## Motor Starting Current

## Motor Starting Current

(OP)

Hello Everyone,

I read in the literature for motors that I_start = (NEMA_CODE x HP)/ sqrt(3)x V

However for a lower voltage at starting, the starting current decreases, but this formula insinuates the opposite. it is kind of confusing.

Any guidance?

Thank you in advance

Samy

I read in the literature for motors that I_start = (NEMA_CODE x HP)/ sqrt(3)x V

However for a lower voltage at starting, the starting current decreases, but this formula insinuates the opposite. it is kind of confusing.

Any guidance?

Thank you in advance

Samy

## RE: Motor Starting Current

https://www.engineeringtoolbox.com/locked-rotor-co..., NEMA Code formula is related to locked rotor kVA/hp. This originates from NEMA Standards MG 1 – 10.37.2.

You can also find that the motor design starting torque and nema code are related. Within some physics constraints this is a design selection.

## RE: Motor Starting Current

Key word here is rated.

When you apply a lower voltage at start, it is NOT the rated voltage. Hence the start current at this lower voltage is reduced proportionately from the rated start current.

Muthu

www.edison.co.in

## RE: Motor Starting Current

## RE: Motor Starting Current

I can see that the concept of starting kva/hp can be misleading to the extent it may be (incorrectly) interpretted to imply that the motor acts like a constant kva device during start.

In fact the starting kva of a given motor is not constant as voltage changes, the starting impedance is. So in this respect, a starting motor acts like a constant impedance device, not a constant kva device.

As edison mentioned the only context in which we can use starting kva/hp to compute starting current is at RATED voltage. To correct to any other voltage, you need to use constant-starting-impedance model (a fractional change in voltage causes a corresponding fractional change in starting current, at least to a first approximation)

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(2B)+(2B)' ?

## RE: Motor Starting Current

@ LionelHutz yes that is correct the denominator is (sqrt(3)x V)

## RE: Motor Starting Current

RATEDvoltage## RE: Motor Starting Current

## RE: Motor Starting Current

Furthermore, the relationship between applied voltage and starting torque is a square law. This means that a 1.0 pu voltage will result in the machine developing 1.0 pu starting torque (and thus accelerating torque). At reduced voltage - like 0.8 pu - the developed torque is 0.8 * 0.8 = 0.64 pu of the RATED starting (and accelerating) torque. This is why having too much line drop (or too little transformer) between the utility and the machine can cause the motor to be unable to accelerate the driven load.

In essence, the correct machine model is indeed a "constant impedance" approach, rather than a "constant kva" approach.

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## RE: Motor Starting Current