Schweitzer Engineering Laboratories SEL-311C - Page 508

13.54
SEL-311C Relay Instruction Manual Date Code 20060320
Testing and Troubleshooting
Test Procedures
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 13.87
so,
Equation 13.88
With a 50QF setting of 0.5 amps, 50QF picks up when I
TEST
is
greater than 0.5 amps.
We calculated |I
A
| = |I
TEST
| = 8.0 amps which satisfies the
supervisory overcurrent conditions described above.
Select V
B
= 67 ∠–120° volts and V
C
= 67 ∠120° volts.
The ground distance elements are supervised by the negative-
sequence directional element.
Step 11. Check the negative-sequence quantities applied and verify that
the 32GF element should assert.
Step 12. Calculate the magnitude and angle of negative-sequence
voltage and current applied for the test quantities listed above.
Step 13. Then calculate the negative-sequence impedance, Z2c, applied
through use of Equation 13.90.
Equation 13.89
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
=
V
2
1
3
---
V
A
a
2
V
B
aV
C
• +• +()• =
I
2
1
3
---
I
A
a
2
I
B
aI
C
• +• +()• =