Motor Starting Current 2

[samy81]

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

 

[FacEngrPE]

Here is a reference from the engineering toolbox

https://www.engineeringtoolbox.com/locked-rotor-code-d_917.html,

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.

 

[edison123]

For a given KVA/HP, the motor rated current is inversely proportional to the rated voltage. Higher the rated voltage, lower the rated run (and the rated start) current. And vice versa.

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

http://www.edison.co.in

 

[LionelHutz]

Did you mean a denominator of (sqrt(3)x V) because the formula is wrong without the brackets.

 

[electricpete]

samy81

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)' ?

 

[samy81]

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

 

[samy81]

Thank you everyone for your replies. I actually came to the same conclusion that this formula can only be used for RATED voltage

 

[LionelHutz]

Yes, the shortcut quick formulas being applied to motors all have limitations.

 

[Gr8blu]

Samy81: Yes - within the constraints of factors which affect impedance, the "basic" short cut formulae can achieve fairly realistic approximations for certain changes. As you have already noted, the relationship between starting (inrush) current and applied line voltage is more-or-less linear, at least when you talk about per unit values. What this really means is that when rated (1.0 pu) volts is applied, the inrush current is also 1.0 pu. If a lesser voltage such as 0.8 pu is applied, then the starting current will be likewise reduced to 0.8 pu.

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.

Converting energy to motion for more than half a century

 

[samy81]

@Gr8blu: The "constant impedance" model is only valid for a "starting" state but when the motor is in "running" state the "constant KVA" model is applicable. I think it is important to notice the difference as this will lead to different behavior vis-à-vis the voltage