Electrical Engg & Ctrl Sys-lm(Ece) 13

Dr.NNCEEEE/IVSEMEE&CSLABLM

1

EC2259 ELECTRICAL ENGINEERING AND CONTROL SYSTEM

LABORATORY

LLAABBOORRAATTOORRYY MMAANNUUAALL

FFOORR IIVV SSEEMMEESSTTEERR BB..EE ((EECCEE))

((FFOORR PPRRIIVVAATTEE CCIIRRCCUULLAATTIIOONN OONNLLYY))

ANNA UNIVERSITY -CHENNAI

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING,

DR.NAVALAR NEDUNCHEZHIYAN COLLEGE OF ENGINEERING,

THOLUDUR 606 303, CUDDALORE DIST.


2

UNIVERSITY PRACTICAL EXAMINATION

ALLOTMENT OF MARKS

Internal assessment : 20 marks

Practical assessment : 80 marks

TOTAL : 100 MARKS

INTERNAL ASSESMENT (20 MARKS)

Staff should maintain the assessment register and the head of the department should monitor it.

SPLIT UP OF INTERNAL MARKS

Observation : 3 marks Record note : 7 marks Model exam : 5 marks Attendance : 5 marks

TOTAL : 20 Marks

UNIVERSITY EXAMINATION

The Exam will be conducted for 100 marks. Then the marks will be converted to 80 marks.

SPLIT UP OF PRACTICAL EXAMINATION MARKS

AIM AND PROCEDURE : 10

CIRCUIT DIAGRAM/PROGRAM : 25

CIRCUIT CONNECTION : 15

TABULATION : 15

CALCULATION/SIMULATION : 15

GRAPH AND RESULT : 10

VIVA VOCE : 10

_____

TOTAL 100


3

_____

LIST OF EXPERIMENTS

1. Open circuit and Load characteristics of separately excited and self excited DC generator.

2. Load test on DC Shunt motor. 3. Swinburnes test and Speed control of DC Shunt motor. 4. Load test on Single phase transformer and open circuit and short circuit test on

single phase transformer. 5. Regulation of three phase alternator by EMF and MMF methods. 6. Load test on three phase Induction motor. 7. No Load and Blocked rotor tests on three phase Induction motor (Determination of

equivalent circuit parameters). 8. Study of DC motor and Induction motor starters. 9. Digital Simulation of Linear systems. 10. Stability analysis of linear systems using MATLAB. 11. Study the effect P, PI, PID controllers using MATLAB. 12. Design of lead and lag compensator. 13. Transfer function of separately excited D.C. generator. 14. Transfer function of armature and field controller D.C. motor.

BEYOND THE SYLLABUS 1. Load test on dc compound motor 2. Determination of transfer function parameters of field controlled dc servo motor 3. Determination of transfer function parameters of armature controlled dc servo motor


4

CONTENTS

Signature of the Lab In Charge

(N.JAYACHITRA)

S.NO

NAME OF EXPERIMENT

PAG

E N

O

1 Open circuit and Load characteristics of separately excited and self excited DC generator.

2 Load test on DC Shunt motor.

3 Swinburnes test and Speed control of DC Shunt motor.

4 Load test on Single phase transformer and open circuit and short circuit test on single phase transformer.

5 Regulation of three phase alternator by EMF and MMF methods.

6 Load test on three phase Induction motor.

7 No Load and Blocked rotor tests on Three phase Induction motor (Determination of equivalent circuit parameters).

8 Study of DC motor and Induction motor starters.

9 Digital Simulation of Linear systems.

10 Stability analysis of Linear systems using MATLAB.

11 Study the effect P, PI, PID controllers using MATLAB.

12 Design of lead and lag compensator.

13 Transfer function of Separately excited D.C. generator.

14 Transfer function of armature and field controller D.C. motor.


5

Exercise Number: 1(A)

Title of the Experiment : OPEN CIRCUIT AND LOAD CHARACTERISTICS OF DC SEPARATELY EXCITED GENERATOR Date of the Experiment :

OBJECTIVE (AIM) OF THE EXPERIMENT:

a) To obtain the open circuit characteristics of a separately excited DC 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 separately excited DC Generator and plot its internal and external characteristics.

a) FACILITIES REQUIRED:

S.NO APPARATUS TYPE RANGE QUANTITY

1 Voltmeter mc (0-300) V 1


3 Ammeter mc (0-20) A 1


5 Rheostat - 250/1.2A 1 6 Rheostat 570/1.2A 1 7 Tachometer Analog - 1

8 Variable Rheostat load - - 1

9 DPSTS - - 1

10 Connecting wires - - Few


6

b) PRECAUTIONS:

c) 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 up to 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.

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.


7

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

original position.

d) EXPRESSION FOR THEORITICAL VALUE:

1. Critical field resistance, Rc = Slope of the tangent drawn to the linear portion of 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

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.


8

e)CIRCUIT DIAGRAM

f)TABULATION:

OPEN CIRCUIT TEST:

S. No Field current

(Amps)

Output voltage at no load (Volts)


9

LOAD TEST:

S. No

Load voltage Load current Field current Ia= IL + IF Eg=Va+IaRa

Volts Amps Amps Amps Volts

g) TO FIND Ra

Circuit Diagram

Tabulation

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

Volts Amps Ohms


10

TO FIND Ra:

1. Connections are given as per the circuit diagram.

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.

h)MODEL GRAPH

i)DISCUSSION QUESTIONS:

1. Why the armature core in d.c machines is constructed with laminated steel sheets instead of solid steel sheets? Lamination highly reduces the eddy current loss and steel sheets provide low reluctance path to magnetic field. 2. Why commutator is employed in d.c.machines? Conduct electricity between rotating armature and fixed brushes, convert alternating emf into unidirectional emf(mechanical rectifier). 3. Distinguish between shunt and series field coil construction? Shunt field coils are wound with wires of small section and have more turns. Series field coils are wound with wires of larger cross section and have less no. of turns.


11

4. How does d.c. motor differ from d.c. generator in construction? Generators are normally placed in closed room and accessed by skilled operators only. Therefore on ventilation point of view they may be constructed with large opening in the frame. Motors have to be installed right in the place of use which may have dust, dampness, inflammable gases, chemicals.etc. to protect the motors against these elements, the motor frames are made either partially closed or totally closed or flame proof. 5. How will you change the direction of rotation of d.c.motor? Either the field direction or direction of current through armature conductor is reversed. RESULT:

Thus the open circuit and load characteristics of separately excited DC generator are

determined.


12

Exercise Number: 01(B)

Title of the Experiment : OPEN CIRCUIT AND LOAD CHARACTERISTICS SELF EXCITED DC SHUNT GENERATOR

Date of the Experiment :


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

a) To conduct the Load test on the given DC Shunt Generator and plot its internal and external characteristics. a) FACILITIES REQUIRED:






5 Rheostat - 250/1.2A 1 6 Rheostat - 570/1.2A 1 7 Tachometer Analog - 1

8 Variable Rheostat load - - 1

9 DPSTS - - 1

10 Connecting wires - - Few

b)EXERCISE:


13

c) 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.

d)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:


14

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.

e) EXPRESSION FOR THEORITICAL VALUE:

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.

f)CIRCUIT DIAGRAM


15


16

g)TABULATION:

OPEN CIRCUIT TEST:

S. No Field current

(Amps)

Output voltage at no load (Volts)

LOAD TEST:

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

Volts Amps Amps Amps Volts


17

h)TO FIND Ra

Circuit Diagram

Tabulation

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

Volts Amps Ohms

TO FIND Ra:

1. Connections are given as per the circuit diagram.

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.

i)MODEL GRAPH


18

j)DISCUSSION QUESTIONS:

1. What is meant by stator voltage control? The speed of the I.M can be changed by changing the stator voltage. Because the torque is proportional to square of the voltage.

2. What are the applications of stator voltage control? Fans, pump drives.

3. What are the advantages of stator voltage control method? 1. The control circuitry is simple. 2. Compact size. 3. Quick response time. 4. There is considerable savings in energy and thus it is economical method.

4. Define base speed? The synchronous speed corresponding to the rated frequency is called the base speed.

RESULT :

Thus the open circuit and load characteristics of self excited DC generator are

determined.


19

Exercise Number: 02

Title of the Experiment: LOAD TEST ON DC SHUNT MOTOR


OBJECTIVE (AIM) OF THE EXPERIMENT :

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

a)FACILITIES 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


20

b)EXERCISE:

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

d)PROCEDURE:

1. Connections are given as per the circuit diagram. 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


21


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


22

f)CIRCUIT DIAGRAM

g)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 %


23

h)MODEL GRAPH:

i)DISCUSSION QUESTIONS

1. What is meant by flux control method? The speed of the dc motor can be controlled by varying the field flux. This method od

speed control can be used for increasing the speed of the motor is inversely proportional to the field flux 2. Under What circumstances does a dc shunt generator fails to generate?

Absence of residual flux, initial flux setup by field may be opposite in direction to residual flux, shunt field circuit resistance may be higher than its critical field resistance, load circuit resistance may be less than its critical load resistance.

3. Define critical field resistance of dc shunt generator? Critical field resistance is defined as the resistance of the field circuit which will

cause the shunt generator just to build up its emf at a specified field.

RESULT:

Thus the load tests of DC shunt motor are determined.


24


Title of the Experiment: SWINBURNES TEST

Date of the Experiment:


To conduct no load test on dc shunt motor and to predetermine efficiency while machine is running as a motor and generator.


S. No APPARATUS TYPE RANGE QUANTITY




3. Rheostat - 250/1.2A 1 4. Tachometer - - 1

5. Connecting wires - - few

b)EXERCISE:

c) 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 starting.


25

d)PROCEDURE:

o Connections are given as per the circuit diagram.

o Close the MCB switch by energizing the circuit. o Start the machine using three-point starter and allow the motor to

run at its rated speed by adjusting the field rheostat. o The corresponding voltmeter and ammeter readings are noted at

no-load. o Bring the field rheostat to its original minimum position and the

MCB switch is switched off. o Now the armature resistance test is conducted as per the circuit

diagram and the corresponding meter readings are noted. o The readings are tabulated and the efficiency of the machine when

it runs as a motor and generator are determined without loading the machine


FOR MOTOR:

Input Power, Pi = VLIL (watts)

Armature current,Ia = IL IF (amps)

Copper loss, Wcu = Ia2Ra (watts)

Total losses, PL = Wc + Wcu (watts)

Constant losses, Wc = ILVL - Ia2Ra (watts)

Output Power, Po = Pin PL (watts)

% Efficiency = output power (Po)/ Input power (Pi)

FOR GENERATOR:

Input Power, PO = VLIL (watts)

Armature current,Ia = IL+ IF (amps)

Copper loss, Wcu = Ia2Ra (watts)

Total losses, PL = Wc + Wcu (watts)


26

Constant losses, Wc = ILVL - Ia2Ra (watts)

Input Power, Pi = Po + PL (watts)

% Efficiency = output power (Po)/ Input power (Pi)

f)CIRCUIT DIAGRAM:


27

g)TABULATION TO FIND OUT CONSTANT LOSS (Wco)

S. No

No-load current

(Io)

Field Current (If)

Terminal Voltage

(V)

No load Armature

current (Iao)

Constant Loss

Wco=VIo- Iao2

Amps Amps Volts Volts Watts

h)CIRCUIT DIAGRAM

TO FIND RA


28

TABULATION TO FIND OUT ARMATURE RESISTANCE

S. No

Armature current

(Ia)

Armature Voltage

(Va)

Aramture Resistance

Ra = Va/Ia

Amps Volts Ohms

TABULATION TO FIND OUT THE EFFICIENCY

RUNNING AS MOTOR

Armature Resistance (Ra) = Rated current (Ir) =

Constant loss (Wco) = Field current (If) =

Terminal Voltage (V) =

S. No

Fraction of load

(x)

Load current

IL= x*Ir

Armature current

Ia =IL+If

Armature loss

Wcu=Ia2Ra

Total loss

Wt

Input power

Wi=VIL

Output power

Wo=Wi-Wt

Efficiency

= Wo/Wi

Amps Amps Amps Watts Watts Watts %


29

RUNNING AS GENERATOR

Armature Resistance (Ra) = Rated current (Ir) =

Constant loss (Wco) = Field current (If) =

Terminal Voltage (V) =

S. No

Fraction of load

(x)

Load current

IL= x*Ir

Armature current

Ia =IL+If

Armature loss

Wcu=Ia2Ra

Total loss

Wt

Output power

Wo=VIL

Input power

Wi=Wo+Wt

Efficiency

= Wo/Wi

Amps Amps Amps Watts Watts Watts %


30

i)DISCUSSION QUESTIONS:

1. What are the principal advantages of rotating field type construction? Relatively small amount of power required for field system can easily supplied to rotating system using slip rings and brushes, more space is available in the stator part of the machine to provide more insulation, it is easy to provide cooling system, stationary system of conductors can easily be braced to prevent deformation. 2. What are the advantages of salient type pole construction used in sy.machines? They allow better ventilation, the pole faces are so shaped radial air gap length increases from pole center to pole tips so flux distortion in air gap is sinusoidal so emf is also sinusoidal. 3. Which type of sy.generators are used in hydroelectric plants and why? As the speed of operation is low, for hydro turbines used in hydroelectric plants, salient pole type sy.generator is used because it allows better ventilation also better than smooth cylindrical type rotor. 4. Why are alternators rated in KVA and not in KW? As load increases I2R loss also increases, as the current is directly related to apparent power delivered by generator, the alternator has only their apparent power in VA/KVA/MVA as their power rating. 5. Why the sy.impedance method of estimating voltage regulation is considered as pessimistic method? Compared to other method, the value of voltage regulation obtained by this method is always higher than the actual value so it is called as pessimistic method.

RESULT:

Thus the Swinburnes tests of DC shunt motor are determined.


31


Title of the Experiment: SPEED CONTROL OF DC SHUNT MOTOR



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

1. Armature control method 2. Field control method





3. Rheostat - 250/1.2A 1 4. Rheostat - 50/5A 1 5. Tachometer Analog - 1


b)EXERCISE:


32

b) 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 starting.

d)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.


33

e)CIRCUIT DIAGRAM:


34

f)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)

g)MODEL GRAPH:


35

h)DISCUSSION QUESTIONS:

1. Why DC series motor should never be started on no-load? When the load current Ia falls to a small value, speed become dangerously high. Hence a DC series motor should never be started without some mechanical load.

2. Define slip. s= (Ns-Nr)/Ns, where, Ns is synchronous speed in rpm Nr is rotor speed in rpm.

3. Define synchronous speed. Three phase balanced power supply is fed to the three stator winding of three

phase induction motor. It creates synchronously rotating magnetic field. The speed of this rotating field is called synchronous speed. Its given by Ns = 120f/p.

4. What are the conditions to be fulfilled by for a dc shunt generator to build back emf? The generator should have residual flux, the field winding should be connected in such a

manner that the flux setup by field in same direction as residual flux, the field resistance should be less than critical field resistance, load circuit resistance should be above critical resistance. 5. Define armature reaction in dc machines? The interaction between the main flux and armature flux cause disturbance called as armature reaction.


36

RESULT:

Thus the speed controls of DC shunt motor are determined.


Title of the Experiment: OPEN CIRCUIT AND SHORT CIRCUIT TEST ON A SINGLE PHASE TRANSFORME



To calculate the equivalent circuit parameter by conducting open circuit and short circuit test on single-phase transformer.

i) To predetermine the efficiency and regulation of the given transformer from equivalent circuit parameter.


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


37


b)EXERCISE:

c)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.

d)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:

1. Connections are given as per the circuit diagram. 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.


38


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 = No load primary current

0 = Phase angle at no load

2. Magnetizing current Im = I0 sin0 Loss compensating current Iw = I0cos 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


39

X01=Z012-R012

2. R02=k*R01 ; X02=k2X01; Z02=k2Z01

Where,

Vsc =Short-circuiting voltage in volts.

f)CIRCUIT DIAGRAM:


40

g)TABULATION:

NO LOAD TEST:

S. No.

Open circuit primary

Voltage (V0)

No load current

(I0)

No load power

(W0)

volts Amps Watts


41

SHORT CIRCUIT TEST:

S. No. Vsc Isc Wsc

Volts Amps Watts

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)/V20 Where,

V20 = Open circuit voltage at secondary

+ = for lagging load

- = for leading load


42

h)MODEL GRAPH:

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


43

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

i)DISCUSSION QUESTIONS

1. Does transformer draw any current when secondary is open? Why?


44

Yes, it (primary) will draw the current from the main supply in order to magnetize the core and to supply for iron and copper losses on no load. There will not be any current in the secondary since secondary is open.

2. List the different methods of speed control applicable to 3 phase slip ring induction motor?

(i) Rotor resistance control (ii) Cascade control (iii)Slip power recovery scheme

3. How speed is achieved by v/f control in 3 phase induction motor? In v/f method, motor can be varied by varying voltage and frequency but we can maintain v/f ratio is constant.

4. Define all day efficiency of a transformer? It is computed on the basis of energy consumed during a certain period, usually a day of

24 hrs. All day efficiency=output in kWh/input in kWh tor 24 hrs.

RESULT:

Thus the open circuit and short circuit test on single phase transformer are

determined.


45


Title of the Experiment: LOAD TEST ON SINGLE PHASE TRANSFORMER



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


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


b)EXERCISE:


46

c)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.

d)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.


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.


47

f)CIRCUIT DIAGRAM:

DISCUSSION QUESTIONS:

1. Mention the difference between core and shell type transformers? In core type, the windings surrounded the core considerably and in shell type the core

surround the windings i.e winding is placed inside the core. 2. What is the purpose of laminating the core in a transformer? To reduce the eddy current loss in the core of the transformer. 3. Give the emf equation of a transformer and define each term?

Emf induced in primary coil E1= 4.44fmN1 volt emf induced in secondary coil E2 =4.44 fmN2.

F - Freq of AC input m - maximum value of flux in the core

N1, N2 - Number of primary & secondary turns. 4. What are the typical uses of auto transformer?

1. To give small boost to a distribution cable to correct for the voltage drop. 2. As induction motor starter.

RESULT:

Thus the load test on single phase transformer is determined.


48

Exercise Number: 05

Title of the Experiment: REGULATION OF THREE PHASE ALTERNATOR BY EMF AND MMF METHOD



To predetermine the voltage regulation of the three-phase alternator by EMF and

MMF methods by conducting open circuit and short circuit tests.


Sl. No. Apparatus Type Range Qty.

1 Voltmeter MI (0-600) V 1

2 Ammeter MC (0-2) A 2

3 Ammeter MI (0-10) A 1

4 Rheostat Wire Wound 570/1.2A 1 5 Rheostat Wire Wound 270/1.2A 1 6 Tachometer Analog 1

7 SPST Switch 1

8 TPST Switch 1

9 Connecting wires 12

b)EXERCISE:


49

c)PRECAUTIONS:

1. The motor field rheostat should be kept in minimum resistance position. 2. The Alternator field rheostat should be in maximum resistance position. 3. Initially all switches are kept in open position

.

d)PROCEDURE:

OPEN CIRCUIT TEST

1

. The connections are made as per the circuit diagram.

2. The D.C. motor is started and made to run at its rated speed of alternator by

adjusting the field rheostat of motor.

3. The alternator field is excited by closing the SPST.

4. By varying the field rheostat the voltage across the alternator is increased upto its

rated voltage.

5. The open circuit test is conducted by varying the field rheostat of the alternator and

the corresponding alternator voltage and field current values are noted.

6. Open circuit characteristics are drawn by taking the field current along X axis and

the phase voltage along Y axis.

SHORT CIRCUIT TEST

1

. The TPSTS is closed and by adjusting the field rheostat of the alternator till the

voltmeter reads zero, ,.

2. The short circuit current and the field current are noted.


50

3. Short circuit characteristics are drawn by taking the filed current along X axis and

the short circuit current along Y axis.

STATOR RESISTANCE TEST

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

2. By adjusting the load, the ammeter and voltmeter readings are tabulated.

3. From the tabulated values the stator resistance of the alternator is calculated.


Let,

E0=no load induced emf in volts

cos= Power Factor Reff=1.6Ra ohms

Zs= E0/Isc ohms

Xs=Zs2- Reff2 ohms 1. E0 at unity power factor

E0= (Vph+IReff)2 + (IXs) 2 volts.

2. E0 at lagging power factor

E0= (Vphcos +IReff)2 + (Vphsin+IXs)2 volts Vph=VL/3 volts

3. E0 at leading power factor


51

E0= (Vphcos +IReff)2 + (Vphsin -IXs)2 volts

4. % Regulation= ( E0-Vph)/ Vph *100

f)CIRCUIT DIAGRAM:


52


53

g)TABULATION:

OPEN CIRCUIT TEST:

Sl. No. Field Current If

in Amps

Open circuit voltage VL

in volts

Phase Voltage VPH

in volts

SHORT CIRCUIT TEST:

EMF METHOD:

MMF METHOD:

Sl. No. Short circuit current ISC in Amps Field Current If in Amps

Sl.

No.

Power

Factor

Eo in volts % Regulation

Lagging Leading Lagging Leading


54

Sl.

No.

Power

Factor

Field Current If in Amps

Eo in volts % Regulation

Lagging Leading Lagging Leading Lagging Leading

h)TO FIND Ra:

TABULATION:


55

Sl. No. Armature Current

in Amps

Armature Voltage

in Volts

Armature resistance

Ra=V/I in Ohms

i)MODEL GRAPHS:

EMF Method:

MMF Method:


56

j)DISCUSSION QUESTIONS:

1. Write down the emf equation for d.c.generator? E= (NZ/60) (P/A) V. P - No. of poles Z - Total no. of conductor - Flux per pole, N-speed in rpm. 2. Why commutator is employed in d.c.machines?

Conduct electricity between rotating armature and fixed brushes, convert alternating emf into unidirectional emf (mechanical rectifier)

3. Distinguish between shunt and series field coil construction? Shunt field coils are wound with wires of small section and have more no. of turns.

Series field coils are wound with wires of larger cross section and have less no. of turns. 4.What is back emf in d.c. motor?

As the motor armature rotates, the system of conductor come across alternate north and south pole magnetic fields causing an emf induced in the conductors. The direction of the emf induced in the conductor is in opposite to current. As this emf always opposes the flow of current in motor operation it is called as back emf. 5.Why MMF method of estimating voltage regulation is considered as optiimisticmethod?

Compared to EMF method, MMF method involves more noof complex calculation steps. Further the OCC is referred twice and SCC is referred once while predetermining the voltage regulation for each load condition. Reference of OCC takes core saturation effect. As this method require more effort, final result is very close to actual value, hence this method is called as optimistic method. RESULT:

Thus the three phase alternators are regulated by EMF & MMF method.


57

Exercise Number: 06

Title of the Experiment: LOAD TEST ON THREE-PHASE SQUIRREL CAGE INDUCTION MOTOR



To determine the performance characteristics of three-phase squirrel cage

induction motor by direct loading method.


Sl. No. Apparatus Type Range Quantity


2 Ammeter MI (0-10)A 1

3 Single element wattmeter LPF 600V,10A 2

4 3 Auto transformer 400V/(0-450)V 1

5 Tachometer Analog 1

6 Connecting wires As per required

b)EXERCISE:

c)PRECAUTIONS:


58

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.

d)PROCEDURE:

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


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 = 2NT/60 watts Where,

N=speed of the motor in rpm

3. % Efficiency = (Output power/Input power)*10


59

f)CIRCUITDIAGRAM:


60

g)TABULATION:

MF=---

Sl.

No.

Line

Voltage

VL

in volts

Line

Current

IL

in amps

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 %

%


61

h)MODEL GRAPH:

Electrical Characteristics:

DISCUSSION QUESTIONS:

1. What are the advantages of cage motor? Since the rotor has low resistance, the copper loss is low and efficiency is very high. On

account of simple construction of rotor it is mechanically robust, initial cost is less, maintenance cost is less, simple starting arrangement. 2. What are the 2 types of 3phase induction motor?

Squirrel cage and slip ring induction motor. 3. Write two extra features of slip ring induction motor?

Rotor has 3phase winding. Extra resistance can be added in rotor circuit for improving PF with the help of three slip rings. 4. Where three point starter and four point starter are recommended?

3 Point starter- For armature speed control of DC motor(below rated speed). 4 point starter- For field speed control of DC motor(above rated speed).

5. What are the methods of starting 3 phase squirrel cage induction motor? DOL starter, primary resistance starter, star-delta starter, auto-transformer starter.

RESULT:

Thus the load test on three phase induction motor are determined.


62

Exercise Number: 07

Title of the Experiment: NO LOAD TEST AND BLOCKED ROTOR TEST ON THREE PHASE INDUCTION MOTOR



To determine the equivalent circuit parameters of three phase induction motor by conducting

i) No load test and ii) Blocked rotor test.


Sl. No. Apparatus Type Range Quantity



3 Ammeter MI (0-5)A 1

4 Ammeter MI (0-10) A 1

5 Single element wattmeter UPF 600V, 5A 2

6 Double element wattmeter LPF 300V, 10A 2

7 3 Auto transformer 400V/(0-450) V 1

8 Connecting wires

b)EXERCISE:


63

c)PRECAUTIONS:

1. The autotransformer should be kept at minimum voltage position.

2. Blocking of the rotor should be done properly.

3. Making sure that only a small voltage is applied across the stator terminal with

rotor blocked.

NO LOAD TEST:

1. The connections are given as per the circuit diagram. 2. The supply is switched ON by closing the TPST switch. 3. The autotransformer is varied until the voltmeter reads the rated voltage of the

machine and the motor starts running at no load. 4. The readings of Voltmeter, ammeter and wattmeter are noted at no load. 5. Then the TPST supply switch is opened.

BLOCKED ROTOR TEST:

1. The connections are given as per the circuit diagram. 2. The rotor of the motor is blocked tightly with the help of brake drum and belt. 3. The supply is switched ON by closing the TPST switch. 4. The autotransformer is adjusted slowly until the voltmeter reads the rated current

of the machine 5. The readings of voltmeter, ammeter and wattmeters are noted.

6. Now release the blocked rotor and open the TPST supply switch.


64


NO LOAD TEST:

1. No load power factor, cos0=W0/(3 V0I0) Where,

W0=No load power in watts

V0=No load voltage in volts

I0=No load current in amps

2. Working component current, Iw=I0 cos0 Amps

3. Magnetizing current,Im= I0 sin0 Amps

4. No load resistance,R0=V0/Iw ohms

5. No load reactance,X0=V0/Im ohms BLOCKED ROTOR TEST:

1. At short circuit power factor,cosSC=WSC/(3VSCISC) Where,

WSC=Short circuit power in watts

VSC=Short circuit voltage in volts

ISC= Short circuit current in Amps

2. Motor equivalent impedance as referred to stator, Z01= VSC/(3ISC) ohms 3. Motor equivalent resistance as referred to stator, R01= Z01 cosSC ohms 4. Motor equivalent reacttance as referred to stator, X01= Z01 sinSC ohms 5. R01 = R01 Reff ohms

6. Reff = 1.6 * Ra ohms

Where,


65

Ra=Armature resistance offered for D.C

7. Equivalent load resistance (RL)= R01 ((1/S)-1)) ohms

Where,

Slip (S)=(Ns-N)/ Ns

Ns=Synchronous Speed in rpm

N=Speed of the rotor in rpm.

8. Effective resistance=R-01+ RL+ X01 ohms

9. Rotor current referred to stator side, I2=(V/3)/Effective resistance 10. Stator current, I1=I0+ I2 Amps


66

f)CIRCUIT DIAGRAM:


67


68

g)TABULATION:

NO LOAD TEST:

Speed of the induction motor: Type of the stator connection:

Multiplication factor:

No load voltage

V0 in volts

No load current

I0 in amps

No load power input P0 in watts

Observed reading

(OR)

Actual reading

AR=OR*MF

MF=

BLOCKED ROTOR TEST:

Type of the stator connection: Multiplication factor:

Line voltage

VSC in volts

Short circuit

current

ISC in amps

Power input PSC in watts

Observed reading

(OR)

Actual reading

AR=OR*MF

MF=


69

Tabulation:

Sl. No.

Armature Current

in Amps

Armature Voltage

in Volts

Armature resistance

Ra=V/I in Ohms


70

h) DISCUSSION QUESTIONS:

1. Why an induction motor never runs at its synchronous speed? If it runs at sy. Speed then there would be no relative speed between the two, hence no

rotor emf, so no rotor current, then no rotor torque to maintain rotation. 2. What are slip rings?

The slip rings are made of copper alloys and are fixed around the shaft insulating it. Through these slip rings and brushes rotor winding can be connected to external circuit. 3. What are the principal advantages of rotating field type construction?

Relatively small amount of power required for field system can easily supplied to rotating system using slip rings and brushes, more space is available in the stator part of the machine to provide more insulation, it is easy to provide cooling system, stationary system of conductors can easily be braced to prevent deformation. 4. In which direction a shaded pole motor runs? The rotor starts rotation in the direction from unshaded part to the shaded part. 5. Why single phase induction motors have low PF? The current through the running winding lags behind the supply voltage by large angle so only single phase induction motors have low PF.

RESULT:

Thus the no load and blocked rotor test of three phase induction motor are determined.


71

Exercise Number: 09

Title of the Experiment: DIGITAL SIMULATIONS OF LINEAR SYSTEMS



To digitally simulate the time response characteristics of a linear system without

non linearities and to verify it manually.

a)FACILITIES REQUIRED

Mat lab Software

b)THEORY

The time characteristics of control systems are specified in terms of time domain specifications. Systems with energy storage elements cannot respond instantaneously and will exhibit transient responses, whenever they are subjected to inputs or disturbances.

The desired performance characteristics of a system of any order may be specified in terms of transient response to a unit step input signal. The transient response characteristics of a control system to a unit step input is specified in terms of the following time domain specifications

Delay time td Rise time tr Peak time tp Maximum overshoot Mp Settling time ts

c)PROCEDURE:

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. 4. From the sink block in the library drag thescope. 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. 9. From the step response obtained note down the rise time, peak time, peak overshoot and settling time.


72

10. For the same transfer function write a matlab program to obtain the step response and verify both the results.

d)PROGRAM

%This is a MATLAB program to find the step response

num=[0 0 25];

den=[1 6 25];

sys = tf (num,den);

step (sys);

e)PLOT


73


1.What is meant by Delay time ? It is the time to required for the response to reach 50% of final value in first attempt

2.What is meant by Rise time? It is the time to required for the response to rise from 10% to 90% of the final value

3.What is meant by over damping? It is the time to required for the response to rise from 0% to 100% of the final value

4.What is meant by peak time? It is the time to required for the response to reach peak time response

5.What is meant by setting time? It is the time to required for the response to reach an stay within a specific tolerance band

RESULT:

Thus the digital simulation of linear system are determined.


74


Title of the Experiment: STABILITY ANALYSIS OF LINEAR SYSTEMS (Bode Plot) Date of the Experiment :


To obtain the bode plot for the given system whose transfer function is given as

G(S)= 242(s+5)

s(s+1)(s2+5s+121)

and to find out whether the system is stable or not.


Mat lab Software

b)THEORY

A Linear Time-Invariant Systems is stable if the following two notions of system stability are satisfied

When the system is excited by Bounded input, the output is also a Bounded output.

In the absence of the input, the output tends towards zero, irrespective of the initial conditions.

The following observations are general considerations regarding system stability and are

If all the roots of the characteristic equation have negative real parts, then the impulse response is bounded and eventually decreases to zero, then system is stable.

If any root of the characteristic equation has a positive real part, then system is unstable.

If the characteristic equation has repeated roots on the j-axis, then system is unstable.

If one are more non-repeated roots of the characteristic equation on the j-axis, then system is unstable.

c)BODE PLOT :

Consider a Single-Input Single-Output system with transfer function

C(s) b0 sm + b1 sm-1 + + bm

=

R(s) a0 sn + a1sn-1 + +an


75

Where m < n.

Rule 1 A system is stable if the phase lag is less than 180 at the frequency for which the gain is unity (one).

Rule 2 A system is stable if the gain is less than one (unity) at the frequency for which the phase lag is 180.

The application of these rules to an actual process requires evaluation of the gain and

phase shift of the system for all frequencies to see if rules 1 and 2 are satisfied. This is

obtained by plotting the gain and phase versus frequency. This plot is called BODE

PLOT. The gain obtained here is open loop gain.

The stability criteria given above represent Limits of Stability. It is well to design a

system with a margin of safety from such limits to allow for variation in components and

other unknown factors. This consideration leads to the revised stability criteria, or more

properly, a Margin of Safety provided to each condition. The exact terminology is in

terms of a Gain Margin and Phase Margin from the limiting values quoted.

If the phase lag is less than 140 at the unity gain frequency, the system is stable. This then, is a 40 Phase Margin from the limiting values of 180.

If the gain is 5dB below unity (or a gain of about 0.56) when the phase lag is 180, the system is stable. This is 5dB Gain Margin.

d)ALGORITHM

1. Write a Program to (or using SIMULINK) obtain the Bode plot for the given system. 2. Access the stability of given system using the plots obtained.

e)PROGRAM

%BODE PLOT OF THE SYSTEM%

%Enter the numerator and denominator of the transfer function

num=[0 0 0 242 1210];

den=[1 6 126 121 0];

sys=tf(num,den)

%Specify the frequency range and enter the command


76

w=logspace(-2,4,1000);

bode(sys,w)

xlabel('Frequency')

ylabel( ' Phase angle in degrees Magnitude of G(s)')

title('Bode Plot of the system 242(s+5)/s(s+1)(s^2+5*s+121)')

%To determine the Gain Margin,Phase Margin, Gain crossover frequency and

%Phase cross over frequency

[ Gm, Pm, Wcp, Wcg ]= margin (sys)

f)PROCEDURE TO OBTAIN BODE PLOT

1. Rewrite the sinusoidal transfer function in the time constant form by replacing s by j

2. Identify the corner frequencies associated with each factor of the transfer

function.

3. Knowing the corner frequencies draw the asymptotic magnitude plot. This plot

consists of straight line segments with line slope changing at each corner frequency by

+20db/decade for a zero and -20db/decade for a pole. For a complex conjugate zero or pole

the slope changes by + 40db/decade.

4. Draw a smooth curve through the corrected points such that it is asymptotic to the

line segments. This gives the actual log-magnitude plot.

5. Draw phase angle curve for each factor and add them algebraically to get the

phase plot.

G)MANUAL CALCULATIONS

i)The sinusoidal transfer function G (j) is obtained by replacing s by j in the given s domain transfer function

G(j)= 242(j +5) j (j +1)( j 2+5 j +121) On comparing the quadratic factor of G(s) with standard form of quadratic


77

factor , and n can be evaluated. s2+5s+121 = s2+2ns + n2 On comparison

n2 = 121 2n= 5 n =11 rad/sec = 0.227 G(j)= 10(1+0.2j) j (1+j)( 1+0.4 j -0.0083 2) ii)CORNER FREQUENCIES

The corner frequencies are c1=1rad/sec c2= 5 rad/sec and c3=11rad/sec Choose a low frequency l such that l< c1 and choose a high frequency h> c3. Let l=0.5 rad/sec and h=100 rad/sec

Term Corner Frequency

rad/sec

Slope db/dec Change in slope db/dec

10

j __ -20 __

1

(1+j) 1 -20 -20-20= - 40

(1+0.2j) 5 20 -40-20 = -20 1

( 1+0.4 j -0.0083 2) 11 -40 -40-20 = -60

Iii) MAGNITUDE PLOTS

Calculate A at l, c1, c2, c3, and h Let A= | G(j)| in db At = l A= 20 log(10/0.5)=26.03db = c1 , A=20log(10/1)=20db


78

= c2 A= -40log(5/1)+20=-7.96 db = c3 A = -20log(11/5) - 7.96 = -14.80 db = h A = -60log(100/11)-14.80 = - 72.3 db These values are plotted in the semilog graph sheet taking frequency along the logarithmic scale and magnitude in db along the linear scale

iv) PHASE PLOT

The phase angle of G(j) as a function of is given by = G(j) = tan-1 0.2 -90 tan-1 tan-1 0.04/(1 0.00832) tan-1 0.2 tan-1 tan-1 {0.04/

(1 0.00832)} = G(j)

0.5 507 26.56 1.15 -112

1 11.3 45 2.31 -126.01

5 45 78.96 14.04 -138

10 63.43 84.29 63.44 -174.3

11 65.5 84.8 85.8 -195.4

20 75.96 87.14 180-19.98=160 -261.18

50 84.3 88.85 180-6=174 -268.55

100 87014 89.43 180-2.9=177.1 -269.3

These values are plotted in the semilog graph sheet taking the same frequency as before along the logarithmic scale and phase angle in degrees along the linear scale.

h)OUTPUT (from simulation)

242 s + 1210

-----------------------------

s^4 + 6 s^3 + 126 s^2 + 121 s


79

Gm = 2.0273

Pm = 41.8270

Wcp = 10.0961

Wcg = 3.6322

OUTPUT (from graph)

gc= gc=3.1rad/sec Phase margin =180+ gc = 180-134 = 46 degrees Gain Margin = 12 db

pc = 10.1 rad/sec


80

i)BODEPLOT


81

j)DISCUSSION QUESTIONS 1. What is polar plot?

As the input frequency is varied from 0 to ,the magnitude of M and phase angle is change and hence the tip of the phase G(jw) trance a locus in the complex plane

2. What is Nyquist contour? In order to investigate ther presence of any zero of q(s)=1+G(s)H(s) in the right of s plane , let us chose a contour which completely enclosed this right half of the s plane .such a contour

3. What is frequency response?

A frequency response is the steady state response of a system when the input to the system is asinusoidal signal.

4. List out the different frequency domain specifications? The frequency domain specification are i) Resonant peak. ii) Resonant frequency.

5. Define resonant Peak? The maximum value of the magnitude of closed loop transfer function is called resonant peak.

RESULT:

Thus the stability analysis of linear system using bode plot are determined.


82


Title of the Experiment: STABILITY ANALYSIS OF LINEAR SYSTEMS (ROOT LOCUS PLOT) Date of the Experiment :


To obtain the Root locus plot for the given system whose transfer function is given as

G(S)= K

s(s+3)(s2+3s+11.25)


Mat lab Software

b)THEORY ROOT LOCUS PLOT :

The characteristic of the transient response of a closed-loop system is related to the location of the closed loop poles. If the system has a variable loop gain, then the location of the closed-loop poles depend on the value of the loop gain chosen. A simple technique known as Root Locus Technique used for studying linear control systems in the investigation of the trajectories of the roots of the characteristic equation.

This technique provides a graphical method of plotting the locus of the roots in the s-plane as a given system parameter is varied over the complete range of values(may be from zero to infinity). The roots corresponding to a particular value of the system parameter can then be located on the locus or the value of the parameter for a desired root location can be determined form the locus. The root locus is a powerful technique as it brings into focus the complete dynamic response of the system . The root locus also provides a measure of sensitivity of roots to the variation in the parameter being considered. This technique is applicable to both single as well as multiple-loop systems.


83

c)PROCEDURE:

1. Write a Program to (or using SIMULINK) obtain the Root locus plot for the given system.

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

d)PROGRAM

%ROOT LOCUS OF THE SYSTEM%

num=[0 0 0 0 1]

den=[1 6 20.25 33.75 0]

sys=tf(num,den)

rlocus(sys)

v=[-10,10,-8,8];

axis(v)

xlabel('Real Axis')

ylabel('Imaginary Axis')

title('Root Locus of the sytem ')

title('Root Locus Plot of the system K/s(s+3)(s^2+3s+11.25))')

e)MANUAL CALCULATIONS

1. Number of poles =4, zeros = 0, number of root locus branches =4. Starting points s=0, -3 & 1.5+ j3.

2. Pole zero plot is as follows

Section between 0 and -3 is part of root locus. One breakway point is between s=0 and s=-3.

3. Angle of asymptotes are 45,135,225 and 315 degrees

4. Centroid = -1.5

5. Three Breakway points are -1.5,-1.5 + j 1.8371


84

6. Intersection with imaginary axis s= + j2.37.

7. Angle of departure -90, +90.

8. Root locus is plotted.

9. Stability for 0< K82.26 system is unstable

f)OUTPUT

num = 0 0 0 0 1

den = 1.0000 6.0000 20.2500 33.7500 0

Transfer function:

1

---------------------------------

s^4 + 6 s^3 + 20.25 s^2 + 33.75 s

g)GRAPH(from Simulation)


85

h)DISCUSSION QUESTIONS

1. Define root locus technique. Its a method of poltting the locus of the root of the characteristic equation in s plane when the gain of the system is varied over the entire range

2. What do you mean by Root-Loci? It is a the locus of the root traced out on the s-plane as its parameter is changed

3. Write the condition for magnitude & angle criterion? G(S)H(S) =1 & G(S)H(S)=1800

4. define the break way point ? its obtain the as root s of characteristic equation after making dK /ds=0

5. Define Resonant frequency? The frequency at which resonant peak occurs is called resonant frequency.

6. What is bandwidth? The bandwidth is the range of frequencies for which the system gain Is more than 3 db. The bandwidth is a measure of the ability of a feedback system to reproduce the input signal , noise rejection characteristics and rise time.

RESULT:

Thus the stability analysis of linear system using locus plot are determined.


86

Exercise Number: 10(C)

Title of the Experiment : STABILITY ANALYSIS OF LINEAR SYSTEMS Date of the Experiment :


To obtain the Nyquist plot for the given system whose transfer function is given as

G(S)= 50

(s+4)(s2+3s+3)

and to find out whether the system is stable or not.


Mat lab Software

b)THEORY

POLAR PLOTS OR NYQUIST PLOTS:

The sinusoidal transfer function G(j) is a complex function is given by G(j) = Re[ G(j)] + j Im[G(j)] or G(j) = G(j) G(j) = M -----------(1) From equation (1), it is seen that G(j) may be represented as a phasor of magnitude M and phase angle

. As the input frequency varies from 0 to , the magnitude M and phase angle changes and hence the tip of the phasor G(j) traces a locus in the complex plane. The locus thus obtained is known as POLAR PLOT.

The major advantage of the polar plot lies in stability study of systems. Nyquist related the stability of a system to the form of these plots. Polar plots are referred as NYQUIST PLOTS.

NYQUIST stability criterion of determining the stability of a closed loop system by investigating the properties of the frequency domain plot of the loop transfer function G(s) H(s).

Nyquist stability criterion provides the information on the absolute stability of a control system as similar to Routh- Hurwitz criterion. Not only giving the absolute stability, but indicates Degree of Stability i.e


87

Relative Stability of a stable system and the degree of instability of an unstable system and indicates how the system stability can be improved. The Nyquist stability citerion is based on a Cauchys Residue Theorem of complex variables which is referred to as the principle of argument.

Let Q(s) be a single valued function that has a finite number of poles in the s-plane. Suppose that an arbitrary closed path q is chosen in the s-plane so that the path does not go through any one of the poles or zeros of Q(s); the corresponding q locus mapped in the Q(s) plane will encircle the origin as many times as the difference between the number of the zeros and the number of poles of Q(s) that are encircled by the s-plane locus q.

The principle of argument is given by

N= Z - P

Where N number of encirclemnts of the origin made by the Q(s) plane locus q. Z number of zeros of Q(s) encircled by the s-plane locus q in the s-plane. P - number of poles of Q(s) encircled by the s-plane locus q in the s-plane.

c)ALGORITHM

1. Write a Program to (or using SIMULINK) obtain the Nyquist plot for the given system. 2. Access the stability of given system using the plots obtained.

d)PROGRAM

%NYQUIST PLOT


num=[0 0 0 50]

den=[1 7 12 12]

sys=tf(num,den)

%Specify the frequency range and enter the command

nyquist(sys)

v=[-3 5 -7 7]

axis(v)

xlabel('Real Axis');

ylabel('Imaginary Axis');


88

title('Nyquist Plot of the sytem 50/(s+4)(s^2+3s+3)')

%To determine the Gain Margin,Phase Margin, Gain crossover frequency and

%phase cross over frequency

[Gm,Pm,Wcp,Wcg]=margin (sys)

v = -3 5 -7 7

Gm = 1.4402

Pm = 11.1642

Wcp = 3.4643

Wcg = 2.9533

e)MANUAL CALCULATIONS:


89

NYQUISTPLOT

f) DISCUSSION QUESTIONS

1. Define Cut-off rate? The slope of the log-magnitude curve near the cut-off is called cut-off rate. The cut-off rate indicates the ability to distinguish the signal from noise.

2. Define Gain Margin? The gain margin ,kg is defined as the reciprocal of the magnitude of the open loop transfer function at phase cross over frequency. Gain margin kg = 1 /|G(jpc) |

3. Define Phase cross over? The frequency at which, the phase of open loop transfer functions is called phase cross over frequency pc.

4. What is phase margin? The phase margin is the amount of phase lag at the gain cross over frequency required to bring system to the verge of instability.

5. Define Gain cross over? The gain cross over frequency gc is the frequency at which the magnitude of the open loop transfer function is unity..

RESULT:

Thus the stability analyses of linear system are determined.


90

Exercise Number: 11

Title of the Experiment: STUDY THE EFFECT P, PI & PID CONTROLLERS USING MATLAB



To design an electronic Proportional Controller.


1. PID designing unit

2. Patch chords

3. Resistors

b)PROCEDURE:

. For the given input and output voltage ranges calculate the gain, KP.

2. Using the formula KP = R2 / R1 and R2 = R3 = 10 K , find R1.

2. Connect the circuit as per the patching diagram.

3. Apply the specified input voltages Ve & Vb ( 0.5 V) to the terminals T2 and

T1 respectively of the P Controller.

4. Note down the output voltage at T3 corresponding to the variation in input

error voltage..


91

c)EXPRESSION FOR THEROTICAL VALUE:

1. Vout= KP Ve + Vb

where KP is proportional gain

Ve is error voltage

Vb is bias voltage

2. Gain = Output Voltage / Input Voltage

3.KP = R2 / R1

d)TABULAR COLUMN:

S.No. Ve (V) Vb (V) Vout (V)

2. DESIGN OF PI CONTROLLER


To design an electronic PI Controller and to study the time response of the

given system.



2. Patch chords

3. Resistors

4. Capacitors

5. CRO and Probes


92

b)EXPRESSION FOR THEROTICAL VALUE:

1.Vout = KP Ve + KI Vedt

Where KP (due to Proportional band, PB) is Vout / Ve = R2 / R 1

Proportional band % is 100/gain where gain is Output / Input

Vout is proportional band % of change in output

Ve is proportional band % of change in input

KI (due to Integral action) isVout /Ve sec-1 =1 / TI

Vout is 0.1% of change in output

Ve is 1% of change in input

2. TI = R2C1

assume C1 = 100F find R2

3. R1 = R2 / KP

c)PROCEDURE:

1. Obtain the design of values of the resistors for the given input and output

voltage ranges.

2. Connect the circuit as per the patching diagram.

3. Apply the specified input voltages Ve &Vb to the terminals T2 and T1

respectively of the P Controller.

4. Note down the output voltage.


93

d)PROCEDURE FOR TIME RESPONSE CHARACTERISTICS:

e)MODEL CALCULATION:

RESULT:

3. DESIGN OF PID CONTROLLER


To design an electronic PID Controller and to study the time response of the

given system.



2. Patch chords

3. Resistors

1. Patch the circuit as per the patching diagram.

2. Set IVPP of square wave and connect it to the controller through error detector.

3. Patch the output terminal of the controller to the input terminal of the process

4. Observe the input & output voltage waveforms in the CRO.

5. Note the time response characteristics ts , tr , tp.


94

4. Capacitors

5. CRO and Probe

b)EXPRESSION FOR THEROTICAL VALUE:

1.Vout = KP Ve + KI Vedt

Where KP (due to Proportional band) is Vout / Ve = R2 / R 1

Proportional band is 100/gain where gain is Output / Input

Vout is proportional band % of change in output

Ve is proportional band % of change in input

KI (due to Integral action) isVout /Ve sec-1 =1 / TI

Vout is 0.1% of change in output

Ve is 1% of change in input

2. TI = R2C1

assume C1 = 100F find R2

3. R1 = R2 / KP

c)PROCEDURE:

1.Obtain the design of values of the resistors for the given input and output

voltage ranges.

2.Connect the circuit as per the patching diagram.

3.Apply the specified input voltages Ve &Vb to the terminals T2 and T1

respectively of the P Controller.

4.Note down the output voltage.


95

d)PROCEDURE FOR TIME RESPONSE CHARACTERISTICS:

1.Patch the circuit as per the patching diagram.

2.Set IVPP of square wave.

3.Observe the input & output voltage waveforms.

4.Note the time response characteristics ts , tr , tp.

e)MODEL CALCULATION:


1. What is the need for a controller? The controller is provided to modify the error signal for better control action

2. What are the different types of controllers? Proportional controller PI controller PD controller PID controller

3. What is proportional controller? It is device that produces a control signal which is proportional to the input error signal.

4. What is PI controller? It is device that produces a control signal consisting of two terms one proportional to error signal and the other proportional to the integral of error signal.

5. What is PD controller? PD controller is a proportional plus derivative controller which produces an output signal consisting of two time -one proportional to error signal and other proportional to the derivative of the signal.

RESULT: Thus the design of electronic PID controller for the given requirements has been done.


96

Exercise Number: 12

Title of the Experiment : DESIGN AND IMPLEMENTATION OF COMPENSATORS



To design a lag compensator in Bode plot (frequency domain) for the system whose transfer function

G(s) = K

s(s+4)(s+80)


Matlab software

b)PROGRAM:

%BODE PLOT OF THE SYSTEM UNCOMPENSATED AND COMPENSATED(LAG)%


num=[0 0 0 9600];

den=[1 84 320 0];

sys=tf(num,den)

bode(sys)

[Gm,Pm,Wcp,Wcg]=margin (sys)

margin (sys)

hold


num=[0 0 0 63.9 30]

den=[0.042 3.52 13.68 1 0];


97

sys=tf(num,den)

bode(sys)

[Gm,Pm,Wcp,Wcg]=margin(sys)

margin(sys)

c)MANUAL CALCULATIONS

i)Calculation of gain

Given Kv = 30 sec -1

Kv = Lt sG(s)H(s)

s 0

Since the system is unity feedback system H(s) = 1

Therefore Kv = Lt sG(s) = Lt K = K

s 0 s 0 s(s+4)(s+80) 4*80

K = 9600

ii)Bode plot of uncompensated system

Let s= j G(j) = 9600 = 30 s(s+4)(s+80) j(1+0.25 j)(1+0.0125 j)

Magnitude plot

The corner frequencies are c1=4rad/sec and c2= 80 rad/sec Choose a low frequency l such that l< c1 and choose a high frequency h> c3. Let l=1 rad/sec and h=100 rad/sec


98

Term Corner Frequency

rad/sec

Slope db/dec Change in slope db/dec

30

j __ -20 __

1

(1+0.25j) 4 -20 -20-20= - 40

1

( 1+0.0125 j) 80 -20 -40-20 = -60

Calculate A at l, c1, c2, and h Let A= | G(j)| in db At = l A= 20 log(30/1)=29.5db = c1 , A=20log(30/4)=18db = c2 A= -40log(80/4)+18= -34 db = h A = -60log(100/80)+(-34) = - 40 db These values are plotted in the semilog graph sheet taking frequency along the logarithmic scale and magnitude in db along the linear scale

The phase angle of G(j) as a function of is given by = G(j) = - tan-1 0.25 -90 tan-1 0.0125

Rad/sec

1 4 10 50 80 100

degrees

-104 -138 -164 -208 -222 -230


99

These values are plotted in the semi log graph sheet taking the same frequency as before along the logarithmic scale and phase angle in degrees along the linear scale.

iii) Determination of phase margin of compensated system

Let, gc = Phase margin of G(j) at gain cross over frequency (gc) And = Phase margin of compensated system. From the bode plot of uncompensated system we found that gc = -168deg Now = 180 + gc = 180-168=12 deg The system requires a phse marginof atleast 33 degrees but the available phase margin is 12 degrees and so lag compensation should be employed to improve the phase margin.

iv) Choose a suitable value for the phase margin of compensated system.

The desired phase margin, o = 33 deg Phase margin of compensated system, o = d + Let initial choice of = 5 deg

o = d + = 33 + 5 =38 degrees v) Determine the new gain crossover frequency

Let gc = New gain cross over frequency and gc = phase of G(j) at gc Now o = 180 + gcn gcn = o 180 = 38 180 = -142 deg From the bode plot the frequency corresponding to a phase of -142deg is 4.7 rad/sec

New gain crossover frequency gcn = 4.7 rad/sec vi) Determine the parameter From the bode plot, the db magnitude at gcn is 16 db Therefore Mag of G(j) = 16 db Also Agcn = 20 log = 10 Agcn


100

=6.3 vii) Determine the transfer function of lag compensator.

The zero of the compensator is placed at a frequency one-tenth of gcn Zero of the lag compensator, zc = 1/T

T = 10/ gcn = 2.13 Pole of the lag compensator pc = 1/ T pc = (6.3*2.13) = 1/13.419

Transfer function of lag compensator Gc(s) = (s+1/T)/(s+1/ T) = 6.3(1+2.13s)/(1+13.419s)

viii) Determine the open loop transfer function of the compensated system

The gain of the compensator is nullified by introducing an attenuator in series with the compensator, as shown in the diagram

Open loop transfer function of compensated system

Gc(s) = 1 * 6.3(1+2.13s) * 30_____

1/6.3 6.3(1+2.13s)(1+13.419s)

30_____s(1+0.25s)(1+0.0125s)


101

6.3 (1+13.419s) s(1+0.25s)(1+0.0125s)

= ________30(1+2.13s)____________

s(1+13.419s)(1+0.25s)(1+0.0125s)

ix) Determine the actual phase margin of compensated system.

Gc(s) = ___30(1+2.13j)____________________ j (1+13.419 j)(1+0.25 j)(1+0.0125 j) Let 0 = Phase of G(j) and gc0 = phase of Gc(j) at = gcn gc0 = tan-1 (2.13 * 4.7) - 90 tan-1(13.419*4.7)

tan-1(0.25 * 4.7) - tan-1(0.0125 * 4.7) = -147degrees

Actual phase margin of compensated system o = 180 + gcn = 180 147 =33 degree

d)OUTPUT:

Transfer function:

9600

--------------------

s^3 + 84 s^2 + 320 s

Gm = 2.8000

Pm = 13.2591

Wcp = 17.8885

Wcg = 10.5470

Current plot heldnum = 0 0 0 63.9000 30.0000

Transfer function:

63.9 s + 30

------------------------------------


102

0.042 s^4 + 3.52 s^3 + 13.68 s^2 + s

Gm =15.7613

Pm = 39.3844

Wcp = 16.9235

Wcg = 3.5782

e)PLOT


103


1.Write the step by step procedure for plotting the magnitude plot and phase plot of a open loop system represented by the transfer function G(s).

Find croner frequency Calculate gain & phase for various Values of w Sketch the plot

Documents

Electrical Engg & Ctrl Sys-lm(Ece) 13