On my last post,i discussed about why we need to determine motor loads and efficiency.So for this post i want to share how to get the motor load value using simple calculation electrical formula.
It is combination from Amperage and Voltage value.Basic formula for Power is P (watt) =I(ampere) x V (volt).Power also measured in Horsepower (hp) unit.For common conversion from electrical Horsepower to Watt is 1hp = 746 watt.
How to calculate Motor Loads
Electrical motor loads are calculated from power in kW and full rated load in kW.Following is the formula how to calculate Motor loads.
We have 1 unit Induction Motor with 30 horsepower (hp),running with 34.9 amps for amperage load and 460 volts 3 phase and power factor value is 0.75 and motor efficiency is 85%
We determine value of power in kW (Pi) using this formula.
We need to determine the full rated load in kW
We can quantify the motor’s part-load by comparing the measured input power under load to the power required when the motor operates at rated capacity using this formula.
First step for actual value of voltage and ampere,i suggest to collected measurement data from motor using the Voltmeter and Clamp on meter.Example as following :-
V red = 462 volt V blue = 461 volt V yellow = 460 volt
V total = (462 + 461 + 460) / 3 = 461 volt
A red = 34.5 amp A blue = 33.8 amp A yellow = 34.2 amp
A total = (34.5+33.8+34.2) / 3 = 34.1 amp
c) Power Factor : 0.75
d) Motor efficiency : 85%
Formula 1 :
Pi = (V xI x PF x 1.732) / 1000
Pi = (461 x 34.1 x 0.75 x 1.732) / 1000
Pi = 20.4 kW
Pir = (HP x 0.7457)/Efficiency
Pir = ( 30 x 0.7457) / 0.85
Pir = 19 kW
LOAD = (Pi / Pir) x 100%
LOAD = (20.4/19) x 100%
Output power as a % of rated power = 107.4%
* From this example value,we can determine our electric motor load in good condition.It is 100% with output power to drive the equipment.It efficiency and not to run with overload.