GE L90 Instruction Manual page 415

5 SETTINGS
DCMA OUTPUT H2 SOURCE
: "4 to 20 mA"
DCMA OUTPUT H2 RANGE
DCMA OUTPUT H2 MIN VAL
DCMA OUTPUT H2 MAX VAL
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 3:
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:
V 0.7
=
min
The base unit for voltage (refer to the FlexElements section in this chapter for additional details) is:
The minimum and maximum voltage values to be monitored (in pu) are:
minimum voltage
The following settings should be entered:
DCMA OUTPUT H3 SOURCE
: "0 to 1 mA"
DCMA OUTPUT H3 RANGE
DCMA OUTPUT H3 MIN VAL
DCMA OUTPUT H3 MAX VAL
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.
GE Multilin
: "SRC 1 Ia RMS"
: "0.000 pu"
: "1.260 pu"
400 kV
×
------------------ - 161.66 kV,
=
3
×
V 0.0664 kV 6024
=
BASE
161.66 kV
= -------------------------- - = 0.404 pu,
400 kV
: "SRC 2 V_1 mag"
: "0.404 pu"
: "0.635 pu"
L90 Line Current Differential System
5.9 TRANSDUCER INPUTS AND OUTPUTS
± × ( ) 6.3 kA ×
0.005 20 4
– =
400 kV
×
V 1.1 ------------------ - 254.03 kV
= =
max
3
400 kV
=
254.03 kV
maximum voltage = -------------------------- -
400 kV
3 .
± × ( ) ×
0.005 1 0 – 254.03 kV
±
0.504 kA
(EQ 5.36)
(EQ 5.37)
= 0.635 pu
(EQ 5.38)
±
= 1.27 kV
5-279
5