conflicting answers between spin up 4-48 and NCEES Power 125

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pepower2012

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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?

 
I think the spin up method works for the NCEES problem....

1mva x .06 = .06

2mva x .045 = .09

Which means transformer 2 can handle 1.5 the load of transformer 2...

1mva + (1.5x1mva) = 2.5mva

Just realized question 125 is kva, but you get the picture.

 
Thanks for trying to help.

I might just gonna stay with MVA method on this particular type of problems for tomorrow. I'm afraid my head my explode otherwise.

 
Guys,

I still can't get irt 100%. Why assumption keeps changing>???

For EX:

for 4-48 in spin up

Gen A has 3MVA 3%

Gen B has 4MVA 2%

SB=2SA

then taking 4 MVa + 4/2 = 6.

As for NCEES, if we follow th exact meathod as above:

Then for NCEES 125

Transformer 1: 1MVA, 4.5%

Transformer 2: 2MVA, 6%

S2=1.5SA

if we take S2 = 2MVA + (2MVA/1.5) = 3.33 MVA not 1.25MVA.

besides, why in the solution for Spin up he assumes the total MVload is 6MVA when applying the formula of: [(MVA1/Z1)/ (MVA1/z1)+(MVA2/Z2)] * MVAload??

this is really confusing me. Please help

 

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