ECE 404
Introduction to Power Systems
HW #7
Chap. 4: 8, 9, 10, 11, 12, 13, & 15
4.8 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,
b) the inductances of each conductor due to both internal and external flux linkages, and
c) the total inductance of the line.
Dcond := 1.5⋅ cm
Dcond = 1.500 ⋅ cm
Ds := 0.5⋅ m
Ds = 0.500 m
a)
Lint :=
1
2
−7 H
⋅ 10
⋅
Lint = 0.050 ⋅
m
mH
km
b)
−1
rp := e
4
⋅
Lx := 2 ⋅ 10
Dcond
rp = 0.584 ⋅ cm
2
⎛ Ds ⎞
⎟
m
⎝ rp ⎠
−7 H
⋅
⋅ ln⎜
Lx = 0.890 ⋅
mH
km
c)
LLoop := 2 ⋅ Lx
C:\JoeLaw\Classes\ECE404
\Homework\HW07\HW07
SolutionV14R2.xmcd
LLoop = 1.78⋅
Page 1 of 9
mH
km
October 1, 2008
ECE 404
Introduction to Power Systems
HW #7
Chap. 4: 8, 9, 10, 11, 12, 13, & 15
4.9 Rework Problem 4.8 with the diameters of each conductor:
a) increased by 20% to 1.8 cm,
b) decreased by 20% to 1.2 cm, without changing the phase spacing.
Compare the results with those of Problem 4.8.
a)
Dcondp20 := 1.2⋅ Dcond
Lint :=
1
2
Dcondp20 = 1.80⋅ cm
−7 H
⋅ 10
⋅
Lint = 0.050 ⋅
m
mH
Same
km
−1
rpp20 := e
4
⋅
Dcondp20
2
⎛ Ds ⎞
⎟
m
⎝ rpp20 ⎠
−7 H
Lxp20 := 2 ⋅ 10
Δp20 :=
⋅
⋅ ln⎜
Lxp20 − Lx
C:\JoeLaw\Classes\ECE404
\Homework\HW07\HW07
SolutionV14R2.xmcd
Lxp20 = 0.853 ⋅
Increased
mH
km
Δp20 = −4.097 ⋅ %
Lx
LLoopp20 := 2 ⋅ Lxp20
ΔLoopp20 :=
rpp20 = 0.701 ⋅ cm
LLoopp20 = 1.71⋅
LLoopp20 − LLoop
LLoop
Page 2 of 9
Decreased
mH
km
ΔLoopp20 = −4.097 ⋅ %
Decreased
October 1, 2008
ECE 404
Introduction to Power Systems
HW #7
Chap. 4: 8, 9, 10, 11, 12, 13, & 15
b)
Dcondm20 := 0.8⋅ Dcond
Lint :=
1
2
−7 H
⋅ 10
⋅
Dcondm20 = 1.20⋅ cm
Lint = 0.050 ⋅
m
mH
Same
km
−1
rpm20 := e
4
⋅
Dcondm20
2
⎛ Ds ⎞
⎟
m
⎝ rpm20 ⎠
−7 H
Lxm20 := 2 ⋅ 10
Δm20 :=
⋅
⋅ ln⎜
Lxm20 − Lx
Lx
LLoopm20 := 2 ⋅ Lxm20
ΔLoopm20 :=
C:\JoeLaw\Classes\ECE404
\Homework\HW07\HW07
SolutionV14R2.xmcd
LLoopm20 − LLoop
LLoop
Page 3 of 9
rpm20 = 0.467 ⋅ cm
Lxm20 = 0.935 ⋅
decreased
mH
km
Δm20 = 5.015 ⋅ %
LLoopm20 = 1.87⋅
Increased
mH
km
ΔLoopm20 = 5.015 ⋅ %
Increased
October 1, 2008
ECE 404
Introduction to Power Systems
HW #7
Chap. 4: 8, 9, 10, 11, 12, 13, & 15
4.10 A 60 Hz three-phase, three-wire overhead line has solid cylindrical conductors arranged in
the form of an equalateral triangle with 4 ft conductor spacing. Conductor diameter is 0.5 in.
Calculate:
a) the positive-sequence inductance in H/m and
b) the positive-sequence reactance in ohms/km
Dcond10 := 0.5⋅ in
Dcond10 = 1.270 ⋅ cm
Ds10 := 4 ⋅ ft
Ds10 = 1.219 m
a)
−1
rp10 := e
4
⋅
Dcond10
⎛ Ds10 ⎞
⎟
m
⎝ rp10 ⎠
−7 H
Lps := 2 ⋅ 10
rp10 = 0.495 ⋅ cm
2
⋅
−6 H
⋅ ln⎜
Lps = 1.10 × 10
Xps := 2 ⋅ π⋅ 60⋅ Hz⋅ Lps
Ω
Xps = 0.415 ⋅
km
⋅
m
b)
C:\JoeLaw\Classes\ECE404
\Homework\HW07\HW07
SolutionV14R2.xmcd
Page 4 of 9
October 1, 2008
ECE 404
Introduction to Power Systems
HW #7
Chap. 4: 8, 9, 10, 11, 12, 13, & 15
4.11 Rework Problem 4.10 with the spacing:
a) increased by 20% to 4.8 ft and
b) decreased by 20% to 3.2 ft.
Compare the results with those of Problem 4.10.
a)
Dsp20 := 1.2⋅ 4 ⋅ ft
Lpsp20 := 2 ⋅ 10
ΔLp20 :=
Dsp20 = 1.463 m
⎛ Dsp20 ⎞
⎟
m
⎝ rp10 ⎠
−7 H
⋅
⋅ ln⎜
Lpsp20 − Lps
Lps
Xpsp20 := 2 ⋅ π⋅ 60⋅ Hz⋅ Lpsp20
ΔXp20 :=
Xpsp20 − Xps
Xps
Lpsp20 = 1.14 × 10
−6 H
⋅
m
ΔLp20 = 3.310 ⋅ %
Increased
Ω
Xpsp20 = 0.429 ⋅
km
ΔXp20 = 3.310 ⋅ %
Increased
b)
Dsm20 := 0.8⋅ 4 ⋅ ft
Lpsm20 := 2 ⋅ 10
ΔLm20 :=
Dsm20 = 0.975 m
⎛ Dsm20 ⎞
⎟
m
⎝ rp10 ⎠
−7 H
⋅
⋅ ln⎜
Lpsm20 − Lps
Lps
Xpsm20 := 2 ⋅ π⋅ 60⋅ Hz⋅ Lpsm20
ΔXm20 :=
C:\JoeLaw\Classes\ECE404
\Homework\HW07\HW07
SolutionV14R2.xmcd
Xpsm20 − Xps
Xps
Page 5 of 9
Lpsm20 = 1.06 × 10
−6 H
ΔLm20 = −4.052 ⋅ %
⋅
m
Decreased
Ω
Xpsm20 = 0.398 ⋅
km
ΔXm20 = −4.052 ⋅ %
Decreased
October 1, 2008
ECE 404
Introduction to Power Systems
HW #7
Chap. 4: 8, 9, 10, 11, 12, 13, & 15
4.12 Calculate the inductive reactance per mile of a single-phase overhead transmission line
operating at 60 Hz, given the conductors to be Partridge and the spacing between centers
to be 20 ft.
From Table A.4
L12 := 4 ⋅ 10
Ds12 = 6.10 m
rp12 := 0.0217⋅ ft
rp12 = 0.6614⋅ cm
⎛ Ds12 ⎞
⎟
m
⎝ rp12 ⎠
−7 H
⋅
Ds12 := 20⋅ ft
⋅ ln⎜
L12 = 2.730 ⋅
mH
km
Ω
X12 = 1.657 ⋅
mi
X12 := 2 ⋅ π⋅ 60⋅ Hz⋅ L12
Note that Table A.4 gives a conductor diameter of 0.642 in.
⎛ 1⎞
⎟
4 0.642 ⋅ in
e ⎝ ⎠⋅
= 0.635 ⋅ cm
−⎜
2
The corresponding radius does not yield the GMR given in the table.
It is a stranded conductor!
C:\JoeLaw\Classes\ECE404
\Homework\HW07\HW07
SolutionV14R2.xmcd
Page 6 of 9
October 1, 2008
ECE 404
Introduction to Power Systems
HW #7
Chap. 4: 8, 9, 10, 11, 12, 13, & 15
4.13 A single-phase overhead transmission line consists of two solid aluminum conductors
having a radius of 2.5 cm, with a spacing of 3.6 m between centers.
a) Determine the total line inductance in mH/m.
b) Given the operating frequency of 60 Hz, find the total inductive reactance of the line in
ohms/km and in ohms/mi.
c) If the spacing is doubled to 7.2 m, how does the reactance change?
rcond := 2.5⋅ cm
rcond = 0.025 ⋅ m
−1
rp13 := e
4
⋅ rcond
rp13 = 1.95⋅ cm
a)
Ds13 := 3.6⋅ m
Ds13 = 3.600 m
⎛ Ds13 ⎞
⎟
m
⎝ rp13 ⎠
−7 H
Lp13 := 4 ⋅ 10
⋅
− 3 mH
⋅ ln⎜
Lp13 = 2.09 × 10
⋅
m
b)
Ω
Xp13 = 0.787 ⋅
km
Xp13 := 2 ⋅ π⋅ 60⋅ Hz⋅ Lp13
Ω
Xp13 = 1.267 ⋅
mi
c)
Ds13c := 7.2⋅ m
Lp13c := 4 ⋅ 10
Ds13c = 7.200 m
⎛ Ds13c ⎞
⎟
m
⎝ rp13 ⎠
−7 H
⋅
⋅ ln⎜
Xp13c := 2 ⋅ π⋅ 60⋅ Hz⋅ Lp13c
ΔXp13 :=
C:\JoeLaw\Classes\ECE404
\Homework\HW07\HW07
SolutionV14R2.xmcd
Xp13c − Xp13
Xp13
Lp13c = 2.37 × 10
− 3 mH
⋅
m
Ω
Xp13c = 0.892 ⋅
km
ΔXp13 = 13.279⋅ %
Page 7 of 9
Increases
October 1, 2008
ECE 404
Introduction to Power Systems
HW #7
Chap. 4: 8, 9, 10, 11, 12, 13, & 15
4.15 Calculate the GMR of a stranded conductor consisting of six outer strands surrounding and
touching one central strand, all strands having the same radius r.
2R
R := 1
D
4R
R = 1.000
2R
D
( 4⋅ R) − ( 2⋅ R)
2
D = 3.464 ⋅ R
R
2R
2
D :=
2R
M := 7
−1
rp15 := e
4
⋅R
rp15 = 0.779 ⋅ R
For the center strand.
d
1, 1
d
1, 3
d
1, 6
:= rp15
:= d
:= d
d
1, 2
1, 2
d
1, 4
1, 2
d
1, 7
:= 2 ⋅ R
:= d
:= d
d
d
1, 2
1, 5
1, 2
:= d
= 2.000 ⋅ R
1, 2
1, 2
7
∏ d1 , k
P :=
1
P = 49.843⋅ R
1
k= 1
For all of the six outer stands
d
2, 1
d
2, 4
:= 2 ⋅ R
:= D
d
d
2, 2
2, 5
:= rp15
:= 4 ⋅ R
d
2, 3
d
2, 6
:= 2 ⋅ R
:= D
d
2, 7
:= 2 ⋅ R
7
P :=
2
∏ d2 , k
P = 299.060 ⋅ R
2
k= 1
n := 3 , 4 .. 7
C:\JoeLaw\Classes\ECE404
\Homework\HW07\HW07
SolutionV14R2.xmcd
P := P
n
2
Page 8 of 9
October 1, 2008
ECE 404
Introduction to Power Systems
M
HW #7
Chap. 4: 8, 9, 10, 11, 12, 13, & 15
2
M
∏ Pm
GMR :=
GMR = 2.177 ⋅ R
m= 1
OR
6
3
rp15 ⋅ ( 2 ⋅ R) = 49.843
2
rp15 ⋅ ( 2 ⋅ R) ⋅ ( 4 ⋅ R) ⋅ ( D) = 299.060
49
⎡r ⋅ ( 2⋅ R) 6⎤ ⋅ ⎡r ⋅ ( 2⋅ R) 3⋅ ( 4⋅ R) ⋅ ( D) 2⎤
⎣ p15
⎦ ⎣ p15
⎦
GMR2 :=
6
GMR2 = 2.177
For fun and learning
AactualCond := 7 ⋅ π⋅ R
2
AactualCond = 21.991⋅ R
2
Look at a Solid Conductor with the same GMR
−1
GMR = e
4
⋅ RsolidCond
RsolidCond :=
GMR
RsolidCond = 2.795 ⋅ R
⎛ − 1⎞
⎜ 4 ⎟
⎝e ⎠
AsolidCond := π⋅ RsolidCond
2
AsolidCond = 24.541⋅ R
2
A solid conductor with the same GMR would have to have a larger
cross-sectional area; i.e., greater cost and heavier.
AsolidCond − AactualCond
AactualCond
C:\JoeLaw\Classes\ECE404
\Homework\HW07\HW07
SolutionV14R2.xmcd
= 11.6⋅ %
Page 9 of 9
October 1, 2008