If one is going down the route of power operated hydraulics then the following calculations are worth remembering !
Sorry it is imperial units ! 1ImpGallon = 4.55L 1Hp = 746Watts, A cubic foot is 6.25 imp gallons
Having a 10cc pump, a 4" diameter pistoned ram with a 28" stroke, an 11Kw 3Ph motor and a honda 12Hp engine, will it work and what will it do ?
Pumps are usually rated in cc per revolution eg 10cc x 2850rpm = 28500/1000 = 28.5/4.55 = 6.26 gal per min
Flow in ImpGals per min x pressure in psi/ 1428 = Hp needed eg 6.26 x 2500 / 1428 = 11 Hp
psi x piston area in Sq inch/ 2240 = tons force eg 2500 x 12.57" / 2240 = 14 Tons This is the extention force, the retraction force will be lower by the area of the piston rod.
piston area x stroke = swept volume- SV SV/ delivery = speed of operation. So 12.57" x 28" = 352 Cu "/ 1728Cu" = 60sec /4.9= 12 seconds for full stroke.
Some figures have been rounded for simplicity..
The above figures were used to show that using the ram and pump, both the electric motor or the honda engine would power the unit giving a force of 14 tons and that its operating speed would be good at 12" travel in 5 seconds. What was it ? A log splitter. Noel.