Schweitzer Engineering Laboratories SEL-311C - Page 507

13.53
Date Code 20060320 Instruction Manual SEL-311C Relay
Testing and Troubleshooting
Test Procedures
Equation 13.82
For a fault on a radial system with no load and when testing a
ground distance element through use of a single current source,
I
A
= I
R
, I
A2
= I
A
/3, and I
A0
= I
A
/3. Equation 13.82 can be
simplified:
Equation 13.83
Select a value for V
A
. Then it is possible to use Equation 13.83
to calculate the magnitude and angle of I
A
required to test the
Zone 2 ground quadrilateral distance element resistive reach.
For the example, select V
A
= 40.0 ∠0°. To simplify the
resistance calculation, select the angle of I
A
= 0°. For the
Zone 2 element, RG2 = 5.00 Ω, ∠Z1ANG = 83.97°, and k01 =
0.726 ∠–3.69°.
Equation 13.84
Because the angle of I
A
equals 0°, simplify the equation above:
Equation 13.85
Canceling the imaginary terms, and rearranging the equation to
calculate |I
A
| yields:
Equation 13.86
The Zone 2 ground distance element is forward. Thus, it is
supervised by the forward directional element 32GF, as well as
the 50L2 and 50GZ2 phase and residual overcurrent elements.
where:
k0 = k01M ∠k01A° for Zone 1, 2, 3, and 4
I
A2
= negative-sequence current flowing in A-phase for the fault
I
A0
= zero-sequence current flowing in A-phase for the fault
R
AG
Im V
A
I
A
k0 I• +
R
()1 Z1ANG∠• ()*()
Im
3
2
---
⎝⎠
⎛⎞
I
A2
I
A0
+()I
A
k0 I
R
• +()1 Z1ANG∠• •
⎝⎠
⎛⎞
*
----------------------------------------------------------------------------------------------------------------------------------------=
R
AG
Im V
A
(I•
A
1k0+()1 Z1ANG)∠• *• ()
Im I
A
(I
A
1k0+()1Z1ANG)*∠• • • ()
--------------------------------------------------------------------------------------------------------=
5.00
Im 40.0 0°∠ (I
A
1 0.726 3.69°–∠+()1 83.97°)∠• *• ()
Im I
A
(I
A
1 0.726 3.69°–∠+()183.97°)*∠• • • ()
--------------------------------------------------------------------------------------------------------------------------------------=
Im 40.0 0°∠ (I
A
1.725 1.55°–∠()183.97°)∠• *• ()
Im I
A
(I
A
1.725 1.55°–∠()183.97°)*∠• • • ()
-----------------------------------------------------------------------------------------------------------------------------=
Im 40.0 0°∠ (I
A
1.725 82.42∠())*• ()
Im I
A
(I
A
1.725 82.42°∠())*• • ()
-------------------------------------------------------------------------------------------=
5.00
40.0 I
A
Im 1.725 82.42°–∠()• •
I
A
2
Im 1.725 82.42°–∠()•
----------------------------------------------------------------------------------=
I
A
40.0
5.0
----------=
8.0 amps=