Lecture 9

Transmission Line Parameters

Reading: 4.1 – 4.6 ; 4.8 – 4.10

Homework 3 – will be posted on the course website

Dr. Lei Wu

Department of Electrical and Computer Engineering

EE 333

POWER SYSTEMS ENGINEERING

Outline

Develop simple model for transmission lines

Line resistance

Line conductance

Line inductance

Line capacitance

Analyze how the geometry of the transmission lines will affect the model parameters

2

Primary Methods for Power Transfer

The most common methods for transfer of electric power are

Overhead ac

Underground ac

Overhead dc

Underground dc

Others

3

Transmission line voltage structure

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)

4

Transmission line

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

Optional shield conductors, which protect against lightning

5

Transmission line

Shieldconductor

Insulator

Phaseconductor

Tower

69kVLine

CompositeInsulatorCrossarm

Compositeinsulator

Steel tower

Twoconductor

bundle

Shield conductor

6

Distribution line

Double circuit69 kV line

Distribution line12.47kV

Wooden tower

Shieldconductor

240V/120Vinsulated line

Transformers

Fuse cutout

Surge arrester

Insulator

7

Transmission line parameter calculation

Characteristics parameters (per unit length)

Series resistance (R)

Shunt conductance (G)

Series inductance (L)

Shunt capacitance (C)

R L

CG

8

Transmission line parameter calculation

Data needed for calculation

Conductor type

Conductor diameter and GMR

Number of conductors per bundle

Bundle spacing

Distance between phases

9

Line resistance

The resistance of the conductor is very important in transmission efficiency evaluation and economic analysis.

The dc resistance of a solid round conductor per unit length (a m or a mile) at a specific temperature is

ρ: conductor resistivity

A: conductor cross-sectional area

L :conductor length

, /dc T TR l Aρ= Ωi

10

Line resistance Example

What is the dc resistance (in Ω/mile) of a 1’’ diameter solid

aluminum wire?

1’’=2.54cm=0.0254meter

1 mile = 1609.344 meter

8min

, 22

1 2.65 10 1609.344 /0.084 /

0.02542

alu umdc T

m m mileR mile

Am

ρ

π

− Ω= = = Ω

i i i

i

8min 2.65 10alu um mρ −= Ωi

11

Line resistance

Stranded conductors are longer than corresponding

transmission line, thus higher resistance

Because ac current tends to flow towards the surface of a

conductor (skin effect), the ac resistance of a line is slightly

higher than the dc resistance.

Resistivity and hence line resistance increase as conductor

temperature increases.

22 1

1T T

T T

T Tρ ρ +

=+

12

Line conductance

Conductance accounts for real power loss between conductors or between conductors and ground. For overhead lines, this power loss is due to leakage currents at

insulators and to corona.

Conductance is usually neglected in power system studies. Losses due to insulator leakage and corona are usually small

compared to conductor I2R loss.

It is a very small component of the shunt admittance.

13

Line inductance

A conductor carrying current produces a magnetic field around

the conductor.

The relationship between current I and magnetic flux linkage λis represented by the inductance

IL

λ=

14

Line inductance

For a solid cylindrical conductor

0 ln2 '

DL

I r

µλπ

= =

0 00.25

ln ln2 2 '

D DIL I I

re r

µ µλπ π−= = =

70where 4 *10 H/m:permeability of free spaceµ π −=

15

Line inductance

For an array of M solid cylindrical conductor

Assume that the sum of the conductor currents is zero

The flux linking conductor k to P

due to current Ik

The flux linking conductor k to P

due to current Im

The total flux linking conductor k in an array of M conductors

0

1, 1

1ln where '

2

M M

k kPi i kki P i ik

I D rD

µλ λπ= →∞ =

= = =∑ ∑

0M

ii

I =∑

16

0 ln2 '

PkkPk k

k

DI

r

µλπ

=

0 ln2

PmkPm m

km

DI

D

µλπ

=

Line inductance

Single-phase two-wire line

0 ln2 '

x xx

x x

DL

I I r

λ λ µπ

= = =

0 ln2 '

y yy

y y

DL

I I r

λ λ µπ

= = =−

20 0 0 0ln ln ln ln

2 ' 2 ' 2 ' ' ' 'x y

x y x y x y

D D D DL L L

r r r r r r

µ µ µ µπ π π π

= + = + = =

17

Line inductance

Three-phase three-wire line

( )

0

0

0 0

1 1 1ln ln ln

2 '

1 1ln ln

2 '

1 1ln ln ln

2 ' 2 '

a a b ca

a b ca

a a aa a

I I Ir D D

I I Ir D

DI I I

r D r

µλπ

µπ

µ µπ π

= + +

= + +

= − =

0 ln2 'a

a

DL

r

µπ

=

18

Line inductance

General formula for calculating inductance

0 ln2

where GMD: geometric mean distance

GMR: geometric mean radius

GMDL

GMR

µπ

=

19

Line inductance

General formula for calculating inductance

0 ln2

xyx

xx

GMDL

GMR

µπ

=

0 ln2

yxy

yy

GMDL

GMR

µπ

=

x yL L L= +

20

Line inductance

GMR

( )( ) ( )2

11 12 1 21 22 2 1 2

ii

ij

where D ' 0.7788

D =distance between conductors i and j

nn n n n nn

i i

GMR D D D D D D D D D

r r

=

= =

⋯ ⋯ ⋯ ⋯

21

Line inductance

For symmetrical configurations

11 12 1n

nGMR D D D= ⋯

( )( )( )3 33 11 12 13 0.7788 2 2 1.4605GMR D D D r r r r= = =

( )( )( )( )4 44 11 12 13 14 0.7788 2 2 2 2

2.0654

GMR D D D D r r r r

r

= =

=

22