Electrician Theory: Delta Star Induction motor question

Question

This is the question I am trying to solve:

A delta connected fully loaded induction motor draws 90A from a 400V supply when started direct online. Show by calculation the current at start up that the motor would draw when connected to a star delta starter in the star position.

I was under the impression with delta star, that when voltage goes up, current goes down, however here the answer to the question shows that the ratio between delta star is 1.732.

With Delta-Star I understand this to be a ratio of 1.732:1 (Delta:Star) So when the phase voltage gets divided by 1.732 the phase current should be increasing by 1.732 times.

This is the solution given.

Why at the final step of the solution do the say 52/sqrt(3) instead of 52*sqrt(3) as I was under the impression that if the phase voltage goes down (is lower in star than in delta) then the current in star should be higher than in delta?

Answer 1

When the motor is at zero RPM it can be seen as a transformer with the stator as primary winding and the rotor as secondary winding.

The secondary winding is shorted hence the startup current is a short circuit current of that transformer. The short circuit current is proportional to the voltage applied, as it is solely determined by the inductance and resistance of the components.

This is different from a motor running close to its synchronous RPM. Running at exactly synchronous RPM is, to stay with that analogy an open secondary winding of that transformer.

That's where the analogy reaches its limit. Close to synchronous RPM lies the nominal RPM and when loaded the motor runs just below the nominal RPM.

When a load is applied the behaviour of the motor is defined by its resistances and inductances and additionally by the behavior of the load. The load of most machinery has a rather flat curve at nominal RPM in comparison to the curve of the motor. Lets say the induction motor moves a conveyor belt. If the conveyor belt moves some material horizontally the torque will increase slowly with the RPM. But near the nominal RPM the M-n curve of the motor is extremely steep. The amplitude of the whole M-n curve however is proportional to the voltage.

Consider a flat curve crossing the steep section of the curve. If the voltage is reduced, the operating point moves faintly to a lower RPM and even more faintly the torque will drop. In total the mechanical power delivered by the motor will be almost the same. However, the power must be delivered by the motor. So if the voltage drops significantly the current has to increase reciprocally.

Those effects at 0 RPM and with typical load near nominal RPM shouldn't be confused.