٧٤

Capacitance of Transmission Line:

The capacitance between the conductors is the charge per unit of potential difference .

mFV

qC /

……………(25)

Capacitance between parallel conductors is a constant depending on the size and spacing of the conductors .

For the length of transmission line less than 80 Km , the effect of capacitance is usually neglected .

The flow of charge is a current , and this current is called the charging current of the line .

The charging current effects the voltage drop along the line , efficiency , power factor and stability of the power system .

Gauss theorem states that at any instant of time , the total electric flux through any closed surface (A) is equal to the total charge enclosed by that surface .

X +

٧٥

2/2

mcoulombx

qD

………..(26)

Where q - charge on the conductor coulomb /m x - distance , from the conductor to the point

where the electric flux density is computed. D - the electric flux density .

The electric field intensity = mediumtheoftypermittiviThe

densityfluxelectricThe

D

E mVx

q/

2

……….(27)

Where - actual permittivity of material .

0/ r

r - relative permittivity ; 0 - permittivity of free space

mF /

1094

11085.8

912

0

1- coulomb E-force on unit charge

Electric flux lines

q- coulomb/meter on line conductor

equipotential Electric surface flux lines

Electric field between two line conductors

٧٦

equipotential surface

1P

1D

q

2D 2P

Therefore , instantaneous voltage drop between 1P and

2P is :

2

1

12

D

D

dxEv

voltsD

Dqdx

x

qD

D 1

2ln22

2

1

…..(28)

Capacitance of a two wire line :

aq bq ar br

D

٧٧

From eq. ( 28)

voltr

Dqqtoduev

a

aaab ln

2

And voltage due to bq , calculated by :

voltr

Dqv

b

bba ln

2

baab vv 1

ln2

ln2

b

b

b

bab r

Dq

r

Dqv

voltD

rqqtoduev bb

bab ln2

By the principle of superposition the voltage drop from conductor (a) to conductor (b) due to charges on both conductors is the sum of the voltage drop caused by each charge alone .

voltD

rq

r

Dqv b

ba

aab

lnln

2

1

…..(29)

For two-wire line aq = - bq , so that : 2

2

ln2

ln2

ba

a

ba

aab

rr

Dq

rr

Dqv

٧٨

volts

rr

Dq

ba

a

ln

meterFarads

rr

Dv

qC

ba

ab

aab /

ln

If rrr ba ( radius of two conductors are same )

mF

rD

C ab /ln

………(30)

mFandIf r /

1094

1,1,

90

KmF

rD

mF

rD

C ab

/ln36

1

/ln36

10 9

…(31)

٧٩

a abC b a n b

naC nbC

abC - capacitance between conductors (a) and (b) .

naC - capacitance between conductor (a) and neutral .

nbC - capacitance between conductor (b) and neutral .

naC = nbC = nC = 2 abC

rD

rD

C n

ln18

1

ln36

12

KmF

rD

C n /ln

0555.0 ……(32)

Capacitive reactance between one conductor and neutral is :

nc Cf

X2

1

٨٠

r

D

ffrD

X c log106.6

2

10ln186

6

Where , 10log303.2ln

KmDfrf

X c /log106.61

log106.6 66

Capacitive reactance Capacitive reactance at 1 meter spacing spacing factor ( aX ) ( dX )

Potential difference between two conductors of a group of charged conductors :

The voltage drop between the two conductors is the sum of the voltage drops due to each charged conductor . bThe voltage drop abD bcD

between (a) to (b) : a acD c amD bmD cmD

m

٨١

volts

D

Dq

D

Dq

D

rq

r

Dq

v

ma

mbm

ca

cbc

ba

bb

a

aba

ab

ln......

...lnlnln

2

1

....…(33) In similar manner :

volts

D

Dq

D

rq

D

Dq

r

Dq

v

ma

mcm

ca

cc

ba

bcb

a

aca

ac

ln......

...lnlnln

2

1

Capacitance of three – phase line:

a- Equilateral spacing : b

D D By using eq. ( 33) a c 0 D

D

Dq

D

rq

r

Dqv cbaab lnlnln

2

1

D

rq

D

Dq

r

Dqv cbaac lnlnln

2

1

D

rqq

r

Dqvv cbaacab ln)(ln2

2

1

Volts

٨٢

But acb qqq

voltsr

Dq

r

Dq

D

rq

r

Dq

D

rq

r

Dqvv

a

aaa

aaacab

ln2

3

ln2

1lnln

2

1

lnln22

1

32

Phasor diagram of voltages cabcab vandvv , : b

30 abab vv abv

30cos2 anab V

v a 30 n bcv

anV2

3 cav

anab Vv 3 c

303 anab Vv

303 ancaac Vvv

303303 ananacab VVvv

3330cos23 anan VV

anV3

٨٣

voltr

DqV

r

DqV

aan

aan

ln2

ln2

33

neutraltomF

rDV

qC

an

an /

ln

2

1rFor

KmF

rD

rD

C n /ln

0555.0

ln18

1

(as in the case of single-phase lines , eq(32)

b) Unsymmetrical spacing but transposed :

1

31D 12D

2

3 23D

31

23

12

121 lnlnln

2

1

D

Dq

D

rq

r

Dqv cbaab ….(34)

٨٤

12

31

23

232 lnlnln

2

1

D

Dq

D

rq

r

Dqv cbaab

23

12

31

313 lnlnln

2

1

D

Dq

D

rq

r

Dqv cbaab

3321 ababab

ab

vvvv

312312

312312

312312

3

3312312

ln

lnln

6

1

DDD

DDDq

DDD

rq

r

DDDq

v

c

ba

ab

3312312

3312312 lnln

2

1

DDD

rq

r

DDDqv baab

If , 3312312 DDDDeq

eqb

eqaab D

rq

r

Dqv lnln

2

1

Similarly ,

eqc

eqaac D

rq

r

Dqv lnln

2

1

٨٥

As in section (a) :

acbanacab qqqandVvv 3

eqa

eqaan D

rq

r

Dqv lnln2

23

1

eqa

eqaan D

rq

r

Dqv lnln

23

12

2

3

3

ln23

1

r

Dqv eq

aan

volts

r

Dqv eqa

an ln2

mF

r

Dv

qC

eqan

an /

ln

2

KmF

r

D

r

DC

eqeqn /

ln

0555.0

ln18

1

…(35)

٨٦

Effect of earth on the capacitance of a line :

The presence of earth will change the electric flux lines and equipotential surfaces considerably , which in effect will change the effective capacitance between the wires,as shown in fig. below :

equipotential surface

Electric fluxlines

h

D

influence of ground on the electric field picture

٨٧

a b

aq bq conductor

h H Ground

h a b

aq bq Images of conductors

D (a) and (b)

H

hq

h

Hq

D

rq

r

Dqv babaab

2ln

2ln

2

1lnln

2

1

2222 4)2( hDhDHbut

22

22

4

2ln

2

4ln

2

1

lnln2

1

hD

hq

h

hDq

D

rq

r

Dqv

ba

baab

٨٨

hD

hDrq

hDr

hDqv baab 2

4ln

4

2ln

2

1 22

22

ba qqbut

)4(

4ln

2

1222

22

hDr

Dhqv aab

2

22 )4(1ln

2

1

hDr

Dqa

)4(1ln2

2

122 hDr

Dqa

volts

hDr

Dqa

)4(1

1lnln

22

mF

hDrDv

qC

ab

aab /

)4(1

1lnln

22

effect of earth term

٨٩

Charging current due to capacitance :

Single-phase line :

ampvcjcj

vI chg (

/

Where , f 2

c-capacitance between lines (Farads) v- phase voltage (volts)

Three – phase line :

)(/

ampsvcjcj

vI chg

Where , v- voltage to neutral (volts) c-capacitance to neutral (Farads)

Capacitance of bundle conductors

Instead of (r) , put bscD , where :

bscD - is the GMR of a bundle conductor, and can be

calculated as in inductance (page 67) .1- For two – conductor bundle :

)()()( 4 22 22dDdDdDD SSS

bSC

٩٠

2- For three – conductor bundle :

3 29 323 3 )()(2

dDdDddDD SSSbSC

3- For four – conductor bundle :

4 3

4 316 434 4

09.1

2)2()2(2

dD

dDdDdddDD

S

SSSbSC

IF conductor is solid , rDS in above three condition

Capacitance of 3-phase line , bundle conductor with equilateral spacing :

KmF

DD

C

bsc

nb /ln

0555.0

And for unsymmetrical spacing but transposed , the capacitance is :

KmF

D

DC

bsc

eqnb /

ln

0555.0

٩١

EX.1 A 3-phase , 50Hz , 132 KV overhead transmission lines has conductor diameter of 4 cm each , are arranged in a horizontal plane as shown in fig . supplies a balanced load , assume the line is completely transposed . Find the capacitance to neutral per phase per Km.

Phase A Phase B Phase C

4m 4m

Solution :

3

CABCABeqm DDDDD

m04.58443

cmrDS 2

KmF

D

DC

S

m

/01.0529.5

0555.0

102

04.5ln

0555.0

ln

0555.0

2

EX.2 A 3-phase , 50Hz , 400 KV overhead transmission lines are arranged in a horizontal plane , each phase has two – strand bundle conductors , the diameter of each strand is 25mm , as shown in the fig. below . Find the capacitance to neutral per phase per Km .

0.3m 25mm

6m 6m Phase A Phase B Phase C

٩٢

Solution :

3CABCABeqm DDDDD

m56.712663

mmrDS 5.12

2

25

dDD SbSC

m0612.03.0105.12 3

KmF

D

DC

bSC

eqnb /0115.0

816.4

0555.0

0612.056.7

ln

0555.0

ln

0555.0

Discussion Questions 1 – Discuss the effect of earth on the capacitance of a line :

2 – Derive in expression for the capacitance to neutral per phase per Km of a single phase overhead transmission line , taking into account the effect of earth .

3 – Derive in expression for the capacitance to neutral per phase per Km of a 3- phase overhead transmission line when conductors are of equilateral spacing .

٩٣

Tutorial Problems

Q1. A 3-phase , 50 Hz , 110 KV , overhead transmission line consists of three solid conductors of 3 cm diameter positioned on the corners of triangle with sides of 2 m , 2.5 m , 3.125 m .

If the conductor of each phase of this line is replaced by three-strand bundle conductor has the same equivalent area of one solid conductor , and the space between the strands of bundle is 0.2 m .

Find the capacitance to neutral per phase per Km for two conditions . Assume the line is transposed.

Q2. A 3-phase , 132 KV , 50 Hz , 100 Km , single circuit bundle conductor transmission line as shown in the figure below . If the diameter of each strand is 1 cm , and the conductors are regularly transposed . Determine , the capacitance to neutral per phase per Km .

6 m

5 m

0.18 m

٩٤

Q3. By using the method of images , prove that the capacitance between conductor and earth , taking into account the effect of earth for 3- phase equilateral spacing (as shown in fig. below) is :

conductor.theofradiustheisr)(and

Where

lnln

2

2

3

31

24

3

42

3

22

1

H

H

HH

H

HH

HH

r

DCan

b

D D

a D c

2H

1H Ground level

3H 4H

a D c

D D

b

٩٥

Performance of Transmission lines

For balance 3-phase system only 1-phase need be considered for analysis .

For purpose of analysis , transmission lines may be classified as :

a- Short line - The length of line is less than 80 Km.b- Medium line - The length of line is between 80 to

250 Km.c- Long line - The length of line is more than 250 Km.

Short Transmission line

The Fig. below , shown the equivalent circuit of a short line .

The effect of shunt capacitance is neglected . The series impedance can be taken a lumped .

rI LI

rV Load

rS VV & - phase voltage (rms value)For short line :

IIII LrS

…………………….(1)

٩٦

rrS IZVV

Eq. (1) can be written in matrix form as :

r

r

S

S

I

VZ

I

V

10

1 …………………(2)

Phasor diagram :

rV - as reference .

- Power angle ( load angle ) , it is the angle between

rS VV & , and is small ( 7to1 for short line ) .

SV IZ IX

s rV IR

III Sr

Phasor diagram per phase , any value taken from it is per phase .

Sr VV ;0

22 )sin()cos( IXVIRVV rrS

٩٧

222 )()()sincos(2 IXIRXRIVVV rrS

Percent voltage regulation of line (V.R) =100

r

ro

V

VV

(at rated current and at a given power factor )

oV - The magnitude of receiving end voltage at no-load.

rV - The magnitude of receiving end voltage at full-load.

With SV - constant .

For short line So VV

100line)short(forV.R

r

rS

V

VV

f

SV

IX o rV a b c IR

s III Sr 90

From phasor diagram :

oaof rS VV

٩٨

is very small ( ocof )

acoaoc rS VV

bcab

tely)(approximasincos XIRI + For lagging P.F.

- For leading P.F.

100

sincos.

rr

rS

V

IXIR

V

VVRV

V.R , depends on I , R ,X and P.F.Condition for zero voltage regulation ( approximately)

0sincos IXIR

V.R becomes zero at leading P.F.

sincos IXIR

X

Rtan

Also from phasor diagram :

IRV

IXV

r

rs

cos

sintan 1

s

٩٩

Medium Transmission line

The effect of shunt capacitance becomes more and more pronounced with the increase in the length of line .For medium length T.L , the shunt capacitance can be considered as lumped , by two type of equivalent circuit :

a) Nominal π circuit .b) Nominal T circuit .

Nominal π circuit

csI crI rI

rV

Z

YVIV

ZIVV

YVI

III

rrr

LrS

rr

crrL

2

2

١٠٠

2

YZVZIV rrr

)3.(..............................

21 ZI

YZV rr

rcrcsLcsS IIIIII

rrS IY

VY

V 22

rrrr IY

VY

ZIYZ

V

2221

21

4

2 YZI

ZYYV rr

)4......(....................

21

41

YZI

YZYV rr

Eq. (3) and (4) can be written in the matrix form as :

)5.(....................

21

41

21

r

r

S

S

I

V

YZYZY

ZYZ

I

V

١٠١

Phasor diagram for nominal π circuit

SV

csI crI IZ IX

rV

sI IR

I csI

rI crI

crI and csI are leads the rV and SV by 90

Angles :

- between sV and rV

- between rV and rI

s - between sV and sI

Nominal T circuit

In the nominal T circuit , the total capacitance of each conductor is concentrated at the center of the line , while the series impedance is split into two equal parts.

١٠٢

SV CV rV

)7........(..........4

12

1

221

2

22

)6.......(..........2

1

2

YZZI

YZV

ZYV

YZI

ZIV

ZI

ZIVV

YVYZ

I

YZ

IVI

YVIIII

rr

rrrr

srrS

rr

rrr

crcrs

Eq. (6) and (7) can be written in the matrix form as :

)8.....(..........

21

41

21

r

r

S

S

I

V

YZY

YZZ

YZ

I

V

١٠٣

Phasor diagram for nominal T circuit

2/ZIr 2/ZIS SV

2

XI S

cI CV 2

RI S

rV 2

XI r

sI 2

RI r

cI

rI

cI - Leads the phasor CV by90

Angles : - between sV and rV

- between rV and rI

s - between sV and sI

Long Transmission line

The line parameters are distributed uniformly over entire length . Assumption of lumped circuit analysis tails in the case of long length T.L.

١٠٤

Fig. below , shows distribution parameter line with series impedance (z) per unit length and shunt admittance (y) per unit length .

rI

rV

( Where z and y per unit length )

From Krichhoff 's voltage law :

)()()( xIxzxVxxV

)(

)()(xzI

x

xVxxV

)()()(

lim0

xzIx

xVxxVx

)9........(....................)()(

or xzIdx

xdV

١٠٥

By Krichhoff 's current law :

)()()( xxVxyxIxxI

Where )( xxVxy is the current flowing in the shunt admittance xy of the element .

)(

)()(xxyV

x

xIxxI

)(lim)()(

lim00

xxyVx

xIxxIxx

)10.......(....................)()(

or xyVdx

xdI

Differentiating eq. (9) and (10) with respect to x :

)11.(....................)()(

)()(

2

2

2

2

dx

xdVy

dx

xId

dx

xdIz

dx

xVd

Substituting values of dx

xdV )( and dx

xdI )( from eq.(9)

and (10) into eq.(11) :

)12.....(..........)()()( 2

2

2

xVxVzydx

xVd

١٠٦

)13.........(..........)()()( 2

2

2

xIxIzydx

xId

yzyz orWhere 2

- Propagation constant , it is complex quantity .

j

Kmradshiftphase

KmfactornAttenuatio

/

/Where

Linear differential equation of the type

solutioncompletitshas,

2

2

ykdz

yd

zkzk ebeay

Where a and b are constant to be evaluatedTherefore , solution of eq. (12) is :

)...(..........)( 14xx ebeaxV

Differentiating eq. (14) ,xx ebea

dx

xdV )(

and substituting the result in eq. (9) to give :

xx ebeaxIz )(

١٠٧

xx ez

be

z

axI )(or

xx e

z

zybe

z

zya

)(

/

1

//xxxx ebea

yze

yz

be

yz

a

)15(...............)(

1 xx

c

ebeaZ

Where yzZc / ohm , is called characteristic impedance , and it is complex quantity .

For lossless line cZ is called Surge impedance and

equal to CL / ohm .

For Over Head Transmission Line , 600400cZ

, and for underground cable 6040cZ

For eq. (14) and (15) , at x = 0 :

rr IxIVxV )(;)(

)(1

bacZcZ

b

cZ

arI

barV

١٠٨

2;

2cZrIrV

bcZrIrVa

xecZrIrVxecZrIrV

xV

22)(

xecZrIxerVxecZrIxerV

2222

)(16.........22 rIcZ

xexerV

xexe

xec

Zr

Ir

VxecZ

rI

rV

cZ

xI 22

1)(

)(17......22

1

rI

cZ

xexer

Vxexe

cZ

In term of hyperbolic functions , eq. (16) and (17) are:

rrc

rcr

IxVxZ

xI

IxZVxxV

)(cosh)(sinh1

)(

)(sinh)(cosh)(

……(18)

To obtain sending end values of voltage and current , we set (x) equal to (l) in eq. (18) :

١٠٩

rrc

S

rcrS

IlVlZ

I

IlZVlV

)(cosh)(sinh1

)(sinh)(cosh

……(19)

Eq. (19) can be written in matrix form :

The equivalent circuit of along line :

rI rI

rV rV

Nominal circuit Equivalent circuit

( approximately ) ( Exactly )

)20.....(..........coshsinh

1

sinhcosh

r

r

c

c

S

S

I

V

llZ

lZl

I

V

١١٠

Nominal circuit is approximate equivalent circuit , generally used for medium length T.L.

The results from nominal circuit are approximate , but the results from equivalent circuit are exact as by the long line equations .

For equivalent circuit :

ZI

ZYVV rrS

21

Compare this equation with equation (19) :

lZZ C sinh)1

lzy

llzlyz

sinhsinh/

)21(........................................sinh

l

lZZ

Where, Z = z l - Total series impedance of the line .

lZY cosh2

1)2

lZ

l

Z

lY

C

sinh

1cosh1cosh2/

l

lll

ZC

sinh

1cosh

2tanhWhere,

2tanh

1

١١١

)22.......(....................2/

2/tanh

22

l

lYY

Where Y = y l - Total shunt admittance of the line . Similar , an equivalent T circuit can be found for T.L.

Transmission efficiency : ( TL )

The ratio of receiving end power to the sending power of transmission line .

)23.....(100cos

cos)(efficiencyonTransmissi

SSS

rrTL IV

IV

Generalized Constants

In any four terminal network , the input voltage and input current can be expressed in term of output voltage and output current . T.L. is a 4-terminal network as shown :

sI rI

A B sV rV

C D 2 – port , 4 terminals network

The network should be : 1- Passive ; 2- Linear ; 3- Bilateral This condition is fully met in T.L.

١١٢

Therefore , equations ( 2,5,8,20 ) can be written in general form as :

)................. 24(

r

r

S

S

I

V

DC

BA

I

V

rrs IBVAV

rrs IDVCI ………………….(25)

The following points may be kept in mind :

1- Constants A,B,C, and D are all Phasors

DACCBBAA

;;; 2- AD

for all symmetrical line .

3- A

- Dimensionless .

B

- Ohms .

C

- Siemens ( mhos ) .

D

- Dimensionless .

4- 1 CBDA

The values of the generalized constants depend upon the particular method adopted for solving a T.L. , as shown in below table .

١١٣

Comparison of ( A,B,C,D ) constant for T.Ls :

A B C D

Short line 1

Z 0 A

Length <80Km(66Kv) ,Capacitance can be ignored , Parameters can be taken as lumped

21

ZY

Z

41

ZYY

A

Mediumline

T2

1ZY

41

ZYZ

Y A

Length 80-250Km ,Capacitance - is line to neutral per Km , Parameters can be taken as lumped ,Z- Total series

impedance of line ,Y – Total shunt admittance of line .

Long line lcosh lZc sinhl

Zc

sinh1

A

Length >250Km ,

ylYzlZ

Y

Z

y

zZ

ZYzlyll

zy

c

;

z–series impedance

per unit length .

= propagation constant . = j = Attenuation constant . = Phase constant ( rad. / unit length )

cZ = Characteristic impedance , for lossless line it is referred to as the surge impedance .

١١٤

Also , can be determined receiving end condition in term of sending end conditions :

)26(...............

S

S

r

r

I

V

AC

BD

I

V

Summary for long line :

ADlZ

C

lZBlA

Y

Z

y

zZ

ZYzlyllzyl

ylYzlZzy

c

c

c

;)(sinh1

)(sinh;)(cosh

;;

rcrS IlZVlV )(sinh)(cosh

rrc

S IlVlZ

I )(cosh)(sinh1

)sin()sinh()cos()cosh()cosh()(cosh bajbajbal )sin()cosh()cos()sinh()sinh()(sinh bajbajbal

)..........!4

)(

!2

)(1()(cosh

42

ll

l

Or

}..........

!5

)(

!3

)(){()(sinh

53

ll

ll

١١٥

:)Series( Cascade Two transmission Lines in

1SI 1rI 2SI 2rI

A1 B1 A2 B2 1SV 1rV 12SV 2rV

C1 D1 C2 D2

222111 22and11 rrSrrS IBVAVIBVAV

222111 22and11 rrSrrS IDVCIIDVCI

2

2

2

2

1

1

1

1

D2C2

B2A2and

D1C1

B1A1

r

r

S

S

r

r

S

S

I

V

I

V

I

V

I

V

2121 and SrSr IIVV

2

2

1

1

D2C2

B2A2

D1C1

B1A1

r

r

S

S

I

V

I

V

2

2

1

1

D1D2B2C1C2D1C1A2

B1D2A1B2B1C2A1A2

r

r

S

S

I

V

I

V

2

2

oo

oo

1

1

DC

BA

r

r

S

S

I

V

I

V

…………………….(27)

١١٦

:Two transmission Lines in Parallel

SI 1SI 1rI rI

A1 B1 SV rV

C1 D1

2SI 2rI

A2 B2 SV rV

C2 D2

21 SSS VVV ; 21 rrr VVV

21 SSS III ; 21 rrr III

)28(.....22and11 21 rrSrrS IBVAVIBVAV )29(.....22and11 2211 rrSrrS IDVCIIDVCI

From eq.(28) :

21 2211 rrrr IBVAIBVA

)21(2

1)(2

12)21(

1

11

12

BBIIB

IBIIB

IBIBAAV

rr

rrr

rrr

١١٧

)30..(..........21

)21(21 BB

AAVIBI rr

r

Put eq.(30) in eq.(28)

21

)21(211

BB

AAVIBBVAV rr

rS

rIBB

BBVr

BB

AABA

21

21

21

)21(11

rS IBB

BBVr

BB

BABAV

21

21

21

1221

21,Further SSS III Substituting values of 1SI and 2SI from eq. (29)

21 2211 rrrrS IDVCIDVCI )(2121 11 rrrr IIDIDVCC

r

rrr ID

BB

AAVIBDDVCC 2

21

)21(2)21(21

rrS IBB

BDDBV

BB

DDAACCI

21

2121

21

)21)(21(21

r

r

oo

oo

S

S

I

V

DC

BA

I

V

………………(31)

١١٨

Experimental Determination of Generalized Constants ( A , B , C , D ) :

1- Open circuit test : 0SI

0rI

A B

000

0

0SSS

S

S ZZI

V

0SV

C D 0rV

)0( 000 rrS IVAV

)0( 000 rrS IVCI

)32.(..........0

00

C

A

I

VZ

S

SS

2- Short circuit test : SSI

rI

SSSSSSSS

SS ZZI

V

SSV

A B 0rV

C D

)0( rrSS VIBV

)0( rrSS VIDI

)33.(..........D

B

I

VZ

SS

SSSS

١١٩

From eq. (32) and (33) :

)34.(..........1

0CDCD

CBDA

D

B

C

AZZ SSS

)35.(........../1

/ 2

0

0 ADACD

CA

ZZ

Z

SSS

S

)36(..........0

0

SSS

S

ZZ

ZA

)37(..........and CBDADA

From equations (34 , 36 and 37 ) can be

determined the constants ),,,( DCBA

.

١٢٠

.L.Power Flow and Power Circle Diagram of T

: er Circle Diagram Receiving End pow

rrs IBVAV

rrs IDVCI

………………….(38)

DACCBBAA

;;;

SSrr VVVV

;0

From eq.(38) :

B

VA

B

V

B

VAVI rSrS

r

The conjugate of rI

is :

B

VA

B

VI rS

r

The complex power at the receiving end ( The volt-amperes delivered to the load ) is given by :

rrrrr QjPIVS

١٢١

B

VA

B

VVQjP rrS

rr

2

……………..(39)

)(cos)(cos

2

B

VA

B

VVP rrS

r

)(sin)(sin

2

B

VA

B

VVQ rrS

r

…………………(40)

Where rP - Real power at receiving end .

rQ - Reactive power at receiving end .

j reactive

power axis rrr QjPS

(Var)

B

AVr2

B

VV rS

real power axis 0 (Watt)

١٢٢

rrrrr IVSS

rr QjP

rrrrrr IVjIV sincos

Var

rr IV K O r Watt

L

B

AVr2

B

VV rS

n d f (x , y)

rrrrr IVS OK

OLcos rrrr IVP

LKsin rrrr IVQ

١٢٣

Coordinates of the center of a receiving end circle :

WattrVB

A....)(x )(cos2Horizontal

VarrVB

A....)(y )(sin2Vertical

………………(41)

)..(circleendreceivingaofRadius (nk) VAB

rVSV

Units of Circle :

1- If voltages are phase to neutral in volts , P and Q ( and coordinates x , y ) are in Watts and Vars per phase.

2- If voltages are line-to-line in volts , the Watts and Vars on the diagram are total three-phase quantities .

3- If voltages are in Kilovolts from line-to-line , the coordinates given by above equation are total in MW , MVar for three-phase .

From circle diagram , maximum power is delivered if :

or

0

١٢٤

)(cos

2

.)(max B

VA

B

VVP rrS

r

df.)(max rP

( For maximum power , the load must draw a large leading current . )

If rV - constant , and SV -Variable , the following receiving-end power circles are obtained :

Q (Var) K3 Load line a K2 K1

b O P(Watt)

r

n

(x , y) 1SV 2SV 3SV

١٢٥

EX. (1) : A 3–phase , 50Hz . medium transmission line , the load at the receiving end is 75MVA at 0.8 power factor lagging with 132KV between lines . If at the sending end , the voltage is

9.65.165 KV , and the power is 68.44 MW at 0.7791 power factor lagging , find the total resistance , inductance and capacitance per phase of line .

Solution : We can be use nominal T circuit method :

SI

2/Z

2/Z

rI

CI

SV

Y

CV

rV

KVphaseVr 21.763

132/

KVVr

021.76

KVphaseVS

9.65.95

3

5.165/

KVj 5.119.94

AIr

9.361.3288.0cos1.328

101323

1075 1

3

6

Aj 9.1965.262

779.0cos5.306779.0105.1653

1044.68 1

3

6

SI

A9.315.306 Aj 46.1623.260

١٢٦

rrC IZ

VV

2

Volt9.361.3282

10021.76 3

Z

SCS IZ

VV

2

9.315.3062

9.361.3282

10021.76108.66.95 33

Z

Z

)4.1623.2609.1965.262(2

1021.7610)5.119.94( 33 jjZ

j

6328

6.31141.666.342

jZ

jZ

Therefore , the resistance per phase of line is = 28 ,

and the reactance per phase of line is = 63

Hf

X L 2.0502

63

2lineofphaseperinductancethe

C

rS

C

C

V

II

V

IY

9.361.3281.666.3410021.76

)9.1965.262()4.1623.260(3

jj

mho9010468.35.86297

35.8657.34 4

١٢٧

CX

Y1

FXf

C

X

C

C

2.11025.0502

1

2

1

1025.0104

1

4

44

EX. (2) A 3-phase , 50 Hz transmission line 160 Km long . delivers a load of 90 MW at 0.85 power factor lagging . The line voltage at the receiving end is 230 KV . The generalized circuit constants for the line are as follows :

01.9000503.0;47.772.85;3.09785.0 CBDA

Calculate :1- The sending end voltage , current , and power factor .2- Efficiency of transmission .

Solution :

Receiving end voltage / phase KVVr 8.1323

230)(

Receiving end current , AI r 4.2828.0102303

10903

6

KVVr

08.132

AI r

9.364.2828.0cos4.282 1

Sending end voltage per phase :

rrs IBVAV

9.364.28247.772.851008.1323.09785.0 3

57.4048.24060103.0130 3

١٢٨

373.156486.182766.680998.129 jj 97.163286.148274 j

KV28.6171.149 Sending end current :

rrs IDVCI

9.364.2823.09785.00108.13201.9000503.0 3

6.3632.27601.98.66 74.16483.22146.1097.65 jj 28.1548.287 j

A19.2854.326

47.3419.2828.6 S

Power factor of sending end = lagging824.047.34coscos S

Sending end power = SSS IV cos3

MW47.120824.054.326171.1493

100powerendSending

powerendReceivingontransmissiofEfficiency

%7410047.120

90

١٢٩

Discussion Questions

1- Draw the phasor diagram of a nominal -circuit of transmission line .

2- Prove that the voltage regulation (V.R) for short line is :

rV

IXIR sincosV.R

Where , I –load current , rV and cos are the receiving end voltage and power factor , R and X are resistance and reactance of line .

3- Prove that the voltage regulation (V.R) for short transmission line , unity power factor is :

rV

IRV.R

Where , I –load current , rV - receiving end voltage , R-resistance of line .

4- State the advantage of bundle conductor lines over single conductor lines .

5- What is the effect of unsymmetrical spacing of conductor in a 3-phase transmission line .

Tutorial Problems

Q1. A 3-phase , 132 KV , 50 Hz transmission line , 200 Km long has constants per phase per Km as follows : resistance = 0.26 Ω , inductive

reactance = j0.38 Ω , shunt susceptance = j 61075.2 mho.The power received is 1.8 MW at 0.8 power factor lagging , calculate : 1) The sending end power . 2) The percentage regulation.

Q2. A 3-phase load of 5 MW at a power factor of 0.9 lagging is to be transmitted over a distance of 100 Km with a receiving end voltage of 66 KV , 50 Hz . Each conductor has a resistance 0f 0.8 Ω /Km and radius of 26.1mm , the conductors have an equilateral spacing of 2 m

١٣٠

. Find the sending end voltage , current and power factor , by using :1) Nominal circuit method . 2) Nominal T circuit method .

Q3, A 3-phase transmission line , 10 Km long consists of three solid copper conductors are arranged at corner of an equilateral triangle of 0.5 m side . Load conditions at receiving end are 5 MW at 0.8 power factor lagging , 11 KV , 50 Hz . The efficiency of transmission is

90.9% and the resistivity of copper is 6107774.1 Ω cm . Calculate :

i) The sending end voltage and power factor .ii) The voltage regulation of the transmission line.

Q4. A 3-phase , 50 Hz , 100 Km long overhead transmission line has the following line constants :

Resistance per phase per Km = 0.153 Ω Inductance per phase per Km = 1.21 mH Capacitance per phase per Km = 0.00958 F

The line supplies a load of 20 MW at 0.9 power factor lagging at a line voltage of 132 KV . Calculate :

1- Sending end voltage and current .2- The voltage regulation of line .3- Efficiency of transmission .

Q5. A 3-phase , 50 Hz , 100 Km transmission line has conductors composed of 12 strands of Aluminum wound around a core of 7steel strands . Each of the aluminum and steel strands has a diameter of 0.28 cm . The resistivity and T constant for the

aluminum are given by 61085.2 Ω m at C20 , and 228respectively . The steel strands carry no current due to the skin effect . If the line conductors are placed at corners of an equilateral triangle of 2 m a side . Determine :

1- The conductor resistance per Km of the line at C402- The line inductance and capacitance . 3- The voltage and current at the sending end , if the receiving end

voltage is 132 KV .

Mechanical Design of Overhead System

: Type of line supports

The supports for an overhead line ( O.H.L) must be capable of carrying the load due to the conductors and insulators ( including the ice and wind loads on the conductors ) .The supports are of the following types :

: Poles -1 a - Wooden poles . b - Reinforced concrete poles . c – Steel pipes poles .

Poles used for high voltage (H.V) distribution and for low voltage (L.V) transmission line up to 33 KV.There are separate types of poles designed for single circuit and double circuit lines , figures below illustrate the various types of poles : FigSingle circuit pole double circuit pole

: Steel Towers -2Steel towers are extremely useful for long distance transmission lines running across open country where long spans are a decided advantage . There are separate types of towers designed for single circuit and double circuit lines , figures below illustrate the types of tower which are used for 66KV and 132 KV :Fig

Single circuit tower Double circuit tower

Figures below illustrate the types of tower which are used for voltage levels above 220 KV :Fig

:and Tension , Sag , Length Calculation of

A. Line Supported at equal levels :

Fig

d- Maximum Sag at mid span , in metres . S- length of section OP.W- Weight/unit length of conductor.l - Span ( distance between supports A and B ).L- Actual length of conductor between A and B

( AOB).

0T - Ultimate strength ( horizontal tension in the section OP at point O in Kg .

- Angle between the horizontal axis at point P with the tangent line at point P .

yX TT , - Horizontal and vertical components of tension in the section OP at point P .

BA TT , - Tension at the supporting points A and B.

dS dy

X

y

T

T

dx

dytan

dx

The part OP (S) under equilibrium :

SWT

TT

y

X

0

)1......(..........tan0T

SW

T

T

dx

dy

X

y

And for the infinitesimal part dS

222 dydxdS 22

1

dx

dy

dx

dS

2

0

1

T

SW

2

0

1

T

SW

dx

dS

2

0

1

T

SW

dSdx

2

0

0

0

1

TSW

TW

dSW

T

dx

Integrating both sides :

Asinh

0

10

T

SW

W

TX

Where , A – Constant of integration .

At X = 0 ( i.e. at point 0 ) , S = 0

0 A

0

10 sinhT

SW

W

TX

)2........(..........sinh

0

0

T

SW

W

TS

Length (L) of the conductor :

2;

2At

LS

lX

0

0

0

0

2sinh

2

2sinh

2

T

lW

W

TL

T

lW

W

TL

:followsasdrepresentebecan2

sinh0

T

lW

........!5

2!3

222sinh

5

0

3

000 T

lW

T

lW

T

lW

T

lW

.etc!52

Ignorting5

0

T

lW

!3

22

23

00

0

T

lW

T

lW

W

TL

6321!3

)3....(....................

241

20

22

T

lWlL