Schweitzer Engineering Laboratories SEL-311C - Page 502

13.48
SEL-311C Relay Instruction Manual Date Code 20060320
Testing and Troubleshooting
Test Procedures
zero-sequence current compensation factor for Zone 1 ground
distance elements and the k0M and k0A settings to define that
factor for the remaining zones.
The reactance the relay ground quadrilateral distance element
measures for a Zone 2 fault is defined by the following
equation:
Equation 13.66
For a fault on a radial system and when testing a ground
distance element through use of a single current source, I
A
= I
R
.
Equation 13.66 can be simplified:
Equation 13.67
When T = 0°, as it does in the example relay settings,
Equation 13.67 can be further simplified through use of the
following identity:
Equation 13.68
If we select a value for V
A
, it then becomes possible to use
Equation 13.68 to calculate the magnitude and angle of I
A
required to test the Zone 2 ground quadrilateral distance
element reactive reach.
For the example, select V
A
= 40.0 ∠0°. To simplify the
reactance calculation, select the angle of I
A
= –90° with respect
to V
A
. For the Zone 2 element, XG2 = 9.36 Ω, ∠Z1ANG =
83.97°, and k0 = 0.726∠–3.69°.
Equation 13.69
The result of computing the imaginary portion of V
A
• I
A
*
when I
A
lags V
A
by 90° is:
Equation 13.70
Substituting this into the equation above yields:
Equation 13.71
where:
k0 = k0M ∠ k0A°
T = Nonhomogeneous system compensation setting
X
AG
Im V
A
I
R
1T°∠• ()*()
Im 1 Z1ANG I
A
k0 I
R
• +()I
R
1T°∠• ()*∠()
---------------------------------------------------------------------------------------------------------------=
X
AG
Im V
A
I
A
1T°∠• ()*()
Im 1 Z1ANG 1 k0+()I
A
(I
A
1T°)∠• • *• • ∠()
--------------------------------------------------------------------------------------------------------------------------=
I
A
2
I
A
() I
A
()*• =
X
AG
Im V
A
I
A
()*()
Im 1 Z1ANG I
A
2
1k0+()• • ∠()
--------------------------------------------------------------------------------------=
9.36
Im 40.0 0° I
A
• 90°∠∠()
I
A
2
Im 1 83.97° 1 0.726 3.69°–∠+()• ∠()•
--------------------------------------------------------------------------------------------------------------=
Im V
A
0° I
A
90°∠• ∠()V
A
I
A
• =
9.36
40.0 I
A
•
I
A
2
1.71()•
---------------------------------=