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6.88
SEL-421 Relay
Instruction Manual
Date Code 20171021
Protection Applications Examples
EHV Parallel 230 kV Underground Cables Example
Convert the power system impedances from primary to secondary so you can
later calculate protection settings.
lists the corresponding secondary
quantities. Convert the impedances to secondary ohms as follows:
Equation 6.52
Equation 6.53
The maximum load current of 777 A primary occurs when the parallel cable is
out of service.
Cable Admittances:
Y
1L1
= Y
1L2
Y
0L1
= Y
0L2
j6.71 • 10
–6
S primary (susceptance)
j6.71 • 10
–6
S primary (susceptance)
Source S Impedances:
Z
1S
= Z
0S
50
87° primary
Source R Impedances:
Z
1R
= Z
0R
35
87° primary
PTR (potential transformer ratio)
230 kV:115 V = 2000
CTR (current transformer ratio)
1000:5 = 200
Phase Rotation
ABC
Table 6.23
Secondary Impedances
Parameter
Value
Cable Impedances:
Z
1L1
= Z
1L2
Z
0L1 (sheath return only)
= Z
0L2 (sheath return only)
Z
0L1 (ground return only)
= Z
0L2 (ground return only)
Z
0L1 (sheath and ground return)
= Z
0L2 (sheath and ground return)
0.48
42.5° secondary
0.95
17.4° secondary
9.14
84.9° secondary
0.96
21.7° secondary
Cable Admittance:
Y
1L1
= Y
1L2
Y
0L1
= Y
0L2
6.71 • 10
–5
S
90° secondary
6.71 • 10
–5
S
90° secondary
Source S Impedances:
Z
1S
= Z
0S
5.0
87° secondary
Source R Impedances:
Z
1R
= Z
0R
3.5
87° secondary
Table 6.22
System Data—230 kV Parallel Underground Cables (Sheet 2 of 2)
Parameter
Value
k
CTR
PTR
------------
200
2000
------------
0.1
=
=
=
Z
1L1 secondary
k Z
1L1 primary
•
=
0.10 4.78
42.5
•
=
0.48
42.5
=
Summary of Contents for SEL-421-4
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