5-254 L30 Line Current Differential System GE Multilin5.9 TRANSDUCER INPUTS AND OUTPUTS 5 SETTINGS5(EQ 5.34)The base unit for current (refer to the FlexElements section in this chapter for additional details) is:(EQ 5.35)The minimum and maximum power values to be monitored (in pu) are:(EQ 5.36)The following settings should be entered:DCMA OUTPUT H2 SOURCE: “SRC 1 Ia RMS”DCMA OUTPUT H2 RANGE: “4 to 20 mA”DCMA OUTPUT H2 MIN VAL: “0.000 pu”DCMA OUTPUT H2 MAX VAL: “1.260 pu”The worst-case error for this application could be calculated by superimposing the following two sources of error:• ±0.5% of the full scale for the analog output module, or• ±0.25% of reading or ±0.1% of rated (whichever is greater) for currents between 0.1 and 2.0 of nominalFor example, at the reading of 4.2 kA, the worst-case error is max(0.0025 × 4.2 kA, 0.001 × 5 kA) + 0.504 kA = 0.515 kA.EXAMPLE: VOLTAGE MONITORINGA positive-sequence voltage on a 400 kV system measured via source 2 is to be monitored by the dcmA H3 output with arange of 0 to 1 mA. The VT secondary setting is 66.4 V, the VT ratio setting is 6024, and the VT connection setting is“Delta”. The voltage should be monitored in the range from 70% to 110% of nominal.The minimum and maximum positive-sequence voltages to be monitored are:(EQ 5.37)The base unit for voltage (refer to the FlexElements section in this chapter for additional details) is:(EQ 5.38)The minimum and maximum voltage values to be monitored (in pu) are:(EQ 5.39)The following settings should be entered:DCMA OUTPUT H3 SOURCE: “SRC 2 V_1 mag”DCMA OUTPUT H3 RANGE: “0 to 1 mA”DCMA OUTPUT H3 MIN VAL: “0.404 pu”DCMA OUTPUT H3 MAX VAL: “0.635 pu”The limit settings differ from the expected 0.7 pu and 1.1 pu because the relay calculates the positive-sequence quantitiesscaled to the phase-to-ground voltages, even if the VTs are connected in “Delta” (refer to the Metering conventions sectionin chapter 6), while at the same time the VT nominal voltage is 1 pu for the settings. Consequently the settings required inthis example differ from naturally expected by the factor of .The worst-case error for this application could be calculated by superimposing the following two sources of error:• ±0.5% of the full scale for the analog output module, or• ±0.5% of readingFor example, under nominal conditions, the positive-sequence reads 230.94 kV and the worst-case error is0.005 x 230.94 kV + 1.27 kV = 2.42 kV.I max 1.5 4.2 kA× 6.3 kA= =I BASE 5 kA=minimum current 0 kA5 kA------------ 0 pu, maximum current 6.3 kA5 kA----------------- 1.26 pu= == =0.005 20 4–( ) 6.3 kA××± 0.504 kA±=V min 0.7 400 kV3-------------------× 161.66 kV, V max 1.1 400 kV3-------------------× 254.03 kV= == =V BASE 0.0664 kV 6024× 400 kV= =minimum voltage 161.66 kV400 kV--------------------------- 0.404 pu, maximum voltage 254.03 kV400 kV--------------------------- 0.635 pu= == =30.005 1 0–( )× 254.03 kV×± 1.27 kV±=
