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.
Before we go further,i want to explain about Power.What is meaning of Power?Of course this term is significant with our discussion topic.Power is rated of energy in electrical circuit.
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.
Example :
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%
STEP 1
We determine value of power in kW (Pi) using this formula.
STEP 2
We need to determine the full rated load in kW
STEP 3
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.
Calculation :-
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 :-
a) VOLTAGE
V red = 462 volt V blue = 461 volt V yellow = 460 volt
V total = (462 + 461 + 460) / 3 = 461 volt
b) AMPERE
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
Formula 2
Pir = (HP x 0.7457)/Efficiency
Pir = ( 30 x 0.7457) / 0.85
Pir = 19 kW
Formula 3
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.
how can I found out operational efficiency of a motor?
If i am going to use the motor with high and low load alternating in very fast time (less than 1 second), but i don’t know how to measure the load, but i can measure the current drawn. how do i calculated motor load for me to select the right size of motor?
Plz tell me how to calculate total load to my box manufacturing plant
My machines specification as follows
Currogation machine 5hp
Pasting machine 3hp
Stitching machine 2 nos 0.5 hp
Slotting machine 3hp
Hi,
Please use formula below :-
P = I x V x 1.732 –> 3 phase power supply
P= Power (watt)
I = Ampere
V= Voltage
(Output power as a % of rated power = 107.4% )
i really don’t understand the above result.
output power is higher than input power?
yeah as i know output power should be smaller than input power
In above formula 2
pir = 26.31
load (pi / pir) x 100% = 77.53%
output power as a% of rated power = 77.53%
Very easy calculate from amps each unit and total it get answer. I beleive you know each unit kw which u state horsepower only.
Board cutting machine 3 hp
I made a blower the fan Wight 55Kg , OD-33.5″, width -6″
So which capecity motor required
And till me formula of this plzz
In what universe is this true:
Formula 2
Pir = (HP x 0.7457)/Efficiency
Pir = ( 30 x 0.7457) / 0.85
Pir = 19 kW
Nice going algebra boy, but it is a multiplied inefficiency not a divided.
HP * 746 WATTS / HP * LOSS = HP KW
HP KW / VOLTAGE = FULL LOAD AMPS
Hence,
30 HP with a 0.85 Efficiency = 19 KW
Divide by the Voltage measured or supplied and get Amps, so in this example with 460 the FLA should be 41.35
pir = 26.31
Formula 2
Pir = (HP x 0.7457)/Efficiency
Pir = ( 30 x 0.7457) / 0.85
Pir = 19 kW
The above calculation is not correct, the answer should 26.318 KW
yes,
output power as a% of rated power = 77.53%
How to know the d.c motor and a.c motor full load calculation formula ???
Hi
I’ve got a 260 KW motor on a soft starter also rated at 260KW it was originally connected to a 330KW gearbox and pump after many years the gearbox failed and can only get a 409KW gearbox to meet my specs will my motor and soft starter handle the load of the gear box I know I’ll loose efficiency but will I lose my motor and starter??
which method is correct for measuring electrical load of motors?
Ampere load by amp clamp meter or KW load by KW clamp meter?