# Operating At Slightly Lower Rated Voltage

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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

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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,

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

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Hi Bob

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

Best regards,

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Hello Marke,

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

Regards.

Bob

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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,

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