academic lab manual department of electrical and

Shri Ram Murti Smarak Engineering Institutions. Page - 1 -

ACADEMIC LAB MANUAL

DEPARTMENT OF

ELECTRICAL AND ELECTRONICS ENGINEERING

VISION

• To produce dynamic competent knowledgeable Electrical and Electronics Engineers who shall lead

a Nation in Technical Education and become a front runner in nurturing the young mind globally by

providing the staunch technical support and research environment.

MISSION

• To offer quality education in Electrical and Electronics Engineering and prepare the students for

professional career and higher studies in Technical Education.

• To create and enhance the skills of the students with futuristic knowledge of new technology

including high moral Ethical in order to become professional in the field of industrial and socio-

economic needs.

• To be center of research and education generating knowledge and technology which lay ground work

in shaping the future in the area of Electrical and Electronics Engineering to achieve the goal of the

institution.

Program Educational Objectives (PEOs) :

PEO1:To produce graduates with a strong foundation in the basic sciences and mathematics to enable that

they become successful and productive engineers, with emphasis on technical competency, and with

attention to teamwork in the broad range of electrical engineering areas to serve the needs of both

private and public sectors.

PEO2:Considering the availability of “renewable and non-renewable “ resources in the country, plan,

estimate and execute power projects to operate generating stations, transformers, electrical

machines, transmission lines and switch gear in an appropriate manner so that the “overall operation

of a power system is economical and efficient”.

PEO3: Graduates able to identify, formulate, analyze, and create engineering solutions using appropriate

current engineering techniques, designing skills and tools to develop novel products and solutions

for the real life problems.

PEO4: To develop the all-round personality and attitude to become good citizens fully aware of national

goals and their role in their achievement with the basic skills to communicate effectively and to

develop the ability to function as members of multidisciplinary teams.

PEO5: Knowledge of Relevant Technologies-The graduates well informed about current technologies

important to electrical engineering, as well as probable future technological advances and contribute

actively to the field by participating in professional societies, publishing, attending conferences and

seeking patents.

Program Outcomes (POs):

Engineering graduates will be able to:

1. Engineering Knowledge: Apply the knowledge of mathematics, science, engineering fundamentals,

and an engineering specialization to the solution of complex engineering problems.

2. Problem Analysis: Identify, formulate, review research literature, and analyze complex engineering

problems reaching substantiated conclusions using first principles of mathematics, natural sciences, and

engineering sciences.

3. 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, and environmental considerations.

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.


5. Modern tool usage: Create, select, and apply appropriate techniques, resources, and modern

engineering and IT tools including prediction and modeling to complex engineering activities with an

understanding of the limitations.

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.

7. Environment and sustainability: Understand the impact of the professional engineering solutions in

societal and environmental contexts, and demonstrate the knowledge of, and need for sustainable

development.

8. Ethics: Apply ethical principles and commit to professional ethics and responsibilities and norms of the

engineering practice.

9. Individual and team work: Function effectively as an individual, and as a member or leader in diverse

teams, and in multidisciplinary settings.

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.

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.

12. Life-long learning: Recognize the need for, and have the preparation and ability to engage in

independent and life-long learning in the broadest context of technological change.

Program Specific Outcomes A graduate of the Electrical and Electronics Engineering Program will demonstrate:

PSO1: To produce graduates who would have developed a strong background in basic electrical and

electronics and ability to use the tools concern, in this fields of specialization. And who would

attain professional competence through life-long learning such as advanced degrees,

professional registration, and other professional activities.

PSO2: To produce graduates who can explore the scientific theories, ideas, methodologies and the

new cutting edge technologies in this field and use this erudition in their professional

development. And to design, develop, analyse and test different types of generation,

transmission, distribution and protection mechanisms in power systems and assess the

performance of electrical and electronics systems; deploy control strategies for power

electronics related and other applications.

PSO3: To produce the graduates so that they are able to utilize of technologies like Smart Grid, PLC,

PMC, process controllers, making PCBs, transducers and HMI and design, install, test, and

maintain power systems and industrial applications. And control the behaviour of electrical

quantities associated with constituents of energy or allied systems.

THE DEPARTMENT IS EQUIPPED WITH THE FOLLOWING LABS:

Electrical Lab

Electrical Machines Lab

Electrical Measurement Lab

Networks Analysis and Synthesis lab

Power Electronics Lab

Control System Lab

Power Systems Lab

Electrical Instrumentation Lab

Project & Simulation Lab

Electric Drive and Control lab

PLC and Automation Lab

FACILITIES AVAILABLE

SRMS TRUST established Tinkering Lab contains world class & all the modern research


equipment like 3D printer, PLC, Digital Storage Oscilloscope (DSO) etc. for ultra-precision

analysis & development of next generation technologies. We aim to promote projects in the

following research areas:

Green computing

Eco-friendly design

Energy Efficient Architecture

PLC and Automation Labfor advance learning.

USP’S

The main goal is to provide the requisite platform to budding engineers and technocrats to sharpen

their reasoning, analytical and communication skills along with requisite technical knowledge so that

they can meet with confidence the future challenges of growing technology.

The course provides the student to excel in designing electrical and electronics and computing

system that are innovative and socially acceptable.

This course helps the student to inculcate the habit of adapting the latest trends of technology and

application in oriental research

The department of EN provides an excellent academic environment for creative and productive work

both for faculty as well as students.

The department has adopted a project based learning system where budding engineer can explore

and innovate.

The department has State-of-the-Art infrastructure and labs.

The students and faculties push the boundaries of possibilities in these labs which creates a constant

buzz in the part of campus where new knowledge is created.

Centre of Excellence in Machines, Network, Circuit Simulation, and PLC Automation.

MoU with several core Industries for placement and onsite training during curriculum.

Incentives to all faculties and students for publication in peer reviewed journals, travel grant for

international conference, seminars, expert lecture etc.

Highly qualifies faculties with extensive research and industry experience in the areas such as power

system, electric drive and control, instrumentation and control, power electronics etc.


KEE-551: POWER SYSTEM LABORATORY – I

Pre-requisites of course: Basic understanding of Scilab/MATLAB/C/C++

Course Outcomes: Knowledge

Level, KL

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

CO1

Use programming tools /Software: Scilab, MATLAB or any C, C++ -

Compiler and formulate a program/simulation model for calculation of

various parameters related to transmission line.

K6

Note: Minimum ten experiments are to be performed from the following list, on a software

platform preferably on Scilab, MATLAB, or any C, C++ - Compiler

1. Calculate the parameters of single-phase transmission line

2. Calculate the parameters of three phase single circuit transmission line

3. Calculate the parameters of three phase double circuit transmission line

4. Determine the ABCD constant for transmission line.

5. Simulate the Ferranti effect in transmission line

6. Calculate the corona loss of transmission line

7. Calculation of sag & tension of transmission line

8. Calculation of string efficiency of insulator of transmission line

9. Calculation for grading of underground cables

10. Simulate the skin effect in the transmission line

11. Calculation of ground clearance of transmission line

12. Calculate the parameters for underground cable.

Spoken Tutorial (MOOCs):

Spoken Tutorial MOOCs, ' Course on Scilab', IIT Bombay (http://spoken-tutorial.org/)

KEE–552: CONTROL SYSTEM LABORATORY

Pre-requisites of course: Basic understanding of Scilab/MATLAB or any equivalent open source software


Note: Minimum 10 experiments are to be performed from the following list:

1. To determine speed-torque characteristics of an AC servomotor.

2. To study i) Synchro Transmitter characteristics. ii) Obtain Synchro Transmitter – Receiver output vs input

characteristics.

3. To determine response of first order and second order systems for step input for various values of constant

‟K‟ using linear simulator unit and compare theoretical and practical results.

4. To study characteristics of positional error detector by angular displacement of two servo potentiometers.

5. To simulate and compare the response of 2nd order system with and without lead, lag, Lead- Lag

compensator / simulate PID controller for transportation lag.

6. To study P, PI and PID temperature controller for an oven and compare their characteristics.

7. To study performance of servo voltage stabilizer at various loads using load bank.

8. To study behavior of separately excited dc motor in open loop and closed loop conditions at various loads.

Software based experiments (Scilab/MATLAB or any equivalent open source software)

9. To determine time domain response of a second order system for step input and obtain performance

parameters.

10. To convert transfer function of a system into state space form and vice-versa.

11. To plot root locus diagram of an open loop transfer function and determine range of gain „k‟ for stability.

12. To plot a Bode diagram of an open loop transfer function.

13. To draw a Nyquist plot of an open loop transfers functions and examine the stability of the closed loop

system.



Reference Books:

1. K.Ogata,“Modern Control Engineering” Prentice Hall of India.

2. Norman S.Nise, “Control System Engineering”, John Wiley & Sons.

3. M.Gopal, “Control Systems: Principles & Design” Tata McGraw Hill.

KEE-553: ELECTRICAL MACHINE-II LABORATORY

Pre-requisites of course: Basic Electrical engineering Lab, Electrical Machine-I Lab.

Note: Minimum 10 experiments are to be performed from the following list:

1. To perform no load and blocked rotor tests on a three phase squirrel cage induction motor and determine

equivalent circuit.

2. To perform load test on a three phase induction motor and draw Torque -speed characteristics

3. To perform no load and blocked rotor tests on a single phase induction motor and determine equivalent

circuit.

4. To study speed control of three phase induction motor by varying supply voltage and by keeping V/f ratio

constant.

5. To perform open circuit and short circuit tests on a three phase alternator.

6. To determine V-curves and inverted V-curves of a three phase synchronous motor.

7. To determine the direct axis reactance (Xd) and quadrature axis reactance (Xq) of synchronous machine.

http://spoken-tutorial.org/


8. To study synchronization of an alternator with the infinite bus by using: (i) dark lamp method (ii) two

bright and one dark lamp method.

9. To determine speed-torque characteristics of three phase slip ring induction motor and study the effect of

including resistance, or capacitance in the rotor circuit.

10.To determine speed-torque characteristics of single phase induction motor and study the effect of voltage

variation.

11.To determine speed-torque characteristics of a three phase induction motor by (i) keeping v/f ratio

constant (ii) increasing frequency at the rated voltage.

12.To draw O.C. and S.C. characteristics of a three phase alternator from the experimental data and

determine voltage regulation at full load, and unity, 0.8 lagging and leading power factors.

13.To determine steady state performance of a three phase induction motor using equivalent circuit.

14. Load Test on Three Phase Alternator.

*The available Experiments from above list may be performed on virtual lab on following virtual lab link:

http://vlab.co.in/


POWER SYSTEM LABORATORY – I

[KEE-551]

EXPERIMENT NO. 1

OBJECT: Calculate the parameters of single-phase transmission line.

SOFTWARE USED: MATLAB

THEORY: Transmission line has four parameters namely resistance, inductance, capacitance and conductance. The

inductance and capacitance are due to the effect of magnetic and electric fields around the conductor. The

resistance of the conductor is best determined from the manufactures data, the inductances and capacitances

can be evaluated using the formula.

FORMULAS: Inductance:

The general formula: L = 0.2 ln (D / r‟) mH / KM

Where, D = Distance between conductor

r' = 0.7788 r

Where, r = radius of conductor

Capacitance:

A general formula for evaluating capacitance per phase in micro farad per km of a transmission line is given

by

C = 0.0556/ ln (D / r) μF/km

PROBLEM STATEMENT:

A two conductor Single Phase line operates at 50 Hz. The diameter of each conductor is 20mm and

spacing between the conductors is 3m.

Calculate (a) The inductance of each conductor per km

(b) The capacitance per phase per km

PROCEDURE:

1. Enter the command window of the MATLAB.

2. Create a new M – file by selecting File - New – M – File

3. Type and save the program in the editor window.

4. Execute the program by either pressing Tools – Run.

5. View the results

PROGRAM:

% SINGLE PHASE TRANSMISSION LINE

D=input(„enter the distance between two conductor D in m:‟);

r=input(„enter the radius of one conductor r in m:‟);

R=0.7788*r;

Y=log(D/R);

L=0.2*Y;

C=(0.0556)/log(D/r);

fprintf(„\n The Inductance per phase per km is %f mh ph/km\n‟,L);

fprintf(„\n The Capacitance per phase is %f F ph/km\n‟,C);

RESULT:

The Inductance per phase per km is 1.19 mh/km

The Capacitance per phase is 9.74×10-9

F/km


EXPERIMENT NO. 2

OBJECT: Calculate the parameters of three phase single circuit transmission line






FORMULAS:

Inductance:

The general formula:

L = 0.2 ln (Dm / Ds) mH / KM

Where, Dm = geometric mean distance (GMD)

Ds = geometric mean radius (GMR)

Single phase 2 wire system

GMD = D

GMR = re-1/4

= r' = 0.7788 r

Where, r = radius of conductor

Three phase – symmetrical spacing

GMD = D

GMR = re-1/4

= r'

Where, r = radius of conductor & GMR = re-1/4

= r' = 0.7788 r

Capacitance: A general formula for evaluating capacitance per phase in micro farad per km of a

transmission line is given by

C = 0.0556/ ln (Deq / r) μF/km

Where, GMD is the “Geometric mean distance” which is same as that defined for inductance under various

cases.

PROBLEM STATEMENT:

A three phase transposed line has its conductors placed at a distance of 11M, 11 M & 22 M. The

conductors have a diameter of 3.625cm Calculate the inductance and capacitance of the transposed

conductors. (a) Determine the inductance and capacitance per phase per kilometer of the above three

lines. (b) Verify the results using the MATLAB program.

PROCEDURE:





5. View the results


PROGRAM:

%3 phase single circuit

D12=input('enter the distance between D12in cm: ');



d=input('enter the value of d: ');

r=d/2;

Ds=0.7788*r;

x=D12*D23*D31;

Deq=nthroot(x,3);

Y=log(Deq/Ds);

inductance=0.2*Y

capacitance=0.0556/(log(Deq/r))

fprintf('\n The inductance per phase per km is %f mH/ph/km \n',inductance);

fprintf('\n The capacitance per phase per km is %f mf/ph/km \n',capacitance);

RESULT:

The inductance per phase per km is 1.377882 mH/ph/km

The capacitance per phase per km is 0.008374 mf/ph/km


EXPERIMENT NO. 3

OBJECT: Calculate the parameters of three phase double circuit transmission line.






FORMULAS: Inductance:

The general formula: L = 0.2 ln (Dm / Ds) mH / KM

Where, Dm = (DAB×DBC×DCA)1/3

DAB=(Dab×Da‟b×Dab‟×Da‟b‟)1/4

DBC=(Dbc×Db‟c×Dbc‟×Db‟c‟)1/4

DCA=(Dca×Dc‟a×Dca‟×Dc‟a‟)1/4

And

Ds = (DSA×DSB×DSC)1/3

DSA=(Daa×Da‟a×Daa‟×Da‟a‟)1/4

DSB=(Dbb×Db‟b×Dbb‟×Db‟b‟)1/4

DSC=(Dcc×Dc‟c×Dcc‟×Dc‟c‟)1/4

Capacitance:

A general formula for evaluating capacitance per phase in micro farad per km of a transmission line is given

by

C = 0.0556/ ln (Dm /Ds) μF/km

PROBLEM STATEMENT:

Figure shows the spacings of a double circuit 3-phase overhead line. The phase sequence is ABC and

the line is completely transposed. The conductor radius in 1·3 cm.

Find :(a)the inductance per phase per kilometer.

(b) The capacitance per phase per km

PROCEDURE:





5. View the results


PROGRAM:


RESULT:


EXPERIMENT NO. 4

OBJECT: Determine the ABCD constant for transmission line.


THEORY: short transmission network

A=1

B=Z

C=0

D=1

Medium line PI network

A=1+((Y*Z)/2)

B=Z

C=Y*(1+((Y*Z)/4))

D=A

Medium line T network

A=1+((Y*Z)/2)

B=Z*(1+((Y*Z)/4))

C=Y

D=A

long transmission line network

A=cosh*sqrt(Z*Y)

B=sqrt(Z/Y)*sinh*sqrt(Z*Y)

C=1/sqrt(Z/Y)*sinh*sqrt(Z*Y)

D=A

PROBLEM STATEMENT:

For short line, length=40

For medium line, length=140

For long line, length=300

Given: Z=(0.2+0.408i)*length;

Y=(0+3.14e-6i)*length

Find ABCD Constant for all transmission lline.

PROCEDURE:





5. View the results

PROGRAM:


RESULT:


EXPERIMENT NO. 5

OBJECT: Simulate the Ferranti effect in transmission line.

SOFTWARE USED: MATLAB SIMULINK

THEORY:

FERRANTI EFFECT: The effect in which the voltage at the receiving end of the transmission line is more than the sending voltage

is known as the Ferranti effect. Such type of effect mainly occurs because of light load or open circuit at the

receiving end.

Ferranti effect is due to the charging current of the line. When an alternating voltage is applied, the current

that flows into the capacitor is called charging current. A charging current is also known as capacitive

current. The charging current increases in the line when the receiving end voltage of the line is larger than

the sending end.

Capacitance and inductance are the main parameters of the lines having a length 240km or above. On such

transmission lines, the capacitance is not concentrated at some definite points. It is distributed uniformly

along the whole length of the line.

When the voltage is applied at the sending end, the current drawn by the capacitance of the line is more than

current associated with the load. Thus, at no load or light load, the voltage at the receiving end is quite large

as compared to the constant voltage at the sending end.

PROCEDURE:

1. Enter the Simulink window of the MATLAB.

2. Create a new model file.

3. Draw and save the model with the help of Simulink library.


5. View the results.

SIMULINK MODEL:


RESULT:


EXPERIMENT NO. 6

OBJECT: Calculate the corona loss of transmission line.


THEORY: The phenomenon of ionization of surrounding air around the conductor due to which luminous glow with

hissing noise is rise is known as the corona effect.

Air acts as a dielectric medium between the transmission lines. In other words, it is an insulator between the

current carrying conductors. If the voltage induces between the conductor is of alternating nature then the

charging current flows between the conductors. And this charging conductor increases the voltage of the

transmission line.

The electric field intensity also increases because of the charging current.

If the intensity of the electric field is less than 30kV, the current induces between the conductor is neglected.

But if the voltage rise beyond the 30kv then the air between the conductors becomes charge and they start

conducting. The sparking occurs between the conductors till the complete breakdown of the insulation

properties of conductors takes place

Corona Formation:

Air is not a perfect insulator, and even under normal conditions, the air contains many free electrons and

ions. When an electric field intensity establishes between the conductors, these ions and free electrons

experience forced upon them. Due to this effect, the ions and free electrons get accelerated and moved in the

opposite direction.

The charged particles during their motion collide with one another and also with the very slow moving

uncharged molecules. Thus, the number of charged particles goes on increasing rapidly. This increase the

conduction of air between the conductors and a breakdown occurs. Thus, the arc establishes between the

conductors.

CORONA POWER LOSS:

The power dissipated in the system due to corona discharges is called corona loss. Accurate estimation of

corona loss is difficult because of its variable nature. It has been found that the corona loss under fair weather

condition is less than under foul weather conditions. The corona loss under appropriate weather conditions is

given below by the Peek‟s formula;

Where Pc – corona power loss

f – frequency of supply in Hz

δ – air density factor

En – r.m.s phase voltage in kV

Eo – disruptive critical voltage per phase in kV

r – radius of the conductor in meters

D – spacing between conductors in meters

It is also to be noticed that for a single –phase line,

En=1/2×line voltage

and for a three phase line,

En = 1/(√3)×line voltage

Peek‟s formula is applicable for decided visual corona. This formula the gives the inaccurate result when the

losses are low, and En/Eo is less than 1.8. It is superseded by Peterson‟s formula given below;

Where,

Pc – corona power loss

f – frequency of supply in Hz


En – voltage per phase

r – radius of the conductor

D – spacing between conductors in meters

Factor F is called the corona loss function. It varies with the ratio (En/Eo). Eo is calculated by the formula

given below,

Where,

Go – maximum value of disruptive critical voltage gradient in V/m.

mo= irregularity factor

PROBLEM STATEMENT:

A sphere , 50Hz , 132KV transmission line consists of conductors of 1.17cm diameter and spaced

equilaterally at the distance of 3cm. The conductors have smooth surface with mo= 0.96. The

barometric pressure is 72cm of Hg and the temp. is 20’C . Determine corona loss per Km per phase

under fair conditions given dielectric strength of air = 33KV & its rms value = 21.22KV/cm( go ).

FORMULA USED :

Air density factor ( ẟ) =

Where b = barometric pressure in cm of Hg

t = temp. in ℃

r =

Under fair condition,

Vd = r ẟ go mo ln ]

PROCEDURE:





5. View the results

PROGRAM:

%single phase transmission line

b = input („enter the value of baraometric pressure ( in cm of Hg)=‟);

D= input („enter the diameter of conductor( in cm)=‟);

t = input („enter the value of temp.( in ℃ )=‟);

V= input („enter the value of line voltage=‟);

mo = input („enter the value of mo=‟);

d = input („enter the value of distance between lines (in cm)=‟);

A = ( 3.92*b )/ (273+t);

r = D/2;

go = 21.21;

Vd = r * A * go * mo *log(d/r);

Vph = V/1.172;

C = r/d;

P = 241 * 0.00001 * ((50+25)/A)*(sqr+(c))*(Vph-Vd)*(Vph-Vd);

fprintf (/n corona loss per Km per phase is %f , p ‟);

RESULT:


EXPERIMENT NO. 7

OBJECT: Calculation of sag & tension of transmission line.


THEORY:

Sag and Tension

For safety purpose, the ground clearance of the conductors at maximum temperature and minimum loading

condition should be maintained. Analysis of the sag and tension is important in the transmission line for the

continuity and quality of electrical services. If the tension of the conductor is increased beyond the limit, it

may get broken, and the power transmission of the system get erupt.

The dip of the conductor between the two level supports is called sag. In other words, the vertical distance

between the highest point of the electrical pole or tower (where the conductor is connected) and the lowest

point of the conductor between the two adjacent level supports is known as sag shown in the figure below.

The horizontal distance between two electrical supports is called the span.

If the weight of a conductor is uniformly distributed along the line, then it is assumed that a freely suspended

conductor shape is a parabola. The shape of sag increases with the increase in the length of the span. For a

small span ( up to 300 meters) parabolic method and large span ( like river crossings) catenary method is

used for the calculation of sag and tension.

Calculation of sag and tension in transmission line depend on the span of the conductor. Span having equal

level supports is called level span, whereas when the level of the supports is not at an equal level is known as

unequal level span.

Tension, T= Ultimate Strength/Safety Factor

Sag,S=Wl2/8T

Where: W= Weight of conductor

l=length of Span

T= tension

PROBLEM STATEMENT:

A 132KV Transmission line has the following data;

Weight of conductor = 680 Kg/Km ;

Length of span = 860m;

Ultimate strength = 3100Kg;

Safety factor = 2;

Calculate the sag and tension .

Formula used :

Tension (T) =

Sag(S) =

Where ,

W = weight of conductor

https://circuitglobe.com/calculation-of-sag-and-tension.html

https://circuitglobe.com/sag-and-tension.html


l = length of span

T = tension

PROCEDURE:





5. View the results

PROGRAM:

%Calculate sag and tension of transmission line.

US = input („enter the value of ultimate strength (in Kg)‟);

SF = input („enter the value of safety factor SF = ‟);

L = input („enter the value of span length (in m)‟);

W = input („enter the value of weight of conductor (in Kg/m)‟);

T = US/SF;

S = ( W*(l*l)) / (8*t);

fprintf („/n tension per Km is %f „,T);

fprintf („/n sag in meter is %f‟, S);

RESULT:


EXPERIMENT NO. – 8

OBJECT : Calculation of string efficiency of insulator of transmission line.

SOFTWARE USED : MATLAB

PROBLEM STATEMENT :

An insulator string ia made of 3 similar insulators. Find the max. voltage that is string can withstand if

the max. voltage per unit is 17.5 KV . Assume k= 0.125.

FORMULA USED :

String efficiency , (ɳ) =

Where,

V = voltage across string

n = no. of disc nearest to conductor

V3= voltage across disc nearest to conductor

V = V1+V2+V3

V1= V3 / k^2 + 3k +1

V2 = V1[1+k]

PROGRAM :

%Calculation of string efficiency of insulator of transmission line .

V3 = input („enter the value of max. voltage per unit ( in volts)‟);

n = input („enter the total number of disc in the string n =‟);

k = input („enter the value of k =‟);

V1 = V3/((k*k)+(3*k)+1);

V2 = V1*(1+K);

V = V1+V2+V3

SE = (V/ (n*V3))*100

fprintf („/n string efficiency is %f‟,SE);

OUTPUT :

V=44.2415KV ; V1 = 12.5842 KV

SE = 84.27% ; V2 = 14.1573 KV


EXPERIMENT NO. – 11

OBJECT : Calculation of ground clearance of transmission line.

SOFTWARE USED : MATLAB

PROBLEM STATEMENT :

An overhead line conductor is supported by the two towers which are at 70cm height above the water

level . The tower are separated from each other by horizontal distance of 300m . the tension in the

conductor is 1500Kg. find the clearance at a point midway between the tower if size of conductor

material is 8.87gm/cm^3.

FORMULA USED :

Weight per meter = area of cross section im m^2 * density in Kg/m^3

W = A * DEN

= 0.95*10^-4 * 8.87*(10^-3/10^-6) = 0.816Kg/m

Sag (S) =

Where,

L= span length

T = tension

Clearance = height – sag

PROGRAM :

%Calculation of ground clearance of transmission line

A= input(„enter the value of area cross-section A=‟);

Den= input („enter the value of density Den =‟);

L = input („enter the value of span length L=‟);

T = input („enter the value of tension T=‟);

h = input („enter the value of height of tower h‟);

W= A*Den;

S = (W*L^2) / ( 8* T);

CL = h-S

fprintf(„Ground clearance is %f,CL)

RESULT:

Ground clearance is 63.88m.


EXPERIMENT NO. -12

OBJECT : Calculation of parameter of underground cables.


PROBLEM STATEMENT :

A Concentric cables has a conductor diameter of 0.6cm & the insulation thickness of 1.4cm . If the

dielectric used has relative permittivity of 5 , calculate the capacitance for 1 Km length of cables ,

where ρ=7.5*10^2 ohm m.

FORMULA USED :

D=d+2t

Where , D= TOTAL DIAMETER WITH SHEATH

d = conductor diameter

capacitance (c) =

insulation resistance =

Ri = (ρ / *ln(D/d)ohm

PROGRAM :

%Calculation of parameter of underground cables.

d = input („enter the value of conductor diameter (in cm)=‟);

input („enter the value of relative permitivity =‟);

input („enter the value of specific resistance ‟);

l = input („enter the value of length (in m)=‟);

t= input („enter the value of insulation thickness (t)=‟);

D = d+2t;

Ri = ( / 2* *l)*log(D/d);

C = (2* *8.85*10^-12* ) / log (D/d);

fprintf („/n insulation resistance is %f‟,Ri);

fprintf („ /n capacitance is %f‟, c);

RESULT:

insulation resistance is 2.0425e+16

capacitance is 0.1603 micro F.


CONTROL SYSTEM LABORATORY

[KEE-552]

EXPERIMENT NO - 1

OBJECT: To study the performance characteristics of a DC motor speed control system.

THEORY: Speed control is a very common requirement in many industrial applications such as rolling

mills, paper factories, etc.Speed control means intentional change of the drive speed to a value required for

performing the specific work process.

In this experiment we will study the speed characteristics of a DC motor (armature controlled ) as open loop

close loop system under varying conditions of gain and we will observe the steady state condition of the

system and hence the speed characteristics which depends upon the system i.e. Km, KA,KT .

At the same time the speed characteristics are observed in the presence of disturbance which is simulated by

the breaking ( eddy current braking ) at different level by varying the system gain i.e. KAis keeping Km and

KT constant.

APPARATUS REQUIRED:

S.No. Name of Equipment Range /Rating Qty Make

CIRCUIT DIAGRAM:

PROCEDURE:

OBSERVATION:

Motor and tachogenerator Characteristics

VR=1 volt

S. No. Ka N Vt Vm

Exp. Ka =

CLOSED LOOP PERFORMANCE:

STEADY STATE ERROR :

VR=1 volt

S.No. Ka SPEED, N VT Theo. Ess=VR-VT Exp.

Ess=


DISTURBANCE REJECTION:

Brake Setting 0 1 2 3 4 5

Open loop speed

Closed loop speed at different gain,

Ka=10

CALCULATION:

Km =

Kt = Vt /ss

T= (Vs/Vt)/(KaKmKt/2f)

Steady state error :Ess=

Transient error :Teff = ( )

RESULT:

PRECAUTIONS-

VIVA-VOICE QUESTION-

1. Why series motor cannot be started on no-load?

2. Which type of motor is used in trains, what is the rating of supply used?

3. Why is the starting current high in a DC motor (or) Why do we use a starter?

4. What happens when a D.C motor is connected across an A.C supply?

5. Why series motor cannot be started on no-load?


EXPERIMENT NO - 2

OBJECT: To study the performance characteristics of an angular position error detector using two

potentiometers.

APPARATUS REQUIRED:


THEORY:

Error detectors detect the difference between the desired input and actual value of the controlled variable or

output and produce a signal proportional to the difference.

The simplest type is a potentiometer type error detector, which produce voltage signal, proportional to the

difference of the input and output signals. The potentiometers may be rotary or linear type, depending on the

type of motion. If input is r and outputc, two identical rotary potentiometers would be connected. The

voltage will depend on the difference r and c would be zero if the positions of the two were identical.

Error detector:

The basic error detector consists of two servo-potentiometers with calibrated dials 1 degree mounted on the

panel. A common Ac/DC (selected by switch) signal is internally connected to these and the potentiometer

output is permanently wired to a unit gain instrumentation amplifier . Theoutput of the instrumentation

amplifier is brought out on the panel. This constitutes the error detector

Demodulator:

This block is needed during the AC operation of the potentiometer. The ACoutput of the potentiometer may

be connected to the demodulator input and output obtained is a phase sensitive DC signal.

The Ke = voltage applied volts/ maximum angular span, radians

CIRCUIT DIAGRAM:

PROCEDURE:

OBSERVATION:

With DC excitation:

Pot 2 fixed at 2 =180 0

S.No. Pot1 position 1 e =2-1 Output Vo volts

With AC excitation:

Pot 2 fixed at 2 =180 0

S.No. Pot1 position 1 e =2-1 Output, Vo (mili volts ) V(Dem.) form DVM , volts

Output, Vo (mili volts)

RESULT:

PRECAUTIONS

a. Viva Questions: -

b. 1) Explain the working of potentiometer?

C. 2) Write the application of potentiometer?


EXPERIMENT NO - 3

OBJECT: To study P, PI and PID temperature controller for an oven and compare the result.

APPARATUS REQUIRED:


THEORY:

In this experiment a plant, i.e. oven is given. This is a thermal system whose transfer function is determined

on the basis of heat supplied to it and the way next dissipates i.e. by conduction, convection or radiation. In

this experiment the lumped parameter has been approximated i.e. the next has been uniformly distributed

through out the oven space the transfer through radiation the conductive and convection heat transfer

= T

= Rate of heat flow

= Constant

T= temp. difference in degree centigrade

Under the assumption of linearity the thermal resistance if defined as

R = T/ = 1/

Similarly the thermal capacitance of the mass is defined by

= cd T / dt

The equation of an oven now become combining the above two equation

= cdT/ dt + 1/R XT

Taking laplace transfer

T(s) / = R/ sCR +1

While experiment the following point to be taken in to account:

a) The temp rise in response to the heat input is not instantaneous. The result in a delay.

b) For steady state error

c) Ess = lim (tref- T) = Tref / (1+AR)

CONTROLLER:

As the experiment already has inbuilt transfer function of an oven we just go for the determination of output

response of the oven with different type of controller used in the industries

A controller is a controlling element, which checks the out put performance of feed back system. If it is out

of desired level. Basically there are four type of controller use in a industry for system performance I

improvement

a) On-off relay: Is a on-off switch and introduce in the system so as to control the on-off time of a

system. it is also referred to as two position controllers, consist of a simple and inexpensive witch /

relay and are therefore used very commonly ij both industrial and domestic control systems. Typical

applications include air conditioner and refrigerators, ovens, heaters.

b) Proportional Controller: It is an amplifier of gain Kp that amplifies the error signal and passes it to

the actuator. The noise drift and bias currents of this amplifier set the lower limit of the input signal

which may be handled reliably and therefore decide the minimum possible value of the error

between the input signal and output. Also the saturation characteristic of this amplifier sets the linear

and non-linear regions of its operation.

c) PI controller:

This controller introduces a pole at the origin, i.e. increases the system type number by unity

The steady state error of the system is therefore reduced


d) PID controller: It increase the damping ratio of the system and therefore improves the dynamic

performance by reducing overshoot. The PID controller helps in reducing the steady state error with

an improvement in the transient response.

CIRCUIT DIAGRAM:

PROCEDURE:

OBSERVATION:

S.No. Time (in second) Temperature (oC)

RESULT:

PRECAUTIONS:

Viva Questions-

1. Write a brief note about PID Controller?

2. Compare the performance of PI and PD controller?

3. Which controller is used for improving the transient response of the system?

4. Which controller is used for improving the steady state response of the system?

5. What is the purpose of PID controller?


EXPERIMENT NO - 4

OBJECT : Measurement of temp. by RTD.

APPARATUS REQUIRED:


THEORY :

RESISTANCE TEMPERATURE DETECTOR (RTD) :

The resistivity of metals increases with an increase in temperature (i.e. the temperature coefficient is

positive), where as in some semiconductors the resistance decreases with an increase in temperature

(i.e. the temperature coefficient is negative).

The resistance thermometer based on the above phenomenon is one of the most accurately

reproducible temperature – sensing device. PT – 100 is unduly used as a R.T.D.

SIGNAL CONDITIONER MODULE :

The AC constant current signal is applied on the R.T.D. to make it operative. The output of the

R.T.D. is directly fed to the input of D.C. differential amplifier and then is fed to a summing

amplifier with a gain and zero adjustment to obtain the output directly in engineering unit of

temperature. The final output of the amplifier is fed to Digital Panel Meter to display the

temperature. Gain adjustment pot is given for the adjustment of amplifier gain and zero pot is given

for the zero adjustment.

VARIABLE RESISTANCE SOURCE :

A 99 to 150 ohm. variable resistance (wire wound potentiometer is provided with the set – up to

calibrate the signal conditioners module for measurement of temperature directly in 0C.

I. Table for Resistance versus Temperature for PT – 100 is given below :

OBSERVATION :

TABLE – I

S.No. Time(sec) Tempearture by RTD Resistance

RESULT:

C.I.1.

C.I.2. PRECAUTIONS:

1. To get best performance you have to put instrument at dust proof and humidity free environment.

2. To get the good performance from the tutor you have to maintain room temperature.

3. Do not try to open the instrument or repair it.

4. Zero error should be removing.

VIVA QUESTIONS-

1. What are the major differences between a thermocouple and an RTD?

2. What are the major advantages of RTD?

3. What are the major disadvantages of RTD ?

4. What is a RTD?


EXPERIMENT NO - 5

OBJECT : Study of Magnetic Amplifier.

APPARATUS REQUIRED:

Sl. No. Name of apparatus Quantity

1 Magnetic amplifier kit one

2 Patch cords five

THEORY :

Amplification is the control of a larger output quantity by the variation of a smaller input quantity. A

magnetic device called magnetic amplifier of magnetic amplifier can perform such amplification. This set up

is design to study the basic characteristics of such amplifiers. The set up consists magnetic amplifier, AC and

DC power supply, two meters for load and control currents and a fixed value RL.

PROCEDURE:

OBSERVATION TABLE:

Mode-1(50 mA)

For Vc Positive direction

Sl. No. Load current IL(mA) Control Current Ic(mA)

For Vc Negative direction


Mode-2(10 mA)

For Vc Positive direction


For Vc Negative direction


RESULT: Characteristics of Magnetic Amplifier for Mode-1 and Mode-2 are shown in Graph Paper.

PRECAUTIONS:

VIVA QUESTIONS-

1. What is an Amplifier?

2. Applications of Magnetic Amplifier


EXPERIMENT NO - 6

OBJECT: To study the performance characteristics of a dc motor angular position control system.

APPARATUS REQUIRED:


THEORY:

In this experiment we are going to study the performance of the closed loop system with proportional

feedback and closed loop system with combined proportional and tachogenerator feed back. This is a second

order system and the analysis of its o/p is to be carried out for its time domain specification i.e. overshoot,

peak time, rise time, damping ratio, under damped frequency and steady state error.

The card is designed to automatically store the time response of the system in a RAM whenever a step input

is given. The stored response is the displayed on the CRO. For this press the MODE switch the unit becomes

ready to capture the step response. Apply the step input now starts the storage. At the end of storage cycle the

mode automatically shifts to display and the response waveform is seen on the CRO.


CIRCUIT DIAGRAM:

PROCEDURE:

D. OBSERVATION:

(A) MANUAL OPERATION OF THE POSITION CONTROL:

KD = 0 ,Tachogenerator Channel Disable KA = 5

S.No.

R

(Deg.)

R

(Deg.)

0

(Deg.)

0

(Deg.)

0 R

(Deg.)

VR

(volt)

Vo

(volt) VR -Vo

(volt)

(B)STEP RESPONSE OF THE POSITION CONTROL WITH OUT TECHOGENERATOR FEED

BACK

KD = 0 VS = 2.5 V

S.No. KA Mp% tr (m sec) tp(m sec) δ ess

Volt

ω

rad/sec

(C) STEP RESPONSE OF THE POSITION CONTROL WITH TECHOGENERATOR FEED BACK

KA = 7 VS = 2.5 V

S.No. KP Mp tr tp δ ess

Volt

ω

rad/sec

CALCULATION:

RESULT:

PRECAUTIONS:

VIVA QUESTIONS -

1. What are the various ways of varying the speed of Dc motors?

2. What is eddy current breaking in Dc speed control?

3. Can a rotary dimmer be used to control Dc motor speed?

4. What are the components of DC position control?

5. How is position control achieved?


EXPERIMENT NO - 7

Object: To obtain torque speed characteristics of AC servo motor.

Theory:

An AC servomotor is basically a two phase induction motor except for certain special design features. The

motor of the servo motor is built with high resistance so that its X/R ratio is small and the toque speed

characteristics are linear. For low resistance, the characteristics are nonlinear. Such a characteristics is

unacceptable in control systems. The motor construction is usually squirrel cage or drag cup type. The

diameter of rotor is kept small in order to reduce inertia to obtain good accelerating characteristics. In servo

applications the voltage applied to the two stator windings are seldom balanced. One of the phases known as

the reference phase is excited by constant voltage and the other phase is known as the control phase is

excited by a voltage of variable magnitude from a servo amplifier and polarity with respect to the voltage

supplied to the reference winding. For low power applications AC servo motors are preferred because they

are light weight, rugged construction.

Circuit Diagram:

Apparatus Required: AC servo motor speed torque unit, multimeter.

AC servomotor specifications: Reference winding voltage: 230v AC Control winding voltage: 230v AC Rated power = 50 watts Moment of

Inertia ( J) = 0.7gm/cm2 Friction co efficient B = 0.021 Speed: 2000rpm

Procedure:

1. Connections are made as per circuit diagram.

2. The load switch is in OFF position, so that DC machine is not connected to auxiliary power supply (12V).

AC servomotor switch is in OFF position.

3. Ensure speed control and load control pot is in minimum position.

4. Switch ON the mains and also AC servomotor. The AC servomotor starts rotating and speed will be

indicated by meter in front panel.

5. The reference winding voltage can be measured.

6. Now switch ON the load control switch and start loading the servomotor.

7. Note down corresponding values of Ia & Ep (& speed).

8. Now new control winding voltage is set by varying position of speed control switch. Again the machine is

loaded & Ia,Ep are noted down.

9. Speed –Torque characteristics are plotted.

Calculation: Power = Eb x Ia Watts Torque = (Eb x Ia x 60 x 1. 0196 x 105 x N gm-cm) / 2


Viva questions:

2. What are the applications of AC servo motor?

3. 2. How is AC servo motor different from normal AC motor?

4. 3. What is the working principle of AC servo motor?


EXPERIMENT NO - 8

OBJECT: To study synchro transmitter and reciver and obtain output Vs input characteristics.

APPARATUS REQUIRED:

Sl. No. NAME OF THE

APPARATUS

TYPE QUANTITY

1. Synchro transmitter- receiver pair kit 1

2 Connecting wires Patch cords As required

Theory: The term synchro is a generic name for a family of inductive devices which works on the principle of a

rotating transformer (Induction motor). Basically they are electro mechanical devices or electromagnetic

transducer which produces an output voltage depending upon angular position of the rotor. A Synchro

system is formed by interconnection of the devices called the synchro transmitter and the synchro control

transformer. They are also called as synchro pair. The synchro pair measures and compares two angular

displacements and its output voltage is approximately linear with angular difference of the axis of both the

shafts. They can be used in the following two ways.

i. To control the angular position of load from a remote place / long distance.

ii. For automatic correction of changes due to disturbance in the angular position of the load.

CIRCUIT DIAGRAM:

PRECAUTIONS:

1. Keep the angular positions of rotors of transmitter and receiver at zero position before starting the

experiment.

2. Handle the angle pointers for both the rotors in a gentle manner.

3. Do not attempt to pull out the angle pointers.

4. Do not short rotor or stator terminals.

PROCEDURE:

1. Connect the mains supply to the synchro Transmitter- receiver system with the help of the given mains

cord.

2. Connect 110V AC supply to the rotor terminals (R1 and R2) of the transmitter only and switch on the

mains supply.

3. Now at zero angular position of rotor of transmitter, note down the voltage between stator winding

terminals i.e., VS1S2, VS2S3 and VS3S1 with the help of given patch cords and tabulate them.

4. Vary the angular positions of rotor of the transmitter in steps by 30 and note down the corresponding

voltages between stator winding terminals in a tabular column.

5. The zero position of rotor and stator coincide with voltage VS1S2 equal to zero. Do not disturb this

condition.

6. Switch off the mains supply of the kit after bringing back the rotor at zero.


7. Plot a graph between angular positions of rotor of transmitter and stator voltages for all three phases.

TABULAR COLUMN:

GRAPH: - Plot a graph of angular position VS voltage for all the three phases.

RESULT: -The synchro Transmitter was studied and waveform of stator winding voltages was

plotted on graph .

To study about synchro transmitter – receiver pair.

APPARATUS REQUIRED:

Sl. No. NAME OF THE APPARATUS TYPE QUANTITY

1. Synchro transmitter- receiver pair kit 1

2 Connecting wires Patch cords As required

CIRCUIT DIAGRAM:


PRECAUTIONS:

1. Keep the angular positions of rotors of transmitter and receiver at zero position before starting the

experiment.

2. Handle the angle pointers for both the rotors in a gentle manner.

3. Do not attempt to pull out the angle pointers.

4. Do not short rotor or stator terminals.

PROCEDURE:

1. Connect the mains supply to the synchro Transmitter- receiver system with the help of the given mains

cord.

2. Connect the stator terminals of transmitter S1, S2, and S3 with stator terminals of receiver S1, S2, and S3

with the help of patch cords respectively.

3. Connect 110V AC supply to the rotor terminals (R1 and R2) of both transmitter and receiver, and then

switch on the mains supply.

4. Now at zero angular position of rotor of transmitter, note down that of receiver and tabulate them.

5. Vary the angular positions of rotor of the transmitter in steps by 30 and note down the corresponding

angular positions of rotor of synchro receiver.

6. It is observed that whenever the rotor of the synchro transmitter is rotated, the rotor of the synchro receiver

follows it both directions of rotations and its positions are linear with the initial error.

7. Switch off the mains supply of the kit after bringing back the rotor of the transmitter at zero.

8. Plot a graph between angular positions of rotor of transmitter and angular positions of rotor of receiver.

TABULAR COLUMN:

RESULT:

Thus the characteristics of synchro transmitter receiver pair were studied.

Viva questions: 1. What is meant by synchro-transmitter receiver pair?

2. Give the application of synchro-transmitter receiver pair?


EXPERIMENT NO - 9

OBJECTIVE: - To determine time domain response of a second order system for unit step input and obtains

performance parameters.

REQUIREMENTS:- Computer system, MATLAB software.

THEORY:-

Let the effect of pole location before finding the unit step response. To find the poles of closed loop tx

function , let us put The following conclusions may be drawn:

(i) Poles are real and unequal if

(ii) Poles are real and equal if

(iii) The poles are complex conjugate if ,where is damping factor.

Matalab program for second order system

t=0:0.1:30

num=[25]

den=[1 2 25]

c1=step(num,den,t)

plot(t,c1)

xlabel('time in sec')

ylabel('c(t)')

title('second order')

RESULT: - Graph of second order system has been plotted using MATLAB.

PRECAUTIONS: -

1. Computer system should be handled carefully.

2. MATLAB Program should be typed carefully.

3. MATLAB Program should be properly saved and run.

4. Results should be observed carefully

Viva Questions: 1) Explain the steady state and transient response?

2) Explain the classification of control system according to the damping ratio?


EXPERIMENT NO - 10

OBJECTIVE: - To plot root locus diagram of an open loop transfer function.


THEORY:-

Root locus:-

In control theory the root locus is the locus of the poles and zeros of a transfer function as the system gain K

is varied on some interval. The root locus is a useful tool for analyzing single input single output (SISO)

linear dynamics systems. A system is stable if all of its poles are in the left-hand side of the s-plane (for

continuous systems) or inside the unit circle of the z-plane (for discrete systems).

Given transfer function:-

G(s) =K(s+1)/s2(s+3.6)

MATLAB Program:-

num=[1 1];

den=conv([1 0 0],[1 3.6]);

g=tf(num,den);

rlocus(g)

grid on

title 'root locus'

RESULT: - Root locus diagram of an open loop transfer function has been plotted using MATLAB.

PRECAUTIONS: - 1. Computer system should be handled carefully.



4. Results should be observed carefully.

GRAPH:-

VIVA QUESTIONS-

1. Explain about MATLAB Software

2) Write steps to plot Root Locus

3) Advantages of using software for analysis.


EXPERIMENT NO – 11

OBJECTIVE: - To plot a Bode diagram of an open loop transfer function.

REQUIREMENTS:- MATLAB software, computer system.

THEORY:-

Frequency Response: The frequency response is the steady state response of a system when the input to the

system is a sinusoidal signal. Frequency response analysis of control system can be carried either nalytically

or graphically. Bode Plot is one of the various graphical techniques available for frequency response

analysis.

Bode plot: The bode plot is a frequency response plot of the transfer function of a system. A bode plot

consists of two graphs. One is plot of the magnitude of a sinusoidal transfer function versus log w . The other

is plot of the phase angle of a sinusoidal transfer function versus logw. The bode diagram of a Transfer

function:-

MATLAB Program:- num =10;

den = [1 1 3 ];

g=tf (num,den);

bode (g)

grid on

title 'bode plot'

RESULT: - Bode diagram of an open loop transfer function has been plotted using MATLAB.





GRAPH:-

-60

-40

-20

0

20

Magnitu

de (

dB

)

10-1

100

101

102

-180

-135

-90

-45

0

Phase (

deg)

bode plot

Frequency (rad/sec)

VIVA QUESTION-

1. What is gain margin and phase margin?

2. What is gain cross over frequency and phase crossover frequency?

3. What are the different types of stability conditions?

4. What are the advantages of frequency response analysis?


EXPERIMENT NO - 12

OBJECTIVE: - To draw a Nyquist plot of an open loop transfer function.


THEORY:-

Frequency Response:

The frequency response is the steady state response of a system when the input to the system is a sinusoidal

signal. Frequency response analysis of control system can be carried either analytically or graphically.

Nyquist plot is one of the various graphical techniques available for frequency response analysis.

Nyquist Stability Criterion: If G(s)H(s) contour in the G(s)H(s) plane corresponding to Nyquist contour in

s-plane encircles the point – 1+j0 in the anti – clockwise direction as many times as the number of right half

s-plain of G(s)H(s). Then the closed loop system is stable.

Given Transfer function:-

MATLAB Program:- num =10;

den = [1 1 3 ];

g=tf (num,den);

nyquist (g)

grid on

title 'nyquist plot'

RESULT: - Nyquist plot of an open loop transfer function has been plotted using MATLAB.





GRAPH:-

-3 -2 -1 0 1 2 3 4 5-6

-4

-2

0

2

4

6

0 dB

-10 dB-6 dB

-4 dB

-2 dB

10 dB6 dB

4 dB

2 dB

nyquist plot

Real Axis

Imagin

ary

Axis

VIVA QUESTION-

1. What is contour?

2. What is the frequency domain analysis?


EXPERIMENT NO - 13

OBJECT: To design Lag, Lead and Lag- Lead compensators using Bode Plot.

RC LEAD COMPENSATING NETWORK

Apparatus Required: Resistors, capacitors, wires, multimeter, and phase- frequency meter.

Theory: If a sinusoidal input is applied to the input of a network and steady state output has a phase lead, then

network is called lead compensator/network. Lead compensator has a zero at s = 1/T and a pole at s = 1/ αT

with zero closer to the origin than pole. This compensator speeds up the transient response and increases the

margin of stability of a system. It also helps to increase the system error constant through to a limited extent.

These compensators are used when fast dynamic response is required.

Effect of Phase Lead Compensation 1. The velocity constant Kv increases.

2. The slope of the magnitude plot reduces at the gain crossover frequency so that relative stability improves

& error decrease due to error is directly proportional to the slope.

3. Phase margin increases.

4. Response become faster.

Advantages of Phase Lead Compensation 1. Due to the presence of phase lead network the speed of the system increases because it shifts gain

crossover frequency to a higher value.

2. Due to the presence of phase lead compensation maximum overshoot of the system decreases.

Disadvantages of Phase Lead Compensation 1. Steady state error is not improved.

Circuit

Diagram:

Derivation of transfer function: Write the above circuit in Laplace form.

Vi(s) = (Z1+Z2)*I(s) (Where I(s) is the current in the circuit and Z1= (R1//C) and Z2 = R2)

Vo(s) = Z2*I(s)

Vo(s)/Vi(s) = Z2 /( Z1+Z2)

After simplification,

GC(S) = (S+ 1/ T ) /(S+ 1/ αT ) where T = R1 C and , α = R2 /( R1 + R2)

Design Equations:

Specifications for the design: Фm =…………. at fm = …………...

1. Sin Фm = (1- α) /(1+α) α<1

2. α = R2 /( R1 + R2)

3. ωm = 1 . /T√ α

4. T = R1 C

Procedure: 1. Derive the transfer function for the Lead network given above.

2. For the given specification, ie for given Фm at given Fm , calculations of R1, R2 and C. are done.

3. Connections are made as per the Lead circuit diagram by the selecting the values found in the above step.

4. Switch ON the mains supply and apply sinusoidal wave by selecting suitable amplitude.


5. The frequency of the signal is varied in steps and at each step note down the corresponding magnitude of

output and phase angle.

6. Draw the frequency response plot and hence find the transfer function & compare it with the design.

Tabular Column:

Input voltage VS = ……….V(volts)

Frequency

(Hz)

Output

VO

(volts)

Ø(degree)

indicated

Gain

(dB)

Typical Lead Characteristics:

Result:

Viva questions: 1. What is lag compensation? Write the frequency response of it.

2. What is the importance of lag network?

RC LAG COMPENSATING NETWORK Apparatus Required: Resistors, capacitors, wires, multimeter, and phase- frequency meter.

Theory: If a sinusoidal input is applied to the input of a network and steady state output has a phase lag, then

network is called lag compensator/network. Lag compensator has a pole at s = 1/ βT and a pole at s = 1/ T

with pole closer to the origin than zero. This compensator improves the steady state behavior of the system

while nearly preserving its

transient response. These compensators are used when low steady state error is required.

Effect of Phase Lag Compensation 1. Gain crossover frequency increases.

2. Bandwidth decreases.

3. Phase margin will be increase.

4. Response will be slower before due to decreasing bandwidth, the rise time and the settling time become

larger.

Advantages of Phase Lag Compensation 1. Phase lag network allows low frequencies and high frequencies are attenuated.

2. Due to the presence of phase lag compensation the steady state accuracy increases.

Disadvantages of Phase Lag Compensation 1. Due to the presence of phase lag compensation the speed of the system decreases.


Circuit Diagram:

Derivation of transfer function: Write the above circuit in Laplace form

Vi(s) = (Z1+Z2)*I(s) (where I(s) is the current in the circuit and Z1= R1 and

Z2=(R2+1/CS)

Vo(s) = Z2*I(s)

Vo(s)/Vi(s) = Z2 / ( Z1+Z2)


GC(S) = (S+ 1/ T ) /(S+ 1/ βT ) where β = ( R1 + R2) / R2 and T = R2 C

Design Equations:

Specifications for the design: Фm =…………. at fm = …………... To find maximum lag angle:

1. Sin Фm = (β -1) /(β +1) β>1

2. β = ( R1 + R2) / R2

3. ωm = 1 . /T√ β

4. T = R2 C

Procedure: 1. Derive the transfer function for the Lag network given above.

2. For the given specifications, ie for given Фm at given fm , calculations of R1, R2 and C. are done.

3. Connections are made as per the Lag circuit diagram by the selecting the values found in the above step.





Tabular Column: Input voltage VS = ……… V( volts)

Frequency

(Hz)

VO(volts)

(rms)

Ø(degree)

indicated

Gain(dB)

20 log(Vo/Vs)


Typical Lag Characteristics:

Result:

Viva questions: 1. What is the need for compensation?

2. What is meant by compensation?

3. What is lag compensation? Write the frequency response of it.

4. What is the importance of lag network?

LAG -LEAD NETWORKS Apparatus Required: Resistors – 10k – 2 nos, capacitors – 0.1μF – 2nos, wires, multimeter, and phase-

frequency meter.

Theory: Lag lead compensator is a combination of a lag compensator and lead compensator. The lag section

has one real pole and one real zero with pole to the right of zero. The lead section also has one real pole and

one zero but zero is to the right of the pole. When both steady state and transient response require

improvement, a lag lead compensator is required.

Advantages of Phase Lag-Lead Compensation 1. Due to the presence of phase lag-lead network the speed of the system increases because it shifts gain

crossover frequency to a higher value.

2. Due to the presence of phase lag-lead network accuracy is improved.

Circuit Diagram:

Procedure: 1. Derive the transfer function for the lag lead network given above.

2. Connections are made as per the Lag lead circuit diagram by the selecting the proper values.





Derivation of transfer function: Write the above circuit in Laplace form.

Vi(s) = (Z1+Z2)*I(s) (Where I(s) is the current in the circuit and Z1= (R1//C1S) and

Z2 = (R2 + 1/C2S)

Vo(s) = Z2*I(s)


Vo(s)/Vi(s) = Z2 /( Z1+Z2)


GC(S) = (S+ 1/ T1 ) (S+ 1/ T2) /(S+ 1/ β T2 )(S+ β/ T1 )

where T1 = R1C1 , T2 = R2C2 , R1C1+ R2C2 + R1C2 = 1/ β T2 + β/ T1

Typical Characteristics:

Result:

Viva questions: 1. What is lag lead Compensation? Write the frequency response of it.

2. What is the importance of lag lead network?

Tabular Column: VS =…………. V(volts)

Frequency

(Hz)

VO

(rms)

Ø(degree)

indicated

Gain(dB)

20 log(Vo/Vs


EXPERIMENT NO – 14

OBJECT: To study the time response of a variety of simulated linear system and to correlate the studies

with theoretical results.

THEORY: In this experiment, we are studying the time domain analysis of a linear first order or second

order system. A designed feedback system must be analyzed in lab for the checking of their transient as well

as steady state performance specification. Because on these specification i.e. Maximum overshoot, rise time,

delay time, settling time, damping ratio, undamped natural frequency, steady state constant etc.the system

behavior depends. The system may have large overshoot or zero or over damped signal all totally depends

upon the for best result or performance & should be between 0.4 to 0.8. The system should be under damped.

The can be checked by varying the forward gain of the system & the effect of the forward gain can be plotted

or analyzed & in turn the system performance is improved. These systems are characterized by two poles an

d up to two zeros. For the purpose of transient response studies zeros are usually are not considered.

DELAY TIME is defined as the time needed for the response to reach 50% of final value.

RISE TIME is the time taken for the response to reach 100% of the final value for the first time.

tr = (- )/wd = tan-1

(1-2)

1/2 /

PEAK TIME is the time taken for the response to reach the first peak of the overshoot

tp = /wn (1-2)

1/2

SETTING TIME is the time required by the system response to reach and stay with in a prescribed

tolerance band

APPARATUS REQUIRED:


CIRCUIT DIAGRAM:

PROCEDURE:

OBSERVATION:

S.No. K Mp tr tP wn

CALCULATION:

Mp = exp (-)/(1-2)

1/2

So that = {(In Mp) 2/

2 +(In Mp)

2}

1/2

ts = 4/wn

RESULT:

PRECAUTIONS:

VIVA QUESTION-

1. Difference between the time response and frequency response.

2. What is the maximum overshoot, damping ratio?


EXPERIMENT NO - 15

OBJECT: To simulates PID controller for transportation lag.

PID controllers are commercially successful and widely used controllers in Industries. For example, in a

typical paper mill there may be about 1500 Controllers and out of these 90% would be PID controllers. The

PID controller consists of proportional controller, integral controller and derivative controller. Depending

upon the application on or more combinations of the controllers are used.(ex: in a liquid control system

where we want zero steady state error, a PI controller can be used and in a temperature control system where

we do not want zero steady state error, a simple P controller can be used.

The equation of the PID controller in time domain is given by,

m(t) = KPe(t) + Ki/Ti ∫ e(t) dt + KdTd de(t) /dt

where KP is a proportional gain Ti is the integral reset time and Td is the derivative time of the PID

controller, m(t) is the output of the controller and e(t) is the error signal given by e(t) = r(t) – c(t).

The characteristics of P, I, and D controllers

A proportional controller (Kp) will have the effect of reducing the rise time and will reduce ,but never

eliminate, the steady-state error. An integral control (Ki) will have the effect of eliminating the steady-state

error, but it may make the transient response worse. A derivative control (Kd) will have the effect of

increasing the stability of the system, reducing the overshoot, and improving the transient response. Effects

of each of controllers Kp, Kd, and Ki on a closed-loop system are summarized in the table shown below.

PROPORTIONAL (P) CONTROLLER:

From the block diagram

C(s)/R(s)=(s+Ki) ωn2/s3 + 2ξ ωns2+ ωn2s+Ki ωn2

The characteristic equation is third order, so system also becomes third order reducing SS error to zero


PD CONTROLLER:

From the block diagram

C(s)/R(s) = ωn2 (1+sTd)/S2 + (2ξ ωn+ ωn2Td)S+ ωn2

Comparing with S2 + 2ξ ωnS+ ωn2, damping ratio increases reducing the peak overshoot in the response

PID CONTROLLER

In PID controller, the error signal is given by

Ea(s)= Kp E(s)+ sTd E(s) + Ki/s E(s)

Procedure:

1. The connections are made as in the diagrams.

2. DC supply from the kit is given.

3. The values of kp, Kd, Ki are adjusted and the waveforms are observed on the CRO.

Result:

Viva questions: 1. What is P-I control?

2. What is P-D control?

3. What is P-I-D control?

4. Why differential control is not used alone?

5. What is the problem with proportional control?


EXPERIMENT NO - 16

Object: MATLAB can also be used to transform the system model from transfer function to state space, and

vice versa.

Transfer Function to State Space

Given a transfer function of the form:

A.


KEE-553 ELECTRICAL MACHINES – II LABORATORY

1. To perform no load and blocked rotor tests on a three phase squirrel cage induction motor and determine

equivalent circuit.

2. To perform load test on a three phase induction motor and draw Torque -speed characteristics

3. To perform no load and blocked rotor tests on a single phase induction motor and determine equivalent

circuit.

4. To study speed control of three phase induction motor by varying supply voltage and by keeping V/f ratio

constant.

5. To perform open circuit and short circuit tests on a three phase alternator.

6. To determine V-curves and inverted V-curves of a three phase synchronous motor.

7. To determine Xd and Xq of a three phase salient pole synchronous machine using the slip test and to draw

the power-angle curve.

8. To study synchronization of an alternator with the infinite bus by using: (i) dark lamp method (ii) two

bright and one dark lamp method.

9. To determine speed-torque characteristics of three phase slip ring induction motor and study the effect of

including resistance, or capacitance in the rotor circuit.

10. To determine steady state performance of a three phase induction motor using equivalent circuit.

11. To draw O.C. and S.C. characteristics of a three phase alternator from the experimental data and


12. To study the speed control of a three phase induction motor using cascade connection

13. To study (i) DOL starter (ii) Star-Delta starter (iii) Auto-Transformer starter

*For Software based experiments (Develop Computer Program in ‘C’ language or use

MATLAB or Equivalent open source software i.e. - Scilab)




EXPERIMENT NO -1

OBJECT: To perform no load and blocked rotor tests on a three phase squirrel cage induction motor and

determine equivalent circuit.

THEORY:

NO LOAD TEST:

The induction motor is made to run at rated voltage and frequency. The no load slip Snl is very small,

therefore r2/Snlis very large as compared to X2 in equivalent circuit of induction motor. In this context,

resultant of parallel branches jX and r2 / Snl is equivalent to jX

For delta connection of the windings

If Vnl= stator voltage at no load

Inl = input current in load

P nl= input power at no load

Znl = 3Vnl/ Inlstator input at no load

Stator no load resistance

R nl = P nl /3 (I nl/3) 2

Xnl = (Z nl 2 - R nl

2

BLOCKED ROTOR TEST:

Blocked rotor test of induction motor is similar to short circuit test on transformers. It is performed on

induction motor to calculate leakage impedance. For performing this test rotor shaft is blocked by external

means. Now voltage is applied to the stator terminal through a three phase variac. The applied voltage is

adjusted till rated current flows in stator windings. In this case

Slip will be Sbr = ( Ns-Nr)/Ns= (Ns-0)/Ns = 1

If Vbr= applied voltage in blocked rotor

Ibr = input current in blocked rotor

Pbr = Input power in blocked rotor

Blocked rotor impedance ,Zbr = Vbr/ (Ibr/3)

Blocked rotor resistance, Rbr = Pbr /3(Ibr/3)2

Blocked rotor reactance,Xbr= (Zbr2-Rbr

2)

Generally for slip ring induction motor x1=x2’

Hence x1=x2’ = Xbr/2

X = Xnl – x1

Rbr = r1+r2‟(X / X2)

2

So r2‟ = (Rbr- r1)*(X2/X)2

where X2 = X + x2‟

APPARTUS REQUIRED :

S.No. Name of Equipment Qty

1

2

3

4

3-Phase IM

Wattmeter

Ammeter

Voltmeter

1

2

1

1


CIRCUIT DIAGRAM:

OBSERVATION TABLE:

NO LOAD TEST:

S.NO. Vnl Inl W1 W2

BLOCKED ROTOR TEST:

S.NO. Vbr Ibr W1 W2

CALCULATION:

RESULT: The No Load and Blocked Rotor tests are performed on the 3 phase induction motor and

parameters of its equivalent circuit are found to be as follows:

PRECAUTIONS:

1. Connections should be tight.

2. Zero error should be removed.


EXPERIMENT NO -2

OBJECT:To perform load test on a three phase induction motor and draw Torque -speed characteristics.

THEORY:

The slip ring induction motor is mechanically loaded by ponney brake mechanical loading arrangement

consisting of CI drum pulley (D = 205mm for 5HP & 3HP) suitable for water cooling, CP wheels for

tightening the belt and measuring the load directly on longitudinal spring balance through which direct load

can be read.

EFFECT ON SPEED: When the motor is on no load, the speed is slightly below the synchronous speed.

The current due to induced emf in the rotor winding is responsible for production of torque required at ni

load. As the load is increased, the rotor speed is slightly reduced. The emf induced in the rotor and hence the

current increases to produce higher torque required.

EFFECT ON SLIP: Slip is expressed as the difference of the speed of the rotor relative to that of the

rotating magnetic field which rotates at synchronous speed.

S = Ns – Nr x 100

Ns

EFFECT ON STATOR CURRENT: Current drawn by the stator is determined by two factors. Its one

component is the magnetizing current required to maintain the rotating field. The second component

produces a field which is equal and opposite to that formed by the rotor current .

EFFECT ON POWER FACTOR: Power factor of an induction motor on no load is very low because of

the high value of magnetizing current. With load the power factor increases because the power component of

the current is increased. Low power factor operation is one of the disadvantage of an induction motor. An

induction motor draws a heavy amount of magnetizing current due to presence of air gap. To reduce this

magnetizing current the air gap is kept small.

Tan ф = 3 (W1-W2) /( W1+W2)

TORQUE: In mechanical loading a brake drum is coupled to the shaft of the rotor 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 of S1 & S2.

Output Power= Torque x Speed

APPARATUS REQUIRED:


1

2

3

4

IM

Wattmeter

Ammeter

Voltmeter

1

2

1

1


CIRCUIT

DIAGRAM:

OBSERVATION:

S.

No.

Applied

Voltage-

Vs

(volts)

Stator

Curren

t- Is

(Amp)

Wattmeter

Reading

(Watt)

Speed

Nr

(rpm)

S1 S2 %

Slip

To

rq

ue

Power

input- Pin

(W1+W2)

(Watt)

Out

put

Power

Po

(Watt)

η

W1 W2

CALCULATION:

RESULT: The variation ofspeed, efficiency, power factor, stator current, torque and slip of 3 phase

induction motor with load is studied and the graph between torque and slip is plotted.

PRECAUTIONS:




EXPERIMENT NO -3

OBJECT: To perform no load and blocked rotor tests on a single phase induction motor and determine

equivalent circuit.

THEORY:

A capacitor start single phase induction motor is provided with a centrifugal switch placed on the rotor shaft

and connected in series with the starting winding. This switch is closed when the motor is at rest and thus the

starting winding is in the circuit and as such the motor can be started as split phase motor. This switch gets

opened and disconnects the starting winding, when the speed of the motor approaches to approx. 75% of the

rated speed.

For performing no load test , the range of the voltmeter should be higher than the rated voltage of the

machine i.e. 300 V and the range of the ammeter should be nearly equal to half of the full load current of the

motor i.e. 5A. As such the range of the wattmeter should be 5A, 250V.

For performing block ed rotor test , the range of the voltmeter could be approx. 40% of the rated voltage i.e.

150V and the range of ammeter more than the full load current of the motor i.e. 10A .

APPARTUS REQUIRED :


1

2

3

1-Phase IM

Ammeter

Voltmeter

1

1

1

CIRCUIT DIAGRAM:

OBSERVATION:

S.

No.

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

resistance

Vo Io Wo Vsc Isc Wsc Vm Im Rdc

CALCULATION:

RESULT:The No Load and Blocked Rotor tests are performed on the single phase induction motor and

parameters of its equivalent circuit are found to be as follows:

PRECAUTIONS:




EXPERIMENT NO -4

OBJECT:To study speed control of three phase induction motor by varying supply voltage and by keeping

V/f ratio constant.

THEORY: The speed of induction motor depends upon slip frequency, slip, and number of poles N= 120

f/P. Varying the supplied stator voltage can vary slip. If the voltage is reduced, torque is reduced as square of

voltage. If stator voltage reduced from V to 0.9 V the torque reduced from T to 0.81 T. Since the torque is

reduced to 81 % ,the rotor speed decreases to N1 i.e. its slip will increase until the increased rotor current will

make up for reduced stator voltage and produce required load at a lower speed N1.

This method of speed control is rarely used for industrial three-phase induction motor because the

requirement of additional costly voltage changing auxiliary equipment. For small induction motor used in

home appliances, the voltage control method of speed control is often used.

APPARATUS REQUIRED:


1

2

3-Phase IM

Voltmeter

1

1

CIRCUIT

DIAGRAM:

OBSERVATION:

S.No. Voltage Speed Slip

RESULT:

PRECAUTIONS:




EXPERIMENT NO -5

OBJECT:To perform open circuit and short circuit tests on a three phase alternator.

THEORY:

The regulation of Alternator is defined as “the rise in terminal voltage” when full-load is removed divided by

rated terminal voltage with speed and excitation of alternator remaining unchanged. The experiment involves

the determination of the following characteristics and parameters:

1. The open -circuit characteristic (the O.C.C).

2. The short-circuit characteristic (the S.C.C).

3. The effective resistance of the armature winding (Ra).

APPARATUS REQUIRED:

S. No. Name of Apparatus Quantity

1.

2.

3.

4.

5.

Alternator (M-G set)

Wattmeter

Ammeter

Voltmeter

Connecting wire

1

3

3

10

CKT. DIAGRAM:


OBSERVATION:

s.no. %VR

RESULT:

PRECAUTIONS:




EXPERIMENT NO -6

OBJECT: To determine V-curves and inverted V-curves of a three phase synchronous motor.

THEORY: When the synchronous machine is operated at constant load, it follows the equation:

EfSin = PeXs

Vf

When Ef varies varies such that EfSin is constant.

Also, Vf Ia Cos = Pe = constant

Ia Cos = Pe = constant

Vf

It means that the projection of the current phasor on Vf must remain constant. The excitation voltage

corresponding to unity power is normal excitation.

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 behavior of the synchronous

motor is described below under three different modes of field excitation

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

UNDER EXCITATION:

When the field current is decreased gradually below the normal excitation the armature current increases and

the operating power factor of motor decreases. The power factor under this condition is lagging. Thus the

synchronous motor draws a lagging current, when it is under excited and equivalent to an inductive load.

OVER EXCITATION:

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

increases and operating power factor increases. How ever the power factor is leading under this condition.

Hence the synchronous motor draws a leading current, when it is over excited and it 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 V is obtained and is known as “V” curves. Such a characteristics curve

plotted between input power factor and the field current for a constant mechanical load on the motor is of the

shape of inverted V and are known as inverted “V” curves.

For increasing the constant mechanical load on the motor the curve bodily shift upwards.

APPARATUS REQUIRED:


1

2

Synchronous motor

voltmeter

1

1


CIRCUIT DIAGRAM:

OBSERVATIONS:

AT NO LOAD –

S.No. W1 W2 Vs Ia If V dc I dc

AT HALF LOAD –

S.No. W1 W2 Vs Ia If V dc I dc

CALCULATION:

RESULT:The variation of armature current with change in field current is observed and hence „V-Curve‟ is

plotted

PRECAUTIONS:




EXPERIMENT NO -7

OBJECT:To determine sub transient direct axis (X”d) and quadrature axis (X”q) reactance‟s of an alternator.

THEORY: When both the field and armature MMF waves are almost stationary with respect to each other,

then the reactance associated with armature are positive sequence reactances. Any two phases of three phase

machine are connected in series and a single phase voltage is impressed across them as shown in figure. The

rotor is at standstill. The impressed voltage is adjusted to pass sufficient current in the two series connected

armature windings. Now the rotor position is adjusted with hand to get maximum deflection of the ammeter

placed in the field winding circuit. Under these conditions, d-axis sub transient impedance Z”d is given by

Z”d = V/(2Imax )

Here V &Imax are the voltmeter & ammeter readings respectively. If the wattmeter records P watts, then

Cosφ = P/ (V Imax )

Sinφ= √ [1- (P/ V Imax )2 ]

d-axis sub transient reactance, Xd” = Z”dSinφ

= V/(2Imax ) √ [1- (P/ V Imax )2 ]

= √ [(VImax)2 – P

2] / Imax

2

Note that induced current in the short circuited field winding is maximum only when field winding axis (i. e.

polar or direct axis) is along the direction of resultant armature m.m.f. as shown in fig. Since the resultant

armature m.m.f. is pulsating in nature, the conditions illustrated in fig. are identical for which Xd” has

defined.

If the rotor shaft is rotated by hand through a half a pole pitch, then peak of resultant armature m.m.f.

coincides the q axis. These conditions are those for which Xq” has been defined. Since the field winding axis

is 900 away from the resultant armature m.m.f. axis, the armature in field winding should record minimum

reading. In view of this, for measuring Xq” the rotor position is adjusted so as to get minimum value of

induced field current. Under these conditions, the instrument reading reading gives,

Xq” = Zq” Sinφ

= V/(2Imin ) √ [1- (P/ V Imin )2 ]

= √ [(VImin)2 – P

2] / Imin

2

When resistance is neglected,

Xd” = V/(2Imax )

Xq” = V/(2Imin )

APPARATUS REQUIRED:

S. No. Name of Apparatus Quantity

1

2

3

Alternator

Ammeter

Voltmeter

1

1

1

CKT. DIAGRAM:

RESULT: The values of direct axis and quadrature axis sub transient reactances are found to be

Xd = Xq =

PRECAUTIONS:




EXPERIMENT NO -8

OBJECT: To study the Synchronization of a three-phase alternator with another three-phase alternator.

THEORY: Synchronization of an alternator is the process of switching on an incoming alternator to the bus

bar, so that it can operate in parallel with another alternator already connected to the bus bar to share the load

on the generating station.

For proper synchronization of alternator three conditions mostly satisfied

a) Terminal voltage of incoming alternator must be equal to the bus bar voltage.

b) The speed of incoming alternator must be such the its frequency = PN/120 equal to the bus bar

frequency.

c) The phase sequence of a incoming alternator voltage must be same as bus bar voltage.

Adjusting the field current of incoming alternator satisfy condition (a).

Adjusting the filed current of a dc motor, which is prime mover satisfy the condition (b).

For condition (c) to be satisfied two bright and one dark lamp method , in this method lamp L1 is connected

between R and R‟ , L2 between Y and B‟ and L3 between Band Y‟.

Where R,Y,B are phase terminal of bus bar and R‟Y‟B‟ are phase terminal of second alternator.

These two sets of star vector all rotate at unequal speed, if the frequency of machines is different. If the

incoming alternator is running faster, this voltage star R‟Y‟B‟ will appear to rotate anti clock wise with

respect to bus bar voltage star RYB at speed corresponding to difference between their frequencies. Lamp

will glow up in particular order. Otherwise lamps will glow up in reverse order. When R‟ coincides with R,

B‟ will coincides B,Y‟ will coincides Y. In that case L1 will be dark, and L2, L3 will be bright.

This is the moment we have to synchronize the incoming alternator.

APPARATUS REQUIRED:


1 Synchronization Panel

1

CIRCUIT:

RESULT:

PRECAUTION:




EXPERIMENT NO -9

OBJECT:To determine speed-torque characteristics of three phase slip ring induction motor and study the

effect of including resistance, or capacitance in the rotor circuit.

THEORY:

The simplest method of starting slip ring induction motor is by means of external resistance in the rotor

circuit. This helps in:

a) Decreasing starting current

b) Increasing starting torque

c) Improving its starting power factor.

In rotor resistance starter, the terminals of the rotor winding are connected to a three phase variable resistor

through slip rings. Resistances are fully in the circuit at starting . The external resistance in the rotor circuit is

gradually cut out, as the motor speeds up and finally the rotor winding is short circuited during normal

running condition.

APPARATUS REQUIRED:


1 Rotor resistance Starter Set up 1

CIRCUIT DIAGRAM:

OBSERVATION:

S.No. Resistance (Ω) Speed (rpm)

RESULT:The speed control of 3 phase induction motor using rotor resistance starter is studied and speed is

found to be -------------- with decrease in rotor resistance.

PRECAUTIONS:




EXPERIMENT NO -10

OBJECT:To determine steady state performance of a three phase induction motor using equivalent circuit.

THEORY: The equivalent circuit of any machine shows the various parameter of the machine such as its

Ohmic losses and also other losses.

The losses are modeled just by inductor and resistor. The copper losses are occurred in the windings so the

winding resistance is taken into account. Also, the winding has inductancefor which there is a voltage drop

due to inductive reactance and also a term called power factor comes into the picture. There are two types of

equivalent circuits in case of a three-phase induction motor-

5. Exact Equivalent Circuit

Here, R1 is the winding resistance of the stator.

X1 is the inductance of the stator winding.

Rc is the core loss component.

XM is the magnetizing reactance of the winding.

R2/s is the power of the rotor, which includes output mechanical power and copper loss of rotor.

If we draw the circuit with referred to the stator then the circuit will look like-

Here all the other parameters are same except-

R2‟ is the rotor winding resistance with referred to stator winding.

X2‟ is the rotor winding inductance with referred to stator winding.

R2(1 – s) / s is the resistance which shows the power which is converted to mechanical power output or

useful power. The power dissipated in that resistor is the useful power output or shaft power.

https://www.electrical4u.com/what-is-inductor-and-inductance-theory-of-inductor/

https://www.electrical4u.com/types-of-resistor/

https://www.electrical4u.com/electrical-resistance-and-laws-of-resistance/

https://www.electrical4u.com/what-is-inductor-and-inductance-theory-of-inductor/

https://www.electrical4u.com/voltage-drop-calculation/

https://www.electrical4u.com/electrical-power-factor/


APPARATUS REQUIRED:

S. No. Name of Apparatus Range Quantity

1

2

3

4

3-Phase IM

Wattmeter

Voltmeter

Ammeter

1

1

1

1

Simulation diagram:

RESULT:

PRECAUTIONS:

1. Write programme carefully.

2. Choose suitable parameters.


EXPERIMENT NO -11

OBJECT: To draw O.C. and S.C. characteristics of a three phase alternator from the experimental data and


THEORY: The variation of terminal voltage from no load to full load of an alternator is called as the

voltage regulation of an alternator.

OPEN CIRCUIT TEST:-

This test is carried out when the alternator is running at rated rpm. and at no load. The field current is varied

and terminal voltage is recorded for about 120 % of the rated value. The characteristics showing the

relationship between field current and emf is known as open circuit curve (OCC).

SHORT CIRCUIT TEST: -

This test is performed when the alternator is running at rated speed. The armature terminals are short

circuited with a very low excitation current. Armature current up to rated value is recorded for various values

of field current. The graph between the armature current and field current is called as the Short circuit

characteristics (SCC). Since the emf generated on open circuit may be regarded, as being responsible for

circulating short circuit current through the synchronous impedance is taken as the ratio of the open circuit

voltage per phase to the short circuit current per phase for a particular field current. The dc resistance of the

stator can be determined with the multimeter but the value of ac resistance can be calculated by multiplying a

factor of 1.3 to calculate the ac resistance. The synchronous reactance can be calculated as Xs = (Z2

s - X2

s)1/2

CALCULATION FOR REGULATION:

The relation ship between terminal voltage and induced emf for a lagging/ leading power factor load is given

by the following expression:-

E = (( V cos + Ia Ra )2 + (V sin + Ia Xs)

2)

1/2

Where, cos = Ra / Zs

% regulation = (E – V)*/V 100 %

CALCULATION FOR EFFICIENCY:-

In open circuit test, the input to the alternator i.ethe power required to drive the alternator is spent as,

friction and windage loss and iron loss. If the field of the alternator is kept unexcited, the input to the

alternator will be equal to the friction and windage loss of the alternator. Input to the alternator can be

calculated by measuring the input to the motor driving the alternator and by knowing the efficiency of the

driving motor.

Input to the alternator = (Input of the driving motor)/ (Efficiency of the driving motor)

Since output of the driving motor is equal to the input to the alternator.

APPARATUS REQUIRED:


1

2

3

Alternator

Ammeter

Voltmeter

1

1

1


CIRCUIT DIAGRAM:

OBSERVATION:

FOR OPEN CIRCUIT TEST

S.No. If (amp) Voc (Volts)

FOR SHORT-CIRCUIT TEST:

CALCULATION:

RESULT: The value of regulation for an alternator is _____________ %.

PRECAUTION:



S.No. If (Amp) Isc


EXPERIMENT NO -12

OBJECT: To study the speed control of a three phase induction motor using cascade connection.

THEORY: Cascading connection is a special method of speed control in which the slip frequency emf

induced in the rotor of the main induction motor is fed to the stator of another induction motor. The main

motor must be a wound rotor motor because its slip rings have to be connected to stator of auxiliary motor;

the slip frequency of main motor is f2 and is equal to s1f1. Auxiliary motor may be a cage motor or a wound

rotor motor. If auxiliary motor is wound rotor motor, its slip rings must be short circuited or connected to a

resistance. This resistance is generally used during starting conditions. During operation the slip rings of

auxiliary motor are short circuited.

Let P1 and P2 be the number of poles of main and auxiliary motors and f1& f2 is the frequencies of supply to

stators of main and auxiliary motors respectively, then

N1 is the synchronous speed of main motor = 120f1/P1

N2 is the synchronous speed of auxiliary motor = 120f2/P2

There are two methods of cascade of two induction motor.

1. Cumulative cascading is the type of cascading the synchronous speed is reduced since here resultant

poles is sum of poles of two induction motor. Resultant Ns = 120 f/(P1+P2)

2. Differential cascading is the type of cascading the synchronous speed is increased since here

resultant poles are difference of poles of two induction motor. Resultant Ns = 120 f/(P1-P2)

APPARATUS REQUIRED:


1

2

IM

Connecting leads

2

l.s.

CIRCUIT:

OBSERVATION:

FOR CUMULATIVE CASCADE CONNECTION:

S.No. Speed of Main motor (rpm) Speed of auxiliary motor

(rpm)

Speed After cascading

(rpm)

RESULT:The speed control of a three phase induction motor using cascade connection is studied and speed

of main motor is --------------- after cascading.

PRECAUTIONS:




EXPERIMENT NO -13

OBJECT: To study (i) DOL starter (ii) Star-Delta starter (iii) Auto-Transformer starter

THEORY:

DOL starter: In the direct on line method of starting cage motors, the cage motor is connected by means of

a starter across the full supply voltage. It consist of a coil operated contactor controlled by the stop and start

pus button which may be installed at convenient places remote from the starter. On pressing the start push

button, the contactor coil energized from two line conductors. The three main contacts and the auxiliary

contact close and the terminal are short-circuited. The motor is thus connected to the supply. When the

pressure is released, it moves back under spring action. Even then the coil remains energized, thus the main

contacts remain close and the motor continues get supplied for this reason, contact is called hold on contact.

When the stop push button is pressed the supply through the contactor coil is disconnected. Science the coil

is de energized; the main contact and auxiliary contact are opened. The supply to motor is disconnected and

the motor stops.

Star-delta Starter:

A star-delta starter is used for cage motor designed to run normally on delta connected stator winding. When

the switch is in start position, the stator windings are connected in star, when the motor picks up speed, say

80 % of its rated value, the change over switch is thrown quickly to the run position which connect the stator

winding in delta, the line current drawn by the motor at starting is reduced to 1/3 as compared to starting

current with the windings connected in delta. At the time of starting when the stator windings are star

connected, each stator phase gets voltage VL/3 where VL is the line voltage. Since the toque develop by an

induction motor is proportional to the square of the applied voltage, star- delta starting reduces the starting

torque to one third that obtain by direct delta starting.

Auto transformer Starter:

In this method, the starting current is limited by using a three-phase autotransformer to reduce the initial

stator applied voltage. The autotransformer is provided with a no. of tapings. In practice starter is connected

to one particular tapping to obtain the most suitable starting voltage. A double through switch is used to

connect the autotransformer in the circuit for starting. When the handle of the switch in the start position, the

primary of the autotransformer is connected to the supply line and the motor is connected to the secondary of

the transformer. When the motor picks up the speed say to about 80% of its rated value, the handle is quickly

moved to the run position. The autotransformer is disconnected from the circuit and the motor is directly

connected to the line and gets its full rated voltage. The handle is held in the run position by the under

voltage relay. In case the supply voltage fails/falls below a certain value, the handle is released and return to

the off position. Over load protection is provided by thermal overload relays.

APPARATUS REQUIRED:


1

2

3

Y-D Starter

DOL Starter

IM

1

1

1


CIRCUIT DIAGRAM:

OBSERVATION:

RESULT:

PRECAUTIONS:



academic lab manual department of electrical and

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