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
