CHAPTER 1

Power Systems Gate Material Vignan University

CHAPTER-1

PERFORMANCE OF TRANSMISSION LINES

1. TRANSMISSION LINE PARAMETERS

Basic Structure of a Power System

The basic structure of a power system is shown in Fig. 1.1.

Fig. 1.1 A typical power system.

It contains a generating plant, a transmission system, a subtransmission system and a

distribution system. These subsystems are interconnected through transformers T1 , T2 and T3 . Let us

consider some typical voltage levels to understand the functioning of the power system. The electric

power is generated at a thermal plant with a typical voltage of 22 kV (voltage levels are usually

specified line-to-line). This voltage is stepped up to 400 kV through transformer T1 for power

transmission. Transformer T2 steps this voltage down to 66 kV to supply power through the sub

transmission line to industrial loads that require bulk power at a higher voltage. Most of the major

industrial customers have their own transformers to step down the 66 kV supply to their desired

levels. The purpose of these change in voltage is to minimize the transmission ine cost for a given

power level. Distribution systems are designed to operate for much lower power levels and are

supplied with medium level voltages.

The power distribution network starts with transformer T3, which steps down the voltage from

66 kV to 11 kV. The distribution system contains loads that are either commercial type (like office

buildings, huge apartment complexes, hotels etc) or residential (domestic) type. Usually the

commercial customers are supplied power at a voltage level of 11 kV whereas the domestic

Dept. of Electrical Engineering


consumers get power supply at 400-440 V. Note that the above figures are given for line-to-line

voltages. Since domestic customers get single-phase supplies, they usually receive 230-250 V at their

inlet points. While a domestic customer with a low power consumption gets a single-phase supply,

both industrial and commercial consumers get three-phase supplies not only because their

consumption is high but also because many of them use three-phase motors. For example, the use of

induction motor is very common amongst industrial customers who run pumps, compressors, rolling

mills etc.

The main components of a power system are generators, transformers and transmission lines.

Necessity of Transmission Lines:- Bulk amount of powers generation could able to produce on

economical basis by employing synchronous generators at remote locations. The bulk amount of

power from the remote generating station could able to be carried out to the load centre by using

suitable network and the network is called transmission network (or) transmission lines.

Necessity of Extra High Voltages for transmission system:

1. The size of the conductor is reduced so that the cost of the conductor is reduced. Ex:- For the

same power, same length and same material the size of the conductor will become 1/n th of

original size as the operating voltage increased by n times.

2. The transmission line loss will be reduced Ex:- For the same power, same length, same

material the copper loss becomes 1/nth to that of original loss as the voltage increased by n

times.

Special Case: As the level of voltage is increasing beyond certain limit, for the same amount of

power to be transmitted, the extra cost required for insulation is more than the saving in conductor

cost, so the selection of operating voltage to transmit the power is the compromise between

in saving of conductor cost and extra cost required for insulation.

Selection of the size of conductor in transmission lines:- In transmission lines there are no

taping (or) consumers in the middle. So the current throughout the transmission line is same from

the generating station to the load centre. Hence the conductor is designed based on current

carrying capacity (or) constant current density. This is up to 220kv transmission line only.

In case of the voltages beyond 220kv, the concept of corona is predominant than that of current

carrying capacity. So the transmission lines are designed based on concept or corona



Types of Conductors:-

Mainly we are having four types of conductors.

1. Solid conductors. 2. Homogeneous Stranded conductors.

3. Composite Stranded conductors. 4. Bundled conductors

Solid Conductors:-

Fig 1.2 Solid conductor

Solid conductors is single piece of conductor

These conductors may be of copper (or) Aluminum

It is having high mechanical strength and tensile strength

Circular solid conductor will be preferred to make electrical field same

throughout the surface of the conductor.

There is no application of solid conductors as transmission lines

Flat (or) solid conductors are used for power transformer winding.

Homogeneous Stranded Conductors:- A number of strands are taken and are twisted

together to increase the current carrying capacity by maintaining the same operating voltage.

Fig 1.3 Stranded conductor

All strands are of same material

Compared to stranded conductor it is having high mechanical strength and

low tensile strength.

Stringing is easy in stranded conductor so transportation is easy

Composite Stranded Conductor:- Strands of different material are twisted together to form

the stranded conductor. Its purpose is to improve the tensile strength

Ex:- ACSR(Aluminium conductor steel reinforced)



As the layer size in increased the number of strands will be increases in the

following manner.

1+6+12+18+24+30

d- is the diameter of each conductor and ‘n’ is the number of layers then

diameter of stranded conductor is

D=(2n-1)d

Generally ACSR conductor is represented by x/y

x← number of Al strands (or) steel strands

y← number of Al strands (or) steel strands

In Power conductors the number of Aluminum ¿ number of steel strands

Bundled Conductors:-

When voltage is above 230KV, corona loss and interference with the communication

lines is more. Corona occurs when the surface potential gradient of a conductor exceeds the

dielectric strength of the surrounding air. This causes ionization of the area near the

conductor. The high voltage surface gradient is reduced by using two or more conductors per

phase in close proximity. This is called conductor bundling. The conductors are bundled in

groups of two, three or four as shown in below Fig. The conductors of a bundle are separated

at regular intervals with spacer dampers that prevent clashing of the conductors and prevent

them from swaying in the wind.

GMR=selfGMD=Ds ,2b=

4√ (D s×d )2=√D s×d

Fig: Two conductors per phase

D s, 3 b=

9√( Ds×d×d )3=3√D s×d×d

Fig: Three conductors per phase



D s, 4 b=

16√ (D s×d×√2d×d )4=1. 094√ Ds×d×d×d

Fig: Four conductors per phase

Parameters of Transmission lines:-

Parameters are:-

Resistance:- It is due to the property of conductor that it opposes the flow of current

Inductance:- Due to magnetic field around the conductor

Capacitance:- Due to electrostatic field around the conductor

Resistance:-

D.C resistance of a wire is given by Rdc=

ρlA

Ω

Where ρ is the resistivity of the wire in Ω - m, l is the length in meter and A is the

cross sectional area in m2. However the resistance of an overhead conductor is not the same

as that given by the above expression. When alternating current flows through a conductor,

the current density is not uniform over the entire cross section but is somewhat higher at the

surface. This is called the skin effect and this makes the ac resistance more than the dc

resistance. Moreover in a stranded conductor, the length of each strand is more that the length

of the composite conductor. Temperature also affects the resistivity of conductors.

R2

R1

=T+t2

T+t1

Where R1 and R2 are resistances at temperatures t1 and t2 respectively and T is a

constant that depends on the conductor material and its conductivity.

Inductance:- Current carrying conductor produces a magnetic field around it the magnetic

flux lines are concentric with their direction specified by Maxwells right hand thumb rule

L= λI



where L is the inductance in Henry, λ is the flux linkage in Weber-turns and I is the phasor

current in Ampere

Internal inductance per unit length is

Lint=12×10−7

Fig:- Cross section of a round conductor

Inductance between any two points outside the conductor

Lext=2×10−7 lnD2

D1 H/mt

Inductance of single phase line

L1=( 12+2×ln

Dr1

)×10−7

H/mt

L1=2×10−7 ln

D

r1'

H/mt r1' =0 .7788 r 1

Loop Inductance

r1=r2=rLloop=L1+L2=4×10−7 ln

D

r 'H/mt

Inductance of 3-phase line with Asymmetrical spacing (Transposed)

L1=2×10−7 ln3√ Dab Dbc Dca

r ' H/mt

GMD=3√Dab Dbc D caFig:- Portion of a transposed line

If lines are of equilateral spacing

Then Dab=Dbc=Dca=D



La=2×10−7 lnD

r 'H/mt

Capacitance:-

Capacitance is due to Electrostatic field around the conductor

C=qv

Where C is the capacitance in Farad, q is charge in coulombs, ν is potential difference

volts.

Capacitance of a single phase circuit:-

C12=πε

ln ( dr )

F/mt

Capacitance to neutral:-

C1n=2 πε

ln ( dr )

F/mt

Capacitance of 3-phase line with Asymmetrical spacing(Transposed):-

Can=2 πε

ln (GMDr )

F/mt

With equal spacing Then Dab=Dbc=Dca=D

GMD=3√Dab Dbc D ca=D


GMD=3√Dab Dbc D ca


Can=2 πε

ln (Dr )

F/mt

Effect of earth on transmission line capacitance:-

Earth surface is considered as a equipotential surface To considered the effect

of earth we are considering image conductors below the ground at a depth equal

to the height of the overhead conductor above the ground with opposite

charge(Method of images)

Cxy=πε

lnDr−ln

H xy

H xx F/mt

Cn=2 πε

lnDr

− lnH xy

H xx F/mt

If conductors are very much above the ground then Hxy=Hxx

With the effect of earth capacitance increases.

For Bundled conductors:-

L1=2×10−7 lnGMDGMR H/mt

Can=2 πε

ln (GMDGMR )

F/mt

S.No Line Description R L XL C Xc

1 Length increases Increases Increases Increases Increases Decreases

2Distance of

separation increasesNo relation Increases Increases Decreases Increases

3Radius of conductor

increasesDecreases Decreases Decreases Increases Decreases

4 Symmetrical spacing No relation decreases Decreases Increases Decreases

5Unsymmetrical

spacingNo relation Increases Increases Decreases Increases



6Effect of earth taken

into accountNo relation No change No change Increases Decreases

7Height of the

conductor increasesNo relation No change No change Decreases Increases

2. PERFORMANCE OF TRANSMISSION LINES:-

Synopsis:-

The important considerations in the operation of Transmission line are voltage drop,

Power loss, Efficiency

The performance of transmission lines depends on the R,L,C and conductor - G

Effect R - Voltage drop- IR and power loss -I2R

Effect L - Voltage drop-2π fl ( quadrature with the current)

Effect C - Charging current=

VXC

=V 2 π fc

Effect G - It is due to leakage current it is predominant only during bad

Whether conditions (so we are neglecting G)

In Overhead lines reactance effects are more

In Underground cables reactance effects are small and capacitance effects are more.

Efficiency:-

η= outputoutput+ losses

%η=Pr

Pr+3 I 2 R×100

(3-phase)

By knowing the efficiency of the system we can understand how much amount of

power loss in the form of heat.

Losses in the network depend on the resistance.

In case of transmission lines R is less (area of cross section of conductor is more) so

power loss in transmission lines is very less (about 2-5%).

In distribution network size of conductor is less so resistance is more losses will be

more, so losses in the distribution system will be more(about 8-10%)



Voltage Regulation:- It is the change of voltage at receiving end from No load to full load

while keeping the sending end voltage constant and also supply frequency made as constant

|V ro|= No load receiving end voltage

|V r|= Full load receiving end voltage

∈=|V ro|−|V r|

As the load at the load end increases from no load to full load the mechanical

input will also increases. So that the the mechanical input is equal to the electrical

output which will result as the synchronous machine is working at synchronous

speed.

However the excitation will remain same so that the sending end voltage is

Constant

Percentage regulation 2.1

The performance of Transmission will be analyzed in % regulation of line

rather than actual regulation

By knowing the regulation of line we can understand the change of voltage

from no load to full load.

Voltage drop =|V s|−|V r|

In transmission lines due to shunt capacitance the regulation of line is not same as

voltage drop. Because |V Ro|∉|V s|

The performance of transmission line will be studied in terms of regulation rather

than the efficiency because of shunt capacitance effect of the line.

Wavelength of transmission line:-

For f =50Hz (where f is the frequency of supply)

fλ=υ


%∈=|V ro|−|V r|

|V r|×100


λ = wavelength i.e the length of the line

υ =velocity of the wave=3×108m/sec

∴ λ= υ

f=3×108

50=6000

km 2.2

Classification of Transmission lines:-

Transmission lines are classified based on the physical length of the line at specified

frequency (50Hz/60Hz)

1. Short Transmission lines length<80KM Capacitance effect is ignored

(R and L are lumped)

2. Medium Transmission lines length (80-250)KM Capacitance effect is considered

(R,L and C are lumped)

Long Transmission lines length>250KM Capacitance effect is

considered

(R,L and C are lumped for steady state)

(R,L and C are lumped for Transient behaviour)

ABCD PARAMETSRS:-

Consider the power system shown in Fig. 2.1. In this the sending and receiving end

voltages are denoted by VS and VR respectively. Also the currents IS and IR are entering and

leaving the network respectively. The sending end voltage and current are then defined in

terms of the ABCD parameters as

2.2

I S=CV R+DI R 2.3


V S=AV R+BI R


Fig. 2.1 Two port representation of a transmission network.

VS ,IS and VR,IR are per phase quantities.

If A=D the network is symmetric

If AD-BC=1 the network is Reciprocal (or) bilateral

A network is said to be Reciprocal the ratio of response to excitation will be

same even though their positions are interchanged.

Open circuit at receiving end:-

From (2.2) we see that

A=V s

V R

|I R=0 2.4

This implies that A is the ratio of sending end voltage to the open circuit receiving end

voltage. This quantity is dimension less.

C=I s

V R

|I R=0 =Open circuit admittance in mhos

C=Y=Wc

2.5


[V S

I S]=[A B

C D ] [V R

I R]

[V R

I R]=[A B

C D ]−1 [V S

I S]

V S

I S

= AC


%∈=|V Ro|−|V R|

|V R|×100

%∈=|V S

A|−|V R|

|V R|×100

Short Circuit at receiving end:-

B=V s

I R

|V R=0Ω

2.6

i.e., B, given in Ohm, is the ratio of sending end voltage and short circuit receiving end

current. In a similar way we can also define

D=I s

I R

|V R=0 (unit less) 2.7

Short circuit impedance in ohms

V S

I S

= BD 2.8

Open circuit and short circuit conditions are ideal cases during load conditions, the

transmission line will have the effect of series impedance and shunt admittance.

The combined effect of the line will be represented as characteristic impedance ZC

Characteristic impedance ZC=2.5×2.8

V S

I S

×V S

I S

= AC

× BD

V S

I S

=√ AC

× BD

ZC=V S

I S

=√ BC

ZC=√ Z sc

Y oc

=√Zsc×Zoc


B=Z=R+ jX


2.2 SHORT LINE

The shunt capacitance for a short line is almost negligible. The series impedance is

assumed to be lumped as shown in Fig. 2.2. If the impedance per km for an l km long line is

z0 = r + jx, then the total impedance of the line is Z = R + jX = lr + jlx. The sending end

voltage and current for this approximation are given by

Fig. 2.2 Short transmission line representation

Z=R+jXL

V S=V R+ZI R 2.9

I S=I R 2.10

The A,B,C,D parameters are given by

A=D=1 , B=Z Ω and C=1 2.11

A=D & AD-BC=1 so short transmission line is symmetric and reciprocal

For short transmission line VRO=VS so voltage regulation is equal to the voltage

drop of line.

Regulation:-

%∈=|V Ro|−|V R|

|V R|×100

At No load condition IR=0 so VRo=VS

%∈=|V S|−|V R|

|V R|×100

%∈=V R+ IZ−V R

V R

×100

%∈= IZV R

×100



From phasor diagram of Short transmission line the approximate value of Sending voltage

will be

V S=V R+ I R R cos φr±I R X sin φr

Where φr is the load power factor + for lagging power factor

_ for leading power factor

%∈=|V S|−|V R|

|V R|×100

%∈=IR R cos φr±IR X sin φr

V R

×100

%∈=%V Rcos φr±%V X sin φr

%V R=I R X

V R

×100

%V X=IR X

V R

×100

Maximum regulation will occur only for lagging power factor load

Condition for maximum regulation:-

Zero regulation will occur only for leading power factor load

Condition for zero regulation:- φr=tan−1 (R / X )

2.2 MEDIUM TRANSMISSION LINES

Medium transmission lines are modelled with lumped shunt admittance. There are two

different representations − nominal-π and nominal-T depending on the nature of the network.

These two are discussed below.

2.2.1 Nominal-π Representation

In this representation the lumped series impedance is placed in the middle while the shunt

admittance is divided into two equal parts and placed at the two ends. The nominal-π

representation is shown in Fig. 2.3. This representation is used for load flow studies, as we shall


φr=tan−1 ( X / R )


see later. Also a long transmission line can be modeled as an equivalent π-network for load flow

studies.

Fig. 2.3 Nominal-π representation

Let us define three currents I1, I2 and I3 as indicated in Fig. 2.3. Applying KCL at nodes

M and N we get

Is I1 I2 I1 I3 IR

I S=Y2

V S+Y2

V R+ I R 2.12

V S=ZI 2+V R=Z [V RY2

+ IR ]+V R

= (YZ

2+1)V R+ZI R

2.13

Substituting (2.13) in (2.12) we get

I S=Y2 [(YZ

2+1)V R+ZI R]+Y

2V R+ I R

=Y (YZ

4+1)V R+(YZ

2+1) I R

2.14

Therefore from (2.13) and (2.14) we get the following ABCD parameters of the nominal-

π representation

A=D=(YZ2

+1)2.15

B=Z Ω 2.16

C=Y (YZ4

+1)mho 2.17

2.2.1 Nominal-T Representation



In this representation the shunt admittance is placed in the middle and the series

impedance is divided into two equal parts and these parts are placed on either side of the shunt

admittance. The nominal-T representation is shown in Fig. 2.4. Let us denote the midpoint

voltage as VM. Then the application of KCL at the midpoint results in

Fig. 2.4 Nominal-T representation.

I S=V M Y+ I R

V S−V M

Z2

=YV M+V M−V R

Z2

Rearranging the above equation can be written as

V M= 2

YZ+4 (V S+V R ) 2.18

Now the receiving end current is given by

I R=V M−V R

Z2 2.19

Substituting the value of VM from (2.16) in (2.17) and rearranging we get

V S=(YZ

2+1)V R+Z(YZ

4+1) I R

2.20

Furthermore the sending end current is

I S=YV M+ I R 2.21

Then substituting the value of VM from (2.18) in (2.21) and solving

I R=YV R+(YZ

2+1) I R

2.22

Then the ABCD parameters of the T-network are



A=D=(YZ2

+1) 2.23

B=Z (YZ4

+1)Ω 2.24

C=Y mho 2.25

A=D & AD-BC=1 so Medium transmission lines are symmetric and reciprocal

For Medium transmission line VRO≠VS so voltage regulation is not equal to the

voltage drop of line.

Direct calculation of ABCD parameters:-

For Series impedance:-

For shunt admittance:-

[V s

I s]=[ 1 0

Y 1 ] [V r

I r]

For medium transmission lines (Nominal Π):-

[V S

I S]=[ 1+ZY /2 ZY (1+ZY /4 ) 1+ZY /2 ] [V r

Ir]

A=1+ZY /2 B=Z

C=Y (1+ZY /4 ) D=1+ZY /2

A=D and AD−BC=1 , So the model is symmetric and reciprocal

For medium transmission lines (Nominal T):-


[V s

I s]=[1 Z

0 1 ] [V r

I r]

[V S

I S]=[ 1 0

Y /2 1 ][1 Z0 1 ] [ 1 0

Y /2 1 ] [V r

I r]


A=1+ZY /2 B=Z (1+ZY /4 )

C=Y D=1+ZY /2

A=D and AD−BC=1 , So the model is symmetric and reciprocal

Source end capacitance:-

A=1. 0 , B=Z , C=Y and D=1+ZY

Load end capacitance:-

A=1+ZY B=Z , C=Y and D=1 .0

Sending end capacitance method is a uncompensation method.Load end capacitance

method is a overcompensated method and the both the method are not symmetric. So the

medium transmission lines are represented in either nominal T (or) nominal II

Transmission lines in Series (or) cascade:-


[V S

I S]=[1 Z /2

0 1 ][1 0Y 1 ][1 Z /2

0 1 ][V r

Ir]

[V S

I S]=[ 1 0

Y 1 ] [1 Z0 1 ] [V r

I r]

[V S

I S]=[1 Z

0 1 ][ 1 0Y 1 ] [V r

I r]


Where

Two transmission lines are in parallel:-

In parallel transmission lines

V s1=V s 2=V s

V r 1=V r 2=V r

I S=I s 1+ I s 2 and I r=I r1+ I r2


[V s

I s]=[ A B

C D ] [V r

I r]

[A BC D ]=[A1 B1

C1 D1] [A2 B2

C2 D2]

[V S

I S]=[A B

C D ] [V r

I r]


Where

[A BC D ]=[

A1B2+ A2 B1

B1+B2

B1 B2

B1+B2

C1+C2+( A1−A2) (D2−D1 )

(B1+B2 )D1 B2+D2 B1

B1+B2

]2.3 LONG LINE MODEL

For accurate modeling of the transmission line we must not assume that the parameters are

lumped but are distributed throughout line. The single-line diagram of a long transmission line is

shown in Fig. 2.5. The length of the line is l. Let us consider a small strip ∆x that is at a distance x

from the receiving end. The voltage and current at the end of the strip are V and I respectively and

the beginning of the strip are V + ∆V and I + ∆I respectively. The voltage drop across the strip is

then ∆V. Since the length of the strip is ∆x, the series impedance and shunt admittance are z ∆x

and y ∆x. It is to be noted here that the total impedance and admittance of the line are

Z=z×l and Y= y×l 2.26

Fig. 2.5 Long transmission line representation

From the circuit of Fig. 2.5 we see that

ΔV =Iz Δx ⇒ ΔVΔx

=Iz 2.27

Again asΔx →0 , from (2.27) we get

dVdx

=Iz 2.28

Now for the current through the strip, applying KCL we get

ΔI=(V + ΔV ) yΔx=Vy Δx+ ΔVy Δx 2.29



The second term of the above equation is the product of two small quantities and therefore

can be neglected. For ∆x → 0 we then have

dIdx

=Vy2.30

Taking derivative with respect to x of both sides of (2.28) we get

ddx ( dV

dx )=zdIdx

Substitution of (2.30) in the above equation results

d2Vdx2

− yzV =02.31

The roots of the above equation are located at √(yz). Hence the solution of (2.31) is of the

form

V=A1 ex √ yz+A2 e− x√ yz2.32

Taking derivative of (2.32) with respect to x we get

dVdx

=A1√ yz ex √yz−A2 √ yz e−x √ yz

2.33

Combining (2.28) with (2.33) we have

I= 1Z ( dV

dx )= A1

√ zy

ex √ yz−A2

√ zy

ex√ yz

2.34

Let us define the following two quantities

ZC=√ zy Ω which is called the characteristic impedance 2.35

γ=√ yz which is called the propagation constant 2.36

Then (2.32) and (2.34) can be written in terms of the characteristic impedance and propagation

constant as

V=A1 eγx+ A2 e−γx 2.37

I=A1

ZC

eγx−A2

ZC

e−γx

2.38

Let us assume that x = 0. Then V = VR and I = IR. From (2.37) and (2.38) we then get



V R=A1+ A2 2.39

I R=1

ZC( A1−A2 )

2.40

Solving (2.39) and (2.40) we get the following values for A1 and A2.

A1=V R+ZC I R

2 and A2=

V R−ZC I R

2

Also note that for l = x we have V = VS and I = IS. Therefore replacing x by l and substituting the

values of A1 and A2 in (2.37) and (2.38) we get

V S=V R+ZC IR

2eγl+

V R−ZC I R

2e−γl

2.41

I S=

V R

ZC

+ I R

2eγl+

V R

ZC

−IR

2e−γl

2.42

Noting that

eγl−e−γl

2=sinh γl

and

eγl−e−γl

2=cosh γl

We can rewrite (2.41) and (2.42) as

V S=V R cosh γl+ZC I R sinh γl 2.43

I S=V Rsinh γl

Zc

+ I Rcosh γl 2.44

The ABCD parameters of the long transmission line can then be written as

A=D=cosh γl 2.43

B=ZC sinh γlΩ 2.44



c=sinh γlZC mho 2.45

A=D & AD-BC=1 so Long transmission lines are symmetric and reciprocal

For long transmission line VRO≠VS so voltage regulation is not equal to the

voltage drop of line.

Propagation constant:-

γ=√ yz

γ=√(R+ jwL )(G+ jwC )

γ=α+ jβ

α =real part = attenuation constant

β = imaginary part= quadrature constant (or) phase constant

Attenuation means fall in the magnitude of voltage and current while it is

travelling in the transmission line from sending end to the receiving end

The inductance and capacitance of transmission line will able to provide the

phase displacement so that the supply voltage and current will be travelled

from the sending end towards the receiving end and it can be understand in

terms β

For lossless transmission line (R=0, G=0)

γ=√( jwL)( jwC )= jw√ LC= jβ

Characteristic Impedance:- When the switch is closed the Transmission line is in a transient

condition and it is dealing with surge waveform so that corresponding impedance is called as

surge impedance (or) characteristic impedance.

ZC=√ zy=√ R+ jwL

G+ jwC

For loss less line Zc=√ jwL

jwC=√ L

C

Surge impedance is the characteristic impedance of a loss free line

Zc(or) Zs=400Ω for Overhead transmission lines , Zs=40Ω for Underground



cables.

During transient condition of the line the transmission line will be neither

inductive nature nor capacitive. So that the nature of the characteristic

impedance is to be treated as resistive.

Characteristic Impedance loading(or) Surge impedance loading:-

It is the loading of the line where the load impedance is equal to characteristic

impedance of the line (or) whenever the transmission line is terminated by a load with an

impedance equal to characteristic impedance of the line. The loading of the line is called

characteristic impedance loading.

When transmission line dealing with surge impedance loading, such a line is

called flat line.

Flat line is the one in which the magnitude of the voltage throughout the line

will be same

Surge impedance loading=

V sV r

Zc

[V s=V r=V ]

SIL

Surge Impedance Loading is the ideal loading capacity.

The most economical loading on transmission line is loading more than the

SIL.

For Underground cables most economical loading is less than the SIL.

Ferranti Effect:-When the transmission line is operating at no-load (or) light load condition.

The receiving end voltage is more than the sending end voltage. This phenomenon is called

Ferranti effect.

A,B,C and D values:-

A=|A|∠α ,

B=|B|∠ β ,

C=|C|∠γ D=|D|∠α


= V 2

Z C


Short transmission lines:-

|A|=1 and α=0o |B|=variable value and β=600-700

|C|=0 and γ =90o |D|=1.0 and α=00

Medium transmission lines:-

|A|=1 and α=1-2o |B|=variable value and β=700-800 |C|=Variable and γ =90o

Long transmission lines:-

|A|=1 and α=2-5o |B|=variable value and β=800-870 |C|=Variable and γ =90o

Power Transfer Equations:- Receiving end

The receiving end power transfer Sr=V r I¿r

The purpose of conjugate is to assign polarity for inductive or capacitive load

reactive powers

Sr=V r(V S−AV r

B )¿

Sr=|V S||V r|

|B|∠( β−δ )−

|A||B|

|V r|2∠ ( β−α )

Pr=|V S||V r|

|B|cos( β−δ )−

|A||B|

|V r|2 cos ( β−α )

Qr=|V S||V r|

|B|sin( β−δ )−

|A||B|

|V r|2 sin ( β−α )

The locus of the above equations is circle. For fixed receiving end voltages they will

be concentric circles

For short transmission lines:-



|A|=1 .0 ,

α=0o ,

|B|=Z ,

∠B=θ ,

Pr=Pr max=|V S||V r|

|Z|cos (θ−δ )−

|A||Z|

|V r|2 cos (θ )

Pr=Pr max=|V S||V r|

|Z|−

|V r|2

|Z|cos (θ )

[ δ=θ ]

Qr=−|V r|

2

|Z|sin θ

For loss less transmission line:-

|A|=1 .0 ,

α=0o ,

|Z|=|X|

θ=900

Pr=|V S||V r|

|Z|sin δ

Pr=Pmax=|V S||V r|

|Z|

[δ=900]Pr=Pmaxsin δ

Qr=|V S||V r|

|X|cos δ−

|V r|2

|X|

In case of short transmission lines δ is very small. So cosδ~1.0

Qr=V r

X [V S−V r ]=V r

XΔV

The real power transfer depends on the power angle and the reactive power depends on the

difference of voltage between the sending end and load end

1) The inductance of a power transmission line increases with (GATE-1992)

a) Decreases in line length

b) Increases in diameter of conductor

c) Increasing in spacing between the phase conductors

d) Increases in load current carried by the conductor



2) A three phase overhead transmission line has its conductors horizontally spaced with spacing

between adjacent conductors equal to ‘d’. If now the conductors of the line are rearranged to

form an equilateral trainable of sides equal to ‘d’ then (GATE-1993)

a) Average capacitance and inductance will increase

b) Average capacitance will increase and inductance will decrease

c) Average capacitance will increase and inductance will increase

d) Surge impedance loading of the line increases

3) The insulation level of a 400KV, EHV over head transmission line is deeded on the basis of

a) Lightning over voltage (GATE-1995)

b) Switching over voltage

c) Corona inception voltage

d) Radio and TV interference

4) For a 500 Hz frequency excitation, a 50km short power line will be modeled as(GATE-1996)

a) Short line

b) Medium line

c) Long line

d) Data insufficient for decisions

5) Consider a long, two-wire line composed of solid round conductors. The radius of both

conductors is 0.25cm and the distance between their centres is 1m. if this distance is doubled,

then the inductance per unit length. (GATE-2002)

a) Doubles c) halves

b) increases but does not double d) decreases but does not halve

6) A long wire composed of a smooth round conductor runs above parallel to the ground

(Assumed to be a large conducting plane). A high voltage exists between the conductor and

the ground. The maximum electric stress occurs at (GATE-2002)

a) The upper surface of the conductor

b) The lower surface of the conductor

c) The ground surface

d) Midway between the conductor and the ground

7) Bundled conductors are mainly used in high voltage overhead transmission lines to

a) Reduce transmission line losses (GATE-2003)

b) Increases mechanical strength of the line

c) Reduce corona



d) Reduce sag

8) The rated voltage of a 3-phase power system is given as (GATE-2004)

a) RMS phase voltage

b) PEAK phase voltage

c) RMS line to line voltage

d) PEAK line to line voltage

9) The phase sequene of the 3-phae system shown in figure is (GATE-2004)

a) RYB b) RBY c) BRY d) YBR

10) The concept of an electricity short, medium and long line is primarily based on the

(GATE-2006)

a) Nominal voltage of the line

b) Physical length of the line

c) Wavelength of the line

d) Power transmitted over the line

Two Marks Questions:

11) For equilateral spacing of conductors of an untransposed 3-phase line, we have (GATE-1996)

a) Balanced receiving end voltage and no communication interference

b) Unbalanced receiving end voltage and no communication interference

c) Balanced receiving end voltage and communication interference

d) Unbalanced receiving end voltage and communication interference

12) For a single phase overhead line having solid copper conductors of diameter 1 cm, spaced 60

cm between centers, the inductance in mH/km is (GATE-1999)

a) 0.05+0.2 In 60

b) 0.2 In 60

c) 0.05+0.2 In (60/0.5)

d) 0.2 In (60/0.5)

13) The conductors of a 10 km long, single phase, two wire likne are separated by a distance of

1.5m. the diameter of each conductor is 1 cm. if the conductors are of copper, the inductance

of the circuit is (GATE-2001)

a) 50.0 mH b) 45.3 mH c) 23.8 mH d) 19.6 mH



14) A 800 kV transmission line is having per phase line inductance of 1.1 mH/km and per phase

line capacitance of 11.78 nF/km. Ignoring the length of the line, its ideal power transformer

capability in MW is (GATE-2004)

a) 1204 MW b) 1504 MW c) 2085 MW d) 2606 MW

15) 230V (phase) 50Hz, three-phase 4-wire, system has a sequence ABC. A unity power-factor

load of 4kW is connected between phase A and neutral N. it is desired to achieve zero neutral

current through the use of a pure capacitor in the other two phases. The value of inductor and

capacitor is (GATE-2007)

a) 72.95 mH in phase C & 139.02µF in Phase B

b) 72.95 mH in phase B & 139.02µF in Phase C

c) 42.12 mH in phase C & 240.79 µF in Phase B

d) 42.12 mH in phase B & 240.79 µF in Phase C

One mark questions

16) The surge impedance of a 400 km long overhead transmission line is 400Ω. For a 200km

length of the same line, the surge impedance will be (GATE-1995)

a) 200 Ω b) 800Ω c) 400Ω d) 100Ω

17) In a 400Kv network, 360 Kv is recorded at a 400kV bus. The reactive power absorbed by a

shunt rated for 50MV AR, 400kV connected at the bus is (GATE-1994)

a) 61.73 MVAR b) 55.56 MVAR c) 45 MVAR d) 40.5

MVAR

18) The reflection coefficient for the transmission line shown in figure at P is (GATE-1998)

a) +1 b) -1 c) 0 d) 0.5

19) Series capacitive compensation in EHV transmission lines is used to (GATE-1998)

a) Reduce the line loading

b) Improve the stability of the system

c) Reduce the voltage profile

d) Improve the protection of the line

20) An overhead line having a surge impedance of 400 Ω is connected in series with an

underground cable having a surge impedance of 100 Ω. If a surge of 50kV travels from the



line towards the cable junctions, the value of the transmitted voltage wave at the junction is

(GATE-1999)

a) 30kV b) 20kV c) 80kV d) -30kV

21) The load carrying capability of long AC transmission line is (GATE-1999)

a) Always limited by the conductor size

b) Limited by stability considerations

c) Reduced at low ambient temperature

d) Decreased by the use of bundled conductors of single conductors

22) Corona losses are minimized when.

a) Conductor size is reduced

b) Smoothness of conductor is reduced

c) Sharp points are provided in the line hardware

d) Current density in conductors is reduced

23) A lossless radial transmission line with surge impedance loading (GATE-2001)

a) Takes negative VAR at sending end and zero VAR at receiving end

b) Takes positive VAR at sending end and zero VAR at receiving end

c) Has flat voltage profile and unity power factor at all points along it

d) Has sending end voltage higher than receiving end voltage and unity power factor at

sending end

24) A 3-phase 11-kV generator feeds power to a constant power unity power factor load of

100MW through a 3-phase transmission line. The line-to-line voltage at the terminals of the

machine is maintained constant at 11 kV. The per unit positive sequence impedance of theline

based on 100MA and 11kV is j0.2. the line to line voltage at the load terminals is measured

to be less than 11 kV. The total reactive power to be injected at the terminals of the load to

increase the line-to-line voltage at the load terminals to 11kV is (GATE-2003)

a) 100MVAR b) 10.1 MVAR c) -100MVAR d) 10.1 MVAR

25) Total instantaneous power supplied by a 3-phase ac supply to a balanced R-L load is

a) Zero (GATE-2004)

b) Constant

c) Pulsating with zero average

d) Pulsating wit non-zero average

26) A 400V, 50Hz, three phase balanced source supplies power to a star connected load whose

rating is 12√3 KVA, 0.8 pf(lag). The rating (in KVAR) of the delta connected (capacitive)

reactive power bank necessary to bring the pf to unity is (GATE-2006)

a) 28.78 b) 21.60 c) 16.60 d) 12.47



27) An extra high voltage transmission line of length 300km can be approximate by a lossless line

having propagation constant β=0.00127 radians per km then the percentage ratio of line

length to wavelength will be given by (GATE-2008)

a) 24.24% b) 12.12% c) 19.05% d) 6.06%

28) For a fixed value of complex power flow in transmission line having a sending end voltage V,

the real power loss will be proportional to (GATE-2009)

a) V b) V2 c) 1

V 2 d) 1V

29) Consider a step voltage wave of magnitude 1 pu travelling along a lossless transmission line

that terminates in a reactor. The voltage magnitude across the reactor at the instant the

travelling at the instant the travelling eave reaches the reactor is (GATE-2010)

a) -1 pu b) 1pu c) 2 pu d) 3 pu

30) Consider two buses connected by an impedance of (0+j5) Ω. The bus 1 voltage is 100angle of

300 V, and bus 2 voltage is 100 angle of 00 V. the real and reactive power supplied by bus1,

respectively, are (GATE-2010)

a) 1000W, 268 V Ar

b) -1000W, -134 V Ar

c) 276.9 W, -56.7 Ar

d) -276.9W, 56.7 V Ar

31) A nuclear power station of 500MW capacity is located at 300km away from a load center.

Select the most suitable power evacuation transmission configuration among the following

options (GATE-2011)

32) For enhancing the power transmission in a long EHV transmission line, the most preferred is

to connect a (GATE-2011)

a) Series inductive compensator in the line

b) Shunt inductive compensator at the receiving end



c) Series capacitive compensator in the line

d) Shunt capacitive compensator at the sending end

Two marks questions

33) A shunt reactor of 100MVAR is operated at 98% of its rated voltage and at 96% of its rated

frequency. The reactive power absorbed by the reactor is (GATE-1998)

a) 98 MVAR b) 104.02 MVAR c) 96.04 MVAR d) 100.04 MVAR

34) A 220 kV, 20 km long, 3 phase transmission line has the following A, B, C, D constants.

A=D=0.96angleof 30, B=55angle of 650 Ω/phase C=0.5x10-4 angle of 900 S/phase. Its correct

charging current per phase is (GATE-1999)

a) 11/√3A b) 11A C) 220A d) 220//√3A

35) A 3-phase, 11kV, 50 Hz, 200kW load has a power factor of 0.8 lag. A delta connected 3-

phase capacitor is used to improve the power factor to unity. The capacitance per-phase of the

capacitor in micro-farads is (GATE-1999)

a) 3.948 b) 1.316 c) 0.439 d) 11.844

b)

36) A transmission line has equal voltages at the two ends, maintained constant by tow sources. A

third source is to be provided to maintain constant voltage (equal to end voltages) at either the

midpoint of the line of at 75% of the distances from the sending end. Then the maximum

power transfer capabilities of the line in the original case and the other two cases respectively

will be in the following ratios (GATE-2000)

a) 1:1:1 b) 1:2:1/0.75 c) 1:2:4 d) 1:4:6

37) The corona loss on a particular system at 50Hz is 1kW/Km per phase. The corona loss at 60hz

would be (GATE-2000)

a) 1kW/km per phase

b) 0.83kW/km per phase

c) 1.2kW/km per phase

d) 1.13kW/km per phase

38) Consider the model shown in figure of a transmission line with a series capacitor at its mid-

point. The maximum voltage on the line is at the location (GATE-2000)



a) P1 b)P2 c) P3 d) P4

39)The ABCD parameters of a 3-phase over head transmission line are A=D=0.9, ∟900 Ω and

C=0.95 x10-3∟900S. At no-load condition a shunt inductive, reaction is connected at the receiving

end of the line to limit the receiving end voltage to be equal to the sending –end voltage. The

ohmic value of the reactor is (GATE-2003)

a) ∞Ω b) 2000 Ω c) 105.26Ω d) 1052.6 Ω

40) A surge of 20kV magnitude travels along a lossless cable towards its junction with two identical

lossless over head transmission lines. The inductance and the capacitance of the cable are 0.4mH

and 0.5µF per km. the inductance and capacitance of the overhead transmission lines are 1.5 mH

and 0.015 µF per km. the magnitude of the voltage at the junction due to surge is (GATE-2003)

a) 36.72kV b) 18.36 kV c) 6.07kV d) 33.93kV

41) A balanced delta connected load of (8+j6) Ω per phase is connected to a 400 V, 50Hz, 3-phase

supply lines. If the input power factor is to be improved to 0.9 by connecting a bank of star

connected capacitors the required KVAR of the bank is (GATE-2003)

a) 42.7 b) 10.2 c) 28.8 d) 39.4

42) A lightning stroke discharges impulse current of 10kA (peak) on a 400kV transmission line

having surge impedance of 250 Ω. The magnitude of transient over-voltage travelling waves in

either direction assuming equal distribution from the point of lightning strike will be

a) 1250kV b) 1650kV c) 2500kV d) 2900kV (GATE-2004)

43) The generalized circuit constants of a 3-phase, 220kV rated voltage, medium length transmission

line are (GATE-2004)

A=D=0.936+j0.016=0.936∟0.980

B=33.5+j138=142.0∟76.40 Ω

C=(-5.18+j914 x 10-6 Ω

If the load at the receiving end is 50MW at 220kV with a power factor of 0.9 lagging, then

magnitude of line to line sending end voltage should be

a) 133.23kV b) 220.00kV c) 230.78kV d) 246.30kV

44) At an industrial sub-station with a 4MW load, a capacitor of 2MVAR is installed to maintain the

load factor at 0.97 lagging. If the capacitor goes out of service, the load power factor becomes

a) 0.85 b) 1.00 c) 0.80 lag d) 0.90 lag (GATE-2005)

45) Two networks are connected in cascade as shown in the figure. With the usual notations the

equivalent A,B,C and D constants are obtained. Given that, C=0.025 ∟450, the value of Z2 is



a) 10∟300 Ω b) 40∟-450 Ω c) 1 ohm d) 0 Ω (GATE-2005)

46) The A,B,C,D constant of a 220kV line are: A=D=0.94∟10, B=130∟730, C=0.001∟900. if the

sending end voltage of the line for a given load delivered at normal voltage is 240kV, the %

voltage regulation of the line is (GATE-2006)

a) 5 b) 9 c) 16 d) 21

47) The total reactance and total susceptance of a lossless overhead EHV line, operating at 50 Hz, are

given by 0.045 pu and 1.2 pu respectively. If the velocity of wave propagation is 3x105 km/s, then

the approximate length of line is (GATE-2007)

a) 122km b) 172 km c) 222km d) 272km

48) A lossless transmission line having Surge impedance Loading (SIL) of 2280MW. A series

capacitive compensation of 30% is emplaced. Then SIL of 2280MW. A series capacitive

compensation of 30% is emplaced. Then SIL of the compensated transmission line will be

(GATE-2008)

a) 1835MW b) 2280MW c) 2725MW d) 3257MW

49) Match the items List-I (To) with the items in List-II (use) and select the correct answer using the

codes given below the lists (GATE-2009)

List I List II

A. Improve power factor 1. Shunt reactor

B. Reduce the current ripples 2. Shunt capacitor

C. Increase the power flow in line 3. Series capacitor

D. Reduce the Ferranti effect 4. Series reactor

(a) a 2, b3, c4, d-1

(b) a 2, b4, c3, d-1

(c) a 4, b3, c1, d-2

(d) a 4, b1, c3, d-2



Miscellaneous questions

50) A factory draws 100kW at 0l.7 p.f. lagging from a 3-phase, 11kV supply. It is desired to raise the

p.f. to 0.95 lagging using series capacitors. Calculate the rating of the capacitor required(GATE-

1997)

51) A 66 kV, 3-phase, 50Hz, 150kM long overhead transmission line is open circuited at the receiving

end. Each conductor has a resistance of 0.25 Ω/km, an inductive reactance of 0.5 Ω/km and a

capacitive admittance to neutral of 0.04x10-4 S/km (GATE-

1999)

a) Draw the nominal π-equivalent circuit and indicate the value of each parameter

b) Calculate the receiving end voltage if the sending end voltage is 66kV

52) A 275kV, 3-phase, 50Hz, 400km lossless line has following parameters: x=0.05 ohms/km, line

charging susceptance y=3.0 micro-Siemens/k (GATE-2000)

a) Calculate the receiving end voltage on open circuit using justifiable assumptions

b) What load at the receiving end will result in a flat voltage profile on the line?

c) If the flat voltage profile is to be achieved at 1.2 times the loading in (b), what will be nature

and quantum of uniformly distributed compensation required?

53) A long lossless transmission line has a unity power factor (UPF) load at the receiving end and an

ac voltage source at the sending end. The parameters of the transmission line are as follows:

characteristics impedance Zc=400 Ω, propagation constant β=1.2x10-3rad/km, and length

l=100km. The equation relating sending and receiving end question is

Vs=Vrcosh(βl)+jZcsinh(βl)IR. Complete the maximum power that can be transferred to the UPF

load at the receiving end if |Vs|=230kV. (GATE-2002)

Practice questions:-

IES 2001

1) The ABCD constants of a 3-phase transmission line are A=D=0.8∟10

A=D=0.8 ∟10

B=170∟850

C=0.002 ∟90.40 mho

The sending end voltages is 400kV. The receiving end voltage under no-load condition is

a) 400kV b) 500kV c) 320kV d) 417kV

2) In a short transmission line,voltage regulation is zero when the power factor angle of the load

at the receiving end side is equal to

a) Tan-1(X/R) b) Tan-1(R/X) c) Tan-1(X/Z) d) Tan-1(R/Z)



3) Consider the following statements:

Surge impedance loading of a transmission line can be increased by

1. Increasing its voltage level

2. Addition of lumped inductance in parallel

3. Addition of lumped capacitance in series

4. Reducing the length of the line

Of these statements

a) 1 and 3 are correct

b) 1 and 4 are correct

c) 2 and 4 are correct

d) 3 and 4 are correct

4) The surge impedance of a 3-phase,400kV transmission line is 400Ω. The surge impedance

loading (SIL) is

a) 400 MW b) 100MW c) 1600MW d) 200MW

5) If a travelling-wave travelling along a loss-free overhead line does not result in any reflection

after it has reached the far end, then the far end of the line is

a) Open circulited

b) Short circuited

c) Terminated into a resistance equal to surge impedance of the line

d) Terminated into a capacitor

6) The active and the reactive power delivered at the receiving end of a short transmission lineof

impedance Z<ψare respectively given by

PR =V S V R

Zcos (φ−δ )−

V 2R

ZCosφand

QR=V S V R

Zsin (φ−δ )−

V 2R

Zsinφ, with

VS and VR being the magnitude of voltage at the sending and receiving ends, δ the power

angle. At the power-limit condition i.e. for maximum PR

a) leading VARs (QR) goes to the load for any values of VS and VR

b) leading VARs (QR) goes to the load ONLY for VS = VR

c) lagging VARs (QR) goes to the load for any values of VS and VR

d) lagging VARs (QR) goes to the load for any values of VS = VR

7) Bundled conductors are used for EHV transmission lines primarily for reducing the

a) Corona loss

b) Surge impedance of the line

c) Voltage drop across the line

d) I2R losses



IES-2002

8) For transmission line which one of the following relations is true?

a) AD-BC=1 b) –AD-BC=1 c) AD-BC=-1 d) AD-BC=0

9) The reflection coefficient for the transmission line shown in figure below at point P is

a) +1 b) 0.5 c) 0 d) -1

10) A 3-phase transmission line has its conductors at the corners of an equilateral triangle with

side 3m. The diameter of each conductor is 1.63cm. the inductance of the line per phase per

km is

a) 1.232 mH b) 1.182 mH c) 1.093mH d) 1.043 mH

11) A cable has inductance of 0.22mH per km and capacitance of 0.202µF per km. The surge

impedance of the cable is

a) 28Ω b) 33Ω c) 42Ω d) 50 Ω

12) For some given transmission line the expression for voltage regulation is given by

¿V s∨−¿V R∨¿

¿V R∨¿¿¿x 100% Hence,

a) This must be a short line

b) This may either be a medium line or a short line

c) This expression is true for any line

d) This may either be a medium line or a long line

13) The capacitance of an overhead transmission line increases with

1) Increase in mutual geometrical mean distance

2) Increase in height of conductors above ground

Select the correct answer from the following

a) Both 1 and 2 are true c) Both 1 and 2 are false

b) Only 1 is true d) Only 2 is true

14) Consider the following statements

Addition of lumped capacitance in parallel to a loss-free transmission line increases

1. Characteristic impedance

2. Propagation constant

3. System stability

4. Charging current



Which of these statements are correct ?

a) 1 and 3 b) 2 and 4 c) 2,3 and 4 d) 1,2 and 4

15) In a certain single-phase a.c. circuit the instantaneous voltage is given by v=Vsin(ωt+300)p.u.

and the instantaneous current is given by i=Isin(ωt-300) p.u. Hence the per unit value of

reactive power is

a) ¼ b) ½ c) √3/4 d) √3/2

IES-2003

16) A 100 km long transmission line is loaded at 110kV. If the loss of line is 15MW and the load

is 150MVA, the resistance of the line is

a) 8.06 ohms per phase

b) 0.806 ohms per phase

c) 0.0806 ohms per phase

d) 80.6 ohms per phase

17) A rectangular voltage wave is impressed on a loss-free over head line, with the far end of the

line being short-circulated. On reaching the end of this line

a) The current wave is reflected back with positive sign, but the voltage wave with negative

sign

b) The current wave is reflected back with negative sign, but the voltage wave with positive

sign

c) Both the current and the voltage waves are reflected with positive sign

d) Both the current and the voltage waves are reflected with negative sign

18) Which one of the following statements is NOT correct for the use of bundled conductors in

transmission lines?

a) Control of voltage gradient

b) Reduction in corona loss

c) Reduction in radio interference

d) Increase in interference with communication lines

19) Ds is the GMR of each sub condutor of a four subconductor bundle conductor and d is the

bundle spacing. What is the GMR of the equivalent single conductor?

a) 1.09√ Ds xd3 b) 1.09√ D3s xd3 c) 1.09

4√ D3s xd3 d) 1.09

4√ Ds xd3

20) Which one of the following statements is correct? Corona loss increases with

a) Decrease in conductor size and increase in supply frequency

b) Increase in both conductor size and supply frequency

c) Decrease in both conductor size and supply frequency



d) Increase in conductor size & decrease in supply frequency

21) For a fixed receiving end and sending end voltage in a transmission system, what is the locus

of the constant power?

a) A straight line b) an ellipse c) a parabola d) a circle

IES - 2005

22) On what basis is the insulation level of a 400kV, EHV overhead transmission line decided?

a) Lightning overvoltage

b) Corona inception voltage

c) Switching overvoltage

d) Radio and TV interference

23) Equivalent π model is quite suitable for analyzing the performance of transmission line of

a) 50kM length b) 150kM Length c) 250 kM length d) all of the above

24) In a 7/30 ACSR conductor why is grease put between steel and aluminium conductors?

a) To reduce corrosion by electrolytic action between zinc (galvanizing agent on steel ) and

aluminum

b) To reduce friction between the stands

c) To reduce leakage of current from aluminium strands to steel stands

d) To eliminate air pockets

25) If X is the inductive reactance /phase and R is the resistance/phase of a short transmission

line, what is the power factor angle of the load for maximum voltage regulation?

a) Cos-1X/R b) tan-1X/R c) Cos-1R/X d) tan-1R/X

26) For a loss-free long transmission line, the conventional line equations become

Vs = (cosβl)VR+(jsinβl)IR and Is=( jsinβlZc

)VR+(cosβl)IR

Which one of the following statements is correct? If the line is terminated at the receiving end

by its natural load impedance Zc, then

a) The voltage is constant in magnitude at all points along the line and, Vs and VR always

remain in phase

b) The voltage is constant in magnitude at all points along the line but V s advances in phase

relative to VR by angle β radius per km

c) The magnitude of the voltage along the line changes in proportion to the line length, and,

Vs and VR always remain in phase

d) The magnitude of the voltage along the line changes in proportion to the line length, Vs

advances in phase relative to VR by β radians per km



IES 2008

27) When is the Ferranti effect on long overhead lines experienced?

a) The line is lightly loaded

b) The line is heavily loaded

c) The line is fully loaded

d) The power factor is unity

28) What is the surge impedance loading of a loss less 400 kV, 3-phase, 50Hz overhead line of

average of surge impedance of 400 ohms?

a) 400 MW b) 400√3 MW c) 400/√3 MW d) 400MW

IES-2009

29) In a transmission line, the mid-point voltage is maintained to V by a compensating device as

shown in the circuit below. What is the real power flow through the line?

a)

V 2

Xsin

δ2 b)

2V 2

Xsin δ

c)

V 2

Xsin δ

d)

2V 2

Xsin

δ2

30) A short transmission line having zero resistance and total series reactance of 0.4 pu is

provided with reactive power compensation at the mid-point of the line such that the mid-

point voltage is held at 0.96pu when the voltage at both ends are 1.0pu. what is the steady

state power transmission limit of such a system?

a) 4.8 pu b) 0.0 pu c) 2.4 pu d) 9.6pu



Solutions:-

1 . Ans:- (c)

Hint:- Inductance per phase of transmission line is Lph=0 .2 ln( d

r ' )mh/KM

Where ' d ' is the space between the two conductors , r'is the effective radius of the

conductor

∴Lph α ln(d )

2 Ans:- (c)

Hint:-

Case(i):- When conductors are horizontally spaced .

Inductance/phase= 0 .2 ln( GMD

Gmr)

Capacitance/phase=

2 πεo

ln (GMDr )

In case (ii)GMD=3√d×2d×d

GMD=3√2×d=1. 26 d

GMR=0 .7788 r

In case (ii):- When conductors are placed in equilateral triangle form

In this case GMD=d

GMR=0 .7788 r

SinceGMR∧r are constant in both the cases



L/ ph α ln(GMD )

C / ph α

1ln(GMD )

∴ From cas(i) and case(ii) the inductance decreases and capacitance increases.

3. Ans:- (b)

Hint:- In any transmission line lightning over voltages are more severe compared to

switching over voltages. The lighting voltages are external voltages. Hence the

insulation is provided for switching voltage.

4. Ans:- (c)

For f =50Hz (where f is the frequency of supply)

fλ=υ

λ = wavelength i.e the length of the line

υ =velocity of the wave=3×108m/sec

∴ λ= υ

f=3×108

50=6000

km

For 6000 Km wave length, the line with more then 160KM length is treated as long

line

For 500Hz supply frequency. λ= υ

f=3×108

500=600

km

Hence with 500Hz supply, a line with more than 16KM length is treated as long line

where line parameters are distributed

∴ For 500Hz supply frequency 50KM line is long line.

5. Ans:- (c)

Hint:- Lph=0 .2 ln( d

r ' )mH/Km



Lph=0 .2 ln( 1

0 .7788×0 .25×10−2)mh/Km

Lph =1.248 mH/Km

If the distance between the conductors is doubled then

Lph=0 .2 ln( 2

0 .7788×0 .25×10−2)mh/Km

Lph =1.38 mH/Km

Hence inductance per unit length increases but not doubles

6. Ans:- (b)

7. Ans:- (c)

Electric field intensity at surface of each conductor

E= V

( Self⋅GMD ) ln( GMDself .GMR )

For bundle conductors increase in self GMD will be predominant compared to

decrease in ln ( GMD

self . GMR )

Hence Electric field intensity at surface of each conductor will be reduced. Hence the

corona loss will be reduced, because chances for ionization of air are reduced.

8. Ans:- (c)

9. Ans:- (b)

Hint:- Phase sequence is verified in opposite direction to the vector rotation. Vectors

rotation is normally considered in anticlockwise direction hence phase sequence is verified

in clock wise direction.

10.Ans:- (c)



Two marks questions:-

11.Ans:- (a)

Hint:- For the transmission line with equilateral spacing of conductors (Inductance/Phase)

is equal

(Current/Phase) is equal

(Flux/Phase) is equal

(Voltage/Phase) is equal

Hence no communication interference and also the receiving end voltages are balanced.

12. Ans:- (c)

Hint:-

Given r=0.5cm, d=60cm

Inductance per conductor La=La . int ernal+ La. external

=( 1

2×10−7+2×10−7 ln( d

r ))H/mt

=( 1

2×10−1+2×10−1 ln( d

r ))mH/Km

From given data La= ( 12×10−1+2×10−1 ln(60

0.5 ))mH/Km

La= (0 . 05+0 . 2 ln(600 .5 ))

mH/Km

13. Ans:- (c)

Sol:- Inductance of 1-Phase circuit with copper conductors

Lab=2 La=2×0 .2 ln( d

r ' )mh/KM



Given, d=1.5m, r=0.5cm, r'=.7788×0 .5×10−2

m

∴Lab=0 . 4 ln( 1. 5

0 . 7788×0. 5×10−2)mH/Km

Lab =2.38 mH/Km

For 10Km length of the line

Lab = 2.38 ×10 mH

Lab =23.8 mH

14. Ans:- (c)

Sol:- Ideal Power transfer capability is nothing but the surge impedance loading of the line

∴surge impedance Zc=√ L /Km

C / Km

=√ 1. 1×10−3

11.68×10−9

= 306.88 Ω

∴Surge impedance loading =V 2

Zc

=

(800×103 )2

306 . 88

= 2085.47 MW

15. Ans:- (b)

Sol:- Load current phase A= 4×103

230∠0°=17 .39

Amp

VPhA= 230∠0 °



VPhB= 230∠−120 °

VPhC= 230∠−240 °

From options

L=72 . 95 mH⇒ X L=22 .91 Ω⇒ I L=10∠−90°

C=72 .95 µF⇒ XC=22 . 9Ω⇒ I C=10∠90°

By assuming inductor in phase B and capacitor in phase C

IBph=10∠−90−120=10∠−210°

ICph=10∠90−240=10∠−150°

(Since VBph=230 ∠−120° and current lags voltage by 90°

in case of inductive load)

∴ For zero neutral current

IAph+ IBph + Icp = 0

17.39∠0°+10∠−210°

+10∠−150°¿0

Hence option (b) is the correct one

Performance of Transmission lines and Voltage Control

One Mark questios:

16. Ans:- (c)

Hin

t:- Surge impedance for a given transmission line is constant and is independent

of length of the transmission line and the frequency of surge. It depends only on

magnitude of inductance/Km and capacitance/Km [∴Zc=√ L/ Km

C / Km ]17. Ans:- (d)



Hint:- 50MVAR,400Kv inductor means for an applied voltage of 400 KV, the

inductor absorbs 50 MVAR reactive power.

Reactive power absorbed by inductor (Q)= 3Vph Iph Sinφ

For a pure inductor φ=90°

Q=3 V ph I phsin 90°=3V ph

V ph

XLph

Q=3 V ph

2

X Lph

At 400Kv, Q=50MVAR

50×106=3×(400

√3×103)

2

X Lph

X Lph=3200Ω

For 360 kV applied voltage the reactive power absorbed by inductor

=

3×(360√3

×103)2

3200

=40.5 MVAR.

18. Ans :- (c)

Hint:- Since load impedance is equal to surge impedance, the voltage & current wave

forms are not going to experience any reflection.

Hence reflection coefficient is zero.

19. Ans :- (b)

Hint:- electrical power supplied by alternator is= EV

Xsin δ

By using series capacitor the effective value of reactance can be reduced.



Hence for a given power transfer, we can operate the alternator at lower values of δ

so that machine will not lose its synchronism during abnormal conditions of

operation.

20. Ans:- (b)

The transmitted (or) refracted voltage

V 2=2 V ( ZL

Z L+ZC)

Here ‘2’ indicates that the voltage V2 is calculating in transient condition

∴V 2=2×50×103×(100100+400 )

V2=20KV

21. Ans:- (b)

Hint:- For a long line with AC transmission , reactance is more and is having more

effect on stability of system.

22. Ans:- (d)

Hint:- current density=

currentunitarea

For a given current, if the current density is reduced means the cross sectional

area of conductor is increased.

But corona loss α (V-Vo)2

Where V= operating voltage

Vo=Critical disruptive voltage

Voα rln ( d

r ) (d= separation between conductors=constant)

Voα r (radius of conductor)

As radius of conductor increases, critical disruptive voltage(Vo) also increases and

hence corona loss decreases.

∴ increase in ‘r’ is more effective than decrease in ln ( d

r )

Note:- Because of bundling radius of the conductor increases so corona loss decreases



23. Ans:- (c)

Hint:- When transmission line dealing with surge impedance loading, then such a line

is called flat line,in which the magnitude of the voltage throughout is purely resistive

in nature. Hence the power factor is unity throughout its length.

24. Ans :- (b)

Sol:-

Given the X1 P.u=j 0.2 p.u

but X1=j0.2¿

(KV b)2

MVAb

= j0 . 2×112

100

= j0.242Ω

To make sending end voltage and receving end voltage equal, the net reactive power

demand at load must be zero.

Let Qc be the reactive power injected at load side and Q r be the reactive power

received from supply.

∴Qr+Qc=Qd

Qr+Qc=0

Qc=-Qr

Qr=

|V s||V r||B|

sin ( β−δ )−|A||B|

|V r|2 sin( β−α )

In this case , β= Line impedance angle=900

|B|=0.242Ω, α=00, |A|=1 .0

Pr=

|V s||V r||B|

cos ( β−δ )−|A||B|

|V r|2 cos ( β−α )

100=

11×110 .242

cos (90−δ )− 1. 00 .242

112cos (90−0)

100=

112

0 .242sin δ

δ = 11.530

Qr=11×110 . 242

sin (90−11. 53 )− 1. 00 .242

112sin(90 )

Qr=-10.1MVAR

∴ The reactive power to be injected =10.1MVAR



25. Ans:- (b)

Sol:- Let the phase voltage be Va=VmSinwt

Vb= VmSin(wt-1200), Vc= VmSin(wt-2400)

Let the phase of load be ‘Φ’, As it is a RL load ‘Φ’ is a lagging angle.

Phase current ia=imSin(wt- Φ),

ib=imSin(wt- 120-Φ), ic=imSin(wt- 240-Φ)

Instantaneous total 3 phase power=( VmSinwt)( imSin(wt- Φ))+

( VmSin(wt-1200))( imSin(wt- 120-Φ))+

(VmSin(wt-2400))( imSin(wt- 240-Φ))

=

=V m Im

2¿ [cos(wt−wt+φ )−cos(2wt−φ)+cos(wt−120−wt+120+φ)−¿ ] [cos(wt−120+wt−120−φ)+cos(wt−240−wt+240+φ )−¿ ]¿

¿¿

=

V m I m

23 cos φ [cos (2 wt−φ )+cos (2 wt−240−φ )+cos (2 wt−480−φ) ]

=

32

V m Imcos φ−[cos (2wt−φ )+cos (2wt−120−φ )+cos (2wt−240−φ )]

=3

V m

√2

I m

√2cos φ−0

= 3 V ph I ph cosφ

26. Ans:- (d)

Sol:- KVAR demand of load at 0.8 power factor =12√3×sin 36 . 86×103

=12.47KVAR

[∵φ=cos−1 0. 8=36 .860 ]

KVAR demand of load at unity power factor =12√3×sin 0=0

∴KVAR to be supplied by capacitor

Bank connected at load=12.47-0=12.47KVAR



27. Ans;- (d)

Sol:- propagation constant for loss less line β=2 πf √LC

Velocity of light υ= 1

√LC

λ= υf= 1

f √LC=2 π

β

λ= 2 π

0. 00127=4947 . 4 Km

∴

Linelengthwavelength

=3004747 . 4

=0 .0606=6 .06%

28. Ans:- (c)

Hint:- P=I2R=( S

3V )2

R= S2 R3V 2

Pα

1

V 2

29. Ans:- (c)

The Reactor is initially open circuit

V2=V+V1=1.0+1.0=2.0 p.u

V1=reflected voltage

V2=Switched voltage

30. Ans:- (a)

Sol:- β=Line impedance angle, δ=30o,α=00, A=D=1.0

(∴ Given line is short line)

Psending

=|DB

||V S|2 cos( β−α )−

|V s||V r||B|

cos ( β+δ )

Qsending

=|DB

||V S|2 sin( β−α )−

|V s||V r||B|

sin ( β+δ )

Psending

=|1 . 05

||100|2cos (90−0)−100×100|5|

cos ( 90+30 )

=1000 W

Qsending

=|1 . 05

||100|2sin(90−0 )−100×100|5|

sin (90+30 )



=268 VAR

31. Ans:- (d)

Sol: SIL=V 2

Zc

SIL==400×400

400=400 Mw

To carrey 500Mw a double circuit 400Kv line is used.

32. Ans:- (c)

Sol:- To enhance the power transmission in a long EHV transmission line, a series

capacitance compensator is used.

P=V 2

X series capacitor will reduce the reactance.

Two Marks Questions:-

33. Ans:- (d)

Sol:- Reactive Power absorbed by reactor=

V 2

X L

Q1=

V12

2 πf 1 L=1000 MVAR

If V2 =0.98V1&f2=0.96f1

Then reactive power absorbed =

V22

2 πf 2 L

Q2=

(0. 98 V 1 )2

2 π (0 . 96 f 1) L

Q2=1 .000416 ( V

12

2 πf 1 L )Q2=1.000416¿ 100MVAR

Q2=100.04MVAR

34. Ans:- (a)

Sol:- Admittance per phase(Y)=C=0.5¿ 10-4∠800S



Charging current per phase = Vph¿ Y

=

220

√3×103×0 .5×10−4

=

11

√3 A

35. Ans:- (b)

Sol:- 3-Phase,11KV,50Hz,200Kw load , at power factor=0.8

KVAR demand of Load (Q1)=

220×103

0. 8×sin (cos−10 . 8 )

∴Q1=150KVAR

KVAR demand of load at UPF=0

So as to operate the load at UPF, we have to supply the 150KVAR by using

capacitor bank

∴ KVAR rating of Δ - connected capacitor bank=

3V ph2

XCph

=150 KVAR

3×(11000 )2

XCph

=150×103

XCph = 2420Ω

12 π fc

=2420Ω

C= 12 π×50×2420

=1 . 3153µF

C¿ 1.316 µF

36. Ans:- (b)

Hint:- Initially without compensator Pmax =

V 2

X

With a compensator at the middle of the line, power transfer capability,

Pmax1=

2V 2

X

If the compensator is kept at the 75% of the line from the sending end, then the

longer transmission line would clearly determine the overall transmission limit.



Pmax2=

V 2

0 .75 X

Pmax: Pmax1 :Pmax1=1:2:

10 .75

37. Ans:- (d)

Sol:- Empirical formula for corona loss is given by

P=241×10−5×( f +25

δ )×√ rd×(V p−V d )2 Kw / ph/km

∴Corona loss, Pα( f +25 )

P2

P1

=(f 2+25 )( f 1+25 )

P2=1Kw×(60+25 )(50+25 )

=1 .13 KW

∴ Corona loss at 60Hz (P2)=1.13KW/Km/Ph

38. Ans:- (a)

The voltage profile for the given line is shown in fig.

39. Ans:- (b)

Sol:- Given A=D=0.9∠00,B=200∠900Ω/C, C=0.95×10-3∠900



[A1 B1

C1 D1 ]=[A BC D ] [1 0

Y 1 ]Vs=A1 Vro

A1=1.0

A+BY=1.0

Y=

0. 1200∠90

; Z=2000Ω

40. Ans:- (d)

Sol:-

In this case Z2=Z3

The Refracted voltage V2=

2VZ2 Z3

Z1 Z2+Z2 Z3+Z3 Z1

=2VZ2

2

Z1 Z2+Z22+Z3 Z1

=

2VZ2

2 Z1+Z2

Z1=characteristic impedance of cable

Z2=Z3=characteristic impedance of overhead line

Z1=√ L/kmC / Km

=√ 0. 4×10−3

0 .5×10−6=28 .28Ω

Z2=√ L/kmC / Km

=√ 1.5×10−3

0.015×10−6=316 .22Ω

V2=2¿20×103×316 .22

(2×28 . 28)+(316 . 22 )

V2=39.93KV

41. Ans:- (b)

Per phase current of the load=400

8+ j6=40∠−36 .860

∴Power factor = cos36.86

=0.8(lagging)



By connecting a bank of star connected capacitors, the reactive power component of

current is changed, where as the active power component of current is unchanged.

Active power component of current = IcosΦ

∴I1cosΦ1= I2cosΦ2

40×0.8=I20.9

I2=35.55A

Φ2=Cos-1(0.9)=25.840

I2=I1+Ic

Ic=I2+I1

=35.55 ∠-25.840 -40 ∠-36.860

Ic=8.5∠900

KVAR of th star connected bank

KVAR demand of load at 0.8 lag power factor

=√3V L I L sin φ1

=√3×400×√3×40×0 .6

=28.800KVAR

KVAR demand of load at 0.9lag power factor

=√3V L I L sin φ2

=√3×400×√3×40×sin 25 .84

=18.593KVAR

∴KVAR to be supplied by star connected capacitor bank

=28.8-18.59

=10.206KVAR

42. Ans:- (a)

Sol:- As the current is distributed equally, current on each side=5000A

Surge impedance of transmission line=205Ω

∴ The magnitude of transient over voltage=250×5000

= 1250KV

43. Ans:- (c)



Sol:-

Power received by load=50MW

Current at receiving end= √3V r I r cos φ

∴ I r=50×106

√3×220×103×0 .9

= 145.79A

Load power factor cosΦ=0.9, Φ=25.840

I r=145 .79∠25 .840

V S=133. 246∠7 . 770KV

Line to line sending voltage= √3×133 . 246∠7 .770KV

=230.78∠7.770 KV

Magnitude of line to line sending end voltage=230.78KV.

44. Ans:- (c)

Sol:- Let the initial power factor angle=Φ1

After connecting a capacitor the power factor angle = Φ2

Given Φ2 =cos-1 0.97

=14.070

KVAR supplied by capacitor= P(Tan Φ1 – Tan Φ2)

2¿106=4¿106

P(Tan Φ1 – Tan 14.07)

Φ1=36.890

cos Φ1=0.8lag

Hence if the capacitor goes out of service the load power factor will become 0.8

lagging

45. Ans:- (b)

Sol:- Tow networks are cascaded


V s=AV r+BIr

[V S

I S ]=[1 Z1

0 1 ] [ 1 0

( 1Z2

) 1 ][V r

Ir ]


C=1/Z2

Given C=0.025∠450

∴ Z2=1/C

Z2=40∠-450Ω

46. Ans:- (c)

Sol:

Under no load condition IR=0

VR(N.L)=

V S

A=240×103

0 .94

VR(N.L)= 255.3KV

% Voltage regulation =

V R (N . L)−V R (F . L)

V R( F .L)×100

=255 . 3−220

220×100

=16%

47. Ans:- (c)

Sol:- Velocity of wave propagation

Velocity wave propagation (

υ )

= 1

√( LKm )( C

Km )

Let ‘l’ be the total length of line

Total reactance of the line=0.045p.u.=2π fL

Total Inductance of line= 0 .045

2π×50



Total susceptance of line=1.2.p.u=2 π fc

Total capacitance of line= 0 .045

2π×50

Inductance/Km = 0 .045

2 π×50×l

Capacitance/Km= 1 .2

2π×50×l

ϑ= 1

√ 0.0452π×50×l

× 1.22 π×50×l

30×105= 1

7 . 4×10−4 l

∴ Length of the line (l)=222km

48. Ans:- (c)

Sol:-

Given SIL=

V 2

ZC =2280MW

Let characteristic impedance(Zc)=√ Zsc

Y oc

=√ 1 . 01 . 0

=1 .0 p .u

=√ impedance /kmadmit tan ce /km

30% Series capacitive compensation is provided

So Zsc=0.7 times original value

Zcnew=√ 0.7×Z sc

Y oc

=√ 0 . 71 . 0

=0 .836 p .u

SILnew=

V 2

ZCnew

= V 2

0. 836 ZC

=22800 . 836 =2727MW



49. Ans:- (b)

Hint:-

(A). Shunt capacitor is used at load end to improve the voltage profile. At the same

time it also improves the power factor at the load side.

(B). Series reactor smoothens the wave shape.

(C). By using series capacitor the net reactance of the line decreases

Power flow in the line=

EVX

Sin δ

Pα1X (So power flow of the line increases)

(D) Ferranti effect occurs during light load and no load condition of a long

transmission

Line due to shunt capacitance. To overcome the shunt capacitance effect we have to

connect the inductor in shunt to prevent the failure of insulation due to over

voltage.

50. Ans:-

Load rating =100 KW at 0.7 power factor (Lagging)

3-phse supply voltage = 11kv

KVAR demand of load at 0.7 P.f is

Q1100×103

0 . 7× sin( cos−1 0 . 7 )=102. 020 KVAR

KVAR demand of load at 0.95 p.f is

Q2100×103

0.95× sin (cos−10 .95 )=32 .868 KVAR

Hence to operate the load at 0.95 power factor

KVAR to be supplied by using series capacitors = (102.020-32.868)KVAR =

69.152KVAR

312XC=69.152KVAR



√3 V L IL cosφ=100×103

√3×11×103×IL×0 . 95=100×103

I L=5. 525 A

Since transmission line is a Y connected one

ILine=Iph

Xc / ph =

69 .152×103

3×(5. 252 )2=755 .125Ω

12 π fC

=755 .125

C/ph=

12 π×50×755 . 125

=4 . 215 μF

Since series capacitors are connected in phase manner

KVAR rating of each capacitor =

69 .1523

=23 .05 KAVR

Rating of capacitor is, 4.215 F, 23.05KAVR, 11KV [GATE – 1997]

51. Ans:

(a) Nominal - equivalent circuit of given transmission line is shown below.

Figure

Total resistance of each conductor, RL = 0.25 x 150 = 37.5 Ω / ph

Total inductive reactance of each conductor, XL = 0.5 x 150 = 75 Ω / ph

Total capacitive admittance to neutral, YC = 0.04 x 10-4 x 150 = 6 x 10-4 S/ph

Y C

2=3×10−4⇒π fC=3×10−4

C=3×10−4

π×50=1 . 9098 μF



V receiving / ph=(66

√3 )×103×− j 3333 .33

(37 . 5+ j75− j 3333. 33 )

¿(66√3 )×103×− j 3333 .33

(37 . 5− j3333 . 33 )V receiving / ph=38 .97 KV

V Re cieving( L−L)=√3×38 . 97=67 . 498 KV

[GATE – 1999]

52. Sol :- (a) By neglecting transmission line resistance for a long line, the nominal -

V s=(275

√3 )XL=0.05 x 400 = 20Ω

Y = 3x400 x 10-6 = 1200 x 10-6 = 1.2 ms

Y2

=1. 2×10−3

2=0 .ms

2 XC=1(Y /2 )

=1666 .67Ω

Receiving end voltage



V rph=V s (− j2 xc )jxL− j2 xC

V rph=275

√3×[− j 1666 .67

− j 1646 .67 ]×103

V rph=160. 7 KV

V r (L−L)=160 . 7 x √3=278 . 34 KV

Sol: (b)For a flat line magnitude of voltage throughout the line length is same. To

have

flat voltage profile the load impedance must be equal to characteristic

impedance of transmission line.

Characteristic impedance ZC= √ LC

=√ impedanceperkmadmit tan ceperkm

=√ 0. 053×10−6

=129Ω

Characteristic impedance loading =

v( L−L )

2

ZC

=

(275×103)2

129

=586.24 MW

(c)1.2 times of surge impedance loading =1.2x586.24 MW

= 703.5 MW

In this case the loading is greater than the surge impedance loading. To achieve

flat voltage profile the compensation must be of capacitive type.[GATE 2000]

53. Ans:-

Sol:- Figure

VS=VR cos h (1) + jZC sinh (1) IR

From standard form VS= AVR + BIR

A=cosh((1) , B= jZC sinh (1)

A= cosh (1.2x10-3x100) = 1.0



B= j 400 sin h (1.2 x 10-3 x 100) = 48.11 ∠900

Since given transmission line is of 100km length, let us consider the line as

short line

VS VR= 230KV

Maximum power transferred Pmax is given as

Prmax =

|V S||V r||B|

−|AB||V r|

2 cos ( β−α )

Qrmax= -|A

B| |V R|

2 sin( β−α)

∴Pr max=230×230×106

48 .11−| 1

48 . 11|(230×103 )2cos (90−0)

Prmax=1102.083MW

∴Qr max=−|148. 11

||230|2 sin( 90−0 )×106

¿−1102.83 MVAR

Total power transferred = (1102.083-j 1102.83) MVA [GATE-2000]

Solutions for Practice questions

1-b 6-a 11-a 16-a 21-d 26-b

2-b 7-a 12-a 17-a 22-c 27-a

3-a 8-a 13-b 18-d 23-c 28-a

4-a 9-c 14-b 19-d 24-a 29-c

5-c 10-a 15-c 20-b 25-b 30-c


CHAPTER 1

Documents

Transcript of CHAPTER 1