Hydraulic Pump Power

[h=3]Hydraulic Pump Power[/h] The ideal hydraulic power to drive a pump depends on the mass flow rate, the liquid density and the differential height
[img]http://www.egpet.net/vb/images/imported/2011/07/4.png[/img]
- either it is the static lift from one height to an other, or the friction head loss component of the system - can be calculated as
[INDENT] [I]P[SUB]h[/SUB] = q ρ g h / (3.6 10[SUP]6[/SUP])(1)[/I]
[I]where [/I]
[I]P[SUB]h[/SUB] = power (kW)[/I]
[I]q = flow capacity (m[SUP]3[/SUP]/h)[/I]
[I]ρ = density of fluid (kg/m[SUP]3[/SUP])[/I]
[I]g = gravity (9.81 m/s[SUP]2[/SUP])[/I]
[I]h = differential head (m) [/I][/INDENT][h=3]Shaft Pump Power[/h] The shaft power - the power required transferred from the motor to the shaft of the pump - depends on the efficiency of the pump and can be calculated as
[INDENT] [I]P[SUB]s[/SUB] = P[SUB]h [/SUB]/ η (2)[/I]
[I]where [/I]
[I]P[SUB]s[/SUB] = shaft power (kW)[/I]
[I]η = pump efficiency[/I]
[/INDENT]