Voltage control Compensation techniques And power factor ...

Voltage control Compensation techniques

And power factor correction.

Prof. N. VISHALI

Dept. of EEE,

JNTUA College of Engineering, Kalikiri

Chittoor District, A P, India

Prof. N. VISHALI, Dept. of EEE, JNTUA College of Engineering, Kalikiri, Chittoor District, A P, India

Outline of Presentation


1. Introduction.

2. Types of compensation techniques.

3. About shunt compensation.

4. Series compensation.

5. Problems on compensation techniques.

6. Approach of compensation problems in gate.

7. Power factor correction.

8. Problems on power factor correction.

Introduction


1. In a power system, voltage at various buses tends to increase or decrease during its daily

operation. To ensure constant voltage to consumers, various techniques are utilized.

2. When the voltage is below the required level, reactive power produced by inductance

needs to be offset by capacitance.

Ex: synchronous condenser, shunt capacitor, series capacitor, tap changing transformer etc.

3. When the voltage is above the required level, reactive power produced by capacitance

needs to be offset by inductance.

Ex: shunt reactor, static VAR compensators etc.

4. Voltage control in an electrical power system is important for proper operation of electrical

power equipment to prevent damage such as overheating of generators and motors, to reduce

transmission losses and to maintain the ability of the system to withstand and prevent voltage

collapse.

Introduction (contd…)


6. We always in practice to reduce reactive power to improve system efficiency. This areacceptable at some level, if system is purely resistively or capacitance it make cause someproblem in Electrical system.7. AC systems supply or consume two kind of power: real power and reactive power. Realpower accomplishes useful work while reactive power supports the voltage that must becontrolled for system reliability.8. Reactive power has a profound effect on the security of power systems because it affectsvoltages throughout the system.9. Voltages can be maintained constant by using compensation techniques. The voltagefluctuations between the sending end and receiving end voltages can be maintained bycontrolling the reactive power in the system. Generally we have two techniques1. Internal control2. External control

Note:We have discussed about internal control in synchronous machines, the excitation of thealternator is controlled.

Types of compensation techniques


In power systems, we have discussed about the external control of voltage fluctuations

those are called compensation techniques. They are:

1. Shunt capacitor

2. Shunt reactor

3. Series capacitor

4. Synchronous condenser

5. Synchronous inductor

6. Synchronous phase modifier.

Among this the first three compensators are static compensators and the next three are

dynamic compensators.

Note:

According to gate syllabus, in this we learn about the static compensator.


Shunt compensation

1. For high voltage transmission line the line capacitance is high and plays a significant role involtage conditions of the receiving end.2. When the line is loaded then the reactive power demand of the load is partially met by thereactive power generated by the line capacitance and the remaining reactive power demandis met by the reactive power flow through the line from sending end to the receiving end.3. By placing shunt capacitor/shunt reactor during the undervoltage/overvoltage conditionsrespectively we can overcome the voltage fluctuations.4. When load is high (more than SIL) then a large reactive power flows from sending end tothe receiving end resulting in large voltage drop in the line.5. To improve the voltage at the receiving end shunt capacitors may be connected at thereceiving end to generate and feed the reactive power to the load so that reactive powerflow through the line and consequently the voltage drop in the line is reduced.

6. To meet the variable reactive power demands requisite number of shunt capacitors areswitched in, in addition to the shunt reactor, which results in adjustable reactive powerabsorption by the combination.


5. Shunt capacitor:a. This method is used to improve the power factor. Whenever an inductive load is connectedto the transmission line, power factor lags because of lagging load current. To compensate, ashunt capacitor is connected which draws current leading the source voltage.b. They also improve the voltage stability and reduce network losses.c. Improving the power factor also means a higher power transmission capability and increasedcontrol of the power flow.

Qc = P (tanΦ1 – tanΦ2)

QR = (VSVR / XL )cosδ – VS * VS /XL

φ1 = angle before compensation.

φ2 = angle after compensation.

Shunt compensation (cont.….)

From the figure the QR = Qc + QL , for UPF the QR = QC

If XC = 1/ Cω be the reactance of the shunt capacitor then the reactive power generated of leading VAR supplied by the capacitor:

QC = VR * VR / XC = VR ^2*Cω

Shunt compensation(cont.…..)

Prof. N. VISHALI, Dept. of EEE, JNTUA College of Engineering, Kalikiri, Chittor District, A P, India

6. Shunt reactor: 1. A shunt reactor is used to absorb the reactive Power. Which means it is used to compensatethe undesirable voltage due to line capacitance (Ferranti effect).2. The sending end voltage is higher than the receiving end voltage. The shunt reactor reducesthe voltage when the receiving end voltage is higher than the sending end voltage. Therefore, itincreases the energy efficiency of the power system.3. It is the most compact device commonly used for reactive power compensation in long high-

voltage transmission lines and in cable systems.4. If XL = Lω be the reactance of the shunt reactor (inductor) then the reactive VAR absorbed

by the shunt reactor:QC = VR*VR/XL = VR^2/Lω

QL = PL tanφ

5. To control the receiving end voltage generallyone shunt rector is installed and switched induring the light load condition.

Series compensation



where K is degree of compensation.The economic degree of compensation lies in the range of 40-70% (K < 1, i.e. 0.4-0.7)

Series compensation (cont.….)

prof N. VISHALI, Dept. of EEE, JNTUA College of Engineering, Kalikiri, Chittor District, A P, India

Series compensation (cont.….)


PROBLEMS

1. A 3-phase 11kv generator feeds power to upf load of 100mw through 3-phase transmission line. The line to line voltage is maintained constant at 11kv. The impedance of line based on 100mva and 11kv is j0.2pu. The line to line voltage at terminal is 11kv. The total reactive power injected at the terminal of load to maintain the line voltage at 11kv is


2. For the system shown below SD1 and SD2 are the complex power demand of buses 1 & 2 respectively. If V2 = 1pu then the VAR rating of capacitor Qc will be ?


Approach of compensation problems in gate


POWER FACTOR CORRECTION

1. The power factor correction means bringing the power factor of an AC circuit nearer to one

by using the equipment which absorbs or supply the reactive power to the circuit.

2. Usually, the power factor correction can be done by using the capacitor and the

synchronous motor in the circuit.

3. The power factor correction will not change the amount of true power, but it will reduce the

apparent power and the total current drawn from the load.

4. The most economical value of power factor lies between 0.9 to 0.95. If the value of power

factor lies below 0.8 (approx.), then it draws more current from the load.

5. The large current increases the losses and requires a large conductor, thus increases the

cost of the system. The loss can be reduced by correcting the power factor of the system.

Power Factor Correction MethodsThe power factor correction methods are mainly classified into two types, i.e., by

using the capacitor or through the synchronous condenser.


Power Factor Correction by using Capacitor Bank:1. In three phase system, the power factor isimproved by connecting the capacitors in star ordelta. The star and delta connected banks areshown in the figure.2. The capacitance requires in star connection ofthree phase transformer is equal to three times thecapacitance requires per phase when the capacitorsare connected in delta.3. Also, the working voltage of the star connectedbank is 1/√3 equal to the delta connected bank. Forthese reasons, the capacitors are connected in thedelta in three phase system for power factorimprovement.4. Delta connection is also better if the capacitors aredesigned for higher working voltage.


Power Factor Correction by Using Synchronous Condenser

1. The power factor can also be correct by installing the specially designed induction motor,

known as the synchronous condenser.

2. The synchronous condenser was running without the mechanical load, and it is connected

in parallel with the load.

3. It absorbs and generates the reactive power (Var) by varying the excitation of the motor

field winding.

4. The synchronous condenser is used for improving the power factor in bulk.

5. The output of the phase modifier can be varied smoothly.

6. The synchronous condenser has some disadvantage like it is costly and their installation,

maintenance and operation are also not easy.


Problems on power factor correction


Home work problems

1. A balanced delta connected load of 8+j6 Ω/ph is connected to 400v, 50hz, 3-Φ

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

connected capacitor the required KVAR of the bank is ……

2. A feeder with reactance of 0.2j pu has a Vs = 1.2pu. If QR = 0.3pu then

approximate voltage drop in feeder will be …… pu.

3. A transmission line has XL = j0.5pu. Calculate the rating and type of compensator

required for maintaining VS = VR = 1pu for the load.

(i) PR = 1pu at UPF. (ii) PR = 1pu at 0.8pf

Models & Performance of Transmission Lines

Prof. N. Visali

Dept. of EEE,

JNTUA College of Engineering, Kalikiri

Chittoor District, A P, India

Prof. N Visali, Dept. of EEE, JNTUA College of Engineering, Kalikiri, Chittoor District, A P, India

Outline of Presentation


Introduction

Types of Transmission lines

1. Short transmission line

2. Medium transmission line

3. Long transmission line

Skin Effect

Proximity Effect

Ferranti Effect

Surge Impedance

Previous GATE questions

1. one mark questions

2. Two mark questions

Work to students

Introduction


• 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

fcVXV

C

2

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

bad Weather 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.

Introduction

Performance of lines is meant the determination of efficiency

and regulation of transmission lines.

% efficiency = Power delivered at the receiving end × 100

Power sent from the sending end


Power delivered at the receiving end + losses

The end of the line where load is connected is called the

receiving end and where source of supply is connected is called

the sending end.


Introduction (contd.)


The regulation of a line is defined as the change in the receiving end

voltage, expressed in per cent of full load voltage, from no load to full load,

keeping the sending end voltage and frequency constant.

Mathematically expressed as,

% regulation = Vr′ - Vr × 100

Vr

where Vr′ is the receiving end voltage under no load condition and Vr the

receiving end voltage under full load condition.

It is to be noted here that Vr′ and Vr are the magnitudes of voltages.



It is to be noted that the electrical power is being transmitted over the

overhead lines at approximately the speed of light. In order to get one full

wave variation of voltage or current on the line, the length of the line for

50 Hz supply will be given by

f . λ = v

where f is frequency of supply,

λ is the wavelength i.e., the length of the line in this case and

v is the velocity of the wave i.e., the velocity of light.



Substituting for f = 50 and v = 3 × 10^8 m/sec.,

λ = v = 3×10^8 = 6 × 10^6 meters

f 50

= 6000 km.

Fig: Voltage distribution of 50HZ supply

From the above result, we can say in order to complete one complete

cycle of sinusoidal wave takes 6000KM physical distance of transmission

line.

Types of transmission lines


Based on the length of the length of the transmission lines it can be

classified as

1. Short Transmission Line ( <80KM)

2. Medium Transmission Line ( >80KM, <160KM)

3. Long Transmission Line ( >160KM)

1. Short Transmission Line (<80KM)


The lines up to about 80 km are termed as short transmission lines

where the effect of shunt capacitance is neglected.

Fig: Short Transmission Line

Short Transmission Line (contd.)


The vector diagram is drawn taking Ir, the receiving end current, as the

reference.

From the vector diagram,

Vs cos φs = Vr cos φr + Ir R

Vs sin φs = Vr sin φr + Ir X

Fig.: vector diagram for short transmission line




Vr

regulation per unit = Vr cos φr + Vx sin φr

where Vr and Vx are the per unit values of resistance and reactance of the line

A four terminal passive network the voltage and current on the

receiving end and sending end are related by the following pair of equations:

Vs = AVr + BIr

Is = CVr + DIr



This means A is the voltage impressed at the sending end per volt at the

receiving end when receiving end is open. It is dimensionless.

B is the voltage impressed at the sending end to have one ampere at the

short circuited receiving end. This is known as transfer impedance in network

theory.



C is the current in amperes into the sending end per volt on the open-

circuited receiving end. It has the dimension of admittance.

D is the current at the sending end for one ampere of current at the

short circuited receiving end.



The sending end voltage and current can be written from the equivalent

network as

Vs = Vr + IrZ

Is = Ir

The constants for short transmission line are when compared to ABCD

parameter equations is given by

A = 1 ; B = Z

C = 0 ; D = 1

Then determine sending end voltage using relation

Vs = AVr + BIr

To determine Vr′ the no load voltage at the receiving end

Vr′= Vs / A , when Ir = 0



From the obtained relations, We can write

To determine % η of transmission, the following relation is made use of



where R is the resistance per phase of the line

2. Medium Transmission Line ( >80KM, <160KM)


It has been mentioned previously that transmission lines with lengths

between 80 km and 160 km are categorized as medium length lines where

the parameters are assumed to be lumped.

The shunt capacitance is either assumed to be concentrated at the

middle of the line or half of the total capacitance is concentrated at each

end of the line.

The two configurations are as follows

nominal-T and

nominal-π

Medium Transmission Line (contd.)


Nominal T In this method, the whole line capacitance is assumed to be concentrated at

the middle point of the line and half the line resistance and reactance are

lumped on its either side as shown in Fig.

Therefore, in this arrangement, full charging current flows over half the line.

In Fig. one phase of 3-phase transmission line is shown as it is advantageous

to work in phase instead of line-to-line values.

Fig: Nominal T



The vector diagram for lagging power factor load is shown in Fig. While

analyzing the medium length lines using nominal-T, it is preferable to take

receiving end current as the reference vector as the calculations become

relatively easier as compared to taking Vr as the reference.

Fig.: vector diagram for nominal T transmission line




Vr

Where,

Vr′ is the receiving end voltage under no load condition

Vr is the receiving end voltage under full load condition

Vr′ can be written as (voltage divider rule)





where P is the 3-phase power delivered at the receiving end, R is

the resistance per phase.



Writing down the voltage and current equation,

Vs = AVr + BIr

Is = CVr + DIr By the relations from phasor diagram we get

Where ,

Z is the total impedance of the transmission line = R+jX

Y is the shunt admittance of the transmission line



In the nominal pi model of a medium transmission line, the series

impedance of the line is concentrated at the centre and half of each

capacitance is placed at the centre of the line. The nominal Pi model of

the line is shown in the diagram below.

Nominal π

Fig.: vector diagram for nominal π transmission line



The vector diagram for lagging power factor load is shown in Fig.

While analyzing the medium length lines using nominal- π

Fig.: vector diagram for nominal π transmission line




Vr

Where,

Vr′ is the receiving end voltage under no load condition

Vr is the receiving end voltage under full load condition

Vr′ can be written as (voltage divider rule)





where P is the 3-phase power delivered at the receiving end, R is the resistance per phase.



Writing down the voltage and current equation,

Vs = AVr + BIr

Is = CVr + DIr By the relations from phasor diagram we get

Where ,

Z is the total impedance of the transmission line = R+jX

Y is the shunt admittance of the transmission line

3. Long Transmission Line (>160KM)


So far electrically short transmission lines less than 160 km in length

have been considered wherein the parameters are assumed to be

lumped.

In case the lines are more than 160 km long, for accurate solutions

the parameters must be taken as distributed uniformly along the length

as a result of which the voltages and currents will vary from point to

point on the line. Consider Fig. below for analysis.

Long Transmission Line (contd.)


Let z = series impedance per unit length

y = shunt admittance per unit length

l = length of the line

Z = zl = total series impedance

Y = yl = total shunt admittance

Fig: Long Transmission Line



Fig: Equivalent circuit for Long Transmission Line

Voltage and current Equations for long transmission line



ABCD parameters for long transmission line

For a lossless line r = 0, g = 0

A pure resistance, and this is known as surge impedance of the line.

This will be explained in later on concepts.


Skin effect is the tendency of an alternating electric current(AC) to become

distributed within a conductor such that the current density is largest near

the surface of the conductor, and decreases with greater depths in the

conductor. The electric current flows mainly at the "skin" of the conductor,

between the outer surface and a level called the skin depth.

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 an alternating

current system, where the current tends to flow with higher density through the

surfaceof the conductors (i.e. skin of the conductor), leaving the core deprived of

necessary number of electrons.

Skin Effect


In fact there even arisesa 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 electrical resistanceof the conductor.

This particular trend of anACtransmission

system to take the surface path for the flow

of current depriving the core is referred to as the skin

Effect in transmission lines.

Skin Effect contd.

Skin Effect contd.


Let us initially consider the solid conductor to be split up into a numberof

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 .

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 surfaceaswell as those in the core.

Whereasthe flux set up by the outer filaments is restricted only to the

surface itself and is unable to link with the inner filaments.

Cause of Skin Effect


Thus the flux linkage of the conductor increases aswe move closer towards

the core and at the same rate increases the inductor as it has a direct

proportionality relationship with flux linkage.

This results in a larger inductive reactancebeing 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

phenomenacalled skin effect in transmission lines.

Skin Effect contd.


Skin Effect contd.


The skin effect in an ac system

depends on

Shape of conductor

Type of material

Diameter ofthe conductors

Operational frequency

Skin Effect contd.


TheACcurrent density J in aconductor decreases exponentially from its value

at the surfaceJS according to the depth d from the surface, as follows:

where δ is called the skin depth.The skin depth is thus defined as the depth

below the surfaceof the conductor at which the current density has fallen to

1/e (about 0.37)of JS. In normal cases it is well approximated as:

where

• ρ =resistivity of the conductor

• ω =angular frequency of current =2π×frequency

• μ =absolute magnetic permeability of theconductor

Skin Effect contd.


Instead of normal conductors/wires A type of cable called litz wire

(from the German Litzendraht, braided wire) is used to mitigate the

skin effect for frequencies of a few kilohertz to about one megahertz.

It consists of a number of insulated wire strands woven together in a

carefully designed pattern, so that the overall magnetic field acts equally

on all the wires and causes the total current to be distributed equally

among them.

With the skin effect having little effect on each of the thin strands, the

bundle does not suffer the same increase in ACresistance that a solid

conductor of the same cross-sectional area would due to the skin effect.

Mitigation of Skin Effect

Skin Effect contd.


Proximity Effect

When the conductors carry the high alternating voltage then the currents

are non-uniformly distributed on the cross-section area of the conductor.

This effect is called proximity effect.

The proximity effect results in the increment of the apparent resistance

of the conductor due to the presence of the other conductors carrying current

in its vicinity.

When two or more conductors are placed near to each other, then their

electromagnetic fields interact with each other. Due to this interaction, the

current in each of them is redistributed such that the greater current density is

concentrated in that part of the strand most remote from the interfering

conductor.


If the conductors carry the current in the same direction, then the

magnetic field of the halves of the conductors which are close to

each other is cancelling each other and hence no current flow

through that halves portion of the conductor. The current is crowded

in the remote half portion of the conductor.

Proximity Effect contd.


When the conductors carry the current in the opposite direction, then

the close part of the conductor carries, the more current and the

magnetic field of the far off half of the conductor cancel each other.

Thus, the current is zero in the remote half of the conductor and

crowded at the nearer part of the conductor.



Factors Affecting the Proximity EffectFrequency – The proximity increases with the increases in the frequency.

Diameter – The proximity effect increases with the increase in the

conductor.

Structure – This effect is more on the solid conductor as compared to the

stranded conductor (i.e., ASCR) because the surface area of the stranded

conductor is smaller than the solid conductor.

Material – If the material is made up of high ferromagnetic material then

the proximity effect is more on their surface.



The proximity effect can be reduced by using the ACSR (Aluminum

Core Steel Reinforced) conductor. In ACSR conductor the steel is

placed at the centre of the conductor and the aluminum conductor is

positioned around steel wire.

The steel increased the strength of the conductor but reduced the

surface area of the conductor. Thus, the current flow mostly in the

outer layer of the conductor and no current is carried in the centre of

the conductor. Thus, reduced the proximity effect on the conductor.


Proximity Effect Mitigation


Ferranti Effect

The effect in which the voltage at the receiving end of the

transmission line is more than the sending voltage is known as the

Ferranti effect. Such type of effect mainly occurs because of light load

or open circuit at the receiving end.

Ferranti effect is due to the charging current of the line. When an

alternating voltage is applied, the current that flows into the capacitor

is called charging current. A charging current is also known as

capacitive current. The charging current increases in the line when the

receiving end voltage of the line is larger than the sending end.


The transmission line generates capacitive reactive volt-amperes in its

shunt capacitance and absorbing reactive volt-amperes in its series

inductance. The load at which the inductive and capacitive reactive volt-

amperes are equal and opposite, such load is called surge impedance load.

It is also called natural load of the transmission line because power is

not dissipated in transmission. In surge impedance loading, the voltage

and current are in the same phase at all the point of the line. When the

surge impedance of the line has terminated the power delivered by it is

called surge impedance loading.

Surge Impedance


Shunt capacitance charges the transmission line when the circuit

breaker at the sending end of the line is close. As shown below

Surge Impedance Contd.


Let V = phase voltage at the receiving end

L = series inductance per phase

XL = series inductance reactance per phase

XC = shunt capacitance reactance per phase

Zo = surge impedance loading per phase

Capacitive volt-amperes (VAr) generated in the line



The series inductance of the line consumes the electrical energy when

the sending and receiving end terminals are closed.

Inductive reactive volt-amperes (VAr) absorbed by the line



Surge impedance loading is also defined as the power load in which

the total reactive power of the lines becomes zero.

The reactive power generated by the shunt capacitance is consumed

by the series inductance of the line.


GATE previous questions (1m)


For a 500HZ frequency excitatioin, a 50KM short power line will be

modeled as

a) Short line

b) Medium line

c) Long line

d) Data insufficiency

Sol. Choice C

λ = v = 3×10^8 = 6 × 10^5 meters= 600 KM

f 500

GATE previous questions (1m) contd.


The concept of electricity short, medium and long line primarily

based on the

a) Nominal voltage of the line

b) Physical length of the line

c) Wavelength of the line

d) Power transmitted over the line

Sol.: choice C

Depending upon the wavelength of the line, charging current

changes and we consider the line parameters as lumped or

distributed. The concept of short, medium and long lines depends on

the parameters (i.e. lumped or distributed).



Total instantaneous power supplied by 3-phase Ac supply to balanced

R-L load is

a) zero

b) constant

c) Pulsating with zero average

d) pulsating with non- zero average

Sol.: choice B , instantaneous phase voltage and currents are

va = Vm sin ωt Ia = Im sin (ωt- Φ)

vb = Vm sin (ωt-120) Ib = Im sin (ωt- Φ-120)

vc = Vm sin (ωt-240) Ic = Im sin (ωt- Φ-240)

Power = va Ia + vb Ib + vc Ic = 3VI Cos φ ( constant )



In a long transmission line withr, l, g and c are the resistance, inductance,

Shunt conductance and capacitance per unit length respectively,

the condition for

Distortion less transmission is

a) rc=lg

b) r=l/c

c) Rg=lc

d) g2 =c/l

Sol.: Choice A



A 220KV, 20Km long, 3-phase transmission line has following ABCD

Parameters A=D=0.96 ∠ 3°,B=55 ∠ 65° ohm/phase,

C=0.5*10-4 ∠ 90° mho/phase. Its correct charging current per phase is

a) 11*sqrt(3) A

b) 11 A

c) 220 A

d) 220/sqrt(3) A

Sol. : choice A

charging current = Vph*Y

=220*103*0.5*10-4/sqrt(3)

= 11/sqrt(3) A


The surge impedance of 400km long overhead transmission line is

400ohms. For a 200 Km length of the same line, the surge impedance will

be (in ohms)

a) 200

b) 800

c) 400

d) 100

Sol.: Choice C

Surge impedance for a given transmission line is constant and is

independent

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

on magnitude of inductance and capacitance.


GATE previous questions (2m)


The ABCD parameters of a 3-phase transmission line are A=D=0.9 ∠ 0°,

B=200 ∠ 90° ohm and C=0.95*10-3 ∠ 90° mho. At no-load condition, a

shunt inductive reactor is connected at the receiving end of the line to

Limit the receiving-end voltage to be equal to sending –end voltage. The

Ohmic value of the reactor is

a) Infinity ohm

b) 2000 ohm

c) 105.26 ohm

d) 1052.6 ohm


Sol.: choice B



At no load condition, a 3-phase, 50HZ, lossless power transmission line has sending-end

And receiving –end voltages of 400Kv and 420kv respectively. Assuming the velocity of

travelling wave to be the Light, the length of the line,in km,is



The generalized circuit constants of 3-phase, 220KV rated, medium

length transmission line are A=D=0.936+j0.016=0.936 ∠ 0.98°,

B=33.5+j138=142.0 ∠ 76.4° ohm

C=(-5.18+j 914) * 10-6 mho

If the load at the receiving end is 50MW at 220KV with a power factor 0.9

Lagging, then magnitude of line to line sending end voltage should be

a) 133.23 KV

b) 220.00 KV

c) 230.78 KV

d) 246.30 KV



Sol.: choice C



For a 400 Km long transmission line, the series impedance is

(0.0+j0.5)/km and the Shunt admittance is (0.0+j5.0)* 10-6 mho/km.

The magnitude of the series impedance (In ohms) of the equivalent pi

circuit of the transmission line is



Sol.:



A three-phase 50 HZ, 400KV transmission line is 300Km long. The line

inductance is1mH/Km per phase and the capacitance is 0.01*10-6F/ Km per

phase. The line is under Open circuit condition at the receiving end and

energized with 400KV at the sending end, the receiving end line voltage in KV

( round off to two decimal places) will be



Sol.:



A lossless transmission line having surge impedance loading (SIL) of 2280MW. ASeries capacitive compensation of 30% is emplaced. Then SIL of the compensated Transmission line will be (in MW)a) 1835b) 2280c) 2725d) 3257

A cable has the following characteristics, L=0.201micro H/m and C= 196.2pico F/m. The velocity of the wave propagation through the cable is (in m/sec)

a) 32b) 159.24c) 0.0312d) 159.24

Work to Students


The total reactance and total susceptance of a lossless overhead EHV

line, operating at 50Hz are given by 0.045 pu and 1.2 pu respectively. If

the velocity of the wave propagation is 3*105 Km/sec, then the

approximate length of line is

A 3-phase, 50HZ generator supplies power of 3MW at 17.32KV to a

balanced 3-phase Inductive load through an overhead line. The per unit

phase line resistance are 0.25 ohm and 3.925 ohm respectively. If the

voltage at the generator terminal is 17.87KV, the Power factor of the load is

Work to Students


The End

Voltage control Compensation techniques And power factor ...

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