I am installing a 600kW 4 pole 3300V squirrel cage induction motor direct-on-line started from a 11000V board via a 1000kVA 11000V/3300V step down transformer. The motor will drive a fan. The question is will the fan accelerate to full speed?

That totally depends upon the fan. As a general rule however, you need a transformer kVA to be LEAST 2 x kW, and in this case it is not. It may still work however, if that fan is the only load on the transformer. it will essentially become something like an autotransformer starter. The overload on the transformer will cause a severe voltage drop on the load side, which will reduce the torque, and thus the current draw, which will act to mitigate the voltage drop as the motor comes up to speed. There is a significant risk of overloading the motor doing it this way however. It will also likely create a significant voltage drop on the 11kV side as well, which may not be acceptable.

That same rough rule also states that with reduced voltage starting, you can often get away with the transformer kVA being 1.1 times the motor kW, so that may be a better solution for you. An electronic soft starter would be a good choice because it is infinitely adjustable as to current and torque settings to match your fan requirements, plus they tend to come with advanced motor protection as well.

Hello sixtus

In effect, by using a transformer that is rated relatively close to the motor rating, you are adding an impedance
in series with the motor. During DOL start, this impedance is going to have quite an influence on the start current and as a result will reduce the start torque of the motor. If the reduction tis too much, there will be insufficient torque to start the fan.

If we assume that i) the LRC of the motor is 600%, and ii) the impedance of the transformer is 4.7%, and iii) there is no significant drop on the 11000 side, then at start, the effective impedance of the motor is 16% of its run impedance.
At run, the motor will have a KVA loading in the order of 700KVA, so the effective impedance of the transformer at that loading will be in the order of 3.3% and the impedance of the motor will be 96.7%
At start, the impedance of the motor will fall to around 16%, so the effective impedance of the motor becomes around 16.1%
The voltage drop in the transformer will be in the order of 3.3/(16.1+3.3) = 17%
I would expect that at a voltage drop of 17% (83% voltage applied to the motor, 69% full voltage torque) the torque would probably be high enough to get the fan to full speed provided that the motor has a reasonable amount of start torque and the fan is well damped during start.

If you have the speed torque and speed current curves of the motor, and the speed torque curve of the motor, it is relatively easy to determine whether the fan will start. Without the curves, there is a lot of guess work and the answer can become meaningless.

Fans typically can take quite some time to start and you may have protection issues with the transforme due to the high overload during start.

Best regards,

Hi Mark or anyone else,

I'm rehashing an old post here but could you please explain this derivation from your last post?

"At start, the impedance of the motor will fall to around 16%, so the effective impedance of the motor becomes around 16.1%"

Should it read:

"At start, the impedance of the motor will fall to around 16.6667% (1/600%), so the effective impedance of the motor becomes around 16.1% (16.66667*96.7%)"

I think this is just a matter of rounding but I've been trying to figure this out for a while and the above seems to make sense.

The lead on questions are then:
1) how do you account for a loaded transformer, i.e., what happens here if the TX was 50% loaded. I guess the motor starting % change is the same but the total voltage drop from nominal needs to be considered too.
2) if the cable has significant length how does this get incorporated to find the v drop at the motor terminals. I suspect we need to normalise to some base somehow.....

Thanks.