7FH2 Generator Cooler and Rated Load

Greetings

While I was reading GEK-107092d, I came across something not very clear. It talks about isolating one cooler (in case it has a failure) of the generator. Fortunately, we haven't had any cooler failures in the past. the quote is:

"If a cooler leak is severe, the defective cooler may be left out of service until repairs can be made. Generators are typically designed to operate at rated power factor at 80% rated load capacity with one hydrogen cooler out of service."

I know that we have to keep monitoring the generator RTDs in case of isolating one cooler.

My question is, what is this rated load? Is it the one on the generator nameplate?

Our generator name plate says the MVA is 228.6 and power factor is 0.85.

By doing small calculations, the rated load will be 155 MW. So 80% would be 124 MWW. Is that correct ?
 
Aptx4869,

If I recall correctly, to determine the possible MW output rating for a synchronous generator nameplate rating given in MVA, one has to multiply the MVA by the rated power factor (from the nameplate). So, 228.6*0.85=193.8 MW, and if you then multiply 193.8 by the 80% (0.80) safety factor for running with one hydrogen cooler out of service you would arrive at 155.04 MW.

It also seems to work the "other" way. If you multiply 228.6 MVA by the 80% safety rating for running with one hydrogen cooler out of service, you get 182.88 MVA. And, if you multiply 182.88 MVA by 0.85 you get 155.45 MW--which is close enough to 155.04 MW for my maths.

But, I'm not a maths wizard. And, I didn't search the World Wide Web for the formula for converting KVA to KW (or MVA to MW). I just used my feeble memory.

In other words, I may be wrong. I would have to refer to my university text, which I don't have with me at this writing.
 
CSA,

Thank you for the reply. I recalculated the MW and reached into the same answer as you had. I seems I made small error in my earlier calculations.

But again, the rated load in the quote, it seems they meant the rated load for the generator. This is what I believe.
 
Aptx4869,

Yes, because the issue is producing electrical power (MVA) without one hydrogen cooler. If necessary, you would have to MANUALLY limit the turbine to keep the electrical power below 155 MW, or 182.88 MVA.

You would be trying to protect the generator, not the turbine. BUT, you would protect the generator by limiting the turbine output (because the generator is driven by the turbine; the generator is "dumb"; it just converts the torque from the turbine to real power (MW)). And you would use the "AVR" to limit the VArs to keep the combination of the watts and the VArs (the MVA) below the limit.

Hope this helps!
 
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