ELE-B7

TRANSMISSION LINES

Development of Line ModelsGoals of this section are

1) Develop a simple model for transmission lines

2) Gain an intuitive feel for how the geometry of the transmission line affects the model parameters

Primary Methods for Power TransferThe most common methods for transfer of

electric power are 1) Overhead AC2) Underground AC3) Overhead DC4) Underground DC

Transmission lines and cablesExtra-high-voltage lines

Voltage: 345 kV, 500 kV, 765 kV

High-voltage linesVoltage: 115 kV, 230 kV

Sub-transmission linesVoltage: 46 kV, 69 kV

Distribution linesVoltage: 2.4 kV to 46 kV, with 15 kV being the most commonly used

High-voltage DC linesVoltage: ±120 kV to ±600 kV

Transmission lines and cablesThree-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; andOptional shield conductors, which protect against lightning

Transmission lines and cables

Shieldconductor

Insulator

Phaseconductor

Tower

69kVLine

CompositeInsulatorCrossarm

Compositeinsulator

Steel tower

Twoconductor

bundle

Shield conductor

Transmission lines and cables

Double circuit69 kV line

Distribution line12.47kV

Wooden tower

Shieldconductor

Distribution Line

Transmission lines and cables

Distribution line

Transformer

240V/120Vinsulated line

Transformers

Fuse cutout

Surge arrester

Insulator

Transmission lines and cables

Span

SagInsulator

Tension Tower

SupportingTower Tension Tower

Definition of Parameters

Aluminum Conductor Steel Reinforced (ACSR);All Aluminum Conductor (AAC); andAll Aluminum Alloy Conductor (AAAC).

ACSR Coductor

Aluminum outer strands2 layers, 30 conductors

Steel core strands,7 conductors

Transmission lines and cables

Transmission lines and cables

Locking Key

Insulator's Head

Expansion Layer

Imbedded Sand

Skirt

Petticoats

Iron Cap

Ball Socket

Compression Loading

Cement

Insulating Glass or Porcelain

BallCorrosion Sleeve for DC Insulators

Steel Pin

Insulators

Transmission lines and cablesInsulator Chain

Line Voltage

Number of Insulators per String

69 kV 4–6

115 kV 7–9

138 kV 8–10

230 kV 12

345 kV 18

500 kV 24

765 kV 30–35

Transmission lines and cablesComposite insulator.

(1) Sheds of alternating diameters prevent bridging by ice, snow andcascading rain.(2) Fiberglass reinforced resin rod.(3) Injection molded rubber (EPDM or Silicone) weather sheds and rodcovering. (4) Forged steel end fitting, galvanized and joined to rod by swaging process.

1 432

Transmission lines and cables

(1) is the clevis ball,(2) is the socket for the clevis,(3) is the yoke plate, and (4) is the suspension clamp. (Source: Sediver)•Figure 4.15 Line post-composite

insulator with yoke holding two conductors.

Transmission lines and cables

Transmission lines and cables

ConductorConductor shieldInsulation

Insulation shieldFiller

Copper screenPVC-sheet

Inductance of a Single WireThe inductance of a magnetic circuit has a constant permeabilitycan be obtained by determining the following:

a) Magnetic field intensity H, from Ampere’s law.b) Magnetic flux density B (B = μH)c) Flux linkage λd) Inductance from flux linkage per ampere

(L= λ/I)

Flux linkages within the wire :

mHI

LId

dxr

Ixdrxd

dxBd

r

xIHB

r

xIHIrxI

xIHIxH

IdlH

r

x

xx

xx

xxxx

enclosed

/102110

21

2

:flux by the linked iscurrent total theof (x/r)fraction only the Since

:islength unit per dflux aldifferenti The2

:conductor magnetic-nonFor 2

:ondistributicurrent uniform Assume2

)2(

7intint

7

0int

4

30

2

2

20

0

2

2

−− ×==⇒×∫ ==

=⎟⎠

⎞⎜⎝

⎛=

=

==

=⇒⎟⎠

⎞⎜⎝

⎛=

=⇒=

=∫ ⋅

λλλ

π

μφλ

φφ

π

μμ

π

ππ

Flux linkages outside of the wire :

mHDD

IL

DDId

dxxIdd

dxBd

xI

xIHB

xIHIxH

IdlH

D

D

x

xx

xx

enclosed

/21ln102

21ln102

102

:conductor theoutsideflux by the linked iscurrent entire theSince

:islength unit per dflux aldifferenti The

1022

104

2)2(

71212

72

112

7

770

⎟⎠

⎞⎜⎝

⎛×==⇒⎟⎠

⎞⎜⎝

⎛××∫ ==

×==

=

×=××==

=⇒=

=∫ ⋅

−−

−

−−

λλλ

φλ

φφ

ππμ

ππ

r x

D1 D2

Total flux linkage:

mHr

DI

L

re

r

DrDe

rD

pp

p

/ln102

r :where

lnI102 lnlnI102lnI102I1021

thenD, D andr D Since linkages.flux external and interanl theof sum theis

D distanceat Ppoint external out toconductor thelinking linkageflux Total

'7

4/1'

'7-4/17-7-7-

21

p

⎟⎟⎠

⎞⎜⎜⎝

⎛×==

=

××=⎟⎠

⎞⎜⎝

⎛+××=⎟

⎠

⎞⎜⎝

⎛××+×=

==

−

−

λ

λ

λ

Total flux linkage, cont.:

conductors M ofarray an in k conductor linkingflux total thegives :where

1ln102

:onmanipulati almathematic someAfter

lnI102

:is Imcurrent todue Ppoint out tok conductor links which and

lnI102

Icurrent todue Ppoint out tok conductor links which linkageflux the: thenzero, is currentsconductor theof sum theand Icurrent

carries mconductor each Assume .conductors solid M ofarray heconsider t Finally,

k

1

7-

m7-

kPm

'k7-

kkPk

m

λ

λ

λ

λ

λ

λ

km

M

mmk

km

Pmkpm

k

Pkkpk

DI

DD

r

D

×∑×=

××=

××=

=

Inductance of single-phase, two-wire

⎟⎟⎠

⎞⎜⎜⎝

⎛×=⇒==

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛×=⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛×=⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛+×=+=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛×==⎟

⎟⎠

⎞⎜⎜⎝

⎛×==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛××−=

=

⎟⎟⎠

⎞⎜⎜⎝

⎛××=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎠

⎞⎜⎝

⎛×−⎟

⎟⎠

⎞⎜⎜⎝

⎛××=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛×+⎟⎟

⎠

⎞⎜⎜⎝

⎛××=

−

−−−

−−

−

'7'y'x'

''7

''

27

''7

'7

'7

'7-

4/1'

'7-

'7-

yx7-

ln104rrr If

ln104ln102ln102

ln102 and conductor,per /ln102

ln102

r :where

ln102

1lnI1lnI102

1lnI1lnI102

:hence and validisrelation previous thezero is currents two theof sum theSince

r

DL

rr

D

rr

D

r

D

r

DLLL

r

DI

LmHr

DI

L

r

DI

similarlyre

r

DI

Dr

DD

yxyxyxyx

yy

yy

xx

xx

yy

xx

xx

xx

xyxxx

λλ

λ

λ

λ

λ

Inductance of three-phase, three-wire with equal phase spacing

⎟⎟⎠

⎞⎜⎜⎝

⎛×=

⎟⎟⎠

⎞⎜⎜⎝

⎛××=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛×−⎟⎟

⎠

⎞⎜⎜⎝

⎛××=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛×++⎟⎟

⎠

⎞⎜⎜⎝

⎛××=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛×+⎟

⎠

⎞⎜⎝

⎛×+⎟⎟

⎠

⎞⎜⎜⎝

⎛××=

'7-

'7-

a'a7-

b'a7-

cb'a7-

ln102

ln102

1lnI1lnI102

1ln)I(1lnI102

1lnI1lnI1lnI102

:hence and validisrelation previous thezero is currents three theof sum theSince

r

DL

r

DI

Dr

DI

r

DDr

a

aa

a

ca

a

λ

λ

λ

λ

Inductance of composite conductors

GMRDGMDDDD

IL

D

DI

DNMDNI

DNMDNI

N

DMI

DNI

NN

k

N

mkmMN

N

k

M

mkm

xx

xyxx

N

k NN

mkm

NMM

mkm

N

K

M

m km

N

m km

N

k

M

m km

N

m km

k

M

m km

N

m kmk

=∏ ∏==∏ ∏=×==

∏

⎟⎟⎠

⎞⎜⎜⎝

⎛∏

⎟⎟⎠

⎞⎜⎜⎝

⎛∏

××=

∑ ⎥⎦

⎤⎢⎣

⎡∑∑ −××∑ ==

⎥⎦

⎤⎢⎣

⎡∑∑ −××==

⎥⎦

⎤⎢⎣

⎡∑∑ −×=

= == =

−

=

=

=−

= ==

−

=

==

−

==

−

2

2

1 1xx

1 '1xy

7

1 /1

1

/1

'17x

1 '1127

1kx

'1127

k

k

'11

7

k

D and D : whereln102

ln102

1ln11ln1102

:isx conductor of linkageflux totalThe

1ln11ln1102

:isk or subconduct of linkageflux theflux, by this linked is Icurrent total theof (1/N)fraction only the Since

1ln1ln102

:is Xconductor ofk or subconduct linking flux totalThe

λ

λ

λλ

φλ

λ

φ

φ

Inductance of unequal phase spacing 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 ingeneral Dab ≠Dac ≠ Dbc

Unless something was done this wouldresult in unbalancedphases

Inductance of unequal phase spacing

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.

D

: whereln102

3 312312eq

7

DDD

DD

LS

eq

=

×= −

D12 D31

D23

Conductor BundlingTo 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.

Inductance of bundled conductors

:follows as calculated is D and D

: whereln102

SL3 312312eq

7

DDD

DD

LSL

eq

=

×= −

( ) ( )dDdD SS ×=×= D 4 2SL

d

d

dd

( ) ( )3 29 3SL D dDddD SS ×=××=

d

d ( ) ( )4 316 4SL 091.1 2D dDdddD SS ×=×××=

Electric field and voltage: Solid cylindrical conductor

The capacitance between conductors in a medium with constant permittivity can be obtained by finding the following:

a) Electric field from Gauss’s law.

b) Voltage between conductors.

c) Capacitance from charge per unit volt (C = q/V)

Gauss law: the total electric flux leaving a closed surface equals the total charge within the volume enclosed by the surface.

Electric field and voltage: Solid cylindrical conductor

1

22

1

2

112 ln

22

2))(2(

DDqdx

xqdxEV

xqEqLLxE

QEds

D

D

D

Dx

xx

enclosed

πεπε

πεεπ

ε

∫ =∫ ==

=⇒=

∫∫ =

Electric field and voltage: Solid cylindrical conductor

∑=

=

=

M

m km

immki

km

immkim

DDqV

DDqV

1

ki

ln2

1

:bygiven is charges all todue V voltage theiion,superposit Using

ln2

πε

πε

Assume each conductor m has a charge qm C/m, the voltage Vkim between conductors k and i due to the charge qm acting alone is:

Capacitance for single phase two wire line

⎟⎠⎞

⎜⎝⎛

=

==

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛==

====

=⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

rD

Cxy

A

rrDVxy

qCxy

DDDDq

DD

qDD

qVxy

yx

yyxx

xyyx

xy

yy

xx

yx

ln

then r,r r if nd

ln

then ,r D and rD D,DD Using

ln2

lnln2

1

yx

yyyxxxyxxy

πε

πε

πεπε

Assume conductor x has a uniform charge q C/m and conductor y has –q. Using the previous last equation with k = x, i = y and m = x, y

X Cxyy

Capacitance for three phase with equal phase spacing

)/ln(2

ln2

1

qqq with and

ln)(ln22

131

3

lnln2

1imilarly

lnln2

1lnlnln2

1

then ,r DD D,DDD Using

lnlnln2

1

acb

yyaacbcaab

rDC

rDqV

Drqq

rDqV

VVVDrq

rDqVac

sDrq

rDq

DDq

Drq

rDqVab

DDq

DDq

DDqVab

an

aan

cbaan

anacab

ca

bacba

ac

bcc

ab

bbb

aa

baa

πεπε

πε

πε

πεπε

πε

=

⎥⎦

⎤⎢⎣

⎡=

−=+

⎥⎦

⎤⎢⎣

⎡++⎟

⎠

⎞⎜⎝

⎛=

=+

⎥⎦

⎤⎢⎣

⎡+=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡++=

=====

⎥⎦

⎤⎢⎣

⎡++=

Capacitance for stranded, unequal phase spacing and bundled conductors

bundleconductor -fourfor 091.1

bundleconductor -for three

bundleconductor -for two

)/ln(2

4 3

3 2

3

rdD

rdD

rdD

DDDD

DDC

sc

sc

sc

acbcabeq

sceq

=

=

=

=

=πε

Line Resistance

-8

-8

Line resistance per unit length is given by

R = where is the resistivityA

Resistivity of Copper = 1.68 10 Ω-m

Resistivity of Aluminum = 2.65 10 Ω-mExample: What is the resistance in Ω / mile of a

ρ ρ

×

×

-8

2

1" diameter solid aluminum wire (at dc)?

2.65 10 Ω-m1609 0.0840.0127m

mRmile mileπ

× Ω= =

×

Line Resistance, cont’dBecause ac current tends to flow towards the surface of a conductor, the resistance of a line at 60 Hz is slightly higher than at dc.Resistivity and hence line resistance increase as conductor temperature increases (changes is about 8% between 25°C and 50°C)Because ACSR conductors are stranded, actual resistance, inductance and capacitance needs to be determined from tables.

Tow-Port Network Model

Vs

Is

VR

IR

Two-port

network

1=−+=+=

BCADDICVIBIAVV

RRs

RRs

Short transmission lines

Vs

Is

VR

IR

Z

oCZBDAII

ZIVV

Rs

RRs

==

===

+=

1

Medium transmission lines

⎟⎠

⎞⎜⎝

⎛ +=

=

+==

⎟⎠

⎞⎜⎝

⎛ ++⎟⎠

⎞⎜⎝

⎛ +=

++=

+⎟⎠

⎞⎜⎝

⎛ +=⎟⎠

⎞⎜⎝

⎛ ++=

41

21

21

41

V of value thesubsitute ,22

21

2

s

YZYC

ZB

YZDA

IYZVYZYI

YVYVII

ZIVYZYVIZVV

RRs

sRRs

RRR

RRs

Long transmission lines

S/m /m

CjGyLjRzωω

+=Ω+=

Long transmission lines, cont.

0)()(

)()()(

)()(:zero approachesx aslimit theTaking

)()()()()()()(

)()(:zero approachesx aslimit theTaking

)()()()()()()(

2

2

2

2

=−

==

=

Δ

=Δ

−Δ+Δ+Δ+=Δ+

=

Δ

=Δ

−Δ+Δ+=Δ+

xzyVdx

xVd

xzyVdx

xdIzdx

xVd

xyVdx

xdI

xyVx

xIxxIxxVxyxIxxI

xzIdx

xdV

xzIx

xVxxVxIxzxVxxV

RRc

RcR

RcRRcR

c

c

xx

xx

xx

IxVxZ

xI

IxZVxxVso

IZVAIZVA

yzZ

ZeAeAxI

xzIeAeAdx

xdVzy

eAeAxV

)cosh()sinh(1)(

)sinh()cosh()(:

2 and

2

ZA-AI(0) I and A A V(0) V Since

impedance. sticcharacteri thecalled is

)(

)()(constantn propagatio thecalled is

)(

21

c

21R21R

21

21

21

γγ

γγ

γγ

γ

γγ

γγ

γγ

+=

+=

−=

+=

==+==

=

−=

=−=

=

+=

−

−

−

Home WorkQ1:

Home WorkQ2: