ANALISIS PERFORMA RELAY DIFFERENTIAL PADA TRANSFORMATOR 1 GIS NEW SENAYAN
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Institusion
Institut Teknologi Perusahaan Listrik Negara
Author
Kristo Siringoringo, Guntur Prip
Mangapul, Juara
Simamora, Yoakim
Mangapul, Juara
Simamora, Yoakim
Subject
Teknik Elektro
Datestamp
2022-10-10 08:10:11
Abstract :
The relay differential is a major protection system on the transformer that protects the transformer from interference. Differential relays protect the transformer when there is a nominal difference in the current flowing on the high voltage side and the low voltage side and work as a sign of delay. In this research, simulation modeling with ETAP 16.0 software and data measurement has been done in GIS New Senayan. When calculating the value according to the theory, the differential relay setting for slope1 is 14% and for slope2 is 28,1% and the setting current1 is 0,1162 A and setting current2 is 0,23263 A. When making a characteristic differential relay, you can use the formula y = Irestrain x (Slope1-endsection) + Idif .. With X-axis = irestrain and Y = I differential. Then when simulated in ETAP 16.0 Software, the researcher gets the value of internal and external short circuit currents. , where when the internal short circuit fault is obtained, the value of the fault current is 2356 A with a differential current of 6.9673A> 0.3 A which makes the differential relay directly secure its working area, while in the external fault the short circuit current is 1108 A with a differential current of 0.2158 A is smaller Id 0.3 A, then the relay does not work to protect its working area
The relay differential is a major protection system on the transformer that protects the transformer from interference. Differential relays protect the transformer when there is a nominal difference in the current flowing on the high voltage side and the low voltage side and work as a sign of delay. In this research, simulation modeling with ETAP 16.0 software and data measurement has been done in GIS New Senayan. When calculating the value according to the theory, the differential relay setting for slope1 is 14% and for slope2 is 28,1% and the setting current1 is 0,1162 A and setting current2 is 0,23263 A. When making a characteristic differential relay, you can use the formula y = Irestrain x (Slope1-endsection) + Idif .. With X-axis = irestrain and Y = I differential. Then when simulated in ETAP 16.0 Software, the researcher gets the value of internal and external short circuit currents. , where when the internal short circuit fault is obtained, the value of the fault current is 2356 A with a differential current of 6.9673A> 0.3 A which makes the differential relay directly secure its working area, while in the external fault the short circuit current is 1108 A with a differential current of 0.2158 A is smaller Id 0.3 A, then the relay does not work to protect its working area
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