5-12 Applications Date Code 20011026
SEL-321/321-1 Instruction Manual
Calculate Z2 for forward faults:
At Bus B:
VF
kV
2
28 6
2000
14 3==
.
.
volts secondary
IF
2
730
200
365==.
A secondary
From above:
ZF
VF
IF
2
2
2
14 3
365
391=
−
=
−
=−
.
.
. Ω
secondary
Calculate Z2 for reverse faults:
At Bus B:
VR
kV
2
29 0
2000
14 5==
.
.
volts secondary
IR
2
286
200
143==.
amps secondary
From above:
ZR
VR
IR
2
2
2
14 5
143
1013=
−
−
==
.
.
. Ω
secondary
Summing the magnitudes of the forward and reverse calculated Z2 as follows:
1013 3 91 14 04...ΩΩΩ+− =
Divide the Z
2
F and Z
2
R plane into three sectors:
14 04
3
468
.
.
Ω
Ω=
Setting for Z2F:
(
)
−+=391 468 077...ΩΩΩ
Setting for Z2R:
()
1013 4 68 545...ΩΩΩ−=
For our example, 50QF = 50QR = 0.5 A (3I2). The maximum load on Line 2 is 450 amps. For
the negative-sequence directional element to operate I
2
> a2 • I
1
.
Therefore, the highest a2 setting is:
I
Iload
2
1
0167
225
0 076
min
.
.
.==
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