ECE 476 – Power System Analysis Fall 2017
Homework 3
Reading: Chapter 4 from textbook.
In-class quiz: Thursday, September 21, 2017
Problem 1. A 60-Hz single-phase, two-wire overhead line has solid cylindrical copper conductors with 1.5 cm diameter. The conductors are arranged in a horizontal configuration with 0.5 m spacing. Calculate in mH/km:
a) The inductance of each conductor due to internal flux linkages only.
The internal inductance of each conductor is given by (4.4.10):
1000 mH
1000 m
1
= 0.05 mH/km per conductor.
Lint = × 10−7 H/m
2
1H
1 km
b) The inductance of each conductor due to both internal and external flux linkages.
The inductance for each conductor due to both internal and external flux linkages is given by (4.4.21):
D
Lx = Ly = 2 × 10−7 ln
r0
! 1000 mH
1000 m
H
0.5
−7
.
= 2 × 10 ln
m
1H
1 km
e−1/4 0.015
2
= 0.8899 mH/km per conductor
c) The total inductance of the line.
The total inductance of the single phase circuit is
L = Lx + Ly = 1.780 mH/km per circuit.
Problem 2. Rework Problem 1 if the diameter of each conductor is:
a) Increased by 20% to 1.8 cm.
1
1000 mH
1000 m
Lint = × 10−7 H/m
= 0.05 mH/km per conductor.
2
1H
1 km
Lx = Ly = 2 × 10
−7
= 2 × 10
−7
ln
ln
D
r0
!
0.5
e−1/4
0.018
2
H
m
1000 mH
1H
= 0.8535 mH/km per conductor.
L = Lx + Ly = 1.707 mH/km per circuit.
1
1000 m
1 km
b) Decreased by 20% to 1.2 cm, without changing the phase spacing.
1
1000 mH
1000 m
−7
Lint = × 10 H/m
= 0.05 mH/km per conductor.
2
1H
1 km
Lx = Ly = 2 × 10−7 ln
= 2 × 10
−7
ln
D
r0
!
0.5
0.012
e−1/4
2
H
m
1000 mH
1H
1000 m
1 km
= 0.9346 mH/km per conductor.
L = Lx + Ly = 1.869 mH/km per circuit.
Compare the results with those of Problem 1.
• Lint is independent of conductor diameter.
• A 20% increase in conductor diameter results in a 4.1% decrease in inductance.
• A 20% decrease in conductor diameter results in a 5% increase in inductance.
Problem 3. A 60-Hz three-phase, three-wire overhead line has solid cylindrical conductors arranged in the form of
an equilateral triangle with 4 ft conductor spacing. Conductor diameter is 0.5 in. Calculate the positive-sequence
inductance in H/m and the positive-sequence inductive reactance in Ω/km.
The line inductance is given by
L = 2 × 10
−7
ln
= 2 × 10−7 ln
D
r0
!
4
e−1/4
0.5
2
1 ft
12 in
= 1.101 × 10−6 H/m per phase.
Then the inductive reactance is
XL = ωL = 2π60 1.101 × 10
−6
Ω
m
1000 m
km
= 0.4153 Ω/km.
Problem 4. Calculate the capacitance-to-neutral in F/m and the admittance-to-neutral in S/km for the three-phase
line in Problem 3 (stated above). Neglect the effect of the earth plane.
The capacitance-to-neutral per line length is
Can =
2π(8.854 × 10−12 )
2π
=
= 1.058 × 10−11 F/m line-to-neutral.
4
ln(D/r)
ln 0.25/12
Then, the admittance-to-neutral per line length is
S
1000 m
−11
Ban = ωCan = 2π60(1.058 × 10 )
= 3.989 × 10−6 S/km line-to-neutral.
m
1 km
Problem 5. Rework Problem 4 if the phase spacing is:
a) Increased by 20% to 4.8 ft.
The capacitance-to-neutral per line length is
Can =
2π
2π(8.854 × 10−12 )
=
= 1.023 × 10−11 F/m line-to-neutral.
4.8
ln(D/r)
ln 0.25/12
Then, the admittance-to-neutral per line length is
S
1000 m
−11
Ban = jωCan = j2π60(1.023 × 10 )
= 3.855 × 10−6 S/km line-to-neutral.
m
1 km
b) Decreased by 20% to 3.2 ft.
The capacitance-to-neutral per line length is
Can =
2π
2π(8.854 × 10−12 )
=
= 1.105 × 10−11 F/m line-to-neutral.
3.2
ln(D/r)
ln 0.25/12
Then, the admittance-to-neutral per line length is
1000 m
S
= 4.166 × 10−6 S/km line-to-neutral.
Ban = ωCan = j2π60(1.105 × 10−11 )
m
1 km
Compare the results with those of Problem 4.
• A 20% increase in phase spacing results in a 3.4% decrease in shunt capacitance and admittance.
• A 20% decrease in phase spacing results in a 4.4% increase in shunt capacitance and admittance.