INTENSIFICATION OF CYLINDER PRESSURE?
I ran the numbers.
With a 2 B 1.125 R cylinder, assuming system pressure (relief valve) at 3000 psi, rod extending to lift a load, and no load ever tending to pull the rod out, I would do a fixed orifice on the closed side, and nothing on the rod side. That is most common, and is best for the rod seal.
16 gpm pump doesn't affect anything, but see the notes below. 16 gpm into the rod end will push out 24 gpm form the closed side. Account for that in the hose and valve design. Edit: I was thinking of the main cylinder. Log lift cyl won’t see 16 gpm anyway.
While this cylinder is lifting at a slow speed, the rest of the 16 gpm pump flow (or the 4 gpm if it is a two stage pump and has unloaded the large section) is going across relief at 3000 psi and is converted to heat, but this is so intermittent, don't worry about it.
Most wood processor applications are so simple there is rarely any issue with just throwing together parts and no one has issues. However, I always check any cylinder application to be sure, especially if it has a small cylinder with large rod, or has over running load.
I use Excel, but here is the theory, for people who want to understand but may have a different application. (Say a lifting crane or winch or something.) I have cut, pasted, and edited from files I have written before, during my work career.
The worst case in your application would be extending the rod, with full pressure into the closed side, no load on the log lift so no load on cylinder rod, and meter out control of the rod side. 3000 psi into closed side will create 4400 psi on the rod/rod seal side.
Not to panic directly. If the cylinder manufacturer designed the cylinder correctly and rated the cylinder operating at 3000 psi, this intensification was accounted for in the seal design, and 4400 psi would be acceptable. Most hydraulic circuits use meter out control so the fluid is always in compression and the load cannot run away, so the tube and seal design accounts for it. However, if the cylinder manufacturer didn't do their job, and designed for 3000 psi on the seal, it may be an issue.
Assuming the load is always pushing the rod back in, and there is no need to control any load pulling the rod outward, a control on the closed side only will be just fine. You could do a fixed orifice (fitting welded closed and drilled through), or adjustable simple needle valve, or two pressure compensated flow controls with reverse free flow checks, mounted in series facing opposite directions.
I would do the orifice, since it is easiest, cheapest, and cannot be adjusted by curious fingers. It will take some trial and error drilling out, and the lift speed and lower speeds will be different and neither one ideal, but this is a log lift and very intermittent use.
Second choice, adjustable high pressure needle valve, no reverse flow check. Will work the same as the fixed orifice except that you can adjust it externally.
Third choice, or first choice if the orifice doesn't give you what you want, is two PCFC. You could adjust those independently. Would add cost though.
Here is the math to determine the 4400 psi.
-Area of a circle is = Pi x R x R.
Since R is half of diameter, it is easier to use = (Pi x D x D) /4.
This reduces to = .7854 x D x D
-Area of the closed side is thus .7854 x B x B. Yours is .7854 x 2 x 2 = 3.141 square inches
-Area of the rod steel is thus .7854 x R x R. Yours is .7854 x 1.125 x 1.125 = .99 square inches.
-Net area for oil on the rod side is 3.141 - .99 of steel = 2.15 square inches.
-'Cylinder area ratio' is a design term, and is closed side area / rod side net area = 3.14 / 2.15 = 1.46. This is useful to relate the flows in and out. With your pump, 16 gpm into the rod side, retracting the cylinder, will push out 16 x 1.46 = 23.4 gpm out of the closed side. This also shows why a 'dump valve' is sometimes used on the closed side of cylinder with large pumps and large rod diameters.
-Force is pressure x area. That is why larger cylinder is higher force.
-In your case, if the closed side has 3000 psi applied to 3.141 sq inches, the extend force is 9423 lbf (lbs force).
-If there is a meter out control on the rod side, the back pressure created will equal the 9423 lbf from the closed side oil pushing, plus or minus any load on the rod. In this case, no load on rod. Rod side has to resist 9423 lbf, so P x Area = 9423, thus Pressure on rod side is 9423 / 2.15 = 4380 psi. That pressure is created by the meter out control device.
-Intensification worst case is when a load is overrunning in rod extend (i.e. tends to pull the rod out further, like a load hanging down from the rod, with the closed end fastened to the ceiling, or a cylinder pulling to lift a conveyor). In that case, the load of the weight would be added to the force of the oil pushing on the closed side. In that situation, the circuit needs to have control on the rod side (load lowering) but a simple orifice or flow control would generate too much pressure, so it would need a counterbalance or load control valve. There are other safety needs in this type of circuit, but that is a whole different topic.
Hope someone else has learned some theory that applies to their application more complicated than log lift.
