em -i lab 2008 reg

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

II Year/IV Semester – BE (EEE)

EE2259 - ELECTRICAL MACHINES LABORATORY – I

Manual cum Observation

(As per Anna University)

2008 Regulation

Academic year 2009 - 2010

1

SYLLABUS

EE2259 - ELECTRICAL MACHINES LABORATORY - I

List of Experiments

1. Open circuit and load characteristics of separately and self excited DC

shunt generators.

2. Load characteristics of DC compound generator with differential and

cumulative connection.

3. Load characteristics of DC shunt and compound motor.

4. Load characteristics of DC series motor.

5. Swinburne’s test and speed control of DC shunt motor.

6. Hopkinson’s test on DC motor – generator set.

7. Load test on single-phase transformer and three phase transformer

connections.

8. Open circuit and short circuit tests on single phase transformer.

9. Sumpner’s test on transformers.

10. Separation of no-load losses in single phase transformer.

2

CONTENTS

S.No. Name of the experiment Page No.

1 Load test on dc shunt motor 6

2 Load test on dc series motor 10

3Open circuit and load characteristics of separately excited dc generator 15

4Open circuit and load characteristics of self-excited shunt generator 22

5 Speed control of dc shunt motor 28

6 Load test on single phase transformer 32

7Open circuit and short circuit test onA single phase transformer 36

8 Swinburne’s test 44

9 Load test on dc compound motor 50

10 Load test on dc compound generator 55

11 Hopkinson’s test 59

12 Three phase transformer connections 65

13 Viva voce questions-DC machines 71

14 Viva voce questions-transformers 79

3

INDEX

S.No.

Date Name of the experimentMark

sStaffInitial

Remarks

LOAD TEST ON DC SHUNT MOTOR

4

CIRCUIT DIAGRAM

Model Graph:

Ex. No: Date:

5

LOAD TEST ON DC SHUNT MOTOR

AIM:

To perform load test on the given D.C shunt motor and to obtain the

following performance characteristics.

1. Torque Vs armature current

2. Speed Vs armature current

3. Speed Vs Torque

4. %Efficiency Vs Power output

APPARATUS REQUIRED:

S.NO

APPARATUS TYPE RANGE QUANTITY

1. Voltmeter mc (0-300) V 1

2. Ammeter mc (0-20) A 1

3. Rheostat Variable 350/1.6A 1

4. Tachometer Analog - 1

5. Connecting wires - - As required

FORMULAE TO BE USED:

1. S=S1 – S2 in Kg.

S =spring balance (Weight in Kg)

S1, S2 = spring balance reading

2. T=S*R*9.81 Nm

S= spring force in Kg.

R=Radius of the brake drum in meter.

T=Torque in Nm.

Tabulation: Radius of the brake drum =

6

S.N

Input voltage

Vin

Input curre

nt

Iin

Spring balance Reading

Speed

N Torque

T

Input power

Pin

Output

power

Pout

Efficiency

S1 S2S1 – S2

volts Amps Kg Kg Kg Rpm Nm Watts Watts %

Model Calculation:

3. Pin =Vin * Iin in watts.

7

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

PRECAUTIONS:

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

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 starters.

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

5. Vary the load up to the rated current.

6. Pour some water on brake drum.

7. For each load setting on brake drum, the input voltage input current,

spring balance readings are noted down and tabulated.

8. From the recorded value, calculate input power, output power and

Efficiency and Characteristics curves were drawn.

RESULT:

The load test on the given D.C shunt motor was conducted and its

performance characteristics were drawn.

CIRCUIT DIAGRAM

8

Load test on DC series motor

Ex.No: Date:

9

LOAD TEST ON DC SERIES MOTOR

AIM:

To perform load test on the given D.C series motor and to obtain the

following performance characteristics.

1. Torque Vs armature current 2.Speed Vs armature current

3. Speed Vs Torque 4.Efficiency Vs Power output

APPARATUS REQUIRED:

S.NO APPARATUS TYPE RANGE QUANTITY




4 Connecting wires As required

FORMULAE USED:

1. S=S1 – S2 in Kg. S =spring balance (Weight in Kg)


2. T=S*R*9.81 Nm

S=spring force in Kg.


T=Torque in Nm.






10

S.

NO

Input voltag

e

Vin

Input current

Iin

Spring balance

ReadingSpeed

N

Torque

T

Input power

Pin

Output power

Pout

%

EffficiencyS1 S2

S1 – S2


MODEL CALCULATION:


11


N= Speed in rpm.

T=Torque in Nm.

5. Percentage Efficiency % = Pout/Pin *100

PRECAUTIONS:

1. The motor should be started with some load.

2. Brake drum should be cooled throughout the experiment.

PROCEDURE:


2. Check whether the initial load is set up in the brake drum.

3. By closing the MCB 230v Dc supply is given to the circuit.

4. The motor is get started by using two point starter.

5. Pour some water to the brake drum and vary the load up to the

rated current and for each load setting on brake drum, the input voltage,

input current, spring balance readings are noted down and tabulated.


efficiency and readings are noted down and tabulated.

7. After taking all the observation reduce the load up to 40% of the

rated current and switch off the MCB.

12

RESULT:

The load test on the given D.C series motor was conducted and its

performance characteristics were drawn.

14

Ex.No: Date:

OPEN CIRCUIT AND LOAD CHARACTERISTICS OF SEPERATELY EXCITED DC GENERATOR

AIM:

To conduct no load and load test on the given separately excited DC shunt generator to plot the following characteristic curves,

i) Open circuit characteristic

ii) Internal characteristics

iii) External characteristics

APPARATUS REQUIRED:

S.NO APPARTUS TYPE RANGE QUANTITY


2. Ammeter

mc (0-20) A 1

mc (0-2) A 1

mc (0-10)A 1

3. Rheostatvariable 350/1.1A 1

variable 530/0.8A 1


5.Variable rheostat load

- - 1

6. DPSTS - - 1

7. SPSTS - - 1



1. Eg = V + Ia Ra in volts.

Eg = Generated voltage in volts, V = Terminal voltage in volts.

Ia =Armature current in Amps. Ra= Armature resistance in ohms.

16

TABULATION:

OPEN CIRCUIT TEST:

S.NO.Field current

Output voltage at no load

Amps volts

LOAD TEST:

S.NO

Load voltage Va

Load current IL

Field current IF

Ia=IL Eg = V + Ia Ra

Volts Amps Amps Amps Volts

17

2. Armature current Ia = IL in Amps.

Where,

IL= Line current in Amps.

Ish = Shunt field current in Amps.

3. Input power = Vin * Iin in watts.

Where,

Vin = Input voltage in volts.

Iin = Input current in Amps.

4. % of Efficiency =Pout/ Pin * 100

Where,

Pin = Input power in watts.

Pout = Output power in watts.

Precautions:

There should be no load at the time of starting.

Auto transformer must be kept at minimum position

PROCEDURE:

NO LOAD TEST:

The connections are given as per the circuit diagram.

Verify whether field rheostats of the generator are kept at maximum

position and field rheostat of motor at minimum position.

Give 230V DC supply to the connected circuit by closing MCB.

Adjusting the excitation of field rheostat of the motor so as to make

the motor to run at rated speed.

The ammeter and voltmeter reading of the generator are noted

when SPSTS switch is open.

Then SPSTS switch is closed and excitation of the generator varied

up to its rated value and the corresponding field current, induced

emf are noted.

18

To Find Ra:

TO FIND Ra:

S.NO

Voltage

(Va)

Current

(Ia)Ra=Va/Ia

Volts Amps Ohms

19

LOAD TEST:

Fix the armature voltage of the generator to the rated voltage by

adjusting the field rheostat of the generator.

Close the DPSTS at the load side of the generator by step by step

note down the corresponding readings of terminal voltage, load

current, and shunt field current up to its rated current.

After the reading are noted the load to its initial position and the

switch of the motor.

TO FIND Ra:

Connections are given as per the circuit diagram.

Gradually vary the rectifier unit knob for the corresponding settings

note down the voltmeter, ammeter readings.

From the tabulated value calculate the armature resistance for each

settings.

RESULT:

Thus the no load and load test on the given separately excited DC

shunt generator were conducted to plot the following characteristic

curves,




20

OPEN CIRCUIT AND LOAD CHARACTERISTICS OF SELF-EXCITED SHUNT GENERATOR

CIRCUIT DIAGARM

21

Ex.No: Date:

OPEN CIRCUIT AND LOAD CHARACTERISTICS OF SELF

EXCITED DC SHUNT GENERATOR

AIM:

To conduct no load and load test on the given self excited DC shunt generator to plot the following characteristic curves

i) Open circuit characteristic ii) Internal characteristics


APPARATUS REQUIRED:

S.NO



2. Ammeter

mc (0-20) A 1

mc (0-2) A 1

mc (0-10)A 1

3. Rheostatvariable 350/1.1A 1

variable 530/0.8A 1


5.Variable rheostatic load

variable 5 KW 1

6. DPSTS - - 1

7. SPSTS - - 1



1. Eg = V + IaRa in volts.

Eg = Generated voltage in volts. V = Terminal voltage in volts.

Ia =Armature current in Amps. Ra= Armature resistance in ohms.

22

TABULATION:

OPEN CIRCUIT TEST:

S.NO. Field currentOutput voltage at

no load

Amps volts

LOAD TEST:

S.NO

Load voltage

V

Load current IL

Field current Ish

Ia=IL+IshEg=Va+IaR

a

Volts Amps Amps Amps Volts

23

2. Armature current Ia = IL + Ish in Amps.

Where,

IL= Line current in Amps.

Ish = Shunt field current in Amps.

3. Input power = Vin * Iin in watts.

Where,

Vin = Input voltage in volts.

Iin = Input current in Amps.

4. % of Efficiency =Pout/ Pin * 100

Where,

Pin = Input power in watts.

Pout = Output power in watts.

PRECAUTIONS:

There should be no load at the time of starting.

Auto transformer must be kept at minimum position

PROCEDURE:

NO LOAD TEST:

The connections are given as per the circuit diagram. Verify whether field rheostats of the generator are kept at maximum

position and field rheostat of motor at minimum position. Give 230V DC supply to the connected circuit by closing MCB. Adjusting the excitation of field rheostat of the motor so as to make

the motor to run at rated speed. The ammeter and voltmeter reading of the generator are noted

when SPSTS switch is open. Then SPSTS switch is closed and excitation of the generator varied

up to its rated value and the corresponding field current, induced emf are noted.

24

To Find Ra:

TO FIND Ra:

S.NO

Voltage

(Va)

Current

(Ia)Ra=Va/Ia

Volts Amps Ohms

25

LOAD TEST:

Fix the armature voltage of the generator to the rated voltage by

adjusting the field rheostat of the generator.

Close the DPSTS at the load side of the generator by step by step

note down the corresponding readings of terminal voltage, load

current, and shunt field current up to its rated current.

After the reading are noted the load to its initial position and the

switch off the motor.

TO FIND Ra:

Connections are given as per the circuit diagram.

Gradually vary the loading rheostat knob for the corresponding

settings note down the voltmeter, ammeter readings.

From the tabulated value calculate the armature resistance for each

setting.

RESULT:

Thus the no load and load test on the given self excited DC shunt

generator were conducted to plot the following characteristic curves,




26

CIRCUIT DIAGRAM:

SPEED CONTROL OF DC SHUNT MOTOR

NAME PLATE DETAILS

MODEL GRAPH:

27

Ex.No: Date:

SPEED CONTROL OF DC SHUNT MOTOR

AIM:

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

1. Armature control method

2. Field control method

APPARATUS REQUIRED:

S.NO





3. Rheostat variable350/1.1A 1

50/5A 1



PROCEDURE:


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

3. Using three-point starter, start the motor.

Armature control method (Below rated speed)

1. Keep the field current constant by adjusting the field rheostat

connected in field Circuit.

2. Vary the rheostat connected in armature circuit for each setting.

Note the corresponding armature voltage, armature current and

tabulate it.

3. From the recorded voltage values draw the graph between speed

and armature voltage.

28

TABULATION:

FIELD CONTROL METHOD:

S.

NO

Va1 = volts Va2 = volts

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

ARMATURE CONTROL METHOD:

S.

NO

If1 = amps If2 = amps

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

29

Field control method:

1. For field control method, keep the armature voltage constant by

adjusting the rheostat connected in armature circuit.

2. Vary the rheostat connected in the field circuit and for each

settings note the corresponding field current and speed and

tabulate it.

3. From the recorded value draw the graph between speed and field

current.

RESULT:

Thus the speed of the given DC shunt motor was achieved by,

1. Armature control method

2. Field control method

30

CIRCUIT DIAGRAM

Load test on Single phase transformer

Model Graph:

31

Ex. No: Date:

LOAD TEST ON SINGLE PHASE TRANSFORMER

AIM:

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

characteristics curves.

APPARATUS REQUIRED:

S.NO

APPARATUS RANGE TYPEQUANTIT

Y

1. Voltmeter (0-300) V MI 2

2. Ammeter (0-5) A MI 2

3. Wattmeter 300V/5A UPF 2

4. Transformer 1KVA/230V - 1

5. Connecting wires - -As

required


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

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

Where,

V02 =Secondary voltage on no load.

V2 = Secondary voltage on load.

Precautions:

Auto transformer must be kept at minimum potential point

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

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.

32

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

TABULATION:

S.

NO

Primary

Voltage

Primary

current

Secondary Voltage

Secondary

current

Input power

Output power

% Efficiency

%

Regulation

Volts Amps Volts Amps Watts Watts % %

MODEL CALCULATION:

33

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.

RESULT:

Thus the load test on single-phase transformer was conducted and

characteristics curves were drawn.

34

CIRCUIT DIAGRAM

Open circuit test on Single phase transformer

TABULATION:

OPEN CIRCUIT TEST:

S.NO.

Open circuit primary Voltage

(V0)

No load current

(I0)

No load power

(W0)

volts Amps Watts

35

Ex.No: Date:

OPEN CIRCUIT AND SHORT CIRCUIT TEST ON

A SINGLE PHASE TRANSFORMER

AIM:

1. To calculate the equivalent circuit parameter by

conducting open circuit and short circuit test on single-phase

transformer.

2. To predetermine the efficiency and regulation of the

given transformer from equivalent circuit parameter.

APPARATUS REQUIRED:

S.NO

ITEM TYPE RANGE QUANTITY

1. Voltmeter MI(0-150)V 1

(0-300)V 1

2. Ammeter MI(0-2)A 1

(0-5)A 1

3. WattmeterUPF 300v/5A 1

LPF 75v/5A 1

4.Auto transformer

-2.7kva,10A,240

V1

5.Connecting wires

- - As required


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

36

Short circuit test on Single phase transformer

TABULATION:

SHORT CIRCUIT TEST:

S.NO.

Short circuit voltage

Vsc

Short circuit current

Isc

Short circuit power

Wsc

Volts Amps Watts

37

2. Magnetizing current Im = I0 sin0

Loss compensating current Iw = I0cos 0

3. No load reactance X0 = V1/ Im

4. No load resistance R0 = V1/Iw

Short circuit test:

1. Z01=Vsc/Isc , R01=Wsc/Isc2 X01=Z01

2-R012

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

Where, Vsc =Short-circuiting voltage in volts.

Isc =Short circuit current in Amps.

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 (Iron loss)

Regulation:

% Regulation = xIsc(R02 cos + X02sin)/V20

Where, V20 = Open circuit voltage at secondary

+ = for lagging load ; - = for leading load

38

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

WattsWatt

s% Watts

Watts

WattsWatt

sWatts % %

39

PRECAUTIONS:

Auto transformer must be kept at minimum potential point.

PROCEDURE:

Open circuit test:


2. To conduct open circuit test, HV winding must be open circuited and LV

winding must be supplied with its rated voltage by using single-phase

Transformer.

3. By closing the DPSTS the rated voltage at normal frequency is

impressed across the primary winding.

4. The ammeter, voltmeter and wattmeter readings are noted and

tabulated.

Short circuit test:

1. To conduct the short circuit test, the HV winding must be shorted and

the rated current must be allowed to flow in the primary winding using

single- phase auto transformer.

2. By closing MCB, the supply is given to the transformer and the rated

current is applied to the primary winding.

40

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

41

RESULT:

Thus the equivalent circuit parameter was calculated by conducting

open circuit and short circuit test on single-phase transformer. And also

the efficiency and regulation of the given transformer from equivalent

circuit parameter was predetermined.

42

SWINBURNE’S TEST

CIRCUIT DIAGRAM

MODEL GRAPH:

43

Ex. No: Date:

SWINBURNE’S TEST

AIM:

To conduct Swinburne’s test on dc shunt motor and to

predetermine efficiency while machine is running as a motor and

generator.

APPARATUS REQUIRED:

S.NO APPARATUS TYPE RANGE QUANTITY

1. Voltmeter MC (0-300) V 1

2. AmmeterMC (0-2) A 1

MC (0-15) A 1

3. RheostatWire

wound350Ώ/1.1A 1

4. Tachometer Digital - 1


FORMULA TO BE USED:

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)*100

44

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

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 curren

t

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

η = Wo/Wi

Amps Amps AmpsWatt

sWatts Watts %

45

FOR GENERATOR:

Input Power, PO = 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)

Input Power, Pi = Po + PL (watts)

% Efficiency = output power (Po)/ Input power (Pi)*100

PROCEDURE:


2. By using three-point starter the motor is started to run at the

rated speed.

3. The motor is switched off using the DPST switch

4. After that the armature resistance test is conducted as per the

circuit diagram and the voltage and current are noted for

various resistive loads.

5. After the observation of reading the load is released gradually.

46

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 curren

t

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

η = Wo/Wi

Amps Amps AmpsWatt

sWatts Watts %

PRECAUTIONS:

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

position.

2. The motor should be run in anti clockwise direction.

3. The motor should be at no load condition throughout the experiment.

47

RESULT:

Thus the efficiency on DC shunt motor was predetermined by

conducting Swinburne’s test while machine is running as a motor

and generator.

CIRCUIT DIAGRAM

Load test on DC compound motor

48

MODEL GRAPH

Ex. No: Date:

LOAD TEST ON DC COMPOUND MOTOR

AIM:

49

To conduct the load test on the given DC Compound motor and to

draw the following characteristics curves,

1.Torque Vs armature current

2.Speed Vs armature current

APPARATUS REQUIRED:

S.NO




3. Rheostat variable 350/1.1A 1


5. Connecting wires - - Few


1. S=S1 – S2 in Kg.

S =spring balance (Weight in Kg)


2. T=S*R*9.81 Nm

S=spring force in Kg.


T=Torque in Nm.


50

S.No

Input voltage

Vin

Input curre

nt

Iin

Spring balance

ReadingSpeed

N

Torque

T

Input power

Pin

Output

power

Pout

Efficiency

S1

S2S1 – S2


M0del Calculation:



51





N= Speed in rpm.

T=Torque in Nm.

5. Percentage Efficiency % = (Pout/Pin) *100

PROCEDURE:


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

3. The motor is started by using four point starters.

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

rated speed.

5. Vary the load up to the rated current.

6. Pour some water on break drum.

7. For each load setting on brake drum, the input voltage, input current,

spring balance readings are noted down and tabulated.

8. Repeat the same procedure for differential mode.


efficiency and characteristics curves were drawn.

PRECAUTIONS:

52

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

position.

2. At the time of starting, the motor should be in no load condition.

RESULT:

The load test on the given D.C compound motor was conducted and

its performance characteristics were drawn.

53

CIRCUIT DIAGRAM:

Load test on DC compound generator

MODEL GRAPH:

54

Ex. No: Date:

LOAD TEST ON DC COMPOUND GENERATOR

AIM:

To conduct the load test on a given DC compound generator and to

draw the characteristics curves.

APPARATUS REQUIRED:

S.NOName of the apparatus

Type Range Quantity

1. Ammeter mc (0-20A) 1

2. Voltmeter mc (0-300)V 1

3 Rheostat Wire wound 300,1.7A 1

.4. Rheostat Wire wound 500,1.2A 1

5. Tachometer Digital - 1

6.Variable Resistive Load

- 5kw 1

55

PRECAUTION:

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

2. The generator field rheostat should be kept at maximum position.

3. At the time of starting, the generator should be in no load condition.

4. The machine should be run at its rated speed throughout the experiment.

PROCEDURE:

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

2. The prime mover is started with the help of the three-point starter and

it is made to run at rated speed when the generator is disconnected from

the load by DPST switch.

TABULATION:

S.no

Differential compound Cumulative compound

Load voltage

(volts)

Load current

(amps)

Load voltage

(volts)

Load current

(amps)

56

3. By varying the generator field rheostat gradually, the rated voltage

(Eg) is obtained.

4. The ammeter and voltmeter readings are tabulated at no load

condition.

5. The ammeter and voltmeter readings are observed for different loads

up to the rated current by closing the DPST switch.

6. After tabulating all the readings, the load is brought in its initial

position.

7. The prime mover is switched off using the DPST Switch after bringing all

the rheostat to initial position.

GRAPH:

The graph is drawn between load voltage Vs load current.

57

RESULT:

Thus the load test on a DC compound generator was conducted and

its characteristics curves were drawn.

CIRCUIT DIAGRAM

Hopkinson’s Test

58

NAME PLATE DETAILS

Ex.No : Date:

HOPKINSON’S TEST

AIM:

59

To conduct Hopkinson’s test to find stray losses and hence to

calculate the efficiency of the machines as a motor and generator.

APPARATUS REQUIRED:

S.NO APPARATUS RANGE TYPE QTY

1 Ammeter (0-10A) MC 3

2 Ammeter (0-1A) MC 2

3 Voltmeter (0-300V) MC 2

4 Rheostat 270Ω/0.8A variable 1

5 Rheostat 770Ω/0.6A variable 1

6 Tachometer - Digital 1

7 Connecting wires -As

required

FORMULAE USED;

For stray loss

1. Total input power = VL . IL

Where,

VL - Line voltage in volts

IL – Line current in Amps

2. Total losses for motor and generator is

= Ia2

g.Ra +Ifg.VL+Ia

2m.Ra +Ifm.VL +Stray loss

TABULATION:

S.NoVL IL Iag Ifg Iam Ifm

Volts Amps Amps Amps Amps Amps

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Calculation of stray loss:

S.NoIL PL Pag Pam VL.Ifg VL.Ifm Ws

Amps Watts Watts Watts Watts Watts Watts

Where,

Ia2g.Ra - Generator armature Cu loss

Ifg. VL – Generator field current loss

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Ia2m.Ra - Motor armature Cu loss

Ifm. VL – Motor field current loss

3. Stray loss of each machine

= (VL . IL - Ia2

g.Ra -Ifg.VL-Ia

2m.Ra -Ifm.VL) / 2

For efficiency as a motor

1. Input power to motor = VL(IL+Ifm+Iag)

Ifm – Motor field current

Iag _ Generator armature current

2. Total losses = (Iag + Ifm)2.Ra + VL.Ifm +Wc

For efficiency as a generator

1. Output power of generator = VL.Iag = Pout

2. Generator field current loss = Ifg. VL

3. Generator armature Cu loss = Ia2g.Ra

4. Input power = Output power + Total losses = Pin

5. % Efficiency = (Pout / Pin) * 100

PROCEDURE:

1. Connections are made as per the circuit diagram.

2. With SPST switch is open condition supply is given by closing DPST

switch.

3. Motor is started using four point starter and speed is brought rated

value by adjusting motor field rheostat.

4. The SPSTS is closed when voltmeter connected across reads zero.

The voltmeter reads double the rated value when polarity of motor

and generator are inter changed.

Determination of Efficiency as a motor:

S.NoPinm Ws Pcu Pfm Pom % η

watts watts watts watts watts %

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Determination of Efficiency as a Generator:

S.NoIa Ws Pog Pcu Pfg Ping % η

amps watts watts watts watts watts %

5. Now the field excitation of the generator is varied and all the meter

readings are noted.

6. From the observed readings, the losses of each machine is

calculated and tabulated.

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RESULT:

Thus Hopkinson’s test on given DC machines were conducted and

efficiency are predetermined and the characteristics were drawn for

generator and motor.

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Ex. No: Date:

THREE PHASE TRANSFORMER CONNECTIONS

AIM:

To study about the different three phase connections of three phase transformers.

APPARATUS REQUIRED:

S.NO Apparatus Type Range Quantity

1. Voltmeter MI (0-600)V 3

2. Voltmeter MI (0-300)V 2

3 Auto Transformer 3 phase 415/0-470 V 1

4. SPST Switch - - 1


1. Star connection:

Line Current= Phase current

Line voltage = Phase Voltage * 3

2. Delta Connection:

Line Current= Phase current * 3

Line voltage = Phase Voltage

3. Transformation Ratio K= V2 / V1

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TABULATION:

S.No

Type of connection

Primary voltage Secondary voltage Transformatiom

ratio(K) = V2/V1Primar

ySecondar

y

Vph

Volts

Vline

Volts

Voltage ratio(V

1)Vph/ Vline

Vph

Volts

Vline

Volts

Voltage ratio(V

2)Vph/ Vline

1

2

3

4

MODEL CALCULATION:

70

PRECAUTION:

1. Three ideal single phase transformers should be used for this study

purpose if a single unit of three phase transformer is not available.

2. High voltage and low voltage sides of the transformer should be

properly used.

PROCEDURE:

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

2. The auto transformer is adjusted to apply the rated primary voltage

of the transformer.

3. The phase voltage, phase current, line voltage and line current are

noted down for both the sides of transformer.

4. For Delta connections and star connected circuits their constraints

are verified respectively.

RESULT:

Thus the different three phase connections of three phase

transformers were studied.

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VIVA VOCE QUESTIONS WITH ANSWERS

DC MACHINES

1. What are all the types of electrical machines?Electrical machines are classified as AC machines and DC machines.

Types of DC machines

1. DC Generator

2. DC Motor

Types of AC machines

1. Transformers

(a) Single phase (b) three phase

2. Alternators

3. Synchronous motor

4. Induction motor

(a) Single phase (b) three phase

2. Mention the different parts of a d.c generator.The different parts of dc generator are

a. Magnetic frame (or) yoke.b. pole core and pole shoesc. pole coil or field coils d. armature windings or conductorse. armature coilsf. commutatorg. Brushes and bearing.

3. State the principle of operation of dc motor?

An electric motor is a machine which converts electric energy into mechanical energy. The motor action is based on the principle that when a current carrying conductor is placed in a magnetic field, it experiences a force whose direction is given by Fleming’s left hand rule and the magnitude is given by

F= BIL Newton.

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4. Mention the two types of armature winding. What is the number of parallel Paths in each case?

Type of winding

lap winding wave winding

No of parallel paths

A= P for lap winding

A= 2 for wave winding

5. What is back emf?

If the voltage is applied and current is passing in the armature, the motor starts rotating. At that time armature conductors also rotates and hence cuts the flux. Due to that an emf is induced in the armature and the direction is found to be opposite to that of the supply this is known as back emf.

6. Write the emf equation of DC generator explaining all terms?

Generator emf, Eg = volts

Where - is the flux per pole in wb

Z- is the total number of armature conductors

N- is the speed of armature is rpm

P- is the no of poles

A- is the number of parallel paths

7. Obtain an expression for armature torque?

Let Ta be the torque developed by the armature of a motor running at N rpm. If Ta is in N-m, the power developed

P = Ta 2 N watts 1.

The electrical energy is converted into mechanical power in the armature

P= EbIa watts 2.

Equation 1) and 2) we get

T a = 0.159 Z Ia (P/A) N-m ;Where, Eb = ZN (P/A) volts.

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8. What is armature reaction? What are its effects?

Armature reaction is the effect of magnetic field set up by armature current on the distribution of flux under main poles of a generator

The armature magnetic field has two effects

i) it demagnetizes or weakens the main flux andii) it cross magnetizes or distorts the main field flux.

9. What is commutation?

The process by which current in the short circuits is reversed while it crosses the MNA is called commutation. The brief period during which coil remains short circuited is known as commutation period Tc

10. What are the use of inter poles in a DC machines?

The uses of inter poles in a DC machines are

i) The commutating or reversing emf generated in the interpoles neutralizes the reactance emf and there by making commutation sparkles

ii) It neutralizes the cross-magnetizing effect of armature reaction. Hence brushes are not to be shifted from the original position.

11. Mention the methods of improved commutation?

The methods of improving commutation are

i) Resistance commutation ii) EMF commutation iii) Interpoles or compoles

12. What is meant by excitation of a dc machines and what are the methods of excitation?

Excitation means giving supply to the field winding to establish flux in the poles and to help the generator to produce an emf in the armature conductors

The field winding may be excited in the following two methods

i) separately excitedii) self excited

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13. Give the types of compound generators?

The three types of compound generator

i) under compoundii) flat compoundiii) Over compounded.

14. What are the characteristics of DC generator?

The characteristics of DC generator are

i) no load or saturation characteristics( Ea/ If)ii) internal characteristics(E/If)

iii) external characteristics( V/If)15. On what operating factor does the speed of a DC motor depend?

Speed volts.

Where Eb is back emf and ф is flux per pole.

16. What are the different methods of speed control in a dc motor?

1. Field Control Methods

2. Armature control Methods

3. Variable voltage Method

17. How do you reverse the direction of a dc motor?

Either the direction of the main field or the direction of current through the armature conductors is to be reversed to reverse the direction of a dc motor

18. How are armature windings classified based on the placement of the coil inside the armature slots?

Single layer winding and Double layer winding

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19. Draw the Speed versus armature current curves for the dc shunt and series motors?

20. Sketch the speed –torque characteristics of dc shunt, series motors and compound motors.

21. A 250V, dc shunt motor takes a line current of 20A. Resistance of shunt field winding is 200Ω and resistance of the armature is 0.3 Ω. Find the armature current and the back e.m.f?

22. Why is commutator employed in d.c machines?

• Conduct electricity between armature and fixed brushes

• Converts altenating emf into unidirectional emf and vice versa

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23. Distinguish between shunt and series field coil constructions.

Shunt field coils are wound with wires of small cross section and have more number of turns. Series field coils are wound with wires of larger cross section and have less number of turns.

24. How does a d.c motor differ from d.c generator in construction?

Generators are normally placed in closed room , accessible only to skilled operators. Therefore on ventilation point of view they may be constructed with large opening in the frame. Motors on the other hand , have to be installed right in the place of use which may have dust, dampness,inflammable gases, chemical fumes etc . To protect the motors against these elements , the motor frames are made either partly closed or totally closed or flame proof etc.

25. What is the function of a no-voltage release coil provided in a dc motor starter?

As long as the supply voltage is on healthy condition the current through the NVR coil produce enough magnetic force of attraction and retain the starter handle in the ON position against spring force. When the supply voltage fails or becomes lower than a prescribed value the electromagnet may not have enough force and the handle will come back to OFF position due to spring force automatically. Thus a no-voltage or under voltage protections given to the motor.

26. Enumerate the factors on which the speed of a dc motor depends.

N = (V-IaRa)/

The speed of dc motor depends on three factors. Flux in the air gap Resistance of the armature circuit Voltage applied to the armature

27. How can one differentiate between long shunt compound generator and short shunt compound generator?

In a short shunt compound generator the shunt field circuit is shorter i.e. across the armature terminals. In a long shunt compound generator the shunt field circuit is connected across the load terminals.

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28. Why is the emf not zero when the field current is reduced to zero in a dc generator ?

Even after the field current/magnetizing force is reduced to zero the machine is left out with some flux as residue. Emf due to this residual flux is available when field current is zero.

29. What are the conditions to be fulfilled for a dc shunt generator to build up emf?

The generator should have residual flux The field winding should be connected in such a manner that the

flux set up by the field winding should be in the same direction as that of residual flux

The field circuit resistance should be less than critical field resistance

Load circuit resistance should be above its critical load resistance

.30. What is the basic difference between dc generator and dc motor

Generator converts mechanical energy into electrical energy. Motor converts electrical energy into mechanical energy. But there is no constructional difference between the two.

31. What are the types of induced emf’s?

1. Dynamically induced emf

2. Statically induced emf

32 Name the different types of dc generators?

1. Separately excited2. Self excited

Self excited generator are further classified as

1. dc shunt generator2. dc series generator3. dc compound generator

33. What are the necessary conditions for the generator to be self excited?

1. Some residual magnetism in the field system in proper direction2. The residual flux and flux produced by the shunt field winding must

aid each other.3. For a series generator, the resistance of external circuit should be

less than the critical resistance.

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34. Why a dc series motor should always be started with load?

When load is heavy, Ia is large and Ф is high. Hence speed is low.

(N ά 1/ Ф). But when a load current falls and hence Ia falls to a small value. Speed becomes dangerously high. Hence a series motor should never be without some mechanical load on it.

35. Why a starter is necessary for a dc motor?

We know Ia=V-Eb/Ra when the motor is at rest, there is no back emf (Eb=0) developed in the armature. If now full supply voltage is applied across the stationary armature, It will draw a very large current because armature resistance is very small (Ia=V/Ra). This excessive current will damage the commentator and brushes. To reduce this high starting current starter is necessary. The main function of starter is to limit the starting current drawn by the motor.

36. What are the merits and demerits of swimburnes test to predetermine the efficiency of a dc machine?

Merits:

1. It is convenient and econ because power required to test a large machine is small ie., only no load input power.

2. The efficiency can be predetermine at any load because constant losses are known.

Demerits:

1. In this motor, the iron losses are assumed to be constant, which is not true case as they change from no load to full load.

2. Due to armature reaction at full load there will be distribution in flux which increase the iron loss.

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TRANSFORMERS

1. What is meant by a transformer?

The transformer is a static piece of apparatus by means of which electrical power is transformed from one alternating current circuit to another with desired change in voltage and current, without any change in the frequency. It works on the principle of mutual induction.

2. What are the advantages of a transformer?

i) Less I2R loss in the transmission line

ii) Less voltage drop in the line

iii) Efficiency of the transmission line is increased

iv) Volume of the conductor required is less.

3. What are the properties of ideal transformer?

i) It has no loss

ii) Its windings have zero resistance.

iii) Leakage flux is zero i.e 100% flux produced by primary links with the secondary

iv) Permeability of core is so high that negligible current is required to establish the flues is it.

4. What are the important parts of a transformer?

Transformer consists of two winding and laminated magnetic core. This core is square or rectangle in shape. It consists of limb and yoke .This core was laminated to reduce the eddy current losses. The two windings are i)primary winding ii)secondary winding

5. Define voltage transformation ratio?

The ratio of secondary induced emf to primary induced emf is called as voltage transformation ratio denoted as K.

E2 N2

= = K

E1 N1

80

6. Define leakage flux.

The entire flux produced by the primary of the transformer does not link with the secondary winding. A part of the primary flux as well as secondary flux complete the path through air and links with the respective flux only. Such a flux is called leakage flux.

7. Why the transformer is called as the constant flux machine?

For any load condition from no load to full load the flux in the core is practically constant, because the load component current always neutralizes the changes in the load. Hence core loss is also constant for all loads and the transformer is called constant flux machine.

8. Write the expression for equivalent resistance and reactance of transformer referred to primary.

Equivalent resistance Ro1 = R1 + R21 = R1 + R2/K2

Equivalent reactance Xo1 = X1 + X11 = X1 + X2/K2

9. Define voltage regulation of a transformer. Give the expression for the same.

The decrease in secondary terminal voltage expressed as a fraction of the no load secondary terminal voltage is called voltage regulation of a transformer.

% voltage regulation=

Where E2 = secondary terminal voltage on no load

V2= secondary terminal voltage on given load

10. What are the losses occurring in a transformer?

i) Core or iron losses or constant losses

ii) Copper losses or variable losses

11. What is meant by core or iron losses?

Core or iron losses are caused as the core gets subjected to an alternating flux.

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12. What is meant by hysteresis losses?

Due to alternating flux set up in the magnetic core of the transformer, it undergoes a cycle of magnetization and demagnetization. Due to hysteresis effect there is loss of energy in this process which is called hysteresis loss.

13. What is meant by eddy current loss?

The induced emf in the core set up,eddy currents in the core and hence the power loss due to the flow of this eddy current is known as eddy current loss

14. What are the methods to reduce the iron losses in a transformer?

The iron losses are minimized by using high grade core material like silicon steel having very low hysteresis loop and by manufacturing the core in the form of laminations.

15. What is meant by copper loss?

The copper losses are the power wasted in the form of I2R due to the ohmic resistances of the primary and secondary windings. The copper loss is found from the short circuit test.

16. What is meant by active or working or iron loss component and magnetizing component?

An active component is the one, which supplied total losses under no load condition. It is also called power component of no load current.

The magnetizing component of no load current is the one, which is required to produce the flux. It is also called wattless component.

17. What is the condition for maximum efficiency of a transformer also Define all day efficiency?

The condition to achieve maximum efficiency is

Copper loss = iron loss i. e Pcu = Pi

All day efficiency is the ratio energy (in kwh) delivered in a 24 hours period to the energy (in kwh) input for the same length of time.

All day efficiency = (for 24 hrs)

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18. Why the OC and SC tests are to be performed in transformer?

The equivalent circuit parameters of a transformer can be determined by conducting the open circuit test (OC) and short circuit (SC) test. The parameters thus determined are effective in determining the regulation and efficiency of a transformer at any load and power factor condition, without actually loading the transformer.

19. What are the advantages of transformer test?

1. The test can be performed and results obtained without much power loss.

2. The parameter can be obtained without actually loading the transformer.

20. what is meant by equivalent circuit and exciting circuit?

Equivalent circuit: The term equivalent circuit of a transformer means the combination of fixed and variable resistance and reactance, which exactly simulates performance and working of the transformer.

Exciting circuit: The equivalent circuit of a transformer on no load consisting of R0&X0 in parallel. This parallel combination is called as exciting circuit.

21. Define efficiency of a transformer?

The efficiency of a transformer is defined as the ratio of the output power(kw) to the input power(kw).

=

22. Why the transformer rating in KVA?

The copper loss of a transformer 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 i.e. it is independent of power factor. So that the rating of transformer is in KVA.

23. What are the two main components of primary current (Io)?

i) Active or core loss component or wattful component (Ic)

ii) Reactive or magnetizing component or wattless component (Im)

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em -i lab 2008 reg

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