Power data

According to the calculations, at equal power the ratio of liquid to be pumped is 9,16 : 1This means that in this example model, in order to move the base of the cone, which is equivalent to the piston of a cylinder of equal power, a pump with a consumption of 9,16 times less. Thus, a hydraulic pump providing 400 bar with a flow rate of 82,04 L/min will print a regime of 73 rpm with a gross power of 250.32*2 = 500.64 kW and a net power of about 15% less.

The power consumption of a hydraulic pump providing 82,04 L/min a 400 bar pressure is calculated by the formula:

Where: P : Power; QFlow rate;p: pressure; ηPump efficiency

The result is that the hydraulic pump needs a power requirement of approx. 64,34 kW to provide such a flow of 82,04 L/min at a pressure of 400 barassuming a efficiency of the 85%.