transformers and large motors

[cvirgil]

I understand that when sizing transformers for large motros you wnat tio limit the starting voltage drop to aroound 10%-12%. This could be checked using a simple calculation knowing the motor starting current and the transformer impedance.

What i'm trying to understand is the "rule of thumb" formula which is VD(%) = (locked rotor current / transformer full load secondary current) x transformer impedance.

Anyone know why and how this works?

 

[ScottyUK]

cvirgil,

At rated current, the transformer terminal voltage drops by a value equal to rated voltage x % impedance. The volt-drop occurs across the leakage reactance and winding resistance of the transformer windings. It is a complex quantity because it has resistive and reactive elements. The actual impedance is calculated from the results of the short-circuit test. The impedance quoted is the percentage of rated voltage required to circulate rated current in the transformer primary with the secondary shorted out

If we re-write the formula as % volt-drop = [Actual Current]/[Rated Current] x % impedance, I'll explain a little.

By substituting rated current for actual current in the above equation, the rule clearly holds. If you substitute the LRC for a motor into the equation, the volt-drop during starting can be calculated.


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If we learn from our mistakes,
I'm getting a great education!

 

[weh3]

It is actually "voltage dip", and it occurs because of the voltage division between the transformer impedance and motor impedance. You can get a pretty good idea of the level by using the impedance in ohms of the transformer (multiply Z% by base Z) and of the motor (divide voltage by locked rotor amps).

Vstarting = Vline x Xm/(Xm+Xt)

William

 

[bigamp]

A good approximation is that the pu voltage dip at the motor bus is motor starting kVA divided by the available 3-phase fault level kVA at the bus. This considers not only the affect of the transformer impedance but also that of the source impedance.

Draw a simple model of the power system and the basis of the approximation will be clear:

Voltage source providing power via source impedance and transformer impedance in series to a motor. The motor impedance is taken as the equivalent R + jX that would draw rated starting current if connected to an infinite bus and recognise that motor starting current is highly reactive (power factor typically 0.2 or less).

Designing for a relative voltage dip of 10% to 12% as you suggest sounds spot on, especially if motors are properly selected (i.e. if they can develop sufficient torque to run the driven equipment to rated speed with say 80% terminal voltage applied).

 

[mc5w]

Look up the starting code letter of your motor in National Electrical Code table 420.7(B). This will tell you how many KVA per horsepower your motor pulls when starting. Multiply this number by the horsepower of your motor, then add in the KVA of all other load on the secondary of your transformer.

Then, use your transformer impedance to get the voltage dip for your transformer. For 5.4% impedance dry transformer and say a Code G motor you could easily need a transformer aroung 2 to 4 times the size of your largest motor depending on how much other load "dilutes" the starting current.

Assuming that the power factor of your load and transformer impedance will give the worst case voltage drop. Since starting current of a motor is highly inductive and the impedance of a typical dry transformer is mostly inductive, this is not a bad approximation. The amount of time that it takes to get the starting power factor of the motor and the power factor of the transformer impedance costs more than what a bigger transformer is worth.

If you are talking about oil filled transformers then you are talking about say 2% impedance for smaller transformers, 5% for medium size, and about 10% for big substation and transmission transformers.