g.h. raisoni college of engineering...

DEPARTMENT OF ELECTRICAL ENGINEERING ivTHsem B.E. electrical

G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR 1

G.H. RAISONI COLLEGE OF ENGINEERING

DEPARTMENT OF ELECTRICAL ENGINEERING

ELECTRICAL MACHINES LABORATORY

ELECTRICAL MACHINES-I/BEMC

INDEX

Sr.

No.

Name of Experiment Page

No.

01 TO PERFORM OPEN CIRCUIT & SHORT CIRCUIT TEST ON A SINGLE-

PHASE TRANSFORMER 2

02 TO PERFORM (a) POLARITY MARKING ON TWO WINDING

TRANSFORMER. (b) CONVERSION OF TWO-WINDING TRANSFORMER

INTO AUTO TRANSFORMER.

6

03 DETERMINATION OF REGULATION & EFFICIENCY OF THREE-PHASE

TRANSFORMER BY DIRECT LOADING. 8

04 TO PERFORM BACK TO BACK TEST ON SINGLE PHASE

TRANSFORMERS. 11

05 TO STUDY SCOTT. CONNECTIONS OF TRANSFORMER. (Study of three

phase to two phase conversion) 13

06 TO PERFORM LOAD TEST ON 3 PHASE SQUIRREL CAGE INDUCTION

MOTOR 15

07 DETERMINATION OF EQUIVALENT CIRCUIT PARAMETERS OF A THREE

PHASE INDUCTION MOTOR BY PERFORMING BLOCKED ROTOR AND

NO LOAD TEST.

18

08 SPEED CONTROL OF 3 PHASE SLIPRING INDUCTION MOTOR BY ROTOR

RESISTANCE CONTROL. 22

09 LOAD TEST ON D.C. SERIES MOTOR WITH MECHANICAL LOAD. 24

10 LOAD TEST ON D.C. SHUNT MOTOR. 26

11 SPEED CONTROL OF D. C. SHUNT MOTOR. BY

1) Armature voltage control method (Below rated speed)

2) Field current control method (Above rated speed)

28

12 TO PERFORM LOAD TEST ON D.C. SHUNT GENERATOR. 31

13 TO FIND REGULATION OF A THREE-PHASE ALTERNATOR BY OPEN

CIRCUIT AND SHORT CIRCUIT TESTS. 34

14 DETERMINE VOLTAGE REGULATION OF 3 PHASE ALTERNATOR BY

DIRECT LOADING 39

15 TO PLOT V & INVERTED V CURVES OF A SYNCHRONOUS MOTOR. 41



EXPERIMENT NO.:1

AIM: TO PERFORM OPEN CIRCUIT & SHORT CIRCUIT TEST ON A SINGLE-PHASE

TRANSFORMER.

APPARATUS: 1) 230/115V, 1KVA Single Phase transformer (1No.)

2) 0-250V, Single Phase dimmerstat (1No.)

FOR O.C. TEST FOR S.C. TEST

3) Voltmeter (0-300 V) (0-150 V) AC. each 1 No. 6) Voltmeter (0-75 V.) AC. 1No.

4) Ammeter (0-1 A) AC, 1 No. 7) Ammeter (0-5 A), (0-10 A) Ac. 1 each.

5) Wattmeter (300V, 1A) 1No. 8) Wattmeter (75 V, 5 A) 1No.

CIRCUIT DIAGRAM: FOR O.C. TEST.

V

A

V

0-150 V0-300 V

0-1 A

M

C

L

V

1- PH

230 VSUPPLY

1 PHASE

DIMMERSTAT

WATTMETER

300 V,1A

230/115 V, 1 PHASE

TRANSFORMER

FOR S.C. TEST.

V

A

A

0-10 A0-75 V

0-5 A

M

C

L

V

1- PH

230 VSUPPLY

1 PHASEDIMMERSTAT

WATTMETER75 V, 5 A

230/115 V,1 PHASE

TRANSFORMER



THEORY: It should cover the following

1) Purpose of O.C. & S.C. test.

1) Brief explanation about connection diagram.

2) Simplified equivalent circuit of a transformer and its parameters.

3) Formulae for efficiency and regulation.

4) Formulae for voltage drop for different power factor loads.

PROCEDURE: FOR O.C. TEST

1) Connect the circuit as shown.

2) Ensure that the dimmerstat Position is at zero.

3) Switch on the single phase AC. Supply.

4) Apply rated voltage of 230V, to the primary side of transformer.

5) Note the ammeter, voltmeter and wattmeter readings.

FOR S.C. TEST


2) Ensure that the dimmerstat position is at ‘0’ (zero).

3) Switch on the single phase AC. Supply.

4) Slowly increase the output voltage of the dimmerstat till the

ammeter on primary side shows rated current of 4.35 amp.

5) Note the ammeter, voltmeter & wattmeter readings.

PRECAUTIONS:

1) All the connections should be perfectly tight.

2) Supply should not be switched ON until& unless the connections are checked by the teacher.

3) Do not bend while taking the readings

4) No loose wires should lie on the work table.

5) Thick wires should be used for current circuit and flexible wires for voltage circuits.

6) The multiplying factor of wattmeter should be correctly used.

OBSERVATIONS:

FOR O.C.TEST (Read on primary side.)

Rated input Voltage

V0

No load current

I0

No load power

W0

230V

FOR S.C. TEST (Read on primary side)

Short circuit voltage

Vsc

Rated primary current

(i.e full load value)

Isc

Short circuit power

Wsc

4.35amp



CALCULATIONS:

FOR O.C. TEST

No load power factor =cosΦo = Wo / (Vo Io)

Magnetising component of Io = Iµ = Io sinΦo amps.

Core loss component of Io = Ic = Io cosΦo amps.

Core loss resistance Ro = Vo/ Ic ohm.

Magnetising reactance Xo = Vo/ Iµ ohms.

Core loss in transformer at any load = Wo

FOR S.C. TEST

Short circuit power factor cosΦsc = Wsc/ (Vsc Isc)

Short circuit impedence Zsc =Vsc / Isc Ω

Short circuit resistance Rsc = Wsc / Isc2 Ω

_________

Short circuit reactance Xsc = √ Zsc2-Rsc

2 Ω

Copper loss in transformer at full load =Wsc watts.

Copper loss in transformer at half full load = Wsc/4 watts.

EFFICIENCIES:

1) At full load and at 0.8 power factor

Full load kVA x103 x cosΦ x 100

η =

Full load KVA x103x cosΦ +core loss +copper loss at full load

2) All half full load and U.P.F.

Half load KVA x 103x cosΦ x100

η =

Half load KVA x 103 x cosΦ + core loss + copper loss at half load

REGULATIONS:

1) At full load and 0.8 power factor lagging.

Voltage drop = Isc (Rsc cosΦ + Xsc sinΦ)

% Regulation = Voltage drop x100

Rated primary voltage (Vo)

2) At full load and 0.8 power factor leading.

Voltage drop = Isc (Rsc cosΦ – Xsc sinΦ)





3) At full load and U.P.F.

Voltage drop = Isc Rsc cosΦ



EQUIVALENT CIRCUIT: Draw simplified equivalent circuit showing calculated values of all parameters on it.

RESULT: - Full load efficiency at 0.8 p.f. =

Full load efficiency at U.P.F. =

Full load regulation at 0.8 lagging p.f. =

Full load regulation at 0.8 leading p.f. =

Full load regulation at U.P.F. =

Ro = ; Xo =

Rsc = ; Xsc =

DISCUSSION: -

Q.1. What is the significance of O.C. & S.C. test?

Q.2. Why h.v. winding is kept open during O.C. test and 1.v. winding is shorted

during S.C. test in case of large transformers?

Q.3. In O.C. test, a voltmeter is connected across secondary winding and still it is

called as O.C. test. Why?

Q.4. What will happen if dc supply instead of ac supply is applied to a transformer?

Q.5. Which is the alternate method for finding efficiency and regulation of a

transformer other than O.C. & S.C. tests ? What are their advantages over each

other?

Q.6. What is the importance of equivalent circuit?

Q.7. Why regulation of transformer is negative for leading p.f. load?

Q.8. “ The wattmeter reading during O.C. test is considered as core loss while

wattmeter reading during S.C. test is considered as copper loss” Justify.

Note: - Answer only 4 questions as told by your teacher

REFERENCES: -

1. A text book on laboratory courses in electrical engineering

--Tarnekar and Kharbanda

2. Electrical technology Volume II – B.L. Theraja



EXPERIMENT NO.:2

AIM : TO PERFORM (a) POLARITY MARKING ON TWO WINDING TRANSFORMER.

(b) CONVERSION OF TWO-WINDING TRANSFORMER

INTO AUTO TRANSFORMER.

APPARATUS : 1) Two winding transformer ( 230 / 115 V, 1 KVA )

2) Voltmeter (0-300 V) 1 No. ( 0-150 V ) 2Nos.

3) Ammeter (0-5 A ) 3 Nos.

4) Loading rheostat ( 5 KW )

5) Single phase dimmerstat ( 2 KVA )

6) Transformer ( Teaser with tapping on primary & secondary )

CIRCUIT DIAGRAM :

V

VV0-300 V

0-150 V

0-150 V

230/115 V, 1 KVATransformer

1 Phase Dimmerstat

P1

P2

S1

S2

230 V 1-phase

A.C. supply

V1V2

V3

Figure (a) Polarity marking on two winding transformer.

V0-300 V

0-600 V

230/115 V, 1 KVA

Transformer1 Phase Dimmerstat

P1

P2

230 V 1-phaseA.C. supply

V1

V2

v

V0-300 V

0-600 V

230/115 V, 1 KVA

Transformer1 Phase Dimmerstat

P1

P2

230 V

1-phaseA.C. supply

V1

V2

v

Fig (b) & (c) Conversion of two-winding transformer into auto-transformer



THEORY : It should cover the following points 1) Explanation of dot and cross marking in general

2) Concept of polarity marking of two mutually coupled coils.

3) Importance of correct polarity in parallel operation of transformers

4 ) Auto transformer

PROCEDURE :

(A) POLARITY MARKING

i) Make the connections as shown in figure (a)

ii ) Connect the primary winding P1 – P2 to supply.

iii) Short circuit the terminals P2 & S2

iv) Connect the voltmeters across primary & secondary windings of transformer & one

voltmeter across P1 and S1

v) Switch on the supply.

vi) By varying the input voltage with the help of dimmerstat take various reading V1 , V2 and V3

for various steps of input voltage.

vii) Analyse the readings and decide about polarity marking of two windings of transformer.

For this assume that a dot is present at terminal P1 of the primary winding.

If V3 = (V1 + V2), the transformer has additive polarity and the other dot should be marked at S2.

If V3 = (V1 - V2), the transformer has subtractive polarity and the other dot should be marked at S1.

(B) AUTO TRANSFORMER

i) Make the connections as shown in figure (b) & (c)

ii) Based on the dot marking, convert the two winding transformer into auto-transformer of

1. Step down type (Fig.b) 2. Step-up type (Fig. c)

iii) Note the corresponding voltmeter readings of the two sides.

PRECAUTIONS: 1) All the connection should be perfectly tight.

2) Supply should not be switched ON unstill & unless the connection are checked by the

teacher.

3) Do not bend while taking the readings.

4) No loose wire should lie on the work-table.

CONCLUSION: a) The given transformer is found to have ____________ polarity . If a dot is marked at P1 on

primary side, the dot on secondary side should be at_____

b) The step-up and step-down modes of auto-transformer were studied.

DISCUSSION: 1. Explain how correct polarity is important in parallel operation of transformers

2. What is inrush current in transformers?

3. What is the effect of magnetic saturation on the excitation current of

transformers?

4. Explain how the input volt-amperes are transformed to secondary side in auto-

transformer

5. Discuss copper saving in auto-transformer

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



EXPERIMENT NO.:3

AIM: DETERMINATION OF REGULATION & EFFICIENCY OF THREE-PHASE

TRANSFORMER BY DIRECT LOADING.

APPARATUS:

Ammeter :- ( 0-10A), (0-20A) AC, each 1 Nos.

Voltmeter :- (0-300 V), (0-600 V) AC each 1 No.

Wattmeter : - Two element type, (300 V, 20 A), (600 V, 10 A)

Dimmerstat : - 3φ AC 10Amp ,1 Nos.

Transformer :- ( 440/220 V, 5 kVA) 3φ

Resistive Load, Connecting Wires etc.

CIRCUIT DIAGRAM:

AA

0-10 A

V

M

V

L

C

M L

CV

R

Y

B

0-300 V

WATTMETER10 A, 600V,

DOUBLE ELEMENT

3PHASE440 V,SUPPLY

A

V

R

B

Y

N

0-20 A0-600 V,

440/220 V,5 KVA,3 PHASE

TRANSFORMER

M L

CV

M L

V C

WATTMETER20 A, 300V,

DOUBLE ELEMENT TYPE

TYPE

3 PHASE 5 KWRESISTIVE LOAD

THEORY: It should cover the following.

1) Brief explanation about connection diagram

2) Efficiency and regulation.

3) Effect of load variation on efficiency and regulation.

4) Advantages/disadvantages of direct loading method.

5) Formula for calculating rated current on both sides.



PROCEDURE: 1) Connect the circuit as shown in figure.

2) Keep load on transformer at off position.

3) Keeping dimmer stat at zero position, switch on 3-φ supply.

4) Now increase dimmer stat voltage for 440 V.

5) Note down the no-load readings.

6) Then increase the load in steps till rated current of the transformer & note down

corresponding readings. Take atleast 8 readings.

7) Calculate efficiency & regulation for each reading.

OBSERVATION TABLE:

No-load secondary voltage E2 = ---------- Volts

Sr.No. I1Amp V1 Volts W1 Watts I 2 Amp V2 Volts W2 Watts % Reg % ŋ 1

2

3

4

5

CALCULATIONS:

O/ Power W2

% η = -------------------- x 100

I/P power W1

No load voltage (E2) – voltage at load (V2)

% Reg = ---------------------------------------------------------- x 100

No load voltage (E2)

GRAPH: Plot the graph output Power Vs efficiency.

PRECAUTIONS: 5) All the connection should be perfectly tight.

6) Supply should not be switched ON unstill & unless the connection are checked by the

teacher.

7) Do not bend while taking the readings.

8) No loose wire should lie on the work-table.

9) Thick wire should be used for current circuit and flexible wires for voltage circuits.

10) The multiplying factor of wattmeter should be correctly noted.

RESULT: The % efficiency & regulation of transformer at full load condition is found as follows.

Percentage efficiency = ------------------%

Percentage regulation = ------------------%



CONCLUSION: Transformer efficiency initially increases as the load on transformer is increased.

After maximum efficiency if we increase the load further, the efficiency of transformer reduces.

Also terminal voltage reduces as the load is increased.

DISCUSSI0N: 1. What is the condition for max efficiency?

2. What is the condition for zero voltage regulation?

3. Which is the other method of finding efficiency and regulation?

4. Draw phasor diagram of transformer at full load and 0.5 p.f. lagging.

5. Draw phasor diagram of transformer at full load and 0.5 p.f. leading.

6. What is the normal nature of output power Vs efficiency curve & why?



EXPERIMENT NO.:4

AIM: TO PERFORM BACK TO BACK TEST ON SINGLE PHASE TRANSFORMERS.

APPARATUS: 1) Two transformers, (1- phase, 1 kVA, 220 /115 V,)

2) Two dimmer stats, (0-270 V, 1- phase, 5 A)

3) Voltmeter, (0-300 V),( 0-75 V)

4) Ammeter, (0-2 A),( 0-10 A)

5) Wattmeter (0-300 V, 2 A),( 0-75 V, 10 A)

6) Connecting wires.

CIRCUIT DIAGRAM:

1Phase

AC

Supply

1Phase

AC

Supply

A

V

v

A

230 V

115 V

230 V

115 V

1 KVA Xmer

S 3

0-2 A

0-300 V

0-75 V0-10 A

C

M

V

L

C

M L

V

300 V,2 A Wattmeter

75 V, 10 A Wattmeter

S 2

S 1

1PhaseDimmerstat

1PhaseDimmerstat

V1

V2

I1

I2

W1

W2

THEORY: It should include the following.

1. Purpose of this test

2. How full load losses are produced in this test without actually loading any

transformer.

3. Explanation about the circuit diagram.

PROCEDURE: 1) Make the connections as shown in circuit diagram.

2) Keep switches S2 & S3 open and the dimmerstats at zero position.

3) Switch ON the supply and check the correctness of polarities of the two transformers.

If V2 = 0 then polarities of connected transformers are correct i.e. connections are back to

back and emf induced in secondaries are in phase opposition but if V2 = 2xKxV1, then

secondary emfs are in phase,in that case change the polarities of any one secondary winding.

4) Note down the readings of V1, I1 and W1

5) Now close switch S2, S3 and increase dimmerstat output voltage gradually so that full

load current flows through secondary windings.

6) Note down V2 , I2 and W2. While doing so , the values shown by V1, I1 and W1 should not

deviate from their earlier readings.



OBSERVATION TABLE:

SR.

No.

Primary

voltage

V1

Primary

current

I2

Primary

power

Iron loss W1

Secondary

voltage

V2

Secondary

Current

I2

Secondary

power

Cu. Loss W2

CALCULATIONS:

Iron loss per transformer Wi = W1 / 2

Copper loss per transformer Wcu = W2 / 2

Out put Power

% Efficiency = ------------------------------ x 100

Output Power + Losses

kVA x Cos Φ

% ŋ = ------------------------------------------------ x 100

kVA x Cos Φ + Iron loss + Cu. Loss

With the help of above equation, calculate efficiency at

1. Full load , UPF

2. Half full load and 0.8 p.f. lagging.

RESULTS: It is found that,

i) % Efficiency at F.L. & unity power factor =

ii) % Efficiency at half full load & 0.8 power factor (lag.) =

DISCUSSION: 1. What is the condition to be satisfied by the two transformers to be tested using this method?

2. What is the main advantage of this test?

3. Other than losses and efficiency, what else can be determined from this test?

4. How are the full load conditions simulated?

5. How are the losses separated?



EXPERIMENT NO.:5

AIM: TO STUDY SCOTT. CONNECTIONS OF TRANSFORMER.

(Study of three phase to two phase conversion)

APPARATUS: Ammeters (0-5 A, AC) 3 Nos ; (0-10 A, AC) 2 Nos.

Voltmeters (0-600 V; AC) 1 Nos, (0-150 V; Ac) 2 Nos.

Transformer 1-phase, 1 kVA, with 86.6% tap on 415V side

Transformer 1-phase, 1 kVA, with 50% tap on 415V side

Dimmer stat (0-440 V, 3 phase) 1 Nos.

Connecting wires, etc.

CIRCUIT DIAGRAM:

0-5 A

S 2

S 1

A

A

A

v

A

0-10 A

0-5 A

0-150 V

0-150 Vv

Primary of main

Xmer

Secondary of

main Xmer

LOAD

LOAD

0-10 A

3 Phase

Supply440 V

v 0-600 V

V 2T

I1T

I 2T

I 1M

V 2M

I 2M

V

TEASER Xmer


1. Explanation and mathematical proof of how a balanced two-phase supply can be obtained by using Scott

connection.

2. Phasor diagram illustrating the phase quadrature between the secondary voltages of the two transformers.

PROCEDURE: 1. Connect the circuit as shown.

2. Use 86.6% tapping for teaser transformer and 50% tapping for main transformer.

3. Keep both loads zero

4. Switch on the 3-ph. supply and take the readings.

5. Vary the loads as per given in observation table and take the readings.



OBSERVATION TABLE:

[A]FOR BALANCED LOAD

[B]FOR UNBALANCED LOAD

CALCULATIONS:

Verify the following calculated values from the measured values.

1. I1T = 1.15 x K x I2T

2. I1M = √[(K x I2M )2 + ( 0.5 x K x I2T )

2]

3. If I V2T I = I V2M I than √[V2T2 + V2M

2] = √2 V2T = √2 V2M

Where K = N2/N1 = transformation ratio

CONCLUSION:

The 3 – Phase to two phase conversion was verified. i.e. 3 – phase system can be converted in to two

phase system using Scott-connections.

DISCUSSION: 1) Is it possible to obtain a 3- phase a.c. supply from 2 – phase a.c. supply by using Scott-connection ?

2) Where dose the Scott-connection find its use?

3) If the two transformers used in Scott. Connection are identical, then how many primary turns of the teaser

transformer are actually used?

4) What is the ratio of number of turns on the primaries of teaser transformer in case of Scott-connection?

5) Are the two transformers connected for Scott-connection coupled magnetically?

6) Do you know any other method of conversion of 3-phese a.c. supply from 2-phese a.c. supply?

1

Teaser transformer Main transformer S.

No.

Load condition

I1T I2T V1T I1M I2M V1M V2M

1.

2.

No load

1kW load on

both

transformers

Teaser transformer Main transformer S.

No.

Load condition

I1T I2T V1T I1M I2M V1M V2M

1.

2.

1kW load on

Teaser

transformer

1kW load on

Main

transformer



EXPERIMENT NO.:6

AIM: TO PERFORM LOAD TEST ON 3 PHASE SQUIRREL CAGE INDUCTION MOTOR.

APPARTUS: 1) 3 Phase induction motor (3 phase, 3.0 HP, 440 V, 1440 rpm)

2) D.C. generator (2.2 kW, 220 V, 1500 rpm.)

3) Ammeter (0-5 A) AC.

4) Voltmeter (0-600 V) AC.

5) Ammeter (0-10 A) D.C.

6) Voltmeter (0-300 V) D.C.

7) Variable resistive load.

8) Wattmeter, double element type ( 440 V, 10 A )

CIRCUIT DIAGRAM:

V

+ -

F

4001.7A

A

AA

D. C.

GENERATOR

R

B

Y0-300 VFF

3 PHASE

SUPPLY

440 V, AC

(LOAD TEST ON 3 PHASE SQUIRREL CAGE INDUCTION MOTOR)

GA

3 PHASE I. M.

V

L

O

A

D

M L

V C

V C

M L

A

+

-

-

WATTMETER

440 V,10 A

0-10 A

A

0-5 A

0-600V

R

B

Y

THEORY: It should cover the following 1) Types of induction motor.

2) Applications of 1- φ & 3-φ induction motors

3) Significance of load test on induction moto

4) Variation of parameters such as speed, p.f., η with load.

5) Write down how the induction motor is loaded in this experiment.




2) Put all the load switches off. Set the generator field rheostat to its maximum.

3) Apply the 3-phase supply to the induction motor with the help of starter and

run it to its normal speed.

4) Reduce the field resistance of D.C. shunt generator so that it generates 170 V.

5) Note the meter readings and the speed with the load at zero.

6) Increase the load in steps keeping the D.C. generator voltage constant at 170 V.

Note the D.C. generator voltage, generator current, motor current, motor voltage,

Power and speed.

7) Take more readings increasing the load gradually till the full load is reached.

8) Calculate performance parameters & plot the different graphs.

OBSERVATION TABLE: Armature resistance of generator Ra= ………Ω

Sr.

No.

MOTOR SIDE

Speed

GENERATOR

SIDE

Vm Im Wm

Volt Amp Watt

N

rpm V dc I dc

Volts Amp

1

2

3

4

5

CALCULATIONS (For each reading):

Power factor of motor = Wm / (√√√√3 Vm. Im)

O/P of generator = Vdc x Idc.

Total losses of generator = I2dc Ra + constant losses.

Constant loss = ½ x wattmeter reading at no load condition.

O/P Power of motor = I/P to generator. = O/P of generator + total losses of generator

MOTOR EFFICIENCY

O/P Power of motor

% η = ---------------------------- x 100

I/P Power of motor

No -N

Slip (s) = ---------- x 100

No

No = No load speed

N = Speed at load



GRAPH: Draw graphs (on same graph paper) of motor output power versus

i) motor efficiency

ii) motor input current

iii) motor power factor

iv) motor speed

RESULT: The speed falls, the power factor (Lagging) improves and the current increases, with an increase in

the output. The efficiency increases and is maximum near full load.

DISCUSSION:

1. Which type of D.C. motor has similar output/speed characteristics?

2. Why should the powerfactor improve with loading?

3. Why powerfactor is poor at no-load?

4. At what load, the efficiency will be maximum?

5. If D.C. supply is connected to the stator winding, what will be the speed of rotating field

And the rotor? Will their be any other effects also?

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



EXPERIMENT NO.:7

AIM: DETERMINATION OF EQUIVALENT CIRCUIT PARAMETERS OF A THREE

PHASE INDUCTION MOTOR BY PERFORMING BLOCKED ROTOR AND

NO LOAD TEST.

APPRATUS: Induction motor (3-ø, 3HP, 440 V, 1440 rpm, 4.5A,)

Ammeters (0-5 A), (0-10 A) AC

Voltmeters (0-600 V) AC

Wattmeters, double element type (5A, 600V), (440 V, 10 A)

Dimmerstat (3- phase, 440 V)

Connecting wires

CIRCUIT DIAGRAM:

(A) FOR BLOCKED ROTOR TEST

0-10

AA

V

M

V

L

C

ROTOR

M L

CV

R

Y

B

0-150 V,

AC

3 Phase.

SUPPLY

STATOR

WATTMETER

10 A, 150V,DOUBLE ELEMENT TYPE

DIMMERSTAT

3 PHASE

N

R

Y

B

N



(B) FOR NO LOAD TEST

0-5 A

A

V

M

V

L

C

ROTOR

M L

CV

R

Y

B

0-600 V,

AC

3 Phase.

SUPPLY

STATOR

WATTMETER

5 A, 600V,DOUBLE ELEMENT TYPE

DIMMERSTAT

3 PHASE

N

R

Y

B

N


1. Purpose of performing these tests.

2. Complete equivalent circuit of 3-phase induction motor and its explanation

3. How the equivalent circuit parameters and machine losses can be obtained from

these tests.

PROCEDURE:


1. Connections are made as per diagram.

2. Keep the dimmerstat position at zero output voltage and hold the rotor shaft so as to disallow its

rotation.

3. Switch on the A.C. supply and gradually increase the motor input voltage till the ammeter

indicates rated current of the motor.

4. Note all the meter readings




1. Connections are made as per the circuit diagram.

2. Keep the dimmerstat position at zero output voltage and the motor at no-load.

3. Switch on the A.C. supply and gradually increase the motor input voltage to a value slightly

greater than its rated value.

4. Note all the meter readings and also the speed.

5. Now reduce the motor input voltage to its rated value and take all the readings.

6. Reduce motor input voltage subsequently in 9 to 10 steps and note the corresponding readings.

7. After finishing both tests, measure the stator winding resistance per phase (R) by using a

multimeter or by ammeter-voltmeter method. This will give the D.C. resistance of stator

winding.

PRECAUTIONS:

1) All connections should be tight.

2) No loose wires should be on the working table.

3) Supply should not be switched on till connections are checked by the teacher.

4) The motor input current should not exceed its rated value.

OBSERVATION TABLE:


Motor

Voltage

Vsc(L-L)

Motor

Current/ph

Isc

Input Power

Psc(3 ph)


CALCULATIONS:

AC resistance of stator winding due to skin effect approximately is

R1= 1.5 x D.C. value of resistance.


Per phase voltage Vsc = Vsc(L-L)/ √3

Per phase power Psc = Psc(3 ph) /3

Per phase equivalent impedance of motor Zsc/ph = Vsc / Isc = (R1 + R2’) +j (X1 + X2’)

Voltage

Applied

Vo(L-L)

No load

Current

Io

Power

Po(3-Ph)

(Po - Io2R1)



Per phase equivalent resistance of motor Rsc/ph = Psc / Isc2

R2’ = Rsc/ph - R1

Per phase equivalent reactance of motor Xsc/ph = √ (Zsc/ph2 - Rsc/ph

2 )

X1 = X2’ = Xsc/ph / 2


Per phase voltage Vo = Vo(L-L)/ √3

Per phase power Po = Po(3 ph) /3

The no-load power drawn from the supply comprises of no-load Cu loss, iron loss (Pc)

and friction/ windage loss (Pf)

Po = Io2R1 + Pc + Pf

The net power input to the stator is (Po - Io2R1) = (Pc + Pf)

Plot net power input versus square of input voltage Vo2

, the intercept of the curve with power axis

will give Pf and Pc.

Shunt resistance Ro = Vo2(rated) / Pc

No-load power-factor Cos Φo = Pc/ ( Vo(rated).Io( at rated Vo) )

Shunt reactance Xo = Vo(rated). / Io( at rated Vo).Sin Φo

EQUIVALENT CIRCUIT: Draw equivalent circuits of 3-phase I. M. at no-load and blocked rotor conditions.

RESULT: Draw complete equivalent circuit of 3-phase I. M. indicating the calculated values of all parameters on it.

DISCUSSION:

1. What are the various losses in an Induction motor?

2. What is the analogy between O.C. & S.C. test of transformer & no load test & Block rotor test of

an Induction motor?

3. For an IM, Pf = 40W/ph , net stator input = 70W/ph, calculate iron loss.

4. What is skin effect?

5. What is the effect of temperature rise on winding resistance?

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



EXPERIMENT NO.:8

AIM: SPEED CONTROL OF 3 PHASE SLIPRING INDUCTION MOTOR BY ROTOR

RESISTANCE CONTROL.

APPARATUS:

1) 3 phase slip-ring Induction motor. 415V, 3 HP, 1500rpm.

2) Voltmeter. (0-600V, AC.)

3) Ammeter. (0-10A, AC.)

4) Rheostat. 50 ohms, 5A, (3Nos.)

5) 3 Phase Dimmer stat.

6) Connecting Wires.

7) Tachometer.

CIRCUIT DIAGRAM: -

R

B Y

U

WV

A

V

50Ohms, 5A

50Ohms, 5A

50Ohms, 5A

0-600 V

0-10 A

STATOR ROTOR3 PHASE

DIMMERSTATE

3 PHASESupply440 V,

R

B

Y


1. Construction and working of a slipring I.M.

2. Application of a slipring I.M.

3. Torque-slip characteristic of I.M. for different values of rotor resistance.

4. Explanation about the circuit connection.

PROCEDURE:

1. Connect the ckt. as shown.

2. Keep the Dimmerstat as zero output voltage & the external rotor resistance at minimum resistance

position.

3. Switch ON the supply & increase the input voltage to stator winding upto its rated value.

4. Measure the speed.

5. Now increase the rotor resistance in steps & note the corresponding values of speed.

6. Draw a graph of rotor resistance versus speed.



OBSERVATION TABLE:

Sr.No. External rotor resistance in Ohms. Speed (rpm)

01

02

03

04

05

06

0Ω

10Ω

20Ω

30Ω

40Ω

50Ω

PRECAUTIONS:

5) All connections should be tight.

6) No loose wires should be on the working table.

7) Supply should not be switched on till connections are checked by the teacher.

8) The motor input current should not exceed its rated value.

GRAPH: Draw a graph of rotor external resistance versus speed.

CONCLUSION: It found that with rise in rotor resistance, there is a corresponding drop in the speed.

DISCUSSION:

1) What will happen if the rotor circuit of a slipring I.M. is kept open & electric supply is given to its stator

winding.

2) What is the constructional difference between a slip-ring & a squirrel cage I.M.

3) Draw & explain the Torque-speed characteristic of a slip ring I.M. for different values of rotor resistance.

4) What is the effect of changing the rotor resistance on the slip at max torque SMT.

5) Can a I.M. rotate at synchronous speed? Justify your answer.

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



EXPERIMENT NO.:9

AIM: LOAD TEST ON D.C. SERIES MOTOR WITH MECHANICAL LOAD.

APPARATUS: D.C.series motor (3H.P., 220 V, 12 A, 1500rpm.) with brake drum attachment.

Voltmeter (0-300 V) D.C.

Ammeter (0-20 A) D.C.

Tachometer.

CIRCUIT DIAGRAM:

BRAKE DRUM

ATTACHAMENT

L A

STARTER

DC SERIES

MOTOR

220 V , DCSUPPLY A

AA

0-300 V V

+

-

-

-

+

+

0-20 A

A

S1

S2


1. Working principle of D.C. series motor

2. Applications of D.C. series motor

3. Purpose of this test

4. Normal variation of efficiency, torque and speed of D.C. series motor with rise in load.


2) Keeping some load on the motor, start itwith the help of starter.

3) At this load, note down the speed and also the forces in springs connected to brake drum.

4) Note voltmeter as well as ammeter reading.

5) Increase the mechanical load in steps by tightening the rope and note all the readings again.

6) Repeat step 5 till the rated current of motor is reached.

7) Calculate torque and efficiency of motor.



PRECAUTIONS: 1) All connections should be perfectly tight and no loose wire should lie on the work table.

2) Before switching ON the dc supply , ensure that the starter’s moving arm is at it’s maximum

resistance position.

3) Do not switch on or operate the D.C. series motor without load

4) Before switching on the D.C.supply, ensure some water inside the drum for cooling purpose.

5) Do not switch on the supply, until and unless the connection are checked by the teacher

OBSERVATION TABLE: Radius of brake drum r = 0.15 m.

CALCULATIONS:

Output torque T = [(F1-F2) x 9.81 x r ] N.m where is radius of brake drum.

Output power Po = 2πNT/60 watts.

Input power Pi = Vm.Im

Po

%η = ---------- x 100

Pi

GRAPH: Plot speed Vs torque and output power Vs efficiency. CONCLUSION: At light load the motor speed is high and it reduces fast with rise in load.

DISCUSSION:

1) What is the significance of back e.m.f. in a d.c. motor?

2) What is the chief advantage of a d.c. series motor?

3) What is the necessity of starter in a d.c. motor?

4) Why is a d.c. series motor used to start heavy loads?

5) Can a d.c. series motor operate on an a.c. supply?

6) Armature core of a dc machine is always laminated, why?

7) Why the D.C. series motor is not operated at zero or light loads?

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

Sr.

No.

Vm

(Volts)

Im

(Amps)

F1

(kg)

F2

(Kkg)

N

(rpm)



EXPERIMENT NO.:10

AIM: LOAD TEST ON D.C. SHUNT MOTOR.

APPARATUS: D.C. Motor ( 3.0 HP, 220 V, 10 A, 1500 rpm.)

D.C. Generator (2.2 kW, 220 V, 1500 rpm.)

Two Voltmeters (0-300 V) D.C

Two Ammeters (0-10 A) D.C.

Two Rheostats (1000 ohms./1.2 A) (400 ohms./ 1.7 A)

Lamp load (1kW., 230 V)

Tachometer (0-1500rpm.)

CIRCUIT DIAGRAM:

+ -

F

4001.7 A

A

AA

D. C. GENERATOR

0-300 V

FF

G

L

OAD

A

+

-

0-10 A

L A

F

STARTER

DC MOTOR

220 V , DCSUPPLY M

A

AA

F

FF

1000

1.2 A

0-300 V

A

VV

+

- -

-

+

+

0-10 A


1) Principle of D.C.shunt motor

2) What is back e.m.f?

3) What is the effect on speed when motor is loaded?

4) How the speed of shunt motor is expressed in terms of back emf & flux (φ) & also in

terms of voltage & armature current

PROCEDURE: 1) Connect the circuit as shown in figure and keep the load zero.

2) Set the field rheostat of motor to zero & field rheostat of generator to maximum.

3) Switch on D.C. supply & start the motor with the help of starter

4) Adjust the field rheostat of motor to obtain rated speed or any suitable speed.

Then don’t disturbed this setting.

5) Adjust the D.C.shunt generator voltage to its rated value(220 V) with the help

of its field rheostat.

6) Note the meter readings and speed at this no-load condition.

7) Now increase the electrical load and adjust the generator terminal voltage

constant at previous value.

8) Note the reading of ammeters, voltmeters & speed.



9) Repeat 6 to 8 above to cover the range of no load to full load of motor.

10) Measure the armature resistance Ra of the motor by multimeter.

PRECAUTIONS: 1) All connections should be perfectly tight and no loose wire should lie on

the work table.

2) Before switching ON the dc supply , ensure that the starter’s moving arm

is at it’s maximum resistance position.

3) Do not switch on the supply, until and unless the connection are checked

by the teacher

4) Avoid error due to parallax while reading the meters.

5) Hold the tachometer with both hands steady and in line with the motor shaft

so that it reads correctly.

OBSERVATIONS:

Motor

side

Generato

r side

Sr.

No.

V m I m Vg I g

Speed

N

rpm.

CALCULATIONS ( for all readings):

Generator output power Pg = Vg.Ig

Generator constant loss Pc = motor constant loss = [Vm(no-load).Im(no-load)]/2

Generator input power = Pg + Ig2.Ra + Pc

Motor output power Pm(o) = Generator input power

Motor input power Pm(i) = Vm.Im

Motor efficiency is

Pm(o)

∴ % η = ----------------- x 100

Pm(i)

GRAPH: Plot on same graph paper

(i) Motor o/p power versus efficiency.

(ii) Motor o/p power versus Motor I/p current.

(iii) Motor o/p power versus speed.

CONCLUSION: With rise in load the motor current increases & the speed decreases slightly.

The efficiency initially increases with the load, reaches to its maximum & then decreases. The

curves obtained experimentally are shown on the graph

DISCUSSIONS: 1. How do you load the motor in this experiment?

2. What was the assumption made in calculations?

3. What is the condition for maximum torque of a D.C. shunt motor?

4. Why is the starter necessary ?

5. What is the power factor of the load used?



EXPERIMENT NO.:11

AIM: SPEED CONTROL OF D. C. SHUNT MOTOR. BY

2) Armature voltage control method (Below rated speed)

3) Field current control method (Above rated speed)

APPRATUS: 1) D.C. Shunt motor – (230 V, 5 A, 1500 rpm, 1KW,)

2) D.C. starter.

3) D.C. Ammeter – (0-2 A)

4) D.C. Voltmeter – (0-300 V)

5) Rheostat 400 Ω, 1.7 A, and 1000 Ω 1.2 A

6) Tachometer

CIRCUIT DIAGRAM:

THEORY: The theory should cover details about following points.

1) Principle of working of dc shunt motor

2) Relation between speed, armature resistance and flux.

3) Explanation of above circuit diagram.

4) Brief explanation about speed control by above two methods.

PROCEDURE: 1) Connect the circuit as shown above.

2) Adjust both rheostats at their minimum resistance position initially.

3) Switch ON the DC supply.

4) Turn the moving arm of starter to it’s minimum resistance position.



(A) FOR ARMATURE VOLTAGE CONTROL

5) Take a set of readings at minimum resistance position of both rheostats.

6) Keeping field current constant, vary the rheostat connected in armature

circuit by increasing it’s value and note down the armature voltage and

corresponding values of speed.

(B) FOR FIELD CURRENT CONTROL

6) Bring the rheostat connected in armature circuit back to minimum resistance

position.

7) Keeping armature voltage constant now increase the resistance of field circuit by

adjusting its rheostat.

8) Measure field current and corresponding values of speed.

9) After taking the required sets of readings adjust both rheostats to their minimum

resistance position and switch OFF the dc supply.

PRECAUTIONS: 1) All connections should be perfectly tight and no loose wire should lie on

the work table.

2) Before switching ON the dc supply , ensure that the starter’s moving arm

is at it’s maximum resistance position.

3) Do not switch on the supply, until and unless the connection are checked

by the teacher


5) Hold the tachometer with both hands steady and in line with the motor shaft

so that it reads correctly.

OBSERVATIONS:

(A) FOR ARMATURE VOLTAGE CONTROL

Field current (If) = Amps (constant)

Sr.

No.

If (Amp) Va (Volts) Speed (N) rpm.

1

2

3

4

5



(B) FOR FIELD CURRENT CONTROL

Armature voltage (Va) = volts (constant)

GRAPH: Plot the following on separate graph papers.

1) N versus Va 2) N versus If

CONCLUSION:

It is observed that the speed of dc shunt motor increases above normal value by field

current control method and decreases below normal value by armature control method.

DISCUSSION:- Q.1.What are the limitations of armature voltage control and field current control

methods?

Q.2. Why both rheostats are kept at minimum resistance position in the starting

condition?

Q.3. What is starter? Why is it required?

Q.4. What is back emf? What is it’s significances?

Q.5. Why is starter required during starting condition & not during running condition?

Q.6. Draw internal and external characteristics of dc shunt motor.

Q.7. What are the applications of dc shunt motor?

Q.8. Why are brushes made form carbon?

Q.9. Why is thin conductor used for field winding? & thick conductor for armature

winding?

Note: Answer any 04 as total by your teacher

REFERENCES:- 1. A text book on laboratory experiments in electrical engg.

Kharbanda & Tarnekar

2. Electrical technology Vol-I I – B.L. Theraja

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

Sr.

No.

If (amp) Va (Volts) Speed (N) rpm.

1

2

3

4

5



EXPERIMENT NO.:12

AIM: - TO PERFORM LOAD TEST ON D.C. SHUNT GENERATOR

APPARATUS: -

M.G. Set, D.C. Generator (1kw, 4.6A, 230V, 1500rpm.self exct. Compound gene.)

D.C.Motor (1kw, 4.6A, 230V, 1500rpm.self exct. Shunt Motor.)

Two Voltmeter (0-300V) D.C.

Two Ammeter (0-10A) D.C.

Two Field Rheostat (400Ω, 1.7A)

Lading Rheostat (1kw,)

Tachometer (0-1500rpm.)

CIRCUIT DIAGRAM: -

+ -

F

4001.7 A

A

AA

D. C. GENERATOR

0-300 V

FF

G

L

O

A

D

A

+

-

0-10 A

L A

F

STARTER

DC MOTOR

220 V , DCSUPPLY M

A

AA

F

FF

10001.2 A

0-300 V

A

VV

+

- -

-

+

+

0-10 A

THEORY: - The theory should cover details about following points

1) What is the working principle of d. c. generator?

2) What is back e.m.f?

3) Why does the voltage drop on load? Does it drop more with more load? If yes, why?

4) What is the effect of changing the excitation current (field current) of the shunt generator?

5) What is the relation between Eb and V?



OBSERVATION TABLE: - Speed = 1500 rpm (constant)

PROCEDURE: -

1) Connection the ckt. As shown in figure.

2) Set the motor field rheostat to its minimum & generator field rheostat to maximum.

3) Start the d.c. motor with the help of starter and adjust the speed to the rated speed of the

generator. Keep the speed constant throughout the experiment with help of motor field

rheostat.

4) Adjust the field rheostat of the d.c. generator to obtain rated terminal voltage at no load

condition. Field rheostat adjusted at this position should nit be disturbed for this set.

5) Change the load current by varying load and note down the corresponding terminal

voltage. Take readings covering the range from no load to little over full load.

6) Plot terminal voltage Vs load current and efficiency Vs output.

CALCULATION: -

O/p of generator

Generator efficiency ηG = -----------------------

I/p of generator

O/p of motor

Motor efficiency ηm = --------------------

I/p of motor

As O/P of motor = I/p of generator

O/p of generator

ηG ηm = -------------------

I/p of motor

O/p of generator

ηG = --------------------

I/p of motor

Sr.

No.

Motor Side

Vm Im

Volts Amp

Generator Side

Vg Ig

Volts Amp

Motor Input

= VmxIm

(Watts)

Generator

Output =

VgxIg (watts)

Efficiency of

generator =

VgIg

------- x 100

VmIm

1/2



GRAPH: -

Vg %

Efficiency

ILOutput of

Generator

RESULT & CONCLUSION: -

The external characteristics of d.c. Shunt generator and efficiency curve are plotted. The

terminal voltage of a d.c. Shunt generator reduces as the load current is increased

DISCUSSION: -

1) What should be done if the d.c. Shunt generator fails to build up?

2) What are the reasons of fall of terminal voltage of a d.c. shunt generator

3) Define armature reaction.

4) What are the different types of losses in generator?

5) What is the e.m.f. Equation of generator?

6) What is the relation between Eb and V?

7) What is the function of commutator?

8) What is the condition for maximum efficiency?



EXPERIMANT NO.:13

AIM: TO FIND REGULATION OF A THREE-PHASE ALTERNATOR BY OPEN CIRCUIT

AND SHORT CIRCUIT TESTS

APPARATUS:

1. Ammeter (0-5A) AC-1No; (0-1A) DC-1 No.

2. Voltmeter (0-300V) AC-1 No.

3. Tachometer – 1 No.

4. Rheostats (400Ω. 1.7A) 1No; 1000Ω. 1.2A 1No.

5. Alternator 3 kVA, 4.2A, 1500 RPM, 3φ

6. D.C. Motor 3 HP, 220V, 1500RPM

7. Connecting wires etc.

CIRCUIT DIAGRAM: [A] OPEN CIRCUIT TEST

A

V

+

-

AFL

STARTER

F

10001.2A

A

AA

D. C. MOTOR

R

NB

Y

0-300 V

AC

0-1A

DC

-

+

+ -

1000

FIELD WDG OF

ALTERNATOR

,1.2A230 V DC

F

FF

FF

D C SUPPLY

230 V M A

ALTERNATOR



[B] SHORT CIRCUIT TEST

ALTERNATOR

A

A

L F A+

-

FF

F

1000

1.2A

AA

A R

B

N

Y

D.C. MOTOR

FIELD WDG

OF ALTERNATOR

+

-

(0-1A)

DC

D.C. SUPPLY -+

F FF

(0 -5 A)

AC

D.C. SUPPLY

230 V

STARTER

M A

THEORY: The theory should cover following points

1. Meanings of regulation and synchronous impedance

2. Details about synchronous impedance method for finding regulation

3. Explanation about above circuit diagrams

4. List of other possible methods for finding regulation

PROCEDURE:

[A] OPEN CIRCUIT TEST


2) Set potential divider to zero output position and motor field rheostat to minimum value.

3) Switch on dc supply and start the motor.

4) Adjust motor speed to synchronous value by motor field rheostat and note the meter readings.

5) Increase the field excitation of alternator and note the corresponding readings.

6) Repeat step 5 till 10% above rated terminal voltage of alternator.

7) Maintain constant rotor speed for all readings.



[B] SHORT CIRCUIT TEST


2) Star the motor with its field rheostat at minimum resistance position and the potential divider set to zero

output.

3) Adjust the motor speed to synchronous value.

4) Increase the alternator field excitation and note ammeter readings.

5) Repeat step 4 for different values of excitations (field current). Take readings up to rated armature

current. Maintain constant speed for all readings

6) Measure the value of armature resistance per phase Ra by multimeter or by ammeter-voltmeter method.

7) Plot the characteristics and find the synchronous impedance.




8) Do not switch on the supply, until and unless the connections are checked by the teacher


10) Hold the tachometer with both hands steady and in line with the motor shaft so that it reads

correctly.

6) Ensure that the winding currents do not exceed their rated values.

OBSERVATIONS:

Alternator armature resistance per phase Ra = ------- Ω Rotor speed = -------------------- RPM (constant).

O.C TEST. S.C.TEST.

Sr. No.

Field current

If

Short circuit

current Isc

Sr. No Field current

If (Amp)

Terminal voltage

Per phase Vo



GRAPH: Plot the readings to draw following graphs. Use same graph paper for both curves.

1. If versus Vo (from OC test) 2. If versus Isc (from SC test)

CALCULATIONS:

OA Vo1

Synchronous impedance Zs = ---------- = ----------- for field current Isc1

OB Isc1

Isc1 is selected over the linear part of OCC, generally it corresponds to rated armature current.

Synchronous reactance Xs = √ (Zs2 - Ra

2)

Where Ra = Armature resistance of alternator (per phase)

Calculate the excitation emf Eo and voltage regulation for full-load and

1. 0.8 lagging p.f.

2. UPF

3. 0.8 leading p.f.

Eo = √[(V cosφ + Ia Ra)2 + (V sin φ + Ia Xs)

2]

+ sign is for lagging pf load.

- sign is for leading pf load.

V = rated terminal voltage per phase of alternator

Eo - V

%Regulation = ------------ x 100

V

PHASOR DIAGRAMS:

Draw phasor diagrams for above three loads and verify the calculated results.

S.N. Zs Zs (av). Xs Xs (av).



RESULT: Regulation of alternator at full load is found to be,

At unity pf = --------------

At 0.8 lagging = ---------------

At 0.8 leading = --------------

Synchronous Impedance varies for different values of excitation.

DISCUSSION: 1. Why OCC looks like B-H curve?

2. Why SCC is a straight line?

3. What is armature reaction effect?

4. What are the causes of voltage drop?

5. When is the regulation negative and why?

6. Can we find regulation of a salient pole machine by this test? Justify your answer.

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



EXPERIMANT NO.:14

AIM: TO DETERMINE VOLTAGE REGULATION OF 3 PHASE ALTERNATOR BY DIRECT LOADING.

APPARATUS:

1) Ammeter (0-5amp.AC) 01 No.

2) Ammeter (0-1amp.DC) 01 No

3) Voltmeter (0.300V AC) 01No

4) Rheostat (400ohms.1.7 Amp) 01 No

5) Rheostat (1000 Ohms.1.2 Amp) 01 No

6) 3 Phase Resistive Load (400volts.10amp ) 01 No

7) Alternator (3 kVA 4.2 amp, 1500rpm) 3 Phase

8) D.C. Motor (Shunt Type) 3 HP, 220V, 1500rpm

CIRCUIT DIAGRAM:

A

+AFL

STARTER

F

1000

1.2A

A

AA

D. C. MOTOR

R

Y

BN

0-1A

DC

-

+

+ -

1000

FIELD WDG OF

ALTERNATOR

230 V DC

F

FF

FF

D C SUPPLY

230 V M A

ALTERNATOR

A

0-5 A

V

0-300 V

R

N

Y

B

1.2 A

L

O

A

D

-

THEORY:

It should cover the following

1. Definition of voltage regulation and its expressions for different power factors of load.

2. Reasons for change in terminal voltage

3. Explanation about circuit diagram

4. List of other methods for finding regulation



PROCEDURE:

1) Connect the circuit as shown in the diagram.

2) Keep load zero, set field potential divider to zero output voltage position.

3) Keep field resistance of motor to its minimum value.

4) Start the motor with the help of starter.

5) With the field rheostat of motor adjust the speed to synchronous value.

6) Switch on DC supply of field (Alternator) and adjust the potential divider so that the voltmeter

reads rated voltage of the alternator. Note this voltage as no load voltage E.

7) Increase the load in steps till rated current of alternator and note the different sets of readings.

8) Keep speed constant during all the readings with the help of motor field rheostat..




13) Do not switch on the supply, until and unless the connection are checked by the teacher


15) Hold the tachometer with both hands steady and in line with the motor shaft so that it reads

correctly.

OBSERVATION: E = Terminal voltage at no load = volts.

Sr.

No.

Load current Terminal voltage

Vt

% Regulation

CALCULATIONS:

E - Vt

% regulation = --------------- x 100

E

RESULT: The regulation at full load and ---- power factor is found to be ---------- %

CONCLUSION: As the load on the alternator increases the regulation also increases

DISCUSSION:

1) Can the terminal voltage rise? Under which load?

2) If the speed of the driving motor falls due to loading what will be the effects?

3) Give the classification of alternator on the basis of rotor and their application?

4) Why the excitation given to alternator is generally DC and not AC?

5) Mention disadvantages of determining the regulation of alternator by direct loading?

6) What is hunting in Alternator?

7) What is the role of damper winding in Alternator?

8) What is chording and write their advantages?

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



EXPERIMENT NO.:15

AIM: TO PLOT V & INVERTED V CURVES OF A SYNCHRONOUS MOTOR.

APPARATUS:

1) Synchronous motor 3 Phase, 3 HP, 440V, 8.2A ,1500 rpm,

2) DC shunt Generator 220V, 9A, 1500 rpm

3) Power factor meter-600V, 10A

4) Voltmeter AC- (0-600V), DC- (0-300V)

5) Ammeter AC- (0-10A)

6) Ammeter DC (0-2A), DC (0-10A)

7) Rheostat-470 ohm, 1.2A,

8) Resistive load bank, tachometer, connecting wires, etc.

CIRCUIT DIAGRAM:

V

+ -

F

400

1.7A

A

AA

D. C.

GENERATOR

R

B

Y0-300 VFF

3 PHASE

SUPPLY

440 V

GA

ALTERNATOR

V

L

O

A

D

M L

V C

V C

M L

A

+

-

0.2 A DC

-

+

+ -

FIELD WDG OF

ALTERNATOR

F FFPF- METER

600 V,10 A

0-10 A

A

0-10A

4

5

6

R

B

Y

A

230 V dc

THEORY:

It should cover the following.

1. Significance of V & inverted V curves of synchronous motor.

2. Phasor diagram of a synchronous motor showing effect of change in excitation

3. Necessary condition for obtaining V & inverted V curves

4. Explanation about circuit diagram

PROCEDURE:

1) Make the connections as shown in circuit diagram.

2) Adjust the field rheostat of DC generator at maximum position, the potential divider at zero output

position and the load at off condition.

3) Switch on the 3-ph. supply, start the synchronous motor and let it run at its rated speed.



4) Switch on the DC supply and adjust the generator field current to a suitable value so that it generates

rated voltage.

5) Increase the alternator field current and note down corresponding power factor and armature current

covering a range from low lagging to low leading power factor through a unity power factor. Note

that armature current is minimum when the p.f. in unity.

6) Repeat step No.5 for some constant load on the Generator.

OBSERVATIONS: [A] AT NO LOAD

[B] AT LOAD

GRAPH: Plot the curves between armature current (Ia) vs field current (If) and power factor

(cosφ) vs field current (If)

CONCLUSION:

1. The variation of armature current (line current) and its power factor due to field current variation at

load and at no load are shown. The armature current is minimum when the PF is unity.

2. As load increases the V curve shifts upward and the inverted V curve shift towards right.

DISCUSSION :

1. With what condition synchronous motor can be used as a synchronous condenser.

2. What are the special applications of an over excited synchronous motor.

3. Explain the effect of change of excitation of a synchronous motor on its armature current.

4. Explain the effect of change of excitation of a synchronous motor on its power factor.

5. With the given excitation a synchronous motor draws a unity PF current . if the mechanical

load is increased what will be the power factor and current for the same excitation.

6. Why V curve shift upwards and inverted V curve shift right as the load increases.

7. Explain the effect of change of excitation of a synchronous generator on its armature current.

8. Explain the effect of change of excitation of a synchronous generator on its power factor.

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

Sr.

No.

If Power Factor

(cosφ)

Ia

Sr.

No.

If Power Factor

(cosφ)

Ia

g.h. raisoni college of engineering...

Documents