Overhead Conductor Overhead Spacer Cable Underground Cable Three-Conductor Cable Service Cables
Power Systems I
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Transmission Lines
ACSR Aluminum Conductor with inner Steel Reinforced strands ACAR Aluminum Conductor with inner Al allow Reinforced strands ACSR/AW Aluminum Conductor with inner Alumoweld Steel Reiforced strands Aluminum - current carrying member Steel - structural support
Power Systems I
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Overhead Conductors
Where conductor close proximity is required Insulating jacket surrounds each conductor Plastic spacers keep conductors from coming in contact with one another
Power Systems I
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Overhead Cable
Power Systems I
Cables
Underground transmission and distribution cables Semiconducting material surrounds the conductor to grade the electric field Plastic jacket provides insulation and protection Neutral strands for an outer shell for protection and return currents
Power Systems I
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Cables
u
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u
ρ = conductor resistivity l = conductor length A = conductor crosssectional area
Rnew = Rold
Rac = 1.02 ⋅ Rdc
ρl Rdc = A
TAl = 228°C
T + told
increased resistance at conductor temperature rises wiring is rated for 65°C, 75°C, or 90°C T + t new ambient temperature is 20°C
Temperature effects
n
skin effect at 60 Hz:
ac resistance
u
n
dc resistance
u
Line resistance
Power Systems I
l
l
Transmission Line Parameters
Power Systems I
φ = ∫ B ⋅ da
Integral of the flux density that is normal to a defined area A
B = µH
Magnetic Flux
Integral of the scalar product of a closed path ie and the magnetic field equals the encircled current
B
A
H
φ
A=
B=
H=
F = ∫ H ⋅ d l = ie
Γ
Γ=
Ampere’s circuital law
Review of Magnetics and Inductance
I
I
I
Power Systems I
φ ∑ ∫ B ⋅ da ∑ ∫ µH ⋅ da ∑ = = =
λ L= I
Inductance
i =1
λ = ∑ φi
N
Flux Linkage
Review of Magnetics and Inductance
infinite straight wire is an approximation of a reasonably long wire
u
u
u
u
u
Image the wire to close at +/- infinity, establishing a kind of “one-turn coil� with the return path at infinity Straight infinitely long wire of radius r Uniform current density in the wire. Total current is Ix Flux lines form concentric circles (i.e. H is tangential) Angular symmetry - it suffices to consider Hx
Assumptions:
u
Conditions:
Power Systems I
l
l
Inductance of a Single Conductor
→ Hx =
µ0 I I → = x x B x 2 2 2π r 2π r
Power Systems I
r
µ0 I r 3 µ0 I x dx = λint = ∫ dλx = 4 ∫ 2π r 0 8π 0
µ0 → Lint = = 0.5 ×10 −7 8π
x2 µ0 I 3 µ0 I x dx xdx → dλx = 2 dφ x = dφ x = Bx dx = 2 4 2π r 2π r r
Ix I = π r 2 π x2
0
Case 1: Points inside of the conductor (x < r)
⋅ dl = I x
l
x
Ix ⇒ H= 2π x
General:
l
∫H
2πx
Inductance of a Single Conductor
D1
λext = ∫ dλx =
D2
µ0 I 2π
µ 0 I D2 1 d x = ln ∫D x D1 2π 1
D2
→ Lext = 2 × 10 −7 ln
µ0 I I x = I → Bx = µ 0 H x = 2π x µ I µ I dφ x = Bx dx = 0 dx → dλx = dφ x = 0 dx 2π x 2π x
Case 2: Points outside of the conductor (x > r)
Power Systems I
l
Inductance of a Single Conductor
D2 D1
r ′ = re −1 4 = DS
D r1
r1
Power Systems I
−7
D D D −7 −7 L = 2 ×10 ln −1 4 = 2 × 10 ln = 2 × 10 ln re r′ DS
−7
1 L1 = 2 × 10 ln −1 4 + ln D r1e r1 = r2 L1 = L2 = L
L1 = L1(int ) + L1( ext ) = 0.5 × 10 −7 + 2 × 10 −7 ln
D r1
conductors of radii r1 and r2, separated by a distance D
L1( ext ) = 2 × 10 −7 ln
l
Inductance of a Single-Phase Line
D
r2
1 ln L22 = 2 ×10 −7 r1′
Power Systems I
1 ln r2′ 1 ln D
λ1 = L11 I1 − L12 I1 I1 = − I 2 → λ2 = − L21 I 2 + L22 I 2
L12 = L21 = −2 ×10 −7 (ln D ) = 2 ×10 −7
L11 = 2 ×10
−7
λ1 = L11 I1 + L12 I 2 λ2 = L21 I1 + L22 I 2
From the 2 conductor case:
Flux Linkage - Self and Mutual Inductances
Power Systems I
λi = 2 ×10 −7
j≠i
n 1 1 I i ln + ∑ I j ′ r D = 1 j i ij
j =1
λi = Lii I i + ∑ Lij I j
n
I1 + I 2 + L + I i + L + I n = 0
General Case:
Total Inductance
j≠i
−7
1 1 1 λa = 2 × 10 I a ln + I b ln + I c ln r′ D D 1 1 λa = 2 × 10 −7 I a ln − I a ln r′ D D −7 λa = 2 × 10 I a ln r′ D L = 0.2 ln DS
I a + Ib + Ic = 0
Symmetrical spacing
Power Systems I
l
Ic
Inductance of Three-Phase Lines
D D
D
Ia
Ib
−7
1 D12 1 ln r′ 1 D32 ln
ln
1 D13 1 ln D23 1 ln r ′
1 1 1 + I c ln λa = 2 ×10 I a ln + I b ln r′ D12 D13 1 1 1 + I c ln λb = 2 ×10 −7 I a ln + I b ln ′ r D21 D23 1 1 1 + I c ln λc = 2 ×10 −7 I a ln + I b ln r′ D31 D32 = LI
Asymmetrical spacing
1 ln ′ r 1 L = 2 × 10 −7 ln D21 1 Power Systems I D31
l
Inductance of Three-Phase Lines
Ic
D31 D23
D12
Ia
Ib
horizontal or vertical configurations are most popular Symmetry is lost - unbalanced conditions
u
c a
a b
b c
a
b
a
c
Each phase occupies each position for the same fraction of the total length of the line
Power Systems I
u
b
c
Average inductance of each phase will be the same
restore balanced conditions by the method of transposition of lines
u
u
The practice of equilateral arrangement of phases is not convenient
position a 1 b 2 c 3
l
l
Transposition
c a
b
D2
Power Systems I
q =Cv
Capacitance
D1
v12 = vD1 − vD2 = − ∫ E ⋅ dl
Electric field
D=εE
Electric field
A
q e = ∫ D ⋅ da
Gauss’s law
R
Review of Electric Fields
D Gaussian Surface
h
q1 D ln 2πε 0 r
q D2 q ln dx = 2πε 0 x 2πε 0 D1 R
h
Power Systems I
q D ln v12 = πε 0 r
q1 D q2 r ln + ln v12 = v12(q1) + v21(q 2 ) = 2πε 0 r 2πε 0 D
q2 r v21(q 2 ) = ln 2πε 0 D
v12 (q1) =
D1
v12 = ∫
D2
Infinite Straight Wire
D
q C= v 2π ε C= D ln r
Infinite wire of radius r
c
C
C
a
n
)
per mile per phase
GMDφ = geometeric mean distance between conductors conductor radius rφ=
0.0389 GMDφ log10 r φ
b
C
C=
Equilateral spacing
Power Systems I
l
Three-Phase Capacitance
12 in
conductor R: 0.3263 Ω/mile GMR: 0.0244 ft Dia.: 0.720 in
44 in
Power Systems I
44 in
)
) = 149.9
Ω mi
) PL SKV
(0.720 in ) ⋅ 121 = 0.03 ft 0.0389 = 0.177 log10 (4.73 0.03)
1 2
X C = 1 (2π 60 ⋅ 0.177
C=
rφ = 12 dia =
4.73 10 0.0244
(60) log
60 = 0.326 + j 0.639 Ω / mi
Z a = (0.3263) + j 0.2794
= 56.8 in = 4.73 ft
GMDφ = 3 d12 d 23 d13 = 3 (45.6 ) (88) (45.6 )
Calculate the resistance, inductive reactance, and capacitive reactance per phase and rated current carrying capacity for the overhead line shown. Assume the line operates at 60 Hz
Example
D12 3 conductors
D12
2 conductors
D13
D12 4 conductors
D13
D14
Commonly used to reduce the electric field strength at the conductor surface Used on overhead lines above 230 kV Conductors are connected in parallel Typical bundled conductor configurations
Power Systems I
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Conductor Bundling
n
i=2
rφ′ = n rφ ⋅ ∏ D1i
Equivalent radius
i=2
GMRφ′ = n GMRφ ⋅ ∏ D1i
n
The use of bundled conductors effects the impedance of the line, the GMRφ , the GMDφ , and the equivalent radius GMDφ : the distance between the center of each bundle is used GMRφ :
Power Systems I
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Conductor Bundling
Power Systems I
10 ft
30 ft
20 in
20 in conductor R: 0.1204 â&#x201E;Ś/mile GMR: 0.0403 ft Dia.: 1.196 in
Calculate the resistance, inductive reactance, and capacitive reactance of the overhead line shown. Assume the line operates at 60 Hz
Example
1 2
Power Systems I
X C = 1 (2π 60 ⋅ 0.177
)
) = 116.85
0.0389 C= = 0.0227 log10 (39.15 0.7568)
rφ = 12 dia =
Ω mi
) PL SKV
(1.196 in ) ⋅ 121 = 0.0498 ft rφ′ = 4 (0.0498) (1.67 ) (1.414 ) (1.67 ) = 0.7568 ft
39.15 10 0.7178
(60) log
60 = 0.0301 + j 0.485 Ω / mi
Z a = (0.0301) + j 0.2794
GMRφ = 4 (0.0403) (1.67 )(1.414 ) (1.67 ) = 0.7178 ft
GMDφ = 3 D12 D23 D13 = 3 (31.6 ) (60 )(31.6 ) = 39.15 ft
R′ = 14 ⋅ 0.1204 = 0.0301 Ω / mi
Example
Voltages are expressed as phase-to-neutral Currents are expressed for one phase The three phase system is reduced to an equivalent single-phase
Line parameters: R, L, C, & G
u
u
depend on the length and the voltage level short, medium, and long length line models
Three types of models
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All lines are made up of distributed series inductance and resistance, and shunt capacitance and conductance
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u
u
Transmission lines are represented by an equivalent circuit with parameters on a per-phase basis
Power Systems I
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Transmission Line Modeling
circuit equations
matrix form
u
u
VS A B VR I = C D I R S
I S = C VR + D I R
VS = A VR + B I R
All transmission line models may be described as a twoport network The ABCD two-port network is the most common representation The network is described by the four constants: A, B, C, & D Network equations:
Power Systems I
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ABCD Two-Port Network
The line length is less than 50 miles (80 km), or The line voltage is not over 69 kV
VS
RL
XL VR
The shunt capacitance and conductance are ignored The line resistance and reactance are treated as lumped parameters
Circuit of the short model
u
u
Modeling of the transmission line parameters
u
u
The short transmission line model may be used when
Power Systems I
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Short Transmission Line Model
Gen.
IS
VS
= VR + I R Z
VS = VR + I R ( R + j ω s L)
IS = IR
Z=R+jωL
Circuit analysis of the short line model
Power Systems I
l
Short Transmission Line Model
VR
Load
IR
Matrix representation:
ABCD values:
u
u
Circuit Equations:
Power Systems I
l
D = A=1
C =0
A=1 B = Z line
VS 1 Z line VR I = 0 1 I R S
IS = IR
VS = VR + Z line I R
Two-Port Representation
R = 0.15 Ω/km L = 1.3263 mH/km
(
381 MVA load at 0.8 lagging pf at 220 kV
)
(
)
Z = 6 + j 20 Ω 220,000∠0° VR = = 127,000∠0° 3 S R ( 3φ ) = 381∠ cos −1 0.8 = 381∠36.9° = 304.8 + j 228.6 MVA
Z = (r + j ω L ) l = 0.15 Ω + j 2 π × 60 ×1.3263 ×10 −3 ⋅ 40
u
Find V, S, V.R., and η at the sending end of the line for
u
40 km, 220 kV transmission line has per phase
Power Systems I
l
l
Short Transmission Line Example
3 VR*
381× 106 ∠ − 36.9° = 1000∠ − 36.9° A = 3 × 127,000∠0°
Power Systems I
250 - 220 304.8 × 100% = 13.6% η = ×100% = 94.4% VR% = 220 322.8
= 322.8 + j 288.6 = 433∠41.8° MVA
S S (3φ ) = 3 ⋅ VS ⋅ I S* = 3 ⋅ (144,330∠4.93°)(1000 ∠ − 36.9°)
VS − LL = 3 ⋅ VS = 250 kV
= 144,330∠4.93°
VS = VR + Z I R = 127,000∠0° + (6 + j 20)(1000∠ − 36.9°)
IR =
S R* ( 3φ )
Short Transmission Line Example
The line length is greater than 50 miles (80 km) The line length is less than 150 miles (250 km)
VS
YC/2
RL
XL
YC/2
VR
Half of the shunt capacitance is considered to be lumped at each end of the line The line resistance and reactance are treated as lumped parameters
Circuit model:
u
u
Modeling of the transmission line parameters
u
u
The medium transmission line model may be used when
Power Systems I
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Medium Transmission Line Model
Gen.
IS
VS
½ YC
IS C
R
R
R
Z line YC 4
R
YC 2
YC 2
Z line YC 2
S
line R
VS = VR + Z line I R + Y2C VR
( ) )V + Z I = (1 + = (I + V ) + V )V + (1 + = Y (1 +
½ YC
Z=R+jωL
Z line YC 2
Circuit analysis of the short line model
Power Systems I
l
)I
Medium Transmission Line Model
R
VR
Load
IR
Matrix representation:
ABCD values:
u
u
Circuit Equations:
Power Systems I
l
C
R
R
R
Z line YC 4
(
A = 1+ C = YC 1 +
Z line YC 2 Z line YC 4
(
)
Z line YC 2
)I
R
Z line VR Z line YC 1 + 2 IR
S
line R
B = Z line D = 1 + Zline2 YC
)
YC 2
YC 2
R
VS 1 + Z line2 YC I = Z line YC Y 1 + S C 4
IS
Z line YC 2
VS = VR + Z line I R + Y2C VR
( ) )V + Z I = (1 + = (I + V ) + V )V + (1 + = Y (1 +
Two-Port Representation
R = 0.036 Ω/km L = 0.80 mH/km
(
270 MVA load at 0.8 lagging pf at 325 kV
)
(
)
325,000∠0° VR = = 187,600∠0° 3 S R (3φ ) = 270∠ cos −1 0.8 = 270∠36.9° = 216 + j162 MVA
Y = ( j ω C ) l = j 2 π × 60 × 0.0112 × 10 −6 ⋅130 = j 0.549 siemens
(
= 4.68 + j 39.2 Ω
Z = (r + j ω L ) l = 0.036 Ω + j 2 π × 60 × 0.8 ×10 −3 ⋅130
u
)
C = 0.0112 uF/km
Find V and S at the sending end of the line for
u
130 km, 345 kV transmission line has per phase
Power Systems I
l
l
Medium Transmission Line Example
3 VR*
270 ×106 ∠ − 36.9° = = 480∠ − 36.9° A 3 ×187,600∠0°
(
)
Power Systems I
= 421.5∠ − 25.58°
(480∠ − 36.9°)(0.989 + j 0.001284)
I S = C VR + D I R = (187,600∠0°) − 3.53 ×10 −7 + j 5.46 ×10 − 4 +
= 199,160∠4.02°
(480∠ − 36.9°)(4.68 + j39.2)
0.989 + j 0.001284 4.68 + j 39.2 ABCD = −7 −4 3 . 53 10 5 . 46 10 0 . 989 0 . 001284 − × + × + j j VS = A VR + B I R = (187,600∠0°)(0.989 + j 0.001284 ) +
IR =
S R* (3φ )
Medium Transmission Line Example
Power Systems I
= 325
345 0.989 + j 0.001284 - 325
×100% = 7.3%
= 218.9 + j124.2 MVA pf = 0.87 VR ( NL ) − VR ( FL ) VS ( FL ) / A − VR ( FL ) VR% = ×100% = × 100% VR ( FL ) VR ( FL )
S S (3φ ) = 3 ⋅ VS ⋅ I S* = 3 ⋅ (199,160∠4.02°)(421 ∠ − 25.58°)
VS − LL = 3 ⋅ VS = 345 kV
Medium Transmission Line Example
The line length is greater than 150 miles (250 km)
u
u
u
Accuracy obtained by using distributed parameters The series impedance per unit length is z The shunt admittance per unit length is y
Modeling of the transmission line parameters
u
The long transmission line model are used when
Power Systems I
l
l
Long Transmission Line Model
V(x + ∆x)
I(x + ∆x) y ∆x ∆x l
y ∆x
I(x) V(x) x
IR VR
Power Systems I
V ( x + ∆x) = V ( x) + z ∆x I ( x) I ( x + ∆x) = I ( x) + y ∆x V ( x + ∆x) V ( x + ∆x) − V ( x) I ( x + ∆x) − I ( x) = z I ( x) = y V ( x + ∆x) ∆x ∆x dV ( x) dI ( x) limit as ∆x → 0 = z I ( x) limit as ∆x → 0 = y V ( x) dx dx
VS
IS
z ∆x
Long Transmission Line Model
VS
I+∆I
V+∆V ∆x
V
I
x
VR
Load
IR
d 2V ( x) dI ( x) d 2 I ( x) dV ( x) =z =y 2 2 dx dx dx dx d 2V ( x) d 2 I ( x) = z ( y V ( x) ) = y ( z I ( x) ) 2 2 dx dx 2 γ = z y propagation constant Power Systems I
Gen.
IS
Long Transmission Line Model
(r + jωL )(g + jωC )
Power Systems I
@x=0⇒
(
VR + I R Z c A1 = 2
(
(
)
VR − I R Z c A2 = 2
)
)
1 dV ( x) γ I ( x) = = A1 eγ x − A2 e −γ x = yz A1 eγ x − A2 e −γ x z dx z 1 Zc = z y I ( x) = A1 eγ x − A2 e −γ x characteristic impedance Zc
γ = α + jβ = z y =
d 2V ( x) 2 V ( x) γ = 2 dx V = A1 eγ x + A2 e −γ x
Long Transmission Line Model
1 e Zc
)
VR +
VR + Z c
−x y z
yz
e
yz
x yz
ex
) y z)I
+e 2
yz
−x y z
− e−x 2
Power Systems I
c
R
R
V ( x) = cosh x y z VR + Z c sinh x y z I R
(
−e 2
+ e−x 2
x yz
yz
( 1 I ( x) = sinh (x y z )V + cosh (x Z
I ( x) =
V ( x) =
ex
VR + Z c I R x y z VR − Z c I R − x y z V ( x) = e e + 2 2 V Z +I V Z −I I ( x) = R c R e x y z − R c R e − x y z 2 2
IR
IR
Long Transmission Line Model
eθ + e −θ cosh θ = 2
eθ − e −θ sinh θ = 2
Hyperbolic Functions
Power Systems I
γ = zy Zc =
z y
cosh (γ l ) Z c sinh (γ l ) ABCD = 1 sinh (γ l ) cosh (γ l ) Zc
1 IS = sinh (γ l ) VR + cosh (γ l ) I R Zc
let x → l VS = cosh (γ l ) VR + Z c sinh (γ l ) I R
Two-Port Representation
Y’/2
′ ′
)
)
(
)
→
→ Z ′ = Z c sinh (γ l )
Y’/2
VR
Y′ 1 cosh (γ l ) − 1 1 γ l tanh = (cosh (γ l ) − 1) = = Z c sinh (γ l ) Z c 2 Z′ 2
Z′Y′ Z′Y′ ′ I S = Y 1 + 4 VR + 1 + 2 I R
(
VS = 1 + Z 2Y VR + Z ′I R
(
Find the values for Z’ and Y’
VS
Represent a long transmission line as a pi-model for circuit analysis Z’ The circuit:
Power Systems I
l
l
l
Pi-Model of a Long Transmission Line
z = 0.045 + j 0.4 Ω/km Y = j 4. 0 uS/km
γ = zy = (0.045 + j 0.4 )(4 × 10 −6 ) = 7.104 × 10 −5 + j 0.001267 Z ′ = Z c sinh (γ l ) = 10.88 + j 98.36 Y′ 1 γ l tanh = = j 0.001008 2 Zc 2
Zc =
0.045 + j 0.4 z = = 316.7 - j17.76 −6 4 × 10 y
Find ABCD for a pi model of the long transmission line
u
250 km, 500 kV transmission line has per phase
Power Systems I
l
l
Long Transmission Line Example
)
′ ′
)
Power Systems I
C = Y ′ 1 + Z 4Y = j 0.00100
(
B = Z ′ = 10.88 + j 98.36
(
Z ′ = 10.88 + j 98.36 Y′ = j 0.001008 2 ′ ′ A = D = 1 + Z 2Y = 0.9504 + j 0.0055
Long Transmission Line Example