Hydraulic Pump Power Calculation.pdf 141z4p

Problem: Find the power required for pumping 130 m3/hr. of water, considering 85m differential head and 60% pump efficiency. Hydraulic Pump Power The ideal hydraulic power to drive a pump depends on the mass flow rate, the liquid density and the differential height, either it is the static lift from one height to another, or the friction head loss component of the system, can be calculated as;

Where Ph q ρ g h

= = = = =

hydraulic power (kW) flow capacity (m3/h) density of fluid (kg/m3) gravity (9.81 m/s2) differential head (m)

Example: Power pumping Water 130 m3/h of water is pumped a head of 85 m. The theoretical pump power can be calculated as;

130 x 1000 x 9.81 x 85 = 108,400,500 Kg.m2/h = 108,400,500 N-m/h.

(1N = 1Kg.m/s2)

1W = 3600N-m/h (or) 1kW = 3,600,000 N-m/h Shaft Pump Power 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;

Where Ps η

= =

shaft power (kW) pump efficiency

1kW = 1.341 HP Answer:

power required is 67.3 HP