ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
G.H. RAISONI COLLEGE OF ENGINEERING
DEPARTMENT OF ELECTRICAL ENGINEERING
ELECTRICAL MACHINES LABORATORY
ELECTRICAL MACHINES-I/BEMC
INDEX
Sr.
No.
01
02
03
04
05
06
07
08
09
Name of Experiment
TO PERFORM OPEN CIRCUIT & SHORT CIRCUIT TEST ON A SINGLEPHASE TRANSFORMER
TO PERFORM (a) POLARITY MARKING ON TWO WINDING
TRANSFORMER. (b) CONVERSION OF TWO-WINDING TRANSFORMER
INTO AUTO TRANSFORMER.
DETERMINATION OF REGULATION & EFFICIENCY OF THREE-PHASE
TRANSFORMER BY DIRECT LOADING.
TO PERFORM BACK TO BACK TEST ON SINGLE PHASE
TRANSFORMERS.
TO STUDY SCOTT. CONNECTIONS OF TRANSFORMER. (Study of three
phase to two phase conversion)
TO PERFORM LOAD TEST ON 3 PHASE SQUIRREL CAGE INDUCTION
MOTOR
DETERMINATION OF EQUIVALENT CIRCUIT PARAMETERS OF A THREE
PHASE INDUCTION MOTOR BY PERFORMING BLOCKED ROTOR AND
NO LOAD TEST.
SPEED CONTROL OF 3 PHASE SLIPRING INDUCTION MOTOR BY ROTOR
RESISTANCE CONTROL.
LOAD TEST ON D.C. SERIES MOTOR WITH MECHANICAL LOAD.
Page
No.
2
6
8
11
13
15
18
22
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.
DETERMINE VOLTAGE REGULATION OF 3 PHASE ALTERNATOR BY
DIRECT LOADING
TO PLOT V & INVERTED V CURVES OF A SYNCHRONOUS MOTOR.
34
14
15
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
1
39
41
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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
3) Voltmeter (0-300 V) (0-150 V) AC. each 1 No.
4) Ammeter (0-1 A) AC, 1 No.
5) Wattmeter (300V, 1A) 1No.
FOR S.C. TEST
6) Voltmeter (0-75 V.) AC. 1No.
7) Ammeter (0-5 A), (0-10 A) Ac. 1 each.
8) Wattmeter (75 V, 5 A) 1No.
CIRCUIT DIAGRAM:
FOR O.C. TEST.
WATTMETER
300 V,1A
M
L
C
V
0-1 A
A
1- PH
230 V
SUPPLY
V
V
0-150 V
0-300 V
1 PHASE
DIMMERSTAT
230/115 V, 1 PHASE
TRANSFORMER
FOR S.C. TEST.
WATTMETER
75 V, 5 A
M
L
C
V
1- PH
230 V
SUPPLY
0-5 A
A
A
V
0-10 A
0-75 V
1 PHASE
DIMMERSTAT
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
230/115 V,
1 PHASE
TRANSFORMER
2
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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
1) Connect the circuit as shown.
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
No load current
No load power
V0
I0
W0
230V
FOR S.C. TEST (Read on primary side)
Short circuit voltage
Rated primary current
Vsc
(i.e full load value)
Isc
4.35amp
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
3
Short circuit power
Wsc
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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-Rsc2 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)
% Regulation =
Voltage drop x100
Rated primary voltage (Vo)
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
4
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
3) At full load and U.P.F.
Voltage drop = Isc Rsc cosΦ
% Regulation =
Voltage drop x100
Rated primary voltage (Vo)
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
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
5
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
EXPERIMENT NO.:2
AIM :
TO PERFORM (a) POLARITY MARKING ON TWO WINDING TRANSFORMER.
(b) CONVERSION OF TWO-WINDING TRANSFORMER
INTO AUTO TRANSFORMER.
APPARATUS : 1)
2)
3)
4)
5)
6)
Two winding transformer ( 230 / 115 V, 1 KVA )
Voltmeter (0-300 V) 1 No. ( 0-150 V ) 2Nos.
Ammeter (0-5 A ) 3 Nos.
Loading rheostat ( 5 KW )
Single phase dimmerstat ( 2 KVA )
Transformer ( Teaser with tapping on primary & secondary )
CIRCUIT DIAGRAM :
0-150 V
V
V3
P1
230 V
1-phase
A.C. supply
0-300 V
S1
V
V
0-150 V
V1
V2
P2
1 Phase Dimmerstat
S2
230/115 V, 1 KVA
Transformer
Figure (a) Polarity marking on two winding transformer.
0-600 V
0-600 V
P1
P1
v
v
230 V
1-phase
A.C. supply
V2
0-300 V
V
V2
230 V
1-phase
A.C. supply
0-300 V
V1
V1
P2
P2
1 Phase Dimmerstat
V
230/115 V, 1 KVA
Transformer
1 Phase Dimmerstat
230/115 V, 1 KVA
Transformer
Fig (b) & (c) Conversion of two-winding transformer into auto-transformer
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
6
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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)
ii )
iii)
iv)
Make the connections as shown in figure (a)
Connect the primary winding P1 – P2 to supply.
Short circuit the terminals P2 & S2
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 autotransformer
5. Discuss copper saving in auto-transformer
---------------------------------------------------------------------------------------------------------
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
7
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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:
0-10 A
0-20 A
0-600 V,
R
M
M
L
L
A
A
V
C
V
V
R
A
C
V
Y
0-300 V
Y
V
B
M
3PHASE
440 V,SUPPLY
C
L
V
C
N
WATTMETER
10 A, 600V,
DOUBLE ELEMENT
TYPE
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.
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
B
L
440/220 V,5 KVA,
3 PHASE
TRANSFORMER
THEORY:
M
8
20 A, 300V,
WATTMETER
DOUBLE ELEMENT
TYPE
3 PHASE 5 KW
RESISTIVE LOAD
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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
1
2
3
4
5
V2 Volts
W2 Watts
% Reg
%ŋ
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 = ------------------%
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
9
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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?
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
10
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
EXPERIMENT NO.:4
AIM:
TO PERFORM BACK TO BACK TEST ON SINGLE PHASE TRANSFORMERS.
APPARATUS: 1)
2)
3)
4)
5)
6)
Two transformers, (1- phase, 1 kVA, 220 /115 V,)
Two dimmer stats, (0-270 V, 1- phase, 5 A)
Voltmeter, (0-300 V),( 0-75 V)
Ammeter, (0-2 A),( 0-10 A)
Wattmeter (0-300 V, 2 A),( 0-75 V, 10 A)
Connecting wires.
CIRCUIT DIAGRAM:
300 V,2 A Wattmeter
0-2 A
M
L
A
W1
I1
1Phase
AC
Supply
C
V
V
V1
S1
0-300 V
1Phase
Dimmerstat
230 V
230 V
1 KVA Xmer
75 V, 10 A Wattmeter
M
L
I2
115 V
A
1Phase
AC
Supply
115 V
0-10 A
C
v
V
0-75 V
V2
W2
S2
THEORY:
1Phase
Dimmerstat
S3
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.
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
11
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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
% Efficiency
% ŋ
Out put Power
= ------------------------------ x 100
Output Power + Losses
kVA x Cos Φ
= ------------------------------------------------ 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.
2.
3.
4.
5.
What is the condition to be satisfied by the two transformers to be tested using this method?
What is the main advantage of this test?
Other than losses and efficiency, what else can be determined from this test?
How are the full load conditions simulated?
How are the losses separated?
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
12
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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
0-10 A
TEASER Xmer
A
A
I 2T
I1T
v
0-150 V
V 2T
LOAD
3 Phase
Supply
440 V
S1
0-5 A
A
I 1M
v
Primary of main
Xmer
0-600 V
V
0-10 A
S2
Secondary of
main Xmer
A
v
I 2M
0-150 V
V 2M
LOAD
THEORY: It should cover the following.
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.
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
13
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
OBSERVATION TABLE:
[A]FOR BALANCED LOAD
S. Load condition
No.
1.
No load
2.
Teaser transformer
I1T
I2T
V1T
I1M
Main transformer
I2M
V1M
V2M
I1M
Main transformer
I2M
V1M
V2M
1kW load on
both
transformers
[B]FOR UNBALANCED LOAD
S. Load condition
No.
1.
1kW load on
Teaser transformer
I1T
I2T
V1T
Teaser
transformer
2.
1kW load on
Main
transformer
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 + V2M2] = √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
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
14
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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:
R
V
L
+
A
A
Y
0-600
V
Y
R
A
FF
A
400
1.7A
C
B
L
3 PHASE I. M.
WATTMETER
440 V,10 A
D. C.
GENERATOR
(LOAD TEST ON 3 PHASE
SQUIRREL CAGE INDUCTION MOTOR)
-
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.
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
0-300 V
V
AA
M
+
G
B
SUPPLY
440 V, AC
V
-
F
C
V
3 PHASE
0-10 A
0-5 A
M
15
L
O
A
D
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
PROCEDURE:
1) Connect the circuit as shown in figure.
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
Vm
Volt
Im
Amp
Wm
Watt
Speed
GENERATOR
SIDE
N
rpm
V dc
Volts
I dc
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 = ---------------------------I/P Power of motor
No -N
Slip (s) = ---------- x 100
No
No = No load speed
N = Speed at load
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
16
x
100
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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?
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G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
17
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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
M
R
V
L
C
R
V
ROTOR
0-150 V,
AC
Y
STATOR
Y
B
V
3 Phase.
SUPPLY
N
C
B
M
L
N
DIMMERSTAT
3 PHASE
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
WATTMETER
10 A, 150V,
DOUBLE ELEMENT TYPE
18
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
(B) FOR NO LOAD TEST
0-5 A
M
A
R
V
L
C
R
V
ROTOR
0-600 V,
AC
Y
STATOR
Y
B
V
3 Phase.
SUPPLY
N
C
B
M
L
N
DIMMERSTAT
3 PHASE
WATTMETER
5 A, 600V,
DOUBLE ELEMENT TYPE
THEORY: It should cover the following.
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:
(A) FOR BLOCKED ROTOR TEST
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
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
19
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
(B) FOR NO LOAD TEST
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:
(A) FOR BLOCKED ROTOR TEST
Motor
Voltage
Vsc(L-L)
Motor
Current/ph
Isc
Input Power
Psc(3 ph)
(B) FOR NO LOAD TEST
Voltage
Applied
Vo(L-L)
No load
Current
Io
Power
Po(3-Ph)
(Po - Io2R1)
CALCULATIONS:
AC resistance of stator winding due to skin effect approximately is
R1= 1.5 x D.C. value of resistance.
(A) FOR BLOCKED ROTOR TEST
Per phase voltage Vsc = Vsc(L-L)/ √3
Per phase power Psc = Psc(3 ph) /3
Per phase equivalent impedance of motor
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
Zsc/ph = Vsc / Isc = (R1 + R2’) +j (X1 + X2’)
20
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
Per phase equivalent resistance of motor
Per phase equivalent reactance of motor
Rsc/ph = Psc / Isc2
R2’ = Rsc/ph - R1
2
2
Xsc/ph = √ (Zsc/ph - Rsc/ph )
X1 = X2’ = Xsc/ph / 2
(B) FOR NO LOAD TEST
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)
The net power input to the stator is
Po = Io2R1 + Pc + Pf
2
(Po - Io R1) = (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
No-load power-factor
Shunt reactance
Ro = Vo2(rated) / Pc
Cos Φo = Pc/ ( Vo(rated).Io( at rated Vo) )
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?
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G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
21
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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: 0-10 A
R
A
3 PHASE
Supply
440 V,
V
R
50Ohms, 5A
U
0-600 V
Y
B
Y
W
V
50Ohms, 5A
B
50Ohms, 5A
3 PHASE
DIMMERSTATE
STATOR
ROTOR
THEORY: It should cover the following.
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.
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
22
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
OBSERVATION TABLE:
Sr.No. External rotor resistance in Ohms.
01
0Ω
02
10Ω
03
20Ω
04
30Ω
05
40Ω
06
50Ω
Speed (rpm)
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.
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G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
23
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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:
STARTER
+
L
+
A
A
-
0-20 A
+
220 V , DC
SUPPLY
S2
V
0-300 V
S1
A
AA
BRAKE DRUM
ATTACHAMENT
-
THEORY:
DC SERIES
MOTOR
It should cover the following
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.
PROCEDURE: 1) Connect the circuit as shown in figure.
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.
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
24
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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.
Sr.
No.
Im
Vm
(Volts) (Amps)
F1
(kg)
F2
(Kkg)
N
(rpm)
CALCULATIONS:
Output torque T = [(F1-F2) x 9.81 x r ] N.m
Output power Po = 2πNT/60 watts.
Input power Pi = Vm.Im
Po
%η = ---------- x 100
Pi
GRAPH:
where is radius of brake drum.
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)
2)
3)
4)
5)
6)
7)
What is the significance of back e.m.f. in a d.c. motor?
What is the chief advantage of a d.c. series motor?
What is the necessity of starter in a d.c. motor?
Why is a d.c. series motor used to start heavy loads?
Can a d.c. series motor operate on an a.c. supply?
Armature core of a dc machine is always laminated, why?
Why the D.C. series motor is not operated at zero or light loads?
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G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
25
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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:
STARTER
+
L
+
A
0-10 A
A
-
F
F
F
+
220 V , DC
SUPPLY
A
A
0-300 V
FF
FF
V
M
-
AA
G
-
AA
A
+
L
O
A
D
400
1.7 A
0-10 A
-
0-300 V
V
1000
1.2 A
-
+
DC MOTOR
D. C. GENERATOR
THEORY: It should cover the following
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.
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
26
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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:
Sr.
No.
Motor
side
Vm Im
Speed
N
rpm.
Generato
r side
Vg I g
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?
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
27
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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)
2)
3)
4)
Connect the circuit as shown above.
Adjust both rheostats at their minimum resistance position initially.
Switch ON the DC supply.
Turn the moving arm of starter to it’s minimum resistance position.
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
28
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
(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
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:
(A) FOR ARMATURE VOLTAGE CONTROL
Field current (If) =
Sr.
If (Amp)
No.
1
2
3
4
5
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
Amps (constant)
Va (Volts)
Speed (N) rpm.
29
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
(B) FOR FIELD CURRENT CONTROL
Armature voltage (Va) =
Sr.
No.
1
2
3
4
5
If (amp)
volts (constant)
Va (Volts)
Speed (N) rpm.
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
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G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
30
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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: STARTER
+
L
+
A
0-10 A
A
-
F
F
F
+
220 V , DC
SUPPLY
A
A
0-300 V
FF
V
FF
G
M
-
AA
0-10 A
-
-
A
1000
1.2 A
+
+
0-300 V
V
-
AA
L
O
A
D
400
1.7 A
DC MOTOR
D. C. GENERATOR
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?
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
31
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
OBSERVATION TABLE: Speed = 1500 rpm (constant)
Sr.
No.
Motor Side
Vm
Volts
Generator Side
Im
Vg
Amp Volts
Ig
Amp
Motor Input
= VmxIm
(Watts)
Generator
Output =
VgxIg (watts)
Efficiency of
generator =
VgIg
------- x 100
VmIm
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
ηG ηm =
O/p of generator
------------------I/p of motor
1/2
O/p of generator
ηG = -------------------I/p of motor
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
32
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
GRAPH: -
Vg
%
Efficiency
Output of
Generator
IL
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?
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
33
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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
+
STARTER
L
F
A
F
D C SUPPLY
230 V
R
A
M
FF
1000
1.2A
V
0-300 V
AC
A
Y
AA
B
N
D. C. MOTOR
ALTERNATOR
F
FIELD WDG OF
ALTERNATOR
0-1A
DC
FF
A
+
1000
,1.2A
230 V DC
+
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
34
-
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
[B] SHORT CIRCUIT TEST
STARTER
+
L
F
A
A
F
M
D.C. SUPPLY
230 V
A
A
FF
B
Y
AA
-
(0 -5 A)
AC
R
N
1000
1.2A
ALTERNATOR
D.C. MOTOR
F
FF
FIELD WDG
OF ALTERNATOR
A
(0-1A)
DC
+
+
D.C. SUPPLY
-
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
1) Connect the circuit as shown.
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.
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
35
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
[B] SHORT CIRCUIT TEST
1) Connect the circuit as shown.
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.
PRECAUTIONS:
6) All connections should be perfectly tight and no loose wire should lie on the work table.
7) Before switching ON the dc supply , ensure that the starter’s moving arm is at it’s maximum
resistance position.
8) Do not switch on the supply, until and unless the connections are checked by the teacher
9) Avoid error due to parallax while reading the meters.
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).
Sr. No
O.C TEST.
Field current Terminal voltage
If (Amp)
Per phase Vo
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
Sr. No.
36
S.C.TEST.
Field current
Short circuit
If
current Isc
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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
Zs = ---------- = ----------for field current Isc1
OB
Isc1
Isc1 is selected over the linear part of OCC, generally it corresponds to rated armature current.
Synchronous impedance
S.N. Zs
Synchronous reactance
Xs = √ (Zs2 - Ra2)
Where Ra = Armature resistance of alternator (per phase)
Zs (av).
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.
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
37
Xs
Xs (av).
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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.
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G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
38
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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:
+
L
STARTER
F
A
0-5 A
A
F
D C SUPPLY
230 V
R
A
0-300 V
A
N
B
AA
1000
N
Y
Y
1.2A
-
L
O
A
D
V
M
FF
R
B
D. C. MOTOR
ALTERNATOR
F
0-1A
DC
A
FF
FIELD WDG OF
ALTERNATOR
+
1000
+
1.2 A
230 V DC
-
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
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
39
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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..
PRECAUTIONS:
11) All connections should be perfectly tight and no loose wire should lie on the work table.
12) Before switching ON the dc supply , ensure that the starter’s moving arm is at it’s maximum
resistance position.
13) Do not switch on the supply, until and unless the connection are checked by the teacher
14) Avoid error due to parallax while reading the meters.
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?
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G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
40
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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:
0-10 A
0-10A
M
R
L
+
A
A
-
4
3 PHASE
SUPPLY
440 V
V
F
C
R
V
5
Y
A
FF
Y
A
G
-
AA
B
M
L
ALTERNATOR
PF- METER
600 V,10 A
D. C.
GENERATOR
FF
F
-
0.2 A DC
L
O
A
D
400
1.7A
C
V
0-300 V
V
B
6
+
FIELD WDG OF
ALTERNATOR
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.
G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR
41
ivTHsem B.E. electrical
DEPARTMENT OF ELECTRICAL ENGINEERING
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.
[A] AT NO LOAD
OBSERVATIONS:
Sr.
No.
If
Power Factor
(cosφ)
Ia
[B] AT LOAD
Sr.
No.
If
Power Factor
(cosφ)
Ia
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.
2.
3.
4.
5.
With what condition synchronous motor can be used as a synchronous condenser.
What are the special applications of an over excited synchronous motor.
Explain the effect of change of excitation of a synchronous motor on its armature current.
Explain the effect of change of excitation of a synchronous motor on its power factor.
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.
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42