Starting current calculation

A

Thread Starter

Ashok

Hi everybody,

Can anybody explain me how to calculate the starting current of an induction motor?

I have a 90 kW 380V 50 Hz induction motor directly coupled with a refrigeration compressor. Starter is star/delta and full load current is 160A. I found the following data on the motor data sheet.
IA/IN = 7.5
MA/MN = 2.4
MK/MN = 3
Mass moment of inertia = 0.52 kgm2

Will the above details help to calculate starting current?

Thanks in advance.

Regards,
Ashok
 
J

Jose Ignacio

Induction motors have a very high starting current, (many times the nominal one) and it depends only on the gemoetry of the motor design. I believe its rated as a factor of the nominal current, and it seems IA/IN=7.5 is the one that matters(because thats a reasonable value for a peak start current). Anyway, for how long it stays on that level will depend on the load.
 
The fast answer is no, never! This is due to the fact that despite available information on complex electrical calculations that can be performed, nothing short of a NASA equivalent series of physics calculations AND experiments can even approximate the actual starting current. Why?
The moment of inertia and other highly varible unknowns. The generally accepted field proven rule is: 225% of the FLA (Full Load Amps) or if not available RLA (Running Load Amps) for a period of 0.5 to 2 seconds duration.
 
Responding to Ashok's query:

Starting-current (also called inrush) is the current drawn by a motor during its runup, or acceleration, to normal operating speed. Its value is in Amperes. Typically it can be from 4 to 8 times the rated current. Ignoring the transients of the first 2 or 3 cycles, then starting-current lasts from the time the motor is first energized (breakaway) until the motor nears full speed... often defined as starting-time.

In your case the nameplate gives the start-to-rated current ratio, Ia/In, as 7.5. This means that if rated voltage were applied to the motor, and its stator winding was delta-connected, then its starting current will be 7.5 x 160 or 1,200 Amperes.

In certain instances, such a current is prohibitive. But, connecting the stator winding as a wye reduces the starting-current to about 58% (1/sqrt3, as indicated by Omer).

Starting-current magnitude is determined only by the motor's design parameters. Neither load torque, nor system inertia, influences its magnitude. However, they are extremely important for calculating starting-time.

Two caveats: 1) Wye-Delta starting limits acceleration duty; and 2) sizing of the motor's feeder conductors requires special attention to derating factors.

To dr.freon... in my experience I never heard of either the 225% rule, nor the 0.2 to 2 sec rule, that you advanced!

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected]) {[email protected]}
 
When the actual load current is equal to the full load current rating of the motor then the starting current can be estimated as FLA x 7.5.

But this is hardly the case since the actual load current is oftentimes less than the design full load current of the motor.

However, if you really want to determine the starting current at any loading condition, then you must have the starting characteristic curve of the motor as well as the load to be able to perform a motor starting analysis.
 
Responding to Anonymous' Thu, Jul 8, 10:49am comments:

Starting-current magnitude is determined by motor parameters. System inertia and net accelerating torque (that developed by motor less that required by load) determines the starting-current duration!

An earlier reply, Mon, Jun 14, 1:49pm, contains detail.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])
 
I also agree with Mr. phil Corso that Starting-current magnitude is determined only by the motor's design parameters. Neither load torque, nor system inertia, influences its magnitude. However, they are extremely important for calculating starting-time.

P.Anand,
[email protected]
 
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