Docs-EEEPSLNotes2013.pdf

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DEPARTMENT OF ELECTRICAL &ELECTRONICS ENGINEERING

COURSE: POWER SYETMS-II

BRANCH: Electrical and Electronics Engineering

CLASS: III/I Sem.

YEAR: 2013-2014

LECTURE NOTES

SHRI VISHNU ENGINEERING COLLEGE FOR WOMEN

VISHNUPUR, BHIMAVARAM-534202


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DEPARTMENT OF EEE - SVECW Page 1

CONTENTS Page No.

UNIT1: Transmission Line Parameters 02

UNITII: Performance of short and Medium Length 15

Transmission lines

UNITII: Performance of long transmission line 27

UNIT IV: Power system transients 32

UNITV: Various factor governing the performance of line 35

UNITV1: Overhead line Insulators 42


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UNIT-I

Transmission Line Parameters


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2. TRANSMISSION LINES

The electric parameters of transmission lines (i.e. resistance, inductance, and capacitance) can be

determined from the specifications for the conductors, and from the geometric arrangements ofthe conductors.

2.1 Transmission Line Resistance

Resistance to d.c. current is given by

AR

dc

where is the resistivity at 20o

Cis the length of the conductor

A is the cross sectional area of the conductor

Because of skin effect, the d.c. resistance is different from ac resistance. The ac resistance is

referred to as effective resistance, and is found from power loss in the conductor

Rpower loss

I 2

The variation of resistance with temperature is linear over the normal temperature range

resistance

( )

R2

R1

T T1 T2 temperature (oC)

Figure 9 Graph of Resistance vs Temperature

( )

( )

( )

( )

R

T T

R

T T

1

1

2

2

0 0


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RT T

T TR2

2

1

1

2.2 Transmission Line Inductive Reactance

Inductance of transmission lines is calculated per phase. It consists of self inductance of the

phase conductor and mutual inductance between the conductors. It is given by:

LGMD

GMR2 10 7 ln [H/m]

where GMR is the geometric mean radius (available from manufacturers tables)GMD is the geometric mean distance (must be calculated for each lineconfiguration)

Geometric Mean Radius: There are magnetic flux lines not only outside of the conductor, but

also inside. GMR is a hypothetical radius that replaces the actual conductor with a hollow

conductor of radius equal to GMR such that the self inductance of the inductor remains the same.

If each phase consists of several conductors, the GMR is given by

1 2

3

n

GMR d d d d d d d d d dn n n n nnn ( .... ).( . .... )......( . ..... )11 12 13 1 21 22 2 1 2

2

where d11=GMR1d22=GMR2

.

.

.

dnn=GMRn

Note: for a solid conductor, GMR = r.e-1/4 , where r is the radius of the conductor.


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Geometric Mean Distance replaces the actual arrangement of conductors by a hypothetical

mean distance such that the mutual inductance of the arrangement remains the same

b c a b

a

m c n

GMD D D D D D D D D Daa ab an ba bb bn ma mb mnmn ( ... ).( ... ).....( ..... )' ' ' ' ' ' ' ' '

'

where Daa is the distance between conductors a and a etc.

Inductance Between Two Single Phase Conductors

r1 r2

D

LD

r1

7

1

2 10 ln' L

D

r2

7

2

2 10 ln'

where r1 is GMR of conductor 1r2 is GMR of conductor 2D is the GMD between the conductors

The total inductance of the line is then

L L LD

r

D

r

D

r r

D

r r

LD

r r

D

r r

T

T

1 2

7

1 2

72

1 2

72

1 2

72

1 2

1 2

7

1 2

2 10 2 10 2 10 21

2

4 10 4 10

ln'

ln'

ln' '

ln' '

ln' '

ln' '

/

If r1 = r2 , then

L Dr

T 4 10 71

ln'


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Example: Find GMD, GMR for each circuit, inductance for each circuit, and total inductance

per meter for two circuits that run parallel to each other. One circuit consists of three 0.25 cm

radius conductors. The second circuit consists of two 0.5 cm radius conductor

9 m

a a

6 m 6 m

b b

6 m circuit B

c

circuit A

Solution:

m = 3, n = 2, m n = 6

GMD D D D B D Daa ab ba bb ca cb' ' ' ' ' '6

where

D D m

D D D m

D m

aa bb

ab ba cb

ca

' '

' ' '

'

9

6 9 117

12 9 15

2 2

2 2

GMD = 10.743 m

Geometric Mean Radius for Circuit A:

m481.0126e1025.0DDDDDDDDDGMR 9 243

41

23cccbcabcbbbaacabaaA

2

Geometric Mean Radius for Circuit B:

m153.06e105.0DDDDGMR 42

2

41

22'a'b'b'b'b'a'a'aB

2

Inductance of circuit A


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LGMD

GMRH mA

A

2 10 2 1010743

04816 2 12 10

7 7 7ln ln

.

.. /

Inductance of circuit B

LGMD

GMRH mB

B

2 10 2 1010 743

01538503 107 7 7ln ln

.

.. /

The total inductance is then

L L L H mT A B 14 715 107. /

The Use of Tables

Since the cables for power transmission lines are usually supplied by U.S. manufacturers, the

tables of cable characteristics are in American Standard System of units and the inductive

reactance is given in /mile.

X fL f GMD

GMRm

X fGMD

GMRm

X fGMD

GMRmile

X fGMD

GMRmile

X fGMR

f GMD mile

L

L

L

L

L

2 2 2 10

4 10

4 10 1609

2 022 10

2 022 101

2 022 10

7

7

7

3

3 3

ln /

ln /

ln /

. ln /

. ln . ln /

Xa Xd

If both, GMR and GMD are in feet, then Xa represents the inductive reactance at 1 ft spacing,

and Xd is called the inductive reactance spacing factor.

Example: Find the inductive reactance per mile of a single phase line operating at 60 Hz. The

conductor used is Partridge, with 20 ft spacings between the conductor centers.

D = 20 ft


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Solution: From the Tables, for Partridge conductor, GMR = 0.0217 ft and inductive reactance at

1 ft spacing Xa= 0.465 /mile. The spacing factor for 20 ft spacing is Xd = 0.3635 /mile. The

inductance of the line is then

X X X mileL a d 0465 03635 08285. . . /

Inductance of Balanced Three Phase Line

Average inductance per phase is given by:

LD

GMR

eq2 10

7ln

where Deq is the geometric mean of the three spacings of the three phase line.

D D D Deq ab ac bc3

Example: A three phase line operated at 60 Hz is arranged as shown. The conductors are ACSR

Drake. Find the inductive reactance per mile.

20ft 20 ft

38 ft

Solution:

For ACSR Drake conductor, GMR = 0.0373 ft

D ft

L H m

X mile

OR

from the tables X mile

eq

L

a

20 20 38 24 8

2 1024 8

0 037313 10

2 60 1609 13 10 0 788

0399

3

7 7

7

.

ln.

./

. /

. /

The spacing factor is calculated for spacing equal the geometric mean distance between the

conductors, that is, X miled 2 022 10 60 24 8 0 3893. ln . . /


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Then the line inductance is X X X mile per phaseline a d 0788. /

Example: Each conductor of the bundled conductor line shown in the figure is 1272 MCM

Pheasant. Find:

a) the inductive reactance in /km and /mile per phase for d = 45 cmb) the p.u. series reactance if the length of the line is 160 km and the base is 100 MVA, 345 kV.

8 m 8 m

Solution:

a) The distances in ft are

d ft

D ft

0 45

0 30481476

8

0 30482625

.

..

..

For Pheasant conductors, GMR = 0.0466 ft.

GMRb for a bundle of conductors is

GMR GMR d ftb 00466 1 476 0 2623. . .

The geometric mean of the phase conductor spacing is

D fteq 26 25 26 25 52 49 33 073 . . . .

The inductance of the line is then

LD

GMRH meq

b

2 10 2 10 330702623

9 674 107 7 7ln ln ..

. /

The inductive reactance is

X fL m km mileL 2 2 60 9 674 10 3 647 10 0 3647 0 58687 4. . / . / . /

b) Base impedance ZV

Sb

b

b

2 2345

1001190

Total impedance of the 160 km line is


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X

XX

Zp u

L

Lp uL

b

160 0 3 647 58 3 5

5835

11900049

. .

.. . .. .

2.3 Transmission Line Capacitive Reactance

Conductors of transmission lines act like plates of a capacitor. The conductors are charged, and

there is a potential difference between the conductors and between the conductors and the

ground. Therefore there is capacitance between the conductors and between the conductors and

the ground. The basic equation for calculation of the capacitance is the definition of the

capacitance as the ratio of the charge and the potential difference between the charged plates:

CQ

VF

where Q is the total charge on the conductors (plates)V is the potential difference between the conductors or a conductor and ground (i.e.

plates)

For transmission lines, we usually want the capacitance per unit length

Cq

VF m/

where q is the charge per unit length in C/m

V is the potential difference between the conductors or a conductor and ground (i.e.

plates)

Capacitance of a Single Phase Line

r1 r2

D

For a two conductor line, the capacitance between the conductors is given by

CD

r r

F m02

1 2

ln

/

where o is the permittivity of free space and is equal to 8.85 10-12 F/m

D is the distance between the conductors, center to centerr1 and r2 are the radii of the two conductors


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Formally, this equation corresponds to the equation for inductance of a two conductor line. The

equation was derived for a solid round conductor and assuming a uniform distribution of charge

along the conductors. The electric field, and therefore the capacitance of stranded conductors is

not the same as for solid conductors, but if the radii of the conductors are much smaller than the

distance between the conductors, the error is very small and an outside radii of the strandedconductors can be used in the equation.

For most single phase lines, r1 = r2 . In this case, half way between the conductors there is a point

where E = 0. This is the neutral point n

a n b

Can Cbn

The capacitance from conductora to point n is Can and is the same as the capacitance fromconductorb to n, Cbn. Can and Cbn are connected in series, therefore C C Can bn ab2 .

1

It follows that

CD

r

an

o2

ln

[F/m]

Since XfC

c

1

2

X

fD

r

f

D

rmc

o

1

22

2 8 62 109

ln

.ln

The capacitive reactance in mile is

Xf

D

r f

D

rmilec

2 862 10 1

1609

1779 109 6

.ln

.ln

Similarly as for inductive reactance, this expression can be split into two terms that are called

capacitive reactance at 1 ft spacing (Xa) and the capacitive reactance spacing factor (Xd).

1 C

C C

C C

C Cab

an bn

an bn

an bn

1

1 1If C Can bn , then C

C

C

Cab

an

an

an

2

2 2

Therefore, C Can ab2


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Xf r f

D milec1779 10 1 1779 106 6.

ln.

ln

Xa Xd

Xa is given in the tables for the standard conductors, Xd is given in the tables for the capacitivereactance spacing factor.

Example: Find the capacitive reactance in M miles for a single phase line operating at 60 Hz.

The conductor used for the line is Partridge, and the spacing is 20 ft.

The outside radius of the Partridge conductor is r in ft06422

0 0268. .

The capacitive reactance is

Xf

D

r fM mileC

1779 10 1779 10 20

0 026801961

6 6.ln

.ln

.. .

OR

From tables X M milea' . .01074

X M miled'

. .00889 for 20 spacing

X X X M mileC a d' ' . .01963

This is the capacitive reactance between the conductor and the neutral. Line-to-line capacitive

reactance is

XX

M mileCL L C

20 0981. .

Capacitance of Balanced Three Phase Line between a phase conductor and neutral is given by

CD

D

F mno

eq

b

2

ln

/

where D D D Deq ab bc ca3 and Dab, Dbc, and Dca are the distances between the centers of the

phase conductors, and Db is the geometric mean radius for the bundled conductors. (in the

expression for Db the outside radius of the conductor is used, rather than the GMR from the

tables.)


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The capacitive reactance to neutral than becomes

Xf

D

Dmilecn

eq

b

1779 106

.ln .

Example:a) A three phase 60 Hz line is arranged as shown. The conductors are ACSR Drake. Find the

capacitive reactance for 1 mile of the line.

b) If the length of the line is 175 miles and the normal operating voltage is 220 kV, find the

capacitive reactance to neutral for the entire length of the line, the charging current for the line,

and the charging reactive power.

20ft 20 ft

38 ft

Solution:

The outside radius for Drake conductors is r in ft1108

20 0462

..

The geometric mean distance for this line is

D fteq 20 20 38 24 83 .

From tables, X M milea' . .00912

Xf

D M mile

X X X M mile

d eq

cn a d

'

' '

.ln

.ln . . .

. .

1779 10 1779 10

6024 8 0 0 952

01864

6 6

This is the capacitive reactance to neutral.

For the length of 175 miles,

XX

Ctotalcn

1751065

Charging current is

I

V

X

k

ACLN

Ctotal

220

3

1065 119


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Reactive power to charge the line is

Q V I k MVAr C LL C3 3 220 119 45 45.

2.5 Transmission Line Losses and Thermal Limits

The power losses of a transmission line are proportional to the value of resistance of the line. The

value of the resistance is determined by the type and length of the conductor. The current in the

line is given by the power being delivered by the transmission line.

P E IR R equiv R cos IP

Eequiv

R

R Rcos

From that,

P I RP

ERloss equiv

R

R R

2

2

cos

Power utilities usually strive to maintain the receiving end voltage constant. The power delivered

by the transmission line is determined by the load connected to the line and cannot be changed

without changing the load. The only term in the above equation that can be regulated is the

power factor. If the power factor can be adjusted to be equal to 1, the power losses will be

minimum.

Efficiency of the transmission line is given by

%

P

P

R

S

100%

Thermal Limits on equipment and conductors depend on the material of the insulation of

conductors. The I2R losses are converted into heat. The heat increases the temperature of the

conductors and the insulation surrounding it. Some equipment can be cooled by introducing

circulation of cooling media, other must depend on natural cooling. If the temperature exceeds

the rated value, the insulation will deteriorate faster and at higher temperatures more immediate

damage will occur.

The power losses increase with the load. It follows that the rated load is given by the temperature

limits. The consequence of exceeding the rated load for short periods of time or by small

amounts is a raised temperature that does not destroy the equipment but shortens its service life.

Many utilities routinely allow short time overloads on their equipment - for example transformers

are often overloaded by up to 15% during peak periods that may last only 15 or 30 minutes.


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UNIT-II

Performance of Short and MediumLength Transmission Lines


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SHORT TRANSMISSION LINES

The transmission lines are categorized as three types

1) Short transmission linethe line length is up to 80 km

2) Medium transmission linethe line length is between 80km to 160 km

3) Long transmission linethe line length is more than 160 km

Whatever may be the category of transmission line, the main aim is to transmit power from one

end to another. Like other electrical system, the transmission network also will have some power

loss and voltage drop during transmitting power from sending end to receiving end. Hence,

performance of transmission line can be determined by its efficiency and voltage regulation.

power sent from sending endline losses = power delivered at receiving end

Voltage regulation of transmission line is measure of change of receiving end voltage from no-load to full load condition.

Every transmission line will have three basic electrical parameters. The conductors of the linewill have resistance, inductance, and capacitance. As the transmission line is a set of conductors

being run from one place to another supported by transmission towers, the parameters are

distributed uniformly along the line.

The electrical power is transmitted over a transmission line with a speed of light that is 3X10 8m sec. Frequency of the power is 50Hz. The wave length of the voltage and current of the power

can be determined by the equation given below,

f. = v where f is power frequency, &labda is wave length and v is the speed of light.


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Hence the wave length of the transmitting power is quite long compared to the generally used

line length of transmission line.

For this reason, the transmission line, with length less than 160 km, the parameters are assumed

to be lumped and not distributed. Such lines are known as electrically short transmission line.

This electrically short transmission lines are again categorized as short transmission line (lengthup to 80 km) and medium transmission line(length between 80 and 160 km). The capacitive

parameter of short transmission line is ignored whereas in case of medium length line the

capacitance is assumed to be lumped at the middle of the line or half of the capacitance may be

considered to be lumped at each ends of the transmission line. Lines with length more than 160

km, the parameters are considered to be distributed over the line. This is called long transmission

line.

ABCD PARAMETERS

A major section of power system engineering deals in the transmission of electrical power from

one particular place (eg. Generating station) to another like substations or distribution units withmaximum efficiency. So its of substantial importance for power system engineers to be thorough

with its mathematical modeling. Thus the entire transmission system can be simplified to a two

port networkfor the sake of easier calculations.

The circuit of a 2 port network is shown in the diagram below. As the name suggests, a 2 port

network consists of an input port PQ and an output port RS. Each port has 2 terminals to connect

itself to the external circuit. Thus it is essentially a 2 port or a 4 terminal circuit, having


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Supply end voltage = VS

and Supply end current = IS

Given to the input port P Q.

And there is the Receiving end Voltage = VR

and Receiving end current = IR

Given to the output port R S.

As shown in the diagram below.

Now the ABCD parameters or the transmission line parameters provide the link between the

supply and receiving end voltages and currents, considering the circuit elements to be linear innature.

Thus the relation between the sending and receiving end specifications are given using ABCD

parameters by the equations below.

VS = A VR+ B IR-(1)

IS = C VR+ D IR-(2)

Now in order to determine the ABCD parameters of transmission line let us impose the requiredcircuit conditions in different cases.

ABCD parameters, when receiving end is open circuited


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The receiving end is open circuited meaning receiving end current IR= 0.

Applying this condition to equation (1) we get.

Thus its implies that on applying open circuit condition to ABCD parameters, we get parameter

A as the ratio of sending end voltage to the open circuit receiving end voltage. Since dimension

wise A is a ratio of voltage to voltage, A is a dimension less parameter.

Applying the same open circuit condition i.e IR= 0 to equation (2)

Thus its implies that on applying open circuit condition to ABCD parameters of transmission

line, we get parameter C as the ratio of sending end current to the open circuit receiving end

voltage. Since dimension wise C is a ratio of current to voltage, its unit is mho.

Thus C is the open circuit conductance and is given by

C = IS VRmho.

ABCD parameters when receiving end is short circuited

Receiving end is short circuited meaning receiving end voltage VR= 0

Applying this condition to equation (1) we get

Thus its implies that on applying short circuit condition to ABCD parameters, we get parameter

B as the ratio of sending end voltage to the short circuit receiving end current. Since dimensionwise B is a ratio of voltage to current, its unit is . Thus B is the short circuit resistance and is


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given by

B = VS IR.

Applying the same short circuit condition i.e VR= 0 to equation (2) we get

Thus its implies that on applying short circuit condition to ABCD parameters, we get parameter

D as the ratio of sending end current to the short circuit receiving end current. Since dimension

wise D is a ratio of current to current, its a dimension less parameter. the ABCD parameters oftransmission line can be tabulated as:-

Parameter Specification UnitA = VS / VRVoltage ratio Unit less

B = VS / IR Short circuit resistance C = IS / VR Open circuit conductance mho

D = IS / IR Current ratio Unit less

SHORT TRANSMISSION LINEThe transmission lines which have length less than 80 km are generally referred as short

transmission lines.

For short length, the shunt capacitance of this type of line is neglected and other parameters like

resistance and inductance of these short lines are lumped, hence the equivalent circuit isrepresented as given below,

Lets draw the vector diagram for this equivalent circuit, taking receiving end current Irasreference. The sending end and receiving end voltages make angle with that reference receiving

end current, of sand r, respectively.


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As the shunt capacitance of the line is neglected, hence sending end current and receiving end

current is same, i.e.

Is = Ir.

Now if we observe the vector diagram carefully, we will get,

Vs is approximately equal to

Vr+ Ir.R.cosr+ Ir.X.sinr

That means,

Vs Vr+ Ir.R.cosr+ Ir.X.sinras the it is assumed that sr

As there is no capacitance, during no load condition the current through the line is considered aszero, hence at no load condition, receiving end voltage is the same as sending end voltage

As per dentition of voltage regulation,

Here, vrand vx are the per unit resistance and reactance of the short transmission line.

Any electrical network generally has two input terminals and two output terminals. If we

consider any complex electrical network in a black box, it will have two input terminals and

output terminals. This network is called twoport network. Two port model of a networksimplifies the network solving technique. Mathematically a two port network can be solved by 2

by 2 matrixes.

A transmission as it is also an electrical network; line can be represented as two port network.

Hence two port network of transmission line can be represented as 2 by 2 matrixes. Here the

concept of ABCD parameters comes. Voltage and currents of the network can represented as ,

Vs= AVr+ BIr(1)

Is= CVr+ DIr(2)

Where A, B, C and D are different constant of the network.

If we put Ir= 0 at equation (1), we get


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Hence, A is the voltage impressed at the sending end per volt at the receiving end when receivingend is open. It is dimension less.

If we put Vr= 0 at equation (1), we get

That indicates it is impedance of the transmission line when the receiving terminals are short

circuited. This parameter is referred as transfer impedance.

C is the current in amperes into the sending end per volt on open circuited receiving end. It has

the dimension of admittance.

D is the current in amperes into the sending end per amp on short circuited receiving end. It is

dimensionless.

Now from equivalent circuit, it is found that,

Vs = Vr+ IrZ and Is = Ir

Comparing these equations with equation 1 and 2 we get,

A = 1, B = Z, C = 0 and D = 1. As we know that the constant A, B, C and D are related for

passive network as

AD BC = 1.

Here, A = 1, B = Z, C = 0 and D = 1

1.1 Z.0 = 1

So the values calculated are correct for short transmission line.


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From above equation (1),

Vs = AVr+ BIr

When Ir= 0 that means receiving end terminals is open circuited and then from the equation 1,

we get receiving end voltage at no load

and as per definition of voltage regulation,

Efficiency of Short Transmission Line

The efficiency of short line as simple as efficiency equation of any other electrical equipment,

that means

MEDIUM TRANSMISSION LINE

The transmission line having its effective length more than 80 km but less than 250 km, is

generally referred to as a medium transmission line. Due to the line length being considerably

high, admittance Y of the network does play a role in calculating the effective circuit parameters,

unlike in the case of short transmission lines. For this reason the modelling of a medium length

transmission line is done using lumped shunt admittance along with the lumped impedance inseries to the circuit.

These lumped parameters of a medium length transmission line can be represented using two

different models, namely.

1) Nominal representation.

2) Nominal T representation.

Lets now go into the detailed discussion of these above mentioned models.

Nominal representation of a medium transmission line


25/46


In case of a nominal representation, the lumped series impedance is placed at the middle of the

circuit where as the shunt admittances are at the ends. As we can see from the diagram of the network below, the total lumped shunt admittance is divided into 2 equal halves, and each half

with value Y 2 is placed at both the sending and the receiving end while the entire circuitimpedance is between the two. The shape of the circuit so formed resembles that of a symbol ,

and for this reason it is known as the nominal representation of a medium transmission line. Itis mainly used for determining the general circuit parameters and performing load flow analysis.

As we can see here, VS and VR is the supply and receiving end voltages respectively, and

Is is the current flowing through the supply end.

IR is the current flowing through the receiving end of the circuit.

I1 and I3 are the values of currents flowing through the admittances. And

I2 is the current through the impedance Z.

Now applying KCL, at node P, we get.

IS = I1 + I2(1)

Similarly applying KCL, to node Q.I2 = I3 + IR(2)

Now substituting equation (2) to equation (1)

IS = I1 + I3 + IR

Now by applying KVL to the circuit,

VS = VR+ Z I2


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Comparing equation (4) and (5) with the standard ABCD parameter equations

VS = A VR+ B IR

IS = C VR+ D IR

We derive the parameters of a medium transmission line as:

Nominal T representation of a medium transmission lineIn the nominal T model of a medium transmission line the lumped shunt admittance is placed in

the middle, while the net series impedance is divided into two equal halves and and placed on

either side of the shunt admittance. The circuit so formed resembles the symbol of a capital T,

and hence is known as the nominal T network of a medium length transmission line and is shown

in the diagram below.

Here also Vs and Vr is the supply and receiving end voltages respectively, and

Is is the current flowing through the supply end. Ir is the current flowing through the receiving

end of the circuit. Let M be a node at the midpoint of the circuit, and the drop at M, be given by

Vm. Applying KVL to the above network we get


27/46


Now the sending end current is

Is = Y VM + IR(9)

Substituting the value of VM to equation (9) we get,

Again comparing Comparing equation (8) and (10) with the standard ABCD parameter equations

VS = A VR+ B IR

IS = C VR+ D IR

The parameters of the T network of a medium transmission line are


28/46


UNIT-III

Performance of Long TransmissionLines


29/46


LONG TRANSMISSION LINE

A power transmission line with its effective length of around 250 Kms or above is referred to as

a long transmission line. Calculations related to circuit parameters (ABCD parameters) of such

a power transmission is not that simple, as was the case for a short or medium transmission line.

The reason being that, the effective circuit length in this case is much higher than what it was forthe former models(long and medium line) and, thus ruling out the approximations considered

there like.

a) Ignoring the shunt admittance of the network, like in a small transmission line model.

b) Considering the circuit impedance and admittance to be lumped and concentrated at a point as

was the case for the medium line model.

Rather, for all practical reasons we should consider the circuit impedance and admittance to be

distributed over the entire circuit length as shown in the figure below.

The calculations of circuit parameters for this reason is going to be slightly more rigorous as we

will see here. For accurate modeling to determine circuit parameters let us consider the circuit of

the long transmission line as shown in the diagram below.


30/46


Here a line of length l > 250km is supplied with a sending end voltage and current of VS and IS

respectively, where as the VRand IRare the values of voltage and current obtained from the

receiving end. Lets us now consider an element of infinitely small length x at a distance x fromthe receiving end as shown in the figure where.

V = value of voltage just before entering the element x.

I = value of current just before entering the element x.

V+V = voltage leaving the element x.

I+I = current leaving the element x.

V = voltage drop across element x.

zx = series impedence of element x

yx = shunt admittance of element x

Where Z = z l and Y = y l are the values of total impedance and admittance of the long

transmission line.

the voltage drop across the infinitely small element x is given by

V = I z x

Or I z = V x

Or I z = dV dx (1)

Now to determine the current I, we apply KCL to node A.

I = (V+V)yx = V yx + V yx

Since the term V yx is the product of 2 infinitely small values, we can ignore it for the sake ofeasier calculation.

we can write dI dx = V y (2)

Now derevating both sides of eq (1) w.r.t x,

d2V d x2= z dI dx

Now substituting dI dx = V y from equation (2)


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d2V d x2 = zyV

or d2V d x2 zyV = 0 (3)

The solution of the above second order differential equation is given by.

V = A1 exyz + A2 e

xyz(4)

Derivating equation (4) w.r.to x.

dV/dx = (yz) A1 exyz (yz)A2 e

xyz(5)

Now comparing equation (1) with equation (5)

Now to go further let us define the characteristic impedance Zcand propagation constant of along transmission line as

Zc= (z/y)

= (yz)

Then the voltage and current equation can be expressed in terms of characteristic impedance and

propagation constant as

V = A1 ex

+ A2 ex

(7)

I = A1/ Zc ex

+ A2 / Zc ex

(8)

Now at x=0, V= VRand I= Ir. Substituting these conditions to equation (7) and (8) respectively.

VR= A1 + A2(9)

IR= A1/ Zc + A2 / Zc(10)

Solving equation (9) and (10),

We get values of A1 and A2 as,

A1 = (VR+ ZCIR) 2

And A1 = (VR ZCIR) 2

Now applying another extreme condition at x=l, we have V = VS and I = IS.

Now to determine VS and IS we substitute x by l and put the values of A1 and A2 in equation (7)

and (8) we get


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VS = (VR+ ZC IR)el

2 + (VR ZC IR)el

/2(11)

IS = (V

RZC + IR)el/2 (VR / ZC IR)e

l/2(12)

By trigonometric and exponential operators we know

sinh l = (el el) 2

And cosh l = (el + el) 2

equation(11) and (12) can be re-written as

VS = VRcosh l + ZC IRsinh l

IS = (VRsinh l)/ZC + IRcosh l

Thus comparing with the general circuit parameters equation, we get the ABCD parameters of a

long transmission line as,

C = sinh l ZC A = cosh l D = cosh l B = ZCsinh l


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UNIT IV

Power Sys tem Trans ients


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Bewley Lattice Diagram

This is a convenient diagram devised by Bewley, which shows at a glance the position and

direction of motion of every incident, reflected, and transmitted wave on the system at every

instant of time. The diagram overcomes the difficulty of otherwise keeping track of the

multiplicity of successive reflections at the various junctions.

Consider a transmission line having a resistance r, an inductance l, a conductance g and acapacitance c, all per unit length.

If is the propagation constant of the transmission line, and

E is the magnitude of the voltage surge at the sending end,

then the magnitude and phase of the wave as it reaches any section distance x from the sending

end is Ex given by.

Ex = E

where

e-.[

e-j [

. e- x = E . e - (+ j ) x = E e - x e - j x

represents the attenuation in the length of line x

represents the phase angle change in the length of line x

Therefore,

. attenuation constant of the line in neper/km

phase angle constant of the line in rad/km.

It is also common for an attenuation factorkto be defined corresponding to the length of a particular

line. i.e. k = e-.O

for a line of length l.

The propagation constant of a line RI D OLQH RI VHULHV LPSHGDQFH z and shunt

admittance y per unit length is given by

= z.y = (r + j l) (g + j c)

Similarly the surge impedance of the line (or characteristic impedance) Zo

Z0

= z= (r + j l)y (g + j c)

When a voltage surge of magnitude unity reaches a junction between two sections with surge

impedances Z1 and Z2, then a part . is transmitted and a part is reflected back. In traversing the

second line, if the attenuation factor is k, then on reaching the termination at the end of the second

line its amplitude would be reduced to N . . The lattice diagram may now be constructed as follows.

Set the ends of the lines at intervals equal to the time of transit of each line. If a suitable time scale is

chosen, then the diagonals on the diagram show the passage of the waves.High Voltage Transient Analysis

In the Bewley lattice diagram, the following properties exist. (1) All waves travel downhill, because time always increases.


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34

(2) The position of any wave at any time can be deduced directly from the diagram.

(3) The total potential at any point, at any instant of time is the superposition of all the

waves which have arrived at that point up until that instant of time, displaced in

position from each other by intervals equal to the difference in their time of arrival.

(4) The history of the wave is easily traced. It is possible to find where it came from

and just what other waves went into its composition.

(5) Attenuation is included, so that the wave arriving at the far end of a line

corresponds to the value entering multiplied by the attenuation factor of the line.

4.3.1 Analysis of an open-circuit line fed from ideal source

Let 2 is the time taken for a wave to travel from one end of the line to the other end of the

line (i.e. single transit time) and kthe corresponding attenuation factor.

Consider a step voltage wave of amplitude unity starting from the generator end at time t =0. Along the line the wave is attenuated and a wave of amplitude kreaches the open end attime 2. At the open end, this wave is reflected without a loss of magnitude or a change ofsign. The wave is again attenuated and at time 2 reaches the generator end with amplitude

k2. In order to keep the generator voltage unchanged, the surge is reflected with a change

of sign (-k2), and after a time 32 reaches the open end being attenuated to -k

3. It is then

reflected without a change of sign and reaches the generator end with amplitude -k4

and

reflected with amplitude +k4. The whole process is now repeated for the wave of amplitude

k4.


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35

UNIT-V

Various Factors Governing thePerformance of Transmission line


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36

SKIN EFFECT

The phenomena arising due to unequal distribution of current over the entire cross section of

the conductor being used for long distance power transmission is referred as the skin effect

in transmission lines. Such a phenomena does not have much role to play in case of a very

short line, but with increase in the effective length of the conductors, skin effect increasesconsiderably. So the modifications in line calculation needs to be done accordingly.

The distribution of current over the entire cross section of the conductor is quite uniform in

case of a dc system. But what we are using in the present era of power system engineering is

predominantly an alternating current system, where the current tends to flow with higher

density through the surface of the conductors (i.e skin of the conductor), leaving the core

deprived of necessary number of electrons. In fact there even arises a condition when

absolutely no current flows through the core, and concentrating the entire amount on the

surface region, thus resulting in an increase in the effective resistance of the conductor. This

particular trend of an ac transmission system to take the surface path for the flow of current

depriving the core is referred to as the skin effect in transmission lines

Why skin effect occurs in transmission lines ?

Having understood the phenomena of skin effect let us now see why this arises in case of an

a.c. system. To have a clear understanding of that look into the cross sectional view of theconductor during the flow of alternating current given in the diagram below.

Let us initially consider the solid conductor to be split up into a number of annular filaments

spaced infinitely small distance apart, such that each filament carries an infinitely small

fraction of the total current.

Like if the total current = I

Lets consider the conductor to be split up into n filament carrying current i such that I = n i .

Now during the flow of an alternating current, the current carrying filaments lying on the core

has a flux linkage with the entire conductor cross section including the filaments of the

surface as well as those in the core. Whereas the flux set up by the outer filaments is

restricted only to the surfaceitself and is unable to link with the inner filaments.Thus the flux

linkage of the conductor increases as we move closer towards the core and at the same rate

increases the inductance as it has a direct proportionality relationship with flux linkage. Thisresults in a larger inductive reactance being induced into the core as compared to the outer

sections of the conductor. The high value of reactance in the inner section results in the

current being distributed in an un uniform manner and forcing the bulk of the current to flow

through the outer surface or skin giving rise to the phenomena called skin effect in

transmission lines.


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37

Factors affecting skin effect in transmission lines.

The skin effect in an ac system depends on a number of factors like:-

1) Shape of conductor.

2) Type of material.

3) Diameter of the conductors.

4) Operational frequency.

FERRANTI EFFECT

In general practice we know, that for all electrical systems current flows from the region of

higher potential to the region of lower potential, to compensate for the potential difference

that exists in the system. In all practical cases the sending end voltage is higher than the

receiving end, so current flows from the source or the supply end to the load. But Sir S.Z.

Ferranti, in the year 1890, came up with an astonishing theory about medium or long distance

transmission lines suggesting that in case of light loading or no load operation of transmission

system, the receiving end voltage often increases beyond the sending end voltage, leading to

a phenomena known as Ferranti effect in power system.

Why Ferranti effect occurs in a transmission line?

A long transmission line can be considered to composed a considerably high amount of

capacitance and inductance distributed across the entire length of the line. Ferranti Effect

occurs when current drawn by the distributed capacitance of the line itself is greater than the

current associated with the load at the receiving end of the line( during light or no load). This

capacitor charging current leads to voltage drop across the line inductance of the transmission

system which is in phase with the sending end voltages. This voltage drop keeps on

increasing additively as we move towards the load end of the line and subsequently the


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38

receiving end voltage tends to get larger than applied voltage leading to the phenomena called

Ferranti effect in power system. It is illustrated with the help of a phasor diagram below.

Thus both the capacitance and inductance effect of transmission line are equally responsible

for this particular phenomena to occur, and hence Ferranti effect is negligible in case of a

short transmission lines as the inductance of such a line is practically considered to benearing zero. In general for a 300 Km line operating at a frequency of 50 Hz, the no load

receiving end voltage has been found to be 5% higher than the sending end voltage.

Now for analysis of Ferranti effect let us consider the phasor diagrame shown above.

Here Vris considered to be the reference phasor, represented by OA.

Thus Vr= Vr(1 + j0)

Capacitance current, Ic= jCVr

Now sending end voltage Vs = Vr+ resistive drop + reactive drop.

= Vr+ IcR + jIcX

= Vr+ Ic (R + jX)

= Vr+jcVr(R + j L) [since X = L]

Now Vs = Vr-2cLVr+ j cRVr

This is represented by the phasor OC.

Now in case of a long transmission line, it has been practically observed that the line

resistance is negligibly small compared to the line reactance, hence we can assume the length

of the phasor Ic R = 0, we can consider the rise in the voltage is only due to OAOC =reactive drop in the line.


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39

Now if we consider c0 and L0 are the values of capacitance and inductance per km of the

transmission line, where l is the length of the line.

Thus capacitive reactance Xc= 1/( l c0)

Since, in case of a long transmission line the capacitance is distributed throughout its length,the average current flowing is,

Ic =1/2 Vr/Xc =

1/2 Vr l c0

Now the inductive reactance of the line = L0 l

Thus the rise in voltage due to line inductance is given by,

IcX =1/2Vr l c0X L0 l

Voltage rise = 1/2 Vr2 l 2 c0L0

From the above equation it is absolutely evident, that the rise in voltage at the receiving end

is directly proportional to the square of the line length, and hence in case of a long

transmission line it keeps increasing with length and even goes beyond the applied sending

end voltage at times, leading to the phenomena called Ferranti effect in power system.

CORONA

Electric-power transmission practically deals in the bulk transfer of electrical energy, from

generating stations situated many kilometers away from the main consumption centers or the

cities. For this reason the long distance transmission cables are of utmost necessity for

effective power transfer, which in-evidently results in huge losses across the system.

Minimizing those has been a major challenge for power engineers of late and to do that one

should have a clear understanding of the type and nature of losses. One of them being the

corona effect in power system, which has a predominant role in reducing the efficiency of

EHV(extra high voltage lines) which we are going to concentrate on, in this article.

What is corona effect in power system and why it occurs?

For corona effect to occur effectively, two factors here are of prime importance as mentioned

below:-

1) Alternating potential difference must be supplied across the line.

2) The spacing of the conductors, must be large enough compared to the line diameter.


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40

Corona Effect in Transmission Line

When an alternating current is made to flow across two conductors of the transmission line

whose spacing is large compared to their diameters, then air surrounding the conductors

(composed of ions) is subjected to di-electric stress. At low values of supply end voltage,

nothing really occurs as the stress is too less to ionize the air outside. But when the potentialdifference is made to increase beyond some threshold value of around 30 kV known as the

critical disruptive voltage, then the field strength increases and then the air surrounding it

experiences stress high enough to be dissociated into ions making the atmosphere conducting.

This results in electric discharge around the conductors due to the flow of these ions, givingrise to a faint luminescent glow, along with the hissing sound accompanied by the liberation

of ozone, which is readily identified due to its characteristic odor. This phenomena of

electrical discharge occurring in transmission line for high values of voltage is known as the

corona effect in power system. If the voltage across the lines is still increased the glow

becomes more and more intense along with hissing noise, inducing very high power loss into

the system which must be accounted for.

Factors affecting corona effect in power system.

As mentioned earlier, the line voltage of the conductor is the main determining factor for

corona in transmission lines, at low values of voltage (lesser than critical disruptive voltage)

the stress on the air is too less to dissociate them, and hence no electrical discharge occurs.

Since with increasing voltage corona effect in a transmission line occurs due to the ionization

of atmospheric air surrounding the cables, it is mainly affected by the conditions of the cable

as well as the physical state of the atmosphere. Let us look into these criterion now with

greater details :-

Atmospheric conditions for corona in transmission lines.

It has been physically proven that the voltage gradient for di-electric breakdown of air isdirectly proportional to the density of air. Hence in a stormy day, due to continuous air flow

the number of ions present surrounding the conductor is far more than normal, and hence its

more likely to have electrical discharge in transmission lines on such a day, compared to a

day with fairly clear weather. The system has to designed taking those extreme situations into

consideration.


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41

Condition of cables for corona in transmission line

This particular phenomena depends highly on the conductors and its physical condition. It has

an inverse proportionality relationship with the diameter of the conductors. i.e. with the

increase in diameter, the effect of corona in power system reduces considerably.

Also the presence of dirt or roughness of the conductor reduces the critical breakdown

voltage, making the conductors more prone to corona losses. Hence in most cities and

industrial areas having high pollution, this factor is of reasonable importance to counter the ill

effects it has on the system.

Spacing between conductors.

As already mentioned, for corona to occur effectively the spacing between the lines should be

much higher compared to its diameter, but if the length is increased beyond a certain limit,

the di-electric stress on the air reduces and consequently the effect of corona reduces as well.

If the spacing is made too large then corona for that region of the transmission line might not

occur at all.


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42

UNIT-VI

OVERHEAD LINE INSULATORS


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43

There are mainly three types of insulator used as overhead insulator likewise

1. Pin Insulator

2. Suspension Insulator

3. Strain Insulator

In addition to that there are other two types of electrical insulator available mainly for lowvoltage application, e.i. Stay Insulator and Shackle Insulator.

Pin Insulator

Pin Insulator is earliest developed overhead insulator, but still popularly used in power

network up to 33KV system. Pin type insulator can be one part, two parts or three parts type,

depending upon application voltage. In 11KV system we generally use one part type insulator

where whole pin insulator is one piece of properly shaped porcelain or glass. As the leakage

path of insulator is through its surface, it is desirable to increase the vertical length of the

insulator surface area for lengthening leakage path. In order to obtain lengthy leakage path,

one, tow or more rain sheds or petticoats are provided on the insulator body. In addition tothat rain shed or petticoats on an insulator serve another purpose. These rain sheds or

petticoats are so designed, that during raining the outer surface of the rain shed becomes wet

but the inner surface remains dry and non-conductive. So there will be discontinuations of

conducting path through the wet pin insulator surface.

In higher voltage like 33KV and 66KV manufacturing of one part porcelain pin insulator

becomes difficult. Because in higher voltage, the thickness of the insulator become more and

a quite thick single piece porcelain insulator can not manufactured practically. In this case we

use multiple part pin insulator, where a number of properly designed porcelain shells are

fixed together by Portland cement to form one complete insulator unit. For 33KV tow parts

and for 66KV three parts pin insulator are generally used.

Designing consideration of Electrical Insulator

The live conductor attached to the top of the pin insulator is at a potential and bottom of theinsulator fixed to supporting structure of earth potential. The insulator has to withstand the

potential stresses between conductor and earth. The shortest distance between conductor and

earth, surrounding the insulator body, along which electrical discharge may take place

through air, is known as flash over distance.

1. When insulator is wet, its outer surface becomes almost conducting. Hence the flash overdistance of insulator is decreased. The design of an electrical insulator should be such that the

decrease of flash over distance is minimum when the insulator is wet. That is why the upper

most petticoat of a pin insulator has umbrella type designed so that it can protect, the rest

lower part of the insulator from rain. The upper surface of top most petticoat is inclined as

less as possible to maintain maximum flash over voltage during raining.

2. To keep the inner side of the insulator dry, the rain sheds are made in order that these rain

sheds should not disturb the voltage distribution they are so designed that their subsurface at

right angle to the electromagnetic lines of force.


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Suspension Insulator

In higher voltage, beyond 33KV, it becomes uneconomical to use pin insulator because size,

weight of the insulator become more. Handling and replacing bigger size single unit insulator

are quite difficult task. For overcoming these difficulties, suspension insulator was

developed.

In suspension insulator numbers of insulators are connected in series to form a string and

the line conductor is carried by the bottom most insulator. Each insulator of a suspension

string is called disc insulator because of their disc like shape.

Advantages of Suspension Insulator

1. Each suspension disc is designed for normal voltage rating 11KV(Higher voltage rating

15KV), so by using different numbers of discs, a suspension string can be made suitable for

any voltage level.

2. If any one of the disc insulators in a suspension string is damaged, it can be replaced much

easily.

3. Mechanical stresses on the suspension insulator is less since the line hanged on a flexible

suspension string.

4. As the current carrying conductors are suspended from supporting structure by suspension

string, the height of the conductor position is always less than the total height of the

supporting structure. Therefore, the conductors may be safe from lightening.


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Disadvantages of Suspension Insulator

1. Suspension insulator string costlier than pin and post type insulator.

2. Suspension string requires more height of supporting structure than that for pin or postinsulator to maintain same ground clearance of current conductor.

3. The amplitude of free swing of conductors is larger in suspension insulator system, hence,

more spacing between conductors should be provided.

Strain insulator

When suspension string is used to sustain extraordinary tensile load of conductor it is referred

as string insulator. When there is a dead end or there is a sharp corner in transmission line,

the line has to sustain a great tensile load of conductor or strain. A strain insulator must

have considerable mechanical strength as well as the necessary electrical insulating

properties.

Shackle Insulator or Spool Insulator

The shackle insulator orspool insulator is usually used in low voltage distribution network.

It can be used both in horizontal and vertical position. The use of such insulator has

decreased recently after increasing the using of underground cable for distribution purpose.

The tapered hole of the spool insulator distributes the load more evenly and minimizes the

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