Selection of hydraulic cylinder
Hydraulic cylinders convert hydraulic energy into mechanical work. Normally they generate a linear movement. For many tasks of hydraulic cylinders you can get a choice:
To avoid that the piston does not drive at full speed against the end stop, a cushioning is required.
Functioning: (1)A damping piece closes, just before reaching the end position, the discharge outlet. (2) The oil is forced to make its way through the small hole with an adjustable throttle screw. The piston thus runs against an oil cushion and gets an end position damping.
The volumetric flow rate Q, which flows into the cylinder, is the cause for a movement of the piston. The piston velocity v depends on the volume flow Q and the effective piston area A.
v = Q v piston speed in m/s
A A volumetric flow rate in l/min
Q effective piston area
The efficiency of a cylinder is determined mainly by friction. Today's seals have low leakage losses, implied that they are new. With increasing operating time, however, the leakage will increase and thus the overall efficiency decreases.
Exercises for this chapter "hydraulic cylinder"
1: The pressure of a hydraulic system is limited to 100 bar by a pressure relief valve. As an actor, a cylinder with d1 = 80 mm and d2 = 35 mm is used. The pump delivers a flow rate Q = 20 liters / min. The efficiency of the cylinder is 0.85.
To calculate: The maximum compressive force F1 of the extending cylinder b) the maximum tensile force F2 of the retracting cylinder c) the piston velocity va during extension and d) piston speed ve during retraction.
Solution for a):
Fcompr. = pe * Apiston * ɳCyl Apiston = ᴫ/4 d12 cm2 = 50,265 cm2
Fcompr. = 1000 N/cm2 * 50,27cm2 * 0,85 = 42,73 kN
Solution for b):
Ftraction = pe * AAnular * ɳCyl AAnular = ᴫ/4 (dKolben2 – dStange2) = ᴫ/4 (64cm2 – 12,25cm2) = 40,64 cm2
Ftraction = 1000 N/cm2 * 40,64 cm2 * 0,85 = 34,54 kN
Solution for c):
va = Qe = 20*10-3m3/60s = 0,066 m/s
Solution for d):
ve = Qe = 20*10-3m3/60s = 0,082 m/s