pressure everywhere including against it's container walls. This is just a simple physical reality of the nature of fluids which are incapable of supporting shear stresses. A more interesting comparison would be to eliminate the rod in the 2nd configuration and just use the fluid as a means of transmitting the forces. What would the pressure be in Cylinder B then?
I agree with Alpha's analysis: the difference is that fluids flow and rods don't. So the pressures have to be equal while the forces do not. For a connecting rod, the opposite is true: forces have to be equal but pressures do not.