Press Cylinder Power Unit: Pressure, Flow, and Heat Balance
A 4.00 in bore, 2.50 in rod, 30 in stroke double-acting cylinder on a 20-ton press must extend against a 20,000 lbf load at 4.0 in/s. The existing power unit has a fixed-displacement 15 gpm pump, a relief valve set at 2,500 psi, a 40 gal reservoir, and no oil cooler. During the extend stroke the cylinder inlet gauge reads 1,850 psi and the return line gauge reads 150 psi. Oil temperature reaches 160 deg F after 45 minutes of cycling, and the operator reports the press slowing on long strokes. The directional valve is a closed-center spool, and all flow passes through the valve before reaching the cylinder. The maintenance log shows no recent pump or relief valve work. The question is whether the power unit is adequate and where the heat is coming from.
1. Calculate the extend area, the pressure required at the cylinder to generate 20,000 lbf at 4.0 in/s, and the flow. Can the 15 gpm pump achieve the speed?
Read the worked answer
Bore area = pi/4 x 4.00^2 = 12.57 in^2; rod area = pi/4 x 2.50^2 = 4.91 in^2; annulus = 12.57 - 4.91 = 7.66 in^2.
The cap pressure must overcome the load plus the return pressure acting on the annulus: p = (20,000 + 150 x 7.66) / 12.57 = (20,000 + 1,149) / 12.57 = 1,682 psi. That is below the 2,500 psi relief setting, so pressure is not the limit.
Flow at 4.0 in/s = 12.57 x 4.0 x 60 / 231 = 13.1 gpm. The 15 gpm pump can supply it; at full pump flow the speed would be 15 x 231 / (12.57 x 60) = 4.59 in/s. The excess 1.9 gpm must go over the relief or be bypassed, so the pump is adequate but inefficient.
2. The gauge reads 1,850 psi. Reconcile that with the calculated pressure and quantify the continuous heat generation while extending.
Read the worked answer
At 1,850 psi, the extra pressure is 1,850 - 1,682 = 168 psi. This is either cylinder friction or a restriction in the valve/lines. The extra force is 168 x 12.57 = 2,112 lbf. The hydraulic power lost to that delta p is 168 x 13.1 / 1714 = 1.28 hp. The 1.9 gpm excess flow at 2,500 psi wastes 1.9 x 2,500 / 1714 = 2.77 hp. Total continuous loss is about 4.05 hp, or 4.05 x 2,545 = 10,300 BTU/hr. With 40 gal (about 300 lb) of oil, that is roughly 34 deg F/hr before cooler losses, which explains 160 deg F in 45 min. A meter-in flow control or a pressure-compensated pump would avoid the excess-flow loss; a cooler alone would mask the problem.
3. For retract at the same 15 gpm and 2,500 psi, calculate speed and available force. What mechanical check matters?
Read the worked answer
Retract uses the annulus: speed = 15 x 231 / (7.66 x 60) = 7.54 in/s, and force at 2,500 psi = 2,500 x 7.66 = 19,150 lbf. Retract is faster and can generate more force than extend, which is normal for a single-rod cylinder. Do not assume the rod can take 19,150 lbf: with a 30 in stroke and a 2.50 in rod, check column buckling and cushioning, and set the relief or a reducing valve to protect the rod. Also check the pilot-operated check or counterbalance valve on the cap side; if it is not set correctly, the retract pressure can spike.