2.12 Line differential protection
LdI> (87L)
2 Protection functions
Technical description
78
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Example 1:Normal situation from relay 1 point of view
Relay1: measured phase current I
L1
=1000A/0°.
Relay2: measured phase current I
L1
=300A/-180°.
CT scaling of relay1 is 1000A / 5A and nominal current is
1000A. CT scaling of relay2 is 1000A / 1A and the nominal
current is 300A. Relay2 sends primary current measurement
information to relay1 and relay1 swaps the angle of received
current by 180 degrees (relay2 phase current I
L1
=300A/-180°
300A/0°).
In BIAS-calculation the measured current amplitude is divided
by the nominal primary current of both ends (might be
different like now).
Relay1: I
PRIMARY MEASURED
/ I
NOMINAL
= 1000A / 1000A = 1
Relay2: I
PRIMARY RECEIVED
/ I
NOMINAL REMOTE
= 300A / 300A = 1
N
d
I
I
0
0
1
0
1
Example 2:Fault situation from relay 1 point of view
Relay1: measured phase current I
L1
=2400A/-30°.
Relay2: measured phase current I
L1
=2100A/-45°.
CT scaling of relay1 is 1000A / 5A and nominal current is
1000A. CT scaling of relay2 is 1000A / 1A and the nominal
current is 300A. Relay2 sends primary current measurement
information to relay1 and relay1 swaps the angle of received
current by 180 degrees (relay2 phase current I
L1
=2100A/-45°
2100A/135°).
In BIAS-calculation the measured current amplitude is divided
by the nominal primary current of both ends (might be
different like now).
Relay1: I
PRIMARY MEASURED
/ I
NOMINAL
= 2400A / 1000A = 2.4
Relay2: I
PRIMARY RECEIVED
/ I
NOMINAL REMOTE
= 2100A / 300A = 7
N
d
I
I
37
.
9
135
7
35
4
.
2
N
b
I
I
1
2
1
1
N
b
I
I
7
.
4
2
7
4
.
2