Consumed Power / Load Study

Hello Jeff Millar

Basically: Yes.

There are other small losses (bearing noise, heat, rotor windage, fan losses etc) but these all come under the basic heading of Efficiency.

Do not forget the Electrical Power Factor of the motor, at variable loads and speeds. If the motor is large, an automatic Power Factor Correction unit may be used to minimise losses.

Kind Regards....

Introducing an alternative:

pump motor is calculated using the following formula:

P=γ Q h/ ŋ 3600

Where:

- P will be computed in W;

- γ is specific weight in N/m³;

- Q is pump flow capacity in m³/h;

- H is discharge height in m;

- ŋ is the motor-pump system total efficiency factor ( 0.50÷0.70 )

Your first sentence is correct but the 2^nd formula is wrong.

Consumed electrical power kW = power input kW = power output kW / motor efficiency = mechanical load kW / motor efficiency.

One thing you need to watch is that motor efficiency varies somewhat with load (actual power output). Motor data sheets usually give efficiency at 100%, 75% and 50% of rated power output.

Cheers........Codey