NCEES 118

[Redskinsdb21]

Hey..Im trying to see different ways to approach this question. I see Vll_p*Il_primary = Vll_s*Il_secondary....on either side of the Xfmr...but shouldnt the line current on the generator side also be able to be calculated from S=VI* ? Meaning shouldnt it be also : I = S/V = 50 MVA/13.2 KV ? I get 3787 amps when doing this? what am I doing wrong? Does this mean the Generator is rated up to 50 MVA but at 13.2 KV its not actually producing an S value of 50 MVA that it can?

~Thanks

 

[Flyer_PE]

Since this is a 3-phase system, you have to divide your answer for current by sqrt3. The rated current for the generator is 2186 Amps.

Also, the problem gives you the transmission line current at 75.93 Amps.

At 132 kV, 75.93 Amps results in a power flow of 17.36 MVA.

17.36 MVA at 13.2 kV gives a line current of 759 Amps

 

[Redskinsdb21]

thanks..i was trying 1-phase equation

 

[asmith01]

You could also look at the problem and solve by the turns ratio for the voltage of the transformer and since you know the current for the tmission line you can then do Isecondary/Iprimary to find the current of the generator.

 

[JillSuzanne]

Why is the impedance in the transformer ignored?

 

[Flyer_PE]

The impedance will affect voltage drop, not the current value.

 

[Swiftman25]

I got the correct answer to 118 but in a different way than the book.

This is how I looked at this problem: generator side as primary and line side as secondary. a = 13.2/(132/root(3)) = 13.2/76.21 = 0.17321.

Know equation: Is/Ip = a; therefore Ip=Is/a = 75.93/0.17321 = 438.38A. HOWEVER, this is amperage in the phases of the delta. LINE current into the delta is root(3) larger,

Therefore Igenerator=root(3)*438.38 = 759.3A. This is close to 760 or answer (A).

 

[Phatso86]

I got the correct answer to 118 but in a different way than the book.

This is how I looked at this problem: generator side as primary and line side as secondary. a = 13.2/(132/root(3)) = 13.2/76.21 = 0.17321.

Know equation: Is/Ip = a; therefore Ip=Is/a = 75.93/0.17321 = 438.38A. HOWEVER, this is amperage in the phases of the delta. LINE current into the delta is root(3) larger,

Therefore Igenerator=root(3)*438.38 = 759.3A. This is close to 760 or answer (A).

I had to read this twice to figure out how it worked out.

I can't believe I got this answer wrong