Starting Current - Auto Transformer Starting

[magneticted]

Hi guys, thanks for your response to my lasst thread regarding "starting current". After further discussions with the utility company, they are now going to use a korndorffer auto transformer with 85% tap.The motor will be rated at 650 kw (6pole) 11kv / 50 hz, and has a starting current value of 252A ( 6 x 42A ). The ulitily company is stating that the starting current must not be greater than 190A. I calculate at 85% tap the starting current will be 182A ( 0.85 x 0.85 x 252 ). Three questions, firstly is this calculation correct, 182A is the value seen at the 11 kv busbar. Secondly, is thier a motor + transformer magnetising current which can influence the calculation and make the 182A value higher. And thirdly as this is a korndoffer type I assume their will be no transient currents occuring at switch over. At the moment we have enough accelerating torque to start the pump on open valve at 85% tap.

 

[rbulsara]

The starting current will reduce proprotional to the voltage reduction (my statement in the last thread was corrected..) when using autotrasnformer. Torque will reduce proportional to square of the voltage reduction.

So 85% tap will give you .85*242=205A, 65% tap will give you 157A. 65% tap is more commonly employed too.

(Just for the note in a star-delta starter the starting current also reduces by 1/3 ,a same as the torque, but it is because of the change in voltage per phase too when in star connection)

 

[Marke]

Hi rbulsara

The current into the motor will reduce proportional to the voltage reduction, the current into the transformer (primary side) will reduce proportional to the voltage reduction squared. VAin = VAout.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[aolalde]

I agree with Marke.

This is an autotransformer, the high voltage terminals are connected to 11 kV line. The secondary (85% tap) is connected to the load (motor). The motor current (secondary load) reduces proportional to the voltage imposed to the motor leads (9.35 kV) or 214.2 Amperes. This load is; kVA=1.732*9.35*214.2 = 3,468.9 kVA.
The primary handles the very same power at 11 kV line voltage, and the line current becomes:

IL =3,468.9/ (1.732* 11.0) = 182 amperes.

This is the current demanded to the utility. And yes the Korndorfer connection provides a “closed transition”.

 

[jraef]

The most important part of what you asked about is if the motor will start the load at that voltage reduction, and you indicated that it will. The other issues are as follows;

Your math is correct on the starting and line current, 182A at the 85% tap.

The magnetising current can indeed be as much as 25% of the FLA, but only for a very brief moment and the utility usually will not have a problem with that. It is typically not considered.

Even with the Korndorfer design, there will still be a transition spike. It is not as bad as an Open Transition spike, but you are still making a jump in voltage, and that always translates into being a torque and current spike, however brief.

The only way around that would be a solid state starter, but then you will not be able to limit the line current as much. However, on a centrifugal pump with a closed outlet, you should be able to start it with 350-400% current limit, so it still should be under that 190A value.

"Our virtues and our failings are inseparable, like force and matter. When they separate, man is no more."
Nikola Tesla

 

[rbulsara]

Thanks for the 'correction' again..aoldale and all..

 

[swgrmfg]

What the utility co. is most likely interested in is the line current, not the motor current. They are different by transformer effect.
The line current for a .85 tap is: .85*.85=.72 p.u. of normal starting current plus about .03 for magnetizing. So .75*252=189A is what you'd expect if there's no voltage drop at the primary side.

Regards.