Hi,

I hve got one 1,0 MW 11 kv slip ring motor which is designed for star connection at 11 k V . However, the supply voltage on my site is only 5,5 k V. I have two options

1. Star connected the motor at 5,5 k V which means that i will only have around 300 k W output shaft power.
or
2. Delta connected the motor at 5,5 k V where I shall have somewhat slightly less than 1,0 mW output shaft power because of the derated voltage, 5,5 k V instaed of 6,35 kV across each winding.

I need only around 600 k W for my application.

Am i following the right path.

Bob

Dear Bob,

You are definitely following right path
1.Out put Kw is propotional to square of voltage.So when you apply only 5.5Kv you get
25% of 1MW that is 250KW

2. If you connect the motor in delta and apply 5.5 kV you get 3 times more out put namely 750KW

At the same time you have to take care of rotor volts which will be 50% in first case and 86% in second case while deciding starting resistances in rotor circuit

subrao

Hi Bob

I agree with Subrao. Good information, and the note on the rotor is certainly one to watch out for.

Best regards,

Hello Marke,

Torque is proportionl to the square of the voltage but does it hold true for power also ?

Regards.

Bob

Hello Bob

Power equals Torque times Speed times a constant, so at a constant speed, the power is directly proportional to the speed.

So yes, at a given speed, power is proportional to the square of the voltage.

Best regards,