hydraulic power pack

The pump HP Formula is HP=GPM X PSI X .000583 Transposing the formula gives

PSI=HP/GPM X .000583

Mr BudT .Pl let us know what is .000583 .

I have always used the formula for input HP = Flow x pressure divided by 1200 to allow for losses.

However, your enquiry seemed to indicate pressure requirement; this is P = Load divided by the Area if used to drive cylinders (rams)

BudT has given the theoretical equation for determining the required power for driving a hydraulic pump. For practical use the pump efficiency should be considered as well.

A gear type pump of good quality should have a total efficieny (includes mechanical and volumetric efficiency) of ETA = 0.92 to 0.95.

Now the required power equation is (metric units):

P = p x Q / (600 x ETA) (Eq.1)

P = min required power consumption at pump shaft in [kW]
Q = flow rate in [Lit/min]
p = pressure at pump discharge port in [bar]
ETA is dimensionless

600 is a conversion factor for transforming units Lit/min to m³/sec and W to kW.

The supply pressure is dependend on the driven equipment and hydraulic components (pipes and valves) in the hydraulic circuit. Transforming Eq.1 you can determine the max. admissible pressure capebility of an existing (assumed) pump-drive (as far as the pump and components are proof for that pressure).

Pumping oil of small or medium flow rates (up to 20 l/min) you should add 20 to 30 % to the calculated Power, to cover variable flow resistances (temperature deviations of the oil and other effects).
Most gear type oil pumps are operated by 4-pole motors (1500 or 1800 rpm). This is ok for oil of viscosity up to ISO VG 46 (15°C Temperature and above). For higher viscosity grades (lower temperature) it is recommended to use motors of lower speed (6 or 8 pole motors). This reduces the churning losses in the pump and would result in a reliable drive of good lifetime.