Sel SEL-DTA2, Инструкция по эксплуатации

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3.8
SEL-DTA2 Instruction Manual
Date Code 20020329
Communications
Command Descriptions
fault with a location of –34.45 miles would not be processed by the
SEL-DTA2; no outputs would be affected.
Settings QUA1–QUA8 control the type of data appearing at the transducer
outputs. Available quantities are shown below:
Currents.
IA, IB, IC, IAB, IBC, ICA, 3I2, IR
Voltages.
VA, VB, VC, VAB, VBC, VCA, 3V2, 3V0,
(SEL-221F/121F) VS
Single-Phase Power.
PA, PB, PC (real), available only with Fast Meter
QA, QB, QC (reactive), available only with Fast Meter
Three-Phase Power.
P (real), Q (reactive)
Refer to your relay instruction manual to determine if the following demand
and peak demand values are available.
Demand.
DA, DB, DC, DR, D3I2, DP, DQ
Peak Demand.
PDA, PDB, PDC, PDR, PD3I2, PDP, PDQ
Fault Location.
FL
In the example setting, QUA1 is set to “VAB,” indicating that Transducer
Channel 1 reflects the voltage measured from Phase A to Phase B.
Settings MIN1–MIN8 and MAX1–MAX8 reflect the measured quantities
corresponding to –5 V (–1 mA or–5 mA) and 5 V (1 mA or 5 mA),
respectively. Values within the set range for a channel are scaled according to
these equations (where n is 1–8):
Using the example settings and a measured value VAB of 231.2 kV, the output
quantities for Analog Channel 1 (with Option 1) would be:
NOTE:
To represent quantities
having both positive and negative
values (i.e. VARs and Watts) such that
negative values give negative analog
outputs and positive values give
positive analog outputs, MIN and MAX
must be symmetrical around zero
(e.g., MIN =
–
200 MW and MAX = +200
MW).
Vout
(Measured Quantity – MINn) • 10 V
MAXn – MINn
--------------------------------------------------------------------------------------- –5 V Option 1 or 5 =
Iout
(Measured Quantity – MINn) • 2 mA
MAXn – MINn
----------------------------------------------------------------------------------------- –1 mA Option 1 =
Iout
(Measured Quantity – MINn) • 10 mA
MAXn – MIN
--------------------------------------------------------------------------------------------n –5 mA Option 5 =
Vout
231.2 – (–250.0)
( )
• 10 V
250.0 – (–250.0)
-------------------------------------------------------------- – 5V Option 1 or 5 =
4.62V =
Iout
231.2 – (–250.0)
( )
• 2 mA
MAX1 – MIN1
---------------------------------------------------------------- – 1mA Option 1 =
0.92 mA =