EM-I LAB MANUAL

LIST OF EXPERIMENTS

1. Open circuit and load characteristics of a separately excited DC Shunt Generator

2. Open circuit and load characteristics of a self excited DC shunt generator

3. Load Characteristics of DC compound generator with cumulative and differential connection

4. Load Characteristics of DC shunt motor

5. Load Characteristics of DC compound motor

6. Load Characteristics of DC series motor

7. Swinburne’s Test

8. Speed control of DC shunt motor

9. Load test on single phase transformer

10. Open circuit and short circuit test on single phase transformer

11. Sumpner’s test on transformers

12. Separation of no load losses in a single phase transformer

13. Three phase transformer connections

EE 1203 ELECTRICAL MACHINES-I LAB

1


2

1. OPEN CIRCUIT AND LOAD CHARACTERISTICS OF SEPARATELY EXCITED DC GENERATOR

AIM:

To conduct open circuit test and load test on a separately excited DC Generator and to draw the following characteristic curves.

A. Open Circuit Characteristics (or) Magnetization CharacteristicsB. Internal CharacteristicsC. External Characteristics Load characteristics andD. Performance Characteristics

APPARATUS REQUIRED:

S.NO APPARATUS RANGE TYPE QUANTITY1. Voltmeter (0-300)V MC 2

2. Ammeter(0-10)A(0-20) A(0-2) A

MCMCMC

111

3. Rheostat 350Ω/1.4 A 24. Tachometer 1

FORMULA USED:

1. Eg = Vt + IaRa Volts Where Eg = Generated voltage in Volts Vt = Terminal Voltage in Volts Ia = Armature Current in Amps. Ra= Armature resistance in Ohms.

2. Input power (Pin)= Vin* Iin Watts. Where Vin= Input voltage in volts Iin= Input current in Amperes.3. Output Power (Pout)= Vt*Ia Watts

Where Vt = Terminal Voltage in Volts. Ia = Armature Current in Amps.

4. Percentage Efficiency = (Pout/Pin)*100

MODEL GRAPH:A B & C D

Eg Eg & VL

If If Ia Po


3

PROCEDURE:OPEN CIRCUIT TEST:

1. Connections are given as per the circuit diagram.2. Verify whether the field rheostat of the motor is kept at minimum position and the field

rheostat of the generator is kept at maximum position. 3. By closing the DPST Switch, 230V DC Supply is given.4. Start the motor by using 3-point starter. 5. The field rheostat of the motor (ie., excitation) is adjusted so as to make the motor to run at

rated speed.6. Emf generated in the generator is noted for various field currents by adjusting the field rheostat

of the generator. Readings are noted up to 125% of the rated voltage.7. Tabulate the corresponding field current (If) & generated voltage (Eg).

Note: The adjustment of the field rheostat is varied in reverse direction, the previous readings and the new one won’t be same due to residual magnetism. So, the adjustment of field rheostat should be in one direction.

8. Draw the open circuited curve from the tabulated values.

LOAD TEST:1. Fix the armature voltage to the rated value by varying the rheostat of the generator.2. DPST Switch of single-phase rheostatic load is closed.3. Apply the load to the generator using Single Phase Resistive Load step by step.4. Vary the load of the generator up to its rated current.5. Tabulate the Input voltage, Input current, terminal Voltage and Armature current for different

load values. 6. While taking each set of readings the field current is maintained constant as that of rated

voltage.(Because due to heating, shunt field resistance is increased).7. Load is gradually decreased and field rheostat is brought to its

original position and supply is switched off.

TO FIND ARMATURE RESISTANCE (Ra):

1. The connections are given as per the circuit diagram.2. By closing the DPST Switch, 230V DC Supply is given.3. Rheostat is varied and the corresponding values of voltage and current are noted down. 4. Calculate DC armature resistance Rdc and multiply by 1.3 to get hot Armature Resistance.

TABULATION:

OPEN CIRCUIT CHARACTERISTICS

If (A)Eg (V)


4

LOAD CHARACTERISTICS

S.No.

InputVoltageVin (V)

InputCurrentIin (A)

TerminalVoltageVt (V)

LoadCurrentIL (A)

FieldCurrentIsh (A)

GeneratedVoltageEg (V)

Input Power

Pin

(watts)

Output Power

Pout

(watts)

= (Pout /Pin) x 100 %

To find RaS.No Voltage in (V) Current in (A) Resistance in ohms

RESULT:

Open circuit and load test is conducted on a separately excited DC generator and the characteristics are drawn.


5


6

2. OPEN CIRCUIT AND LOAD CHARACTERISTICS OF SELF EXCITED DC GENERATOR

AIM:

To conduct open circuit test and load test on a self excited DC Generator and to draw the following characteristic curves.

A. Open Circuit Characteristics (or) Magnetization CharacteristicsB. Internal CharacteristicsC. External Characteristics Load characteristics andD. Performance Characteristics

APPARATUS REQUIRED:


2. Ammeter(0-10)A(0-20) A(0-2) A

MCMCMC

111


FORMULA USED:

1. Eg = Vt + IaRa Volts Where Eg = Generated voltage in Volts Vt = Terminal Voltage in Volts Ia = Armature Current in Amps. Ra= Armature resistance in Ohms.

2. Input power (Pin)= Vin* Iin Watts. Where Vin= Input voltage in volts Iin= Input current in Amperes.3. Output Power (Pout)= Vt*Ia Watts

Where Vt = Terminal Voltage in Volts. Ia = Armature Current in Amps.

4. Percentage Efficiency = (Pout/Pin)*100

MODEL GRAPH:A B & C D

Eg Eg & VL

If If Ia Po


7

PROCEDURE:OPEN CIRCUIT TEST:

1) Connections are given as per the circuit diagram.2) Verify whether the field rheostat of the motor is kept at minimum position and the field

rheostat of the generator is kept at maximum position. 3) By closing the DPST Switch, 230V DC Supply is given.4) Start the motor by using 3-point starter. 5) The field rheostat of the motor (ie., excitation) is adjusted so as to make the motor to run at

rated speed.6) .Emf generated due to residual magnetism is noted.7) Emf generated in the generator is noted for various field currents by adjusting the field rheostat

of the generator. Readings are noted up to 125% of the rated voltage.8) Tabulate the corresponding field current (If) & generated voltage (Eg).

a. Note: The adjustment of the field rheostat is varied in reverse direction, the previous readings and the new one won’t be same due to residual magnetism. So, the adjustment of field rheostat should be in one direction.

9) Draw the open circuited curve from the tabulated values.

LOAD TEST:1) Fix the armature voltage to the rated value by varying the rheostat of the generator.2) DPST Switch of single-phase rheostatic load is closed.3) Apply the load to the generator using Single Phase Resistive Load step by step.4) Vary the load of the generator up to its rated current.5) Tabulate the Input voltage, Input current, terminal Voltage and Armature current for different

load values. 6) While taking each set of readings the field current is maintained constant as that of rated

voltage.(Because due to heating, shunt field resistance is increased).7) Load is gradually decreased and field rheostat is brought to its

original position and supply is switched off.

TO FIND ARMATURE RESISTANCE (Ra):Repeat the procedure and circuit diagram from the previous experiment

TABULATION:

OPEN CIRCUIT CHARACTERISTICS

If (A)Eg (V)


8

LOAD CHARACTERISTICS

S.No.

InputVoltageVin (V)

InputCurrentIin (A)

TerminalVoltageVt (V)

LoadCurrentIL (A)

FieldCurrentIsh (A)

GeneratedVoltageEg (V)

Input Power

Pin

(watts)

Output Power

Pout

(watts)

= (Pout /Pin) x 100 %

RESULT:

Open circuit and load test is conducted on a self excited DC generator and the characteristics are drawn.


9


10

3. LOAD CHARACTERISTICS OF DC COMPOUND GENERATOR WITH CUMULATIVE AND DIFFERENTIAL CONNECTION

AIM:To conduct a load test on a DC compound generator and to draw the load characteristics in

both cumulative and differential mode of operation.

APPARATUS REQUIRED:


2. Ammeter(0-20)A(0-2) A(0-10) A

MCMCMC

111

3. Rheostat 350Ω/1.5 A 2

4. Tachometer 1

FORMULA USED:

1. For a long shunt compound motor, the generated voltageEg= V + Ia(Ra+Rse) VoltsWhere Ia = Armature current in Amps Ra = Armature resistance in Ohms

Rse = Series field resistance in Ohms V = Terminal voltage in Volts

2. For a short shunt compound motor, the generated voltageEg= V + Ia*Ra+IL*Rse VoltsWhere Ia = Armature current in Amps Ra = Armature resistance in Ohms

Rse = Series field resistance in Ohms IL = Load Current in Amps.

V = Terminal voltage in Volts

MODEL GRAPH:

V & Eg Egcumulative

VVEg diff

IL


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PROCEDURE:

1. Connections are given as per the circuit diagram.2. Verify whether the field rheostat of the motor is kept at minimum position and the field rheostat of the generator is kept at maximum position . 3. By closing the DPST Switch, 230V DC Supply is given.4. Start the motor by using 4-point starter. 5. The field rheostat of the motor is adjusted so as to make the motor to run at rated speed.6. Vary the field rheostat of the generator to obtain the rated voltage.7. Note the no load readings. 8. The load side DPST switch is closed.9. Load is applied to the motor up to 125% of rated value by using rheostatic load and the corresponding terminal voltage & load current values are tabulated.10. While taking each set of readings the field current is maintained constant as

that of rated voltage.(Because due to heating, shunt field resistance is increased).11. The series field of the generator is interchanged and the same procedure is

repeated for all modes of operation.

NOTE: If the shunt field is connected across both the armature and series field, it is long shunt and if the shunt field is connected across the armature only then it is called short shunt.TABULATION

CUMULATIVE – LONG SHUNTS.No Line

VoltageVin (V)

Line Current

IL(A)

Field Current

Ish(A)

Armature Current

Is=IL+Ish (A)

Generated voltage Eg in (V)

CUMULATIVE –SHORT SHUNTS.No Line

VoltageVin (V)

Line Current

IL(A)

Field Current

Ish(A)

Armature Current

Is=IL+Ish (A)

Generated voltage Eg in (V)

DIFFERENTIAL – LONG SHUNTDIFFERENTIAL-SHORT SHUNTDraw similar tabulation as cumulative

RESULT:The load test is conducted and the load characteristics of the given DC compound generator are

drawn.


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4. LOAD CHARACTERISTICS OF DC SHUNT MOTOR

AIM:

To conduct load test on the given on DC shunt motor and to draw the following characteristic curves.

A. Armature current Vs Torque (Electrical Characteristics)B. Speed Vs Torque (Mechanical Characteristics)C. Performance characteristics

a. Output power Vs Speedb. Output power Vs Currentc. Output power Vs Torqued. Output power Vs Efficiency

APPARATUS REQUIRED:

S.NO APPARATUS RANGE TYPE QUANTITY1. Voltmeter (0-300)V MC 22. Ammeter (0-10) A MC 1


FORMULA USED:

1.F=[F1-F2] Kg.Where F = Force in Kg.

F1, F2= Spring Balance readings in Kg.

2. Torque (T) = 9.81 *F*RWhere R= Radius of the Brake Drum in metre.

3.Input Power (Pin) =Vin * Iin WattsVin =Input voltage in AmperesIin = Input current in Amperes

4.Outpower (Pout ) =2** N*T/60 wattsWhere N= Speed of the motor in rpm

T= Torque in Nm.

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


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14

MODEL GRAPH:

A B C

T N

T Ia N

I T Pout

PROCEDURE:

1.Connections are given as per the circuit diagram.2.Verify whether the field rheostat of the motor is kept at minimum position.3.By closing the DPST Switch, 230V DC Supply is given.4.Start the motor by using 3 point starter.5.The field rheostat of the motor(i.e., excitation) is adjusted so as to make the motor to run at

rated speed. After adjusting the rheostat to the rated speed it should not be altered.6.At no load condition ,the input voltage ,current, speed are noted. 7.Load is applied to the motor upto 125% of rated value by using Brake Drum and the corresponding reading are tabulated.8.Load is gradually decreased and field rheostat is brought to its original position and supply is switched off.9.From the tabulated values the performance characteristic curves are drawn.

TABULATION:

S.No

Input Voltage

Vin

Input Current

Iin

Spring Balance Reading

Speed N

TorqueT

InputPower

Pin

Output Power

Pout

Efficinecy Pout * 100 Pin

F1 F2 F1~F2

Unit Volts Amps Kg kg Kg RPM Nm Watts Watts %

RESULT:

Thus the load test on a DC shunt motor is conducted and the performance characteristic curves are drawn.


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5. LOAD CHARACTERISTICS OF DC COMPOUND MOTOR

AIM:

To conduct load test on DC compound motor and to draw the following characteristic curves

A. Armature current Vs SpeedB. Speed Vs Torque C. Performance characteristics

o Output power Vs Speedo Output power Vs Currento Output power Vs Torqueo Output power Vs Efficiency

APPARATUS REQUIRED:


2. Ammeter (0-10)A MC 1

3. Rheostat 350/1.4 A 14. Tachometer 1

FORMULA USED:

1. F=(F1~F2) Kg.Where, F = Force in Kg. F1,F2 = Spring Balance readings in Kg.

2. T = 9.81* F *RWhere, T = Torque in N-m

R = Radius of the Brake Drum in meter.

3. Pin = Vin*Iin Watts.Where, Pin = Input power in watts.

Vin = Input voltage in volts. Iin = Input current in Amperes.

4. Pout = 2**N*T/60 watts.Where Pout = Output power in watts.

N = Speed of the motor in rpm

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


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MODEL GRAPH:

N A N B Ia CT

N

Ia T Pout

PROCEDURE:

1. Connections are given as per the circuit diagram.2. Verify whether the field rheostat of the motor is kept at minimum position.3. By closing the DPST switch, 230V DC supply is given.4. Start the motor by using 4-point starter.5. The field rheostat of the motor (i.e., excitation) is adjusted so as to make the motor to run at

rated speed. After adjusting the rheostat to the rated speed it should not be altered.6. At no load condition, the input voltage, current, speed are noted.7. Load is applied to the motor up to 125% f rated value by using Brake Drum and the

corresponding reading are tabulated.8. Load is gradually decreased and field rheostat is brought to its original position and supply is

switched OFF.9. From the tabulated values the performance characteristic curves are drawn.

TABULATION:

S.No Input Voltage

Vin

Input Current

Iin


Speed N

TorqueT

InputPower

Pin

Output

Power Pout

Efficinecy Pout * 100PinF1 F2 F1~F2

Unit Volts Amps Kg kg Kg RPM Nm Watts Watts %

RESULT:

Thus the load test on a DC Compound motor is conducted and the characteristic curves are drawn.


18

6. LOAD CHARACTERISTICS OF DC SERIES MOTOR

AIM:To conduct load test on a DC series motor and to draw the performance characteristic curves.

A. Armature current Vs Torque (Electrical Characteristics)B. Speed Vs Torque (Mechanical Characteristics)C. Performance characteristics

a. Output power Vs Speedb. Output power Vs Currentc. Output power Vs Torqued. Output power Vs Efficiency

APPARATUS REQUIRED:


2. Ammeter(0-20)A(0-2) A(0-10) A

MCMCMC

111

3. Rheostat 350Ω/1.5 A 2

4. Thread5. Tachometer 1

FORMULA USED:

1.F=[F1-F2] Kg.Where F = Force in Kg.

F1, F2= Spring Balance readings in Kg.

2. Torque (T) = 9.81 *F*RWhere R= Radius of the Brake Drum in metre.

3.Input Power (Pin) =Vin * Iin WattsVin =Input voltage in AmperesIin = Input current in Amperes

4.Outpower (Pout ) =2** N*T/60 wattsWhere N= Speed of the motor in rpm

T= Torque in Nm.5.Percentage Efficiency = (Pout / Pin )*100


19


20


21

MODEL GRAPH:

A B C

T I T N

N

I T Pout

PROCEDURE:

1.Connections are given as per the circuit diagram.2.Verify whether the field rheostat of the motor is kept at minimum position and ensure that some load is applied to the brake drum.3.By closing the DPST Switch, 230V DC Supply is given.4.Start the motor by using 3 point starter.5.The excitation of the field rheostat of the motor is adjusted so as to make the motor to run at rated speed. After adjusting the rheostat to the rated speed it should not be altered.6.At no load condition, the input voltage, current, speed are noted. 7.Load is applied to the motor upto 125% of rated value by using Brake Drum and the corresponding reading are tabulated.8.Load is gradually decreased and field rheostat is brought to its minimum position and supply is switched off.9.From the tabulated values, torque, input power and the percentage efficiency are calculated & the performance characteristic curves are drawn.

TABULATION:

S.No Input Voltage

Vin

Input Current

Iin


Speed N

TorqueT

InputPower

Pin

Output Power

Pout

Efficinecy Pout * 100

PinF1 F2 F1~F2Unit Volts Amps Kg kg Kg RPM Nm Watts Watts %

RESULT:The load test is conducted and the performance characteristics curves are drawn for the given

DC series motor.


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7. SWINBURNE’S TEST

AIM:

To conduct the Swinburne’s test (or) loss summation test on the given DC shunt motor and to predetermine the efficiency of a DC machine operating as

1. a motor2. a generator

APPARATUS REQUIRED:


2. Ammeter(0-5)A(0-2) A

MCMC

11


FORMULA USED:

1. No load input power Pino=VLo* ILo Watts Where VLo= No load input voltage in volts ILo= No load input current in Amps2. Armature Current Iao= ILo- Ifo Amps Where Ifo= No load field current in Amps ILo= No load input current in Amps3. Armature copper loss Wcuo= Iao

2Ra WattsWhere Iao= No load armature current in Amps Ra= Armature resistance in Ohms

4. Constant losses of the DC shunt machine Wc= Pino-Wcuo Watts Where Pino= No load input power in Watts

Wcuo=Armature copper loss in WattsMODEL GRAPH:

Gen GenMot Mot

X Pout


24

PROCEDURE:

1.Connections are given as per the circuit diagram.2.Verify whether the armature rheostat of the motor is kept at maximum position and the field rheostat is kept at minimum position .3.By closing the DPST Switch, 230V DC Supply is given.4.Start the motor by using 3 point starter.5.The field rheostat of the motor(i.e., excitation) is adjusted so as to make the motor to run at rated speed. After adjusting the rheostat to the rated speed it should not be altered.6.At no load condition, the input voltage, current, speed are noted.7.After taking the readings, switch off the supply.8.From the recorded values, the no load losses are calculated.9.The efficiency of the DC shunt machine is predetermined for various load conditions and the graph is drawn between efficiency Vs output power for the machine as generator and as motor.

TABULATION:

NO LOAD READING OF THE MACHINE AS A MOTOR

S.No No Load VoltageVin

No Load Current Iin No Load Field

Current If

No Load Armature Current

Ia=Iin-If

Unit Volts Amps Amps Amps

EFFICIENCY AS A MOTORS.No Fraction

of LoadLoad

Current IL

=12*x

Input Power

Pin=Vin*IL

Armature CurrentIa=IL-If

Copper Loss

Wcu=Ia2Ra

Total Loss PL=Wc+Wcu

Output Power

Pout=Pin-PL

Efficinecy Pout * 100

Pin

Unit X Amps Watts Amps Watts Watts Watts %

EFFICIENCY AS A GENERATORS.No Fraction

of LoadLoad

Current IL=12*x

Input Power

Pout=VL*IL

Armature CurrentIa=IL+If

Copper Loss

Wcu=Ia2Ra

Total Loss PL=Wc+Wcu

Output Power

Pin=Pout+PL

Efficinecy Pout * 100Pin

Unit x Amps Watts Amps Watts Watts Watts %

RESULT: Thus the swinburne’s test is conducted on the given DC shunt motor and the performance(Efficiency)characteristics are obtained for the machine running as a motor and as a generator.


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8. SPEED CONTROL OF DC SHUNT MOTOR

AIM:To Control the speed of the given DC Shunt motor by the following methods.

1. Armature control method2. Field control method.

APPARATUS REQUIRED:


2. Ammeter(0-5)A(0-2) A

MCMC

11

3. Rheostat 350Ω/1.4 A55 Ω/ 4.6 A

11

4. Tachometer 1

FORMULA:Back EMF Eb= Va- IaRa

MODEL GRAPH:1.

2.

N Eb

If N

PROCEDURE: A) ARMATURE CONTROL METHOD

1.Connections are given as per the circuit diagram.2.Verify whether the armature rheostat of the motor is kept at maximum position and the field rheostat is kept at minimum position .3.By closing the DPST Switch, 230V DC Supply is given.4.Start the motor by using 3 point starter.5.For armature control method, keep the field current constant by adjusting the field rheostat

connected in the field circuit. 6.Vary the rheostat connected in the armature circuit and note the corresponding armature voltage, armature current and speed and tabulate it.


27

B) FIELD CONTROL METHOD

1.For field control method keep the armature voltage as constant by adjusting the rheostat connected to the armature circuit.2.Vary the rheostat connected in the field circuit and for each setting note the corresponding field current and speed and tabulate it.

TABULATION:

FIELD CONTROL METHOD: Va=constant

S.No ArmatureVoltageVa (volts)

Field CurrentIf (amps)

SpeedN (rpm)

ARMATURE CONTROL METHOD: I f =cons tant

S . N O I f (Amps) V a ( vo l t s ) Ia (Amps) E b ( vo l t s ) N(rpm)

RESULT:The speed control of DC Shunt motor is conducted and the curves are drawn.


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9.LOAD TEST ON SINGLE PHASE TRANSFORMER

AIM:To conduct load test on the given single phase transformer and to draw the performance

characteristics curves.1. Efficiency Vs Output Power 2. Regulation Vs Output Power.

APPARATUS REQUIRED:

S.NO APPARATUS RANGE TYPE QUANTITY1. Voltmeter (0-150)V

(0-300) VMIMI

11

2. Ammeter(0-10)A(0-20) A

MIMI

11

3. Wattmeter 150V,20A 300V,10A

UPFUPF

11

4. Auto transformer 240 V, 2.7 KVA,10A

1

FORMULA USED:

1.Percentage Regulation = (Vo2-V2) /Vo2*100 Where Vo2 = Secondary voltage on no load Vo = Secondary voltage at a particular load

2. Power factor = Pout/V2*I2

Where Pout = Secondary wattmeter readings in Watts V2 = Secondary voltage in Volts I2 = Secondary current in Amps

3. Percentage efficiency = Pout/Pin*100 Where Pout = Secondary wattmeter readings in Watts Pin = Primary wattmeter readings in Watts. MODEL GRAPH:

% reg

Pout


29


30

PROCEDURE

1.Connections are given as per the circuit diagram. 2.Verify whether the autotransformer is kept at zero voltage position.

3.By closing the DPST switch, 230V,1,50HZ AC supply is given to the transformer.4.At no load, the readings from the meters are noted down.5.The load is applied to the transformer in steps upto 125% of the rated value of the primary Current by using rheostatic load..6.The corresponding values from the meters are tabulated for different loads. 7.Then the load is removed gradually, auto transformer is brought to its minimum position and the supply is switched off.8.From the recorded values, the regulation, power factor and efficiency are calculated.

TABULATION:

S.No

Primary VoltageV1 (V)

Primary Current

I1 (A)

Primary Wattmete

rW1 (W)

Secondary

VoltageV2 (V)

Secondary

CurrentI2 (A)

Secondary

Wattmeter

W2 (W)

PowerFactor Cos

%Regulation

%

%

RESULT Thus the load test on the single phase transformer is conducted and the characteristic curves are drawn. The tested transformer attains maximum overall efficiency of______________ at an output power of


31

OPEN CIRCUIT TEST


32

10. OPEN CIRCUIT AND SHORT CIRCUIT TEST ON SINGLE PHASE TRANSFORMER

AIM:To conduct open circuit and short circuit test on single phase transformer and hence

to determine i. The equivalent circuit parameters

ii. The percentage efficiency and the regulation at any desired load.

APPARATUS REQUIRED:


(0-75) VMIMI

11

2. Ammeter (0-10) A(0-2.5) A

MIMI

11

3. Wattmeter 150 V, 20A150 V, 20A

LPFUPF

11

4. Auto transformer 240V,2.7KVA,10A 1

FORMULA USED:I. To determine the equivalent circuit parameters:

1. No load power factor Coso = Wo/VoIo

Where Vo = Open circuit voltage in Volts Io = Open circuit current in Amps Wo = No load power in Watts (Iron loss)

2. Working component of no load current Iw = IoCoso 3. Magnetizing component of no load current I = IoSino 4. Ro = Vo/Iw

5. Xo = Vo/ I

6. Equivalent impedance of transformer w.r.t HV side Z02=Vsc/Isc7. Equivalent resistance of transformer w.r.t HV side R02 = Wsc/Isc2

8. Equivalent reactance of transformer w.r.t HV side X02 = Zo22-Ro2

2 9. Transformation ratio K = V2/V1

10.Equivalent resistance of transformer w.r.t LV side R01 = R02/K2

11. Equivalent reactance of transformer w.r.t LV side X01 = X02/K2

II. To Predetermine Percentage Efficiency: Q = KVA of the given transformer

x = Fraction of loadTotal losses = x2*Wsc + WoOutput = x*Q*CosInput = Output+ LossesPercentage Efficiency = (Output/Input)*100


33

SHORT CIRCUIT TEST:


34

III. To Calculate Percentage Regulation:For Lagging power factor,

Percentage Regulation = (x*Isc/Vo)*(Ro2 Cos + Xo2 Sin) * 100For Leading power factor,

Percentage Regulation = (x*Isc/Vo)*(Ro2 Cos - Xo2 Sin) * 100

MODEL GRAPH + % reg

1 10.8 0.8

lead pf lag pf

Pout X - % regEQUIVALENT CIRCUIT:

Re’ Xe’ Re Xe

V1 R0 X0 V2’ V1’ R0’ X0’ V2

PROCEDURE

OPEN CIRCUIT TEST:1.Connections are given as per the circuit diagram.2.Verify whether the autotransformer is kept at zero voltage position.3.To conduct no load test, high voltage windings must be open circuited and the rated voltage b should be applied to the low voltage winding .4. By closing the DPST switch, 230V, 1, 50Hz, AC supply is given to the transformer.5. By adjusting the auto transformer, apply the rated voltage across the primary windings and the corresponding readings from ammeter, voltmeter and wattmeter are noted.6.Bring the autotransformer to the minimum position and remove the supply. 7.From the noted values, the exciting branch parameters (Ro & Xo) of the equivalent circuit are calculated.

SHORT CIRCUIT TEST:1.Connections are given as per the circuit diagram.2.Verify whether the autotransformer is kept at zero voltage position.3.To conduct short circuit test, low voltage windings must be short-circuited.4.By closing DPST switch, 230V, 1, 50HZ AC supply is given to the autotransformer.


35

5. By adjusting the auto transformer, apply the rated current across the primary windings and the corresponding readings from ammeter, voltmeter and wattmeter are noted.6.Bring the autotransformer to the minimum position and remove the supply. 7. From the noted values, the exciting branch parameters (Ro1 & Xo1) of the equivalent circuit are calculated.

Predetermination Of Efficiency And Regulation: From the calculated values of equivalent circuit parameters, the regulation is calculated for

various values of assumed power factor. For various values of assumed power factor the efficiency is calculated and the graph is

plotted.

TABULATION:Open Circu i t Tes tNo load Voltage Vo in Volts

No Load Current Io in Amps

Wattmeter Reading Wo in Watts

Shor t C ircu i t Tes tPrimary Voltage Vsc in Volts

Primary Current Isc in Amps

Wattmeter Reading Wsc in Watts

De terminat ion Of Percentage Regu la t ionCos Sin

1-Cos2Percentage Regulation

Lagging (+) Leading (-)0,.2,.4,.6,.8,1

Predeterminat ion Of Eff ic iency For power factors of 0 .8 and 1 .0S.No Fraction

of LoadOutput PowerPo(XQcos)

Copper Loss

Wcu=x2Wsc

Total Loss Wc=Wo+x2Wsc

Input PowerOutput+losses

Efficinecy

Unit x Amps Watts Watts Watts %

RESULT:Thus the open circuit test and short circuit test is conducted on a single-phase transformer and,

i. The equivalent circuit parameters are found out. The equivalent circuit parameters of the given transformer are

Ro = Xo=R01= R02=ii. The performance curves of the transformer are determined at various loads and power factors.


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11. SUMPNER’S TEST ON TRANSFORMERS

AIM: To conduct Sumpner’s test on two similar single-phase transformers. To draw equivalent circuit of each transformer with reference to high voltage and low

voltage sides. To predetermine the percentage efficiency and percentage regulation of each transformer at

any load and to draw the performance characteristics.

APPARATUS REQUIRED:


(0-600) V(0-75) V

MIMIMI

111

2. Ammeter(0-10)A(0-5) A

MIMI

11

3. Wattmeter 150V,5A 75V,10A

LPFUPF

11

4. Auto transformer 240 V, 2.7 KVA,10A 25. SPST Switch 1

FORMULA USED:1. Open circuit parameters of each transformer

Vo = V1

Io = I1/2Wo = W1/2

2. Short circuit parameters of each transformerVsc = V2/2Isc = I2

Wsc = W2/2 3. No load power factor Coso = Wo/VoIo

Where Vo = Open circuit voltage in Volts Io = Open circuit current in Amps Wo = No load power in Watts (Iron loss)

4.Working component of no load current Iw = IoCoso 5.Magnetizing component of no load current I = IoSino 6.Ro = Vo/Iw

7.Xo = Vo/ I

8.Equivalent impedance of transformer w.r.t HV side Z02=Vsc/Isc9.Equivalent resistance of transformer w.r.t HV side R02 = Wsc/Isc2

10.Equivalent reactance of transformer w.r.t HV side X02 = Zo22-Ro22 11.Transformation ratio K = V2/V1

12.Equivalent resistance of transformer w.r.t LV side R01 = R02/K2

13.Equivalent reactance of transformer w.r.t LV side X01 = X02/K2


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TO PREDETERMINE PERCENTAGE EFFICIENCY:

Q = KVA of the given transformerx = Fraction of loadTotal losses = x2*Wsc + WoOutput = x*Q*CosInput = Output+ LossesPercentage Efficiency = (Output/Input)*100

TO CALCULATE PERCENTAGE REGULATION:

For Lagging power factor,Percentage Regulation = (x*I2/V1)*(Ro2 Cos + Xo2 Sin) * 100

For Leading power factor,Percentage Regulation = (x*I2/V1)*(Ro2 Cos - Xo2 Sin) * 100

MODEL GRAPH & EQUIVALENT CIRCUIT:(As in the previous experiment.)

PROCEDURE1.Connections were given as per the circuit diagram.2.Verify whether the auto transformer is kept at zero voltage position.3.Keeping the switch S as open, 230V,1,50HZ AC supply is given to the auto transformer by closing the DPST switch.4.Auto transformer1 is adjusted and rated voltage is applied to LV side of test transformer.5.If the voltmeter connected across the switch reads zero, then switch S is closed.6.The auto transformer2 is adjusted such that the ammeter reads the required secondary rated current.7.The corresponding readings of ammeter, voltmeter and wattmeter are noted.8. Bring the auto transformer to the minimum position and remove the supply. 9.The wattmeter in the secondary gives the total copper loss for both the transformer and

wattmeter reading in the primary gives the core loss for both the transformers

TABULATION:BACK TO BACK TEST OF 2 TRANSFORMERS

S.No V1 I1 W1 V2 I2 W2

Unit Volts Amps Watts Volts Amps Watts

PREDETERMINATION OF EFFICIENCY

Cos Fraction X2*Wsc Total losses Output InputPower= Efficiency


39

of Load

Wo+x2Wsc PowerXQCos

Output+ Losses

X Watts Watts Watts Watts %

TO DETERMINE PERCENTAGE REGULATION

S.No Fraction of Load x Cos Sin Percentage RegulationLagging Leading

RESULT:The sumpner’s test is conducted on the given 2 identical single phase transformer. The

equivalent circuit parameters of the given transformer are Ro = Xo=R01=R02=

The performance curves of the transformer are determined at various loads and power factors.


40

12. SEPARATION OF NO LOAD LOSSES IN A SINGLE PHASE TRANSFORMER

AIM: To separate the iron losses in a single phase transformer into its components

a. Hysterisis losses andb. Eddy current losses

APPARATUS REQUIRED:S.NO APPARATUS RANGE TYPE QUANTITY

1. Voltmeter (0-300)V MI 12. Ammeter (0-1)A MI 13. Wattmeter 300V, 5A LPF 1

4. Rheostat 350/1.4 A55/4.6 A

22

5. Tachometer 1

FORMULA USED:1. Frequency f = PN/120 HZ

2. Total iron loss Wi = Wh + We

Where Wh= Hysteresis loss We = Eddy current loss

3. Hysteresis loss Wh = Ch*fWhere Ch = Intercept value of characteristic curve on y axis.

4. Eddy current loss We = Ce*f2

Where Ce = Slope of the characteristic curve

5. Total iron loss Wi = Wh + We

Where Wi = Ch*f + Ce*f2

Wi/f = Ch + Ce*f

MODEL GRAPH

Wi / f

Ce = slope

Ch

f

41

PROCEDURE:

42

1. Connections are given as per the circuit diagram.2. Verify whether the armature rheostat of the motor is kept at maximum position and the field rheostat is kept at minimum position.3. By closing the DPST Switch, 230 V DC Supply is given.4. Start the motor by using the 4 Point Starter.5. The field rheostat of the alternator is adjusted to have a rated voltage.6. Keeping the field rheostat of the motor constant, the armature rheostat is varied and the following speed, voltage and power are noted.7. The procedure is repeated for different speeds and the readings are noted.

TABULATION:

SEPARATION OF NO LOAD LOSSES

S.No Speed N Voltage V Iron loss Wi

Frequencyf=pn/120

Wi/f We Wh

Unit RPM Volts Watts HZ W/Hz Watts Watts

RESULT:

Thus the no load loss of the single-phase transformer is separated with the help of alternator.

43

44

13. THREE PHASE TRANSFORMER CONNECTIONS

AIM:To make and study scott (open delta or V-V) connections using 3 phase

transformers.

APPARATUS REQUIRED:


(0-600) VMIMI

1

2. Auto transformer 1

PROCEDURE:

1. Connections are given as per the circuit diagram.2. Verify whether the autotransformer is kept at zero voltage position.3. By closing the DPST switch, 230V, 1, 50HZ AC supply is given to the transformer.4. The voltages are verified in the primary and the secondary side for scott connections.

TABULATION:

VERIFICATION OF SCOTT CONNECTIONS

S.No Type of connection

Primary side line voltage Vpin

Volts

Secondary side line voltage Vsin Volts

1. Scott

RESULT:Thus the Scott connection is verified.

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EM-I LAB MANUAL