5-244 L60 Line Phase Comparison Relay GE Multilin5.8 TRANSDUCER INPUTS/OUTPUTS 5 SETTINGS5For 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 3:A 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.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±=
