This difference in these areas makes hydraulic cylinder forces the piston side are greater than the rod side, the flows are uneven consumed or discharged one side over the other if you keep the speed as input and output, and finally can be operated as pressure boosters.
necessary pressure on the piston side (P2) to move the load is equal to the pressure to overcome the load on the cylinder rod pressure to balance the back pressure (P3) generated in the chamber piston rod side.
This counter is basically due to the restriction that the throttle valve offers flow of hydraulic fluid to pass through it.
For the analysis we consider the previous figure, where the charge is positive and the cylinder out.
The power to be exercised by the cylinder is equal to the shit but opposite sign.
On the other hand, the pressure drop at the throttle valve are determined by:
ΔP1 is the loss of pressure from the throttle-return (value usually very low) and ΔP2 is the pressure drop across the choke. In this last expression we have:
and substituting in the equation of force:
But
Substituting we have:
From here:
P3 Solving the general equation, which is the value of pressure that we find, as is the back pressure generated in the piston rod side due to the regulation of the discharge flow, the formula is:
Noting the equation we can deduce that the value of the counter is dependent on load on the cylinder, so that if the load increases, the counter decreases. This makes sense because the pressure on the side of the cylinder piston would be using to beat the charge, ie taking advantage to do the job, if the load can consume the entire value of the pressure on the piston side, then back stem side would be zero, which means no movement, but a completely static condition.
Applying this same logic but reducing the value of the load (Fc) on the cylinder deduce that the counter increases, reaching its maximum value for the case where the cylinder out without charge.
This maximum value is just the value of the pressure factor piston and cylinder areas as we can see, the pressure is multiplied in that factor.
In fact, returning to the introductory fails, pressure in the chamber of the cylinder rod side reached values \u200b\u200bvery close to 300 bar (30 MPa), higher pressure by a factor of 2 to the maximum allowed by the standard for hoses SAE 100 R 2, ie outside the field of security supported by the provision that provides as a safety factor 4 as a minimum.
Applying the same reasoning as above, but to find the value of the pressure side when the cylinder piston retracts, we reach the following expression:
For the return of the cylinder, what happens is a pressure reduction ratio!.
This is the reason why only burst hoses rod side. The solution was simply used to the rod side hoses or exceed the SAE 100 R 11 and that is precisely what was done and the problem disappeared.
This short analysis is not accurate, because certain conditions were disregarded real and drum performance, strength and dead load on the cylinder, however it has enough validity to demonstrate the fails because the values \u200b\u200brecorded in the computer with the usual industrial gauges are almost the same as those calculated.