valliammai engineering college semester/ee6461-electrical... · open circuit and short circuit test...

VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur – 603 203.

DEPARTMENT OF ELECTRICAL & ELECTRONICS

ENGINEERING

LAB MANUAL

NAME :

REGISTER NUMBER :

BRANCH /SECTION : ECE

SEMESTER : IV SEM

SUBJECT CODE : EE6461

SUBJECT : ELECTRICAL ENGINEERING AND

CONTROL SYSTEM LABORATORY

Prepared by 1. Dr.R.Arivalahan/ Assoc. Prof.

2. Ms.P.Bency/ Asst. Prof.(O.G)

3. Ms.P.Dhivya/Asst. Prof.(O.G)

4. Ms.K.Durgadevi/ Asst. Prof.(O.G)

5. Ms.G.Shanthi/Asst. Prof.(O.G)

6. Mr.S.Venkatesh/Asst. Prof.(O.G)

VALLIAMMAI ENGINEERING COLLEGE

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

IV Semester - Electronics And Communication Engineering

EE6461 - ELECTRICAL ENGINEERING AND

CONTROL SYSTEM LABORATORY

Manual

(2013 Regulation)

Academic Year

2017-2018 EVEN SEMESTER

Prepared by,

1. Dr.R.Arivalahan/ Assoc. Prof.

2. Ms.P.Bency/ Asst. Prof.(O.G)

3. Ms.P.Dhivya/Asst. Prof.(O.G)

4. Ms.K.Durgadevi/ Asst. Prof.(O.G)

5. Ms.G.Shanthi/Asst. Prof.(O.G)

6. Mr.S.Venkatesh/Asst. Prof.(O.G)

General Instructions to students for EEE Lab courses

Be punctual to the lab class.

Attend the laboratory classes wearing the prescribed uniform and shoes.

Avoid wearing any metallic rings, straps or bangles as they are likely to prove dangerous at

times.

Girls should put their plait inside their overcoat

Boys students should tuck in their uniform to avoid the loose cloth getting into contact with

rotating machines.

Acquire a good knowledge of the surrounding of your worktable. Know where the various live

points are situated in your table.

In case of any unwanted things happening, immediately switch off the mains in the worktable.

This must be done when there is a power break during the experiment being carried out.

Before entering into the lab class, you must be well prepared for the experiment that you

are going to do on that day.

You must bring the related text book which may deal with the relevant experiment.

Get the circuit diagram approved.

Prepare the list of equipments and components required for the experiment and get the indent

approved.

Plan well the disposition of the various equipments on the worktable so that the experiment

can be carried out.

Make connections as per the approved circuit diagram and get the same verified. After getting

the approval only supply must be switched on.

For the purpose of speed measurement in rotating machines, keep the tachometer in the

extended shaft. Avoid using the brake drum side.

Get the reading verified. Then inform the technician so that supply to the worktable can be

switched off.

You must get the observation note corrected within two days from the date of completion of

experiment. Write the answer for all the discussion questions in the observation note. If not,

marks for concerned observation will be proportionately reduced.

Submit the record note book for the experiment completed in the next class.

If you miss any practical class due to unavoidable reasons, intimate the staff in charge and do

the missed experiment in the repetition class.

Such of those students who fail to put in a minimum of 75% attendance in the laboratory class

will run the risk of not being allowed for the University Practical Examination. They will have

to repeat the lab course in subsequent semester after paying prescribed fee.

Being machines lab students should be aware at operating voltage levels and ensure for

any loose connection and short circuits.

Regarding control system part, as students will be working in MATLAB software. And to

be usual each of them is requested to maintain a separate folder for future reference.

EE6461 ELECTRICAL ENGINEERING AND CONTROL SYSTEM LABORATORY

L T P C

0 0 3 2

OBJECTIVES:

To provide hands on experience with generators and motors.

To Understand the working of DC/AC motors and generators

To study the characteristics of transducers

To learn the use of transformer

To understand the behavior of linear system through simulation

To gain knowledge of controllers

LIST OF EXPERIMENTS:

1. Study of DC & AC motor starters

2. Study of three phase circuits

3. Speed Control of DC shunt motor

4. Load Test on DC shunt motor

5. OCC & Load Characteristics of DC shunt generator

6. Transfer Function of separately excited D.C.Generator.

7. Regulation of three phase alternator

8. Open Circuit and Short Circuit test on single phase transformer to draw its equivalent

circuit

9. Load test on single-phase transformer

10. Load test on single phase and three-phase Induction motor

11. Measurement of passive elements using Bridge Networks.

12. Study of transducers and characterization.

13. Digital simulation of linear systems.

14. Stability Analysis of Linear system using MATLAB or equivalent Software.

15. Study the effect of P, PI, PID controllers using MATLAB or equivalent Software.

16. Design of Lead and Lag compensator.

TOTAL: 45 PERIODS

INDEX

SL.NO DATE TITLE MARKS SIGN

CYCLE I

1

2

3

4

5

6

7

CYCLE II

1

2

3

4

5

6

7

3

4

LEARNING OBJECTIVES

1. To expose the students to the basic operations of electrical machines and help them

to develop experimental skills.

2. To study the concepts, performance characteristics, time and frequency response of

linear systems.

3. To study the effects of controllers.

Course Outcome

1. Evaluate the electrical characteristics of electric machines.

2. Design the electronic instruments to perform control system analysis and measurement.

3. Design some electrical and mechanical systems.

4. Compare root locus method and frequency domain design methods.

5. Analyze the characteristics of DC and AC machines.

6. Analyze the standard methods to determine accurate modeling/simulation parameters for

various general-purpose electrical machines and transformers.

PRECAUTIONS:

1. Remove the fuse carriers before wiring and start wiring as per the circuit diagram.

2. Check the positions of the various rheostats as specified.

3. The switch is kept open at the time of starting the experiments.

4. Selection of fuse wires

* 120% of rated current.

5. Replace the fuse carriers with appropriate fuse wires after the circuit connections are

checked by the staff-in-charge.

5

Exp. No. :

Date :

SPEED CONTROL OF DC SHUNT MOTOR

AIM:

To control the speed of the given DC shunt motor by,

1. Armature control method

2. Field control method

APPARATUS REQUIRED:

S.NO APPARATUS TYPE RANGE QUANTITY

1. Voltmeter mc (0-300) V 1

2. Ammeter mc (0-2) A 1

3. Rheostat - 500/.2A 1

4. Rheostat - 50/3.5A 1

5. Tachometer Analog - 1

6. Connecting wires - - few

THEORY: An electric motor is a machine, which converts electrical energy in to mechanical energy.

Its action is based on the principle that when a current carrying conductor is placed in a magnetic

field, it experiences a mechanical force whose direction is given by Fleming’s left hand rule and

whose magnitude is given by F=BIL Newton.

Speed of a motor is given by,

N = V-IaRa (A) =k (V-IaRa/ )

z p Hence, speed can be controlled by varying

i) Flux/pole (field control)

ii) Resistance Ra of armature circuit (Rheostat control)

iii) Applied voltage V (Voltage control)

SPEED CONTROL OF SHUNT MOTOR:

Speed of DC shunt motor can be controlled by using the following methods,

i) Armature control method:

This method is used when speeds below the no load speed are required. In this case, the

voltage across the armature is varied in order to control the speed of DC shunt motor.

ii) Field control method:

In this method, the speed above rated speed can be obtained. In this case, the flux of a DC

motor can be changed by changing Ish in order to control the speed of a DC shunt motor.

6

CIRCUIT DIAGRAM:

15A

NAME PLATE DETAILS:

Volts 230V

Amps 19A

Speed 1500

Fuse calculations No Load : 25% of rated Current 25 X 19 = 4.75 A ~ 5A 100 Full Load 125% of full load 125 X 19 = 23.75A ~ 25A 100

7

PRECAUTIONS:

1. Ensure that all the connections are tight.

2. Select the meters of proper range.

3. Ensures the rating of fuses.

4. The starter arm must be in OFF position at the time of startin

PROCEDURE:

1. Connections are given as per the circuit diagram.

2. Switch on the 230V DC supply by closing the MCB.

3. Start the motor, using three-point starter.

Armature control method (Below rated speed)

1. By keeping the field current constant, the armature rheostat is adjusted in steps and note

down the corresponding armature voltage, and armature current.

2. The readings are tabulated and a graph of speed Vs armature voltage is drawn.

3. The same procedure may be repeated for some other field current.

Field control method (Above the rated speed)

1. By keeping the armature voltage as constant, the field rheostat is adjusted in steps and

note down the corresponding field current and speed.

2. The readings are tabulated and a graph of speed Vs field current is drawn.

3. The same procedure may be repeated for some other armature voltage.

8

TABULATION:

ARMATURE CONTROL METHOD:

S. No If1 = If2 =

Va (v) N (rpm) Va (v) N (rpm)

FIELD CONTROL METHOD:

S. No Va1 = Va2 =

If (A) N (rpm) If (A) N (rpm)

9

MODEL GRAPH:

DISCUSSION QUESTIONS:

1. What are the factors affecting the DC motor speed?

2. Define Base speed

3. What are the methods available to control the Dc motor speed?

4. What is the function of commutator?

5. What is the different types DC motor?

6. Mention the use of Interpole.

7. What are the advantages of armature controlled method over field control method?

8. Define Back emf.

9. What type of starter used in DC shunt motor?

10. Mention the use of Equalizer ring.

RESULT:

10

Exp. No. :

Date :

LOAD TEST ON DC SHUNT MOTOR

AIM:

To conduct the load test on DC shunt motor and to draw the following characteristics

curves,

1. Torque Vs armature current

2. Speed Vs armature current

3. Speed Vs Torque

4. %Efficiency Vs Power output

APPARATUS REQUIRED:


1. Voltmeter mc (0-300) V 1

2. Ammeter mc (0-20) A 1

3. Rheostat - 250/1.2A 1

4. Tachometer Analog - 1

5. Connecting wires - - Few

FORMULAE TO BE USED:

1. S=S1 – S2 in Kg.

S =Spring balance (Weight in Kg)

S1, S2 = Spring balance readings in Kgs.

2. T=S*R*9.81 N mt

S=Spring force in Kg.

R=Radius of the brake drum in meter.

T=Torque in N mt.

3. Pin =Vin * Iin in watts.

Vin = Voltage input in volts.

Iin =Input current in Amps.

Pin =Input power in watts.

4. Pout = (2NT) /60 in watts.

Pout = output power in watts.

N= Speed in rpm.

T=Torque in Nm.

5. Percentage Efficiency = Pout/Pin *100

11

CIRCUIT DIAGRAM

MACHINE SPECIFICATIONS:

Motor

KW / HP

Volts

Amps

Speed

THEORY:

An electric motor is a machine, which converts electrical energy in to mechanical energy.

Its action is based on the principle that when a current carrying conductor is placed in a magnetic

field, it experiences a mechanical force whose direction is given by Fleming’s left hand rule and

whose magnitude is given by F=BIL Newton.

DC SHUNT MOTOR:

In DC shunt motor, the field winding is connected in parallel with the armature. The field

winding has a large number of turns and smaller cross sectional area. Since, the field current is

small, the field power loss is also small.

12

CHARACTERISTICS OF DC SHUNT MOTOR:

The performance of Dc shunt motor can be judged from its characteristics curves known

as motor characteristics. By conducting the load test on DC shunt motor, the following

characteristics are obtained,

1. Torque Vs armature current characteristics

2. Speed Vs armature current characteristics

3. Speed Vs Torque characteristics

PRECAUTIONS:


2. Select the meters of proper range.

3. Ensures the rating of fuses.

4. The starter arm must be in OFF position at the time of starting.

5. The motor should be loaded up to rated current of the machine.

PROCEDURE:


2. By closing the MCB, 230v DC supply is given to the circuit.

3. The motor is started by using three point starter.

4. Adjust the field rheostat of the motor so as to make the motor to run at rated speed.

5. Vary the load insteps and at each step note down the input voltage, input current,

spring balance readings till the rated current.

6. The readings are tabulated and a graph of,

Torque Vs armature current characteristics

Speed Vs armature current characteristics

Speed Vs Torque characteristics

TABULATION: Circumference of the Brake drum =

Radius of the brake drum =

S.

No

Input

voltage

Input

current

Spring balance

Reading

Speed

N

Torque

T

Input

power

Pin

Output

power

Pout

Efficiency

S1 S2 S1 – S2

volts Amps Kg Kg Kg Rpm Nm Watts Watts %

MODEL GRAPH:

12

13


1. What are the different types of DC motor?

2. What are the different types of starters used for DC motors?

3. Why starter is necessary in DC motors?

4. List out the parts in 3 Point starter.

5. Can you use 4 point starter in DC shunt motor?

6. Define commutation.

7. How to minimize the commutation effect in DC machine?

8. Compare armature and wave winding in DC machine.

9. Mention the two advantages of Yoke.

10. Why shunt motor is also called as constant speed motor ?

RESULT:

14

Exp. No. :

Date :

OPEN CIRCUIT AND LOAD CHARACTERISTICS OF DC

SHUNT GENERATOR

AIM:

a) To obtain the open circuit characteristics of a DC Shunt Generator at rated speed and

to determine (i) Critical field resistance at rated speed (ii)Critical Speed (iii) Voltage

built up by the generator at rated speed and (iv) to plot the OCC characteristic curves

b) To conduct the Load test on the given DC Shunt Generator and plot its internal and

external characteristics.

APPARATUS REQUIRED:


1 Voltmeter mc (0-300) V 1

2 Ammeter mc (0-20) A 1


4 Rheostat - 500 Ω 2A 1

5 Rheostat - 1000Ω 0.8 1

6 Tachometer Analog - 1

7 Variable Rheostat load - - 1

8 Connecting wires - - Few

FORMULAE TO BE USED: 1. Critical field resistance ,Rc = Slope of the tangent drawn to the linear portion of the

OCC and passing through origin.

Rc = …………………. Ohms.

2. Critical Speed Nc = Rsh * Nrated / R

3. Voltage build up Egm =Voltage corresponding to the point at which the OCC at rated

Speed and Rsh line intersect.

Egm =……………….. volts.

4. Eg = V + IaRa in volts. Where,

Eg = Generated voltage in volts.

V = Terminal voltage in volts.

Ia = Armature current in Amps.

Ra = Armature resistance in ohms.

5. Armature current Ia = IL+ Ish in Amps. Where,

IL = Load current in Amps.

Ish = Shunt field current in Amps.

15

CIRCUIT DIAGRAM

NAME PLATE DETAILS:

Volts 230V

Amps 19A

Speed 1500


16

THEORY:

An electric generator is a machine, which converts mechanical energy in to

electrical energy. The energy conversion is based on the principle of the

production of dynamically induced emf, whenever a conductor cuts the magnetic flux;

dynamically induced emf is produced in it according to Faraday’s laws of

electromagnetic induction. This emf causes a current to flow if the conductor circuit is

closed.

Induced emf direction can be found by Flemings Right hand rule.

Self excited Generator:

Separately -excited generators are those whose field magnets are energized

from the generator itself.

Characteristics:

The following characteristics are drawn by conducting the open circuit and load test on

DC separately excited generator

i) No load saturation characteristics (E Vs If)

ii) Internal characteristics (E Vs Ia) iii) External characteristics (V Vs Il)

TABULATION:

OPEN CIRCUIT TEST:

S. No Field current

(Amps)

Output voltage at no load

(Volts)

17

LOAD TEST:

S. No Load current Load voltage Field current Ia= IL + IF Eg=Va+IaRa

Amps volts Amps Amps Volts

18

PRECAUTIONS:

1. All the connections should be tight

2. Ensure both SPST and DPST switches are in open position.

3. Check whether the load is under off position.

4. Keep the motor field rheostat at minimum resistance position during starting.

5. Keep the motor armature rheostat at maximum resistance position at starting.

6. Keep the generator field rheostat at maximum resistance position during starting.

PROCEDURE:

OPEN CIRCUIT TEST:

1. The connections are given as per the circuit diagram.

2. Verify whether field rheostat of the generator is kept at maximum position and field

rheostat of motor at minimum position.

3. Switch ON the DC supply.

4. Adjust the excitation of field rheostat of the motor so as to make the motor to run at rated

speed.

5. The ammeter and voltmeter readings of the generator are noted with SPSTS switch in

opened position.

6. Close the SPSTS switch and excitation of the generator is varied insteps by adjusting the

field rheostat of the generator, at each step the readings of field current and induced emf

are noted upto its rated generator voltage.

7. The readings are tabulated and a graph of open circuit characteristics is drawn between

generated voltage Vs field current.

LOAD TEST:

1. Fix the armature voltage of the generator to the rated voltage by adjusting the field

rheostat of the generator.

2. Close the DPSTS at the load side of the generator and increase the load in steps till the

rated armature current and at each step the readings of terminal voltage, load current, and

shunt field current are noted.

3. Finally reduce the load insteps and bring the generator and field rheostat to its original

position.

19

TO FIND Ra

Circuit Diagram

NAME PLATE DETAILS:

Volts 230V

Amps 19A

Speed 1500

Fuse calculations Full Load 125% of full load 125 X 19 = 23.75A ~ 25A 100

20

Tabulation

S. No Current (Ia) Voltage (Va) Ra=Va/Ia

Amps volts Ohms

Avg Ra = Ω

TO FIND Ra:


2. Gradually vary the loading rheostat insteps and at each step note down the corresponding

voltmeter and ammeter readings.

3. From the tabulated value calculate the average armature resistance.

21

MODEL GRAPH


1. What are the reasons for the decrease in terminal voltage when the shunt generator is

loaded?

2. What is the difference between internal and external characteristics of DC shunt

generator?

3. What is the purpose of providing interpole in series with the armature of DC shunt

generator?

4. What are the different types of DC generator?

5. Define fuse and how will you choose the fuse rating of the machine?

6. Why OCC is often called magnetization characteristics?

7. What are the different types of excitations available for DC generator?

8. What do you mean by critical speed of a DC generator?

9. Define Armature Reaction.

10. List the applications of DC shunt generator?

RESULT :

22

Exp. No. :

Date :

TRANSFER FUNCTION OF SEPERATELY EXCITED DC GENERATOR

AIM:

To determine the transfer function of separately excited dc generator.

APPARATUS REQUIRED:


1 Voltmeter mc (0-300) V 1



4 Rheostat - 500 Ω 2A 1

5 Rheostat - 1000Ω 0.8 1

6 Tachometer Analog - 1

7 Variable Rheostat load - - 1

8 Connecting wires - - Few

23

CIRCUIT DIAGRAM:

NAME PLATE DETAILS:

Volts 230V

Amps 19A

Speed 1500

24


25

THEORY:

The transfer function for DC generator is defined as ratio of Laplace Transform

Of output V1(t) to Laplace Transform of Input Vf(t).

Transfer function = V1(t) / Vf(t)

The KVL to field circuit is

Vf(t) = If(t)Rf + Lf (dIf(t) /dt) ………………… 1

The armature induced emf Ea(t) is

Ea(t) If(t)

= Kf If(t) ………………… 2

where Kf is proportionality constant

The KVL to armature circuit is given by

Ea(t) = Ia(t) (Ra +RL) +La (dIa /dt) ………………… 3

The load voltage

V1(t) = Ia(t) RL ………………… 4

Ia(t) = IL(t)

Taking Laplace transform for equations 1,2,3 & 4 we get

Vf(s) = If(s) Rf + sLf If(s) ………………… 5

Ea(s) = Kf If(s) ………………… 6

Ea(s) = Ia(s) (Ra +RL) + sLa Ia(s) ………………… 7

V1(s) = Ia(s) RL ………………… 8

From the above equations we get

Ea(s) = Kf Vf(s) / (Rf + sLf ) ………………… 9

From 7 & 9

Kf Vf(s) / (Rf + sLf ) = Ia(s) (Ra +RL) + sLa Ia(s) ………………… 10

From equation 10

Ia(s) = Kf Vf(s) / (Rf + sLf ) [(Ra +RL) + sLa ] ………………… 11

V1(s) = Kf Vf(s) R1 / (Rf + sLf ) [(Ra +RL) + sLa ]

V1(s) / Vf(s) = Kf R1 / (Rf + sLf ) [(Ra +RL) + sLa ]

G(s) = V1(s) / Vf(s) = Kf R1 / (Rf + sLf ) [(Ra +RL) + sLa ………………… 12

G(s) = V1(s) / Vf(s) = (Kf RL / Lf La ) / (s + Rf / Lf) (s + (Ra +RL) /La )

Where Lf / Rf = field time constant, La / Ra = armature time constant

La / (Ra +RL) = total time constant

The above equation is known as the load transfer function of separately excited D.C generator.

PRECAUTIONS: (Not to be included in the Record)

1. Remove the fuse carriers before wiring and Start wiring as per the circuit diagram.

2. Check the positions of the various rheostats as specified.

3. The SPST switch is kept open at the time of starting the experiments.

4. Fuse calculations: As this is a no-load test the required fuse ratings are 20% of the rated

current.

5. Replace the fuse carriers with appropriate fuse wires after the circuit connections are checked

by the staff-in-charge.

PROCEDURE: (To Find K)

1. The circuit connections are made as per the circuit diagram in the shown figure 4.1.

2. Keeping the motor field rheostat in its minimum position, generator field rheostat in

maximum position and the starter in its OFF position, the main supply is switched ON to the

circuit.

3. The motor is started using the 3-point starter by slowly and carefully moving the starter

handle from its OFF to ON position.

4. The motor is brought to its rated speed by adjusting its rheostat and checked with the help of

a tachometer.

5. With the SPST switch open, the residual voltage is noted.

6. Now the SPST switch is closed and the generator field rheostat is gradually decreased in

steps and at each step the field current (If) and the corresponding induced EMF (Eg) are

recorded in the tabular column. This procedure is continued until the generator voltage

reaches 120% of its rated value.

7. After the experiment is completed the various rheostats are brought back to their original

position in sequence and then main supply is switched OFF.

22

23

CIRCUIT DIAGRAM:

TO DETERMINE RF AND LF OF A GENERATOR

NAME PLATE DETAILS:

Volts 230V

Amps 19A

Speed 1500 rpm

24


Procedure: (To Find Ra & La)

1. The circuit connections are made as per the circuit diagram shown in figure 4.2.

2. Keeping autotransformer in minimum position, Main is switched ON.

3. Slowly adjust the variac and apply a small voltage (say 20V) to the armature winding.

4. Note down voltmeter, ammeter and wattmeter readings.

5. Bring the variac to minimum position and switch OFF the main supply.

Procedure: (To Find Rf & Lf)

1. The circuit connections are made as per the circuit diagram shown in figure 4.3.

2. Keeping autotransformer in minimum position, Main is switched ON.

3. Slowly adjust the variac and apply a small voltage (say 60V) to the field winding.

4. Note down voltmeter, ammeter and wattmeter readings.

5. Bring the variac to minimum position and switch OFF the main supply

25

f

( )

f f

=

, Ra

( (

& La in

) = )

standar

/

d equations

+

+

( +

)

( )

= /

F

f f

f

1. The open circuit characteristic is drawn to scale as shown in model graph.

2. A tangent is drawn to the linear portion of this OCC.

3. The slope of tangent is found using the relation Kf = Ea / If

4. The inductance and resistance of the field winding are calculated as follows:

W1=I 2R

Rf =W1 / I 2

Zf=Vf / If

2

2 1/2

Xf = (Zf – Rf )

Lf = Xf / 2пf

5. The inductance of the armature winding is calculated using the equation

W2=Ia2Ra

Ra =W2 / Ia2

Za=Va / Ia

Xa = (Za2

– Ra2)1/2

La = Xa / 2пf

6. The transfer function of the given separately excited D.C shunt generator is then evaluated by

substituting the values K , R , L the

or no Load

TABULAR COLUMN :

Tabulation 1: OCC test

S.NO If (A) Eg (V)

26

Tabulation 2: To find Rf and Lf

Sl.No. W1 If Vf

Tabulation 3: To find Ra and La

Sl.No. W2 Ia Va

Model Graph

MODEL CALCULATION:

RESULT:

Thus the Transfer Function of Separately Excited D.C Shunt Generator is determined &

is given by

G(s)NL =

G(s)L =

27

Expt No:

Date:

REGULATION OF THREE PHASE ALTERNATOR BY EMF AND MMF METHODS

AIM:

To predetermine the regulation of three phase alternator by EMF and MMF method and

also to draw the vector diagrams.

Name plate details:

3Alternator DC Shunt motor

415v 5A 1500 rpm 5kva 220V 28A 1500 rpm

Fuse rating:

125 % of current (Full load current)

For dc shunt motor.

For alternator

28

Apparatus required:

s. no Name of the apparatus Type Range Quantity

1 Voltmeter MI (0-600)V 1

2 Ammeter MI (0-5)A 1

3 Ammeter MC (0-2)A 1

4 Rheostat - 500Ω 2A 1

5 Rheostat - 250Ω1.5A 1

6 Connecting wire - - few

Formulae used:

Emf method: Armature resistance Ra = 1.6 Rdc where - Rdc is the resistance in DC supply.

Synchronous impedance Zs = Open circuit voltage (E1 (ph))/short circuit current (Isc)

Synchronous impedance Xs = (Zs2-Ra

2)

Open circuit voltage Eo = ((Vrated cos + Ia Ra) 2

+ (Vrated sin +IaXs)2)(For lagging power

factor)

Open circuit voltage Eo = ((Vrated cos + Ia Ra)

2+(Vrated sin - IaXs)

2) (For leading power

factor)

Open circuit voltage Eo = ((Vrated cos + Ia Ra)2+( IaXs)

2) (For unity power factor)

Percentage regulation = (Eo-Vrated /Vrated)*100(For both EMF and MMF methods)

Precaution:

i. The motor field rheostat should be kept in the minimum resistance position.

ii. The alternator field potential divider should be in the maximum voltage position. iii.

Initially all switches are in open position.

Procedure for both emf and MMF method:

1. Note down the nameplate details of motor and alternator.

2. Connections are made as per the circuit diagram.

3. Give the supply by closing the dust switch.

4. Using the three point starter, start the motor to run at the synchronous speed by varying the

motor filed rheostat.

5. Conduct an open circuit test by varying the potential divider for various values of field

current and tabulate the corresponding open circuit voltage readings.

6. Conduct a short circuit test by closing the TPST switch and adjust the potential divider to set

the rated armature current, tabulate the corresponding field current.

7. Conduct a stator resistance test by giving connection as per the circuit diagram and

tabulate the voltage and current readings for various resistive loads.

29

Procedure to draw the graph for EMF method:

1. Draw the open circuit characteristics curve (generator voltage per phase Vs field current)

2. Draw the short circuit characteristics curve (short circuit current Vs field current)

3. From the graph find the open circuit voltage per phase (E1 (ph)) for the rated short circuit

current (Isc).

4. By using respective formulae find the Zs, Xs, Eo and percentage regulation.

Open circuit test:

S.NO Field current(If) Open circuit line

voltage (VOL)

Open circuit phase

voltage(Vo(ph))=

VOL /

Amps Volts Volts

Circuit Diagram

30

Short circuit test:

S.No

Field current(If)

Short Circuit Current

(120 to 150 % of rated

current ) (Isc)

Amps Amps

Tabulation to find out the armature resistance (ra):

S.No Armature current

(I) Armature voltage

(V) Armature Resistance

Ra=V/I

Amps Volts Ohms

Procedure to draw the graph for MMF method:

1. Draw the open circuit characteristics curve (generator voltage per phase Vs field current

2, Draw the short circuit characteristics curve (short circuit current Vs field current)

3. Draw the line OL to represent If’ which gives the rated generated voltage (V).

4. Draw the line LA at an angle (90Φ) to represent If” which gives the rated full load

current.(Isc) on short circuit [(90Φ) for lagging power factor and (90- Φ) for leading

power factor].

5. Join the points O and A and find the field current (If) measuring the distance OA that

gives the open circuit voltage (E0) from the open circuit characteristics.

6. Find the percentage regulation by using suitable formula.

Procedure to draw the vector diagram:

1. Draw the line OA that represents the rated voltage V.

2. Draw the line OB to represent the rated current Ia, which makes an angle Φ (it may

lags/leads in phase) with the voltage.

3. Draw the line AC to represent IRa drop, which is parallel to current axis (OB)

4. Draw the perpendicular line CD with the line AC (IRa drop) to represent IXs drop.

5. Join the points D and A to represent the IZs drop.

6. Join the points O and D and measure the length OD by voltage scale to find open circuit

voltage Eo.

7. Find the percentage regulation by using suitable using formulae.

31

RESULTANT TABULATION FOR REGULATION OF THREE PHASE ALTERNATOR BY EMF AND

MMF METHODS

S.N

o

Power

factor

EMF method MMF method

Lagging Leading unity Lagging Leading unity

1.

2.

3.

4.

5.

6.

7.

MODEL CALCULATION:

RESULT:

32

Expt No:

Date :

OPEN CIRCUIT AND SHORT CIRCUIT TEST ON A

SINGLE PHASE TRANSFORMER

AIM:

i) To calculate the equivalent circuit parameter by conducting open circuit and short

circuit test on single-phase transformer.

ii) To predetermine the efficiency and regulation of the given transformer from

equivalent circuit parameter.

APPARATUS REQUIRED:

S.NO ITEM TYPE RANGE QUANTITY

1. Voltmeter MI (0-150)V 1

2. Voltmeter M (0-300)V 1

3. Ammeter MI (0-2)A 1

4. Ammeter MI (0-5)A 1

5. Wattmeter UPF 300V/5A 1

6. Wattmeter LPF 75V/5A 1

7. Auto transformer - 2.7KVA,10A,270V 1



Open circuit test:

1. W0 = I0 V0 cos 0 watts. Where,

cos 0 = (W0 / I0 V0 ) W0 = Real power consumed

V0 = Primary no load voltage in volts

I0

0

=

=

No load primary current

Phase angle at no load

2. Magnetizing current Im = I0 sin0

Loss compensating current Iw = I0 cos 0

3. No load reactance X0 = V0/ Im

4. No load resistance R0 = Vo/Iw

Short Circuit Test:

1. Z01=Vsc/Isc , R01=Wsc / Isc2

2 2

X01=Z01 -R01

2. R02=k*R01 ; X02=k

2X01; Z02=k

2Z01

Where,

Vsc =Short-circuiting voltage in volts.

33

CIRCUIT DIAGRAM

O.C. TEST:

S.C. TEST:

34

S. No.

V

sc

I

s

c

W

sc

Volts

Amps Watts

TABULATION:

NO LOAD TEST:

S. No.

Open circuit

primary

Voltage (V0)

No load current

(I0)

No load power

(W0)

volts Amps Watts

SHORT CIRCUIT TEST:

35

Efficiency: 1. Efficiency = Output power/input power

= Output power/(Output power + total losses)

2. Output power =(x *kva* cos) 3. Total loss = copper loss + iron loss

X = Assumed fraction of load

KVA = Rating of the transformer

Cos = Assumed power factor

Wsc = Short circuit wattmeter reading (cu loss)

W0 = Open circuit wattmeter reading (Fe loss)

Regulation: % Regulation = xIsc(R02 cos + X02sin)/V2

Where,

V20

= Open circuit voltage at secondary + = for lagging load

- = for leading load

THEORY:

A Transformer is a static (or stationary) piece of apparatus by means of which electric

power in one circuit is transformed in to electric power of the same frequency in another circuit.

Principle of operation involved in a transformer is mutual induction principle.

OC and SC TEST:

OC and SC tests are carried out in order to determine the equivalent circuit constants

and efficiency of a transformer. These tests are very economical and convenient, because they

furnish the required information without actually loading the transformer.

OC Test: The purpose of this test is to determine the load or core loss and no load current I0 which is

helpful to find X0 and R0.

SC Test:

This is an economical test for determining equivalent impedance (Z01 or Z02), leakage

reactance (X01 or X02) and total resistance (R01 or R02), cu loss at full load.

MODEL GRAPH:

34

PRECAUTIONS: 1. Ensure that all the connections are tight.

2. Secondary side of the transformer should be loaded upto rated current of the machine.

3. Calculate the multiplication factors for wattmeters.

4. Calculate the full load current of the transformer.

5. The transformer should not be loaded beyond 125% of the rated full load current.

PROCEDURE:

Open circuit test: 1. Connections are given as per the circuit diagram.

2. Keeping the autotransformer at its zero output position ,Switch ON the

3. A.C.supply.

4. 3.Adjust the auto transformer output to get the rated voltage across the LV

5. winding of the transformer.

6. Note down all the meter readings.

7. 5.Bring the autotransformer to zero output position and switch OFF the supply.

Short circuit test:


2. Keeping the autotransformer at its zero output position, Switch ON the AC supply.

3. Gradually increase the output voltage of autotransformer till the rated full load current

flow through the transformer windings.

4. Note down all the meter readings.

5. Bring the autotransformer to zero output position and switch OFF the supply.

35

TABULATION FOR PERCENTAGE EFFICIENCY:

S.

No.

Iron

loss

Cu

loss

Load

x

Total loss

=

W0+x2Wsc

Output power

Input power %

Efficiency

UPF 0.8PF UPF 0.8PF UPF 0.8PF

Watts Watts % Watts Watts Watts Watts Watts % %

TABULATION FOR PERCENTAGE REGULATION:

S.

No.

Power

Factor

cos

125% of load

x =1.25

100% of load

x = 1.00

50% of load

x = 0.5

25% of load

x = 0.25

Lagging Leading Lagging Leading Lagging Leading Lagging Leading

DISCUSSION QUESTIONS

1. Why is the open circuit test on a transformer usually carried out on the low voltage side

and short circuit test performed on the high voltage side?

2. Why are iron losses constant at all loads in a transformer?

3. Draw the equivalent circuit of single phase transformer.

4. Compare transformer and auto transformer.

5. Define all day efficiency

6. List the applications of Step up transformer.

7. Why transformer ratings are expressed in KVA?

8. What are the losses occur in transformer?

9. Why transformer efficiency is high compare to Dynamic machine?

10. Distinguish between core type and shell type transformer

RESULT:

36

Expt No:

Date :

LOAD TEST ON SINGLE PHASE TRANSFORMER

AIM:

curves.

To conduct the load test on single-phase transformer and to draw the characteristics

APPARATUS REQUIRED:

S.No. ITEM RANGE TYPE QUANTITY

1. Voltmeter (0-300) V MI 1

2. Voltmeter (0-150) V MI 1

3. Ammeter (0-5) A MI 1

4. Ammeter (0-20)A MI 1

5. Wattmeter 300V/5A UPF 2

6. Transformer 1KVA - 1



1. Percentage efficiency = (Output power/Input power) *100

2. Percentage Regulation =(V02 –V2)/ V02 * 100

Where,

V02 =Secondary voltage on no load.

V2 = Secondary voltage on load.

THEORY:

A Transformer is a static (or stationary) piece of apparatus by means of which electric

power in one circuit is transformed in to electric power of the same frequency in another circuit.

Principle of operation involved in a transformer is mutual induction principle.

When the load is connected to the secondary of the transformer, the load current flow

through the secondary and the load then the transformer is said to be loaded. By conducting the

load test on transformer the performance of the transformer is determined by drawn the

following curves,

1. Output power Vs % Efficiency

2. Output power Vs % Regulation

37

CIRCUIT DIAGRAM:

NAME PLATE DETAILS:

Fuse calculations Full Load 125% of full load 125 X 9 = 12A 100

Specifiicion Primary secondary

Rated volt 115V 230V

Rated current 9A 5A

Rated power 1kva 1kva

Loading rating 5 kva,230v

38

PRECAUTIONS:


2. Secondary side of the transformer should be loaded upto rated current of the machine.

3. Calculate the multiplication factors for wattmeters.

PROCEDURE:

1. Give the connections as per the circuit diagram.

2. By closing MCB, 230v single phase, 50Hz AC supply is given to the transformer.

3. Note the readings from all meters on no load.

4. Now load is applied to the transformer in step by step-up to the rated primary current

and the corresponding readings are noted.

5. From the recorded values calculate the percentage efficiency and percentage

regulation and draw the required graph.

39

TABULATION:

S. No

Primary

Voltage

Primary

current

Secondary

Voltage

Secondary

current

Input

power

Output

power

%

%

Regulation

Volts Amps Volts Amps Watts Watts % %

Model Graph:

40

DISCUSSION QUESTIONS

1. Why transformer is a static machine?

2. What is transformer?

3. Why transformer ratings are expressed in KVA?

4. What are the losses occur in transformer?

5. Why transformer efficiency is high compare to Dynamic machine?

6. Distinguish between core type and shell type transformer.

7. List the properties of ideal transformer.

8. What are the different types of transformer?

9. How to minimize the core lose in transformer?

10. Draw the equivalent circuit of single phase transformer.

RESULT:

41

Expt No:

Date :

LOAD TEST ON SINGLE PHASE INDUCTION MOTOR

AIM:

To determine the performance characteristics of Single-phase induction motor by direct loading

method.

APPARATUS REQUIRED:

Sl.no Apparatus Type Range Qty.

1 Voltmeter MI (0-300) V 1


3 Single element wattmeter UPF 300V,10A 1

4 1Ø Auto transformer 1Ø 230V/(0-270)V 1

5 Connecting wires

FORMULAE USED:

1. Torque,T=9.81*Reff*(S1-S2) N-m

Where,

Reff =Effective radius of the brake drum in m.

S1,S2=Spring balance readings in Kg.

2. Output power,Po = 2πNT/60 watts

Where,

N=speed of the motor in rpm

3. % Efficiency = (Output power/Input power)*100

PRECAUTIONS: 1. At the time of starting, the motor should be in the no load condition.

2. Initially DPST switch is in open position.

PROCEDURE:

1. The connections are made as per the circuit diagram.

2. The supply is given by closing the DPSTS.

3. The motor is started using DOL starter.

4. The voltmeter, ammeter, wattmeter, speed and spring balance readings are noted for no load.

5. The load is increased in steps and at each increase in step of the load the readings of voltmeter,

ammeter, wattmeter, speed and spring balance are noted.

42

NAME PLATE DETAILS:

Volts 415V

Amps 7.5A

Speed 1430

Fuse calculations Full Load 125% of full load 125 X 7.5 = 10A 100

43

TABULATION: MF-

Sl.

No. Line

Voltage VL

(volts)

Line

Current IL

(amps)

Input

Power PL

(watts)

Speed

N

Rpm

Spring balance

readings Torque

T

N-m

Output

Power

watts

η

% Slip

% PF

S1

Kg S2

Kg S1 - S2

in Kg

MODEL GRAPH

44


1. Why single phase induction motor is not self starting?

2. What is the use of auxiliary winding?

3. What are the different types of single phase induction motor?

4. Briefly explain the double field revolving theory.

5. What is meant by shaded pole induction motor?

6. What is the motor used in mixie?

RESULT:

45

Expt No:

Date :

LOAD TEST ON THREE-PHASE SQUIRREL CAGE

INDUCTION MOTOR

AIM:

To determine the performance characteristics of three-phase squirrel cage induction

motor by direct loading method.

APPARATUS REQUIRED:

Sl. No. Apparatus Type Range Quantity

1 Voltmeter MI (0-600) V 1


3 Single element wattmeter LPF 600V,10A 2

4 3Ø Auto transformer 400V/(0-450)V 1

5 Tachometer Analog 1

6 Connecting wires As per required

FORMULA USED:

1. Torque,T=9.81*Reff*(S1-S2) N-m

Where,

Reff =Effective radius of the brake drum in m.

S1, S2= Spring balance readings in Kg.

2. Output power,Po = 2∏NT/60 watts

Where,

N=speed of the motor in rpm

3. % Efficiency = (Output power/Input power)*100

4.Slip% = (Ns-N) /Ns X 100

5. cosø =W(31/2 NLIL)

PRECAUTION: 1. At the time of starting, the motor should be in the no load condition.

2. Initially switch is in open position and the autotransformer should be kept at minimum voltage

position.

CIRCUITDIAGRAM:

NAME PLATE DETAILS:

Volts 415V

Amps 7.5A

Speed 1430

Fuse calculations Full Load 125% of full load 125 X 7.5 = 10A 100

46

Starter connection

Star –delta starter

PROCEDURE:


2. The supply is given by closing the TPST switch.

3. The motor is started by using star delta starter.

4. The voltmeter, ammeter, wattmeter, speed and spring balance readings are noted for no load

condition.

5. The load is increased in steps at each increase in step of the load the readings of voltmeter,

ammeter, wattmeter, speed and spring balance are noted.

47

TABULATION:

MF=---

Sl.

No.

Line

Current

IL

in amps

Line

Voltage

VL

in

volts

Input

Power

PL

in

watts

Speed

N in

Rpm

Spring balance

readings

Torque

T

Ouput

Power

Effi.

Slip

PF

S1

in

Kg

S2

in

Kg

S1 -

S2

in Kg

N-m

watts

%

%

Electrical Characteristics:

Scale X axis 1 cm =150 Y axis 1 cm =10

WHEATSTONE BRIDGE

AIM:

To measure the given medium resistance using Wheatstone bridge.

APPARATUS REQUIRED:

S.No. Name of the Trainer Kit/

Components

Quantity

1. Wheatstone bridge trainer 1

2. Unknown Resistors specimen 5 different values

3. Connecting wires Few

4. DMM 1

5. CRO 1

THEORY:

Wheatstone bridge trainer consists of basic bridge circuit as screen printed on front panel with a

built in 1 kHz oscillator and an isolation transformer. The arm AC and AD consists of a 1K resistor.

Arms BD consists of variable resistor. The unknown resistor (Rx) whose value is to be determined is

connected across the terminal BC .The resistor R2 is varied suitably to obtain the bridge balance

condition. The DMM is used to determine the balanced output voltage of the bridge circuit.

For bridge balance,

For the galvanometer current to be zero the following conditions also exists

xx

RR

EII

11 and

E = EMF of the supply, combining the above equations we obtain

The unknown resistance. If three of the resistances are known, the fourth may be determined.

PROCEDURE:

1. Connect the unknown resistor in the arm marked Rx.

2. Connect the DMM across the terminal CD and switch on the trainer kit.

3. Vary R2 to obtain the bridge balance condition.

4. Find the value of the unknown resistance Rx using DMM after removing wires.

5. Compare the practical value with the theoretical value of unknown resistance Rx calculated using

the formula.

PANEL DIAGRAM

TABULATION:

Sl.No R1 (Ω) R2 (Ω ) R3 (Ω ) Rx(Ω )

(Actual)

Rx(Ω )

(Observed)

Percentage

Error

1

2

3

4

5

MODEL CALCULATION:

RESULT:

Review Questions

1. What are the applications of Wheatstone bridge?

2. What are standard arm and ratio arm in Wheatstone bridge?

3. What are the detectors used for DC Bridge?

4. What do you meant by sensitivity?

5. Why Wheatstone bridge cannot be used to measure low resistances?

CIRCUIT DIAGRAM

Ex. No:

Date:

2(a). MAXWELL’S BRIDGE

AIM:

To measure the unknown inductance and Q factor of a given coil.

APPARATUS REQUIRED :

S.No Name of the Trainer Kit/ Components Quantity

1. Maxwell’s inductance- capacitance bridge

trainer kit

1

2. Unknown inductance specimen 3 different values

3. Connecting wires Few

4. Head phone/ CRO 1

THEORY :

In this bridge, an inductance is measured by comparison with a standard variable capacitance.

The connection at the balanced condition is given in the circuit diagram.

Let L1 = Unknown Inductance.

R1 = effective resistance of Inductor L1.

R2, R3 and R4 = Known non-inductive resistances.

C4 = Variable standard Capacitor.

writing the equation for balance condition,

3244

411

1RR

RCj

RLjR

separating the real and imaginary terms, we have

Thus we have two variables R4 and C4 which appear in one of the two balance equations and hence the

two equations are independent. The expression for Q factor is given by

441

1 RCR

LQ

FORMULA USED:

Phasor Diagram

PANEL DIAGRAM

Procedure:


2. Connect the unknown inductance in the arm marked Lx .

3. Switch on the trainer kit.

4. Observe the sine wave at secondary of isolation transformer on CRO.

5. Vary R4 and C4 from minimum position in the clockwise direction to obtain the bridge balance

condition.

6. Connect the CRO between ground and the output point to check the bridge balance.

TABULATION:

Sl.

No.

R1

(Ω)

R3

(Ω)

C

(µF)

Lx

(mH)

Actual

Lx

(mH)

Observed

Quality

factor

Q

MODEL CALCULATION:

RESULT:

REVIEW QUESTIONS

1. What are the sources of errors in AC bridges?

2. List the various detectors used for AC Bridges.

3. Define Q factor of an inductor. Write the equations for inductor Q factor with RL series and parallel

equivalent circuits.

4. Why Maxwell's inductance bridge is suitable for medium Q coils?

5. State merits and limitations of Maxwell's bridge when used for measurement of unknown inductance.

DIGITAL SIMULATION OF LINEAR SYSTEMS

a)Digital Simulation Of First Order Systems

AIM:

To digitally simulate the time response characteristics of a first order system.

APPARATUS REQUIRED:

A PC with MATLAB package

THEORY:

STUDY OF BASIC MATLAB COMMANDS:

The name MATLAB stands for MATRIX LABORATORY. MATLAB was originally written

to provide easy access to matrix software developed by the LINPACK and EISPACK projects.

Today, MATLAB engines incorporate the LAPACK and BLAS libraries, embedding the state of

the art in software for matrix computation. It has evolved over a period of years with input from

many users. In university environments, it is the standard instructional tool for introductory and

advanced courses in MATHEMATICS, ENGINEERING, AND SCIENCE. In industry,

MATLAB is the tool of choice for high-productivity research, development, and analysis.

MATLAB is a high-performance language for technical computing. It integrates computation,

visualization, and programming in an easy-to-use environment where problems and solutions are

expressed in familiar mathematical notation. Typical uses include,

Math and computation

Algorithm development

Data acquisition Modeling, simulation, and prototyping

Data analysis, exploration, and visualization

Scientific and engineering graphics

Application development, including graphical user interface building

It is an interactive system whose basic data element is an array that does not require

dimensioning. This allows you to solve many technical computing problems, especially those

with matrix and vector formulations, in a fraction of the time it would take to write a program in

a scalar non-interactive language such as C or Fortran. It also features a family of add-on

application-specific solutions called toolboxes. Very important to most users of MATLAB,

toolboxes allow you to learn and apply specialized technology. Toolboxes are comprehensive

collections of MATLAB functions (M-files) that extend the MATLAB environment to solve

particular classes of problems. Areas in which toolboxes are available include SIGNAL

PROCESSING, CONTROL SYSTEMS, NEURAL NETWORKS, FUZZY LOGIC,

WAVELETS, SIMULATION, AND MANY OTHERS.

Some practical examples of first order systems are RL and RC circuits.

48

PROCEDURE:

To write a program in MATLAB

1.Start the MATLAB.

2.Type edit on MATLAB prompt and hit enter or follow File NewM file option from

main menu bar or on new icon in main toolbar.

3.Now an editor/ debugger window will open. This is where you write, edit, create and

save your own program in file called M-file.

4. Execute the program by clicking Debug Run option from main menu bar or on new

icon in main toolbar.

To build a simulink model in MATLAB To build a SIMULINK model to obtain step response / sine response of a first order

system, the following procedure is followed:

1. In MATLAB software open a new model in SIMULINK library browser.

2. From the continuous block in the library drag the transfer function block.

3. From the source block in the library drag the step input/ sine input.

4. From the sink block in the library drag the scope.

5. From the math operations block in the library drag the summing point.

6. Connect all to form a system and give unity feedback to the system.

7. For changing the parameters of the blocks connected double click the respective

block.

8. Start simulation and observe the results in scope. (Use a mux from the signal

routing block to view more than one graph in the scope)

BLOCK DIAGRAM:

Step response of a first order system:

49

Sine response of a first order system:

MATLAB (m-file) program to obtain the step response and sine response

MATLAB PROGRAM FOR STEP RESPONSE OF FIRST ORDER SYSTEM:

clear all;

t=0:0.2:15;

n=[0 1];

d1=[2.5 1];

c1=step(n,d1,t);

d2=[1 1];

c2=step(n,d2,t);

d3=[0.5 1];

c3=step(n,d3,t);

plot(t,c1,'g',t,c2,'r',t,c3,'b');

grid on;

title('step response of first order system');

xlabel('time in sec');

ylabel('c(t)');

gtext('t=2.5');

gtext('t=1');

gtext('t=0.5');

MATLAB PROGRAM FOR SINE RESPONSE OF FIRST ORDER SYSTEM:

clear all;

t=0:0.2:15;

u=sin(t);

n=[0 1];

d1=[2.5 1];

g1=tf(n,d1);

c1=lsim(g1,u,t);

50

d2=[1 1];

g2=tf(n,d2);

c2=lsim(g2,u,t);

d3=[0.2 1];

g3=tf(n,d3);

c3=lsim(g3,u,t);

plot(t,c1,'k',t,c2,'b',t,c3,'g');

grid on;

title('sine response of first order system');


ylabel('c(t)');

gtext('t=2.5');

gtext('t=1');

gtext('t=0.2');

Output graphs:

RESULT:

51

B)Digital Simulation Of Second Order Systems

Aim:

To digitally simulate the time response characteristics of second -order system

APPARATUS REQUIRED:

A PC with MATLAB package

THEORY:

STUDY OF BASIC MATLAB COMMANDS: The name MATLAB stands for MATRIX LABORATORY. MATLAB was originally written

to provide easy access to matrix software developed by the LINPACK and EISPACK projects.

Today, MATLAB engines incorporate the LAPACK and BLAS libraries, embedding the state of

the art in software for matrix computation. It has evolved over a period of years with input from

many users. In university environments, it is the standard instructional tool for introductory and

advanced courses in MATHEMATICS, ENGINEERING, AND SCIENCE. In industry,

MATLAB is the tool of choice for high-productivity research, development, and analysis.

MATLAB is a high-performance language for technical computing. It integrates computation,

visualization, and programming in an easy-to-use environment where problems and solutions are

expressed in familiar mathematical notation. Typical uses include,

Math and computation

Algorithm development

Data acquisition Modeling, simulation, and prototyping

Data analysis, exploration, and visualization

Scientific and engineering graphics

Application development, including graphical user interface building

It is an interactive system whose basic data element is an array that does not require

dimensioning. This allows you to solve many technical computing problems, especially those

with matrix and vector formulations, in a fraction of the time it would take to write a program in

a scalar non-interactive language such as C or Fortran. It also features a family of add-on

application-specific solutions called toolboxes. Very important to most users of MATLAB,

toolboxes allow you to learn and apply specialized technology. Toolboxes are comprehensive

collections of MATLAB functions (M-files) that extend the MATLAB environment to solve

particular classes of problems. Areas in which toolboxes are available include SIGNAL

PROCESSING, CONTROL SYSTEMS, NEURAL NETWORKS, FUZZY LOGIC,

WAVELETS, SIMULATION, AND MANY OTHERS.

PROCEDURE:

To write a program in MATLAB

1.Start the MATLAB.

2.Type edit on MATLAB prompt and hit enter or follow File NewM file option from

main menu bar or on new icon in main toolbar.

52

3.Now an editor/ debugger window will open. This is where you write, edit, create and

save your own program in file called M-file.

4. Execute the program by clicking Debug Run option from main menu bar or on new

icon in main toolbar.

To build a simulink model in MATLAB

1. Start MATLAB.

2. Start Simulink.

3. Open the libraries that contain the blocks you will need. These usually will include the

sources, sinks, math and continuous function block and possibly other.

4. Open a new simulink window.

5. Drag the needed blocks from the library folders to that new untitled simulink window.

You must give it a name using the Save As menu command under the File menu heading.

The assigned filename is automatically appended with an .mdl extension.

6. Arrange these blocks in orderly way corresponding to the equations to be solved.

7. Interconnect the blocks by dragging the cursor from the output of one block to the input

of another block.

8. Double click on any block having parameters that must be established and set these

parameters.

9. It is necessary to specify a stop time for the simulation, this is done by clicking on the

simulation parameters entry on the simulation > parameters entry on the simulation

toolbar.

10. Now we are ready to simulate our block diagram. Press start icon to start the simulation.

After simulation is done, double click the scope block to display the output. Click the

autoscale icon in the display window to scale the axis as per variable range.

53

PROGRAM:

MATLAB (m-file) program to obtain the step response and sine response

MATLAB PROGRAM FOR STEP RESPONSE OF SECOND ORDER SYSTEM:

clear all;

t=0:0.2:50;

k=20;

m=25;

b1=22.36;

b2=22.36;

w=[sqrt(k/m)];

x=[0 w*w];

z=(b1+b2)/(2*sqrt(k*m));

d1=[1 2*z*w w*w];

c1=step(x,d1,t);

b1=11.18; b2=11.18;

z=(b1+b2)/(2*sqrt(k*m));

d2=[1 2*z*w w*w];

c2=step(x,d2,t);

b1=0; b2=0;

z=(b1+b2)/(2*sqrt(k*m));

d3=[1 2*z*w w*w];

c3=step(x,d3,t);

b1=100; b2=150;

z=(b1+b2)/(2*sqrt(k*m));

d4=[1 2*z*w w*w];

c4=step(x,d4,t);

plot(t,c1,t,c2,t,c3,t,c4);

grid on;

title('time response of second order system for step input');


ylabel('c(t)');

gtext('underdamped z<1');

gtext('overdamped z>1');

gtext('critically damped z=1');

gtext('undamped z=0');

54

MATLAB PROGRAM FOR SINE RESPONSE OF SECOND ORDER SYSTEM:

clear all;

t=0:0.2:50;

k=20;

m=25;

b1=22.36;

b2=22.36;

u=sin(t);

w=[sqrt(k/m)]; x=[0

w*w];

z=(b1+b2)/(2*sqrt(k*m));

d1=[1 2*z*w w*w];

g1=tf(x,d1);

c1=lsim(g1,u,t);

b1=11.18;

b2=11.18;

z=(b1+b2)/(2*sqrt(k*m));

d2=[1 2*z*w w*w];

g2=tf(x,d2);

c2=lsim(g2,u,t);

b1=0; b2=0;

z=(b1+b2)/(2*sqrt(k*m));

d3=[1 2*z*w w*w];

g3=tf(x,d3);

c3=lsim(g3,u,t);

b1=100; b2=150;

z=(b1+b2)/(2*sqrt(k*m));

d4=[1 2*z*w w*w];

g4=tf(x,d4);

c4=lsim(g4,u,t);

plot(t,c1,t,c2,t,c3,t,c4);

grid on;

title('time response of second order system for sine input');


ylabel('c(t)');

gtext('underdamped z<1');

gtext('overdamped z>1');

gtext('critically damped z=1');

gtext('undamped z=0'); 55

Time response of second order system for step input

Crictical damped

system

1

s2+1.7888s+0.8

Underdamped system

Step

1

s2+0.8944s+.8

1

s2+0.8

Undamped system

Mux

Scope

1

s2+10s+0.8

Overdamped system

56

Time response of second order system for sinusoidal input

Crictical damped

system

1

s2+1.7888s+0.8

Underdamped system

Sine Wave

1

s2+0.8944s+.8

1

s2+0.8

Undamped system

Mux

Scope

1

s2+10s+0.8

Overdamped system

OUTPUT GRAPHS:

RESULT:

57

57

to ob

( )

tain

=

he bode p ( )

)(

lot for

)

t

( e given sys

te

t

m usin

ot for

( )

the

=

iven uni ( )

)(

ty feed

)

g

(

ba

For

=

magni

(

) = (

+ 2) 10)(

80 ) (

0( +

+ 40

2 ) 1600

) 1

4(1 +

(1

+

+

40 (1 +

)

)

10)

)(1 + 0

400(

) (1

5

)

Expt No.:

Date:

STABILITY ANALYSIS OF LINEAR SYSTEMS

(a) Stability analysis of linear systems using bode plot

AIM:

To write a program he given system

To access the stability of th g the plots obtained,

APPARATUS REQUIRED:

System with MATLAB software.

THEORY:

Let us draw the bode pl ck control system has

Step 1:

tude plot:

=

Step 2:

0. = ( + 0.025 .1

The corner frequencies are ωc1 =2 rad/sec, ωc2 =10 rad/sec, ωc3 =40 rad/sec.

58

4 )

0 5 )

)

)

= 2

= 4

0

2

= 2

0

= 0 d

= (

∗

) +

=

= -2 0

= −

0

og 5 + 0 = -13

− 13

.

= 6

0

− 37

Term Corner frequency Slope (dB) Change in

slope(db/dec)

( (1 + .

1

(1 + 0.1

1

(1 + 0.025

-

2

10

40

-40

20

-20

-20

-

-20

-40

-60

Step 3:

To find the points of magnitude plot choose a low frequency ω L such that ωL < ωc1,

choose a high frequency ωh such that ωh > ωc2. Let ωL =0.2 rad/sec and ωh = 50 rad/sec. ωL =0.2 rad/ ωc1 =2 rad/sec, ωc2 =10 rad/sec, ωc3 =40 rad/sec ωh = 50 rad/sec.

At ω= ωL =0.2

dB .

At ω= ωc1 =

0 B

At ω= ωc2 =10

Similarly

At ω= ωc3

=4

At ω= ωh

=5

Step 4:

0 l 9 dB

4 .9 = −37.9 dB

0 .9 = −43.7 dB

In a graph sheet mark the frequencies on X-axis and range of dB magnitude on Y-axis

after choosing proper scale.

Step 5:

Mark all the points obtained in step on the graph and join the points by its curve,Mark the

slope and every point of the graph

59

=

0

0

0

Tabulation:

ω A in dB

0.2

2

10

40

50

40

0

-13.9

-37.9

-43.7

For Phase plot:

−180 − tan .54 − tan .1 − tan .025

ω in deg

0.5

1

2

5

25

10

40

60

-170

-161

-139

-155

-195

-160

-294

-222

By Plotting the above points we get, Gain Margin= 25.5db and phase margin = 330

PROGRAM 1:

clear all;

clc;

Num=800*[1 2];

D1=[1 0 0];

D2=[0 1 10];

D3=[0 1 40];

Den=[conv(D1,conv(D2,D3))];

sys=tf(Num,Den);

Bode(sys);

margin(sys);

allmargin(sys);

grid on;

PROGRAM 2: clear all;

clc;

Num=800*[1-2];

D1=[1 0 0];

D2=[0 1 10];

60

D3=[0 1 40];

Den=[conv(D1,conv(D2,D3))];

sys=tf(Num,Den);

Bode(sys);

margin(sys);

allmargin(sys);

grid on;

PROCEDURE: 1. Solve the given problem manually and obtain the gain margin and phase margin for

the given system.

2. Type edit on MATLAB prompt and hit enter or follow File newM-file option

from the main menu bar or click an icon in main toolbar.

3. Now an editor /debugger window will open. This where you write , edit, create , run

and save the program for the given system to obtain the bode plot in file called M-

files.

4. Access the stability of the given system using the plot obtained by using the

following conditions

a) if both Kg and γ are positive the system is stable.

b) If any one of the two or both are negative the system is unstable.

c) If Kg and γ are zero the given system is limitedly stable.

OUTPUT GRAPH

to ob

( )

tain

=

he bode p ( )

)(

lot for

)

t

( e given sys

te

t

m usin

LOC

( )

US f

=

r the giv ( )

)(

en unit

)

(

y

locus

asym

of asy

on real a

totes:

mptote=6

xis

= ( )

0o

angle of asymptote=300

entroid:

=

) (

)

(b) Stability Analysis Of Linear Systems Using Root Locus

AIM:

To write a program he given system

To access the stability of th g the plots obtained,

APPARATUS REQUIRED:

System with MATLAB software.

THEORY:

Let us draw the ROOT o feedback control system has

1. To locate poles and zeros:

Poles = 0,0,-10,-40 (0 is double pole)

Zeros = -2

Number of poles n=3

Number of zeros m=1

Number of separate root loci =4

2. To find the root

3. To find angle of p

When q=0 angle

When q=1 angle of asymptote=180o

When q=2 o

4. To find C

( =

=-16

5. To find break away and break in point:

S=0 is the actual break away point

6. To find crossing point on imaginary axis:

The characteristic equation is given by S2(S+10)(S+40)+800(S+2)=0

Put S=jω

The characteristic equation becomes

(jω)2(jω+10)( jω+40)+800(jω+2)=0

61

Equate real and imaginary part equal to zero. We get ω=±4j which is crossing point

on imaginary axis

7. Verify manually obtained plot with the MATLAB software and access the stability of

given system using plots obtained.


clc;

Num=800*[1 2];

d1=[1 0 0];

d2=[0 1 10];

d3=[0 1 40];

den=[conv(d1,conv(d2,d3))];

G=tf(Num,den);

rlocus(G);

margin(G);

allmargin(G);

grid on;

title('root locus for the system g(s)=800(S+2)/S^2(S+10)(S+40’);


clc;

Num=800*[1-2];

d1=[1 0 0];

d2=[0 1 10];

d3=[0 1 40];

den=[conv(d1,conv(d2,d3))];

G=tf(Num,den);

rlocus(G);

margin(G);

allmargin(G);

grid on;

title('root locus for the system g(s)=800(S+2)/S^2(S+10)(S+40)');

62

63

OUTPUT GRAPH

RESULT:

64

Expt No:

Date :

AIM:

STUDY OF D.C MOTOR STARTERS

To study the different kinds of D.C motor starters

EQUIPMENT AND APPARATUS REQUIRED :

Sl No. Name of the apparatus Quantity

1 Two Point starter 1

2 Three Point starter 1

3 Four Point starter 1

THEORY :

The value of the armature current in a D.C shunt motor is given by Ia = ( V – Eb )/ Ra

Where V = applied voltage.

Ra = armature resistance.

E b = Back .e.m.f .

In practice the value of the armature resistance is of the order of 1 ohms and at the instant of

starting the value of the back e.m.f is zero volts. Therefore under starting conditions the value of

the armature current is very high. This high inrush current at the time of starting may damage the

motor. To protect the motor from such dangerous current the D.C motors are always started

using starters.

The types of D.C motor starters are

i) Two point starters

ii) Three point starters

iii) Four point starters.

The functions of the starters are

i) It protects the from dangerous high speed.

ii) It protects the motor from overloads.

i) TWO POINT STARTERS: ( refer fig 1)

It is used for starting D.C. series motors which has the problem of over speeding due to the loss

of load from its shaft. Here for starting the motor the control arm is moved in clock-wise

direction from its OFF position to the ON position against the spring tension. The control arm is

held in the ON position by the electromagnet E. The exciting coil of the hold-on electromagnet

E is connected in series with the armature circuit. If the motor loses its load, current decreases

and hence the strength of the electromagnet also decreases. The control arm returns to the OFF

position due to the spring tension, Thus preventing the motor from over speeding. The starter

also returns to the OFF position

when the supply voltage decreases appreciably. L and F are the two points of the starter which

are connected with the motor terminals.

66

ii) THREE POINT STARTER: ( refer fig 2 )

It is used for starting the shunt or compound motor. The coil of the hold on electromagnet

E is connected in series with the shunt field coil. In the case of disconnection in the field circuit

the control arm will return to its OFF position due to spring tension. This is necessary because

the shunt motor will over speed if it loses excitation. The starter also returns to the OFF position

in case of low voltage supply or complete failure of the supply. This protection is therefore is

called No Volt Release

( NVR).

Over load protection:

When the motor is over loaded it draws a heavy current. This heavy current also flows

through the exciting coil of the over load electromagnet ( OLR). The electromagnet then pulls

an iron piece upwar6.ds which short circuits the coils of the NVR coil. The hold on magnet gets

de-energized and therefore the starter arm returns to the OFF position, thus protecting the motor

against overload. L, A and F are the three terminals of the three point starter.

iii) FOUR POINT STARTER:

The connection diagram of the four point starter is shown in fig 3. In a four point starter

arm touches the starting resistance, the current from the supply is divided into three paths. One

through the starting resistance and the armature, one through the field circuit, and one through

the NVR coil. A protective resistance is connected in series with the NVR coil. Since in a four

point starter the NVR coil is independent of the of the field ckt connection , the d.c motor may

over speed if there is a break in the field circuit. A D.C motor can be stopped by opening the

main switch. The steps of the starting resistance are so designed that the armature current will

remain within the certain limits and will not change the torque developed by the motor to a great

extent.

67

Expt No:

Date :

STUDY OF INDUCTION MOTOR STARTERS

AUTO –TRANSFORMER STARTING

An auto transformer starter consists of an auto transformer and a switch as shown in the

fig. When the switch S is put on START position, a reduced voltage is applied across the motor

terminals. When the motor picks up speed, say to 80 per cent of its mornal speed, the switch is

put to RUN position. Then the auto-transformer is cut out of the circuit and full rated voltage

gets applied across the motor terminals.

(Ref. To text book for fig)

The circuit dia in the fig is for a manual auto-transformer starter. This can be made push button

operated automatic controlled starter so that the contacts switch over from start to run position as

the motor speed picks up to 80% of its speed. Over-load protection relay has not been shown in

the figure. The switch S is air-break type for small motors and oil break type for large motors.

Auto transformer may have more than one tapping to enable the user select any suitable starting

voltage depending upon the conditions.

Series resistors or reactors can be used to cause voltage drop in them and thereby allow low

voltage to be applied across the motor terminals at starting. These are cut out of the circuit as the

motor picks up speed.

STAR- DELTA METHOD OF STARTING:

The startor phase windings are first connected in star and full voltage is connected across its free

terminals. As the motor picks up speed, the windings are disconnected through a switch and they

are reconnected in delta across the supply terminals. The current drawn by the motor from the

lines is reduced to as compared to the current it would have drawn if connected in delta.The

motor windings, first in star and then in delta the line current drawn by the motor at starting is

reduced to one third as compared to starting current with the windings delta-connected.

In making connections for star-delta starting, care should be taken such that sequence of supply

connections to the winding terminals does not change while changing from star connection to

delta connection. Otherwise the motor will start rotating in the opposite direction, when

connections are changed from star to delta. Star-delta starters are available for manual operation

using push button control. An automatic star – delta starter used time delay relays(T.D.R)

through which star to delta connections take place automatically with some pre-fixed time delay.

The delay time of the T.D.R is fixed keeping in view the starting time of the motor.

(Ref. To text book for fig)

69

FULL VOLTAGE OR DIRECT –ON-LINE STARTING

When full voltage is connected across the stator terminals of an induction motor, large

current is drawn by the windings. This is because, at starting the induction motor

71

behaves as a short circuited transformer with its secondary, i.e. the rotor separated from the

primary, i.e. the stator by a small air-gap.

At starting when the rotor is at standstill, emf is induced in the rotor circuit exactly

similar to the emf induced in the secondary winding of a transformer. This induced emf of the

rotor will circulate a very large current through its windings. The primary will draw very large

current from the supply mains to balance the rotor ampere-turns. To limit the stator and rotor

currents at starting to a safe value, it may be necessary to reduce the stator supply voltage to a

low value. If induction motors are started direct-on-line such a heavy starting current of short

duration may not cause harm to the motor since the construction of induction motors are rugged.

Other motors and equipment connected to the supply lines will receive reduced voltage. In

industrial installations, however, if a number of large motors are started by this method, the

voltage drop will be very high and may be really objectionable for the other types of loads

connected to the system. The amount of voltage drop will not only be dependent on the size of

the motor but also on factors like the capacity of the power supply system, the size and length of

the line leading to the motors etc. Indian Electricity Rule restricts direct on line starting of 3

phase induction motors above 5 hp.

RESULT:

Thus the construction and working of different starters for starting D.C series, shunt,

compound and three phase induction motors are studied.

72

Expt. No:

Date:

Aim:

STUDY THE EFFECT OF P, PI, PID CONTROLLERS USING MAT LAB To Study the effect of P, PI, PID controllers using Mat lab.

1 Choice of the Controller type

In so far were described proportional, integrative and derivative modes of the controllers

and a rational behind their use was explained. However, excerpt for a few tips, an attention was

not given to a question when to use different types of controllers. The rest of this section will

give some answers on that particular topic.

1.1 On-off Controller

On-off controller is the simplest controller and it has some important advantages. It is

economical, simple to design and it does not require any parameter tuning. If oscillations will

hamper the operation of the system and if controller parameter tuning is to be avoided, on-off

controller is a good solution. In addition, if actuators work in only two modes (on and off), then

it is almost always only controller that can be used with such actuators. That is a reason why on-

off controllers are often used in home appliances (refrigerators, washers etc.) and in process

industry when control quality requirements are not high (temperature control in buildings etc.).

Additional advantage of on-off controllers is that they in general do not require any maintenance.

1.2 P Controller

When P controller is used, large gain is needed to improve steady state error. Stable

system do not have a problems when large gain is used. Such systems are systems with one

energy storage (1st order capacitive systems). If constant steady state error can be accepted with

such processes, than P controller can be used. Small steady state errors can be accepted if sensor

will give measured value with error or if importance of measured value is not too great anyway.

Example of such system is liquid level control in tanks when exact approximate level of liquid

suffice for the proper plant operation. Also, in cascade control sometime it is not important if

there is an error inside inner loop, so P controller can a good solution in such cases.

Derivative mode is not required if the process itself is fast or if the control system as

whole does not have to be fast in response. Processes of 1st order react immediately on the

reference signal change, so it is not necessary to predict error (introduce D mode) or compensate

for the steady state error (introduce I mode) if it is possible to achieve satisfactory steady state

error using only P controller.

1.3 PD controller

It is well known that thermal processes with good thermal insulation act almost as

integrators. Since insulation is good and thermal losses are small, the most significant art of the

energy that is led to the system is used temperature rise. Those processes allow 18 for large gains

so that integral mode in the controller is not needed. These processes can be described as

different connections of thermal energy storages. Thermal energy is shifted from one storage into

another. In general, with such processes there is present a process dynamics with large inertia.

73

Since dynamics is slow, derivative mode is required for control of such processes. Integral mode

would only already slow dynamics make more slowly. The other reason for using PPD

controllers in such systems is that is possible to measure temperature with low level of noise in

the measured signal.

PD controller is often used in control of moving objects such are flying and underwater

vehicles, ships, rockets etc. One of the reason is in stabilizing effect of PD controller on sudden

changes in heading variable y(t). Often a "rate gyro" for velocity measurement is used as sensor

of heading change of moving object.

1.3 PI controller

PI controllers are the most often type used today in industry. A control without D mode is

used when:

a) fast response of the system is not required

b) large disturbances and noise are present during operation of the process

c) there is only one energy storage in process (capacitive or inductive)

d) there are large transport delays in the system

If there are large transport delays present in the controlled process, error prediction is

required. However, D mode cannot be used for prediction because every information is delayed

till the moment when a change in controlled variable is recorded. In such cases it is better to

predict the output signal using mathematical model of the process in broader sense (process +

actuator). The controller structures that can be used are, for example, Otto-Smith predictor

(controller), PIP controller or so called Internal Model Controller (IMC).

An interesting feature of IMC is that when the model of the process is precise

(A = AM and B = BM), then a feedback signal eM = y – yM is equal to disturbance:

It follows that a control signal is not influenced by the reference signal and control

systems behaves as open loop. A usual problems with stability that arrise when closed loop

systems are used are then avoided. Control system with IMC controller will be stable and if IMC

and process are stable. With the exact model of process IMC is actually a feedforward controller

and can designed as such, but, unlike feedforward controllers, it can compensate for unmeasured

disturbances because feedback signal is equal to disturbance, which allows suitable tuning of the

reference value of the controller.

If model of the process is not exact5 (AM = A, BM =B), then feedback signal eM will

contain not only disturbance d but a modeling error, also. Thus, a feedback will

have its usual role, and stability problem can arise. This requires for parameters6 to be tuned

again so the stability is not lost.

74

1.4 PID controller Derivative mode improves stability of the system and enables increase in gain K and

decrease in integral time constant Ti, which increases speed of the controller response. PID

controller is used when dealing with higher order capacitive processes (processes with more than

one energy storage) when their dynamic is not similar to the dynamics of an integrator (like in

many thermal processes). PID controller is often used in industry, but also in the control of

mobile objects (course and trajectory following included) when stability and precise reference

following are required. Conventional autopilot are for the most part PID type controllers.

1.5 Topology of PID controllers

Problem of topology (structure) of controller arises when:

• designing control system (defining structure and controller parameters)

• tuning parameters of the given controller

There are a number of different PID controller structures. Different manufacturers design

controllers in different manner. However, two topologies are the most often case:

• parallel (non-interactive)

• serial (interactive)

75

Parallel structure is most often in textbooks, so it is often called "ideal" or "textbook

type". This non-interactive structure because proportional, integral and derivative mode are

independent on each other. Parallel structure is still very rare in the market. The reason for that is

mostly historical. First controllers were pneumatic and it was very difficult to build parallel

structure using pneumatic components. Due to certain conservatism in process industry most of

the controller used there are still in serial structure, although it is relatively simple to realize

parallel structure controller using electronics. In other areas, where tradition is not so strong,

parallel structure can be found more often.

1.5.1 Parallel PID topology A parallel connection of proportional, derivative and integral element is called parallel or non-

interactive structure of PID controller. Parallel structure is shown in Fig.

It can be seen that P, I and D channels react on the error signal and that they are unbundled. This

is basic structure of PID controller most often found in textbooks. There are other non-interactive

structures.

RESULT:

Thus the effect of P, PI, PID controllers using Mat lab are studied.

76

77

Expt No:

Date:

DESIGN AND IMPLEMENTATION OF COMPENSATORS

AIM:

To study the compensation of the second order process by using lead – Lag Compensator

EQUIPMENT REQUIRED: 1. LEAD – Lag network system kit

2. Capacitors – 0.1μF

3. Decade Resistance Box

4. CRO

DESIGN:

Lead – Lag network using operational amplifier:

An electronic lead –lag network using operational Amplifier is shown in Fig. 4.1.

The transfer function for this circuit can be obtained as follows:

Let

Z1 = R1 || C1

Z2 = R2 || C2

The second op-amp acts as a sign inverter with a variable gain to compensate for the magnitude.

The transfer function of the entire system is given by G(j).

R R (1 R C s) G(s)

4 2 1 1

R3 R1 (1 R2 C2 s)

78

1

1

TABULATION:

SI NO AMPLITUDE

(V)

FREQUENCY

(HZ)

PHASE

SHIFT

()

R1

(OHMS)

C1 (F) R2 (OHMS)

C2 (F)

We have

R R (1 T 2

2 )

G( j ) 2 4 1

R R (1 T 2

2 ) 1 3 2

Where

T1 = R1C1

T2 = R2C2 and

Thus the steady state output is

2 2

Yss (t ) R2 R4 (1 T1 )

Sin(t tan 1

T1 tan

T2 ) for an input Esint.

R R (1 T 2

2 ) 1 3 2

From this expression, we find that if T1>T2, then tan 1

T1 tan T2 >0. Thus if T1>T2, then the

network is a lead network. If T1 <T2, the network is a lag network.

Determination of values for angle compensation:

Frequency of sine wave (f)= 20Hz.

Phase angle to be compensated =14.50

79

1 1 tan (2fT1 ) tan (2fT2 )

Let T1 0.1sec

1 1

14.5 tan (2 * 20 * 0.1) tan (2 * 20 * T2 )

T2 0.023sec

Hence the values of T1 and T2 are chosen from which the values of R1, R2, C1, and C2 can be

determined. For Example,

T1 = R1C1 = 0.1;

If C1 = 0.1μF, R1=1MΩ T2 = 0.023sec

If C2 = 0.1 μF, R2 =230KΩ.

These values produce a phase lead of 14.5o, which is the desired compensation angle.

PROCEDURE:

1. Switch ON the power to the instrument.

2. Connect the individual blocks using patch chords bypassing the compensating

network and gain as shown in fig. 4.2.

3. Give a sinusoidal input as the set value to the error detector.

4. Measure the amplitude and frequency of the input signal.

5. Measure the amplitude and phase difference of the output signal with respect to the

input signal using DSO.

6. Using the technique explained previously, calculate the values of R1, R2, C2, and C1

to compensate for the phase shift of the output signal.

7. Connects the components at the points provided.

8. Now include the compensation block in the forward path before the process using

patch chords as shown in fig.4.2.

9. Observe the compensated wave form through DSO.

CALCULATION: (frequency = Hz)

CALCUALTION : (frequency = Hz)

RESULT:

Thus the compensator is designed for the given process and the performance of

the compensated system is found to work satisfactorily.

Frequency : Hz R1 =

R2 =

C1 =

C2 =

Frequency :

Hz R1 =

R2 =

C1 =

C2 =

80

ADDITIONAL EXPERIMENTS

Exp No :

Date :

Swinburns Test on Dc Shunt motor

To predetermine the efficiency o the D.C. machine when it act as

(i) Motor

(ii) Generator

APPARATUS REQUIRED:-

Sl.No. Name of the apparatus Range Type Quantity

1. Ammeter (0 -5) A MC 1

2. Ammeter (0 - 2) A MC 1

3. Voltmeter (0 - 300)V MC 1

4. Rheostat 400, 1.1 A Wire wound 1

5. Tachometer Digital 1

PRECAUTION:

1. The field rheostat should be kept at minimum resistance position.

2. There should be no load at the time of starting the experiment.

PROCEDURE:


2. The DPST switch is closed.

3. The motor is started with the help of three point starter.

4. The field rheostat of the motor is adjusted to bring the motor speed to the rated

value.

5. The no load current, voltage and shunt field current are noted.

CIRCUIT DIAGRAM:

NAME PLATE DETAILS:

Volts 230V

Amps 19A

Speed 1500


TABULAR COLOUMN

Voltage, V (volts) Field current, If

(A)

No load current, I0

(A)

For motor

Line

Current,

IL

(A)

Field

current

If

(A)

Ia =

IL -If

(A)

WCu 2

= Ia

Ra

Constant

Loss

(watts)

Total

Loss

(watts)

Input

Power

(watts)

Output

Power

(watts)

Efficiency

%

For generator

Line

Current,

IL

(A)

Field

current

If

(A)

Ia =

IL+If

(A)

WCu 2

= Ia

Ra

Constant

Loss

(watts)

Total

Loss

(watts)

Input

Power

(watts)

Output

Power

(watts)

Efficiency

%

Measurement of Ra:

Voltage (v) Current(A) Armature resistance

Ra(ohms)

Model

Graph

FORMULA

USED:

Constant loss Wc = V Io – (Ia –If)2

Ra

Ra – Resistance of armature

For Motor

1. Armature Current Ia = IL – If 2

2. Armature Copper Loss Wcu = Ia Ra

3. Total loss Wt = Wc + Wcu

4. Input power Pi = VIL

5. Output Power Po = Pi – Wt

6. Efficiency (Output Power / Input Power)

For Generator

1. Armature Current Ia = IL + If 2

2. Armature Copper Loss Wcu = Ia Ra

3. Total loss Wt = Wc + Wcu

4. Output power Po = VIL

5. Input Power Pi = Po + Wt

6. Efficiency (Output Power / Input Po

RESULT:

Thus the efficiency of the DC machine has been predetermined and

characteristics were drawn.

Ex. No:

Date:

7.(b) DETERMINATION OF TRANSFER FUNCTION OF DC MOTOR

AIM:

1. To determine the transfer function of an armature controlled DC motor.

2. To determine the transfer function of an field controlled DC motor.

APPARATUS REQUIRED:

Name Range Qty Type

Ammeter

Voltmeter

Auto transformer

Rheostat

Tachometer

Stopwatch

(0-5A),(0-2A),(0-10A),

(0-100mA)

(0-300V),(0-300V)

(0-300V),(0-150V)

1Ф,230V/(0-270V),5A

400Ω,1.1A/50 Ω,3.5A/250

Ω,1.5A

Each 1

Each1

Each1

1

Each1

1

1

MC

MC

MI

THEORY:

TRANSFER FUNCTION OF ARMATURE CONTROLLED DC MOTOR

The differential equations governing the armature controlled |DC motor

speed control system are

………………………..1

..……………………..2

…………………………3

…………………………4

On taking Laplace transform of the system differential equations with zero initial conditions we get

………………………5

TO MEASURE FIELD RESISTANCE Rf:

………………………………6

………………………………7

………………………………8

on equating equation 6 and 7

………………………………9

Equation 5 can be written as

………………………………10

Substitute Eb(s) and Ia(s) from eqn 8,9 respectively in equation 10

The required transfer function is

Where La / Ra =Ta= electrical time constant

J /B =Tm =mechanical time constant

TRANSFER FUNCTION OF FIELD CONTROLLED DC MOTOR

The differential equations governing the field controlled DC motor speed control system are,

………………………………11

………………………………12

………………………………13

Equation 12 and 13

………………………………14

TO MEASURE ARMATURE INDUCTANCE(La):

TO MEASURE FIELD INDUCTANCE( LF):

TO MEASURE Ka

………………………………15

The equation 4 becomes

………………………………16

On substituting If(s) from equation 7 and 8, we get

………………………………17

………………………………18

………………………………19

Where

Motor gain constant Km = Ktf /Rfb

Field time constant Tf= Lf/Rf

Mechanical time constant Tm = J/B

PROCEDURE:

To find armature resistance Ra:

1. Connections were given as per the circuit diagram.

2. By varying the loading rheostat take down the readings on ammeter and voltmeter.

3. Calculate the value of armature resistance by using the formula Ra = Va / Ia.

To find armature resistance La:

1. Connections were given as per the circuit diagram.

2. By varying the AE positions values are noted.

3. The ratio of voltage and current gives the impedance Za of the armature reading. Inductance

La is calculated as follows.

To find armature ka:


2. Keep the rheostat in minimum position.

3. Switch on the power supply.

4. By gradually increasing the rheostat, increase the motor to its rated speed.

5. By applying the load note down the readings of voltmeter and ammeter.

6. Repeat the steps 4 to 5 times.

To find kb:


2. By observing the precautions switch on the supply.

3. Note down the current and speed values.

4. Calculate Eb and 𝛚.

TO FIND Kb

TABULATION:

To find Ra:

Va(V) Ia(A) Ra(Ω)

To find Rf:

Vf(V) If(A) Rf(Ω)

To find La:

Va(V) Ia(A) Za(Ω) La(Ω)

To find Lf:

Vf(V) If(A) Zf(Ω) Lf(Ω)

ARMATURE CONTROLLED DC MOTOR:

Va Ia N T ω Kb Kt Eb

FIELD CONTROLLED DC MOTOR:

Va Ia N Tm ω Eb Km T Ktf Tf

MODEL GRAPH:

MODEL CALCULATION

RESULT:

VIVA questions for the experiments

SPEED CONTROL OF DC SHUNT MOTOR

VIVA QUESTIONS: 1. Which method of speed control is used for controlling the speed of the motor above its rated

speed? Give reason.

2. Which method of speed control is used for controlling the speed of the motor below its rated

speed? Give reason.

3. Explain the reasons for the shape of the graphs obtained.

4. State any method to control the speed of a D.C series motor?

LOAD TEST ON DC SHUNT MOTOR

VIVA QUESTIONS:

1. What is the need for a starter? 2. Name the different types of starters for DC motors.

3. Why a DC shunt motor is called a constant Speed motor?

4. State few applications of DC shunt motor.

5. What is the role of commutator in a DC motor.

6. What is the effect of armature reaction on the performance of DC motor?

7. What happen when the field circuit gets opened when a DC shunt motor is running?

8. How to reverse the direction of rotation of DC shunt motor?

OPEN CIRCUIT CHARACTERISTICS OF SELF EXCITED

DC SHUNT GENERATOR

VIVA QUESTIONS: 1. Define critical field resistance and critical speed.

2. State the conditions to be satisfied by a DC shunt generator to build-up voltage.

3. What is residual flux and what happens to the generator induces EMF when residual flux is

zero?

4. What is the purpose of SPST switch connected in the field circuit of the generator?

5. Why the speed must be maintained constant throughout the experiment?

LOAD TEST ON SELF EXCITED DC SHUNT GENERATOR

VIVA QUESTIONS:

1. What is a prime mover?

2. Why the speed of generator should be maintained constant during the experiment?

3. Why does the terminal voltage fall as the load on the generator is increased?

4. What is armature reaction and what are its effects on the performance of DC generator?

OPEN CIRCUIT CHARACTERISTICS OF A SEPARATELY

EXCITED DC SHUNT GENERATOR

VIVA QUESTIONS:

1. Define critical field resistance and critical speed.



zero?



TRANSFER FUNCTION OF SEPERATELY EXCITED DC SHUNT GENERATOR

VIVA QUESTIONS 1. What is the need for transfer function of a system?

2. What are the types of transfer function?

3. Define feedback.

4. Mention few applications of Separately Excited DC generator.

5. What are the basic elements used for modeling mechanical rotational system?

PREDETERMINATION OF REGULATION BY EMFAND MMF METHOD

VIVA QUESTIONS:

1. Define regulation. 2. What is meant by pessimistic method?

3. Which method is called as optimistic method?

4. What are the advantages of EMF and MMF method?

5. Name some other methods used to predetermine the regulation.

O.C AND S.C TESTS ON A SINGLE PHASE TRANSFORMER

VIVA QUESTIONS: 1. Why O.C test is conducted on the L.V side and S.C test on the H.V side?

2. Define regulation in a transformer.

3. Why the regulation graph is not passing through the origin?

4. State the condition for maximum efficiency?

5. What is the regulation of an Ideal transformer?

6. At what fraction of full load does the efficiency of the given transformer is maximum?

LOAD TEST ON A SINGLE PHASE TRANSFORMER

VIVA QUESTIONS: 1. Define Regulation of a Transformer.

2. What is the effect of load p.f on regulation of Transformer?

3. What is the condition for maximum efficiency?

4. Determine the percentage load at which maximum efficiency occurred for the given

single-phase transformer?

5. What is the effect of change in frequency on the efficiency of the transformer?

82

OPEN CIRCUIT CHARACTERISTICS OF A SEPARATELY

EXCITED DC SHUNT GENERATOR

VIVA QUESTIONS:




zero?






3. Define feedback.




VIVA QUESTIONS: 1. Define regulation. 2. What is meant by pessimistic method?


















OPEN CIRCUIT CHARACTERISTICS OF A SEPARATELY EXCITED DC SHUNT GENERATOR

VIVA QUESTIONS:




zero?






3. Define feedback.




VIVA QUESTIONS:

1. Define regulation. 2. What is meant by pessimistic method?


















82

LOAD TEST ON SQUIRREL CAGE INDUCTION MOTOR

VIVA QUESTIONS: 1. What is squirrel cage induction motor?

2. What is Skewing?

3. What is cogging?

4. What is crawling?

5. What is the no load current of an induction motor?

6. Distinguish between rotating transformer and static transformer?

7. Define slip.

NO LOAD AND BLOCKED ROTOR TEST ON THREE PHASE INDUCTION MOTOR

VIVA QUESTIONS: 1. How will you measure three phase power using two watt meters?

2. What is the necessity to have starter for three phase induction motor?

3. How mechanical load is represented in the equivalent circuit of induction motor?

4. Define synchronous speed.

5. Why induction motors cannot run at synchronous speed?

DIGITAL SIMULATION OF LINEAR SYSTEMS

VIVA QUESTIONS: 1) Define linear system.

2) What is impulse signal?

3) What is MIMO?

4) What is time invariant system?

5) What is step signal?

VIVA QUESTIONS: STABILITY ANALYSIS OF LINEAR SYSTEM

1) What is root locus?

2) What is bode plot?

3) What is Nyquist plot?

4) What is phase margin?

5) What is gain margin?

STUDY OF D.C MOTOR STARTER

VIVA QUESTIONS

1. Differentiate two point and three point starter

2. What is the need for starter in electrical technology?

3. Differentiate four point and three point starter

4. What are the types of starter?

5. What are the protective devices used in starters?

STUDY OF INDUCTION MOTOR STARTERS

VIVA QUESTIONS:

1. Differentiate star – delta and auto transformer starter

2. What is the need for starter in electrical technology?

3. Differentiate auto transformer and DOL starter

4. What are the types of AC starters?

5. What are the protective devices used in starters?

DESIGN AND IMPLEMENTATION OF COMPENSATORS

VIVA QUESTIONS: 1. What is compensation?

2. What is compensator?

3. When lag/lead/lag-lead compensators are employed?

4. What are the types of compensator?

5. Differentiate lag and lead compensators?

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