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https://tgadvisers.com/steam-turbine-and-generator-reduced-minimum-load-considerations/ If you got 20k, have at it. https://www.epri.com/research/products/000000003002013589

Phase to phase IR? Is it 6 leads generator? If not, you should be getting zero megohms between phases.

Under powered steam turbine will have water condensation, scaling of blades leading to lower life.

I rotate large electric machines up to 20 tons in our shop. I use two 20 ton hoists mounted independently in the same cross travel bridge.

Greg - 5 to 10 minutes is my plan. It's a time vs KW decision. MintJulep - Ok.

"If I reduce the KVA that should equate to a reduction in amps, correct?" Correct. "if there is an amp reduction, shouldn't I be able to calculate an efficiency increase?" The reduced current means less I^2R loss in the generator, cables, transformer, which is indeed a power saving. You need...

I am looking at 3300/600 = 5.5 rpm per sec. That's linear acceleration, right?

govt department advised correctly. 24x7 capacitor with no other inductive load will drop your pf like a stone.

Littleinch Yes, if the acceleration time is increased from 5 to 10 minutes for 0 to 3300 RPM, the power required drops to half i.e. 14.34 KW. I am looking at a linear acceleration which the VFD will be programmed to do. Both the drive cage induction motor and VFD are rated for 132 KW. I...

Thanks guys. But when the load is running at 3300 RPM, the motor has to deliver that peak power of 28.68 KW, right?

Driving motor has enough power and speed range up to 3300 RPM and the whole set up is via a big VFD.

Thanks, Greg. I did do some math (had to recall some college days equations) and got exactly double of that i.e. 28.68 KW. Wonder where I am going wrong. Got moment inertia I = 72 kg-sq m right Torque = Moment of inertia x Angular acceleration = 82.97 N-m Power = torque x angular speed =...

What will be the power needed to rotate a uniform cylindrical rotor weighing 1600 kgs dia 600 mm speed 3300 RPM 5 minutes acceleration time from zero speed to 3300 RPM Any particular equation to calculate this? Will acceleration time play a role in power needed? The cylindrical rotor will be...

In a given motor frame size, 2 to 4 speeds can be designed with corresponding power rating. You need to look for the same frame with the same speed and HP that you want. No standards will specify the power and speed ratings for frame sizes.

Is that a stator - rotor rub? Aluminum die-cast rotor? Speed?

"We are bit sure that the problem is in rotor laminations because a similar motor with the same symposium was deconstructed and found a deformation in it’s lamination." What kind of deformation? Post some pictures and complete motor nameplate.

"The VFD acts as the filter". How? why not go for a 690 V motor of same size? standard VFD and reduced cable size.

Looks like the top mcb is the output with output voltmeter connected. The connector terminal blocks seem to be unfit for load output. Zero nameplate details. OP has to engage to more

Great info, Gianoli. 33,500 A ? :oops:

Who is making 15,000 amps breakers? The maximum I am aware of is 5,000 amps VCB. Large generators have the breakers on the HV side of the generator step-up transformers.