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Id
X+Z
Fig.K-2 Internal fault in Id-Ir plane including out-flow current
In order to ensure that the GRL100 relay will operate correctly in this case, the point shown on the
plot must fall within the operating zone.
According to this requirement, DIFI2 can be calculated as follows.
X +Z > X+2Y+Z−2DIFI2
DIFI2 > Y
This means that DIFI2 must be larger than the amount of out-flowing current.
CASE 1. The case with the distance protection applied as a main protection in addition to the
current differential protection for the special case
Therefore the condition shown in Table K-1 and Table K-2 should be replaced by that in Table
K-5 and Table K-6 respectively for "CASE 1".
Table K-5 CT Requirement defined by V
Td [ms]
35
50
75
100
150
Td [ms]
V
≧ I
35
k
50
V
≧ I
k
75
V
≧ I
k
100
V
≧ I
k
150
V
≧ I
k
V
: Knee point voltage [V]
k
R
: Secondary CT resistance [ohms]
ct
R
: Actual secondary burden [ohms]
2
I
: Maximum secondary fault current
fmax
I
: Maximum secondary fault current at the zone 1 reach point
f_z1_max
I
: Maximum secondary fault current for a close-up reverse fault
f_rev_max
I
: Maximum secondary load current
Lmax
Id = Ir - 2DIFI2
Id = 1/6Ir+5/6DIFI1
2DIFI2
X+2Y+Z
Requirement 1
V
≧ I
(R
+ R
)×3
k
fmax
ct
2
V
≧ I
(R
+ R
)×3
k
fmax
ct
2
V
≧ I
(R
+ R
)×4
k
fmax
ct
2
V
≧ I
(R
+ R
)×4
k
fmax
ct
2
V
≧ I
(R
+ R
)×8
k
fmax
ct
2
Requirement 3
(Rct+ R
)×2
V
f_rev_max
2
(Rct+ R
)×3
V
f_rev_max
2
(Rct+ R
)×6
V
f_rev_max
2
(Rct+ R
)×6
V
f_rev_max
2
(Rct+ R
)×6
V
f_rev_max
2
404
Ir
(K-2)
(Special case)
k
Requirement 2
V
≧ I
(R
+ R
k
f_z1_max
ct
V
≧ I
(R
+ R
k
f_z1_max
ct
V
≧ I
(R
+ R
k
f_z1_max
ct
V
≧ I
(R
+ R
k
f_z1_max
ct
V
≧ I
(R
+ R
k
f_z1_max
ct
Requirement 4
>
Max{I
+ I
/2, I
}
×(R
k
Lmax
fmin
fmaxout
>
× (R
Max{I
+ I
/2, I
}
k
Lmax
fmin
fmaxout
>
× (R
Max{I
+ I
/2, I
}
k
Lmax
fmin
fmaxout
>
× (R
Max{I
+ I
/2, I
}
k
Lmax
fmin
fmaxout
>
Max{I
+ I
/2, I
}
×(R
k
Lmax
fmin
fmaxout
6 F 2 S 0 8 5 0
)×6
2
)×7
2
)×8
2
)×8
2
)×8
2
+ R
)×14.4
ct
2
+ R
)×20
ct
2
+ R
)×28.8
ct
2
+ R
)×36.8
ct
2
+ R
)×50.4
ct
2
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