labmannual 2nd cycle

8/22/2019 LabMannual 2nd Cycle

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Experiment No.1

Synchronization of a given Alternator with infinite bus

MOTIVATION

In modern power systems, several generators have to be run in parallel for reliableand economic operation. Maintenance of the units can be optimally scheduled and

additional units can always be added as and when required with ease. Hence in large

power stations a study of the load sharing among the generators running in parallel is of

great significance. Unlike the case of D.C generators where there is no constraints on thespeed, in case of alternators operating in parallel, the frequency of operation is fixed. For

alternators running in parallel load sharing can be affected by changing the energy inputs

to the alternators. For a given prime mover setting, changes in excitation will simply

affect the reactive power, while the sharing of the active power will depend on the speedload characteristics of the prime movers. Thus, it is essential to study the factors affecting

the sharing of load between two alternators in parallel.

OBJECTIVES

(a) Synchronization of a given alternator with infinite bus.(b) To study the effect of speed controller of Prime mover on real power sharing.

(c) To study the effect of excitation on reactive power sharing.

THEORY

In order that alternators may operate satisfactorily in parallel, their prime movers musthave drooping speed load characteristics. All types of prime movers, steam engines,

steam turbines, water wheels, etc. which operate with fixed governor settings, for

example with fixed tension of governor springs, must decrease in speed slightly in order

to carry increase in load since the governors of the prime movers depend for their actionupon a change in speed . When an increase in load is applied to a prime mover, it slows

down. This causes the governor to act and admit more steam or water as the case may be.

The prime mover continues to slow down until equilibrium is established between thepower it receives and its output plus its losses. It follows that, if a prime mover must

operate at a fixed speed, the only way that it can be made to deliver more or less power is

to change the governor setting, usually by changing the tension of the governor spring, insuch a way as to increase or decrease the power given to the prime mover at the fixed

speed. The only way the synchronous generator can be made to deliver more power at the

same speed is to increase the input from its prime mover by changing the governorsetting.

Before two three-phase systems can be synchronized the following conditions

should be met:

(a) Frequency of both the systems should be the same.


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(b) Voltages must be equal.

(c) Phase sequences of both the systems should be same.

Voltage magnitude can be checked by means of voltmeters and the rest,simultaneously, by means of synchroscope.

A simple form of synchroscope consists of three bulbs which may be connected intwo ways. One is known as two lamps bright, one lamp dark method and the other is

known as all lamps dark method. First method is preferred over the second.

Figure 1.1(a) shows the speed load characteristics of each of the prime moversdriving the alternators. For clarity the change of speed is exaggerated. The speed load

curves of the prime movers (turbines) of the alternators are determined by their

governors. If the alternators are motor driven, the speed load characteristics depend

on the motor speed load characteristics.The prime mover governors are adjusted. Figure 1.1(a) shows that the no load

frequency f0 of the two alternators is the same. Let frbe the rated frequency at which

the system is operating. At this frequency, oa is the load on the alternator 1 and ob is

the load on the alternator 2.The kilowatt output of a synchronous generator operating in parallel in a system

of constant frequency and constant voltage cannot be changed by changing itsexcitation. As the generator speed does not change, the governor and its prime mover

cannot act to alter the generator output. A change in the excitation alone merely

changes the power factor at which the synchronous generator operates with out

changing the power output.The output of a synchronous generator with fixed terminal voltage, fixed

excitation and fixed frequency can be changed by changing its governor setting

(speed load characteristic) in such a way as to give it more steam at the same speed insteam turbine or more water in a water in a water wheel.


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Assume that it is intended to increase the system load and that the increase in load

should be taken by unit 1 with unit 2 retaining its earlier load as per Fig 1.1(a). In

such a situation it is necessary to raise the speed load characteristic of machine 1 asshown in Fig. 1.1(b). The system frequency and load terminal voltage are however to

be maintained constant at the rated values. If for a given system load (= oa + ob

shown in Fig.1.1(a)) it is intended to increase the load on machine 1 and decrease theload on machine 2 then it is necessary to raise the speed load characteristic of the

machine 1, and lower the speed load characteristic of the machine 2, keeping in mind

constant system frequency. Conceptually the different speed load characteristics can

be achieved in a d.c. motor alternator set by changing the speed of the d.c. motorworking as a prime mover to the alternator.

PRE-EXPERIMENTAL QUIZ

1. What do you mean by synchronization?2. What is the effect of improper synchronization?

3. While following the synchronization procedure, is it necessary to follow certainorder?

4. What are the different methods of synchronization? Which one do you think is the

best way and why?

5. Why is it necessary to run various alternators in parallel?6. What will happen if the prime movers have flat speed load characteristics?

7. What is the effect of changing excitations in two alternators running in parallel?

8. How are the real load and reactive load sharings affected?9. How would you ensure balanced loading?

EQUIPMENT AND COMPONENTS(a) Two synchronous machines coupled to d.c. shunt motors (prime movers).

(b) d.c. and a.c. ammetrers and voltmeters.

(c) Wattmeters.(d) Synchronizing panel with lamps.

(e) Tachometer

(f) Loading rheostats and field rheostats.

PROCEDURE, CONNECTION DIAGRAM AND EXPERIMENTATION

Test 1. Synchronization of a given alternator with infinite bus.

a) Make the connections as per diagram shown in Fig 1.2.b) Close the switch SW1. Drive the d.c. machine M1 as prime mover. Adjust R1 to

run the alternator G1 at its rated speed corresponding to the rated frequency. Adjust

the alternator excitation to provide rated terminal voltage at no load.

c) Check the conditions for synchronization with infinite bus as follows


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Frequency of both the systems should be the same. Voltages must be equal. Phase sequences of both the systems should be same.Voltage magnitude can be checked by means of voltmeters and the rest,

simultaneously, by means of synchroscope using two lamps bright, one lamp dark

method.d) Close the switch SW2, when conditions for synchronization are satisfied.

Test 2. To study the effect of speed controller of Prime mover on real power sharing.

a) Make the connections as per diagram shown in Fig 1.2.

b) The procedure for synchronization remains the same.c) Change the speed of prime mover and observe the real power exchange with

infinite bus.

d) Note down the reading as per the table given below.

S.No.

Ammeter

reading (A1)(Amps)

Ammeter

reading (A2)(Amps)

Wattmeter

(W1)(Watts)

Wattmeter

(W2)(Watts)

TerminalVoltage

(Alternator)

(Volts)


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Test 3. To study the effect of excitation on reactive power sharing.

a) Make the connections as per diagram shown in Fig 1.3.b) The procedure for synchronization remains the same.

c) Change the excitation of synchronous machine and observe the reactive power

exchange with infinite bus.d) Note down the reading as per the table given below.

S.No.Terminal Voltage

(Alternator)

(Volts)

Excitation

Current

(A3)(Amps)

Armaturecurrent (A2)

(Amps)

Wattmeter(W1)

(VAR)


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Experiment No. 2

SteadyState power limit of a transmission line

MOTIVATION:

Normal steady-state conditions are used as a basis for designing an electrical

system and determining its main technical and economical characteristics. Power-Angle

curves are not only important in determining steady-state power limit of a system, but

they are also useful in determining the transient power limits of a system.

OBJECTIVES:

i) To draw the Power-angle diagram for a transmission line.

ii) To study the change in power transfer for variations in sending-end and receiving-end

voltages.

THEORY:

The equations for the real and reactive powers in terms of the nodal voltages and

the driving-point and transfer admittance can be established. Such equations are found

very useful in power system analysis work. The simple circuit of Fig. 2.1 can be analyzed

in detail and the results extended to the general case of a network with n nodes.

In Fig. 2.1,

12R11SS YVYVI

22R21SR YVYVI

11Y , 22Y are the driving-point admittances and Y,Y12 are the transfer admittances.

The complex power is defined as


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PS jQS =*SV IS

Where QS is positive when the reactive power is delivered to the system. The

expressions for real power are then: 2S S 11 11 S R 12 12 2P V Y sin V V Y sin

121221RS22222RR sinYVVsinYVP

where

12112

1111 90

1212 90

2222 90

1Vs Vs

2r Vr V

12Yij Yij

1221

The expressions for reactive powers are

121212RS11112SS cosYVVcosYVQ

121221RS22222RR cosYVVcosYVQ

For a lossless system 11, 12 etc. are zero.

For a lossless short-line representation, having a reactance X, the real power at the

two ends will be the same and equal to

12RS

RS sinX

VVPP

and the reactive powers are given by

2

S S R

S 12

V V Vcos

X XQ


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12RS

2R cos

X

VV

X

V

The above equations show that the transfer of real power between the sending and

receiving end is directly proportional at the two ends. Keeping the magnitudes of the

voltages constant, a plot between the power and the phase angle , known as the power

angle diagram can be drawn. The maximum power transfer occurs when = 90 i.e. the

steady-state power limit is

X

VVP RSm

The power that can be transmitted over a transmission system is limited by

thermal considerations, economics, steady state power limits and transient power limits.

Two typical cases of stability arise: (1) the line supplies a load to the receiving

end which has no generation in which case the maximum power transfer depends on the

load characteristics (2) the line is a part of the network with synchronous generators at

each end, so loss of synchronism is possible. The second is usually the more important

and common case.

The power in a static impedance load of constant power factor is proportional to

the square of the voltage. As the load power is increased, the voltage drops slightly as

shown in Fig. 2.3, then more rapidly until maximum power is reached. Thereafter the

voltage and power fall still further, but the system can operate stably at these lower

voltages. A system normally contains motor as well as static (lighting and heating) loads,

and with induction motors the reactive power requirements increase as the voltage falls,

and beyond the maximum reactive power point as the motors will stall.

The simplest criterion for synchronous stability is [ p/ > 0]. The criterion for

load instability seen in induction motors is derived from the VQ characteristics of the

induction motors. Below a certain voltage the reactive power consumed increases with

decrease in voltage until dQ/dv when the voltage collapses.


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PRE-EXPERIMENTAL QUIZ

1. What is a power-angle curve?

2. Explain briefly the significance of Power-Angle curve in stability studies.

3. What is synchronizing-power co-efficient? What should be the condition on

synchronizing power co-efficient for system stability?

4. Formulate the synchronous and load stability criteria for a simple electrical system.

5. What is the effect of X/R ratio of the transmission line on the Power Angle curve?

6. What is a Receiving-end power chart? How is it helpful in determining steady-state

power limit of the transmission line?

7. What is the maximum value for from synchronous stability point of view?

8. What could be the difference between normal, faulted, and the post-fault power-angle

characteristics of a system?

MATERIALS AND EQUIPMENTS

Transmission-line model, 3-phase phase-shifting transformer, 3-phase and 1-

phase variacs, Voltmeters, Ammeters and Wattmeters of suitable ratings.

PROCEDURE, CONNECTION DIAGRAM, EXPERIMENTATION AND

PRECAUTIONS

1. The connections are given as in Fig. 2.4.

2. Set the value of the transmission line reactance at X1, Keeping the switch S

open, adjust the 3-phase and 1-phase variacs to obtain 100V in voltmeters V S and

VR.

3. Adjust the 3 phase phase-shifting transformer to obtain zero voltage across the

switch S. In our case we are using the 3-phase induction motor for getting the

phase shift, and then zero voltage can be reached by adjusting the rotor winding

position manually.

4. Close S and check the readings of the ammeters and wattmeters. They should be

at zero.


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5. Note the phase shifting transformer rotor angle as zero value. Increase the phase

angle up to 180 in steps of 10 with the help of the phase shifting transformer.

6. Note the ammeter and wattmeter readings for each value of. The values of VS

and VR should be held constant. The current in the ammeters should not exceed

the rating of the transmission line model or the phase shifter.

7. Repeat the experiment for different values of Vs / VRby keeping the other three

quantities VR/ Vs, X and as constant.

PRECAUTIONS

1. The Voltmeter across the switch S should have double the range of VS(or VR).

2. Preferably use low power-factor wattmeters.

3. If an iron-cored choke is used in the transmission line model, the choke should not

saturate.

DATA SHEET VS= VR=

Line reactance =

S.No 12 WS WR IS IR cosS

(calculated)

cosR

(calculated)

QS QR

Data-Processing and Analysis

i) Plot the curves (a) WS vs and (b) WRvs for different values of line reactances.

ii) Find out the steady-state power limit from the Power angle curve.

iii) Find out P/ at = 30 from the power-angle curve.

iv) Find out P/V with V = 5V at = 30.


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v) Plot a) QS vs and b) QR vs for different values of line reactances, QS and QR may

be determined by finding the power factors at the sending and receiving ends.

vi) Find out Q/V and Q/ at = 30.

POST-EXPERIMENTAL QUESTIONS, DISCUSSIONS AND CRITICAL

EVALUATION

1. Find out the steady-state power limits from the power angle curve determined

experimentally. Compare it with the theoretical value. Explain any difference.

2. How does to the reactance of the transmission line affect, the steady-state power

limit?

3. What conclusion can you draw about the system from the experimental values of

P/ and P/V at = 30?

4. What conclusions can you draw about the system from the experimental values of

Q/ and Q/V at = 30?

5. If the generator power is kept constant, what do you expect the form of variation

between VS (or VR) with .? Can you arrive at the critical condition for steady-state

power limit from the above relation?

6. Enumerate the various methods of increasing the steady-state power limit.

7. Will series-capacitor compensation aid in improving the system steady-state power

limit? If so, explain how it improves?

8. What are the various problems associated with series compensated lines?

9. What is the expression for steady-state stability limit for the following electrical

system in terms of sending end voltage, line reactance, and p.f of the load?.

A synchronous generator supplying power to a static impedance load through a

transmission line at a given power factor.

SUGGESTIONS FOR FURTHER STUDY

1. Draw the variation of receiving end power vs receiving end voltage for different

p.fs of the load for synchronous system supplying, a static load and there from

determine the steady-state power limit of the system.

2. Study the power-angle curve for the following systems.


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i) Synchronous generator supplying power to a static load

through a transmission line.

ii) Synchronous generator supplying power to an induction

motor through a transmission line.

3. Experimentally determine the variation of Vs. VS (or VR) for a constant power

output.

4. A phase-shifter is included in one of the lines of a double-circuit system. Compare the

power transfer capability of such a system with a double-circuit system having no

phase-shifter.

REFERENCES

1. Transient processes in Electrical power systems by V. Venikov

2. Elements of Power system analysis by Stevenson.

3. Electric power transmission by Woodruff.

4. Transmission lines by Zaborsky.

Y11 = y1 + y2

Y22 = y1 + y3

Y12 = Y21= - y1

Fig 2.1 Simple Transmission line model


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Fig 2.2 One line diagram of Transmission line model

Fig 2.3 PV curve for different power factor

Fig 2.4 Connection Diagram for finding power limit


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Experiment No.3

a) Study of a power directional relay

MOTIVATION:

A 4-pole induction cup construction is widely used for realizing almost

any steady state relay characteristic with identical components except for the coil

windings. Power directional relay is realized using 4 pole induction cup construction.

These relays are employed for the protection of parallel feeders, ring feeder and feeder

fed at both the ends. It is extremely important to learn constructional features andoperating characteristic of such relay.

OBJECTIVIES:

1. To study construction and principle of operation of a 4 pole induction cup type

power directional relay.

2. To determine operating characteristics of such relay.

THEORY:

Fig. 3.1 shows the construction of a 4-pole induction cup relay. It has a magnetic core

possessing four pole which face inwards. Two diametrically opposite poles carry current

winding and the other pair of poles carries voltage winding. The moving element of the

relay, in the form of a cylindrical rotor (cup) is arranged in the space between the pole

tips. The moving contact is attached to the cup. The turning of the rotor is restricted by a

spring (not shown in the diagram.)

The current I when passed through the current coil create flux I in the working gap.

Under the applied voltage V, a current Iv flows through the voltage coil to develop the

flux V. Iv lags the voltage V by an angle v, i.e.


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v = tan1

(Xv / Rv)

where Xv and Rv are the reactance and resistance of the voltage winding.

Hence two fluxes V and I shifted relative to each other by an angle of 90 degrees actupon the rotor. Inter action of eddy currents and fluxes produce a driving torque T acting

upon the rotor, where

T = K1 V I Sin

Where the angle between V and I. (See fig.3.2)

For an unsaturated magnetic system it may be assumed that the fluxes are proportional to

the currents, which create them.

i.e. I I and V IV V

Hence the driving torque takes the form

T = KVI Sin = KVI Sin (90 )

Where is the angle by which current phasor I lags the voltage phasor.

= 90v, termed the internal angle of the relay. Finally the expression for driving

torque becomes

T = KVI Cos ( + )

If the torque developed in the relay exceeds the resisting torque created by spring and

friction in the pivots, and, further more has a positive sign (direction), the relay operates.


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The relay develops maximum torque for given V and I, if = - i.e. when

Cos ( + ) = 1. The phase angle between V and I for which relay develops maximum

torque is known as maximum torque angle (MTA). Fig.3., shows the operating

characteristic of directional relay.

The high operating seed is obtained by forming the induction disc into an

inverted cup, so that its inertia is greatly reduced and by designing the pole system to

give maximum torque per volt ampere input. In the 4-pole unit almost all the eddy

currents induced in the cup by one pair of poles appear directly under the other pair of

poles , so that the torque /VA is about three times that of the induction disc even when

the later is equipped with a C shaped magnet. The greater efficiency of the induction cup

relay is due to the arrangement of its magnetic circuit to minimize magnetic leakage and

to reduce the resistance of the rotor induced current paths.

By designing to avoid magnetic saturation, the operating characteristic of

the relay can be made linear and accurate over a very wide range, with pick up and reset

values close together; this simplifies application and testing.

This relay unit is particularly suited to role as a directional or phase

comparison unit, not only on account of its sensitivity and speed but because it has a

steady non vibrating torque.

PRE EXPERIMENTAL QUIZ:

1. What will the torque equation of an induction type directional relay are if one flux

is direct current and the other is alternating current?

2. How would you realize a directional over current relay using an over current and

directional relay units?

3. What do you understand by polarizing quantity of a directional relay?


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4. Discuss the principle of operation of O.C directional relays?

APPARATUS:

Power directional relay

3 phase variac

3 phase induction regulator

Ammeters

Voltmeters

Current Transformer

Potential Transformer

Rheostats

1 phase variac

PROCEDURE:

1. Connect the circuit as shown in the Fig. 3.4.

2. Adjust the position of the induction regulator to make phase difference between V

and I zero.

3. Set current at a low value and obtain the magnitude of V required for the

operation of relay when the phase angle is varied in the range zero to 360degrees

4. Repeat the experiment for different values of I.

DATA SHEET:

Current in mA S.No in degrees Voltage in Volts


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DATA PROCESSING:

Plot a graph between V and for given current I. Obtain MTA of this relay.

POST-EXPERIMENTAL QUIZ:

1. What do you understand by 30 and 90-degree connection schemes of power

directional relays?

2. A 50 Hz, single phase directional relay of the current voltage type has voltage coil

whose impedance is 230+560j . What is the MTA of this relay?

It is desired to modify this relay so that it will develop a maximum positive torque

at 45 degree power factor lagging. Suggest the desired modifications.

3. Give some applications of the power directional relay?

4. Suggest an electronic circuit for realizing a power directional relay.

REFERENCES:

1. M. Titarenko and Noskov-Duketsky, Protective Relaying in Electric Power

Systems Peace Publishers, Moscow.

2. C.R.Mason, The Art and Science of Protective Relaying Wiley Eastern 1977.

3. A.R. Van Warrington, Protective Relays, theory and practice, Vol. I, Chapmanand Hall 1969.


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Fig. 3.1: Four pole induction cup relay

V

I

I

IV

V

V

Fig. 3.2: Phasor diagram showing the voltage, currentand fluxes for power directional relay


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Non

OperatingZone

V

Iv

= 90

MTA

I

V

I

MTA = Maximumtorque angle

Operating

Zone

Fig. 3.3: Operating and non-operating zones of apower directional relay

Dc

Voltage

source

3 induction

regulator

3phase

variac

R

Y

B

A2

B2

C2

415 V

3

supply c2

b2

V

-

+

30 V

65

7

8

9

10

Relay

ARheostat

1

variacV

Ph

N

230 V

1

supply

Fig. 3.4: Circuit diagram for studying characteristics of powerdirectional rela


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b) NEGATIVE PHASE SEQUENCE RELAY

MOTIVATION:

Negative phase sequence currents flow in the armature windings of an

alternator due to unbalanced loading and unsymmetrical faults at the terminals of the

machine. The negative sequence component of armature current produces reaction

magnetic field, which rotates in direction opposite to the rotating D.C field system, which

cuts the rotor at twice the rotational velocity, there by inducing double frequency currents

in the field system and in the rotor body. The resulting eddy currents are very large andcause severe heating of the rotor. So severe is this effect that a single phase load equal

to the normal three phase rated current can quickly heat the brass rotor slot wedges to

the softening point.

Short time heating is of interest during system fault conditions; the heat

dissipated during such periods is negligible and the heat generated can be considered to

be entirely retained with in the thermal capacity of the rotor. Hence it is extremely

necessary to provide negative phase sequence protection to the generator.

OBJECTIVES:

1. To study the construction of a typical negative phase sequence relay.

2. To experimentally verify the calibration of the typical negative phase sequence

relay.

THEORY:

The negative sequence component can be detected by the use of a filter

network. Many circuits for this purpose have been evolved. The filter network employed

in Type CAN relay is shown in Fig.(attached sheets).


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The R phase CT is loaded with a R-C circuit, where as the B phase CT

energizes a resistor, the value of resistance is equal impedance of R-C circuit. The power

factor of the R-C circuit is 0.5. The R phase voltage drop, therefore lags the current I Rby

60 degrees. The B phase voltage drop is in phase w.r.t. IB. With positive sequence

currents R and B phase voltages are in phase opposition and so the voltage across XY is

zero, where as the negative sequence currents results in condition shown in the phasor

diagram Fig.(attached sheets), and produces a substantial combined voltage between the

points X and Y. A relay connected to these points will therefore respond to the negative

sequence component. Since the source of unbalance is in the system and will affect all

generators in the vicinity, these should not be disconnected unless the condition remains

uncorrected to such a time that there is a danger to the generator being damaged. The

protection should have a time delay characteristic which is as near as is practicable to the

heating characteristic of the machine, thereby providing as much time as possible for the

operation staff to locate and isolate the fault before shut down becomes necessary. Full

advantage will not be taken f the time so provided unless the operator is warned at an

early stage; the protection should therefore contain an alarm feature which operated at a

setting equal to or slightly lower that of the tripping element, and is delayed by only a

few seconds, to avoid unnecessary alarms being given for faults that are cleared in the

usual way.

PRE EXPERIMENTAL QUIZ:

1. Show that the negative phase sequence component of current for line-to-line fault

is equal to 1/ 3 times the fault current.

2. Will this relay respond to zero sequence current?

3. Name the type of fault, in which negative sequence component of the fault current

is zero.

4. Why are the negative phase sequence relays not made instantaneous in operation?

5. The secondary winding of a /Y connected transformer is subjected to line to

ground fault. Name sequence components of the line current percent in the delta

side of this transformer. Assume grounded neutral.


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APPARATUS:

1. Negative phase sequence relay.

2. Current transformers 2 nos.

3. Rheostats

4. Ammeters

5. Stop Watch

PROCEDURE:

1. Make connections as shown in the circuit diagram (Fig.3.5) choosing CTs,

Rheostats and ammeters of the appropriate rating.

2. Choose minimum plug setting on the relay.

3. Create fault between R-Y, closing switches 1 & 2and increase the current till relay

operates. Note down the operating current.

4. Repeat the step 3 for line-to-line faults between Y & B ad B & R.

5. Observe the operating time of the alarm and trip circuit for different values of

fault currents (more than the pick current).

DATA SHEET:

a. VERIFICATION OF CALIBRATION

Nature of fault Relay Setting Ammeter Reading Negative sequencecurrent =

1/ 3 Ammeter

Reading

R-Y

Y-B

B-R

POST EXPERIMENTAL QUIZ:

1. Is it possible to use negative phase sequence relay for detecting single phasing

of a 3- phase Induction motor? Explain.


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2. How much phase difference exists in the negative sequence components of the

currents between primary and secondary currents of /Y connected 3 phase

transformer?

3. Suggest an alternative negative sequence filter network insensitive to zero

sequence component of the currents.

4. Suggest a negative phase sequence filter network using operational amplifiers.

5. The desired operating characteristics of negative phase sequence relay is I22

t = k,

where I2 is negative sequence component (p.u of continuous mean rating CMR), t

is the time in second and K is the constant proportional to the thermal capacity of

the generator. What is the approximate value of K for turbo alternator and that for

hydro generator?

6. What should be the values of RR and RB if the capacitor is of 4f and system

frequency is 50 Hz?

REFERENCES:

1. C. R. Mason, The art and science of protective relaying , Wiley Eastern.

2. B. Ravindranath & M. Chander, Power system Protection and switchgear,

Wiley Eastern.

3. A. R. Van C. Warrington, Protective Relays: The theory and practice,

Chapman and Hall (Vol I. and Vol II.)


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Experiment No. 4

Voltage control using Var Compensation through

synchronous condenser

Motivation:

Owing to inevitable resistance and reactance of all conducting system (whether

overhead transmission lines or underground cables), there will be a drop of voltage along

the system when current is flowing and further this drop will vary with the current andthe power factor. In short transmission line, it is possible to design an economical system

in which the voltage drop does not exceed values, which are outside the capacities of the

automatic voltage regulators, which control the generator terminal voltage.

With very long transmission lines, the line capacitance becomes

significant, this having the effect not only of adding its own component to

the total line drop, but also of causing wide variations of power factor withsuch a system cannot be kept at the low percent value as in the short lines,

necessitates that the control of the receiving end voltage shall be affected byan entirely different method.

Several methods are used to achieve the voltage control and one such method is to

connect the idle running synchronous motor, called a synchronous condenser across the

line at receiving end. A detailed study of the synchronous condenser performance and itsVAR compensation capability is necessary for deciding the compensation to be provided

on a power system for an efficient and smooth voltage control.

Objective:

1. To determine the change in voltage, voltage regulation, line power factor and

transmission efficiency for varying loads with and without VAR compensation.2. To determine the VAR supplied by a synchronous condenser for maintaining zero

regulation for a given transmission line for varying loads.

Theory:

A synchronous condenser is synchronous motor operating on load. It the propertyof such a motor that it takes lagging KVA, when the field current is below a certain

value, and leading KVA, when the field current is above this value. The efficiencies ofthese machines are very high. The real power it takes will be small, just its losses. It istherefore justifiable to assume that the current taken by the machine leads the applied

voltage by 900. For simplicity let us consider a line having resistance R and inductive

reactance X only, and work in terms of the voltage to neutral Ep, which we will assume is

to be the same at both ends of the line.

Let I1= Load current at a p.f. of cos


26/42

And Im= synchronous condenser current

It is convenient to split I1 into in phase and quadrature components, i.e.,

Ia= I1 cos , the in phase component

Ib= I1 sin , the quadrature component

The three currents Ia, Ib, and Im produce resistance and reactance drop per phase of IaR,IaX, IbR, IbX and ImR, ImX; the resistance drops being in phase with and reactive drop inquadrature leading with the corresponding currents. The phasor diagram is as shown in

the Fig. 4.1. Note that since Im and Ib are in direct opposition and that Im must be greater

than Ib, the four drops due to these two currents simplify to(Im-Ib)R in phase with Im.

and (Im-Ib )X in quadrature leading Im.

Since the lengths OA and OF are to be equal, we have from Fig. 4.1

OF2=(OA+AB-DF)

2+ (BC+CD)

2

Ep2=[Ep+ IaR-(Im- Ib )X]

2+[ IaX+(Im- Ib)R ]

2

This is an equation with only one known, viz, Im and the solution therefore gives thenecessary wattless leading current to be taken by the synchronous condenser. Finally

Wattless KVA (leading) capacity of the condenser

=1000

3

1000

3mmp EIIE

where E is the line voltage.

Generally the load delivered by a transmission line is not a constant and it is thus possible

to prepare a family of curve from which one can tell the VAR required for any load at

any power factor normally expected to be supplied using a given transmission line.

Pre-experimental Quiz:

1. Reactance power in a single ac circuit is a product ofa) Voltage times the current in the circuit.

b) Voltage times the current times the cosine of the angle between the

voltage and current.c) Voltage times current times the sine of the angle between the voltage and

control.

2. In a lightly loaded circuit of such a length that the capacitive reactance isappreciable, the receiving end voltage

a) Is always less that the sending end voltage

b) May exceed the sending end voltage

c) Is always equal to the sending end voltage3. Why does an ac transmission line require a VAR input?

4. On a long high voltage transmission line under heavy load condition VAR

compensation can be provided by installing


27/42

a) Series inductive reactance

b) Series capacitors

c) Shunt inductive reactors5. If any synchronous motor (properly synchronized to the supply) is running on no

load and is having negligible loss then

a) The stator current will be very highb) The stator current will be zero

c) The stator current will be very small

d) The back emf will be more than the supply voltage.6. The minimum armature current of the synchronous condenser corresponds to

a) 0.8 lagging p.f.

b) zero p.f.

c) 0.8 leading p.f.d) Unity p.f.

7. A synchronous motor can be operated at desired power factor by varying the

excitation of the motor

a) Trueb) False

8. A synchronous capacitor is nothing but a synchronous motor running on no loadwith over excitation

a) True

b) False

Material and Equipment:

3-phase variac, transmission line model, synchronous machine, wattmeter, voltmeters,ammeters, RLC load units.

Procedure, Connection Diagram and Experimentation:

(i). Voltage Regulation of RL load without VAR compensation:

1. Connect the circuit as per the connection diagram shown in Fig.4.2.

2. Switch on the 3 phase supply and adjust the 3 phase variac to set the sending

end voltage (Vs) at 400 V.3. Keeping the sending end voltage constant, with varying loads in steps at the

receiving end, measure receiving end voltage (Vr), line current (Iline), load

current (Iload) and load power (Wload ).

4. switch of the supply after reducing the load and variac position.

(ii). Voltage Regulation of RLC load without VAR compensation:


2. Switch on the 3 phase supply and adjust the 3 phase variac to set the sendingend voltage (Vs) at 400 V.


28/42

3. Keeping the sending end voltage constant, with varying loads in steps at the

receiving end, measure receiving end voltage (Vr), line current (Iline), load

current (Iload) and load power (Wload ).4. switch of the supply after reducing the load and variac position.

(iii). Voltage Regulation of RL load with VAR compensation:


2. Switch on the 3 phase supply and adjust the 3 phase variac to start the

machine connected at the receiving end of the transmission line as inductionmotor.

3. Adjust the variac to get the rated speed of the machine for synchronizing the

machine to the mains and let it run as a synchronous motor.

4. Switch on the DC supply for the field excitation, starting from zero, increasethe excitation current till line current becomes minimum value corresponding

to the sending end voltage (say 400V).

5. Adjust the excitation gradually and make the receiving end voltage equal to

the sending end voltage. Note the VAR taken by the synchronous condenser(including sign) at the various steps in the process.

6. Keep the receiving end voltage at this value (say 400 V) switch on a load ofabout 1A at u.p.f.

7. Adjust the excitation of the synchronous condenser and the variac to make

both the sending and receiving end voltages 400V and the VAR taken by the

condenser. Also record Vr, Iline, Iload, IM and Wload.8. Repeat the steps 5 and 7 for different values of loads at u.p.f.

Precautions:

1. Increase the load in small increments preferably 1-2 % of system capacity.

Data Sheet:

(i) RL load without VAR compensation:

Sl. no Vs Vr Iline Iload Wload

1

2

3

4


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(ii) RLC load without VAR compensation:

Sl. no Vs Vr Iline Iload Wload

1

2

3

4

(iii) RL load with VAR compensation:

Sl.

noVs Vr Iline Iload IM Wload

VAR

supplied

1

2

3

4

Data Processing and Analysis:

1. For each type of load, draw the curve VAR versus power taking care of the

sign.2. Compare the experimental results with the theoretical results for a known load

and p.f.

Post Experimental Quiz:

1. When synchronous condensers are used to provide VARs, leading VARs areproduced by

a) Increasing the field current

b) Reducing the field currentc) Increasing the speed of the machine

2. a) With a 100 MVA generator operating at 85% p.f. lagging, how muchMVARs are produced?b) To what MW load should the machine be limited so that its MVA rating

will not be exceeded?

3. VARs characterized by the fact that they always flow

a) From point of low voltage to high voltage.b) Without affect by the voltage at the line terminals.

c) From point of high voltage to low voltage.


30/42

4. Can the synchronous condenser deliver the rated output both on leading and

lagging power factors?

5. Can the synchronous condenser be operated at zero p.f.?6. What are the other methods of voltage control employed in power system?

References:

1. H. Cotton and H. Barber, The Transmission and Distribution of ElectricalEnergy, ELBS and the English University Press Ltd. London, 1970.

2. A. T. Starr, Generation, Transmission and Utilization of Electrical Power,

STR ISAAC PITMAN & SONS, Ltd, London,1962.

3. B. M. Weedy, Electric Power System, 3rd, John Wiley & Sons, New York,1979.

4. O. I. Elgerd, Electric Energy System Theory: An Introduction, McGraw

Hill, 1971.

Fig. 4.1 Phasor Diagram

XII )(

OA

F

B

bm

RII bm )(

XIa

mI

)( bm II

aI

lIbI

pE

RIa

Sending end

)( pE

Receiving end )( pE

C

D


31/42

Fig. 4.2 Connection Diagram for RL load without VAR compensation

Fig. 4.3 Connection Diagram for RLC load without VAR compensation

3 phaseWattmeter (

L)

R, LLoad Units

I load

I line

3 phase

AC50 Hz

supply

3PhaseV

ariac

Vs Vr

Transmis

sion

Linemod

el

3 phaseWattmeter ( W

L)

R, L, C

Load Units

Iload

I line

3 phase

AC

50 Hz

supply

3Phase

Variac Vs

Vr

Transm

ission

Linem

odel


32/42

Fig. 4.4 Connection Diagram for RL load with VAR compensation

3 phaseWattmeter

L)

R, L

Load Units

Iload

Iline

IM

M L

VAR Measurement

3 phase

AC

50 Hz

supply

Synchronous

Condense

Fiel

220 VDC

ff

R1

+

-3Phase

VsVr

Transmission

Linemodel

VC


33/42

A Micro Computer Controlled Static Var Compensator

For Power System Laboratory

List of Components Supplied:

1. Variac 3 Phase 0 400V, 50 HZ, 2A 1 No.

2. Fuses 0 2A 3 Nos.

3. Indicating Lamps 0 230V 3 Nos.

4. 3 Phase Transformer 400V/130V, 5A star/star 1 No.

5. Transmission line 3 Phase, with the following details 1 set.

a) Line length each of equivalent 100 Km length P1 line

unit referred to 110V, 5A, representing an equivalent

400KV, 400MVA, 3 phase transmission line.b) Each P1-line unit inductance 3.2mh, 5A

consists of : Capacitance at each end 15MF,

400V

c) Total length of line 300Km referred to 110V, 5A.

6. Compensator Unit

It consists of the following details:-

1 No.

i. Fixed capacitors - 60F, 400V connected in delta

ii. Air gap type inductors 110V, 3A, connected in delta

with anti parallel thyristors

iii. Thyristors 16A, 1000PIV

iv. Ammeter to measure current through inductor 0 5A.

3 Nos. (1 set)

3 Nos. (1 set)

6 Nos.

1 No.

7. A.C. Ammeter to measure the load current 0 6A 1 No.

8. A.C. Voltmeter to measure load voltage 0 150V 1 No.

9. A.C. Single phase Wattmeter to measure load power 1 No.

10 3 phase variable load (Resistive), Connected in star, loads are

varied from 0 to 5.5A in steps of 0.5A.

1 No.

11. Control Card with necessary thyristor triggering circuits. 1 No.

12. A single board Micro Computer 8085 MP based Details are given

in the Users Mannual

1 No.


34/42

Table 1

Visual Display

Meaning of theprompt

Addressfield

prompt

Initialnumerical

values indata field

Assignedkey for

futuredisplay

Press thekey for

nextdisplay

Loop count LC 02h 0 NEXT

Initial alpha lower

byte

ALPI 00h 1

Initial alpha higher

byte

ALP2 26h 2

Proportional constant PC 01h 3

Derivative constant DC 01h 4

Reference voltage REFV 72h 5

Multiplying factor PF 30h 6

Summated error

voltage

IE 00h 7

Integral constant IC 08h 8

Initial error voltage KP 00h 9

8 bit alpha ALPA 09h A

Program mode PR 01h B


35/42

Table 2

Experimental Results

S.E. Voltage Fixed At 100V

Without SVC With SVC

Receivingend

voltage(line to

line)(V)

Loadcurrent

(A)

3-phasepower

transferredto

(W)

Receivingend

voltage(line to

line)(V)

Loadcurrent

(A)

3-phasepower

transferredto load

(W)

Inductorcurrent

(A)

TO RUN THE STATIC VAR COMPENSATOR EQUIPMENT

1. Make necessary connections

2. press Go EXEC 0800H

3. visual display appear as per Table 1, one after another each time

after pressing the next Key.

4. after the program mode is displayed, if the next Key is pressed, the

display goes blank and the reactive power compensation program is

brought into operation.

(CS 01 for PD control, and CS 02 for PID control).


36/42

R

Y

B400V,

3-P

haseSupply

Switch

Fuse

Fuse

Fuse

Lamp

Lamp

Lamp

N

OR

OY

OB

0 - 400V

2A, 3Ph

2R

2B

2Y

400V/120V

5A, Y/YN

Variac Transformer

Sending End

R3

Y3

B3

N3

N3

N3

300Km Long PI - 4 No. Unit

A

0 - 5A

0 - 150V

W

3-PhaseVariableRe

sistiveload

0-5.5

Ainsteps

at0.5

A

R4

Y4

B4

N

Receiving

EndR4

Y4

B4

N

0.5A0.5A

R

Y B

R

YB

A AY

Generator

SwitchOFF

ON

Var Compensator Unit

V

SE RE

R

R

BB

Fig.1 : Basic Scheme

300 Km Long PI 3 Nos


37/42

G0

G1

G2

D0

D1

D2

PC4

AD/C Ch6

+5V

GND

8255

8253

P

T

E

R

M

IN

A

L

M

IC

R

O

C

O

M

P

U

T

E

R

1

5

0

V

1

5

0

V

1

5

0

VR4

R4

Y4

Y4

B4

B4

20

20

20

20

20

20

0

0

0

G1

G4

G3

K1

G2

K2

K3

K5

K4

K6

G5

G6

SP

S1

S2

BLACK

BLACK

BLUE

BLUE

GREY

GREY

K

GTh1

K

KG Th2

K

G

G

G

G Th3

Th4

KTh5

Th6

K

+

o

+

o

o

o

+

+

o

o

+

+

OR N

2

3

0

V

2

3

0

V

2

3

0

V

2

3

0

V

,

M

a

in

s

RS

YS

BS

1

0

0

K

m

L

in

e

1

0

0

K

m

L

in

e

1

0

0

K

m

L

in

e

R4

Y4

B4

27 2

2

27

22

272211,1KV

11

,1KV 3

Th1T

h2

11,1

KVT

h3

Th

4

R6

Y6

B6

60 f

Th6

Th5

Fig.2(a): Control Circuit Connection Diagram of Static Var Compensator Unit


38/42

Step by Step Operating Instructions of SVC

1. Connect the links of S.E. and R.E.

2. Switch on the 3-Phase 400V ON/OFF switch to the ON position, keeping the

voltage adjust Variac to a minimum.

3. The three RED, YELLOW and BLUE Lamp glows.

4. Gradually increase the Variac voltage and maintain the sending ending voltage to

about 100V.

5. Connect the R, Y and B points of the R.E. to the 3-phase load box (Resistive).

6. Keeping the S.E. voltage at about 100V, always note the R.E. voltages from no

load to full load up to 5A in steps and in each case also note the Wattmeterreadings. [current coil 10A, voltage scale 300V].

7. Next make the SVC Switch to the ON position and connect the Micro-Computer

to the Jack position when the Microprocessor display SBS 1.

8. Press Go 0800H EXEC KEY :- Different displays appear one after another as

shown in Table 1.

9. Adjust reference voltage to about 79H, and CS-01 or CS-02 for either PD or PID

Control.

10. Maintain the S.E. voltage at 100V and note the R.E. voltages. R.E. currents,

wattmeter readings etc. from no load to full load of 5A in steps of 0.5A.

11. Observe how the R.E. voltage, remains constant and more Power is transferred

when the SVC is switched ON.

12. Vary LC, PROP.CONS, DERV. CONS, and Integral CONS. and observe the

performance of the SVC in controlling the R.E. voltage.


39/42

5. FAULT STUDIES

The description of the fault studies software and the relevant data file is available in the

accompanying sheets. With the given software and the sample study system, the

following studies should be carried out.

1. Comparison of the severity of different types of faults: Choose a bus for the

application of a solid fault. Do the studies with 3-phase, LG, LL and LLG faults

and rank the faults in the order of severity.

2. Effect of transformer connection in fault studies: For a fault on either side of a

transformer study the effect of changing transformer connections on the fault

currents and voltages.3. Effect of network changes on fault level (only 3 phase fault): Consider the fault

at a bus and study the effect of adding/removing a line, adding/removing a

generator on the fault level.

4. Effect of neutral connections in fault studies: For a fault at a bus, change

appropriate neutral connections (one at a time) and see the effect on fault current

and voltages. This study is to be done for all faults involving ground.

Note: Draw sequence diagrams for all studies respectively.


40/42

General Instructions:1) Change the path in the MATLAB command window and set it to the destination

on your computer where the fault analysis folder is stored.

2) You need to use only the command window for your lab purposes.

Instructions related to using the package:

The program can be run in two ways:

1) Entering the data in an interactive manner

2) Changing the data in text files which may/may not be present in the directory

when you are first accessing it.

Running the program:

1) enter faultmain in the command window prompt

2) the program asks you for the following entries:a. number of buses

b. number of generatorsc. number of lines

d. number of transformers

3) in the next stage, you are given the option of choosing eitherdata entry orprogram execution also known asfault analysis

4) if the following text files (gendata.txt, faultdata.txt, linedata.txt and

transformerdata.txt) are not present in the directory, you will have to choose the

data entry option. Once you have entered the data, the text files are created andyou can use these to correct any errors you might have made while entering the

values or else you can proceed with the program execution.

5) While entering reactances, do not include i along with the numerical value. Forexample, enter 0.03 and not 0.03i

6) To execute the program follow steps 1 and 2 and after that chooseprogram

execution

7) The files are described as below. You need to open them in WORDPAD to viewthem correctly.

a. faultdata.txt: This file describes the fault specifications. The file contains the

following elements described column wisei.faultbus: this is the bus number of the faulted bus

ii.faulttype: this describes the type of fault 1-three phase 2-SLG 3-LL 4-LLG

iii.fault impedance: this stores data regarding the fault impedanceb. gendata.txt: This describes the generator specifications from left to right column

wise

i. Generator_bus: This is the bus number of the generatorii. Positive sequence reactance of the generator

iii. Zero sequence reactance

iv. Grounding reactance


41/42

c.linedata.txt: This describes the line specifications. Read from left to right, the

different columns are:

i. starting bus of the lineii. ending bus of the line

iii. positive sequence reactance

iv. zero sequence reactancev. real part of the off nominal tap

vi. imaginary part of the off nominal tap

d. transformerdata.txt: The transformer specifications read from left to right are:i.HV bus

ii.HV bus Connection type: 0-star grounded 1-star ungrounded 2-delta

iii.LV bus

iv.LV bus connection typev.Positive sequence reactance

vi.Zero sequence reactance

The output.txt file is created once your program has completed execution. Use word pad

to open it and view your results.

Single line diagram of the system


42/42

DATA FORMAT

Generator data:Bus No +ve seq zer oseq groundi ng r eact ance

6. 00 0. 10 0. 05 0. 00

7. 00 0. 10 0. 05 0. 00

Transformer dataHVbus HVbusconnect i on LVbus LVbusconnect i on +veseq zeroseq

1. 00 0. 00 6. 00 2. 00 0. 10 0. 10

2. 00 0. 00 7. 00 2. 00 0. 10 0. 10

Line data

Fr ombus Tobus +veseq zer oseq r eal par t i mag. par t

1. 00 2. 00 0. 06 0. 18 1. 00 0. 00

2. 00 5. 00 0. 12 0. 36 1. 00 0. 00

5. 00 4. 00 0. 24 0. 72 1. 00 0. 00

4. 00 3. 00 0. 03 0. 12 1. 00 0. 00

3. 00 1. 00 0. 24 0. 72 1. 00 0. 00

2. 00 3. 00 0. 18 0. 54 1. 00 0. 00

Fault data

Faul t bus f aul t t ype f aul t i mpedance

5. 00 1. 00 0. 00

labmannual 2nd cycle

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