ECE 404
Introduction to Power Systems
j :=
HW #9
−1
Chapter 4: 34, 35, & 39
Chapter 4: 2
pu := 1
ε 0 = 8.854 × 10
− 12 F
⋅
MVA := MW
m
MVAR := MW
4.34 Calculate the capacitance-to-neutral in F/m and the admittance-to-neutral in S/km for the
three-phase line in Problem 4.10. Neglect the effect of the earth plane.
4.10 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.
Deq := 4 ⋅ ft
r :=
0.5
2
Can :=
⋅ in
Deq = 1.219 m
r = 0.635 ⋅ cm
2 ⋅ π⋅ ε 0
− 11 F
Can = 1.058 × 10
⎛ Deq ⎞
⎟
ln⎜
⎝ r ⎠
Yan := j ⋅ 2 ⋅ π⋅ 60⋅ Hz⋅ Can
C:\JoeLaw\Classes\ECE404
\Homework\HW09\HW09
Solution.xmcd
Yan = 3.989j × 10
Page 1 of 6
⋅
m
−6 S
⋅
km
October 10, 2008
ECE 404
Introduction to Power Systems
HW #9
Chapter 4: 34, 35, & 39
Chapter 4: 2
4.35 Rework Problem 4.34 if the phase spacing is:
a.
b.
increased by 20% to 4.8 ft and
decreased by 20% to 3.2 ft.
Compare the results with those of Problem 4.34.
Deqa := 4.8⋅ ft
a.
Cana :=
Deqa = 1.463 m
2 ⋅ π⋅ ε 0
⎛ Deqa ⎞
⎟
ln⎜
⎝ r ⎠
− 11 F
Cana = 1.023 × 10
Cana − Can
Can
Yana := j ⋅ 2 ⋅ π⋅ 60⋅ Hz⋅ Cana
Yan
b.
Deqb := 3.2⋅ ft
Canb :=
2 ⋅ π⋅ ε 0
⎛ Deqb ⎞
⎟
ln⎜
⎝ r ⎠
⋅
km
Deqb = 0.975 m
Canb = 1.105 × 10
Canb − Can
− 11 F
⋅
m
= 4.432 ⋅ %
−6 S
Yanb = 4.166j × 10
Yanb − Yan
Yan
C:\JoeLaw\Classes\ECE404
\Homework\HW09\HW09
Solution.xmcd
−6 S
= −3.352 ⋅ %
Can
Yanb := j ⋅ 2 ⋅ π⋅ 60⋅ Hz⋅ Canb
m
= −3.352 ⋅ %
Yana = 3.855j × 10
Yana − Yan
⋅
Page 2 of 6
⋅
km
= 4.432 ⋅ %
October 10, 2008
ECE 404
Introduction to Power Systems
HW #9
Chapter 4: 34, 35, & 39
Chapter 4: 2
4.39 Calculate the capacitance-to-neutral in F/m and the admittance-to-neutral in S/km for the
three-phase line in Problem 4.18. Also calculate the line-charging current in kA/phase if the
line is 100 km in length and is operated at 230 kV. Neglect the effect of the earth plane.
4.18 A 230-kV, 60 Hz, three-phase completely transposed overhead line has one
ACSR 954-kcmil conductor per phase and flat horizontal phase spacing,
with 8 m between adjacent conductors.
D12 := 8 ⋅ m
Deq39 :=
r39 :=
3
D13 := 16⋅ m
D12⋅ D13⋅ D23
1.196
Can39 :=
2
D23 := 8 ⋅ m
⋅ in
Deq39 = 10.079 m
r39 = 1.519 ⋅ cm
2 ⋅ π⋅ ε 0
Can39 = 8.562 × 10
⎛ Deq39 ⎞
ln⎜
⎟
⎝ r39 ⎠
C:\JoeLaw\Classes\ECE404
\Homework\HW09\HW09
Solution.xmcd
⋅
m
−6 S
Yan39 := j ⋅ 2 ⋅ π⋅ 60⋅ Hz⋅ Can39
Ichg := Yan39⋅ 100 ⋅ km⋅
− 12 F
Yan39 = 3.228j × 10
⋅
km
230 ⋅ kV
Ichg = 0.0429j ⋅ kA
3
Page 3 of 6
October 10, 2008
ECE 404
Introduction to Power Systems
HW #9
Chapter 4: 34, 35, & 39
Chapter 4: 2
5.2 A 150-km, 230-kV, 60 Hz, three-phase line has a positive-sequence impedance
z=0.08 + j0.48 Ω/km and a positive-sequence shunt admittance
y = j 3.33 *10^-6 S/km.
At full-load, the line delivers 250 MW at 0.99 pf lagging and at 220 kV.
Using the nominal π circuit, calculate:
a.
b.
c.
the ABCD parameters,
the sending-end voltage and current, and
the percent voltage regulation.
length 2 := 150 ⋅ km
length 2 = 150 ⋅ km
z := ( 0.08 + j ⋅ 0.48) ⋅
Ω
km
−6 S
y := j ⋅ 3.33⋅ 10
a.
A2 := ⎛⎜ 1 +
⎝
⋅
km
Y⋅ Z ⎞
2
⎟
⎠
Z := z⋅ length 2
Z = ( 12 + 72j) Ω
Y := y ⋅ length 2
Y = 4.995j × 10
−4
⋅S
−3
A2 = 0.982 + 2.997j × 10
( )
A2 = 0.982
B2 := Z
arg A2 = 0.175 ⋅ deg
B2 = ( 12 + 72j) Ω
( )
B2 = 72.993 Ω
C2 := Y⋅ ⎛⎜ 1 +
⎝
Y⋅ Z ⎞
4
⎟
⎠
arg B2 = 80.538⋅ deg
(
C2 = −7.485 × 10
−7
−4
C2 = 4.95 × 10
D2 := A2
( )
arg C2 = 90.087⋅ deg
−3
D2 = 0.982 + 2.997j × 10
D2 = 0.982
C:\JoeLaw\Classes\ECE404
\Homework\HW09\HW09
Solution.xmcd
⋅S
)⋅S
−4
+ 4.95j × 10
Page 4 of 6
( )
arg D2 = 0.175 ⋅ deg
October 10, 2008
ECE 404
Introduction to Power Systems
b.
250 j ⋅ acos( 0.99)
S2 :=
⋅e
⋅ MVA
0.99
HW #9
Chapter 4: 34, 35, & 39
Chapter 4: 2
S2 = ( 250 + 35.623j ) ⋅ MVA
( )
S2 = 252.525 ⋅ MVA
VRFL :=
220 j ⋅ 0⋅ deg
⋅e
⋅ kV
3
⎯
S2
IR :=
3 ⋅ VRFL
arg S2 = 8.11⋅ deg
VRFL = 127.017 ⋅ kV
IR = ( 656.08 − 93.486j ) A
( )
IR = 662.707 A
Vs := A2 ⋅ VRFL + B2 ⋅ IR
Vs = ( 139.337 + 46.497j ) ⋅ kV
Vs = 146.89⋅ kV
Is := C2 ⋅ VRFL + D2 ⋅ IR
( )
arg Vs = 18.454⋅ deg
Is = ( 644.467 − 26.964j ) A
Is = 645.031 A
⎯
Ss := 3 ⋅ Vs⋅ Is
arg IR = −8.11⋅ deg
( )
arg Is = −2.396 ⋅ deg
Ss = ( 265.633 + 101.168j) ⋅ MVA
( )
Ss = 284.246 ⋅ MVA
c.
IRNL := 0 ⋅ A
arg Ss = 20.85 ⋅ deg
IRNL = 0
Vs = A2 ⋅ VRNL + B2 ⋅ IRNL
Vs
VRNL :=
A2
VRNL = ( 142.032 + 46.915j ) ⋅ kV
(
)
VRNL = 149.579 ⋅ kV arg VRNL = 18.279⋅ deg
VRFL = 127.017 ⋅ kV
VR :=
VRNL − VRFL
VRFL
C:\JoeLaw\Classes\ECE404
\Homework\HW09\HW09
Solution.xmcd
VR = 17.763⋅ %
Page 5 of 6
October 10, 2008
ECE 404
Introduction to Power Systems
HW #9
Chapter 4: 34, 35, & 39
Chapter 4: 2
For Fun
IsNL := C2 ⋅ VRNL + D2 ⋅ IRNL
IsNL = ( −23.329 + 70.272j ) A
IsNL = 74.043 A
⎯
SsNL := 3 ⋅ Vs⋅ IsNL
(
)
arg IsNL = 108.366 ⋅ deg
SsNL = ( 0.05 − 32.629j ) ⋅ MVA
(
)
SsNL = 32.629⋅ MVA arg SsNL = −89.912⋅ deg
C:\JoeLaw\Classes\ECE404
\Homework\HW09\HW09
Solution.xmcd
Page 6 of 6
October 10, 2008