GE Multilin L30 Line Current Differential System 9-5
9 APPLICATION OF SETTINGS 9.2 CURRENT DIFFERENTIAL (87L) SETTINGS
9
9.2.6 CT TAP
If the CT ratios at the line terminals are different, the
CURRENT DIFF CT TAP 1(2) setting must be used to correct the ratios to
a common base. In this case, a user should modify the CURRENT DIFF BREAK PT and CURRENT DIFF PICKUP settings
because the local current phasor is used as a reference to determine which differential equation is used, based on the
value of local and remote currents. If the setting is not modified, the responses of individual relays, especially during an
external fault, can be asymmetrical, as one relay can be below the breakpoint and the other above the breakpoint. There
are two methods to overcome this potential problem:
1. Set
CURRENT DIFF RESTRAINT 1 and CURRENT DIFF RESTRAINT 2 to the same value (e.g. 40% or 50%). This converts the
relay characteristics from dual slope into single slope and the breakpoint becomes immaterial. Next, adjust differential
pickup at all terminals according to CT ratios, referencing the desired pickup to the line primary current (see below).
2. Set the breakpoints in each relay individually in accordance with the local CT ratio and the CT TAP setting. Next, adjust
the differential pickup setting according to the terminal CT ratios. The slope value must be identical at all terminals.
Consider a two-terminal configuration with the following CT ratios for relays 1 and 2.
(EQ 9.4)
Consequently, we have the following CT tap value for relays 1 and 2.
(EQ 9.5)
To achieve maximum differential sensitivity, the minimum pickup is set as 0.2 pu at the terminal with the higher CT primary
current; in this case 2000:5 for relay 2. The other terminal pickup is adjusted accordingly. The pickup values are set as fol-
lows:
(EQ 9.6)
Choosing relay 1 as a reference with a breakpoint of 5.0, the break point at relay 2 is chosen as follows:
(EQ 9.7)
Use the following equality the verify the calculated breakpoint:
(EQ 9.8)
Therefore, we have a breakpoint of 5.0 for relay 1 and 2.5 for relay 2.
Now, consider a three-terminal configuration with the following CT ratios for relays 1, 2, and 3.
(EQ 9.9)
Consequently, we have the following CT tap value for relays 1, 2, and 3.
(EQ 9.10)
In this case, the relay channels communicate as follows:
• For relay 1, channel 1 communicates to relay 2 and channel 2 communicates to relay 3
• For relay 2, channel 1 communicates to relay 1 and channel 2 communicates to relay 3
CT
ratio
relay 1()1000 5⁄=
CT
ratio
relay 2()2000 5⁄=
CT
tap
relay 1()2.0=
CT
tap
relay 2()0.5=
Pickup relay 1()0.4=
Pickup relay 2()0.2=
Breakpoint relay 2()Breakpoint relay 1()
CT
ratio
relay 1()
CT
ratio
relay 2()
----------------------------------------
×=
5.0
1000 5⁄
2000 5⁄
--------------------
×= 2.5=
Breakpoint relay 1()CT
ratio
relay 1()× Breakpoint relay 2()CT
ratio
relay 2()×=
CT
ratio
relay 1()1000 5⁄=
CT
ratio
relay 2()2000 5⁄=
CT
ratio
relay 3()500 5⁄=
CT
tap1
relay 1()2.00= CT
tap2
relay 1()0.50=
CT
tap1
relay 2()0.50 CT
tap2
relay 2()0.25==
CT
tap1
relay 3()2.00 CT
tap2
relay 3()4.00==
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