GE B30 UR Series

5-200 B30 Bus Differential System GE Multilin
5.9 TRANSDUCER INPUTS AND OUTPUTS 5 SETTINGS
5
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 nominal
For 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 MONITORING
A positive-sequence voltage on a 400 kV system measured via source 2 is to be monitored by the DCmA H3 output with a
range 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 5.24)
The base unit for voltage (refer to the FlexElements section in this chapter for additional details) is:
(EQ 5.25)
The minimum and maximum voltage values to be monitored (in pu) are:
(EQ 5.26)
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 quantities
scaled to the phase-to-ground voltages, even if the VTs are connected in “Delta” (refer to the Metering conventions section
in chapter 6), while at the same time the VT nominal voltage is 1 pu for the settings. Consequently the settings required in
this 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 reading
For example, under nominal conditions, the positive-sequence reads 230.94 kV and the worst-case error is
0.005 x 230.94 kV + 1.27 kV = 2.42 kV.
0.005 20 4–()6.3 kA××± 0.504 kA±=
V
min
0.7
400 kV
3
-------------------
× 161.66 kV, V
max
1.1
400 kV
3
-------------------
× 254.03 kV====
V
BASE
0.0664 kV 6024× 400 kV==
minimum voltage
161.66 kV
400 kV
---------------------------
0.404 pu, maximum voltage
254.03 kV
400 kV
---------------------------
0.635 pu====
3
0.005 1 0–()× 254.03 kV×± 1.27 kV±=