Help regarding AC induction motor starting current

Thread Starter

Sattamassagana

Joined May 29, 2011
3
Hello,

I am a bit troubled over calculating the starting current for a motor I want control.
I have here a 0.15KW 3 phase induction motor. It will be wired in Delta configuration.

Specs:

Nominal speed: 880rpm
Nominal current: 0.72A
Starting to nominal current ratio: 2.1A
Nominal torque: 1.63Nm
efficiency: 50%
Nominal power factor: cosφ=0.58
Starting to nominal current ratio: 2.1A
Starting to nominal torque ratio: 2.2Nm

On the nameplate I saw that A(Amps) is rated at 1.25A when delta wired @ 230V.

Also, the motor is fitted with a worm gear unit so that

Output speed: 9 rpm
Output torque: 72Nm


What I' d like to know is


1) Do the Amps rated on the nameplate refer to the Full load current?



2) If yes, am I correct with calculating the starting current?


(Starting to nominal current ratio) * Full Load Current= 2.62A


3) Does the worm gear unit reduces or increases the duration of the starting current, and is it possible to quantify it?



Thank you in advance
 

PackratKing

Joined Jul 13, 2008
847
The nameplate on the motor should list " Locked-rotor Amps " or LRA

Since all 3-phase motors start " with a bang " your start current should be a very short pulse, likely only 3 times load current, that is if you could find a meter short of an oscilloscope, capable of measuring such a short pulse.
The worm drive should not affect the start parameter.
 
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