motor inrush reduction 2

[electrolitic]

I connect a 3,5MVA transformer, Z=7%, 11.0kV/3.3kV to a 2,5 MW induction motor, Is=6 In. Circuit breakers at primary and secondary. Can I suppose around 30% of reduction voltage at starting, Is reduction to aprox. 4,0 In as it was a reator starting (70%) without additional breakers and bump during transition? Other loads, if any, to be connected just after the main motor starting.
Is this assumption correct? If yes, is it normally applied?

 

[waross]

I get 30% with a simple calculation, BUT, with a 30% reduction in voltage, the current will not be enough to cause a 30% reduction in voltage.
I'd guesstimate 15% reduction in voltage and see if I can live with the numbers.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jraef]

I interpret that you are asking if a 30% reduction in voltage through a Reactor Starter will bring the starting current down to 400% from 600% if starteed Across-the-Line (DOL).

The answer is, it's not quite that simple.

Reduced voltage starting reduces the starting torque, and current, by the square of the voltage reduction. So at 70% voltage through the reactor during start, the torque and current will be at 49% of normal. 49% of 600% is 294% (300%), So Is = 3.0 In. But with the reduction in current comes a reduction in starting torque Ts, which would normally be 160% of Tn, so your accelerating torque will be 49% of that as well or roughly 80% of Tn. If that allows your motor to accelerate to 90% speed or higher, no problem. But if not, when you transition to full voltage you will get a transition spike that can theoretically still be 6.0 In anyway, albeit for much less time.

So it depends on the application and load acceleration torque/speed profile.


"Will work for salami"

 

[waross]

Hi Jeff. I took the issue to be using the 3.5 MVA transformer as the starting reactance. ie: what will happen if the motor is started DOL on a 3.5 KVA transformer.
Perhaps the OP will clarify this for us.
electrolitic: I should have mentioned that the impedance of the transformer is applicable only to short circuit calculations. At normal loads and at a given power factor the Regulation is applicable. Under heavy loading the resistance and induction of the load in series with the resistance and the impedance of the transformer is applicable. In most instances the current of motor will different than that predicted by a simple calculation based on transformer impedance alone. The only case where the simple calculation is accurate is when the motor X/R ratio equals the transformer X/R ratio.
BUT
The motor X/R ratio changes with both speed and loading.
Hence I suggested a simplified approximation that describes a worst case that will probably never be reached.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electrolitic]

Waross
You interpreted my question very well. It is not an auto trasformer starting a motor where the reduced voltage level is already defined by th AT tap selection. It is a voltage drop caused by the motor starting current overloading the transformer reducing the motor starting current. It is a recurrent calculation, I suppose. I don't know how to do it.
I agree with your explanation about transformer regulation and impedance but I don't need an accurate value. It is just the understanding of the process.
Thanks

 

[waross]

I'll say it differently in case I wasn't clear.
Using the 7% impedance voltage in a simple calculation yields a 30% voltage drop.
But a reduction in voltage causes a proportional reduction in current, so use 1/2 the impedance voltage or 3.5%.
This is not exact but is possibly the worst case limit.
For an exact value you must know the effective resistance of the motor. This is not a parameter that I see often (or ever).
Once the motor starts to turn, everything starts changing but the current will be dropping and the voltage will be rising so you are moving away from the worst case scenario.
Hopes this helps.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

Motor acts like constant impedance during start.
Zmotor_start = Vstart/Istart

It remains to convert the motor impedance and transformer impedance to a common base (the hard part if you haven't done it in awhile) and then the easy part is to solve the voltage divider problem to find voltage at motor
Vmotor = 3.3 * Zms/[Zms+Zt]
Assume both impedances have the same phase angle (close to inductive) for simplicity.

Reduction in starting current is proportional to reduction in motor voltage.

=====================================
(2B)+(2B)' ?

 

[LionelHutz]

Reduced voltage starting reduces the starting torque, and current, by the square of the voltage reduction. So at 70% voltage through the reactor during start, the torque and current will be at 49% of normal.

The current reduces proportionally with the voltage reduction. AT 70% voltage you'd have 70% current and 49% torque.