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Motor starting current

Motor starting current

Motor starting current

(OP)
What could be the difference between starting current of a motor (induction) when started on load ( let 100% of full load)and same motor at no load (uncoupled). Is starting current will remain the same and only starting current time will reduce.
I want to test higher rated motor on a motor test bench rated for maximum 90 kw motors.

RE: Motor starting current

The answer is in the motor model.

Why the stator takes 5 or 6 times more while starting?

Because rotor's impedance equals zero in the begining and stator has short circuit current.

If 100% full load exist, notting will change in electrical diagram, so current will same.

But, as mechanical view;

Moment model of motor, related (I) moment of stability. So that to reach desing speed, it takes more time.

Result: current will same, but time will longer.

The longer the time, makes to long time for current.

Therefore wires or isolation can burn.

All engineers know that if you have heavy load,

You start first triangle stator coils (delta connection)to improve impedance of stator and less current
(x=y=z)
(a=b=c)

Then star connection to get efficent from motor.
(x y z)
" " "
(a b c)

RE: Motor starting current

The cause of high starting current:
The starting current will be high (of the order of 4 to 6 times of rated current, depending on the motor rating) if it's on no-load or on load. The current will come down as the motor picks up speed, and settles down at a stable state. When the motor is ON-LOAD, because of the higher inertia more time is needed to attain this state. After that, a steady current will be drawn- the current will be more if the load is more.

Usually, the no-load current will be 2-10% of the full load current- again, this will depend on the motor-rating.

I've one suggestion to make in Merbas's posting above:

Usually, for heavy loads, we use the STAR-DELTA transformation, and not delta-star. If you've three windings A1-A2, B1-B2 and C1-C2 where '1' indicates start of winding and '2' indicates the end,
for Star connection, you connect A2-B2-C2 together. The 3-phase power is applied across A1, B1 andd C1. This will cause less current to flow through the windings.
Once you pick up speed, you quickly change over from star to delta.

For delta, you connect A2-B1 together, B2-C1 together and C2-A1 together. The 3-phase supply is connected across the three junctions. This allows more current with the same 3-phase voltage.

Regards.

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