transmission lines ste 2013
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EE 369
POWER SYSTEM ANALYSIS
Lecture 5
Development of Transmission Line ModelsTom Overbye and Ross Baldick
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Development of Line Models
Goals of this section are:
1) develop a simple model for transmission
lines, and
2) gain an intuitive feel for how the geometry of
the transmission line affects the model
parameters.
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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 dcThe analysis will be developed for ac lines.
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Magnetics Review
Magnetomotive force: symbol F, measured inampere-turns, which is the current enclosed by aclosed path,
Magnetic field intensity: symbol H, measured in
ampere-turns/meter: The existence of a current in a wire gives rise to an
associated magnetic field.
The stronger the current, the more intense is themagnetic field H.
Flux density: symbol B, measured in webers/m2
or teslas or gauss (1 Wb /m2 = 1T = 10,000G): Magnetic field intensity is associated with a magnetic
flux density.5
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Magnetics Review
Magnetic flux: symbol measured in webers,which is the integral of flux density over asurface.
Flux linkages measured in weber-turns. If the magnetic flux is varying (due to a changing
current) then a voltage will be induced in aconductor that depends on how much magnetic fluxis enclosed (linked) by the loops of the conductor,according to Faradays law.
Inductance: symbol L, measured in henrys:
The ratio of flux linkages to the current in a coil.
,
,
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Magnetics Review
Amperes circuital law relates magnetomotive
force (the enclosed current) and magnetic
field intensity:
e
= mmf = magnetomotive 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
F
H l
H
l
l
= Algebraic sum of current linked by 7
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Line Integrals
Line integrals are a generalization of standard
integration along, for example, the x-axis.Integration along the
x-axis
Integration along a
general path, which
may be closed
Amperes law is most useful in cases of symmetry,
such as a circular path of radiusxaround an infinitely
long wire, so that H and dl are parallel, |H| is constant,
and |dl| integrates to equal the circumference 2x.
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Flux Density
Assuming no permanent magnetism, magneticfield intensity and flux density are related by the
permeability of the medium.
0
0
= magnetic field intensity (amp-turns/meter)
= flux density (Tesla [T] or Gauss [G])(1T = 10,000G)
For a linear magnetic material:
= where is the called the permeability
=
= permeability of frees
r
H
B
B H
-7pace = 4 10 H m
= relative permeability 1 for airr
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Magnetic Flux
2
Magnetic flux and flux density
magnetic flux (webers)
= flux density (webers/m or tesla)
Definition of flux passing through a surface is
=
= vector with direction normal to the surfaceIf flux
A
A
d
d
B
B a
a
density B is uniform and perpendicular to anarea A then
= BA10
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Magnetic Fields from Single Wire
Assume we have an infinitely long wire with
current ofI =1000A.
Consider a square, located between 4 and 5
meters from the wire and such that the square
and the wire are in the same plane.
How much magnetic flux passes through the
square?
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Magnetic Fields from Single Wire
Magnetic flux passing through the square?
Easiest way to solve the problem is to take
advantage of symmetry.
As an integration path, well choose a circle
with radiusx, withxvarying from 4 to 5
meters, with the wire at the center, so the
path encloses the current I. 12
Direction of H is given
by the Right-hand Rule
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Single Line Example, contd
40
5 0
4
70
5
22
2 10 2T Gauss
(1 meter)
25 5
ln 2 10 ln2 4 4
4.46 10 Wb
A
Id xH I H
x
B Hx x
IdA dx
xI
I
H l
B
For reference,the earths
magnetic field is
about 0.6 Gauss
(Central US)
13
H is perpendicular
to surface of square
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Flux linkages and Faradays law
i=1
Flux linkages are defined from Faraday's law
d= , where = voltage, = flux linkages
d
The flux linkages tell how much flux is linking anturn coil:
=
If flux links every coil then
N
i
V Vt
N
N
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Inductance
For a linear magnetic system; that is, one
where B = H,
we can define the inductance, L, to be the
constant of proportionality relating the
current and the flux linkage: = L I,
where L has units of Henrys (H).
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Substation Bus
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Inductance Example
Calculate the inductance of an N turn coil woundtightly on a torodial iron core that has a radius ofR and a cross-sectional area ofA. Assume
1) all flux is within the coil
2) all flux links each turn3) Radius of each turn is negligible compared to R
Circular path of radius R
within iron core
encloses all N turnsand hence links NI.
Since radius of each turn
is negligible compared to R,
all circular paths within
the iron core have radius
approximately equal to R. 17
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Inductance Example, contd
0
0
20
2 (path length is 2 )
2
2
H2
e
r
r
r
I d
NI H R R
NIH B H HR
A B N LI
NINAB NA R
N AL
R
H l
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Inductance of a Single Wire
To develop models of transmission lines, we firstneed to determine the inductance of a single,infinitely long wire. To do this we need todetermine the wires total flux linkage, including:
1. flux linkages outside of the wire2. flux linkages within the wire
Well assume that the current density within thewire is uniform and that the wire is solid with a
radius ofr.In fact, current density is non-uniform, andconductor is stranded, so our calculations will beapproximate.
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Flux Linkages outside of the wire
We'll think of the wire as a single loop "closed" by
a return wire that is infinitely far away. Therefore
= since there is = 1 turn. The flux linking
a length of wire outside it to a distance of
N
R
0A
from
the wire center is:
d length
2
R
r
Idx
x
B a20
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Flux Linkages outside, contd
0A
00
d length2
Since length = we'll deal with per unit length values,
assumed to be per meter.
lnmeter 2 2
Note, this quantity still goes to infinity as
R
r
R
r
Idx
x
I Rdx I
x r
R
B a
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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
1the skin depth = where isff
the frequency in Hz
and is the conductivity in siemens/meter.
0.066 mFor copper skin depth 0.33 inch at 60Hz.
For derivation we'll assume a uniform current density.
f
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Flux linkages inside, contdWire cross section
x
r
2
2
2
Current enclosed within distance
from center
2 2
e
ex
xx I I
r
I Ix
H x r
2
inside 2 2A 0
3
40
However, situation is not as simple as outside wire
case since flux only links part of wire (need Biot-Savart law
to derive): d (length) d2
(length) d (length)2
r
r
Ix xx
r r
Ix
xr
B a
0
.8
r
I
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Line Total Flux & Inductance0 0
0
0
(per meter) ln2 8
(per meter) ln2 4
(per meter) ln2 4
Note, this value still goes to infinity as we let
go to infinity.Note that inductance depends on
r
Total
rTotal
r
R
I Ir
RI
r
RLr
R
logarithm
of ratio of lengths.
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Inductance Simplification
0 0 4
0 4
Inductance expression can be simplified usingtwo exponential identities:
ln( )=ln + ln ln ln ln ln( )
ln ln ln ln2 4 2
ln ln2
r
r
a
r
aab a b a b a e
bR
L R r er
L R re
0
4r
ln2 '
Where ' 0.78 for 1r
Rr
r r e r
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Two Conductor Line Inductance
Key problem with the previous derivation is weassumed no return path for the current. Nowconsider the case of two wires, each carrying thesame current I, but in opposite directions; assume
the wires are separated by distance D.
D
Creates counter-
clockwise field
Creates a
clockwise field
To determine the
inductance of each
conductor we integrate
as before. Howevernow we get some
field cancellation.
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Two Conductor Case, contd
D D
Direction of integration
R
Key Point: Flux linkage due to currents in each
conductor tend to cancel out.
Use superposition to get total flux linkage.
0 0left
For distance , greater than 2 , from left line
ln ln2 ' 2
R D
R R DI I
r D
Left Current Right Current
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Two Conductor Inductance
0left
0
0
0
0
Simplifying (with equal and opposite currents)
ln ln2 '
ln ln ' ln( ) ln2
ln ln2 '
ln as2 '
ln H/m2 '
left
R R DI
r D
I R r R D D
D RI
r R D
DI Rr
DL
r
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Many-Conductor Case
Now assume we now have kconductors, each with
current ik, arranged in some specified geometry.
Wed like to find flux linkages of each conductor.
Each conductors fluxlinkage, k, depends upon
its own current and the
current in all the other
conductors.
To derive the flux linkage for conductor 1, 1, well be integrating fromconductor 1 (at origin) to the right along thex-axis.
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Many-Conductor Case, contd
At point b the netcontribution to 1from ik, 1k, is zero.
Wed 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 of1k.
Point a is at distance d1k from conductor k.
Rk is the
distance
from con-
ductor k
to pointc.
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Many-Conductor Case, contd
0 1 21 1 2'
12 11
0
1 2' 12 11
01 1 2 2
1 1 2
0
1
ln ln ln ,2
1 1 1ln ln ln
2
ln ln ln2
As goes to infinity so the second
term from above can be written =2
nn
n
n n
n n
n
n
jj
RR Ri i id dr
i i id dr
i R i R i R
R R R R
i
1lnR
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Many-Conductor Case, contd
1
01 1 2'
12 11
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 1ln ln ln ,
2
System has self and mutual inductan
n
jj
nn
n n
i
i i id dr
L i L i L i
ce.
However, the mutual inductance can be canceled for
balanced 3 systems with symmetry. 32
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Symmetric Line Spacing 69 kV
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Line Inductance Example
Calculate the reactance for a balanced 3, 60Hztransmission 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 1ln( ) ln( ) ln( )2 '
a a b ci i ir D D
Since system is assumed
balanced
a b ci i i
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Line Inductance Example, contd
0a
0
70
3
6
Substituting , obtain:1 1
ln ln2 '
ln .2 '
4 10 5ln ln
2 ' 2 9.67 10
1.25 10 H/m.
Again note logarithm of ratio of distance between
p
a b c
a a
a
a
i i i
i ir D
Dir
DL
r
hases to the size of the conductor. 35
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Line Inductance Example, contd
6
6a
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
3.79
(this is the total per phase)
The reason we did NOT have mutual inductance
was because
aL
X
X
of the symmetric conductor spacing36
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Conductor Bundling
To increase the capacity of high voltage transmissionlines 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.
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Bundled Conductor Flux LinkagesFor the line shown on the left,
define dijas the distance betweenconductors iandj.
We can then determine kfor conductor k.
Assuming of the phase current flows
in each of the four conductors in
a given phase bundle, then for conductor 1:
18
12 13 14
0115 16 17
19 1,10 1,11 1,12
1 1 1 1ln ln ln ln
4 '
1 1 1 1ln ln ln ln2 4
1 1 1 1ln ln ln ln
4
a
b
c
i
r d d d
id d d d
i
d d d d
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Bundled Conductors, contd
1
412 13 14
01 1
415 16 17 18
14
19 1,10 1,11 1,12
Simplifying
1ln
( ' )
1ln
2( )
1ln
( )
a
b
c
i
r d d d
i
d d d d
i
d d d d
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Bundled Conductors, contd
14
12 13 14
1
12 1
1
1415 16 17 18 2 3 4
1 19 1
geometric mean radius (GMR) of bundle
( ' ) for our example
( ' ) in generalgeometric mean distance (GMD) of
conductor 1 to phase b.
( )
(
b
bb
b
b b b ab
c
R
r d d d
r d dD
d d d d D D D D
D d d
14
,10 1,11 1,12 2 3 4) c c c acd d D D D D
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Inductance of Bundle
01
0 01
01
If and
Then
1 1ln ln
2
ln 4 ln2 2
4 ln , which is the2
self-inductance of wire 1.
ab ac bc a b c
a a
b
ab b
b
D D D D i i i
i i
R DD D
I IR R
DLR
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Inductance of Bundle, contd
01
But remember each bundle has conductorsin parallel (4 in this example).
So, there are four inductances in parallel:
/ ln .2
Again note that inductance depends on the
logarithm of t
ab
b
DL L bR
he ratio of distance between phasesto the size of bundle of conductors.
Inductance decreases with decreasing distance between
phases and increasing bundle size. 42
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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 ofr= 1.24 cm. Except now assume
each phase has 4 conductors in a square bundle,
spaced 0.25 meters apart. What is the new inductanceper meter?
2 3
13 4
70
1.24 10 m ' 9.67 10 m
9.67 10 0.25 0.25 ( 2 0.25)
0.12 m (ten times bigger than !)
5ln 7.46 10 H/m
2 0.12
Bundling reduces inductance.
b
a
r r
R
r
L
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T i i T C fi ti
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Transmission Tower Configurations
The problem with the line analysis weve done
so far is we have assumed a symmetrical towerconfiguration.
Such a tower configuration is seldom practical.
Typical Transmission Tower
Configuration
Therefore ingeneral Dab
Dac Dbc
Unless something
was done this would
result in unbalanced
Phases.
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Transposition
To keep system balanced, over the length ofa 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. 45
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Line Transposition Example
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Line Transposition Example
47
Transposition Impact on Flux
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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 1ln ln ln
3 2 '
1 1 1 1ln ln ln
3 2 '
1 1 1 1ln ln ln
3 2 '
a b c
a b c
a b c
I I Ir d d
I I Ir d d
I I Ir d d
a phase in
position 1
a phase in
position 3
a phase in
position 2
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Transposition Impact, contd
13
13
12 13 230
a
13
12 13 23
Recognizing that
1(ln ln ln ) ln( )
3
We can simplify so1 1
ln ln'
2 1ln
a b
c
a b c abc
I Ir d d d
Id d d
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Inductance of Transposed Line
1
312 13 23
0 0a
70
Define the geometric mean distance (GMD)
Then for a balanced 3 system ( - - )
1 1ln ln ln2 ' 2 '
Hence
ln 2 10 ln H/m2 ' '
Again, logarithm of ratio
m
a b c
ma a a
m
m ma
D d d d
I I I
DI I Ir D r
D DLr r
of distance between phases
to conductor size. 50
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Inductance with Bundling
0a
70
If the line is bundled with a geometric meanradius, , then
ln
2
ln 2 10 ln H/m2
b
ma
b
m ma
b b
R
DI
RD D
LR R
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Inductance Example
Calculate the per phase inductance andreactance of a balanced 3, 60 Hz, 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.
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Inductance Example
13
12 13 231/ 3
4
13
12 13
1/ 3
0
7
,
(10 (2 10) 10) 12.6m,
'= 0.0078m,
( ' ) ,
( ' 0.3 0.3) 0.0888m,
ln2
9.9 10 H/m,
2 (1600m/mile) = 0.6 /mile.
r
m
b
ma
b
a a
D d d d
r r e
R r d d
r
DLR
X fL
53
Review of Electric Fields
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Review of Electric Fields
A
To develop a model for transmission line capacitancewe first need to review some electric field concepts.
Gauss's law relating electric flux to enclosed charge):
d = (integrate over closed surface)eq D a
2
where
= electric flux density, coulombs/m
d = differential area da, with normal to surface
A = total closed surface,
= total charge in coulombs enclosedeq
D
a
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Gausss Law ExampleSimilar to Amperes Circuital law, Gausss Law ismost useful for cases with symmetry.
Example: Calculate D about an infinitely longwire that has a charge density ofqcoulombs/meter.
Since D comesradially out,
integrate over the
cylinder bounding
the wire.D is perpendicular
to ends of cylinder.A
d 2
where radially directed unit vector
2
eD Rh q qh
q
R
r r
D a
D a a
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Electric Fields
The electric field, E, is related to the electric fluxdensity, D, by
D = E
where
E = electric field (volts/m) = permittivity in farads/m (F/m)
= o r
o = permittivity of free space (8.85410
-12
F/m) r = relative permittivity or the dielectricconstant
(1 for dry air, 2 to 6 for most dielectrics)
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Voltage Difference
P
P
The voltage difference between any two
points P and P is defined as an integral
V ,
where the integral is along any path
from point P to point P .
d
E l
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Voltage Difference
In previous example, , with radial.2
Consider points P and P , located radial distance and
from the wire and collinear with the wire.
Define to be the radial distance from the wir
o
q
R
R R
R
r r
E a a
e
on the path from points P to P , so2
Voltage difference between P and P (assuming = ) :
V ln2 2
o
o
R
Ro o
qd dR
R
Rq qdR
R R
E l
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Voltage Difference, contd
V ln2 2
So, if is positive then those points closer to the
charge have a higher voltage.
The voltage between two points (in volts)
is equal to the amount of ene
Repeating:
rg
R
Ro o
Rq qdRR R
q
y (in joules)
required to move a 1 coulomb charge
against the electric field between the two points.
Voltage is infinite if we pick one of the points to be
infinitely far away. 59
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Multi-Conductor Case
1
Now assume we have parallel conductors,each with a charge density of coulombs/m.
The voltage difference between our two points,
P and P , is now determined by superposition
1V ln
2
i
ni
iii
nq
Rq
R
where is the radial distance from point P
to conductor , and the distance from P to .
i
i
R
i R i
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Multi-Conductor Case, contd
=1
1 1
11
11 1
1
If we assume that 0 then rewriting
1 1 1V ln ln
2 2
We then subtract ln 0
1 1 1V ln ln2 2
As we move P to infinity, ln 0
n
ii
n n
i i i
ii in
ii
n ni
i iii i
i
q
q q RR
q R
Rq qR R
R
R
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Absolute Voltage Defined
1
Since the second term goes to zero as P goes toinfinity, we can now define the voltage of a
point w.r.t. a reference voltage at infinity:
1 1V ln2
This equation holds for any point as long a
n
iii
qR
s
it is not inside one of the wires!Since charge will mostly be on the surface
of a conductor, the voltage inside will equal
the voltage at the surface of the wire. 62
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Three Conductor Case
A
BC
Assume we have threeinfinitely 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 1ln ln ln2
ln2
a a b c
aa
V q q qr D D
q DV
r
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Line Capacitance
1 11
For a single capacitor, capacitance is defined as
But for a multiple conductor case we need to
use matrix relationships since the charge onconductor may be a function of
i i i
j
n
q C V
i V
q C
q
1 1
1
n
n nn n
C V
C C V
q C V
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Line Capacitance, contd
We will not be considering thecases with mutual capacitance. To eliminate
mutual capacitance we'll again assume we have
a uniformly transposed line, using similar argumentsto the case of inductance. For the previous
three conductor example:
2Since = Cln
aa a
a
qq V CDV
r
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Bundled Conductor Capacitance
1
1
12
Similar to the case for determining lineinductance when there are bundled conductors,
we use the original capacitance equation just
substituting an equiva
Note for the ca
lent radius
( )
p
n
c nb
n
R rd d
acitance equation we use rather
than ' which was used for in the inductance
equation
b
r
r R
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Line Capacitance, contd
1
13
1
12
-12o
For the case of uniformly transposed lines we
use the same GMR, , as before.
2
ln
where
( ) (note NOT ')
in air 8.854 10 F/m
n
m
mcb
m ab ac bc
c nb
D
CD
R
D d d d
R rd d r r
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Line Capacitance Example
Calculate the per phase capacitance and susceptanceof a balanced 3, 60 Hz, transmission line withhorizontal phase spacing of 10m using three conductorbundling with a spacing between conductors in thebundle of 0.3m. Assume the line is uniformlytransposed and the conductors have a a 1cm radius.
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Line Capacitance Example, contd1
3
13
12
11
11
8
(0.01 0.3 0.3) 0.0963 m
(10 10 20) 12.6 m
2 8.854 10 1.141 10 F/m12.6ln
0.0963
1 1
2 60 1.141 10 F/m
2.33 10 -m (not / m)
cb
m
c
R
D
C
X C
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Line Conductors
Typical transmission lines use multi-strand
conductors
ACSR (aluminum conductor steel reinforced)
conductors are most common. A typical Al. toSt. ratio is about 4 to 1.
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Line Conductors, contd
Total conductor area is given in circular mils.One circular mil is the area of a circle with adiameter of 0.001, and so has area 0.00052 square inches
Example: what is the area of a solid, 1diameter circular wire?Answer: 1000 kcmil (kilo circular mils)
Because conductors are stranded, theequivalent radius must be provided by themanufacturer. In tables this value is knownas the GMR and is usually expressed in feet.
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Line Resistance
-8
-8
Line resistance per unit length is given by
= where is the resistivityA
Resistivity of Copper = 1.68 10 -m
Resistivity of Aluminum = 2.65 10 -m
Example: What is the resistance in / mile of a
R
-8
2 2
1" diameter solid aluminum wire (at dc)?2.65 10 -m m
1609 0.084mile mile(0.0127) m
R
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Line Resistance, contd
Because ac current tends to flow towards thesurface of a conductor, the resistance of a lineat 60 Hz is slightly higher than at dc.
Resistivity and hence line resistance increase as
conductor temperature increases (changes isabout 8% between 25C and 50C)
Because ACSR conductors are stranded, actualresistance, inductance, and capacitance needsto be determined from tables.
73
ACSR Table Data (Similar to Table A 4)
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ACSR Table Data (Similar to Table A.4)
Inductance and Capacitance
assume a Dm of 1 ft.GMR is equivalent to r
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ACSR Data, contd
7
3
3 3
2 4 10 ln 1609 /mile
12.02 10 ln ln
12.02 10 ln 2.02 10 ln
mL
m
m
DX f L f
GMR
f DGMR
f f DGMR
Term from table,
depending on conductor type,
but assuming a one foot spacing
Term independent
of conductor, but
with spacing Dm
in feet.75
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ACSR Data, Cont.
0
6
To use the phase to neutral capacitance from table
21-m where
2 ln
11.779 10 ln -mile (table is in M -mile)
1 1 1
1.779 ln 1.779 ln M -mile
Cm
m
m
X CDf C
r
D
f r
Df r f
Term from table,
depending on conductor type,
but assuming a one foot spacing
Term independent
of conductor, but
with spacing Dm
in feet.76
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Dove Example
7
0.0313 feet
Outside Diameter = 0.07725 feet (radius = 0.03863)
Assuming a one foot spacing at 60 Hz
12 60 2 10 1609 ln /mile0.0313
0.420 /mile, which matches the table
For the capacitance
a
a
C
GMR
X
X
X
6 41 11.779 10 ln 9.65 10 -mile
f r
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Additional Transmission Topics
Multi-circuit lines: Multiple lines often share acommon transmission right-of-way. This DOES causemutual inductance and capacitance, but is oftenignored in system analysis.
Cables: There are about 3000 miles of underground accables 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 provideinsulation
inductance is lower
capacitance is higher, limiting cable length
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Additional Transmission topics
Ground wires: Transmission lines are usuallyprotected from lightning strikes with a ground
wire. This topmost wire (or wires) helps to
attenuate the transient voltages/currents thatarise 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!
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Additional Transmission topics
Shunt conductance: Usually ignored. A smallcurrent may flow through contaminants oninsulators.
DC Transmission: Because of the large fixed
cost necessary to convert ac to dc and then backto ac, dc transmission is only practical forseveral specialized applications
long distance overhead power transfer (> 400 miles)
long cable power transfer such as underwater
providing an asynchronous means of joiningdifferent power systems (such as the Eastern andERCOT grids).
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EE 369
POWER SYSTEM ANALYSIS
Lecture 7
Transmission Line Models
Tom Overbye and Ross Baldick
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A
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Announcements
For lectures 7 to 9 read Chapter 5.
HW 5 is problems 4.32, 4.33, 4.36, 4.38, 5.7,
5.16, 5.18; is due Thursday, September 29.
Midterm September 29 will cover through
material in Homework 4:
Bring in page of notes, calculator, writing
implements.
Homework 6 is 5.14, 5.23, 5.26, 5.27, 5.28,
5.33, 5.37, 5.43; due October 6.
82
T i i Li M d l
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Transmission Line ModelsPrevious lectures have covered how to calculate
the distributedseries inductance, shuntcapacitance, and series resistance oftransmission lines: That is, we have calculated the inductance L,
capacitance C, and resistance rper unit length, We can also think of the shunt conductance g per
unit length,
Each infinitesimal length dxof transmission lineconsists of a series impedance rdx + jLdxand a
shunt admittance gdx + jCdx, In this section we will use these distributed
parameters to develop the transmission linemodels used in power system analysis.
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Transmission Line Equivalent Circuit
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Transmission Line Equivalent Circuit
Our current model of an infinitesimal lengthof transmission line is shown below:
For operation at frequency , let
and (with usually equal 0)
z r j L
y g j C g
Units on
z and y are
per unit
length!
dx dx
dx
84
D i ti f V I R l ti hi
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Derivation of V, I Relationships
We can then derive the following relationships:
d ( ) d
d ( ( ) d ) d ( )
d ( ) d ( )( ) ( )
d d
V I x z x
I V x V y x V x y dx
V x I xz I x yV x
x x
dx dx
dx
85
Setting up a Second Order Equation
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g p q
2
2
2
2
d ( ) d ( )( ) ( )d d
We can rewrite these two, first order differential
equations as a single second order equationd ( ) d ( )
( )dd
d ( ) ( ) 0d
V x I xz I x yV xx x
V x I xz zyV x
xx
V x zyV xx
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V I R l ti hi td
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V, I Relationships, contd
Define the propagation constant as
where
the attenuation constant
the phase constant
yz j
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E ti f V lt
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Equation for Voltage
1 2
1 2 1 2
1 1 2 2 1 2
1 2
1 2
The general equation for is( )
Which can be rewritten as
( ) ( )( ) ( )( )2 2
Let and . Then
( ) ( ) ( )2 2
cosh( ) sinh( )
x x
x x x x
x x x x
VV x k e k e
e e e eV x k k k k
K k k K k k
e e e eV x K K
K x K x
88
R l H b li F ti
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Real Hyperbolic Functions
dcosh( ) dsinh( )sinh( ) cosh( )
d d
x xx x
x x
For real , the cosh and sinh functions have
the following form:
x
cosh( )x sinh( )x
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C l H b li F ti
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Complex Hyperbolic Functions
For complex
cosh( ) cosh cos sinh sin
sinh( ) sinh cos cosh sin
x j
x j
x j
90
D t i i Li V lt
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Determining Line Voltage
The voltage along the line is determined based uponthe current/voltage relationships at the terminals.
Assuming we know and at one end (say the
"receiving end" with and where = 0) we canR R
V I
V I x
1 2determine the constants and , and hence the
voltage at any point on the line.
K K
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Determining Line Voltage contd
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Determining Line Voltage, contd
1 2
1 2
1
1 2
2
( ) cosh( ) sinh( )
(0) cosh(0) sinh(0)
Since cosh(0) 1 & sinh(0) 0
( )( ) sinh( ) cosh( )
( ) cosh( ) sinh( )
where characterist
R
R
R RR
R R c
c
V x K x K x
V V K K
K V
dV xzI x K x K xdx
zI I z zK I
yyz
V x V x I Z x
zZ
y
ic impedance
92
Determining Line Current
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Determining Line Current
By similar reasoning we can determine ( )
( ) cosh( ) sinh( )
where is the distance along the line from thereceiving end.
Define transmission efficiency as ;
that is, efficiency means
RR
c
out
in
I xV
I x I x xZ
x
P
P
the real power out (delivered)
divided by the real power in.
93
Transmission Line Example
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Transmission Line Example
6 6
Assume we have a 765 kV transmission line with
a receiving end voltage of 765 kV(line to line),
a receiving end power 2000 1000 MVA and
= 0.0201 + 0.535 = 0.535 87.8 mile
= 7.75 10 = 7.75 10 90
RS j
z j
y j
3
S.0mile
Then
2.036 10 88.9 / mile
262.7 -1.1c
zy
zZ
y
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Transmission Line Example contd
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Transmission Line Example, cont d
*6
3
Do per phase analysis, using single phase power
and line to neutral voltages. Then
765 441.7 0 kV3
(2000 1000) 101688 26.6 A
3 441.7 0 10
( ) cosh( ) sinh( )
441,700 0 cosh(
R
R
R R c
V
jI
V x V x I Z x
3
3
2.036 10 88.9 )
443,440 27.7 sinh( 2.036 10 88.9 )
x
x
95
Transmission Line Example contd
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Transmission Line Example, cont d
Receiving end Sending end
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Lossless Transmission Lines
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Lossless Transmission Lines
For a lossless line the characteristic impedance, ,
is known as the surge impedance.
(a real value)
If a lossless line is terminated in impedance then:
Then so we get.
c
c
c
R
c R
R c R
Z
j l lZ
j c c
Z
VZ
I
I Z V
..
97
Lossless Transmission Lines
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Lossless Transmission Lines( ) cosh( ) sinh( ),
cosh sinh ,
(cosh sinh ).
( ) cosh( ) sinh( )
cosh sinh ,
(cosh sinh )
( )That is, for every location , .
( )
R R c
R R
R
RR
c
R R
c c
R
c
c
V x V x I Z x
V x V x
V x x
VI x I x x
ZV V
x xZ Z
V x xZ
V xx Z
I x
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Lossless Transmission Lines
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Lossless Transmission Lines
2 2
2
Since the line is lossless this implies that for
every location , ( ( ) ( )*)
( ( ) ( )*) ( | ( ) | ) | ( ) |
is constant so, ( ) and ( )
( )Define to be the "surge impedance loading"
c c c
R R
c
x V x I x
Z I x I x Z I x Z I x
V x V I x I
V x
Z
(SIL). If load power P > SIL then line
consumes vars; otherwise line
generates vars.
99
Transmission Matrix Model
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Transmission Matrix Model
Oftentimes were only interested in the terminalcharacteristics of the transmission line. Therefore
we can model it as a black box:
VS VR
+ +
- -
IS IRTransmissionLine
S R
S R
WithV VA B
I IC D
100
Transmission Matrix Model contd
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Transmission Matrix Model, cont d
With
Use voltage/current relationships to solve for , , ,
cosh sinh
cosh sinh
cosh sinh1
sinh cosh
S R
S R
S R c R
RS R
c
c
c
V VA BI IC D
A B C D
V V l Z I l V
I I l lZ
l Z lA B
l lC DZ
T
101
Equivalent Circuit Model
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Equivalent Circuit ModelWe will try to represent as a equivalent circuit
Next well use the T matrix values to derive the
parametersZ'and Y'that match the behaviorof the equivalent circuit to that of the T matrix.
We do this by first finding the relationship
between sending and receiving end for the
equivalent circuit.102
Equivalent Circuit Parameters
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Equivalent Circuit Parameters'
' 2
' '1 '
2
' '2 2
' ' ' '' 1 1
4 2
' '1 '
2
' ' ' '' 1 1
4 2
S RR R
S R R
S S R R
S R R
S R
S R
V V Y
V IZ
Z YV V Z I
Y YI V V I
Z Y Z YI Y V I
Z YZ
V V
Z Y Z YI IY
103
Equivalent circuit parameters
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Equivalent circuit parameters
We now need to solve for ' and '.
Solve for ' using element:
sinh '
Then using we can solve for '
' '= cosh 1
2
' cosh 1 1 tanh2 sinh 2
C
c c
Z Y
Z B
B Z l Z
A Y
Z YA l
Y l lZ l Z
104
Simplified Parameters
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Simplified Parameters
These values can be simplified as follows:
' sinh sinh
sinh with (recalling )
' 1tanh tanh
2 2 2tanh
2 with2
2
C
c
z l zlZ Z l l
y l zl
lZ Z zl zyl
Y l y l yl l
Z z l yll
YY yl
l
105
Simplified Parameters
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Simplified ParametersFor medium lines make the following approximations:
sinh' (assumes 1)
' tanh( / 2)
(assumes 1)2 2 / 2
50 miles 0.998 0.02 1.001 0.01
100 miles 0.993 0.09 1.004 0.04
20
lZ Z
l
Y Y l
l
sinhl tanh(l/2)Length
l l/2
0 miles 0.972 0.35 1.014 0.18 106
Three Line Models
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Three Line Models(longer than 200 miles)
tanhsinh ' 2use ' ,2 2
2
(between 50 and 200 miles)
use and2
(less than 50 miles)use (i.e., assume is zero)
ll Y Y
Z Zll
YZ
Z Y
Long Line Model
Medium Line Model
Short Line Model
107
Power Transfer in Short Lines
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Power Transfer in Short Lines
Often we'd like to know the maximum power thatcould be transferred through a short transmission line
V1 V2
+ +
- -
I1 I2TransmissionLine with
ImpedanceZS
12S
21
1
** 1 2
12 1 1 1
1 1 2 2 2
21 1 2
12 12
with , ,
( )
Z
Z Z
V VS V I V
Z
V V V V Z Z
V V VS
Z Z
108
Power Transfer in Lossless Lines
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Power Transfer in Lossless Lines
21 1 2
12 12 12
12 12
1 212 12
If we assume a line is lossless with impedance and
are just interested in real power transfer then:
90 (90 )
Since - cos(90 ) sin , we get
sin
Hence the maxi
jX
V V VP jQ
Z Z
V VP
X
1 212
mum power transfer is
Max V VPX
109
Limits Affecting Max. Power Transfer
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Thermal limits limit is due to heating of conductor and hence
depends heavily on ambient conditions.
For many lines, sagging is the limiting constraint.
Newer conductors limit can limit sag. Forexample, in 2004 ORNL working with 3Mannounced lines with a core consisting ofceramic Nextel fibers. These lines can operate at
200 degrees C.
Trees grow, and will eventually hit lines if theyare planted under the line.
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Tree Trimming: Before
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Tree Trimming: Before
111
Tree Trimming: After
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Tree Trimming: After
112
Other Limits Affecting Power Transfer
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g
Angle limits while the maximum power transfer occurs when
line angle difference is 90 degrees, actual limit issubstantially less due to multiple lines in the
system
Voltage stability limits
as power transfers increases, reactive losses
increase as I2
X. As reactive power increases thevoltage falls, resulting in a potentially cascadingvoltage collapse.