ECE 476 Power System Analysis Lecture 5:Transmission Line Parameters Prof. Tom Overbye Dept. of...

ECE 476 Power System Analysis

Lecture 5:Transmission Line Parameters

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

overbye@illinois.edu

Special Guest Lecturer: TA Won Jang

Announcements

• Please read Chapter 4• HW 2 is 2.44, 2.48, 2.49, 2.51

• It does not need to be turned in, but will be covered by an in-class quiz on Sept 10

• San Diego Gas & Electric is on campus for the ECE Career Fair on 9/9) (ARC Gym) and then for interviews on 9/10

2

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

3

Primary Methods for Power Transfer

• The most common methods for transfer of electric power are – Overhead ac– Underground ac– Overhead dc– Underground dc– other

4

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

H l

H

l

l

Algebraic sum of current linked by

5

Line Integrals

•Line integrals are a generalization of traditional integration

Integration along thex-axis

Integration along ageneral path, whichmay be closed

Ampere’s law is most useful in cases of symmetry, such as with an infinitely long line

6

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

B

B H

bility 1 for air

7

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

=

Ad

d

BA

B a

a

8

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 givenby the “Right-hand” Rule

Easiest way to solve the problem is to take advantage of symmetry. For an integration path we’ll choose acircle with a radius of x.

9

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

10

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 0left

For distance Rp, greater than 2R, from left line

ln ln2 ' 2

Rp Rp RI I

r R

Left Current Right Current11

Two Conductor Inductance

0left

0

0

0

0

Simplifying (with equal and opposite currents)

ln ln2 '

ln ln ' ln( ) ln2

ln ln2 '

ln as Rp2 '

ln H/m 2 'left

Rp Rp RI

r R

I Rp r Rp R R

R RpI

r Rp R

RI

r

RL

r

12

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, k, depends upon

its own current and the

current in all the other

conductors.To derive 1 we’ll be integrating from conductor 1 (at origin)

to the right along the x-axis. 13

Many-Conductor Case, cont’d

At point b the net

contribution to l1from ik , l1k, 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 l1k. Point a is at distance d1k from conductor k.

Rk is the

distance

from con-

ductor k

to point

c.

14

Many-Conductor Case, cont’d

0 1 21 1 2'

12 11

01 1 2'

12 11

01 1 2 2

1 1 2

0

1

ln ln ln2

1 1 1ln ln ln

2

ln ln ln2

As R goes to infinity R so the second

term from above can be written =2

nn

n

nn

n n

nn

jj

RR Ri i i

d dr

i i id dr

i R i R i R

R R

i

1ln R

15

Many-Conductor Case, cont’d

1

01 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 1ln ln ln

2

System has self and mutual inducta

n

jj

nn

n n

i

i i id dr

L i L i L i

nce. However

the mutual inductance can be canceled for

balanced 3 systems with symmetry.16

Symmetric Line Spacing – 69 kV

17

Birds Do Not Sit on the Conductors

18

Line Inductance Example

Calculate the reactance for a balanced 3, 60Hz

transmission line with a conductor geometry of an

equilateral triangle with D = 5m, r = 1.24cm (Rookconductor) and a length of 5 miles.

0 1 1 1ln( ) ln( ) ln( )

2 'a a b ci i ir D D

a

Since system is assumed

balanced

i b ci i

19

Line Inductance Example, cont’d

a

0a

0

70

3

6

Substituting

i

Hence

1 1ln ln

2 '

ln2 '

4 10 5ln ln

2 ' 2 9.67 10

1.25 10 H/m

b c

a a

a

a

i i

i ir D

Di

r

DL

r

20

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.

21

Bundled Conductor Pictures

The AEP Wyoming-JacksonFerry 765 kV line uses

6-bundle conductors.Conductors in a bundle areat the same voltage!

Photo Source: BPA and American Electric Power22

Bundled Conductor Flux Linkages

For the line shown on the left,

define dij as the distance bet-

ween conductors i and j. We

can then determine l for each

18

12 13 14

01

15 16 17

19 1,10 1,11 1,12

1 1 1 1ln ln ln ln

4 '

1 1 1 1ln ln ln ln

2 4

1 1 1 1ln ln ln ln

4

a

b

c

ir d d d

id d d d

id d d d

23

Bundled Conductors, cont’d

14

12 13 14

01 1

415 16 17 18

14

19 1,10 1,11 1,12

Simplifying

1ln

( ' )

1ln

2 ( )

1ln

( )

a

b

c

ir d d d

id d d d

id d d d

24

Bundled Conductors, cont’d

14

12 13 14

1

12 1

1

14

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

bb

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

14

,10 1,11 1,12 2 3 4) c c c acd d D D D D

25

Inductance of Bundle

a

01

0 01

01

If D and i

Then

1 1ln ln

2

ln 4 ln2 2

4 ln2

ab ac bc b c

a ab

ab b

b

D D D i i

i iR D

D DI I

R R

DL

R

26

Inductance of Bundle, cont’d

0a 1

But remember each bundle has b conductors

in parallel (4 in this example). So

L / ln2 b

DL b

R

27

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

13 4

b

70a

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!)

5L ln 7.46 10 H/m

2 0.12

r r

28

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

29

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.

30

Line Transposition Example

31

Line Transposition Example

32

Transposition Impact on Flux Linkages

0a

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”

33

Transposition Impact, cont’d

13

13

12 13 230a

13

12 13 23

Recognizing that

1(ln ln ln ) ln( )

3We can simplify so

1 1ln ln

'

2 1ln

a b

c

a b c abc

I Ir d d d

Id d d

34

Inductance of Transposed Line

13

m 12 13 23

0 0a

70

Define the geometric mean distance (GMD)

D

Then for a balanced 3 system ( - - )

1 1ln ln ln

2 ' 2 '

Hence

ln 2 10 ln H/m2 ' '

a b c

ma a a

m

m ma

d d d

I I I

DI I I

r D r

D DL

r r

35

Inductance with Bundling

b

0a

70

If the line is bundled with a geometric mean

radius, R , then

ln2

ln 2 10 ln H/m2

ma

b

m ma

b b

DI

R

D DL

R R

Inductance Example

Calculate the per phase inductance and reactance 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.

Answer: Dm = 12.6 m, Rb= 0.0889 m Inductance = 9.9 x 10-7 H/m, Reactance = 0.6 /Mile