pepower2012
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- Jul 29, 2012
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for 4-48 in spin up
Gen A has 3MVA 3%
Gen B has 4MVA 2%
A&B synchronized and operating in parallel, What is the total load of the set up without overload any Gen?
The answer was 3M * 2%= 6, which is half of 4MVA *3, then we can assume that Gen A has the half of load capacity of Gen B, there for 4 + 4/2 = 6MVA
Then for NCEES 125
Transformer 1: 1MVA, 4.5%
Transformer 2: 2MVA, 6%
What is the total load that can be served without overload either transformer?
The answer was using PU method and it came out to be 2500. I had problem understand why. And the more important note was that this two methods can't be used on each other. So if you use the PU method for 4-48, you will get 9MV, not 6MVA. And if you use Spin up method for the 125, it doesn't work either! Am I missing something here?
Gen A has 3MVA 3%
Gen B has 4MVA 2%
A&B synchronized and operating in parallel, What is the total load of the set up without overload any Gen?
The answer was 3M * 2%= 6, which is half of 4MVA *3, then we can assume that Gen A has the half of load capacity of Gen B, there for 4 + 4/2 = 6MVA
Then for NCEES 125
Transformer 1: 1MVA, 4.5%
Transformer 2: 2MVA, 6%
What is the total load that can be served without overload either transformer?
The answer was using PU method and it came out to be 2500. I had problem understand why. And the more important note was that this two methods can't be used on each other. So if you use the PU method for 4-48, you will get 9MV, not 6MVA. And if you use Spin up method for the 125, it doesn't work either! Am I missing something here?