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DEPARTMENT

OF

ELECTRICAL AND ELECTRONICS ENGINEERING

II B.TECH II SEMESTER

REGULATION/LAB CODE: R16/EE407ES

LABORATORY MANUAL

ELECTRICAL MACHINES-II

Electrical Machines-II Laboratory

Department of Electrical & Electronics Engineering

DEPARTMENT

OF

ELECTRICAL AND ELECTRONICS ENGINEERING

VISION

To become a renowned department imparting both technical and non-technical

skills to the students by implementing new engineering pedagogy’s and research to

produce competent new age electrical engineers.

MISSION

To transform the students into motivated and knowledgeable new age electrical

engineers.

To advance the quality of education to produce world class technocrats with an

ability to adapt to the academically challenging environment.

To provide a progressive environment for learning through organized teaching

methodologies, contemporary curriculum and research in the thrust areas of

electrical engineering.



Program Educational Objectives (PEO’s):

PEO 1: Apply knowledge and skills to provide solutions to Electrical and Electronics

Engineering problems in industry and governmental organizations or to

enhance student learning in educational institutions

PEO 2: Work as a team with a sense of ethics and professionalism, and communicate

effectively to manage cross-cultural and multidisciplinary teams To provide

experience in selecting and using variety of electrical instruments &

accessories.

PEO 3: Update their knowledge continuously through lifelong learning that

contributes to personal, global and organizational growth



PO 1: Engineering knowledge: Apply the knowledge of mathematics, science,

engineering fundamentals and an engineering specialization to the

solution of complex engineering problems.

PO 2: Problem analysis: Identify, formulate, review research literature, and

analyze complex engineering problems reaching substantiated

conclusions using first principles of mathematics, natural science and

engineering sciences.

PO3: Design/development of solutions: design solutions for complex

engineering problems and design system components or processes that

meet the specified needs with appropriate consideration for the public

health and safety, and the cultural, societal and environmental

considerations.

PO 4: Conduct investigations of complex problems: use research based

knowledge and research methods including design of experiments,

analysis and interpretation of data, and synthesis of the information to

provide valid conclusions.

PO 5: Modern tool usage: create, select and apply appropriate techniques,

resources and modern engineering and IT tools including prediction and

modelling to complex engineering activities with an understanding of the

limitations.

Program Outcomes(PO’s):

A graduate of the Electrical and Electronics Engineering Program will

demonstrate:



PO 6: The engineer and society: apply reasoning informed by the contextual

knowledge to assess societal, health, safety, legal and cultural issues and

the consequent responsibilities relevant to the professional engineering

practice.

PO 7: Environment sustainability: understand the impact of the professional

engineering solutions in the societal and environmental contexts, and

demonstrate the knowledge of, and need for sustainable development.

PO 8: Ethics: apply ethical principles and commit to professional ethics and

responsibilities and norms of the engineering practice.

PO 9: Individual and team work: function effectively as an individual and as a

member or leader in diverse teams, and in multidisciplinary settings.

PO 10: Communication: communicate effectively on complex engineering

activities with the engineering community and with society at large,

such as, being able to comprehend and write effective reports and

design documentation, make effective presentations, and give and

receive clear instructions.

PO 11: Project management and finance: demonstrate knowledge and

understanding of the engineering and management principles and apply

these to one’s own work, as a member and leader in a team, to manage

projects and in multidisciplinary environments.

PO 12: Lifelong learning: recognize the need for, and have the preparation and

ability to engage in independent and lifelong learning in the broader

context of technological change.



Program Specific Outcomes(PSO’s):

PSO-1: Apply the engineering fundamental knowledge to identify, formulate,

design and investigate complex engineering problems of electric circuits,

power electronics, electrical machines and power systems and to succeed in

competitive exams like GATE, IES, GRE, OEFL, GMAT, etc.

PSO-2: Apply appropriate techniques and modern engineering hardware and

software tools in power systems and power electronics to engage in life-long

learning and to get an employment in the field of Electrical and Electronics

Engineering.

PSO-3: Understand the impact of engineering solutions in societal and

environmental context, commit to professional ethics and communicate

effectively.



Course Outcomes (CO’s):

Upon completion of this course, the student will be able to:

CO1: Explain principles of operation of DC motors. Identify the various terms

associated with /DC motors.

CO2: Summarize National Electric Code (NEC) regulations governing the

installation of transformers and AC/DC motors.

CO3: Calculate motor horsepower, speed, slip, efficiency, power factor, and

torque for Electrical machines properties under various loads

CO4: Analyze Machine circuit Connections.



KG REDDY COLLEGE OF ENGINEERING AND TECHNOLOGY

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

II B TECH II SEMESTER

ELECTRICAL MACHINES LAB-II

S.No. Name of the Experiment

1. O.C. and S.C. Tests on Single phase Transformer

2. Sumpner’s test on a pair of Single phase Transformers

3. Brake Test on Three phase Induction Motor

4. No-Load and Blocked rotor tests on three phase induction motor

5. Regulation of a three phase alternator by synchronous impedance

and m.m.f. methods

6. ‘V’ and ‘Inverted V’ curves of a three phase synchronous motor

7. Equivalent Circuit of a single phase induction motor

8. Determination of Xd and Xq of a salient pole synchronous machine

9. Parallel operation of Single Phase Transformers

10. Scott connection of Transformers



EXPERIMENT-1

O.C. and S.C. Tests on Single phase Transformer

AIM: 1. To predetermine the efficiency and regulation of the given single phase

Transformer at different power factors.

2. To draw equivalent circuit referred to the primary.

APPARATUS:

THEORY:

OPEN CIRCUIT TEST:

The main purpose of conducting this test is to determine core losses and no load

parameters Ro and X0 . While conducting this test one winding of the transformer usually high

voltage winding is kept open and the low voltage winding is engaged with measuring instruments

like voltmeter and ammeter.

From open circuit test data

Rcl=Vo/Ic

Xml=Vo/Im

SHORT CIRCUIT TEST:

This test is an economical method for determining the following

1. Equivalent quantities like Ro1, R02, X01, X02, Z01 , Z02 as referred to the winding in which

the instruments are connected.

2. Copper losses at full load can be determined which is useful for finding the efficiency.

3. Regulation of transformer can also be calculated

In this test one winding usually the low voltage winding is short circuited by a thick gauge of wire

and measuring instruments are connected in HV winding.

From short circuit test data:

S.No: Apparatus Type Rating Quantity

1. Ammeter Moving iron 0-10A, 0-2A 2

2. Voltmeter Moving iron 0-300V 1

3. Wattmeter Dynamometer 5A/150V,LPF 1

4. Wattmeter Dynamometer 20A/150V,UPF 1



Z2=Vsc/Isc

R2=Wsc/Isc2

X2=(Z22-R2

2 )1/2

Referred to LV side

R1= R2*(N1/N2)2

X1=X2*(N1/N2)2

Efficiency curve:

Efficiency at any load (X times full load) and at a given power factor can be calculated as follows.

Output at X times full load =X*rated KVA*PF

Iron loss=Wo

Copper loss at x times full load =X2 *full load copper loss

%Efficiency = output*100/(output+losses)

Efficiency at different assumed loads for a given power factor are calculated and tabulated.

Regulation curve:

Percentage regulation= rCos ± X sin

r, percentage resistance=I*R1*100/V

x, percentage reactance=I*X1*100/V

I, rated LV side current

V, rated LV side voltage

Positive sign for lagging power factor and negative sign for leading power factor percentage

regulation for full load for different power factors are calculated and results are tabulated.



CIRCUIT DIAGRAM:

OPEN CIRCUIT DIAGRAM:

SHORT CIRCUIT DIAGRAM:

PROCEDURE:

Open circuit test:

1. Connections are made as per the circuit diagram

2. HV side is kept open and rated voltage is applied to the low voltage winding by adjusting

the autotransformer

3. The meter readings are noted down and tabulated

Short circuit test:

1. Connections are made as per the circuit diagram

2. LV side is short circuited and adjusting the autotransformer, rated current is send on the

HV side



3. The meter readings are noted down and tabulated

TABULAR FORMS:

Open circuit test:

Vo

(V)

Io

(A)

Wo

(W) Coso=Wo/(Vo*Io)

Ic=Io* Coso

(A)

Im=Io*Sino

(A)

Short circuit test:

Vsc(V) Isc(A) Wsc(Watts)

PRECAUTIONS:

1. Wattmeter connections must be done as per the rating of the transformer

2. LPF wattmeter to be used for open circuit test

RESULT:

VIVA VOCE QUESTIONS:

1. What is the operating principle of transformer?

2. Why OC and SC tests are convenient and very economical?

3. What is the main purpose of OC and SC test?

4. Why iron losses are negligible during SC test of a transformer?

5. What is the primary reason to conduct OC test only on the Low voltage winding of the

transformer?



EXPERIMENT-2

Sumpner’s test on a pair of Single phase Transformers

AIM :

1. To perform Sumpner’s (Back to Back) test on two identical transformers.

2. Determine the efficiency at 1/4 , 1/2, 3/4, full load and 1.25 times the full load and at 0.85

p.f. lagging.

3. Plot efficiency Vs output characteristic.

APPARATUS :

S. No. Name Type Range Quantity

1.

2.

3.

4.

5.

6.

7.

8.

Ammeter

Ammeter

Voltmeter

Voltmeter

Voltmeter

Wattmeter

Wattmeter

Single phase variac

MI

MI

MI

MI

MI

Dynamometer

Dynamometer

Fully variable

2.5/5 A

15/30 A

300 V

600 V

30 V

2.5 A, 200 V

15 A, 75 V

230/0-270 V, 15 A

1

1

1

1

1

1

1

1

THEORY

This test needs two identical transformers. The primary windings of these transformers are

connected in parallel and supplied at rated voltage and frequency, while the two secondaries are

connected in phase opposition. Thus the voltage across the two secondaries is zero, when the

primary windings are energized. As such, this test is also called back to back test. In this test, iron

losses occur in the cores and full load copper losses occur in the windings of the two transformer.

Current flowing in the two secondaries is rated full load current of the transformer. Thus, heat run

test can be conducted on the transformer can be estimated. The current drawn by the primaries is

twice the no load current of each transformer. The wattmeter W1 connected in the circuit of the

primaries measures the total core losses of both the transformers.

Thus, iron losses of each transformer = 1/2 W0

Where, W0 is the reading of wattmeter, W1, when rated voltage is applied to the primaries

of the transformers.



Similarly, wattmeter W2 connected in the secondary circuit measures the total full load

copper losses of the two transformers.

Hence, full load copper losses of each transformer = 1/2 Wc

Where, Wc is the reading wattmeter W2, when full load current is flowing in the secondary

circuit. A low voltage, hardly 8 to 10 percent of the rated value is applied across the secondaries

for full load current to flow.

(a) Efficiency at full load :

Let the output in KVA of each transformer be P0

Total losses of each transformer under full load operation = 1/2 W0 + 1/2 Wc

P0 X 1000 X cos φ

Percentage efficiency at full load, f = X 100

P0 X 1000 X cos φ + 1/2 W0 + 1/2 Wc

(b) Efficiency at half full load :

Power output of each transformer at half full load = 1/2 P0

Iron losses at half the full load = 1/2 W0 (constant)

Copper losses at half the full load = (1/2 )2 [1/2 Wc ] = 1/8 Wc

1/2 P0 X 1000 X cos φ

Thus, efficiency at half full load, 1/2f = X 100

1/2 P0 X 1000 X cos φ + 1/2 W0 + 1/8 Wc



CIRCUIT DIAGRAM:

PROCEDURE

1. Connect the circuit as per circuit diagram.

2. Ensure that switches S2 and S3 are open.

3. Energize the primaries by closing the switch S1.

4. Observe the reading of voltmeter V1, which should be zero for correct connection of the

secondaries. In case, the voltmeter reads twice the rated voltage of each transformer,

open the switch S1 and interchange the connections at the secondary terminals of one of

the transformer. Close the switch S1 again and verify that the voltmeter V1 now reads

zero. Important caution : Even if the voltmeter V1 reads zero at the first instance, it is

advisable to check the reading of Voltmeter V1 by interchanging the connections at the

secondary terminals of one of the transformer

5. Adjust the setting of the variac, to give nearly zero output voltage.

6. Replace the voltmeter V1, by a low range voltmeter.

7. Close the switch S3 and then S2.

8. Adjust the output voltage of the variac, so that the current flowing in the secondaries is

full load secondary current of each transformer.

9. Record the readings of all the instruments connected in the primary and secondary

circuit. Only one set of reading is sufficient to calculate the efficiency at different loads.

10. Switch off the supply to primary and secondary circuits.



TABULAR COLUMN:

S. No. Primary side Secondary side

V0 I0 W0 Vsc Isc Wsc

RESULT:

VIVA QUESTIONS

1. Why two transformers, and that too identical, are needed in this test ?

2. Discuss various losses occurring in a transformer along with their variation with respect

to loading condition and the parts in which these occur.

3. If the iron losses and copper losses at full load single phase, 30 KVA, 1100/250 V, 50

Hz transformer are 300 watts and 400 watts respectively. Find out these losses at 3/4 of

the full load.

4. Based on the data of the above question, find out the load at which the efficiency of the

transformer will be maximum and also calculated the maximum efficiency.

5. Using the data obtained in this test, find out the equivalent resistance and reactance of

each transformer referred to secondary.



EXPERIMENT-3

Brake Test on Three phase Induction Motor

AIM :

1. Perform load test on 3-phase induction motor

2. Compute Torque, output power, input power, efficiency, input power factor and slip for

every load setting.

APPARATUS:

THEORY:

The load test on induction motor is performed to compute its complete performance

i.e. torque, slip, efficiency, power factor etc. During this test, the motor is operated at rated

voltage and frequency and normally loaded mechanically be brake and pulley arrangement from

the observed data, the performance can be calculated, following the steps given below.

SLIP :

The speed of the rotor, Nr droops slightly as the load on the motor is increased. The

synchronous speed, Ns of the rotating magnetic field is calculated field is calculated, based on the

number of poles, P and the supply frequency, i.e.

Synchronous speed, Ns = 𝟏𝟐𝟎𝐅

𝑷

Then, slip, S = 𝐍𝐬−𝐍𝐫

𝐍𝐬

Normally, the range of slip at full load is from 2 to 5 per cent.


1. Ammeter MI 0-10/20 A 1

2. Voltmeter MI 0-300/600 V 1

3. Wattmeter Dynamometer 10/20 A, 2

4. 3-phase variac Fully variable 200/400 V 1

5. Tachometer Digital 15 A, 400/0-

400 V, 0-2000

rpm

1



TORQUE :

Mechanical loading is the most common type of method employed in laboratories.

A brake drum is coupled to the shaft of the motor and the load is applied by tightening the belt,

provided on the brake drum. The net force exerted at the brake drum in kg is obtained from the

readings S1 and S2 of the spring balances i.e.

Net force exerted, W = (S1 – S2) kg

Then, load torque, T = W X d/2 kg –m

= W X d/2 X 9.8 Nw-m

where, d- effective diameter of the brake drum in meters.

OUTPUT POWER, P0 :

The output power in watts developed by the motor is given by,

Output power, P0 = 𝟐 𝐍 𝐓

𝟔𝟎 watts

where, N is the speed of the motor in r.p.m.

INPUT POWER :

Input power is measured by the two wattmeters, properly connected in the circuit

i.e.

Input power = (W1 + W2) watts

Where, W1 and W2 are the readings of the two watttmeters.

INPUT POWER FACTOR :

Input power factor can also be calculated from the readings of two wattmeters for

balanced load. If φ is the power factor angle, then

tan φ = 3𝐖𝟏−𝐖𝟐

𝐖𝟏+𝐖𝟐

Knowing the power factor angle, φ, from the above, power factor, cos φ can be

calculated. It may be noted clearly at this stage, that the power factor of the induction motor is

very low at no load, hardly 0.1 to 0.25 lagging. As such, one of the wattmeter will record a

negative reading, till the power factor is less than 0.5, which may be measured by reversing the

connections of either the current coil or pressure coil of this wattmeter.

EFFICIENCY :

Percentage efficiency of the motor, =𝐎𝐮𝐭𝐩𝐮𝐭 𝐩𝐨𝐰𝐞𝐫

𝐢𝐧𝐩𝐮𝐭 𝐩𝐨𝐰𝐞𝐫 X 100



Full load efficiency of 3 phase induction motor lies in the range of 82 % (for small

motors) to 92 % (for very large motors).

CIRCUIT DIAGRAM:

PROCEDURE:


2. Ensure that the motor is unloaded and the variac is set at zero output voltage.

3. Switch-on 3 phase ac mains and start the motor at reduced applied voltage. Increase the

applied voltage, till its rated value.

4. Observe the direction of rotation of the motor. In case, it is reverse, change the phase

sequence of the applied voltage.

5. Take-down the readings of all the meters and the speed under no load running.

6. Increase the load on the motor gradually by turning of the hand wheels, thus tighting the

belt. Record the readings of all the meters and the speed at every setting of the load.

Observations may be continued upto the full load current rating of the motor.

7. Reduce the load on the motor and finally unload it completely.

8. Switch-off the supply to stop the motor.

9. Note-down the efficiency diameter of the brake drum.

TABULAR COLUMN:

S. No. Line

Voltage

Input

current

W1 W2 S1 S2 speed



PRECAUTIONS:-

1. Ensure that the starter arm is at extreme left position.

2. Avoid loose connections

3. Note down the readings form the meters without any parallax error

RESULT:

CONCLUSION:

VIVA VOCE QUESTIONS:

1. Why does the speed of 3 phase induction motor falls slightly, when the load on the motor

is increased?

2. Why one of the wattmeters reads negative, when the motor is lightly loaded ?

3. Can this experiment be conducted with one wattmeter only ? If so, draw the modified

circuit diagram?

4. Why it is essential to start the induction motor at reduced voltage ?

S.No. Current Input

power

Torque Output

power

Slip Power

factor

Efficiency



EXPERIMENT-4

NO LOAD AND BLOCK ROTOR TEST

AIM :

1. Perform no load and block rotor test on 3 phase induction motor.

2. Find out, input current, power factor, slip, torque and efficiency, corresponding to full

load, using the above circle diagram.

3. Compute (i) Max power (ii) Max torque (iii) Starting torque and best power factor,

utilizing the above circle diagram.

APPARATUS:


1.

2.

3.

4.

Ammeter

Voltmeter

Wattmeter

3 phase variac

MI

MI

Dynamometer

Fully variable

0-10/20 A

1-300/600 V

10/20 A, 200/400 V

15 A, 400/0-400 V

1

1

2

1

THEORY

To draw the circle diagram of a 3 phase induction motor, following data is essential.

No load current, I0 and its power factor angle, φ0

Short circuit current, Isc` corresponding to rated voltage and its power factor angle,

φsc

NO LOAD TEST:

To obtain no load current and its power factor angle, φ0, no load test is performed at

rated voltage and frequency. Let the readings of ammeter, voltmeter, and two wattmeters

connected in the circuit be, I0, V0, W01 and W02 respectively during no load test. Then,

tan φ = 3𝐖𝟏−𝐖𝟐

𝐖𝟏+𝐖𝟐

Hence, no load power factor angle, φ0, can be calculated from the readings of two

wattmeters. No load current, I0 has been directly measured by the ammeter.

BLOCK ROTOR TEST:



To obtain short circuit current and its power factor angle, block rotor test is

performed on the motor. In this test, rotor is not allowed to move (blocked either by tightening the

belt, in case provided or by hand) and reduced voltage (25 to 30 percent of the rated voltage) of

rated frequency is applied to the stator winding. This test is performed with rated current flowing

in the stator winding. Let the readings of ammeter, voltmeter and two wattmeters be, Isc, VSC,

WSC1 and WSC2 respectively under block rotor condition. Then,

tan φ = 3𝐖𝐬𝐜𝟏−𝐖𝐬𝐜𝟐

𝐖𝐬𝐜𝟏+𝐖𝐬𝐜𝟐

Thus, short circuit power factor angle, φsc can be calculated from the above

equation.

Short circuit current, Isx observed during the block-rotor test corresponds to reduced

applied voltage, VSC, which should be converted to rated voltage of the motor for plotting the

circle diagram. The relation between the short circuit current and the applied voltage is

approximately a straight line. Thus, short circuit current, IS C corresponding to rated voltage, V of

the motor is given by,

Short circuit current, Isc` = 𝐈𝐒𝐂

𝐕𝐒𝐂× 𝑽

It may be remembered, that the power factor of the power factor of the motor is

quite low at no load as well as under blocked rotor condition. Thus, one of the wattmeter

connected in the circuit will give negative reading in both the test, which may be recorded by

reversing the terminals of the pressure coil or the current coil.



CIRCUIT DIAGRAM

PROCEDURE

No Load Test:


2. Ensure that the motor is unloaded and the variac is set at zero position.

3. Switch-on the supply and increase the voltage gradually, till the rated voltage of the motor.

Thus the motor runs at rated speed under no load.

4. Record the reading of all the meters connected in the circuit.

5. Switch-off the ac supply to stop the motor.

Block Rotor Test:

6. Readjust the variac at zero position

7. Change the range of all the instruments for block rotor test as suggest in the discussion on

circuit diagram.

8. Block the rotor either by tightening the belt firmly or by hand.

9. Switch-on the ac supply and apply reduced voltage, so that the input current drawn by the

motor under blocked rotor condition is equal to the full load current of the motor.

10. Record the readings of all the meters, connected in the circuit.

11. Switch-off the ac supply fed to the motor.

12. Measure the resistance per phase of the stator winding, following ohm;s law concept.



TABULAR COLUMN:

No load test Block rotor test

S. No. V0 I0 W01 W02 VSX ISC WSC1 WSC2

RESULT:

QUESTIONS

1. Find out various losses occurring under full load condition, using the circle diagram

drawn.

2. Mark clearly the stable and unstable region on the circle diagram, drawn on the basis of

experimental data.

3. Why the no load power factor is quite small in case of 3-phase induction motor ? Mention

its normal range.

4. Discuss the fact that the input power factor of the motor increases with increase in load.

Draw an approximate curve to indicate this increase in power factor with load.

5. Why the no load current of an induction motor is so high, as compared to a transformer of

identical rating ?

6. Utilizing the information obtained in this experiment, find out the parameters of the

equivalent circuit of 3 –phase induction motor.



EXPERIMEMT-5

REGULATION BY SYNCHRONOUS IMPEDANCE METHOD

AIM :

1. Perform no load and short circuit tests on a 3-phase alternator.

2. Measure the resistance of the stator winding of alternator.

3. Find out regulation of alternator at full load and at (i) unity p.f. (ii) 0.85 p.f. lagging (iii)

0.85 p.f leading, using synchronous impedance method.

APPARATUS:

S.No. Name Type Range Quantity

1.

2.

3.

4.

5.

Ammeter

Ammeter

Voltmeter

Rheostat

Techometer

MC

MI

MI

Single tube

digital

0-1/2 A

0-10/20 A

0-300/600 V

370Ω, 1.7 A

0-2000 rpm

1

1

1

2

1

NAME PLATES DETAILS:

THEORY

To find out the regulation of alternator by synchronous impedance method,

following characteristics and data has to be obtained experimentally,

1. open circuit characteristic at synchronous speed.

2. short circuit characteristic at synchronous speed.

3. ac resistance of the stator winding, per phase i.e. Ra.

The open circuit and short circuit characteristics of a 3 phase alternator, plotted on

the phase basis. To find out the synchronous impedance from these characteristics, open circuit



voltage, E1 and short circuit current, I1 (preferably full load current), corresponding to a particular

value of field current is obtained. Then , synchronous impedance per phase is given by,

Synchronous impedance, Za = 𝑬𝟏

𝑰𝟏

Then, Synchronous reactance, Xz = Z2S - R

2a

The phasor diagram of the alternator, supplying full load current of Ia ampere,

lagging the terminal voltage V by an angle φ. The open circuit voltage E of the alternator is given

by, E = V + Ia Ra + Ia Xa (Phasor sum)

The diagram has been drawn with the current as the reference phasor and is self

explanatory. The open circuit voltage as finally obtained from the phasor diagram, corresponding

to this loading condition is E volts. Then the regulation of the alternator under the above loading

condition is given by,

Regulation = 𝑬−𝑽

𝑽X 100 percent

An approximate expression for the open circuit voltage can be establish referring to

the phasor diagram.

Open circuit voltage, E = OD2 + DC2

= (OF + FD)2 + (DB + BC)2

or E = (V cos φ + Ia Ra )2 + (V sin φ + Ia Xa)

2 ( for lagging p.f. load)



The above expression is for lagging power factor load. In case, alternator is

operating at leading power factor, open circuit voltage, E can be found out in a similar way and is

given by,

E = (V cos φ + Ia Ra)2 - (V sin φ + Ia Xa)

2 (for leading p.f. load)

The value of regulation obtained by this method is higher than obtained from as

actual load test; as such it is called the pessimistic method.

CIRCUIT DIAGRAM

PROCEDURE


2. Adjust the position of rheostat, R1 for maximum possible current in the field circuit of

dc motor, to

3. Ensure (i) low starting speed (ii) high starting torque.

4. Set the position of rheostat, R2 for minimum current in the field circuit of alternator, to

ensure low value of generated emf at starting.

5. Switch on the dc mains, feeding the dc motor and the field circuit of alternator.

6. Start the dc motor, using the starter properly. Various resistance steps of the starter

should be cur out slowly, so that the motor does not draw high current during starting.

7. Set the speed of the motor and hence the alternator at its rated value by varying

rheostat, R1, provided in the field circuit of the motor.

8. Note-down the open circuit voltage of the alternator and the field current.



9. Repeat step 7 for various value of field current (can be obtained by varying the

rheostat, R2 provided in the field circuit of alternator). Observation should be

continued, till the open circuit voltage is 25 to 30 percent higher than its rated value.

10. Set the position of rheostat, R2 again for minimum possible current in the field circuit

of alternator.

11. Short-circuit the stator winding of the alternator, by closing the switch, provided for

this purpose in the circuit diagram.

12. Note-down the short circuit current and the field current.

13. Repeat step 11, for various values of field current, till the short circuit current

becomes equal to the full load current of alternator.

14. Readjust the setting of rheostats R1 and R2 to their initial positions and then switch-

off the dc supply to stop the dc motor.

15. Measure the dc resistance of the stator winding by usual voltmeter-ammeter method.

To obtain ac resistance, skin effect must be taken into account. As such, ac resistance

may be taken approximately 1.3 times the dc resistance measured.

TABULAR COLOUMN:

Open Circuit Test Short Circuit Test

S. No. If E If Isc

RESULT:



QUESTIONS :

1. Write down and discuss the approximate expression for the regulation of

alternator (i) for lagging load (ii) for resistive load (iii) for leading load.

2. Is the value of synchronous impedance constant at various value of field

excitation ? If not explain why ?

3. Find out the value of synchronous impedance, for various value of field current

and plot a curve on the same graph, where o.c. & s.c. characteristics have been

drawn.

4. Using the same o.c. and s.c. characteristics, find out the regulation of alternator

at full load and at 0.85 p.f. lagging, using the mmf method. Draw the

corresponding phasor diagram and explain the method involved.

5. Discuss in details, why the value of regulation obtained in synchronous

impedance method is higher than that obtained from an actual load test.



EXPERIMENT-6

V AND ‘INVERTED V’ CURVES OF SYNCHRONOUS MOTOR

AIM :

a. To study the effect of variation of field current upon the stator current an power factor

with synchronous motor running at no load, hence to draw V and inverted V curves of

the motor.

b. To repeat the above, with synchronous motor loaded to half the full load and 3/4 the full

load.

APPARATUS:

S.No. Name Type Range Quantity

1.

2.

3.

4.

5.

6.

Ammeter

Ammeter

Wattmeter

Voltmeter

Ammeter

Voltmeter

MI

MC

Dynamometer

MI

MC

MC

0-10/20 A

0-5/10 A

10/20 A, 200/400 V

0-300/600 V

0-10/20 A

0-300 V

1

1

2

1

1

1

NAME PALTE DETAILS:



THEORY

With constant mechanical load on the synchronous motor, the variation of field current

changes the armature current drawn by the motor and also its operating power factor. As such, the

behaviour of the synchronous motor is described below under three different modes of field

excitation.

1. Normal excitation :

The armature current is minimum at a particular value of field current, which is called the

normal field excitation. The operating power factor of the motor is unity at this excitation and thus

the motor is equivalent to a resistive type of load.

2. Under excitation :

When the field current is decreased gradually below the normal excitation, the armature

current increases and the operating power factor of the motor decreases. The power factor under

this condition is lagging. Thus, the synchronous motor draws a lagging current, when it is under

excited and is equivalent to an inductive load.

3. Over Excitation :

When the field current is increased gradually beyond the normal excitation, the armature

current again increases and the operating power factor decreases. However, the power factor is

leading under this condition. Hence, the synchronous motor draws a leading current, when it is

over excited and is equivalent to a capacitive load.

If the above variation of field current and the corresponding armature current are plotted

for a constant mechanical load, a curve of the shape of ‘V’ is obtained. Such a characteristic of

synchronous motor is commonly called as ‘V’ curve of the motor. The characteristic curve plotted

between input power factor and the field current for a constant mechanical load on the motor are

of the shape of inverted ‘V’ and are known as inverted ‘V’ curves.



CIRCUIT DIAGRAM

PROCEDURE


2. Switch-on the ac supply feeding to 3-phase synchronous motor and start the motor, using

the starter.

3. Observe the direction of rotation of the motor, in case, it is rotating in opposite direction,

stop the motor and reverse the phase sequence. Start the motor again, using the starter.

Ensure that the motor is running on no load.

4. In this case, field winding is excited automatically with the help of exciter, provided on the

shaft of the main motor.

5. Set the rheostat in the field circuit of the motor to the position of normal excitation. Under

this



condition, armature will drew minimum current from the mains. Note down the readings

of all the meters connected in the circuit.

6. Reduce the excitation in steps and note down the corresponding armature current and

reading of both the wattmeters. Excitation may be reduced, till the current in the armature

winding is of rated value. Under this condition, armature current should increase on

reducing the excitation .

7. Again, adjust the rheostat in the field circuit to normal excitation. Now increase the

excitation in steps and note-down the readings of all the meters at each setting of increased

excitation. Excitation may be increased, till the behaviour of the motor is normal. At large

excitation, the motor will try to fall out of step.

8. Adjust the voltage of the dc generator coupled to synchronous motor to rated value by

varying the field current of the generator.

9. Load the dc generator to half the full load and maintain it constant throughout the next part

of the experiment.

10. Repeat step 5, 6, and 7 sequentially under this condition of loading.

11. Increase the load on the generator to 3/4th of full load, keeping its voltage constant.

Maintain this load constant throughout the next part of the experiment.

12. Repeat step 5, 6, and 7 sequentially for this increased load on the motor.

13. Remove the load on the dc generator gradually.

14. Switch-off the supply to the motor to stop it.

TABULAR COLUMN:

S.

No.

V If Ia W1 W2 Vdc Idc Cos



RESULT:

QUESTIONS

1. What are the basic differences between a 3-phase synchronous motor and 3-phase

induction motor ?

2. what are the various methods of starting a 3-phase synchronous motor ?

3. what is the power factor of the motor at normal excitation ?

4. What is the nature of power factor, when a synchronous motor is operated (i) under

excited (ii) over excited ?

5. Is it possible to operate a synchronous motor on any other speed than the synchronous

speed ?



EXPERIMENT-7

NO LOAD AND BLOCKED ROTOR TEST ON SINGLE PHASE INDUCTION

MOTOR

AIM :

1. Perform no load and blocked-rotor test on single phase induction motor.

2. Determine the parameters of the equivalent circuit drawn on the basis of double-

revolving field theory.

APPARATUS


1.

2.

3.

4.

5.

6.

7.

8.

Ammeter

Voltmeter

Wattmeter

Auto

transformer

Ammeter

Voltmeter

Lamp bank load

Techometer

MI

MI

Dynamometer

-

MC

MC

Resistive

digital

0-5 A/10 A

0-150/300 V

5/10 A, 200/400

V

8 A, 230/0-270

V

0-10 V

0-30 V

10 A, 250 V

0-2000 rpm

1

1

1

1

1

1

1

1

NAME-PLATE DETAILS:



THEORY

No load and block-rotor test are performed on single phase induction motor to

determine its parameters of the equivalent circuit. The approximate equivalent circuit of a single

phase induction motor, drawn on the basis of double revolving field theory, in which the iron loss

component has been neglected. The motor consists of a stator winding, represented by its

resistance R1 and leakage reactance X1 and two imaginary rotors, generally called as forward and

backward rotors. Each rotor has been assigned half the actual rotor value in stator terms. Existing

branch has been shown with exciting reactance only, with one-half the total magnetizing reactance

assigned to each rotor. If the forward rotor operates at a slip of s, then the backward rotor has a

slip of (2-s). The complete parameters of this equivalent circuit can be calculated following the

steps suggested below, using the informations obtained in the no load and block rotor test.



CIRCUIT DIAGRAM

PROCEDURE

No load test:

1. Connect the circuit as per circuit diagram, with the meter ranges suggested above.

2. Switch-on the ac supply to the circuit.

3. Adjust the voltage applied to the circuit to a low value and then increase it to the rated

voltage of the motor.

4. Centrifugal switch will get opened automatically at approximately 75 percent of the rated

speed, disconnecting the starting winding from the main winding.

5. Record the applied voltage, V0, the no load current, I0 and the no load power input, P0.

6. Reduce the applied voltage in suitable steps and record the no load current and power for

various values of applied voltages

7. Switch-off the supply to stop the motor.

Block-rotor test :

1. Connection are basically the same for block-rotor test except that the meters are replaced

with proper ranges suggested already and the starting winding is disconnected from the

circuit.

2. Adjust the variac in the circuit, such that its output voltage is quite low approximately 5 to

10 percent of the rated voltage of the motor.

3. Switch-on the ac to the circuit.

4. Adjust the applied voltage, such that the current drawn by the motor is full load rated

current. Record applied voltage, input current and power.

5. Switch-off the ac main to stop the motor.



TABULAR COLUMN:

(a) No Load Test (b) Block Rotor Test (c) Measurement of

resistance

S. No V0 I0 W0 Vsc Isc Wsc Vm Im Rdc

RESULT:

QUESTIONS

1. Why single phase induction motor do not produce the starting torque of their own ?

2. Compare the performance of single phase induction motor with that of a 3-phase induction

motor with respect to efficiency and operating power factor.

3. Compare the parameters of the auxiliary winding with that of main winding of a single

phase induction motor.

4. What is the main purpose in using a capacitor along with single phase induction motor and

in which circuit it is connected ?

5. Why a switch is normally provided on the rotor of a single phase induction motor ? What

is the basic operating principle of this switch .



EXPERIMENT-8

STEADY STATE REACTANCES (Xd, Xq)/Slip test

AIM :

a) To measure direct-axis synchronous reactance of synchronous machine.

b) To measure quadrature-axis synchronous reactance by slip test.

APPARATUS:


1.

2.

3.

4.

5.

6.

Ammeter

Ammeter

Voltmeter

Voltmeter

3-phase variac

Rheostat

MI

MC

MI

MI

-

single tube

0-15 A

0-2 A

0-150 V

0-600 V

400/0-400 V, 15

A

272 Ω, 1.7 A

1

1

1

1

1

2

NAME PLATE DETAILS:



THEORY

Direct-axis synchronous reactance and quadrature- axis synchronous reactance are the

steady state reactances of the synchronous machine. These reactances an be measured by

performing, open circuit, short-circuit tests and the slip test on a synchronous machine.

(i) Direct-axis synchronous reactances, Xd

The direct-axis synchronous reactance of synchronous machine in per unit is equal to the

ratio of field current, Ifsc at rated armature current from the short circuit test, to the field current, Ifo

at rated voltage on the air gap line i.e.

Direct-axis synchronous reactance, Xd = Ifsc

Ifo per unit

Thus direct-axis synchronous reactance can be found out by performing open circuit and

short circuit test on an alternator.

(ii) Quadrature-axis synchronous reactance, Xq by slip test :

For the slip test, the alternator should be driven at a speed, slightly less than the

synchronous speed, with its field circuit open. 3 phase balanced reduced voltage of rated

frequency is applied to armature (stator) terminals of the synchronous machine. Applied voltage is

to be adjusted, so that the current drawn by the stator winding is full load rated current. Under

these conditions of operation, the variation of the current drawn by the stator winding, voltage

across the stator winding and the voltage across the field winding. The wave shapes of stator

current and stator voltage clearly indicate that these are changing between minimum and

maximum values. When the crest of the stator mmf wave coincides with the direct axis of the

rotating field, the induced emf in the open field is zero, the voltage across the stator terminals will

be maximum and the current drawn by the stator winding is minimum. Thus approximate value of

direct-axis synchronous reactances, Xds is given by,

Emax = XdS

Imin



When the crest of stator mmf wave coincides with the quadrature-axis of the rotating field, the

induced emf in the open circuit field is maximum, the voltage across the stator terminals will be

minimum and the current drawn by the stator winding is maximum. Hence, approximate value of

quadrature-axis synchronous reactance, Xqs is given by,

Xqs = Emin

Imax

For best result, these values are not taken as final values. The most accurate

method for determining the direct-axis synchronous reactances, Xd is the one, that has already

been described in (i) above. The most accurate value of quadrature-axis synchronous reactance,

Xq can now be found out using the above information i.e. Xds, Xqs and Xd.

Xqs

Quadrature-axis synchronous reactance, Xq =

Xds

Emin Imin

Xd = Xd per unit

Imax Emax



Hence the accurate value of Xq can be found out by recording minimum and maximum

values of the above quantities. Accurate results can be obtained, if the osillograms are taken

during experimentation for stator current, stator voltage and injected voltage across the field.

It may be noted clearly that for synchronous machine, Xd is greater than Xq i.e. Xd > Xq.

CIRCUIT DIAGRAM:

PROCEDURE:

Open circuit test


2. Ensure that the external resistance in the field circuit of dc motor acting as a

primover for alternator is minimum and the external resistance in the field circuit

of alternator is maximum.

3. Switch on dc supply to dc motor and the field of alternator.

4. Start the dc motor with the help of starter. The starter arm should be moved slowly,

till the speed of the motor builds up and finally all the resistance steps are cut out

and the starter arm is held in on position by the magnet of no volt release.

5. Adjust the speed of the dc motor to rated speed of the alternator by varying the

external resistance in the field circuit of the motor.

6. Record the field current of the alternator and its open circuit voltage per phase.



7. Increase the field current of alternator in steps by decreasing the resistance and

record the field current and open circuit voltage of alternator for various values of

field current.

8. Field current of alternator is increased, till the open circuit voltage of the alternator

is 25 to 30 percent higher than the rated voltage of the alternator.

9. Decrease the field current of alternator to minimum by inserting the rheostat fully

in the field circuit.

Short circuit test:

10. With the dc motor running at rated speed and with minimum field current of

alternator, close the switches, thus short-circuiting the stator winding of alternator.

11. Record the field current of alternator and the short circuit current.

12. Increase the field current of alternator in steps, till the rated full load short-circuit

current. Record the readings of ammeters in both the circuits at every step. 4 to 5

observations are sufficient, as short circuit characteristic is a straight line.

13. Decrease the field current of alternator to minimum and also decrease the speed of

dc motor by field rheostat of the motor.

14. Switch off the dc supply to dc motor as well as to alternator field.

Slip Test

1. Connect the circuit of alternator as per circuit diagram, keeping the connection of the dc

motor same.

2. Ensure that the resistance in the field circuit of dc motor is minimum.

3. Switch on the dc supply to the motor.

4. Repeat steps 4 described in open circuit test.

5. Adjust the speed of the dc motor slightly less than the synchronous speed of the alternator

by varying the resistance in the field circuit of the motor. Slip should be extremely low,

preferably less than 4 percent.

6. Ensure that the setting of 3 phase variac is at zero position.

7. Switch on 3 phase ac supply to the stator winding of alternator.

8. Ensure that the direction of rotation of alteration, when run by the dc motor and when run

as a 3 phase induction motor at reduced voltage (alternator provided with damper winding

can be run as 3-phase induction motor) is the same.

9. Adjust the voltage applied to the stator winding, till the current in the stator winding is

approximately full load rated value.



10. Under these conditions, the current in the stator winding, the applied voltage to the stator

winding and the induced voltage in the open field circuit will fluctuate from minimum

values to maximum values, which may be recorded by the meters included in the circuit.

For better results, oscillogram may be take of stator current, applied voltage and induced

voltage in the field circuit.

11. Reduce the applied voltage to the stator winding of alternator and switch off 3 phase ac

supply.

12. Decrease the speed of dc motor and switch off dc supply.

TABULAR COLUMN:

Open Circuit Test Short Circuit Test Slip Test

S. No. If V0 If Isc Imin Imax Vmin Vmax

RESULT:



QUESTIONS

1. Define and discuss with suitable diagrams, the direct-axis and quadrature-axis

synchronous reactances, of synchronous machine.

2. Express direct-axis and quardrature-axis synchronous reactance in terms of leakage

reactance of stator winding per phase and the armature reaction reactance along the direct-

axis and quadrature-axis.

3. What is normally the range of steady state reactances of large rating synchronous machine

?

4. Why Xq is less then Xd in salient pole alternators, where as they are approximately equal

in non-salient pole alternators ?

5. Why synchronous machines are built with high values of steady state reactances ?



EXPERIMENT-9

PARALLEL OPERATION OF SINGLE PHASE TRANSFORMER

AIM :

1. To separate the two transformers in parallel.

2. To study the load sharing by each transformer.

APPARATUS:


1.

2.

3.

4.

5.

6.

Ammeter

Ammeter

Voltmeter

Wattmeter

Wattmeter

Load

MI

MI

MI

Dynamometer

Dynamometer

Inductive

0-15 A

0-30 A

0-300/600 V

200 V, 15 A

200 V, 30 A

250 V, 7.5 Kw

2

1

1

2

1

1

NAME PLATE DETAILS:



THEORY:

Parallel operation of transformers is frequently necessary in the power system network,

which consist of a number of transformers installed at generating stations, substations etc. When

operating two or more transformers in parallel (on the primary as well as secondary sides), their

satisfactory performance require that the following condition must be satisfied.

(a) For single phase transformers

(i) The same polarity

(ii) The same voltage ratio.

(b) For three phase transformers

(i) The same polarity

(ii) Zero-relative phase displacement

(iii) Same phase sequence

(iv) Same voltage ratio.

In addition to the above essential requirements, the transformers to be operated in parallel

should have the following for better load sharing and operating power factor.

(i) Equal per unit impedances

(ii) Equal ratios of resistance to reactance.

CIRCUIT DIAGRAM



PROCEDURE:

Polarity check


2. Switch-on the supply to the primary circuit, where the primaries of both the

transformers are connected in parallel.

3. The voltmeter connected in the secondary circuit of the transformers will read either

zero or twice the secondary terminal voltage of each transformer.

(i) If the voltmeter reads zero-connect a1 to a`1 and a2 to a`2 for the two

secondaries to be in parallel

(ii) In case, the voltmeter reads twice the secondary terminal voltage, then connect

a1 to a`2 and a2 to a`1 for parallel operation of the two transformers.

parallel operation:

1. Connect the circuit as per the circuit diagram.

2. Ensure that the two secondaries have been connected properly.

3. Close the switch S1 to energize both the primaries. Ensure that the switch S2 is kept open. In

case the voltage ratio of the two transformers are unequal, there will be a circulating

current, which may be recorded.

4. Close the switch S2. Adjust a particular load on the secondaries and record the readings of

all the instruments connected in the circuit.

5. Repeat step 3 for various values of load current upto the rated capacity of the two

transformers operating in parallel.

6. Switch off the load slowly. Open the switch S2 and then switch off the supply to the

primaries of the transformers.



TABULAR COLUMN

S. No. VL IL WL I1 W1 I2 W2

RESULT:

QUESTIONS

1. Discuss the essentiality of parallel operation of transformers in a practical system

supplying Power to a wide area.

2. What is meant by circulating current with regard to parallel operation of

transformers ? How

3. Explain the additional essential and desirable conditions that should be satisfied

for parallel operation of 3 phase transformer compared to single phase

transformers.



EXPERIMENT-10

SCOTT-CONNECTION (Three phase to two phase conversion)

AIM :

a) To obtain a balanced two phase supply form a 3 phase balanced system.

b) To study the nature of currents on the primary side with balanced loading on the

two phases.

APPARATUS:


1.

2.

3.

4.

Voltmeter

Ammeters

3 phase Variac

Lamp bank load

MI

MI

-

Resistive

0-300/600V

0-20 A

400/0-400 V, 25

A

250 V, 3 Kw

1

5

1

2

THEORY

Three phase to two phase conversion or vice versa is essential under the following

circumstances.

(i) To supply power to two phase electric furnaces

(ii) To supply power to two-phase apparatus from a 3-phase source

(iii) To interlink three phase system and two phase systems.

(iv) To supply power to three phase apparatus from a two phase source.

The common type of connection which can achieve the above conversion is normally called

scott-connection.

Transformer A -50 percent tapping and is called the main transformer.

Transformer B – 86.6 percent tapping and is called the teasure transformer. The phasor

diagram of voltages across the primaries and secondaries. The voltage across the primary, CO of

the teasure transformer will be 86.6 percent of the voltage across the primary AB of main

transformer. The neutral point of the three phase system will be on the teasure transformer, such

that the voltage between O and N is 28.8 percent of the applied voltage. Thus the neutral point

divided the teasure primary winding, CO in the ratio of 1:2.’



The voltages across the two secondaries a1 a2 and b1 b2 should be same in magnitude but in

phase quadrature, which may be verified experimentally by recording the voltage across the two

secondaries Va1a2, Vb1b2 and the voltage across a2 b2 with a1 and b1 connected together. The

voltage Va1a2 and Vb1b2 will be in phase quadrature, if the following relationship holds good

between the three voltages.

Va2b2 = V2a1a2 + V2

b1b2

The behaviour of the above circuit can be studied experimentally, under the following

different conditions of loading.

(i) Equal loading on the two secondaries at unity power factor :

If the two secondaries of main and teasure transformers carry equal currents at unity power

factor (resistive load), the current flowing in the primary windings on three phase side will also

will be equal and that too at unity power factor. This fact may be verified experimentally.

(ii) Equal loading on the two secondaries at 0-8 p.f. lagging :

Load the two secondaries with equal current but with inductive load at 0.8 p.f. lagging. Then

the currents on the primary side will also be balanced and that too at 0.8 p.f. lagging, a fact which

may be verified experimentally.

(iii) Unequal loading on the two secondaries with different power factors.

If both the current and power factor are different in the two secondaries of the transformer

used for scott-connection, then the current on the primary side will also be unbalanced, again a

fact which can be verified experimentally.



CIRCUIT DIAGRAM :

PROCEDURE


2. Ensure that the switches, S1 and S2 are open.

3. Adjust the 3 phase variac for minimum voltage in its output circuit.

4. Switch on the ac supply and apply rated voltage across the primaries of the transformers.

5. Record voltage V1, V2 and V3 and verify that the output is a balanced two phase supply.

6. Switch off the ac supply and remove the dotted connection of the two secondaries and

the voltmeter,

7. V3. Adjust the variac to minimum output voltage.

8. Switch on the ac supply again. Adjust the output voltage of the variac as per the rated

voltage of the primaries of the transformers.

9. Close the switches S1 and S2 to load both the secondaries. Adjust equal loading of both

the secondaries. Record the readings of all the meters connected in the circuit (primaries

as well as secondaries).

10. Repeat step 8 for various equal loading condition on the two secondaries.

11. Repeat step 8 for various unequal loading conditions on the two secondaries.

12. Switch off the load from both the seondaires and adjust the variac, so that its output

voltage is minimum.

13. Switch off the ac supply.



TABULAR COLUMN:

For balanced two phase supply Under loaded conditions

S. No. V1 V2 V3 S. No. I2m I2T I1 I2 I3 V1 V2

RESULT:

VIVA QUESTIONS

1. Is it possible to obtain 3 phase balanced ac system from a 2 phase balanced ac system

using Scott connection?

2. Why is it essential that 86.6 percent tapping must be there in teasure transformer?

3. What tapping should be available on the main transformer and why?

4. Comment about the iron losses occurring in main and teasure transformer, specially from

the consideration of their equality or inequality.

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