transmission lines ste 2013

7/27/2019 Transmission Lines Ste 2013

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EE 369

POWER SYSTEM ANALYSIS

Lecture 5

Development of Transmission Line ModelsTom Overbye and Ross Baldick

1


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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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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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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

43

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

74


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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

77


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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

78

dd l


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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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dd l


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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

81

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.

83

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


86/113

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

86

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

87

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

89

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

91

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

94

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

96

Lossless Transmission Lines


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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



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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

98



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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


102/113

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


105/113


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



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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.

110

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.

transmission lines ste 2013

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