Schweitzer Engineering Laboratories SEL-311C - Page 503

13.49
Date Code 20060320 Instruction Manual SEL-311C Relay
Testing and Troubleshooting
Test Procedures
Solving for the magnitude of I
A
yields:
Equation 13.72
The Zone 2 ground distance element is forward reaching. Thus,
it is supervised by the forward directional element 32GF, as
well as the 50L2 and 50GZ2 phase and residual overcurrent
elements. Applied phase current must exceed the 50L2 setting,
applied residual current must exceed the 50GZ2 setting, and
applied 3I
2
must exceed the 50QF setting.
The 50L overcurrent elements operate based upon the
magnitude of the phase current. Through use of the current
connections shown in Figure 13.5, we can see that the
magnitude of I
A
is equal to the magnitude of the applied test
current.
With a 50L2 setting of 0.50 amps, 50L2 picks up when I
TEST
is
greater than 0.50 amps.
The 50GZ overcurrent elements operate based upon the
magnitude of the residual current. Through use of the current
connections shown in Figure 13.5, we can see that the
magnitude of I
R
is equal to the magnitude of the applied test
current.
With a 50GZ2 setting of 0.50 amps, 50GZ2 picks up when
I
TEST
is greater than 0.50 amps.
The 50QF negative-sequence overcurrent element operates
based upon the magnitude of 3I
2
applied. Through use of the
current connections shown in Figure 13.5, calculate the
magnitude of 3I
2
applied based upon the magnitude of I
TEST
.
Equation 13.73
so:
Equation 13.74
With a 50QF setting of 0.5 amps, 50QF picks up when I
TEST
is
greater than 0.5 amps.
We calculated |IA| = |I
TEST
| = 2.49 amps which satisfies the
supervisory overcurrent conditions described above.
Step 11. Select V
B
= 67 ∠–120° volts and V
C
= 67 ∠120°volts.
The ground distance elements are supervised by the ground
directional element.
Step 12. Check the negative-sequence quantities applied and verify that
the 32GF element asserts.
I
A
40.0
9.39()1.71()•
-------------------------------------=
2.49 amps=
I
2
1
3
---
I
A
a
2
I
B
aI
C
• +• +()• =
3I•
2
I
A
a
2
I
B
aI
C
• +• +()=
I
A
I
TEST
and I
B
I
C
0===
3I
2
• I
TEST
=