Schweitzer Engineering Laboratories SEL-311L, Instruction Manual, Page: 727 / 734

Date Code 20010625 Example Calculations for 87L Settings J-5
SEL-311L Instruction Manual
L
INE
C
HARGING
C
URRENT
C
ALCULATION
E
XAMPLES
Example 1: 500 kV OH Transmission Line: 100 Miles of Single 1113 MCM Conductor
N L
V • C • f • • 2 I
−
π =
[A per-phase per mile]
Equation J.7
( )
l
r / D log
0388 . 0
C
EQ 10
=
[µF]
Equation J.8
where:
D
EQ
1
= Equivalent spacing between three conductors: [D
EQ
= (D
AB
• D
BC
• D
CA
)
1/3
] (ft)
r = Conductor radius (ft) (1113 MCM ACSR, 0.6465 in = 0.0539 ft)
l = Conductor length (miles) (100 miles)
C = Shunt capacitance (Equation J.8: 0.01265
µ
F/mi • 100 mi = 1.265
µ
F)
f = system frequency (60 Hz)
V
L-N
= system phase-neutral voltage (500 /
√
3 kV)
1
Assuming a conductor spacing of 50 ft between the conductors of a horizontal line
configuration (D
EQ
= 63 ft).
@ V = 500 kV /
√
3, I
1CHARGING
= 137.6 A primary for a line length = 100 miles
@ V = 2886 V, I
2CHARGING
= 1.37 A primary for the same line length, with 1% unbalance
If the current transformer ratio for this 500 kV application is 400:1 (2000/5), the phase charging
currents are 0.334 A secondary, and the negative-sequence charging current (3I
2
) is 0.01A
secondary. Make setting 87L2P = 0.5 for excellent security and sensitivity.
Example 2: 15 kV Underground Cable: 5 Miles
l
G
k • n • 0169 . 0
C
=
[µF]
Equation J.9
where:
C = cable capacitance (see Equation J.9, 0.2345 • (5.28) • (5 mi) = 6.19
µ
F)
n = number of conductors (3)
k = cable dielectric constant (3.7)
V
L-N
= system phase-neutral voltage (15 /
√
3 kV)
G = cable geometric factor (shaped differs from circular, etc.
Our example assumes a G = 0.8)
l = cable length in 1,000s of feet (5 mi • 5.28 kft/mi)
@ V = 8.66 kV, I
1CHARGING
= 60.6 A primary for a 15-mile long cable
@ V = 86.6 V, I
2CHARGING
= 0.60 A primary for the same cable length