ELE-B7
TRANSMISSION LINES
Development of Line Models
†
Goals of this section are
1)
Develop a simple model for transmission
lines
2)
Gain an intuitive feel for how the geometry of
the transmission line affects the model
parameters
Primary Methods for Power Transfer
The most common methods for transfer of
electric power are
1)
Overhead AC
2)
Underground AC
3)
Overhead DC
4)
Underground DC
Transmission lines and cables
†
Extra-high-voltage lines
†
†
115 kV, 230 kV
Voltage: 46 kV, 69 kV
Distribution lines
„
†
Voltage:
Sub-transmission lines
„
†
345 kV, 500 kV, 765 kV
High-voltage lines
†
†
Voltage:
Voltage: 2.4 kV to 46 kV, with 15 kV being the most
commonly used
High-voltage DC lines
„
Voltage: ±120 kV to ±600 kV
Transmission lines and cables
†
†
†
†
†
Three-phase conductors,
which carry the electric
current;
Insulators, which support
and electrically isolate the
conductors;
Tower, which holds the
insulators and conductors;
Foundation and grounding;
and
Optional shield conductors,
which protect against
lightning
Transmission lines and cables
Shield conductor
Shield
conductor
Insulator
Composite
insulator
Phase
conductor
Steel tower
Tower
69kV
Line
Crossarm
Two
conductor
bundle
Composite
Insulator
Transmission lines and cables
Shield
conductor
Double circuit
69 kV line
Distribution line
12.47kV
Wooden tower
Distribution Line
Transmission lines and cables
Distribution line
Insulator
Surge arrester
Fuse cutout
Transformers
240V/120V
insulated line
Transformer
Transmission lines and cables
Span
Tension Tower
Supporting
Tower
Insulator
Sag
Definition of Parameters
Tension Tower
Transmission lines and cables
†
†
†
Aluminum Conductor
Steel Reinforced
(ACSR);
All Aluminum
Conductor (AAC);
and
All Aluminum Alloy
Conductor (AAAC).
†
ACSR Coductor
Aluminum outer strands
2 layers, 30 conductors
Steel core strands,
7 conductors
Transmission lines and cables
Insulators
Locking Key
Insulator's Head
Expansion Layer
Iron Cap
Ball Socket
Compression
Loading
Imbedded Sand
Cement
Insulating Glass
or Porcelain
Skirt
Petticoats
Corrosion Sleeve
for DC Insulators
Steel Pin
Ball
Transmission lines and cables
Insulator Chain
Line Voltage
Number of Insulators per
String
69 kV
4–6
115 kV
7–9
138 kV
8–10
230 kV
12
345 kV
18
500 kV
24
765 kV
30–35
Transmission lines and cables
Composite insulator.
†
(1) Sheds of alternating diameters prevent bridging by ice, snow and
cascading rain.
†
(2) Fiberglass reinforced resin rod.
†
(3) Injection molded rubber (EPDM or Silicone) weather sheds and rod
covering.
†
(4) Forged steel end fitting, galvanized and joined to rod by swaging process.
1
2
3
4
Transmission lines and cables
†
†
†
†
•Figure 4.15 Line post-composite
insulator with yoke holding two
conductors.
(1) is the clevis
ball,
(2) is the socket
for the clevis,
(3) is the yoke
plate, and
(4) is the
suspension clamp.
(Source: Sediver)
Transmission lines and cables
Transmission lines and cables
PVC-sheet
Copper screen
Filler
Insulation
Insulation shield
Conductor shield
Conductor
Inductance of a Single Wire
The inductance of a magnetic circuit has a constant permeability
can be obtained by determining the following:
a)
b)
c)
d)
Magnetic field intensity H, from Ampere’s law.
Magnetic flux density B (B = μH)
Flux linkage λ
Inductance from flux linkage per ampere
(L= λ/I)
Flux linkages within the wire :
∫ H ⋅dl = I enclosed
Ix
2πx
Assume uniform current distribution :
H x (2πx) = I x ⇒ H x =
2
xI
⎛x⎞
Ix = ⎜ ⎟ I ⇒ Hx =
⎝r⎠
2πr 2
For non - magnetic conductor :
μ xI
Bx = μ0 H x = 0
2πr 2
The differential flux dφ per unit length is :
dφ = Bx dx
Since only the fraction (x/r) 2 of the total current is linked by the flux :
2
μ0 x 3 I
⎛x⎞
dλ = ⎜ ⎟ dφ =
dx
4
r
⎝ ⎠
2πr
r
λint = ∫ dλ =
0
λ
1
1
× 10 − 7 I ⇒ Lint = int = × 10 − 7 H / m
2
I
2
Flux linkages outside of the wire :
∫ H ⋅dl = I enclosed
H x (2πx) = I ⇒ H x =
I
2πx
Bx = μ0 H x = 4 × π × 10 − 7
I
= 2 × 10 − 7
2πx
The differential flux dφ per unit length is :
I
x
dφ = Bx dx
Since the entire current is linked by the flux outside the conductor :
dλ = dφ = 2 × 10 − 7
D2
r
I
dx
x
λ
⎛ D1 ⎞
⎛ D1 ⎞
⎟ ⇒ L12 = 12 = 2 × 10 − 7 ln⎜
⎟H / m
I
⎝ D2 ⎠
⎝ D2 ⎠
λ12 = ∫ dλ = 2 × 10 − 7 I × ln⎜
D1
D2
D1
x
Total flux linkage:
Total flux linkage λp linking the conductor out to external point P at distance D
is the sum of the interanl and external flux linkages. Since D1 = r and D 2 = D, then
λp =
1
D⎞
D
⎛
⎛D⎞
× 10-7 I + 2 × 10-7 I × ln⎜ ⎟ = 2 × 10-7 I × ⎜ ln e1 / 4 + ln ⎟ = 2 × 10-7 I × ln
r ⎠
2
⎝
⎝ r ⎠
r'
where : r ' = e −1 / 4 r
Lp =
λp
I
= 2 × 10
−7
⎛D⎞
ln⎜⎜ ⎟⎟ H / m
⎝ r' ⎠
Total flux linkage, cont.:
Finally, consider the array of M solid conductors. Assume each conductor m carries
current I m and the sum of the conductor currents is zero, then :
the flux linkage λkPk which links conductor k out to point P due to current I k
D
λkpk = 2 × 10-7 I k × ln Pk
r 'k
and λkPm which links conductor k out to point P due to current Im is :
D
λkpm = 2 × 10-7 I m × ln Pm
Dkm
After some mathematical manipulation :
λk = 2 × 10
-7 M
∑ I m × ln
m =1
1
Dkm
where : λk gives the total flux linking
conductor k in an array of M conductors
Inductance of single-phase, two-wire
Since the sum of the two currents is zero the previous relation is valid and hence :
λ x = 2 × 10
⎛ 1 ⎞⎞
⎛ 1 ⎞
⎟⎟
⎜
⎟
+ I y × ln ⎜
I × ln ⎜
⎟
⎜ x
⎜ D xy ⎟ ⎟
⎝ D xx ⎠
⎠⎠
⎝
⎝
⎛
-7 ⎜
⎛
⎞
⎞
⎟ − I × ln ⎛⎜ 1 ⎞⎟ ⎟
⎟
⎜
⎝ D ⎠ ⎟⎠
⎠
⎝
⎛ D ⎞
⎟
λ x = 2 × 10 - 7 I × ln ⎜
⎜ r' ⎟
⎝ x⎠
⎛ 1
⎜ r'
⎝ x
λ x = 2 × 10 - 7 ⎜ I × ln ⎜
where : r ' x = e −1 / 4 rx similarly
λ y = − 2 × 10
Lx =
-7
⎛ D
I × ln ⎜
⎜ r'
⎝ y
⎞
⎟
⎟
⎠
⎛ D
= 2 × 10 − 7 ln ⎜
⎜ r'
Ix
⎝ x
λx
⎛
⎞
λ
⎟ H / m per conductor, and L y = y = 2 × 10 − 7 ln ⎜ D
⎟
⎜ r'
Iy
⎠
⎝ y
⎛ D
D
+
L = L x + L y = 2 × 10 − 7 ln ⎜
⎜ r'
'
⎝ x ry
2
⎞
⎛
⎟ = 2 × 10 − 7 ln ⎜ D
⎟
⎜ r' r'
⎠
⎝ x y
⎛D⎞
If r ' x = r ' y = r ' ⇒ L = 4 × 10 − 7 ln ⎜⎜ ⎟⎟
⎝ r' ⎠
⎛
⎞
⎟ = 4 × 10 − 7 ln ⎜
⎜
⎟
⎜
⎠
⎝
⎞
⎟
⎟
⎠
⎞
⎟
⎟
' ' ⎟
r xr y ⎠
D
Inductance of three-phase, three-wire
with equal phase spacing
Since the sum of the three currents is zero the previous relation is valid and hence :
⎛
⎛1⎞
⎛ 1 ⎞⎞
⎛1⎞
⎟⎟ + I b × ln⎜ ⎟ + Ic × ln⎜ ⎟ ⎟
⎜
'
⎝ D ⎠ ⎟⎠
⎝D⎠
⎝r ⎠
⎝
⎛
⎛1⎞
⎛ 1 ⎞⎞
= 2 × 10-7 ⎜ Ia × ln⎜⎜ ⎟⎟ + (I b + I c ) × ln⎜ ⎟ ⎟
⎜
⎝ D ⎠ ⎟⎠
⎝ r' ⎠
⎝
⎛
⎛1⎞
⎛ 1 ⎞⎞
= 2 × 10-7 ⎜ Ia × ln⎜⎜ ⎟⎟ − Ia × ln⎜ ⎟ ⎟
⎜
'
⎝ D ⎠ ⎟⎠
⎝r ⎠
⎝
λa = 2 × 10-7 ⎜ Ia × ln⎜⎜
λa
λa
⎛D⎞
⎟
'⎟
⎝r ⎠
⎛D⎞
La = 2 × 10-7 ln⎜⎜ ⎟⎟
⎝ r' ⎠
λa = 2 × 10-7 I a × ln⎜⎜
Inductance of composite conductors
The total flux φk linking subconductor k of conductor X is :
⎡I N
1
1 ⎤
I M
−
∑ ln
∑ ln
⎥
⎣ N m =1 Dkm M m =1' Dkm ⎦
Since only the fraction (1/N) of the total current I is linked by this flux,
φk = 2 × 10 − 7 ⎢
the flux linkage of λk subconductor k is :
⎡ 1
φ
λk = k = 2 × 10 − 7 I × ⎢
N
1
1 M
1 ⎤
ln
ln
−
∑
∑
⎥
2
N
⎣ N m =1 Dkm NM m =1' Dkm ⎦
The total flux linkage of conductor x is :
N
N ⎡ 1
N
1
1
M
1 ⎤
−
λx = ∑ λk =2 × 10 − 7 I × ∑ ⎢
∑ ln
∑ ln
⎥
2
k =1
K =1 ⎣ N m =1 Dkm NM m =1' Dkm ⎦
1 / NM
⎛ M
⎞
⎜ ∏ Dkm ⎟
⎟
N ⎜
⎠
λx = 2 × 10 − 7 I × ln ∏ ⎝ m =1'
1/ N 2
k =1 ⎛ N
⎞
⎜ ∏ Dkm ⎟
⎜
⎟
⎝ m =1
⎠
Dxy
λ
Lx = x = 2 × 10 − 7 ln
I
Dxx
N
M
N
N
where : D xy = MN ∏ ∏ Dkm = GMD and D xx = N 2 ∏ ∏ Dkm = GMR
k =1 m =1'
k =1 m =1
Inductance of unequal phase spacing
†The
problem with the line analysis we’ve done
so far is we have assumed a symmetrical tower
configuration. Such a tower figuration is
seldom practical.
Therefore in
general Dab ≠
Dac ≠ Dbc
Typical Transmission Tower Configuration
Unless something
was done this would
result in unbalanced
phases
Inductance of unequal phase
spacing
†
To keep system balanced, over the length of a transmission
line the conductors are rotated so each phase occupies each
position on tower for an equal distance. This is known as
transposition.
D12
L = 2 × 10
−7
ln
Deq
DS
D eq = 3 D12 D23 D31
D31
D23
where :
Conductor Bundling
To increase the capacity of high voltage transmission
lines it is very common to use a number of conductors per
phase. This is known as conductor bundling. Typical
values are two conductors for 345 kV lines, three for 500
kV and four for 765 kV.
Inductance of bundled conductors
L = 2 × 10
−7
ln
Deq
DSL
where :
D eq = 3 D12 D23 D31 and DSL is calculated as follows :
d
d
d
DSL = 4 (DS × d )2 =
d
(DS × d )
DSL =
d
16
d
(
DSL = 9 (DS × d × d )3 = 3 DS × d 2
(DS × d × d × d 2 )
4
(
= 1.0914 DS × d 3
)
)
Electric field and voltage: Solid
cylindrical conductor
The capacitance between conductors in a medium with
constant permittivity can be obtained by finding the
following:
a) Electric field from Gauss’s law.
b) Voltage between conductors.
c) Capacitance from charge per unit volt (C = q/V)
Gauss law: the total electric flux leaving a closed surface
equals the total charge within the volume enclosed by the
surface.
Electric field and voltage: Solid
cylindrical conductor
∫∫ Eds =
Qenclosed
ε
E x (2πx)( L) =
qL
ε
⇒ Ex =
D2
D2 q
D1
D1 2πεx
V12 = ∫ E x dx = ∫
q
2πεx
dx =
q
2πε
ln
D2
D1
Electric field and voltage: Solid
cylindrical conductor
Assume each conductor m has
a charge qm C/m, the voltage
Vkim between conductors k and i
due to the charge qm acting
alone is:
qm
D
ln im
2πε
D km
Using superposit iion, the voltage Vki due to all charges is given by :
V kim =
V ki =
1
M
∑ q m ln
2πε m =1
Dim
D km
Capacitance for single phase two wire
line
Assume conductor x has a uniform charge q C/m
and conductor y has –q. Using the previous last
equation with k = x, i = y and m = x, y
D yx Dxy
D yy ⎤
D yx
1 ⎡
q
⎢q ln
⎥=
− q ln
ln
2πε ⎢⎣
Dxy ⎥⎦ 2πε Dxx D yy
Dxx
Using D xy = D yx = D, D xx = rx and D yy = ry , then
Vxy =
Cxy =
q
=
Vxy
πε
⎛
⎜ D
ln⎜
⎜ rx ry
⎝
And if rx = ry = r, then
Cxy =
πε
⎛D⎞
ln⎜ ⎟
⎝ r ⎠
⎞
⎟
⎟
⎟
⎠
X
Cxy
y
Capacitance for three phase with equal
phase spacing
Dba
D
D ⎤
1 ⎡
+ qb ln bb + qc ln bc ⎥
⎢qa ln
Daa
Dab
Dac ⎦
2πε ⎣
Using D ab = D ca = D cb = D, D aa = D yy = r , then
Vab =
D
r
D⎤
1 ⎡
1
q
q
q
+
+
=
ln
ln
ln
a
b
c
r
D
D ⎥⎦ 2πε
2πε ⎢⎣
similarly
Vab =
D
r⎤
1 ⎡
qa ln + qc ln ⎥
⎢
r
D⎦
2πε ⎣
Vab + Vac = 3Van
Vac =
D
r⎤
⎞⎡
⎟ ⎢2qa ln + (qb + qc ) ln ⎥
r
D⎦
⎠⎣
and with q b + q c = −q a
Van =
1⎛ 1
⎜
3 ⎝ 2πε
D⎤
1 ⎡
qa ln ⎥
⎢
r⎦
2πε ⎣
2πε
Can =
ln( D / r )
Van =
D
r⎤
⎡
q
q
+
ln
ln
a
b
⎢
r
D ⎥⎦
⎣
Capacitance for stranded, unequal
phase spacing and bundled conductors
C=
2πε
ln( Deq / Dsc )
Deq = 3 Dab Dbc Dac
Dsc = rd
for two - conductor bundle
3
Dsc = rd 2
4
for three - conductor bundle
Dsc = 1.091 rd 3
for four - conductor bundle
Line Resistance
Line resistance per unit length is given by
R =
ρ
where ρ is the resistivity
A
Resistivity of Copper = 1.68 × 10-8 Ω-m
Resistivity of Aluminum = 2.65 × 10-8 Ω-m
Example: What is the resistance in Ω / mile of a
1" diameter solid aluminum wire (at dc)?
2.65 × 10-8 Ω-m
m
Ω
1609
R =
= 0.084
2
mile
mile
π × 0.0127m
Line Resistance, cont’d
†
†
†
Because ac current tends to flow towards the
surface of a conductor, the resistance of a line
at 60 Hz is slightly higher than at dc.
Resistivity and hence line resistance increase
as conductor temperature increases (changes is
about 8% between 25°C and 50°C)
Because ACSR conductors are stranded,
actual resistance, inductance and capacitance
needs to be determined from tables.
Tow-Port Network Model
IR
Is
Two-port
Vs
network
Vs = AVR + BI R
I s = CVR + DI R
AD − BC = 1
VR
Short transmission lines
IR
Is
Vs
Z
Vs = VR + ZI R
Is = IR
A = D =1
B=Z
C =o
VR
Medium transmission lines
V Y ⎞ ⎛ YZ ⎞
⎛
Vs = VR + Z ⎜ I R + R ⎟ = ⎜1 +
⎟VR + ZI R
2
2
⎝
⎠ ⎝
⎠
V Y VY
I s = I R + R + s , subsitute the value of Vs
2
2
⎛ YZ ⎞
⎛ YZ ⎞
I s = Y ⎜1 +
⎟VR + ⎜1 +
⎟I R
4
2
⎝
⎠
⎝
⎠
A = D =1+
YZ
2
B=Z
⎛ YZ ⎞
C = Y ⎜1 +
⎟
4
⎝
⎠
Long transmission lines
z = R + jωL Ω/m
y = G + jωC S/m
Long transmission lines, cont.
V ( x + Δx ) = V ( x ) + ( zΔx) I ( x )
V ( x + Δx ) − V ( x)
= zI ( x)
Δx
Taking the limit as Δx approaches zero :
dV ( x )
= zI ( x )
dx
I ( x + Δx ) = I ( x) + ( yΔx )V ( x + Δx )
I ( x + Δx ) − I ( x )
= yV ( x )
Δx
Taking the limit as Δx approaches zero :
dI ( x )
= yV ( x )
dx
2
d V ( x)
dx 2
d 2V ( x)
dx
2
=z
dI ( x )
= zyV ( x)
dx
− zyV ( x) = 0
V ( x) = A1eγx + A2e −γx
γ = zy
is called the propagation constant
dV ( x)
= γA1eγx − γA2e −γx = zI ( x)
dx
A1eγx − A2e −γx
I ( x) =
Zc
Zc =
z
y
is called the characteristic impedance.
Since VR = V(0) = A1 +A 2 and I R = I(0) =
V + Zc I R
V − Zc I R
A1 = R
and A2 = R
2
2
so :
V ( x) = cosh(γx)VR + Z c sinh(γx) I R
I ( x) =
1
sinh(γx)VR + cosh(γx) I R
Zc
A1 - A 2
Zc
Q1:
Home Work
Q2:
Home Work