DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGINEERING

EC2259

ELECTRICAL ENGINEERING and CONTROL SYSTEM LABORATORY

IV Sem ECE

SSN COLLEGE OF ENGINEERING KALAVAKKAM- 603 110

LAB MANUAL

Prepared By

Dr Mrunal Deshpande

Associate Professor

EEE

IV Semester - Electronics and Communication Engineering

List of Experiments

1. Open circuit and load characteristics of separately excited D.C. generator. 2. Open circuit and load characteristics of self excited D.C. generator. 3. Load test on D.C. shunt motor. 4. Swinburne’s test on D.C. Shunt motor and Speed control of D.C. shunt

motor. 5. Load test on single phase transformer 6. Open circuit and short circuit test on single phase transformer 7. Load test on three phase induction motor. 8. No load and blocked rotor tests on three phase induction motor

(Determination of equivalent circuit parameters) 9. Regulation of three phase alternator by EMF and MMF methods. 10. Transfer Function of separately excited D.C. Generator. 11. Transfer Function of armature and Field Controller D.C. Motor 12. Study of D.C. motor and induction motor starters. 13. Digital simulation of linear systems. 14. Stability Analysis of Linear system using MAT lab. 15. Study the effect of P, PI, PID controllers using MAT lab. 16. Design of Lead and Lag compensator.

SSN College of Engineering, Kalavakkam

Department of Electrical and Electronics

Engineering

EC2259 – Electrical Engineering and Control

System Laboratory

INDEX

CYCLE-I

(Fill in the order by which done) Ex. No.

Date Experiment Name Page No

Staff signature

1.

2.

3.

4.

5.

6.

7.

8.

CYCLE-II

9.

10.

11.

12.

13.

14.

15.

16.

GENERAL INSTRUCTIONS TO STUDENTS FOR EEE LABORATORY COURSES

SAFETY: 1 You are doing experiments with the help of electrical power. You have to be very

careful. You must clearly know the supply system to your worktable in particular and the entire laboratory in general.

2 Incase of any thing goes wrong observation, you have to IMMEDIATELY SWITCH OFF the supply to the worktable.

3 You have to tuck in your shirts and you have to wear an overcoat. 4 Wearing loose garments inside the lab is strictly prohibited. 5 You have to wear shoes compulsorily.

ATTENDANCE: 1 Every time you come to the laboratory class, you have to come with your record note

book, observation notebook, calculators etc. 2 You have to attend the lab classes at the stroke of the bell in the laboratory. 3 You have to give your attendance. You have to submit your records and have to show

the day's experiment's circuit diagram and get it attested. You have to occupy the respective worktable of the machine. Collect required meters etc as per indent slip.

MAKING CONNECTIONS: 1 Start giving connections as per the circuit diagram from one side of the circuit. 2 Series circuits are to be only given first, with the help of connecting wires. 3 Make parallel connections, namely voltmeters last using silk wires. 4 Terminals of the meters should not be used as junction points. There should not be any

loose connection. Thoroughly check the connections and choose correct rheostat positions before starting. Also keep all meters in horizontal position to read the readings conveniently. Call staff to check the connections.

DOING EXPERIMENTS: 1 Switch ON the supply only in presence of staff member. 2 The experiments which include starters, move the handle of the starter slowly from

OFF to ON position. 3 Start the experiment as per the procedure. First check for the direction of rotation in

case of machines and deflection of meters. OBSERVATION:

1 Enter all readings in the tabulation. Note down the multiplication factor of any meter immediately.

2 During load test on motors, the needle of the spring balances may be vibrating. Arrest gently the vibrations and take the reading.

3 Pour water in the brake drum at the time of loading. CALCULATION:

1 Calculate all required quantities and enter in the tabulation. Show model calculations for any one set of readings only.Units are VERY, VERY IMPORTANT. Draw the necessary graphs. Write the result. Show it to the staff for getting signature.

RECORD: 1 As the name Implies, it is a record: PERMANENT RECORD for reference. Write

neatly; Draw circuit diagrams neatly and label correctly. 2 Enter readings in the tabulation. 3 UNITS are to be written for various quantities. 4 Draw Graph. Complete the record before you come for next lab class.

Additional Instructions Avoid wearing any loose metallic rings, straps or bangles, as they are likely to prove

dangerous at times. Before entering into the laboratory class, you must be well prepared for the experiment

that you are going to do on that day. You must bring the related textbook, which may deal with the relevant experiment. Get the circuit diagram approved with correct meter & fuse ratings Get the reading verified. After the experiment 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.

If you miss any practical class due to unavoidable reasons, intimate the staff in charge and complete that experiment in the repetition class.

The 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.

Girls should put their plait inside their overcoat 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. The same must be done when there is a power breakdown when the experiment is being carried out.

Avoid carrying too many instruments at a time. Avoid using water hydrant for electrical fires.

Exp.No: Date:

1. Open Circuit Characteristics (O.C.C) & Load Characteristics of D.C Separately Excited Shunt Generator.

AIM: To determine the no load magnetization or open circuit characteristic of a separately excited dc shunt generator, compare it with self excited one, and hence to deduce the O.C.C. for various speed, 1000 and 1250 rpm. To determine the critical field resistance and critical speed. To determine the external and internal (load) characteristics of the separately excited DC shunt generator by actually loading the machine. APPARA TUS REQUIRED: S.No Name of the apparatus Range Type Quantity

1 2 3 4 5 6 7 8

THEORY : O.C.C In any D.C generator generated emf is directly proportional to the exciting field current i.e. flux. O.C.C shows the relation between the no-load generated emf in the armature, 'Eo' and the field or exciting current 'If’ at a given fixed speed. It should be noted that OCC for a higher speed would be above this curve and for a lower speed, below it. (ii) Load Charactristics: Let us consider a dc generator giving it’s rated no load voltage 'Eo' for a certain constant field current. If there were no armature reaction and armature voltage drop, then this voltage would have remained constant as shown in model graph by the dotted horizontal line I. But when the generator is loaded, the voltage falls due to these two causes, thereby giving slightly drooping characteristics. If we subtract from Eo, the value of voltage drop due to armature reaction for different loads, then we get the value of E, the emf actually induced in the armature under load conditions. Curve II is plotted in this way and is known as the INTERNAL CHARACTERISTIC. If we subtract from E the armature drop laRa, we get terminal voltage V. The reverse procedure may also be adopted to obtain Internal characteristic (ie) adding laRa to V. Curve III represents the EXTERNAL CHARACTERISTIC. Comparison of self and separately excited generator gen characteristics: Since the field current depends on terminal voltage in self excited generator, it’s load characteristics would be more drooping than the other one. In O.C.C. there is no much appreciable difference between the two types.

CIRCUIT DIAGRAM:

R2

Name Plate Details:

Fuse Rating Calculation: D.C. Shunt Motor: O.C.C D.C Generator: O.C.C Load Test: Load Test:

D.C. Shunt Motor D.C Shunt Generator Capacity: Capacity: Voltage: Voltage: Current: Current: Speed: Speed: Field Current: Field Current:

+

D P S T S

AL

V M A

AA

-

220 V D.C.

L F A

Af

SPSTS

G VRL

+ -

DPSTS

+

- z

zz

+

-

z

zz

3 Point Starter

A

AA

OBSERVATION: O.C.C Test

Sl.no If (Amps)

E 0

(Volts) E0 At different speeds

750rpm 1500rpm

RANGE FIXING:

The current drawn by the shunt motor on no-load is 15 to 20% of full load current.

The current drawn by the shunt motor on load must not exceed 120% of full load current.

Motor Field circuit rheostat rating is _____ Ω ; ____ A (the current rating should be

slightly higher than the rated current)

Generator load Current IL = _______ A ∴ The range of ammeter AL is (0- )A

The rated field current is _____ A

∴ The range of ammeter Af is (0- )A

Generator Field circuit rheostat rating is _____ Ω ; ____ A (the current rating should be slightly higher than the rated current and ohmic value should be as high as possible). Rated voltage of Generator V = _______ Volts ∴ The range of voltmeter V is (0- ) Volts

PROCEDURE: O.C.C.Test:

1. Connections are given as shown in the circuit diagram. 2. After the connections are checked, Keeping R1 minimum resistance position and R2 in

maximum resistance position, the supply switch is closed. 3. The motor is started with the help of a dc 3 point starter.

4. The field rheostat R1 of the motor is adjusted to make it run at the rated speed. Keeping the SPST switch open, Emf generated due to residual magnetism is noted from voltmeter. (when .field current is zero.)

5. Then SPST switch is closed. By varying the field rheostat R2 in the field circuit of generator, different values of generated emf Eo and field current If are noted from voltmeter and ammeter and tabulated.

Load Test: 1. Keeping the generator side DPST open, the field rheostat in the generator side is

adjusted for the rated voltage of the generator which is seen in the voltmeter. 2. Now the DPST switch is closed and the resistive load is put up on the generator step by

step. The terminal voltage, armature and load current values are noted down for each step from the respective meters.

3. Note that while taking each set of readings, the field current is maintained constant as that for rated voltage [because due to heating, shunt field resistance is increased]

Measurement of Ra and Rsh : 1. Connections are given as shown in the circuit diagram. 2. Keep the resistances in maximum position and close the supply switch and take

Minimum three readings. Load Test:

Speed: .............. RPM; No load voltage: ............. volts

Sl. No.

Terminal voltage

Load current IL = Ia

If Eg = V + IaRa

MODEL GRAPHS: E0 Draw Rc line, such that it is tangent to the initial portion of O.C.C. at rated speed and passes through origin. The value of critical field resistance, Rc = the slope of Rc line

Critical speed, Nc = speedRatedNhereNAC

BCRR −× ;

MODEL CALCULATION: O.C.C

E0 ∝ N

So, for different speeds, O.C.C. can be deduced from the O.C.C.at rated speed.

N1 = E1 N2 E2

Load test: For separate excitation Ia = IL So, induced emf on load, Eg = V + IaRa GRAPH: Open circuit characteristic is drawn by taking field current 'If' along x axis and generated voltage Eo along y axis. To find critical field resistance: From the origin a tangent is drawn to OCC at the linear portion. The slope of the tangent will give the critical field resistance. Calculate E0 at different speeds and draw O.C.C. for diff speeds in a separate graph. External characteristic is drawn taking load current 'IL' along x axis and terminal voltage 'V' along yaxis. Internal characteristic is drawn by adding laRa drop to external characteristic curve, 'la' along x axis and 'E' along y axis.

Rsh Line

Eg

If

Rc line

O.C.C. Line

V, Eg

IL, Ia

V vs IL

Eg vs Ia

C

B

A

Measurement of Rsh of Generator Measurement of Ra for Generator OBSERVATION:

A + -

V +

-

DPSTS

+

-

220 V D.C

Z

ZZ

A + -

V G +

-

A

AA

DPSTS

+

-

RPS

Measurement of Ra:

Sl. No.

V (Volts)

I (Amps)

Ra = V/I (Ohms)

Measurement of Rsh:

Sl. No.

V (Volts)

I (Amps)

Rsh = V/I (Ohms)

RESULT: QUESTIONS:

1. What is the principle of operation of DC generator?

2. Where the field winding is placed in DC generator? Why?

3. List out the factors involved in voltage build up of a DC shunt generator.

4. Explain briefly the function of a commutator in a DC machine. Can commutator action

be performed by a solid state device.

5. Define and explain critical field circuit resistance and critical speed of DC shunt

generator.

6. What is armature reaction?

7. What is internal & external characteristics of DC shunt generator?

8. What are the applications of DC shunt generator?

REFERENCES:

1. I.J.Nagrath & D.P.Kothari “ Electrical Machines.” 2. Albert E. Clayton & H.N. Honcock “ The performance and Design of Direct Current

Machines.” 3. K. Murugash Kumar “ D.C. Machines & Transformers”

Ex.No : Date:

2.Open Circuit Characteristics (O.C.C) & Load Characteristics of Self Excited D.C Shunt Generator.

AIM:

To determine the no load magnetization or open circuit characteristic of self excited dc shunt generator for various speed, 1000 and 1250 rpm.

To determine the critical field resistance and critical speed. To determine the external and internal (load) characteristics of the self excited DC shunt generator by actually loading the machine. APPARA TUS REQUIRED: S.No Name of the apparatus Range Type Quantity

1 2 3 4 5 6

THEORY : (i) O.C.C & (ii) Load Charactristics: Refer the expiment “separately excited shunt generator” Comparison of self and separately excited generator gen characteristics: Since the field current depends on terminal voltage in self excited generator, it’s load characteristics would be more drooping than the other one. In O.C.C. there is no much appreciable difference between the two types. RANGE FIXING: The current drawn by the shunt motor on no-load is 15 to 20% of full load current. The current drawn by the shunt motor on load must not exceed 120% of full load current.

Motor Field circuit rheostat rating is _____ Ω ; ____ A (the current rating should be slightly higher than the rated current)

Generator load Current IL = _______ A ∴ The range of ammeter AL is (0- )A

The rated field current is _____ A

∴ The range of ammeter Af is (0- )A

Generator Field circuit rheostat rating is _____ Ω ; ____ A (the current rating should be slightly higher than the rated current and ohmic value should be as high as possible). Rated voltage of Generator V = _______ Volts ∴ The range of voltmeter V is (0- )V

CIRCUIT DIAGRAM:

Name Plate Details:

Fuse Rating Calculation: D.C. Shunt Motor: D.C Generator: O.C.C O.C.C Load Test: Load Test:

D.C. Shunt Motor D.C Shunt Generator Capacity: Capacity:

Voltage: Voltage:

Current: Current:

Speed:

Speed: Field Current: Field Current:

+

D P S T S

AL

V M

A

AA

-

220 V D.C.

L F A

SPSTS

G VRL

+ -

DPSTS

+

- z

zz

3 Point Starter

Af

A

AA +

- z

zz

OBSERVATION: O.C.C Test

Sl.no If (Amps)

E 0

(Volts) E0 At different speeds

750rpm 1500rpm

Load Test:

Speed: .............. RPM; No load voltage: ............. volts

Sl. No.

Terminal voltage

Load current IL

If Ia =IL + If Eg = V + IaRa

MODEL GRAPHS:

Rsh Line

E0

If

Rc line

O.C.C. Line

V, Eg

IL, Ia

V vs IL

Eg vs Ia

C

B

A

Draw Rc line, such that it is tangent to the initial portion of O.C.C. at rated speed and passes through origin. The value of critical field resistance, Rc = the slope of Rc line

Critical speed, Nc = speedRatedNhereNAC

BCRR −× ;

PROCEDURE: O.C.C.Test:

1. Connections are given as shown in the circuit diagram. 2. After the connections are checked, Keeping R1 minimum resistance position and R2 in

maximum resistance position, the supply switch is closed. 3. The motor is started with the help of a dc 3 point starter. 4. The field rheostat R1 of the motor is adjusted to make it run at the rated speed.

Keeping the SPST switch open, Emf generated due to residual magnetism is noted from voltmeter. (when .field current is zero.)

5. Then SPST switch is closed. By varying the field rheostat R2 in the generator side, different values of generated emf Eo and field current If are noted from voltmeter and ammeter and tabulated.

Load Test: 1. Keeping the generator side DPST open, the field rheostat in the generator side is

adjusted for the rated voltage of the generator which is seen in the voltmeter. 2. Now the DPST switch is closed and the resistive load is put up on the generator step by

step. The terminal voltage, armature and load current values are noted down for each step from the respective meters.

3. Note that while taking each set of readings, the field current is maintained constant as that for rated voltage [because due to heating, shunt field resistance is increased]

Measurement of Ra and Rsh : 1. Connections are given as shown in the circuit diagram. 2. Keep the resistances in maximum position and close the supply switch and take

Minimum three readings. Measurement of Rsh of Generator :

A + -

V +

-

AAA 3 P

+

-

220 V D.C

Z

ZZ

Measurement of Ra for Generator: OBSERVATION:

Model Calculation: O.C.C

E0 ∝ N

So, for different speeds, O.C.C. can be deduced from the O.C.C.at rated speed.

N1 = E1 N2 E2

A + -

V G +

-

A

AA

DPSTS

+

-

RPS

Measurement of Ra:

Sl. No.

V (Volts)

I (Amps)

Ra = V/I (Ohms)

Measurement of Rsh:

Sl. No.

V (Volts)

I (Amps)

Rsh = V/I (Ohms)

Load test: For separate excitation Ia = IL

For self excitation Ia = IL + If

So, induced emf on load, Eg = V + IaRa

GRAPH: Open circuit characteristic is drawn by taking field current 'If' along x axis and generated voltage Eo along y axis. To find critical field resistance: From the origin a tangent is drawn to OCC at the linear portion. The slope of the tangent will give the critical field resistance. Calculate E0 at different speeds and draw O.C.C. for diff speeds in a separate graph. External characteristic is drawn taking load current 'IL' along x axis and terminal voltage 'V' along yaxis. Internal characteristic is drawn by adding laRa drop to external characteristic curve, 'la' along x axis and 'E' along y axis. Result: QUESTIONS:

1. What is the principle of operation of DC generator?

2. Where the field winding is placed in DC generator? Why?

3. List out the factors involved in voltage build up of a DC shunt generator.

4. Explain briefly the function of a commutator in a DC machine. Can commutator action

be performed by a solid state device.

5. Define and explain critical field circuit resistance and critical speed of DC shunt

generator.

6. What is armature reaction?

7. What is internal & external characteristics of DC shunt generator?

8. What are the applications of DC shunt generator?

* * * * *

Ex.No: Date:

3. Load test on D.C. Shunt Motor AIM: To conduct the load test on a given dc shunt motor and draw the performance curves. APPARATUS REQUIRED: S.No Name of the apparatus Range Type Quantity

1

2

3

4

5

THEORY: Principle of operation of a D.C. motor: To operate DC machine as motor, field windings are excited with DC source and North and South Pole magnetic fields established in the air gap by the stator poles, a DC current has to be passed through the armature terminals. Current carrying armature conductors cause additional magnetic field to be set up. Interaction of this magnetic field with the main magnetic field set up by the stator poles cause a force to be developed around the conductors. The force developed in all the conductors is unidirectional (anti-clockwise) due to commutator action. In case, clockwise direction is required either the direction of the main field or the direction of the current through the armature conductor is to be reversed. As the armature rotates, the system of conductors in the armature come across alternate North and South pole magnetic fields. Therefore an emf is induced in the conductors. This is called as Back emf. The plot of speed, torque, efficiency and input current (vs) output power are called the performance characteristics of a motor. The speed characteristics is almost a horizontal line and the speed regulation is good. Hence a d.c shunt motor is generally referred to as a constant speed motor and are used in applications such as lathes, centrifugal pump, machine tools, blower fans and etc. The torque varies directly with the output load and the armature current because the field is approximately constant over the load range .As compared to other motors a shunt motor has a low means that a motor draws more current from the main supply than a series motor or a compound motor at the time of starting. The small input current on no load goes to meet the various losses occurring within the machine.

The efficiency curve with is usually the same for all motors It is advantageous to have an efficiency curve which is fairly flat over a wide range of loads and having maximum efficiency at full loads.

The rated field current is _____ A

Field circuit rheostat rating is _____ Ω ; ____ A (the current rating should be slightly higher than the rated current)

Rated voltage of motor V = _______ Volts

∴ The range of voltmeter V is (0- ) Volts

FORMULAE: Torque, (T) = (S1-S2)Rx9.81 Nm Where S1, S2 -spring balance readings (kg) R- Radius of the brake drum (m) Input power, Pi = VI watts Output power, Po = 2IINT/60watts Percentage efficiency = (Output power / Input power) x 100 PROCEDURE:

1. The circuit connections are given as shown in the circuit diagram. Ensure that there is no load on the brake drum.

2. The supply is given by closing the DPST switch, keeping R1 in minimum resistance position.

3. The motor is started using the dc 3 point starter, and it is allowed to run at rated speed by adjusting the rheostat R1 which is connected to the field circuit. After setting the speed rheostat position should not be altered.

4. At no load condition, the input voltage, current and speed are noted using voltmeter, ammeter and tachometer. For load spring balance reading S1 & S2 are noted.

5. Now the load on the brake drum is increased, gradually and the corresponding voltmeter, ammeter readings and speed, spring balance readings are noted down and tabulated.

6. Then the load is gradually decreased and field rheostat is brought to the minimum position and the supply is switched off.

7. Then torque, input power, output power and percentage efficiency are calculated by using the formulae and tabulated.

MODEL CALCULATION: Circumference of the brake drum = cms

Radius, r = mt.

Torque applied on the shaft of the motor, T = (F1 ~ F2) r 9.81 Nm

Output power, Po = WattsNT

60

2π

Input Power, Pi = V X I Watts

% Efficiency, η = 100×i

o

P

P

Circuit Diagram: Name Plate Details: . Fuse rating calculation: MODEL GRAPHS:

D.C. Shunt Motor

Capacity: Speed:

Voltage:

Current: Field Current:

+

D P S T S

A

V M

A

AA 220 V D.C.

L F A +

F

FF

3 Point Starter

−

+

− #

F1 F2

Brake drum

−

P0

N, T, η, I

T I

P0

N

Ia

T T

Ia

N

T

N

OBSERVATION:

Radius of brake drum, r = __________ mts.

Sl. No.

V I

Spring Balance Speed

N Torque

T Output

Power P0 Input

Power Pi Efficienc

y η F1 Kg F2 Kg

GRAPH: The performance characteristic curves are drawn as i. Output power Vs speed ii- Output power Vs current iii- Output power Vs Torque and iv, Output power Vs Efficiency. RESULT: QUESTIONS:

1. Explain the principle of operation of a DC motor 2. How does the back emf in a d.c. motor make the motor self regulating? 3. How can we reverse the direction of a d.c. shunt motor? 4. Justify the statement - “D.C. shunt motor will run almost at constant speed” 5. What is the need for starters for starting a d.c. machine? 6. What are the various D.C. starters? 7. What are the different types of d.c. motors? 8. What is the expression for torque developed ?

9. Write the condition for maximum power developed by d.c. motor.

10. Write the voltage equation of a d.c. shunt motor. 11. What is the relationship between back emf, speed of rotation and flux per pole?

REFERENCES:

1. I.J.Nagrath & D.P.Kothari “ Electrical Machines.”

2. Albert E. Clayton & H.N. Honcock “ The performance and Design of Direct Current

Machines.”

3. K. Murugash Kumar “ D.C. Machines & Transformers”

Ex.No: Date :

4a. Swinburne Test on D.C.Shunt Motor

AIM: To predetermine the efficiency of a dc shunt machine (motor and generator) by performing Swinburne test. . APPARATUS REQUIRED: S.No Name of the apparatus Range Type Quantity

1

2

3

4

5

THEORY: Swinburne test is performed on a DC machine to predetermine its efficiency. The efficiency is obtained without loading the machine. The DC motor is run on no load and the obtained data is used to calculate efficiency for both DC motor and generator. As the machine is not loaded the losses taking place are minimum and machine is not heated. RANGE FIXING: The current drawn by the shunt motor on no-load is 15 to 20% of full load current. Current drawn by the motor on no-load (approx.) I0 = _______ A ∴ The range of ammeter AL is (0- )A The rated field current is _____ A

CIRCUIT DIAGRAM: NAME PLATE DETAILS: . FUSE RATING CALCULATION: MEASUREMENT OF R A

∴ The range of ammeter Af is (0- )A & Field circuit rheostat rating is _____ Ω ; ____ A (the current rating should be slightly higher than the rated current) The no-load armature current is Iao = I0- If

Armature resistance Ra is low. ∴ The ohmic value of armature circuit rheostat required = 220 / Iao = ___ Select the nearest higher ohmic and current value. ∴ Armature circuit rheostat rating is _____ Ω ; ____A. Rated voltage of motor V = _______ Volts ∴ The range of voltmeter V is (0- ) Volts

D.C. Shunt Motor

Capacity: Speed:

Voltage:

Current: Field Current:

220V D.C

+

−

D P S T S

Af

V M A

AA Z

ZZ

+

+

−

−

PROCEDURE:

1. Apply rated voltage to the armature of the motor

2. Run the motor at the rated speed.

3.Observe the meter readings Measure the armature resistance of the machine MODEL CALCULATIONS : Constant losses, Wc = (No-load input to the motor − Armature copper loss at N.L) = VI0 − Iao

2Ra where Iao = I0 − I f

Stray losses, Ws = Wc − Shunt field current loss = Wc − VI f

Predetermination of η as generator: Assume load current as IL Amp. Armature current Ia = IL + If

Variable loss or Armature copper loss Wv = Ia2 Ra

Total losses, W = Wc + Wv

Output power, Po = VIL

Input power, Pi = Po + W Efficiency, η = Po / Pi

Predetermination of η as motor: Assume motor line current as IL Amp. Armature current , Ia = IL − I f

Variable loss or Armature copper loss Wv = Ia2 Ra

Total losses, W = Wc + Wv

Input power, Pi = VIL

Output power, Po = Pi − W Efficiency, η = Po / Pi

Predetermination of maximum η: For η to be maximum, Wv = Wc i.e. Ia2 Ra = Wc

Ia =

With this value of Ia, maximum efficiency of the d.c.machine when it runs as generator and motor can be computed as above

OBSERVATION: 1. No Load Test

Predetermination of efficiency of Generator

S No Assumed IL

Ia Wv Total Loss

Po (watts)

Pi (watts) Efficiency

Predetermination of efficiency of Motor

S No Assumed IL

Ia Wv Total Loss

Po (watts)

Pi (watts) Efficiency

S No

V (Volts)

I0

(amps) If

(amps)

MODEL GRAPHS: RESULT: Discussion questions: 1. What are stray losses and constant losses? 2. What are variable losses? 3. What is the condition for maximum efficiency? 4. How eddy current losses can be minimized? 5. How the different losses in a d.c.shunt machine depend on load?

Ex.No: Date :

4b. Speed Control of D.C.Shunt Motor

AIM: To determine the speed control characteristics of a dc shunt motor by i. Armature control method and ii. Field Control method. APPARATUS REQUIRED: S.No Name of the apparatus Range Type Quantity

1

2

3

4

5

THEORY: We know that back emf is given by the following equation. Eb = V – Ia Ra = ΦZNP / A N = ( V – Ia Ra / ΦZ ) * A/P rps. N = K ( V – Ia Ra / Φ ) rps Simply, The speed of D.C. shunt motor = N α V/ Φ Where V = applied voltage (or armature voltage), Φ = flux. ( field current) So the speed is directly proportional to armature voltage and inversely proportional to field current. It is clear that the variation of speed is possible in three different ways:

(i) VARIATION OF FIELD (FLUX) CONTROL METHOD

A more economical method of speed control is by rheostat adjustment of the field current. If the field flux is reduced, the speed increases. Using this method, speed can be increased above rated speed. In non- inter poles, it is in the ratio 2:1 and in inter poles maximum to minimum speed of 6:1 is fairly common. But it is not possible to reduce the speed below the rated speed.

The disadvantage of this method are i) Output torque gets reduced & ii) Higher speed results in poor commutation.

(ii) ARMATURE CONTROL METHOD:

In this method, armature current is controlled by means of a rheostat in the armature circuit. For a given armature current the larger the resistance of the controller, the smaller will be the voltage across the armature and hence the speed will lower. This method is used when Speed below the no-load speed is required. The main disadvantages of this type of speed control are

1. The losses are more and hence it is relatively more costly. 2. The efficiency is low. 3. The speed cannot be increased.

CIRCUIT DIAGRAM: Name plate details: . Fuse rating calculation:

D.C. Shunt Motor

Capacity: Speed:

Voltage:

Current: Field Current:

220V D.C

+

−

D P S T S

Af

V M A

AA Z

ZZ

+

+

−

−

Model graphs:

Armature control Field control OBSERVATION:

Armature control:

Sl.No. If1 = ------- A If2 = -------- A If3 = --------- A Va Speed Va Speed Va Speed

Va

N

If1(rated)

If2<If1

If3<If2

Va1(rated)

Va2<Va1

Va3<Va2

N

If

Field control:

Sl.No. Va1 = ------- A Va2 = -------- A Va3 = --------- A If Speed If Speed If Speed

PROCEDURE: ARMATURE CONTROL METHOD:

1. Connections are given as shown in the circuit diagram and initially the field rheostat R1 is kept in minimum position and armature rheostat R2 is kept in maximum position.

2. Supply is given and the motor is started. The armature resistance is cutdown and speed is measured.

3. If the motor runs below the rated speed, adjust the field rheostat and motor is brought to its rated speed, The field current, 'If' is kept constant.

4. Now the armature voltage (ie potential drop across the armature) is varied by varying the armature rheostat and corresponding voltmeter, ammeter (la) readings and speed are noted.

5. The same procedure is repeated for another value of constant field current 'If' FIELD CONTROL METHOD:

1. The motor is run at the rated speed as usual. Armature voltage and hence armature current is kept constant.

2. The field current is varied and the value of the field current 'If' and speed are noted in the tabular column.

3. The same procedure is repeated for another constant value of armature current.

GRAPH: ARMATURE CONTROL METHOD: Graph is drawn between speed and armature voltage taking armature voltage along x axis and speed along y axis. FIELD CONTROL METHOD: Graph is drawn between speed and field current, taking speed along y axis and field current along x axis. RESULT: Questions: 1. Which speed control will give the speed greater than the rated speed and which one will give

less than the rated speed? State also the reason. 2. What are the factors that decide the speed of a dc machine? 3. What are the various methods of speed control in dc series motor? 4. Write few merits and demerits of the rheostatic control. 5. What will happen if the field winding of a running dc motor is opened?

Ex.No: Date:

5. Load Test on SINGLE PHASE TRANSFORMER AIM: To conduct load test on a single phase transformer to determine its efficiency and find the variation of secondary terminal voltage with respect to the load current. APPARATUS REQUIRED: S.No Name of the apparatus Range Type Quantity

1

2

3

4

5

THEORY:

1. EFFICIENCY: Transformer is a static device by which electric power is transformed from one circuit to another circuit at the same frequency .Hence in transformer there are no friction or winding losses as in the case electric motors. For small transformer direct loading method can be employed where input & output power can be measured .The efficiency of a good transformer will be in the range of 95 to 98% and motors have 70 to 80%

2. REGULATION:

Regulation is an impedance parameter by which the performance of a transformer can be assessed The lesser the value, the better the transformer because a good transformer should keep its secondary terminal voltage as constant as possible under all conditions of load.

When a transformer is supplied with a constant primary voltage, the secondary voltage decreases (assuming lagging p.f loads) because of its internal resistance and leakage reactance, Let the change in secondary terminal voltage from no load to full load is (VNL – VFL). % Regulation down = (VNL – VFL) / VNL. X 100 If this change is divided by VNL then the regulation is known as .REGULATION DOWN' If this change is divided by VFL ' then it is known as 'REGULATION UP' % Regulation up = (VNL – VFL) / VFL. X 100

Normally, we are interested in finding regulation 'down'

The loads can be Resistive, Inductive, Capacitive or combinations of these loads are possible, The regulation characteristics curve will be as shown in the model graph. where it is drawn between power factor and % regulation, But for resistive loads, it is enough to draw the same between output power and % regulation, RANGE FIXING:

Rated primary current, I 1= Rated capacity in VA/Primary voltage,V1

Rated secondary current, I 2 = Rated capacity in VA/Secondary voltage , V 2

The load used is resistive in nature ∴ The range of Ap, Vp, Wp are …………A, ……………V, …………W respectively The range of As, Vs, Ws are ……………A, …………….V, …………..W respectively Fuse Rating:

120% of rated current ∴ The fuse rating is ____ A

PROCEDURE: 1. Excite the transformer to its rated voltage on no load.

2. Observe the meter readings at no load.

3. Gradually load the transformer and note the meter readings for each loading.

4. Load the transformer to its rated capacity i.e. till it draws rated current from the supply.

Note that applied voltage to the primary side should be kept at its rated voltage on Loading

CIRCUIT DIAGRAM:

MODEL GRAPH:

Single phase transformer Specifications

Capacity: Primary Voltage: Secondary Voltage

Single phase variac Specifications

Output voltage: Current rating:

MODEL CALCULATION:

1.Output power = Ws 2. Input Power = Wp

% Efficiency, η = (Ws/Wp) *100

% Regulation=(Vs0 − V s)/V s0× 100 (where Vs0 – no load secondary rated terminal voltage OBSERVATION:

MF= MF=

RESULT: Discussion Questions: 1. Define regulation of transformer. 2. Draw the phasor diagram of transformer for lagging, leading and unity power factor load condition. 3. What is the condition for maximum efficiency of transformer?

Sl. No.

Vp Volts

Ip Amps

Wp watts Vs

Volts

Is

Amps

Ws watts %Efficiency

η %

Regulation Reading Power Reading Power

Ex.No: Date:

6. OC and SC Test on SINGLE PHASE TRANSFORMER AIM: To conduct open circuit and short circuit test on a single phase transformer to determine equivalent circuit parameters and also to determine its efficiency and regulation at desired load. APPARATUS REQUIRED: S.No Name of the apparatus Range Type Quantity

1

2

3

4

5

THEORY: Open Circuit Test: In this test the high voltage side of the transformer is left open. Primary side is connected across the normal rated voltage. Normal flux is hence set up in the core and the transformer has only iron loss and no copper loss as under no load current is negligible. Hence power consumed during open circuit is only for feeding iron loss or core loss and it remains constant irrespective of load conditions. The voltmeter, ammeter and watt meter readings are noted down and the current Iu and Iw are calculated. Short Circuit Test: This is an economical method for determining the equivalent impedance (Z01 or Z02), Leakage reactance (X01 or X02) and total resistance (R01 or R02) of the transformer as referred to the winding in which the measuring instruments are placed and the full load copper loss. The low voltage winding is generally short circuited and rated full load current is made to flow in the circuit by applying a small percentage of normal voltage by means of an auto transformer. Because of this, core losses are very small with the result that wattmeter reading represents full load copper loss for the whole transformer. Equivalent Circuit: The equivalent circuit consists of no load parameters R0 and X0 which are found out by performing open circuit test on single phase transformer. The winding resistance and leakage reactance are obtained from short circuit test. Once these parameters are known, the transformer efficiency, regulation can be obtained for various load conditions and power factor.

It is evident that copper loss depends on current and iron loss on voltage. Hence total transformer loss depends on volt-ampere (VA) and not on phase angle between voltage and current. (ie) it is independent of load power factor. That is why rating of transformer is in kVA and not in kw. RANGE FIXING : Oc Test: Full load capacity in VA Primary voltage V1 Secondary voltage V2 Let both O.C. and S.C. test be conducted on primary side. In O.C. test the current drawn by the transformer is about 5 – 10% of full load primary current. Ammeter range is (0 - )A The rated primary voltage is applied. Voltmeter range (0 - )V Wattmeter: The current rating and voltage rating of Wattmeter must be near to the value calculated above. Under O.C. condition the reactive power drawn is more and the active power drawn is less. So power factor under no-load condition is very low.

LPF wattmeter can be used The range of wattmeter is ……… V, ……. A, LPF. S .C. T e s t : The voltage applied to the transformer primary to circulate rated full load current is about 5 to 10% of rated primary voltage. The voltmeter range is (0 - )V and Ammeter range is (0 - )A The active power drawn by the transformer on S.C. condition is more and reactive power drawn is less. UPF wattmeter can be used. Range of wattmeter is ………V, ……….A, UPF.

CIRCUIT DIAGRAM

PROCEDURE: OPEN CIRCUIT TEST: 1. The connections are given as shown in the circuit diagram. 2. Keeping the secondary at no load (ie left open), apply the rated primary voltage by closing the DPST switch. 3. Note down the readings of ammeter, voltmeter and wattmeter. 4. From these values, the exciting branch parameters (Ro, Xo) of the equivalent circuit are calculated using the ,formulae appearing in the theory. SHORT CIRCUIT TEST: 1. The connections are made as shown in the circuit diagram. 2. Keeping the secondary winding shorted, the voltage applied to the primary by auto transformer is adjusted for rated primary current. 3. The corresponding voltmeter and wattmeter readings are taken. Take care to see that current does not exceed rated value. 4. From these values, RO1, XO1 of the transformer are determined using the formulae appearing in the theory and marked all parameters on the equivalent circuit. Then efficiency and regulation are found out using the formulae and entered in the tabulation.

OBSERVATION

For Efficiency and Regulation: S NO

% of load x

Cu loss= Wc=x2Wsc

(watts)

Total loss=

Wi+Wsc (watts)

Cos Ø=1

Cos Ø=1

Cos Ø=1

P0 Pi η

P0

Pi

η

P0

Pi η

1 0

2 20

3 40

4 60

5 80

6 100

7 120

S NO

Cos Ø=1 Sin Ø=1 % Regulation Lagging pf Leading pf

1 0

2 0.2

3 0.4

4 0.6

5 0.8

6 1

CALCULATIONS: i. Equivalent Circuit Both OC and SC test are performed on primary side of the transformer. OC Test

,

,

,

SC Test:

ii. Efficiency: Let load be x% of full load KVA and cosØ the power factor. Then Power output= Po= x*(FL KVA)* cosØ*100 Copper Loss= x2*Wsc Total Loss= W=Wsc+Wi Power input=Pi=Po+W Efficiency=Po/Pi iii Regulation :

% !"# $

% 100

I2=rated secondary current V2=rated secondary voltage +ve sine for lagging power factor -ve Sine for leading power factor iii Maximum efficiency For maximum efficiency: cu loss= iron loss ( I2 is obtained from above formula. Power output= Po= Total Loss = W= 2Wi Efficiency=Po/Pi

MODEL GRAPHS Equivalent Circuit

RESULT Q u e s t i o n s : 1. What are the two components of transformer no load current? 2. Write the emf equation of transformer. 3. Draw the no load phasor diagram for the transformer. 4. Name the instrument transformers.

Ex.No: Date:

7. Load Test on THREE PHASE INDUCTION MOTOR AIM: To conduct load test on a three phase induction motor to determine its power and power factor for balanced inductive load by two single element watt meter and obtain the load characteristics. APPARATUS REQUIRED: S.No Name of the apparatus Range Type Quantity

1

2

3

4

5

Range Fixing: The current rating of the motor is A. The range of the ammeter is (0- ) A. The rated line voltage of the motor is V. The range of the voltmeter is (0- )V. The current rating and voltage rating of Wattmeter are to be nearer to the value calculated above. The Induction motor can be loaded up to 120% of Full load capacity. The power factor of Induction motor on Full load is around 0.8. So, UPF wattmeter can be chosen. Range of wattmeter is ………V, ……….A, UPF.

CIRCUIT DIAGRAM:

Induction Motor Specifications

Capacity: Voltage: Line current: Speed: Connection:

PROCEDURE: 1. Give connection as per circuit diagram and check for the motor is on no-load

condition.

2. Close the TPSTS on the supply side.

3. Bring the starter handle to START (or) Y position. Allow the motor to pick up speed.

4. Now bring the starter handle to RUN (or) D position.

5. Apply load in convenient steps of Amps, till 120% of rated load and observe all

meters readings, force applied on brake drum and speed. Note: On no-load and light loading, one of the wattmeter will give negative reading. Observe these readings as minus reading by interchanging the terminals of current or pressure coil of that wattmeter. MODEL CALCULATION: Circumference of the brake drum = _____ cms Radius of the brake drum, r = _____ m Torque applied on the shaft of the rotor, T = (F1 ~ F2) *r *9.81 Nm Output power, Po = 2π NT/60 watts Input power Pi = W1 + W2 % Efficiency =Po/Pi *100 Power factor of motor =Pi/√3VI % Slip S =NS−N/Ns *100; N s = Synchronous speed Check also the power factor cos φ from the two wattmeter readings : where

φ = tan−1[√3 (W 2 −W1 )/(W 2 +W1 ) ; W2 – Higher reading watt meter, W1 – Lower reading watt meter.)

OSERVATION

MODEL GRAPH:

S. No.

V Volts

I Amps

Wp watts Speed N rpm

Force kg Torque

Slip Input

watts

Output watts

Pf

%Efficiency η

Reading

Power

F1 F2

RESULT: Discussion Questions: 1. Name the types of 3-f Induction motor. 2. Which type of 3-f Induction motor has higher starting torque? 3. Define: slip speed, slip 4. Why power factor of Induction motor is low on no-load condition? 5. Mention the applications of 3-f Induction motors?

Ex.No: Date: 8.NO LOAD AND BLOCKED ROTOR TEST ON THREE PHASE INDUC TION MOTOR

AIM: To draw equivalent circuit of the given 3-φ squirrel cage induction motor and hence predetermine the performance characteristics of the induction motor. APPARATUS REQUIRED: S.No Name of the apparatus Range Type Quantity

1

2

3

4

5

PROCEDURE:

(i) No load test:

1. Make the circuit connections as per the circuit diagram. 2. By varying the variac start the motor at no load. 3. Apply rated voltage to the motor and note down all the meter readings.

(ii) Blocked rotor test:

1. Make the circuit connections as per the circuit diagram. 2. Add some load initially so that the motor is blocked from rotating. 3. By varying the variac circulate rated current in the motor stator circuit. 4. Note down all the meter readings.

CIRCUIT DIAGRAM No Load Test Blocked Rotor Test

Name Plate Details Power: (KW) Rated Voltage: (V) Rated Current: (A) Rated Speed: rpm Frequency: (Hz)

OBSERVATION:

1. Open Circuit Test 2. Short Circuit Test

Vo Volts Io Amps Wo watts

RESULT: Discussion Questions: 1. List out the method of starting of 3-f induction motor. 2. What are the data needed for constructing circle diagram of an induction motor? 3. What should be minimum dimension for the diameter of the circle diagram? 4.What is the limitation on capacity of induction motor for using DOL starters? 5. What is an induction generator?

Vs Volts Is Amps Ws watts

Ex.No: Date: 9. REGULATION OF 3- φ ALTERNATOR- EMF &MMF (PREDETERMINATION) AIM: To conduct open circuit and short circuit test on a three phase alternator to predetermine its efficiency. APPARATUS REQUIRED: S.No Name of the apparatus Range Type Quantity

1

2

3

4

5

PRECAUTIONS: Motor field rheostat should be at the minimum resistance position while switching on switching off the DPSTS. The rheostat on the field circuit of generator should be kept at minimum potential (output) position before closing the TPSTS. RANGE FIXING: On both O.C. and S.C. tests the power delivered by the alternator is zero. The power drawn by the M-G set from source is equal to the no load power requirement of MG set, which is approximately 30-40% of full load current of D.C. motor. Fuse rating on D.C. side is ……………….A Fuse rating on A.C. side is based on the rated current of alternator is …………. The rated field current of motor is _____ A ∴ Field circuit rheostat rating of motor is _____ Ω ; ____ A (the current rating should be slightly higher than the rated current)

The rated field current of alternator is _____ A ∴ The range of ammeter Af is (0- ) A ∴ Field circuit rheostat rating of alternator is _____ Ω ; ____ A (the current rating should be slightly higher than the rated current) The current rating of alternator is _____ A ∴ The range of ammeter A is (0- )A. The rated Line voltage of alternator is ______ V The range of voltmeter V is (0- )V Note: Voltmeter is connected between phase and neutral. PROCEDURE: For O.C.C Test: 1.Give the connections as per the circuit diagram. 2.Close the DPSTS on the supply side and run the MG set at rated speed.

3.Keeping TPSTS open on alternator side and vary the If in convenient steps, note down the values of If & V (Phase voltage) till rated field current of alternator.

4. Bring the alternator field rheostat to original position. For Load Test: 1.Keeping TPSTS closed and increase If for rated current in the armature circuit of alternator. 2.Note down the field and line current of alternator.

3. Measure the stator resistance per phase (d.c. value) with the circuit given.

CIRCUIT DIAGRAM :

Alternator Specifications

Capacity:

Voltage:

Line Current:

Field current:

Speed:

Motor Specifications

Capacity:

Voltage:

Line Current:

Field current:

Speed:

MODEL GRAPHS OBSERVATION: SC Test OC Test

Rdc= ohms EMF Method

S No

Power factor

cosØ

SinØ

E0 % Regulation Leading PF Lagging Pf Leading PF Lagging

Pf 1 0

2 0.2

3 0.4

4 0.6

5 0.8

6 1

S No V vollts If amps

I (rated current) amps

If amps

MMF Method

S

No

Power

factor cosØ

E0 Total field current If % Regulation Leading

PF Lagging

Pf Leading

PF Lagging

Pf Leading

PF Lagging

Pf 2 0.2

3 0.4

4 0.6

5 0.8

6 1

MODEL CALCULATION:

(i) EMF method:

Stator resistance / phase a.c. value, Ra = 1.3 X Rdc; Ra = ………… . Plot the O.C.C. and S.C.C. on common If Synchronous impedance / phase Zs = OC voltage , V ph/ SC current , I sc [Note: Zs should be calculated in the linear range of O.C.C.] Synchronous reactance / phase X s = √(Zs 2 − Ra 2 ) Full Load regulation: I= Full load current V = Rated terminal voltage phase value Cosf - Load power factor (assumed) No load induced emf Eo = √(V Cosφ + I Ra ) 2+(V Sin φ ± I X s) 2

+ve sign for lagging power factor load. -ve sing for leading power factor load.

% Regulation =(E0 – V)/V* 100

(ii)MMF method:

If1 = Field current which corresponds to rated phase voltage (V) from OCC (neglecting IRa drop) If2 = Field current which corresponds to rated current from SCC Total field current If is : I f = √(If1

2+If22±2If1If2cos(90-Ø))

Where f is power factor angle (assumed)

+ve sign for lagging p.f. loads -ve sign for leading p.f. loads

Eo = no load phase voltage from OCC curve corresponds to the total field current If calculated above.

% regulation =(Eo – V/V) *100 RESULT: Discussion Questions:

1. Write down the emf equation of 3-f generator.

2. Define regulation of alternator.

3. Is regulation of alternator can be +ve, -ve and zero? If so under what conditions?

4. For which nature of load power factor the regulation of alternator is maximum?

5. How S.C.C. is a straight line?

6. Draw phasor diagrams for different nature of loads.

7. What are pessimistic and optimistic methods?

EXPT. NO.: DATE:

10.TRANSFER FUNCTION OF ARMATURE AND FIELD

CONTROLLED DC MOTOR AIM: To determine the transfer function of an armature controlled dc servomotor. APPARATUS REQUIRED: S.No Name of the apparatus Range Type Quantity

1

2

3

4

5

6.

THEORY: Transfer function is defined as the ratio of Laplace transform of the output variable to the Laplace transform of input variable at zero initial conditions. Armature controlled DC shunt motor

In this system, Ra = Resistance of armature in Ω La= Inductance of armature windings in H Ia = Armature current in A If = Field current in A e = Applied armature voltage in V eb = back emf in V Tm = Torque developed by the motor in Nm

J = Equivalent moment of inertia of motor and load referred to motor shaft in kgm2

B= Equivalent viscous friction coefficient of inertia of motor and load referred to motor shaft in Nm/(rad/s)

In Servo applications, DC motors are generally used in the linear range of the magnetization curve. Therefore, the air gap flux φ is proportional to the field current. φ α If

φ = Kf If ,where Kf is a constant. ----------------------------- (1)

The torque Tm developed by the motor is proportional to the product of the armature current and air gap flux. Tm α φ Ia

Tm =Ki φ Ia = Ki Kf If Ia , where Ki is a constant --------------(2) In the armature controlled DC motor, the field current is kept constant. So the above equation can be written as Tm = Kt Ia , Where Kt is known as motor torque constant.------ (3) The motor back emf being proportional to speed is given by

eb α dθ/dt,

eb = Kb dθ/dt, where Kb is the back emf constant.----------------(4)

The differential equation of the armature circuit is

e = IaRa + La dIa/dt + eb ----------------------------------------- (5) The torque equation is

Tm = Jd2θ/dt2 + B dθ/dt ------------------------------------------ (6)

Equating equations (3) and (6)

Jd2θ/dt2 + B dθ/dt = Kt Ia ---------------------------------------(7) Taking Laplace transforms for the equations (4) to (7), we get

Eb(s) = Kb s θ(s) -------------------------------------------- (8)

(s La + Ra ) Ia(s) = E(s) – Eb(s). ------------------------------- (9)

( J s2+ B s) θ(s) = Tm (s) = Kt Ia(s) ---------------------------- (10)

From equations (8) to (10) , the transfer function of the system is obtained as

Block diagram Using the above equations, the block diagram for the armature controlled DC motor is given below: E(s) + θ (s)

ωωωω(s) - Eb(s) For field control Transfer Function the air gap flux φ is proportional to the field current. φ α If

φ = Kf If ,where Kf is a constant. -------------------------------- (1)

The torque Tm developed by the motor is proportional to the product of the armature current and air gap flux. Tm α φ Ia

Tm =K′φ Ia = K′ Kf If Ia = Km Kf If , where Ki is a constant ----(2) Appling Kirchhoff’s voltage law to the field circuit, we have Lf dIf/dt + RIf = ef ------------------------------------------------- (3) Now the shaft torque Tm is used for driving the load against the inertia and frictional torque. Hence, Tm = Jd2θ/dt2 + B dθ/dt ------------------------------------------- (4) Taking Laplace transforms of equations (2) to (4), we get Tm(s) = KmKf If (s) ----------------------------------------------- (5)

1/[Ra+sLa] K t 1/s[Js+B]

s Kb

Ef(s) = (s Lf + Rf) If(s) -------------------------------------------- (6)

Tm(s) = (J s2+ B s) θ(s) ------------------------------------------- (7)

Solving equations (5) to (7), we get the transfer function of the system as

CIRCUIT DIAGRAM To determine Kt and Kb PROCEDURE:

i)No-Load test to determine Kb and Kf. 1. Initially keep all the switches in the off position. 2. Keep all the voltage adjustment knobs in the minimum position. 3. Give connections. 4. Switch on the power and the SPST switch. 5. Set the field voltage to the rated value. 6. Adjust the armature voltage and note the armature voltage, current and speed. 7. Calculate the back emf eb and plot the graph between back emf and ω 8. Determine Kb from graph. 9. For Kf plot graph between torque and rated and 80% of field current corresponding to same

armature current. 10. From the graph obtain Kf.

ii)Load test to determine Kt

1. Initially keep all the switches in the off position. 2. Keep all the voltage adjustment knobs in the minimum position. 3. Give connections. 4. Switch on the power and the SPST switch. 5. Adjust the field rheostat to obtain rated speed. Note the corresponding current as rated field

current. 6. Conduct the load test for rated field current and 80% of rated field current. 7. Apply load and note the armature voltage, current and spring balance readings. 8. Calculate torque and plot the graph between torque and armature current. 9. Determine Kt from graph. 10. Ra. La, Rf and Lf values are noted from machine manual.

iii)Retardation test to determine J and F

1. Total losses can be divided into constant and variable losses. The constant losses include frictional and inertia losses.

Total losses in a circuit= VI-Ia2Ra-VIf

Losses = V(Ia+If)-Ia2Ra-VIf Energy = Losses*t=1/2[J(w1

2-w22)]

N1=1600 rpm, N2=1400 rpm and w=2πN/60. J=(Losses*2*t)/( (w1

2-w22)]

N1=N*e-t1/Tm; N2=N*e-t1/Tm

PROCEDURE:

1. Give connections as per circuit diagram 2. Close the SPST and measure V, Ia and If

3. Run motor at 1600 rpm and open the switch suddenly. 4. Using stop watch, note down time taken for motor speed to fall down from 1600 rpm to 225

rpm. 5. Moment of inertia J is then calculated from the relation between loss, time and w. 6. The friction constant f is obtained using exponential relation between speed, time and time

constant.

Tabulation to determine Kb and Kt

I f

amps Armature Voltage Va

(V)

Armature Current I a

(A)

Speed N (rpm)

Eb= Va-I a Ra

(V)

ωωωω = 2ππππN/60 (rad/sec)

Torque=EbIa/ωωωω (N-m)

Rated I f

80% of I f

- For retardation Test: From Load Test:

Retardation test for

Range of speed (rpm) Time (sec)

J only

Both J and F

Ia If Va Torque

MODEL GRAPH To find Kt To find K b To find K mK f Torque Eb( V) Torque (Nm) (Nm) Ia(A) ω(rad/sec) I f (A) MODEL CALCULATION

Ra = ……..Ohms

Za = …….. Ohms

La = √(Za2 –Ra

2 ) / 2πf = ……. H

f = 50 Hz

J = 0.074 kg/m2, B = 0.001Nm/rad/sec

From Graph,

Kt = Torque constant = ∆T / ∆ Ia = ………… Nm / A

Kb = Back emf constant = ∆Eb / ∆ ω = ………. V/(rad/s)

Rf = ……..Ohms

Zf = …….. Ohms

Lf = √(Zf2 –Rf

2 ) / 2πf = ……. H

From Graph, KmKf = ∆T / ∆If = ………… Nm / A

RESULT:

Ex.No: Date:

11.DETERMINATION OF TRANSFER FUNCTION OF SEPRATELY EXITED DC GENERATOR

AIM: To determine the transfer function of separately exited generator. APPARATUS: S.No Name of the apparatus Range Type Quantity

1

2

3

4

5

THEORY: The operation of a separately excited DC generator assumes that armature resistance is driven at constant speed (by means of a prime mover) and field excitation If is adjusted to given rated voltage at no load and is then held constant at this value throughout operation is considered. The armature circuit is governed by the equation: V = Ea - Ia Ra Inspite of the field excitation , Ea drops off with load owing to demagnetizing effect of armature reaction. In case of self excited generator , a part of the supply voltage itself is used for excitation of field. Transfer function : Let Rf = Resistance of field circuit ; W Lf = Inductance of field circuit ; Henry Eg = Generated voltage ; Volts Vf = Excitation voltage ; Voltages Ra = Armature resistance ; Ohms Lf = Armature Inductance; Henry

Now for a generator Eg α Φ ------------------------ (1) Flux is directly proportional to field current If through the windings. Eg α If ------------------------ (2) Let Kg be the generator constant in V/A Eg = Kg If --------------------- (3) Applying Kirchoff’s law to field circuit, Ef(t) = Rf If(t) + Lf dIf(t)/dt -------------(4) Taking Laplace transform of equation (4) Ef(s) = Rf If(s) + Lf s If(s), which implies If(s) / Ef(s) =1 / [Rf+s Lf]----------------(5) Eg (s)= Kg If (s)-------------------(6) From (5) & (6) Eg(s) / Ef(s) = Kg / [Rf+s Lf] ---------- (7) We know that Eg = Ra IL + La dIL / dt -------------------- (8) Taking laplace transform of equation (8) Eg(s) = (Ra + La s) IL(s), From this IL(s) / Ef(s) = Kg / [Rf+s Lf][ (Ra + La s)] ----------(9) The transfer function of a separately excited DC generator is IL(s) / Vf(s) = Kg / [Rf+s Lf][ (Ra + La s)] ----------(10) can be represented in block diagram format as shown below.

IL(s) -Load current ; Vf(s) - Excitation voltage ; Kg - Induced emf constant in V/Amp ; Rf - Field resistance ; Lf -Field Inductance ; Ra -Armature resistance ; La -Armature Inductance ; Kg - can be obtained by conducting open circuit test. ; Rf , Lf , Ra , La - can be found by using V –A method CIRCUIT DIAGRAM: Determination of Kg: Tabulation: The values of Rf, Lf Ra and La will be provided. MODEL Graph: To find kg

S NO E0 volts If amps

PROCEDURE: Determination of Kg : 2. The connection are made as shown in Fig (I). 3. The DPST switch is closed. 4. The motor is started by moving the starter handle from OFF to ON position smoothly. 5. The motor is brought to the rated speed by adjusting the motor field rheostat. The motor drives the generator at a rated speed. 6. Note down the field current If and open circuit voltage Eo . By adjusting Rf , the field current is increased in convenient steps up to the rated field current. 7. In each step the readings of Eo and If are noted. Throughout the experiment the speed is maintained constant. 8. A plot of Eo Versus If is drawn by taking If on X-axis and Eo on Y- axis. 9. A tangent to the linear portion of the curve is drawn through the origin. 10. The slope of this line , Eo Versus If gives Kg . RESULT: Discussion Questions: 1. The transfer function defined by equation (10) does not hold under field saturated condition. Justify this statement. 2.State the value of field time constant of the generator. What does this signify? 3.Derive the transfer function of separately excited generator . 4. Define OCC. 5. What is Auto Transformer? Why it is used instead of ordinary transformer. 6. Differentiate separately excited and self excited generator. 7. What is the purpose of 3 –point starter? 8. How do you fix the Fuse rating for a particular circuit. 9. Differentiate MI and MC instrument. 10.What is the use of transfer function. 11.What are the different types of modeling techniques?

Ex.No: Date:

12. DIGITAL SIMULATION OF LINEAR SYSTEMS AIM: To digitally simulate the time response characteristics of Linear SISO systems using state variable formulation.

APPARATU REQUIRED: A PC with MATLAB package.

THEORY: State Variable approach is a more general mathematical representation of a system, which,

along with the output, yields information about the state of the system variables at some predetermined points along the flow of signals. It is a direct time-domain approach, which provides a basis for modern control theory and system optimization.

SISO (single input single output) linear systems can be easily defined with transfer function

analysis. The transfer function approach can be linked easily with the state variable approach. The state model of a linear-time invariant system is given by the following equations:

X(t) = A X(t) + B U(t) State equation

Y(t) = C X(t) + D U(t) Output equation Where A = n x n system matrix, B = n x m input matrix, C= p x n output matrix and D = p x m transmission matrix, PROGRAMME: . OPEN LOOP RESPONSE (FIRST ORDER SYSTEM) T.F=4/(s+2) Response of system to Step and Impulse input n=[4]; n=[4]; d=[1 2]; d=[1 2];

sys=tf(n,d); sys=tf(n,d); step(sys) impulse(sys) SIMULINK

Step Input Open Loop –I Order

Impulse Input Open Loop –I Order Close Loop Response n=[4]; d=[1 2]; sys=tf(n,d); sys=feedback(sys,1,-1) step(sys) impulse(sys) SIMULINK

Step Input Close Loop –I Order

Impulse Input Close Loop –I Order

SECOND ORDER SYSTEM TF= 4/s2+6s+16 Open Loop Response n=[4]; d=[1 6 16]; sys=tf(n,d); step(sys) impulse(sys) SIMULINK

Step Input Open Loop –II Order

Impulse Input Open Loop –II Order Close Loop Response n=[4]; d=[1 6 16]; sys=tf(n,d); sys=feedback(sys,1,-1) step(sys) impulse(sys)

SIMULINK

Step Input Close Loop –I Order

Impulse Input Close Loop –II Order STATE SPACE EQUATION A=[-6 -16; 1 0]; B=[1; 0]; C=[ 0 4]; D=[0]; Sys=ss(A,B,C,D); step(sys) impulse(sys) RESULT:

Ex.No: Date:

13. STABILITY ANALYSIS OF LINEAR SYSTEM USING MATLA B

AIM:

To analyze the stability of linear system using Bode plot/ Root locus.

APPARATU REQUIRED:

A PC with MATLAB package.

THEORY:

A Linear Time-Invariant Systems is stable when the following two conditions for stability are satisfied

When the system is excited by a bounded input, the output is also bounded. In the absence of the input, the output tends towards zero, irrespective of the initial

conditions.

PROCEDURE: 1. Write a Program to obtain the Bode plot / Root locus / Nyquist plot for the given system.

2. Determine the stability of given system using the plots obtained.

PROGRAM :

Consider a transfer function TF=

)*.,-)*.,-

Open a M file of MAT lab and enter the above transfer function. For root locus:

n=[10]; d=[0.0045 0.22 1 0]; sys=tf(n,d) rlocus(sys)

For Bode Plot

n=[10]; d=[0.0045 0.22 1 0]; sys=tf(n,d) bode(sys)

RESULT: Ex.No: Date:

14. STUDY THE EFFECT OF P, PI, PID CONTROLLERS USING MAT LAB.

AIM: To design a P, PI, PID controller using “Ziegler – Nichols” Tuning with process control

simulator. APPARATU REQUIRED:

A PC with MATLAB package. THEORY: The transient response of a practical control system often exhibits damped oscillation before reaching steady state value. In specifying the transient response characteristics of control systems to unit step input, it is common to specify the following

i) Delay Time(Td) ii) Rise time(Tr) iii) Peal time( Tp) iv) Max. overshoot (Mp) v) Settling time( Ts)

Proportional control:

The output of the controller is proportional to input U(t) = Kp e(t)

E(t) = error signal U(t) controller out[put Kp = proportional constant

• It amplifies the error signal and increases loop gain. Hence steady state tracking accuracy , disturbance signal rejection and relative stability are improved.

• Its drawbacks are low sensitivity to parameter variation and it produces constant steady state error.

Proportional + Integral Control:

The output of the PI controller is given by t

U(t) = Kp [ e(t) + (1/Ti )∫ e(t) dt ]

0 where Kp is the proportionality constant and Ti is called the integral time.

• This controller is also called RESET controller. • It introduces a zero in the system and increases the order by 1. • The type number of open loop system is increased by 1 • It eliminates steady state error. Damping ratio remains same. • Increase in order decreases the stability of system.

Proportional + Integral Control + Differential Cont rol: The output of a PID controller is given by t

U(t) = Kp [ e(t) + (1/Ti )∫ e(t) dt + Td de(t)/dt ] 0

The PID controller introduces a zero in the system and increases the damping. This reduces peak overshoot and reduces rise time. Due to increase in damping, ultimately peak overshoot reduces.

The stability of the system improves. In PID controller, all effects are combined. Proportional control stabilizes gain but produces steady state error. Integral control eliminates error. Derivative controller reduces rate of change of error.

TUNING OF PID CONTROLLERS

Proportional-integral-differential (PID) controllers are commonly employed in process control industries. Hence we shall present various techniques of tuning PID controllers to achieve certain performance index for systems dynamic response. The technique to be adopted for determining the proportional, integral and derivative constants of the controller depends upon the dynamic response of the plant.

In presenting the various tuning techniques we shall assume the basic control configuration, wherein the controller input is the error between the desired output and the actual output. This error is manipulated by the controller (PID) to produce a command signal for the plant according to the relationship.

U(s)=Kp (1+(1/sτi)+sτd)

Where Kp= proportional gain constant τI= integral time constant. τd= Derivative time constant.

General Block Diagram

Block Diagram for P Controller

Block Diagram For PI Controller

Block Diagram Of PID Controller

Block Diagram of Closed Loop Control Using PID Controller PROGRAMME: clear all close all G=tf(4500,[1 361.2 0]); Kc=184.1; M1=feedback(Kc*G,1); t=[0:0.0001:0.04]; y1=step(M1,t);

PID(KP,Ti,Td )

M(s)

Transfer Function

C(s)

C(s)

R(s) E(s)

KD=0.324; D1=tf([KD Kc],1); M2=feedback(D1*G,1); y2=step(M2,t); figure(1); plot(t,y1,t,y2);grid;xlabel('t');ylabel('y(t)'); text(0.005,1.4,'With P control') text(0.003,1.07,'With PD control') Kc=14.728;KI=147.28; D2=tf([Kc KI],[1 0]); M3=feedback(D2*G,1); y3=step(M3,t); figure(2); plot(t,y1,t,y3);grid; xlabel('t');ylabel('y(t)'); text(0.005,1.5,'With P control') text(0.02,1.15,'With PI control')

RESULT:

Ex.No: Date:

15. DESIGN oF LEAD AND LAG COMPENSATOR AIM: To design a lead and lag compensator network for the process given in the Process Control

Simulator.

APPARATU REQUIRED:

A PC with MATLAB package. THEORY:

Practical feedback control systems are often required to satisfy design specification in the transient as well as steady state regions. This is not possible by selecting good quality components alone (due to basic limitations and characteristics of these components). Cascade compensation is most commonly used for this purpose and design of compensation networks figures prominently in any course in automatic control systems.

In general, there are two situations in which compensation is required. In the first case the

system is absolutely unstable and the compensation is required to stabilize it as well as to achieve a specified performance. In the second case the system is stable but the compensation is required to obtain the desired performance. The systems which are of type 2 or higher are usually unstable. For these systems, lead compensator is required, because the lead compensator increases the margin of stability. For type 1 and type 0 systems stable operation is always possible. If the gain is sufficiently reduced, in such cases, any of three components viz. Lag, Lead, Lag – Lead must be used to obtain the desired performance. The simulation of this behavior of the Lead – Lag Compensator can be done with the module (VLLN – OI).

An electronic Lead - lag network using Operational amplifiers is given figure 1.

C2

C1 R2 R4

- -

R1 + R3 +

Fig.1 LEAD -LAG NETWORK USING OPERATIONAL -AMPLIFIER

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

Let Z1 = R1 ׀׀ C1

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(jω) = ( R4 R2/ R3 R1 ) (1+R1C1s) / (1+R2 C2 s)

1+T1√ ) ( R4 R2/ R3 R1 ) = ׀G(jω)׀2ω2) / ( √1+T2

2ω2 )

where T1 = R1 C1 ; T2 = R2 C2

φ = angle G(jω) = - tan-1(T1ω) – tan-1(T2ω).

Thus steady state output is

For an input =X sinωt,

Yss(t) =X (R4 R2/ R3 R1) (( √1+T12ω2) / ( √1+T2

2ω2 ))sin(wt – tan-1 T1ω –tan-1 T2ω )

From this expression, we find that if T1 > T2, then tan-1 T1ω –tan-1 T2ω > 0.

Thus if T1 > T2 , then the network is a LEAD NETWORK.

If T1 < T2 , then the network is a LAG NETWORK.

DETERMINATION OF VALUES FOR ANGLE COMPENSATION:

Frequency of sine wave = 20 Hz

Angle to be compensated = 70º

φ= tan-1 (2π f * T1) –tan-1 (2π f * T2)

T1 = 10, then substituting in above equation

70 - tan-1 (2 * π * 20 * 10) –tan-1 (2 * π * 20 *T2)

solving for T2

T2 = 0.003 .

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

determined .

For example, T1 =R1 C1 = 10 ; If C1= 1µF, then R1 = 10 MΩ.

T2 = 0.003 = R2 C2 then C2 =1 µF, and hence R2 = 3 MΩ.

These values produce a phase lead of 70º which is the desired compensation angle.

Nominal Value for R1 =1 MΩ C1 = 0.1 µF

R2 =20 KΩ C2 = 0.01 µF

PROCEDURE:

For the given transfer function of the system design theoretically the lead and lag compensator.

Then using M file of MAT lab write a programme to design the same and compare the values

obtained.

PROGRAMME:

Program for lag compensator design(Bode Plot) clear all;close all den=[1 1 0]; G=tf(10,den); figure(1); margin(G); w=logspace(-1,1,100); [mag ph]=bode(G,w); ph=reshape(ph,100,1); mag=reshape(mag,100,1); Phi=input('Enter phase angle Phi :') wg=interp1(ph,w,Phi) beta=interp1(ph,mag,Phi) wcu=input('Enter upper corner frequency :') tau=1/wcu; D=tf([tau 1],[beta*tau 1]); figure(2); margin(D*G); w=[1:0.1:10]; [mag,ph,w]=bode(feedback(G,1)); magdB=20*log10(mag); wb=interp1q(-magdB,w,3) [mag1,ph1,w]=bode(feedback(D*G,1)); mag1dB=20*log10(mag1); wb1=interp1q(-mag1dB,w,3) figure(3) Gc=D*G; ngrid;nichols(G,Gc); M=feedback(D*G,1); ltiview('step',M)

Program for Lead Compensator design (Bode plot) clear all;close all den=[1 1 0]; G=tf(10,den); figure(1); margin(G); phiM=input('Enter required phase lead :'); alpha=(1-sin(phiM*pi/180))/(1+sin(phiM*pi/180)); w=logspace(-1,1,100); [mag,ph]=bode(G,w); magdB=20*log10(mag); magdB=reshape(magdB,100,1); wm=interp1(magdB,w,-20*log10(1/sqrt(alpha))); tau=1/(wm*sqrt(alpha)); D=tf([tau 1],[alpha*tau 1]) figure(2); margin(D*G); w=[1:0.1:10]; [mag,ph,w]=bode(feedback(G,1)); magdB=20*log10(mag); wb=interp1q(-magdB,w,3) [mag1,ph1,w]=bode(feedback(D*G,1)); mag1dB=20*log10(mag1); wb1=interp1q(-mag1dB,w,3) figure(3) Gc=D*G; ngrid;nichols(G,Gc); pause [theta,dB]=ginput(1); rp=10.^(dB./20); theta=theta*pi/180; [x,y]=pol2cart(theta,rp); Mr=((x.^2+y.^2)./(((1+x).^2)+y.^2)).^0.5; Mr=20*log10(Mr) M=feedback(D*G,1); ltiview('step',M)

Program for the design of compensators (Root locus) close all clear all G=tf(1,[1 2 0]); figure(1) rlocus(G); axis equal; axis([-7 2 -5 5]); %spchart(gca,[0.5 1],4); sgrid beta=input('Enter trial value of beta:') D=tf([1 2.9],[1 beta]); figure(2) rlocus(D*G) axis equal;axis([-7 2 -5 5]); %spchart(gca,[0.5 1],4); sgrid [KK polesCL]=rlocfind(D*G) Gv=tf([1 0],1)*KK*D*G; Kv=dcgain(Gv) M=feedback(KK*D*G,1); ltiview('step',M)

RESULT: