Schweitzer Engineering Laboratories SEL-311C - Page 477

13.23
Date Code 20060320 Instruction Manual SEL-311C Relay
Testing and Troubleshooting
Test Procedures
Equation 13.2
Coincidentally, these voltage and current signals could represent a forward
single line-ground fault on a system with a 90° impedance angle. Normally, a
forward fault results in a negative Z2c value.
Here is an example for a reverse single line-ground fault.
∠Z1ANG = 90°
V
2
= 10 ∠180° volts, secondary
I
2
= 2 ∠ 90° amps, secondary
Equation 13.3
Notice that the result of Equation 13.1 is positive for a reverse fault and
negative for a forward fault. This result is consistent with actual behavior on
the power system.
The relay determines fault direction by comparing the result of Equation 13.1
to forward and reverse impedance thresholds that are dependent upon the Z2F
and Z2R relay settings and the magnitude of negative-sequence voltage
divided by the magnitude of negative-sequence current.
When Z2c is less than the forward Z2 threshold, Z2FT, the fault is in the
forward direction, so the relay sets the F32Q and F32QG elements, if other
supervisory conditions permit. When Z2c is greater than the reverse Z2
threshold, Z2RT, the fault is in the reverse direction, so the relay sets the R32Q
and R32QG elements, if other supervisory conditions permit.
The forward directional threshold is calculated through use of Equation 13.4
when Z2F is negative or Equation 13.5 when Z2F is positive.
When Z2F ≤ 0:
Equation 13.4
When Z2F > 0:
Equation 13.5
Where:
Equation 13.6
The reverse directional threshold is calculated through use of Equation 13.7
when Z2R is positive or Equation 13.8 when Z2R is negative.
Z2c
Re 10 180° 190°∠ 290°–∠• ()• ∠ *[]
2
2
--------------------------------------------------------------------------------------------=
–20 ohms
4
-----------------------=
–5 ohms=
Z2c
Re 10 180° 190°∠ 2+ 90°∠• ()• ∠ *[]
2
2
----------------------------------------------------------------------------------------------=
20 ohms
4
--------------------=
5 ohms=
Z2FT 0.75 Z2F 0.25 Z2m• –• =
Z2FT 1.25 Z2F 0.25 Z2m• –• =
Z2m
V
2
I
2
---------=