Engineering designers, after determining the centrifugal pump flow head, need to determine another important parameter of the pump: pump motor power. Usually we can determine the motor type and motor power according to the flow rate and the head. We can derive the technical formula of the pump motor according to the principle of energy conservation.
The effective work done by the pump is W=Mgh (send a certain weight of medium to a certain height h, h is the lift)
- M is the mass of water m = ρV (ρ is the density of the medium, the volume of the V medium) V = Qt (Q represents the flow rate of the pump, t represents the pump working time)
So the effective work done by the pump W=ρQtgH
The effective power of the pump P=W/t=ρQgH
The shaft power (actual output power) of the pump is P1=ρQgH/η
——η indicates the efficiency of the pump
Actual motor power P2 = γP1
—— γ represents the safety margin of the motor (the value range of γ is 1.1-1.3, generally 1.2)
If the medium we are playing is water, then the motor power calculation formula is P2=(1.2QgH)/(3600*η)
The unit of flow Q is: m3/h The unit of head H is: m
Need to pay attention to is: P2 calculated according to the formula, not necessarily the motor power, such as 33.56kw, then we choose the motor to choose 37kw, if it is 30.56kw, then we choose 30kw motor.
Through the above formula, we can know the following situations.
1. Regardless of the manufacturer, the actual effective work is fixed in the case of flow and head determination.
2. The pump power consumption does not look at the pump motor power, depending on the shaft power of the pump. The same is a 30kw motor, the shaft power of a family is 20.56kw, one is 25.18kw, obviously 20.56kw to save energy. The key to the shaft power is the efficiency of the pump.
