Schweitzer Engineering Laboratories SEL-311C - Page 484

13.30
SEL-311C Relay Instruction Manual Date Code 20060320
Testing and Troubleshooting
Test Procedures
Equation 13.21
Using a single-phase current source simplifies the I
2
calculation.
Equation 13.22
Assume that you apply the following test voltages:
V
A
= 49.0 V ∠0°
V
B
= 67.0 V ∠–120°
V
C
= 67.0 V ∠120°
Equation 13.23
Step 10. Determine the test angle of A-phase current from the Z1ANG
relay setting. For Equation 13.1 to yield a positive result, I
A
,
hence I
2
, should lag V
2
by the angle of Z1ANG. For
Equation 13.1 to yield a negative result, I
A
should lead V
2
by
(180° – ∠Z1ANG°).
Step 11. When you apply the voltage signals shown above, V
2
is 180°
out of phase from V
A
. Take this into account, and calculate the
angle of I
A
with respect to the angle of V
A
. Equation 13.1
yields a positive result when I
A
leads V
A
by
(180° – ∠Z1ANG°). Equation 13.1 yields a negative result
when I
A
lags V
A
by ∠Z1ANG°.
We are testing for positive values of Z2c, so I
A
should lead V
A
by (180°–∠Z1ANG°). Assuming that V
A
= 49.0 ∠0° volts, the
angle of I
A
for this test should be 96°.
Step 12. Calculate the magnitude of A-phase current where Z2c equals
Z2RT or Z2FT through use of Equation 13.24:
Equation 13.24
For Z2R = 5.45 Ω:
Equation 13.25
Step 13. Calculate Z2m:
V
2
1
3
---
V
A
a
2
V
B
aV
C
• +• +()• =
I
2
1
3
---
I
A
a
2
I
B
aI
C
• +• +()• =
I
B
I
C
= 0 amps=
I
2
1
3
---
I
A
()• =
V
2
1
3
---
49.0 0° 1240° 67 120°–1120° 67 120°∠• ∠+∠• ∠+∠()volts• =
1
3
---
49.0 0° 67 120° 67 120– °∠+∠+∠()volts• =
6.0 180°volts∠=
I
A
3
V
2
Z
2
---------
⎝⎠
⎛⎞
• =
I
A
3
6.0 volts
5.45 ohms
-------------------------
⎝⎠
⎛⎞
• =
3.3 am
p
s=