Question

The diagram shows one phase of a 3-phase generation and distribution system. We would like to determine the 12 kV short circuit (maximum) current at the indicated location attributed to the generation, in the event of a fault.

For this question, we work in the per unit pu system.

a. We pick S(base)=100 MVA, and V(base) = 12 kV. Then I(base) and Z(base) are fixed. The 12 kV of the transformer secondary is line-neutral voltage since this is a grounded neutral voltage. Converting the 12 kV to line-line voltage, show that I(base) = 4,811 A.
b. The distribution line impedance is referenced to the 12 kV line-neutral voltage. Since impedance = (power/voltage2), show that
Z(transmission line) = 10.6 pu on a 100 MVA base.
c. Taking Z(transformer) = 0.075 pu referenced to 7.5 MVA, show that Z(transformer) =1.0 pu referenced to the 100 MVA base.
d. We can now convert I(base) to the short-circuit (maximum) current at the fault location. This is I(base)/(total Z pu) = 415 A.
e. 100 MVA was chosen as S(base) but any value for S(base) could have been chosen. Why?
This is a common type of calculation in the pu system. Note that the 60 kV of the transformer primary did not matter in this calculation, because the fault was in the 12 kV line

Distribution Line Z=j15.3112 Fault location Generation 60 kV 100 MVA Transformer 60/12 kV 7.5 MVA X= 7.5%

Answer 1

4811. 25 413.69 A d) Now, I base Ans. Toral Zpu (I+10.63) of generator is not given Note: Per-unit impedance the question. the value of of e) Yes, any value for spase can be chosen, because as Shase will change corresponding. per unit impedance the equipment such as Generator, Transformer, Distribution line etc. will change accordingly as per fomulae shown in equation Also, corresponding bases such as Cunnent Base and Impedance Base value will also change occor " dingly