Power system slide notes

3/5/2007 459 Chapter 4 Transmission lines and cables 1
ECE 459
Power System Analysis
Professor G. Huang
Chapter 3
Parameters of Transmission Lines and cables

Lecture 4

4.1 Construction

Transmission lines and cables
• Extra-high-voltage lines
• Voltage: 345 kV, 500 kV, 765 kV
– Interconnection between systems
• High-voltage lines
• Voltage: 115 kV, 230 kV
– Interconnection between substations, power plants
• Sub-transmission lines
– Voltage: 46 kV, 69 kV
– Interconnection between substations and large industrial customers
• Distribution lines
– Voltage: 2.4 kV to 46 kV, with 15 kV being the most commonly used
– Supplies residential and commercial customers
• High-voltage DC lines
– Voltage: ±120 kV to ±600 kV
– Interconnection between regions (e.g., Oregon-California)

• 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

Shield
conductor
Insulator
Phase
conductor
Tower
69kV
Line
Composite
InsulatorCrossarm
Composite
insulator
Steel tower
Two
conductor
bundle
Shield conductor

Double circuit
69 kV line
Distribution line
12.47kV
Wooden tower
Shield
conductor
Distribution Line

Distribution line
Transformer
240V/120V
insulated line
Transformers
Fuse cutout
Surge arrester
Insulator

Span
Sag
Insulator
Tension Tower
Supporting
Tower Tension Tower
Definition of Parameters

Residential transformer
vault or pedestal
Fuse cutout
Open
disconnect
switchCable
Supply 1
Supply 2
Fuse
Transformer
Connection diagram residential
Concert conduit with cables in a manhole

4.2 Components of
transmission lines

Frequently used towers
• Lattice tower, used for 220 kV and above;
• Guyed lattice tower, 345 kV and above;
• Tapered steel tube with cross-arm, 230 kV and
below;
• Concrete tower, for distribution and sub-
transmission; and
• Wood tower, for distribution to 220 kV

Lattice
Tower

Guyed Lattice Tower

Tapered Steel Tube Tower

• 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

Locking Key
Insulator's Head
Expansion Layer
Imbedded Sand
Skirt
Petticoats
Iron Cap
Ball Socket
Compression
Loading
Cement
Insulating Glass
or Porcelain
Ball
Corrosion Sleeve
for DC Insulators
Steel Pin
Insulators

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

Composite insulator.
• (1) Sheds of alternating diameters prevent bridging by ice, snow and
cascading rain.
• (2) Fiberglass reinforced resin rod.
• (3) Injection molded EPDM rubber weather sheds and rod covering.
• (4) Forged steel end fitting, galvanized and joined to rod by swaging
process. (Source: Sediver, Inc., York, SC)
1 432

• (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)•Figure 4.15 Line post-composite
insulator with yoke holding two
conductors.

4.3 Cables

Figure 4.16 Single-phase high-voltage cable with solid dielectric

Conductor
Conductor shield
PEX-
Insulation
Insulation shield
Filler
Copper screen
PVC-sheet
Figure 4.17 Three-phase distribution cable with solid dielectric.

4.4 Transmission Line Electrical
Parameters

Cross-Section Diameter Resistance (ohms/mile)
dc ac at 60 Hz
Aluminu
m Steel 25°C 50°C 75°C
100°
C
--- 2776. 1407
.
1521
.
84x.1818 19x.109
1
2.00
0
.546 4 3219 81.6 .0338 .0395 .0421 .0452 .0482 .066
7
Joree 2515. 1274
.
1344
.
76x.1819 19x.084
9
1.88
0
.425 4 2749 61.7 .0365 .0418 .0450 .0482 .0516 .062
1
Thrasher 2312. 1171
.
1235
.
76x.1744 19x.081
4
1.80
2
.407 4 2526 57.3 .0397 .0446 .0482 .0518 .0554 .059
5
Kiwi 2167. 1098
.
1146
.
72x.1735 7x.1157 1.73
5
.347 4 2303 49.8 .0424 .0473 .0511 .0550 .0589 .057
0
Bluebird 2156. 1092
.
1181
.
84x.1602 19x.096
1
1.76
2
.480 4 2511 60.3 .0426 .0466 .0505 .0544 .0584 .058
8
Chukar 1781. 902. 976. 84x.1456 19x.087
4
1.60
2
.437 4 2074 51.0 .0516 .0549 .0598 .0646 .0695 .053
4
Falcon 1590. 806. 908. 54x.1716 19x.103
0
1.54
5
.515 3 2044 54.5 .0578 .0602 .0657 .0712 .0767 .052
1
Code
Al
(mm
2)
Total
(mm
2)
Con
d.
(in.)
Core
(in.)
GM
R
(ft)
25°
C
Weight
(lbs
per
1000
ft)
Stren-
gth
(kips)
Stranding
Table 4.1 ACSR cable technical data

Transmission line parameter
• Resistance Ohm/mile Table 4.1
• Inductance Henry/mile Equation
• Capacitance Farad/mile Equation

Data from the Table 4.1
• Name Bird name Kiwi
• Conductor Diameter
• Resistance at 50C or 75C
• Geometrical Mean Radius (GMR)

Data needed for calculation
• Transmission line length
• Number of conductor / Bundle
• Conductor diameter and GMR
• Distance between phases
• Conductor height

Typical transmission
line
• Horizontal arrangement
• Two ground conductors
• Two conductors per bundle
28'-35'
12'-17'
64'-130'
22'-26'
27'

Resistance
– Stranding increase resistance
– Increases by temperature
– Increases by skin effect
– AC resistance higher than DC
– Accurate value from Table 4.1

Capacitance
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
==
c
AN
A
AN
r
V
Q
C
GMD
ln
2 0επ
cequ rdr =
3 2
cequ rdr =
4 3
09.1 cequ rdr =
Two-conductor bundle
Three-conductor bundle
Four-conductor bundle:
3
ACBCAB DDDGMD =

Inductance
4.31 Transposed line
3
ACBCAB DDDGMD =
cd GMRGMR =
3 2
GMRGMR cd=
4 3
GMR09.1GMR cd=
LA_inductance
μo
2 π⋅
ln
GMD
GMRc
⎛
⎜
⎝
⎞
⎟
⎠
⋅:= XA ω
μo
2 π⋅
ln
GMD
GMRc
⎛
⎜
⎝
⎞
⎟
⎠
⋅
⎛
⎜
⎝
⎞
⎟
⎠
⋅
⎡
⎢
⎣
⎤
⎥
⎦
:=

Inductance/Capacitance
3
ACBCAB DDDGMD =
XA ω
μo
2 π⋅
ln
GMD
GMRc
⎛
⎜
⎝
⎞
⎟
⎠
⋅
⎛
⎜
⎝
⎞
⎟
⎠
⋅
⎡
⎢
⎣
⎤
⎥
⎦
:=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
==
c
AN
A
AN
r
V
Q
C
GMD
ln
2 0επ

Inductance/Capacitance
4.31 Transposed line3
ACBCAB DDDGMD =
cd GMRGMR =
3 2
GMRGMR cd=
4 3
GMR09.1GMR cd=
cequ rdr =
3 2
cequ rdr =
4 3
09.1 cequ rdr =
m
H
104 7
0
−
= πμ
m
F
36
10 9
0
π
ε
−
=

Numerical exercise
Transmission lines Parameter calculation

28'-35'
12'-17'
64'-130'
22'-26'
27'
D

Vline 500kV:= LLine 206mi:= f 60Hz:= ω 2 π⋅ f⋅:=
d 18in:= n 2:= D 32ft:=
Bluebird conductor
rc
1.762in
2
:= GMRc 0.0588ft:= R75 0.0544
Ω
mi
:=
εo
10
9−
36 π⋅
F
m
⋅:= μo 4 π⋅ 10
7−
⋅
H
m
⋅:=

Transmission Line Parameters
GMR d GMRc⋅:= GMR 0.091m=
requ d rc⋅:= requ 0.101m=
GMD
3
D D⋅ 2⋅ D⋅:= GMD 12.289m=
XL ω
μo
2 π⋅
⋅ ln
GMD
GMR
⎛
⎜
⎝
⎞
⎟
⎠
:= XL 0.596
Ω
mi
=
XLine XL LLine⋅:= XLine 122.754Ω=

RLine
R75
n
LLine⋅:= RLine 5.603Ω=
ZLine RLine j XLine⋅+:= ZLine 5.603 122.754j+ Ω=
CLine
2 π⋅ εo⋅
ln
GMD
requ
⎛
⎜
⎝
⎞
⎟
⎠
:= CLine 18.627
nF
mi
=
XCS
1−
ω
CLine
2
⋅ LLine⋅
:= XCS 1.383− kΩ=

Equivalent circuit

MATCAD
XCL XCS:= XCL 1.383− kΩ=
ZCS j XCS⋅:= ZCS 1.383j− kΩ= ZCL j XCL⋅:= ZCL 1.383j− kΩ=
Vnet1_ln
Xnet1
Inet1_short RLine
XLine
IL_short
XCS
IL_shortICS_short XCL

Figure 4.50 Equivalent network of a short line.

RLine
LLine
CLine
/ 2 CLine
/ 2
Figure 4.49 Equivalent network of a medium-length line.

Figure 4.49 Equivalent network of a medium-length line.
j X
VS VR
IS R
C/2C/2
Ics I ICR
Ir

Numerical exercise
Transmission lines Loading

• Steps of calculation (4.5.1.3)
• Given data:
• The power is variable, selected value is:
• Load current calculation:
VL_ll Vline:= VL_ll 500kV= pfL 0.8:= lagging( )
PL 400M W⋅:=
PL1 PL( )
PL
3
:= VL_n
VL_ll
3
:=
SL1 PL( )
PL1 PL( )
pfL
e
j acos pfL( )⋅
⋅:= SL1 PL( ) 133.333 100j+ M V⋅ A⋅=
IL PL( )
SL1 PL( )
VL_n
⎛
⎜
⎝
⎞
⎟
⎠
⎯
:= IL PL( ) 461.88 346.41j− A=

• Capacitive current:
• Line current
• Supply voltage:
ICL
VL_n
ZCL
:= ICL 208.798jA=
ILine PL( ) ICL IL PL( )+:= ILine PL( ) 461.88 137.612j− A=
VS_n PL( ) VL_n ILine PL( ) ZLine⋅+:= VS_n PL( ) 308.156 55.926j+ kV=

• Capacitive current:
• Supply/network current
ICS PL( )
VS_n PL( )
ZCS
:= ICS PL( ) 40.451− 222.888j+ A=
Inet1 PL( ) ILine PL( ) ICS PL( )+:= Inet1 PL( ) 421.429 85.276j+ A=

Sending
end
Receiving
end
···
Figure 4.48 Equivalent network for a long transmission line.

Derivation of Transmission Line Models

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
5) other

Magnetics Review
• Ampere’s circuital law:
e
F = mmf = magnetomtive force (amp-turns)
= magnetic field intensity (amp-turns/meter)
d = Vector differential path length (meters)
= Line integral about closed path
(d is tangent to path)
I =
eF d I
Γ
Γ
= =
Γ
∫
∫
H l
H
l
l
Algebraic sum of current linked by Γ

Line Integrals
•Line integrals are a generalization of
traditional integration
Integration along the
x-axis
Integration along a
general path, which
may be closed
Ampere’s law is most useful in cases of symmetry,
such as with an infinitely long line

Magnetic Flux Density
•Magnetic fields are usually measured in
terms of flux density
0
-7
0
= flux density (Tesla [T] or Gauss [G])
(1T = 10,000G)
For a linear a linear magnetic material
= where is the called the permeability
=
= permeability of freespace = 4 10
= relative permea
r
r
H m
μ μ
μ μ μ
μ π
μ
×
B
B H
bility 1 for air≈

Magnetic Flux
Total flux passing through a surface A is
=
= vector with direction normal to the surface
If flux density B is uniform and perpendicular to an
area A then
=
A
d
d
BA
φ
φ
∫ B a
a

Magnetic Fields from Single
Wire
•Assume we have an infinitely long wire with
current of 1000A. How much magnetic flux
passes through a 1 meter square, located
between 4 and 5 meters from the wire?
Direction of H is given
by the “Right-hand” Rule
Easiest way to solve the problem is to take advantage
of symmetry. For an integration path we’ll choose a
circle with a radius of x.

Single Line Example, cont’d
0
5 0
0 4
7
0
5
4
2
2
2
5 5
ln 2 10 ln
2 4 4
4.46 10 Wb
2 10 2
B T Gauss
x
A
I
xH I H
x
B H
I
H dA dx
x
I
I
x
π
π
μ
μ
φ μ
π
φ μ
π
φ
−
−
−
= → =
=
= =
= = ×
= ×
×
= =
∫ ∫
For reference, the earth’s
magnetic field is about
0.6 Gauss (Central US)

Flux linkages and Faraday’s law
N
i=1
Flux linkages are defined from Faraday's law
d
V = where V = voltage, = flux linkages
The flux linkages tell how much flux is linking an
N turn coil:
=
If all flux links every coil then
i
dt
N
λ
λ
λ φ
λ φ=
∑

Inductance
• For a linear magnetic system, that is one
where
• B = μ H
• we can define the inductance, L, to be
• the constant relating the current and the flux
• linkage
• λ = L i
• where L has units of Henrys (H)

Inductance Example
•Calculate the inductance of an N turn coil
wound tightly on a torodial iron core that has
a radius of R and a cross-sectional area of A.
Assume
•1) all flux is within the coil
•2) all flux links each turn

Inductance Example, cont’d
0
0
2
0
2 (path length is 2 R)
H
2
2
H
2
e
r
r
r
I d
NI H R
NI
B H H
R
AB N LI
NI
NAB NA
R
N A
L
R
π π
μ μ μ
π
φ λ φ
λ μ μ
π
μ μ
π
Γ
=
=
= = =
= = =
= =
=
∫ H l

Inductance of a Single Wire
•To development models of transmission lines,
we first need to determine the inductance of a
single, infinitely long wire. To do this we need
to determine the wire’s total flux linkage,
including
– 1. flux linkages outside of the wire
– 2. flux linkages within the wire
•We’ll assume that the current density within
the wire is uniform and that the wire has a
radius of r.

Flux Linkages outside of the wire
R
0A r
We'll think of the wire as a single loop closed at
infinity. Therefore = since N = 1. The flux linking
the wire out to a distance of R from the wire center is
d length
2
I
dx
x
λ φ
φ μ
π
= =∫ ∫B a

Flux Linkages outside, cont’d
R
0A r
R 0
0r
d length
2
Since length = we'll deal with per unit length values,
assumed to be per meter.
ln
2 2
Note, this quantity still goes to infinity as R
I
dx
x
I R
dx I
meter x r
λ φ μ
π
μλ μ
π π
= = =
∞
= =
→ ∞
∫ ∫
∫
B a

Flux linkages inside of wire
Current inside conductor tends to travel on the outside
of the conductor due to the skin effect. The pentration
of the current into the conductor is approximated using
1
the skin depth = where f is
fπ μσ
the frequency in Hz
and is the conductivity in mhos/meter.
0.066 m
For copper skin depth 0.33 inch at 60HZ.
f
For derivation we'll assume a uniform current density.
σ
≈ ≈

Flux linkages inside, cont’d
Wire cross section
x
r
2
2
2
Current enclosed within distance
x of center I
2 2
e
e
x
x
I
r
I Ix
H
x rπ π
= =
= =
2 3
0
inside 2 2 40 0
Flux only links part of current
2 82
r r rIx x Ix
dx dx I
r r r
μ μμ
λ μ
π ππ
= = =∫ ∫

Line Total Flux & Inductance
0 0
0
0
(per meter) ln
2 8
(per meter) ln
2 4
L(per meter) ln
2 4
Note, this value still goes to infinity as we integrate
R out to infinity
r
Total
r
Total
r
R
I I
r
R
I
r
R
r
μ μ μ
λ
π π
μ μ
λ
π
μ μ
π
= +
⎛ ⎞= +⎜ ⎟
⎝ ⎠
⎛ ⎞= +⎜ ⎟
⎝ ⎠

Inductance Simplification
0 0 4
0 4
Inductance expression can be simplified using
two exponential identities:
a
ln(ab)=ln a + ln b ln ln ln ln( )
b
ln ln ln ln
2 4 2
ln ln
2
r
r
a
r
a b a e
R
L R r e
r
L R re
μ
μ
μ μμ
π π
μ
π
−
−
= − =
⎛ ⎞⎛ ⎞⎛ ⎞= + = − +⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠
⎛ ⎞⎛ ⎞
= −⎜ ⎜ ⎟
⎝ ⎠⎝ ⎠
0
4
r
ln
2 '
Where r' 0.78 for 1
r
R
r
re r
μ
μ
π
μ
−
=⎟
≈ =

Two Conductor Line Inductance
•Key problem with the previous derivation is
we assumed no return path for the current.
Now consider the case of two wires, each
carrying the same current I, but in opposite
directions; assume the wires are separated by
distance R. R
Creates counter-
clockwise field
Creates a
clockwise field
To determine the
inductance of each
conductor we integrate
as before. However
now we get some
field cancellation

Two Conductor Case, cont’d
R
R
Direction of integration
Rp
Key Point: As we integrate for the left line, at distance 2R from
the left line the net flux linked due to the Right line is zero!
Use superposition to get total flux linkage.
0 0
left
For distance Rp, greater than 2R, from left line
ln ln
2 ' 2
Rp Rp R
I I
r R
μ μ
λ
π π
−⎛ ⎞= − ⎜ ⎟
⎝ ⎠
Left Current Right Current

Two Conductor Inductance
( )
0
left
0
0
0
0
Simplifying (with equal and opposite currents)
ln ln
2 '
ln ln ' ln( ) ln
2
ln ln
2 '
ln as Rp
2 '
ln H/m
2 '
left
Rp Rp R
I
r R
I Rp r Rp R R
R Rp
I
r Rp R
R
I
r
R
L
r
μ
λ
π
μ
π
μ
π
μ
π
μ
π
−⎛ ⎞⎛ ⎞= − ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
= − − − +
⎛ ⎞
= +⎜ ⎟−⎝ ⎠
⎛ ⎞= → ∞⎜ ⎟
⎝ ⎠
⎛ ⎞= ⎜ ⎟
⎝ ⎠

Many-Conductor Case
Now assume we now have k conductors, each with
current ik, arranged in some specified geometry.
We’d like to find flux linkages of each conductor.
Each conductor’s flux
linkage, λk, depends upon
its own current and the
current in all the other
conductors.
To derive λ1 we’ll be integrating from conductor 1 (at origin)
to the right along the x-axis.

Many-Conductor Case, cont’d
At point b the net
contribution to λ1
from ik , λ1k, is
zero.
We’d like to integrate the flux crossing
between b to c. But the flux crossing
between a and c is easier to calculate and
provides a very good approximation of λ1k.
Point a is at distance d1k from conductor k.
Rk is the
distance
from con-
ductor k
to point
c.

[ ]
0 1 2
1 1 2'
12 11
0
1 1 2'
12 11
0
1 1 2 2
1 1 2
0
1
ln ln ln
2
1 1 1
ln ln ln
2
ln ln ln
2
As R goes to infinity R so the second
term from above can be written =
2
n
n
n
n
n
n n
n
n
j
j
RR R
i i i
d dr
i i i
d dr
i R i R i R
R R
i
μ
λ
π
μ
λ
π
μ
π
μ
π =
⎡ ⎤
= + + +⎢ ⎥
⎣ ⎦
⎡ ⎤
= + + + +⎢ ⎥
⎣ ⎦
+ + +
= =
L
L
L
1ln R
⎛ ⎞
⎜ ⎟
⎝ ⎠
∑

1
0
1 1 2'
12 11
1 11 1 12 2 1
Therefore if 0, which is true in a balanced
three phase system, then the second term is zero and
1 1 1
ln ln ln
2
System has self and mutual inducta
n
j
j
n
n
n n
i
i i i
d dr
L i L i L i
μ
λ
π
λ
=
=
⎡ ⎤
= + + +⎢ ⎥
⎣ ⎦
= + +
∑
L
L
nce. However
the mutual inductance can be canceled for
balanced 3 systems with symmetry.φ

Symmetric Line Spacing – 69 kV

Birds Do Not Sit on the Conductors

Line Inductance Example
Calculate the reactance for a balanced 3φ, 60Hz
transmission line with a conductor geometry of an
equilateral triangle with D = 5m, r = 1.24cm (Rook
conductor) and a length of 5 miles.
0 1 1 1
ln( ) ln( ) ln( )
2 '
a a b ci i i
r D D
μ
λ
π
⎡ ⎤= + +⎢ ⎥⎣ ⎦
a
Since system is assumed
balanced
i b ci i= − −

Line Inductance Example, cont’d
a
0
a
0
7
0
3
6
Substituting
i
Hence
1 1
ln ln
2 '
ln
2 '
4 10 5
ln ln
2 ' 2 9.67 10
1.25 10 H/m
b c
a a
a
a
i i
i i
r D
D
i
r
D
L
r
μ
λ
π
μ
π
μ π
π π
−
−
−
= − −
⎡ ⎤⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
⎛ ⎞= ⎜ ⎟
⎝ ⎠
×⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠×
= ×

Line Inductance Example, cont’d
6
6
a
4
Total for 5 mile line
1.25 10 H/m
Converting to reactance
2 60 1.25 10
4.71 10 /m
0.768 /mile
X 3.79
(this is the total per phase)
The reason we did NOT have mutual inductance
was because
aL
X π
−
−
−
= ×
= × × ×
= × Ω
= Ω
= Ω
of the symmetric conductor spacing

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.

Bundled Conductor Flux
Linkages
For the line shown on the left,
define dij as the distance between
conductors i and j. We can then
determine λ for each
18
12 13 14
0
1
15 16 17
19 1,10 1,11 1,12
1 1 1 1
ln ln ln ln
4 '
1 1 1 1
ln ln ln ln
2 4
1 1 1 1
ln ln ln ln
4
a
b
c
i
r d d d
i
d d d d
i
d d d d
μ
λ
π
⎡ ⎤⎛ ⎞
+ + + +⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥
⎢ ⎥⎛ ⎞
⎢ ⎥= + + + +⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠
⎢ ⎥
⎛ ⎞⎢ ⎥
+ + +⎜ ⎟⎢ ⎥
⎢ ⎥⎝ ⎠⎣ ⎦

Bundled Conductors, cont’d
1
4
12 13 14
0
1 1
4
15 16 17 18
1
4
19 1,10 1,11 1,12
Simplifying
1
ln
( ' )
1
ln
2 ( )
1
ln
( )
a
b
c
i
r d d d
i
d d d d
i
d d d d
μ
λ
π
⎡ ⎤⎛ ⎞
⎢ ⎥⎜ ⎟ +
⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥
⎢ ⎥⎛ ⎞
⎢ ⎥⎜ ⎟= +⎢ ⎥⎜ ⎟
⎢ ⎝ ⎠ ⎥
⎢ ⎥
⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Bundled Conductors, cont’d
1
4
12 13 14
1
12 1
1
1
4
15 16 17 18 2 3 4
1 19 1
geometric mean radius (GMR) of bundle
( ' ) for our example
( ' ) in general
geometric mean distance (GMD) of
conductor 1 to phase b.
( )
(
b
b
b
b
b b b ab
c
R
r d d d
r d d
D
d d d d D D D D
D d d
=
=
= ≈ ≈ ≈ ≈
=
K
1
4
,10 1,11 1,12 2 3 4) c c c acd d D D D D≈ ≈ ≈ ≈

Inductance of Bundle
a
0
1
0 0
1
0
1
If D and i
Then
1 1
ln ln
2
ln 4 ln
2 2
4 ln
2
ab ac bc b c
a a
b
a
b b
b
D D D i i
i i
R D
D D
I I
R R
D
L
R
μ
λ
π
μ μ
π π
μ
π
= = = = − −
⎡ ⎤⎛ ⎞ ⎛ ⎞= − ⎜ ⎟⎢ ⎥⎜ ⎟
⎝ ⎠⎝ ⎠⎣ ⎦
⎛ ⎞ ⎛ ⎞
= =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞
= × × ⎜ ⎟
⎝ ⎠

Inductance of Bundle, cont’d
0
a 1
But remember each bundle has b conductors
in parallel (4 in this example). So
L / ln
2 b
D
L b
R
μ
π
⎛ ⎞
= = ⎜ ⎟
⎝ ⎠

Bundle Inductance Example
0.25 M0.25 M
0.25 M
Consider the previous example of the three phases
symmetrically spaced 5 meters apart using wire
with a radius of r = 1.24 cm. Except now assume
each phase has 4 conductors in a square bundle,
spaced 0.25 meters apart. What is the new inductance
per meter?
( )
2 3
1
3 4
b
70
a
1.24 10 m ' 9.67 10 m
R 9.67 10 0.25 0.25 2 0.25
0.12 m (ten times bigger!)
5
L ln 7.46 10 H/m
2 0.12
r r
μ
π
− −
−
−
= × = ×
= × × × × ×
=
= = ×

Transmission Tower
Configurations
•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.
Typical Transmission Tower
Configuration
Therefore in
general Dab ≠
Dac ≠ Dbc
Unless something
was done this would
result in unbalanced
phases

Transposition
• 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.
Aerial or side view of conductor positions over the length
of the transmission line.

Transposition

Transposition Impact on Flux
Linkages
0
a
12 13
0
13 23
0
23 12
For a uniformly transposed line we can
calculate the flux linkage for phase "a"
1 1 1 1
ln ln ln
3 2 '
1 1 1 1
ln ln ln
3 2 '
1 1 1 1
ln ln ln
3 2 '
a b c
a b c
a b c
I I I
r d d
I I I
r d d
I I I
r d d
μ
λ
π
μ
π
μ
π
⎡ ⎤
= + +⎢ ⎥
⎣ ⎦
⎡ ⎤
+
+ + +⎢ ⎥
⎣ ⎦
⎡ ⎤
+ +⎢ ⎥
⎣ ⎦
“a” phase in
position “1”
“a” phase in
position “3”
“a” phase in
position “2”

Substation Bus

Transposition Impact, cont’d
( )
( )
1
3
1
3
12 13 230
a
1
3
12 13 23
Recognizing that
1
(ln ln ln ) ln( )
3
We can simplify so
1 1
ln ln
'
2 1
ln
a b
c
a b c abc
I I
r d d d
I
d d d
μ
λ
π
+ + =
⎡ ⎤+ +⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎢ ⎥⎣ ⎦

Inductance of Transposed Line
( )
1
3
m 12 13 23
0 0
a
70
Define the geometric mean distance (GMD)
D
Then for a balanced 3 system ( - - )
1 1
ln ln ln
2 ' 2 '
Hence
ln 2 10 ln H/m
2 ' '
a b c
m
a a a
m
m m
a
d d d
I I I
D
I I I
r D r
D D
L
r r
φ
μ μ
λ
π π
μ
π
−
=
⎡ ⎤
= − =⎢ ⎥
⎣ ⎦
= = ×

Inductance with Bundling
b
0
a
70
If the line is bundled with a geometric mean
radius, R , then
ln
2
ln 2 10 ln H/m
2
m
a
b
m m
a
b b
D
I
R
D D
L
R R
μ
λ
π
μ
π
−
=
= = ×

Line Transposition Example

Inductance Example
• Calculate the per phase inductance and
reactance of a balanced 3φ, 60 Hz,
transmission line with horizontal phase
spacing of 10m using three conductor
bundling with a spacing between conductors
in the bundle of 0.3m. Assume the line is
uniformly transposed and the conductors
have a 1cm radius.
Answer: Inductance = 9.9 x 10-7 H/m, Reactance = 0.6 Ω/Mile

Review of Electric Fields
eA
2
To develop a model for line capacitance we
first need to review some electric field concepts.
Gauss's law:
d = q (integrate over closed surface)
where
= electric flux density, coulombs/m
d = differential
∫ D a
D
a
2
e
area da, with normal to surface
A = total closed surface area, m
q = total charge in coulombs enclosed

Gauss’s Law Example
•Similar to Ampere’s Circuital law, Gauss’s
Law is most useful for cases with symmetry.
•Example: Calculate D about an infinitely long
wire that has a charge density of q
coulombs/meter.
Since D comes
radially out inte-
grate over the
cylinder bounding
the wireeA
d 2 q
where radially directed unit vector
2
D Rh qh
q
R
π
π
= = =
=
∫
r r
D a
D a a

Electric Fields
•The electric field, E, is related to the electric flux
density, D, by
• D = ε E; where
• E = electric field (volts/m)
• ε = permittivity in farads/m (F/m)
• ε = εo εr
• εo = permittivity of free space (8.854×10-12 F/m)
• εr = relative permittivity or the dielectric constant
(≈1 for dry air, 2 to 6 for most dielectrics)

Voltage Difference
P
P
The voltage difference between any two
points P and P is defined as an integral
V
In previous example the voltage difference between
points P and P , located radial distance R and R
f
dβ
α
α β
βα
α β α β
−∫ E l
R
R
rom the wire is (assuming = )
V ln
2 2
o
o o
Rq q
dR
R R
β
α
α
βα
β
ε ε
πε πε
= − =∫

Voltage Difference, cont’d
R
R
With
V ln
2 2
if q is positive then those points closer in have
a higher voltage. Voltage is defined as the energy
(in Joules) required to move a 1 coulomb charge
against an ele
o o
Rq q
dR
R R
β
α
α
βα
βπε πε
= − =∫
ctric field (Joules/Coulomb). Voltage
is infinite if we pick infinity as the reference point

Multi-Conductor Case
i
1
Now assume we have n parallel conductors,
each with a charge density of q coulombs/m.
The voltage difference between our two points,
P and P , is now determined by superposition
1
V ln
2
n
i
i
ii
R
q
R
α β
α
βα
βπε =
=
where is the radial distance from point P
to conductor i, and the distance from P to i.
i
i
R
R
α α
β β
∑

Multi-Conductor Case, cont’dn
i
i=1
1 1
1
1
11 1
1
If we assume that q 0 then rewriting
1 1 1
V ln ln
2 2
We then subtract ln 0
1 1 1
V ln ln
2 2
As we more P to infinity, ln 0
n n
i i i
ii i
n
i
i
n n
i
i i
ii i
i
q q R
R
q R
R
q q
R R
R
R
βα α
β
α
α
βα
β α
α
α
α
πε πε
πε πε
= =
=
= =
=
= +
=
= +
→
∑
∑ ∑
∑
∑ ∑

Absolute Voltage Defined
1
Since the second term goes to zero as P goes to
infinity, we can now define the voltage of a
point w.r.t. a reference voltage at infinity:
1 1
V ln
2
This equation holds for any point as long a
n
i
ii
q
R
α
β
βπε =
= ∑
s
it is not inside one of the wires!

Three Conductor Case
A
BC
Assume we have three
infinitely long conductors,
A, B, & C, each with radius r
and distance D from the
other two conductors.
Assume charge densities such
that qa + qb + qc = 0
1 1 1 1
ln ln ln
2
ln
2
a a b c
a
a
V q q q
r D D
q D
V
r
πε
πε
⎡ ⎤= + +⎢ ⎥⎣ ⎦
=

Line Capacitance
j
1 11 1
For a single line capacitance is defined as
But for a multiple conductor case we need to
use matrix relationships since the charge on
conductor i may be a function of V
i i i
n
n
q C V
q C C
q
=
⎡ ⎤
⎢ ⎥ =
⎢ ⎥
⎢ ⎥⎣ ⎦
L
M M L M
1
1n nn n
V
C C V
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
=q C V
M
L

Line Capacitance, cont’d
In ECE 476 we will not be considering theses
cases with mutual capacitance. To eliminate
mutual capacitance we'll again assume we have
a uniformly transposed line. For the previous
three conductor exam
a
a a
ple:
q 2
ince q = C
ln
a
a
V V
S V C
DV
r
πε
=
→ = =

Bundled Conductor Capacitance
1
1
c
b 12
Similar to what we did for determining line
inductance when there are n bundled conductors,
we use the original capacitance equation just
substituting an equivalent r
Note fo
adius
r t
( )
he
R n
nrd d= L
b
capacitance equation we use r rather
than r' which was used for R in the inductance
equation

Line Capacitance, cont’d
[ ]
1
m
1
3
m
1
c
b 12
-12
o
For the case of uniformly transposed lines we
use the same GMR, D , as before.
2
ln
where
D
R ( ) (note r NOT r')
ε in air 8.854 10 F/m
n
m
c
b
ab ac bc
n
C
D
R
d d d
rd d
πε
ε
=
=
=
= = ×
L

Line Capacitance Example
•Calculate the per phase capacitance and
susceptance
of a balanced 3φ, 60 Hz, transmission line
with horizontal phase spacing of 10m using
three conductor bundling with a spacing
between conductors in the bundle of 0.3m.
Assume the line is uniformly transposed and
the conductors have a 1cm radius.

Line Capacitance Example,
cont’d
1
3
1
3
m
12
11
c 11
8
(0.01 0.3 0.3) 0.0963 m
D (10 10 20) 12.6 m
2 8.854 10
1.141 10 F/m
12.6
ln
0.0963
1 1
X
2 60 1.141 10 F/m
2.33 10 -m (not / m)
c
bR
C
C
π
ω π
−
−
−
= × × =
= × × =
× ×
= = ×
= =
× ×
= × Ω Ω

Line Conductors
• Typical transmission lines use multi-strand
conductors
• ACSR (aluminum conductor steel
reinforced)
conductors are most common. A typical Al.
to St. ratio is about 4 to 1.

Line Conductors, cont’d
• Total conductor area is given in circular
mils. One circular mil is the area of a circle
with a diameter of 0.001 = π × 0.00052
square inches
• Example: what is the the area of a solid, 1”
diameter circular wire?
Answer: 1000 kcmil (kilo circular mils)
• Because conductors are stranded, the
equivalent radius must be provided by the
manufacturer. In tables this value is known
as the GMR and is usually expressed in feet.

Line Resistance
-8
-8
Line resistance per unit length is given by
R = where is the resistivity
A
Resistivity of Copper = 1.68 10 Ω-m
Resistivity of Aluminum = 2.65 10 Ω-m
Example: What is the resistance in Ω / mile of a
ρ
ρ
×
×
-8
2
1" diameter solid aluminum wire (at dc)?
2.65 10 Ω-m
1609 0.084
0.0127m
m
R
mile mileπ
× Ω
= =
×

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.

ACSR Table Data (Similar to Table
A.4)
Inductance and Capacitance
assume a Dm of 1 ft.
GMR is equivalent to r’

ACSR Data, cont’d
7
L
3
3 3
X 2 4 10 ln 1609 /mile
1
2.02 10 ln ln
1
2.02 10 ln 2.02 10 ln
m
m
m
D
f L f
GMR
f D
GMR
f f D
GMR
π π −
−
− −
= = × × Ω
⎡ ⎤= × +⎢ ⎥⎣ ⎦
= × + ×
Term from table assuming
a one foot spacing
Term independent
of conductor with
Dm in feet.

ACSR Data, Cont.
0
C
6
To use the phase to neutral capacitance from table
21
X -m where
2 ln
1
1.779 10 ln -mile (table is in M -mile)
1 1 1
1.779 ln 1.779 ln M -mile
m
m
m
C
Df C
r
D
f r
D
f r f
πε
π
= Ω =
= × × Ω Ω
= × × + × × Ω
Term from table assuming
a one foot spacing
Term independent
of conductor with
Dm in feet.

Dove Example
7
0.0313 feet
Outside Diameter = 0.07725 feet (radius = 0.03863)
Assuming a one foot spacing at 60 Hz
1
2 60 2 10 1609 ln Ω/mile
0.0313
0.420 Ω/mile, which matches the table
For the capacitance
a
a
C
GMR
X
X
X
π −
=
= × × × ×
=
6 41 1
1.779 10 ln 9.65 10 Ω-mile
f r
= × × = ×

Additional Transmission Topics
• Multi-circuit lines: Multiple lines often share a
common transmission right-of-way. This DOES
cause mutual inductance and capacitance, but is
often ignored in system analysis.
• Cables: There are about 3000 miles of
underground ac cables in U.S. Cables are
primarily used in urban areas. In a cable the
conductors are tightly spaced, (< 1ft) with oil
impregnated paper commonly used to provide
insulation
– inductance is lower
– capacitance is higher, limiting cable length

Additional Transmission topics
• Ground wires: Transmission lines are
usually protected from lightning strikes with
a ground wire. This topmost wire (or wires)
helps to attenuate the transient
voltages/currents that arise during a lighting
strike. The ground wire is typically grounded
at each pole.
• Corona discharge: Due to high electric
fields around lines, the air molecules become
ionized. This causes a crackling sound and
may cause the line to glow!

Additional Transmission topics
• Shunt conductance: Usually ignored. A small current
may flow through contaminants on insulators.
• DC Transmission: Because of the large fixed cost
necessary to convert ac to dc and then back to ac, dc
transmission is only practical for several specialized
applications
– long distance overhead power transfer (> 400 miles)
– long cable power transfer such as underwater
– providing an asynchronous means of joining different
power systems (such as the Eastern and Western grids).

Bonus question #2 (AC vs DC)
Due on March 8
• Give technical analysis why overhead
HVDC lines need to be more than 400
miles to be cost effective? What is the
technical bottleneck?
• Why now engineers are suggesting DC lines
for distribution lines around 1KV? Are the
concerns for HVDC an issue here? Why?

Transmission Siting
• Building new transmission lines can be
extremely difficult, particularly if a new
right-of-way is needed
• The siting of new lines is currently a state
responsibility, which can cause difficulties
if a line spans more than one state
– the exception federal power marketing administrations
and TVA, which have their own siting authority

Summary
• How inductance is derived? How capacitance is
derived? What is corona?
• What is GMR? Why GMR? What is GMD? Why
GMD?
• Why delta construction has zero mutual
inductance? Why flat instruction has nonzero
mutual inductance?
• What is bundling? Why bundling?
• What is transposition? Why transposition?

Power system slide notes

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