GANGGUAN TAK SIMETRI

TIPE GANGGUAN

Shunt faults are more severe than series faults. Balanced faults are simpler to calculate than unbalanced faults. Simultaneous faults, involving two or more faults that occur simultaneously, are usually considered to be the most difficult fault analysis problem. The probability of having a simultaneous fault is much less than the shunt fault.

• Most of the faults that occur on power systems are not the balanced (i.e., symmetrical) threephase faults but the unbalanced (i.e., unsymmetrical) faults, specifically the single line-to-ground (SLG) faults. For example, Reference [5] gives the typical frequency of occurrence for the threephase, SLG, line-to-line, and double line-to-ground (DLG) faults as 5%, 70%, 15%, and 10%, respectively.

• In general, the three-phase fault is considered to be the most severe one. However, it is possible that the SLG fault may be more severe than the three-phase fault under two circumstances: (1) the generators involved in the fault have solidly grounded neutrals or low-impedance neutral impedances and (2) it occurs on the wye-grounded side of delta–wye-grounded transformer banks. The line-to-line fault current is about 86.6% of the three-phase fault current.

• Faults can be categorized as shunt faults (short circuits), series faults (open conductor), and simultaneous faults (having more than one fault occurring at the same time). Unbalanced faults can be easily solved by using the symmetrical components of an unbalanced system of currents or voltages. Therefore, an unbalanced system can be converted to three fictitious networks: the positive-sequence (the only one that has a driving voltage), the negative-sequence, and the zero-sequence networks interconnected to each other in a particular fashion depending on the fault type involved.

ASUMSI-ASUMSI

SINGLE LINE TO GROUND FOULT

Contoh

Gangguan satu fasa ketanah pada bus 1

LINE TO LINE FOULT

Contoh

Gangguan antar fasa pada bus 3

DOBLE LINE TO GROUND FOULT

Contoh

Gangguan antar fasa ke tanah pada bus 4

Y11= 1/x =1/j0.05=-j20 Y12 = 0 tidak terhubung Y22 =1/x = 1/(j0.10 + j0.15) = -j4

TUGAS 8

Assume faults at the middle of the transmission line TL23