S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
1
CONTENTS
Sl.No.
NAME OF EXPERIMENT
Page
No.
CYCLE 1
1
Brake Test on Squirrel Cage Induction Motor
4
2
No-load & Blocked-rotor Tests on 3-phase Squirrel Cage
Induction Motor
13
3
Slip Test on 3-phase Salient Pole Synchronous Machine
25
4
Voltage Regulation on Alternator
34
5
No-load & Blocked-rotor Tests on Slip Ring Induction Motor
48
CYCLE 2
6
Induction Machine as Generator & Motor
59
7
No-load & Blocked-rotor Tests on Pole Changing Induction
Motor
66
8
No-load & Blocked-rotor Tests on Single Phase Induction Motor
81
9
V-Curves of Synchronous Machines
88
10
Speed Control of Induction Motor by Variable Frequency Method
96
MODEL QUESTIONS
100
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
2
INSTRUCTIONS
1.
WEAR SHOES COMPULSORILY
2. SHIRTS SHOULD BE TUCKED IN
3. GIRLS SHOULD PROTECT THEIR HAIR
4. DO NOT ALLOW CHAINS TO HANG
5. DO NOT LEAN OVER ROTATING MACHINERY
6. ENERGIZE THE CIRCUIT ONLY AFTER GETTING APPROVAL FROM THE FACULTYIN-CHARGE
7. MAKE SURE THAT THE CORRECT SWITCH HAS BEEN SWITCHED ON/OFF
BEFORE/AFTER THE EXPERIMENT.
MAKING CONNECTIONS
• Make sure that the supply is OFF.
•
Meters should be positioned properly.
•
Do not connect more than one wire to each terminal of ammeters & voltmeters.
•
Make series connections before parallel connections.
•
All the connections should be tight.
•
Get the connections checked before switching ON.
•
Check the position of rheostats, autotransformers, switches before switching ON.
•
Never exceed the permissible values of current or voltage.
•
While conducting brake test, pour water on the brake drum to avoid overheating.
•
Show the readings to the faculty-in-charge before switching off.
ROUGH RECORD
1. Write Name of the experiment with number & date, aim, apparatus required, neat
circuit diagram, tabulations, sample calculations (for one set of readings showing the
substitution of the values) and results. No need to write principle or procedure.
2. Take at least six sets of readings, if possible. Each student in a group should do sample
calculations for different sets.
3. Get signature of the faculty-in-charge after completing the rough record.
FAIR RECORD
1. Write the name of the experiment on the top of the right side in capital letters
2. Experiment Number & date should be written at the top.
3. Each record should contain the following on the right side
•
Aim of the experiment
•
Apparatus required
•
Principle
•
Procedure
•
Sample Calculation (on the left side if possible; if calculations are too long, write on
right side so that no pages on the right side are left blank)
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
•
3
Result (at the end)
4. On left side
•
Neat circuit diagram with PEN
•
Name plate details/specifications
•
Tabulations
•
Sample Calculation (on the left side if possible; if calculations are too long, write on
the right side so that no pages on the right side are left blank)
•
Graph (draw with PEN if possible; use different colors for different graphs on the
same graph sheet).
Do experiment TODAY; submit Rough Record in the NEXT CLASS & Fair Record in the THIRD
CLASS.
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
4
==================================================================
Experiment No. 1
BRAKE TEST ON SQUIRREL CAGE INDUCTION MOTOR
==================================================================
AIM: Conduct the brake test on 3 phase squirrel cage induction motor and plot the following
performance characteristics.
a) Electrical characteristics – Speed, line current, torque, power factor, efficiency &
% slip Vs output power
b) Mechanical characteristics – Speed Vs Torque
Also find the additional kVAR required to improve the power factor to 0.95 at various
loads.
APPARATUS:
S.No.
1.
2.
3.
Name of the apparatus
Voltmeter
Ammeter
Wattmeter
4.
5.
Tachometer
TPDT switch
Type
MI
MI
Dynamometer
type
Range
(0-500V)
(0-10A)
500V, 10A, UPF
Quantity
1
1
2
1
1
PRINCIPLE:
The two types of 3-phase induction motors are i) squirrel cage induction motor and ii)
slip-ring induction motor. Three-phase squirrel cage induction motor is generally preferred
because it is rugged in construction, requires less maintenance and is economical as
compared to 3-phase slip ring induction motor.
When the stator winding is connected to three phase ac supply, a rotating magnetic
field is established in the air gap which rotates at synchronous speed. Initially, rotor is
stationary. Due to relative speed between the rotating magnetic field and stationary rotor
conductors, an emf is induced in the rotor. As the rotor circuit is closed, currents will
circulate through them. According to Lenz’s law, these induced currents will flow in such a
direction so as to oppose the cause producing it. Here the cause is relative speed. In order to
reduce the relative speed, the currents in the rotor produce a torque tending to rotate the rotor
in the same direction of rotating field.
At synchronous speed of the rotor, the relative speed is zero, no emf and no torque is
developed, rotor tends to stop, hence rotor can not attain synchronous speed. Motor runs at a
speed slightly less than synchronous speed.
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
PROCEDURE:
Make the connections as shown diagram.
Precautions: i) Keep TPST switch open
ii) Keep TPDT in position 1 (Star connection)
5
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
.
6
iii) Keep belt on brake drum in loose position (motor on no load)
Switch on the 3 phase supply while the motor is on no load. When the motor gains
speed, change the TPDT switch to delta position (position 2). By tightening the brakedrum,
increase the load on the motor upto rated value (=7.5A). Note down the speed, spring balance
readings, voltmeter, ammeter and wattmeter readings. Now decrease the load in steps upto no
load and note down the readings each time. If any of the wattmeter readings shows negative
on no load or light loads, switch of the supply & interchange the terminals of pressure
coils/current coils (not both) of that wattmeter. Now, again starting the motor (follow above
procedure for starting), take readings. Switch off the supply. Measure the radius of the brake
drum.
TABULATION:
SAMPLE CALCULATION (Set No. ____)
V = ______V,
N =_______ rpm,
I =_______ A,
W1 = _______W,
S1 = ________Kg,
W2 = ________ W,
S2 =________Kg
Radius of brake drum R = 0.135 m
Synchronous speed Ns = 1500rpm
Input Power W= W1 + W2 = _________ watts
Power factor, cosΦ1= cos(tan −1
3 × (W1 − W2 )
) =__________
(W1 + W2 )
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
Percentage slip, s = =
7
Ns − N
× 100 =_________%
Ns
Torque T = R × ( S1 − S2 ) × g = _________ N-m
Output power =
Efficiency, η =
2π NT
= _________W
60
output
×100 = _________%
input
Additional kVAR required to improve the power factor to 0.95 (cosΦ2) =
= W (tan Φ1 − tan Φ 2 ) ×10−3 =__________
Value of capacitance required to improve the power factor =
=
MODEL GRAPHS
kVAR × 1000
=
ωV 2
kVAR × 1000
=________μF
2 × π × f ×V 2
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
Speed
8
Speed Vs Torque Characteristic
0
Torque
RESULT:
i)
Brake test conducted on 3 phase squirrel cage induction motor
Performance characteristics plotted
ii)
Additional kVAR required and the value of capacitance to improve the power
iii)
factor for each load to 0.95 determined.
==================================================================
Do you know?
1. Why are starters needed for induction motors ?
Equivalent circuit of the induction motor at starting (S=1) is shown below (magnetizing
V
current neglected). Starting current, I SC =
.
Ro1 + jX o1
Ro1
ISC
Xo1
V
If rated voltage is applied, large starting current (5 to 8 times full load current) will flow.
This causes appreciable voltage drop in the line and may affect other equipments
connected to the same line. Also, if a large current flows for a long time it may overheat
the motor and damage the insulation. In such case, reduced voltage starting must be
used.
2. What are the different types of starters used for 3 phase induction motors?
Direct On Line (DOL) starter
i)
Stator impedance (resistance/reactance) starter
ii)
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
iii)
iv)
v)
9
Autotransformer starter
Star-Delta Starter
Rotor resistance starter (Only for slip ring induction motors)
3. The no load current for an induction motor is larger than that in a transformer of
same VA rating. Give reason.
Because of the presence of air-gap in the induction motor, for same flux, the
magnetizing current (flux = mmf / reluctance) is far larger. Also, in addition,
induction motor has to supply mechanical losses (friction and windage losses) on no
load which is not present in transformer.
4. Why is an induction motor not capable of running at synchronous speed ?
When the motor speed reaches synchronous speed, no rotor emf, no rotor current
and hence no torque is produced. Hence, the induction motor never attains
synchronous speed.
5. Explain star-delta starter.
A star-delta method of starting is employed to provide reduced voltage at start. In
this method, the normal connection of the stator windings is delta while running. If
these windings are connected in star at start, the phase voltage is reduced ( = V 3 ),
resulting in less current at starting. As the motor approaches full speed, the windings
will be connected in delta.
6. What you mean by synchronous speed? What is the synchronous speed of an
induction motor whose rated speed is 1440rpm ?
Synchronous speed Ns is the speed of the rotating magnetic field in a poly-phase
120 f
induction motor. N s =
. Synchronous speed will be slightly greater than the
P
rated speed. For 50Hz supply, possible synchronous speeds are 3000rpm (2 poles),
1500rpm (4 poles), 1000rpm (6poles), 750 rpm (8 poles) etc. If rated speed is 1440
rpm, synchronous speed is 1500rpm.
7. The rotor core loss of an induction motor under running condition is usually
neglected. Why?
During running condition, rotor frequency is equal to slip frequency (=sf) which
is very small. Hence rotor core loss will be very small which can be neglected.
Note: The rotor core loss is not constant for all load conditions. As the load
increases, slip increases, hence rotor frequency and rotor core loss increases.
8. What is the normal value of full-load slip during running condition?
2 to 8%
9. How can the direction of rotation of a 3 phase induction motor be reversed ?
By interchanging any two supply leads connected to the motor.
What is the speed of the stator and rotor magnetic fields with respect to
10.
stator?
Both are rotating at synchronous speed Ns with respect to stator.
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
10
11. Compare between squirrel cage and slip ring induction motors.
SCIM
1. Its rotor consists of copper/aluminium
bars permanently short circuited at both
ends
2. Starting torque is poor
3. Separate starting methods are needed in
order to reduce the staring current
4. Its rotor can adjust to any number of
stator poles. So different speeds can be
obtained by different arrangements of stator
winding.
5. It has fewer components and hence less
labor since it has no rotor winding. So it is
cheaper
6. Better efficiency
7. Better cooling (larger space to provide
fan blades on rotor)
8. Less maintenance
SRIM
1. Its rotor consists of 3 phase windings
which is then connected to external
resistance through slip rings and brushes
2. Better starting torque can be achieved
by inserting external resistance in the
rotor circuit.
3. Rotor resistance starting can be used
4. Rotor and stator are wound for same
number of poles
5. Slip rings, brushes, starting resistance
etc. increases the cost
6. Lower efficiency
7. Cooling not efficient
8. More maintenance
12. At no load, one of the two wattmeters connected in the input side of the motor is
negative. Why ?
The no load power factor of an induction motor is always less than 0.5 because
the no load current is mainly used for magnetizing the core.
13. What is time harmonics and space harmonics ?
If the supply currents are non-sinusoidal, it contains harmonics. These harmonic
currents (time harmonics) produce rotating fields in the air gap. The time harmonic
currents and their rotating fields produce parasitic torques in the machine.
Even if the supply currents are sinusoidal, air gap flux may be non-sinusoidal due
to winding arrangement, slotting, air gap irregularity etc. Hence, air gap flux
contains harmonics and these harmonics are called space harmonics. The space
harmonics also produce parasitic torques in the machine.
Time harmonics produce no significant effects on the operation of the induction
motor. Effects of space harmonics are crawling and cogging.
14. What is meant by cogging and crawling?
In the case of squirrel cage induction motors, with certain relationships between
the number of poles and stator and rotor slots, peculiar behavior may be observed
when the machine is started. With the number of stator slots S1 = number of rotor
slots S2, the induction motor may refuse to start at all. This phenomenon is known as
cogging. With other ratios, S2/S1, the motor may exhibit tendency to run stably at low
speed, e.g. one seventh of the normal speed. This is known as crawling.
15. Is the rotational losses (stator core loss + rotor core loss + friction & windage loss)
constant in an induction motor?
11
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
If the induction motor is connected to a supply of fixed voltage and frequency, the
stator core loss is fixed. At no load, the machine will operate close to synchronous
speed. Therefore, the rotor frequency f2 is very small and hence rotor core loss is very
small. At a lower speed, f2 increases and so does the rotor core loss. The total core
losses thus increase as the speed falls. On the other hand, at no load, friction and
windage losses are maximum and as speed falls these losses decrease. Therefore, if a
machine operates from a constant-voltage and constant-frequency source, the sum of
core losses and friction and windage losses remains essentially constant at all
operating conditions.
16. Draw the IEEE-recommended equivalent circuit?
The resistance Rc is omitted and the core loss is lumped with the windage and
friction losses. The magnetizing reactance Xm can not be moved to the machine
terminals.
I2'
I1
R1
X1
Im
Xm
V1
X2'
R2'/s
17. Why the power factor of an induction motor is low at starting?
The rotor frequency and rotor reactance are high under starting conditions and
therefore, rotor currents lag the rotor emf by a large angle. This results in low power
R2
R
≈ 2 , R2<<X2. Hence power factor is low)
factor at starting. ( p. f . =
2
2
X2
R2 + X 2
18. The starting torque of a squirrel cage induction motor can not be altered, when the
applied voltage is constant. Why?
The starting torque of a squirrel cage induction motor can not be increased as
there is no provision for inserting resistance in the rotor circuit.
19. What type of protection is provided in the starter meant for 3 phase induction
motors?
Overload and under-voltage protection.
20. What is the standard direction of an induction motor?
Counter clockwise when looking from non-drive end of the motor.
21. What is single phasing?
Single phasing is a fault condition in which a 3 phase motor is operating with one
line open. The 3 phase motor will not start with one line open. If the motor is running
when single phasing occurs, it will continue to run as long as the shaft load is less
than 80% rated load and the remaining single phase voltage is normal; rotation of the
rotor produces a quadrature field that maintains rotation. If single phasing occurs
while operating at or near rated load, the increase in phase current will cause rapid
heating of the windings, and therefore protective devices, must be provided to trip the
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
12
machines from the supply lines, or severe damage to stator and rotor winding may
occur.
22. When the applied rated voltage per phase is reduced to one-half, what will be the
starting torque of a squirrel cage induction motor in terms of its starting torque
with full voltage?
One fourth of starting torque with full rated voltage.
23. Draw the diagram of a direct-on-line starter?
On pressing the START push button S1, the contactor coil C is energized from two
line conductors L1 and L2. The three main contacts M and the auxiliary contact A
close and the terminals a and b are short-circuited. The motor is thus connected to the
supply. When the pressure on S1 is released, it moves back under spring action. Even
then the coil C remains energized through ab. Thus, the main contacts M remain
closed and the motor continues to get supply.
When the STOP push button S2 is pressed, the supply through the contactor coil is
disconnected. Since the coil C is de-energised, the main contacts M and auxiliary
contact A are opened. The supply to motor is disconnected and the motor stops.
Undervoltage protection : When the voltage falls below a certain value, or in the
event of failure of supply during motor operation, the coil C is de-energised. The
motor is then disconnected from the supply.
Overload protection: In case of an overload, one or all the overload coils (OLC) are
energized. The normally closed contact D is opened and the contactor coil C is deenergised to disconnect the supply to the motor.
L1
L2
L3
Fuse
C
a A
M M
M
b
S2
Stop
Start
S1
D
OLC
S3
Motor
Remote
Stop
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
================================================================
Experiment No. 2
NO-LOAD AND BLOCKED-ROTOR TESTS ON A 3 PHASE
SQUIRREL CAGE INDUCTION MOTOR
================================================================
AIM: i) To conduct no load and blocked rotor tests on 3 phase squirrel cage induction
motor
ii)To determine the equivalent circuit parameters and hence predetermine the
performance at full load from the equivalent circuit and
iii)To draw the circle diagram and hence predetermine the performance
characteristics from circle diagram.
APPARATUS:
S.No.
Name of the
apparatus
1.
Voltmeter
2.
3.
4.
Ammeter
5.
6.
7.
Wattmeter
8.
9.
Type
Range
Quantity
MI
MI
MC
MI
MI
MC
Dynamometer
0-500V
0-150V
0-30V
0-5A
0-10A
0-10A
500/250/125V,
5/10A,UPF
150V,10A,LPF
9Ω,8.5A
1
1
1
1
1
1
2
Dynamometer
Wire wound
Rheostat
1
1
PRINCIPLE:
The performance characteristics of induction motors can be determined
approximately by graphical method such as circle diagram. This is applicable both for the
squirrel cage and slip ring induction motors. From the approximate equivalent circuit,
I2 ' =
V
R '
( R1 + 2 ) 2 + ( X 1 + X 2 ') 2
s
sin Φ =
=
V
sin Φ
X1 + X 2 '
where
X1 + X 2 '
R '
( R1 + 2 ) 2 + ( R1 + R2 ') 2
s
If the leakage reactances X1 and X2’ are assumed to remain constant regardless of load,
and the applied voltage V is constant, the above equation represents the polar equation of
a circle with diameter
V
(1 − s )
) and Φ,
. By changing the load RL (where RL = R2 '
X1 + X 2 '
s
the value of the current I2’ changes. The locus of the current, however, lies on a circle
(Figure 1).
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
Figure 1
14
Figure 2
Thus in the case of induction motors, the locus of the current due to load lies on a
circle and the diagram is known as a circle diagram. If no load current taken by the motor
is also to be accounted for to obtain the stator current, the diagram can then be shown as
in figure 2. The stator current I1 is then the phasor sum of I2’ and Io.
No load and blocked rotor tests are conducted for determining the equivalent
circuit parameters, for predetermining the efficiency at any load and to draw the circle
diagram. No-load test is conducted at rated voltage keeping the motor on no-load. Since
the no-load current is only 20-40% of the full load current, the I2R losses can be
neglected. Input power is equal to constant iron, friction and windage losses of the motor.
In blocked rotor test, rotor is blocked and a reduced voltage is applied to the stator
through a 3-phase autotransformer. Due to low voltage and no rotation, core and
mechanical losses are neglected. Input power is equal to copper loss only.
PROCEDURE:
A) NO LOAD TEST
Make the connections as shown in figure.
Precautions : i) Keep the autotransformer in minimum voltage position
ii) Keep belt on brake drum in loose position (motor on no load)
Switch on the 3 phase supply. Adjust the autotransformer and apply rated voltage
to the stator. Since the power factor of the induction motor under no load condition is
generally less than 0.5, one wattmeter will show negative reading. Then switch off the
supply and interchange the connections of the pressure coil (or current coil) of that
wattmeter. Note down the ammeter, voltmeter and wattmeter readings at rated supply
voltage. Switch off the supply.
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
B) BLOCKED ROTOR TEST
Make the connections as shown in figure.
Precautions : i) Keep the autotransformer in minimum voltage position
ii) Rotor is blocked by tightening the belt on the brake drum.
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
16
Switch on the 3 phase supply. Adjust the autotransformer so that rated current (to
get full load copper loss) flows in the ammeter. Note down voltmeter, ammeter and
wattmeter readings. (If any of the wattmeter reads negative, switch off the supply and
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
interchange the connections of the pressure coil (or current coil) of that wattmeter and
continue the above procedure). Switch off the supply.
C) STATOR RESISTANCE MEASUREMENT
Make the connections as shown in figure.
+
10A
9Ω, 8.5A
+
0-10A MC
-
+
28V
DC
V
0-30V MC
-
S1
A
R
-
10A
S3
S2
Precautions : Keep the rheostat in maximum resistance position
Switch on 28V d.c. supply. Note down voltmeter and ammeter readings for
different positions of rheostat. (Note: Resistance/phase =
3
x Delta resistance)
2
Procedure to draw the circle diagram: (Do not write in fair record)
1. Draw the lines by taking the current (I) in X-axis, voltage (V) in Y-axis. (V & I
are line values)
2. From the No-load test find out the current Io and draw the OA vector with the
magnitude of Io from the origin by suitable current scale, which lags the voltage
Woc
(Y-axis) V by an angle Φo where Φ o = cos −1 (
).
3Voc I oc
3. From the current Isc find out the ISN (short circuit current corresponding to the
V
normal voltage) through the formula I SN = I sc ( rated ) , draw the OB vector with the
Vsc
magnitude of ISN from the origin by the same current scale, which lags the voltage
Wsc
(Y-axis) V by an angle ΦSC where Φ SC = cos −1 (
).
3Vsc I sc
4. Join the points B and A to get the output line.
5. Draw the parallel line for the X-axis from point A and for the Y-axis from point B
upto the X-axis (point E), let both the lines intersects at point D.
6. Then draw the bisector for the output line and extend it to the line AD let the point
of intersection be C.
7. By keeping the point C as center draw a semi circle with radius CA.
8. Let EB be the line of total loss [EB = ED + DB where ED = constant loss and DB
= variable loss]
9. In the line DB locate the point G to separate the stator and rotor copper losses
by using the formula,
Rotor Copper loss I 2 '2 R2 '
R '
= 2 where R1= stator
=
2
Stator Copper loss I 2 ' R1
R1
resistance per phase and R2= rotor resistance per phase.
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
BG R2 '
Rotor Copper loss
=
=
.
BD Ro1 Stator Copper loss + Rotor Copper loss
10. To get the torque line, join the points A and G.
11. To find the full load quantities, draw line BK (=Full load output/power scale).
Now, draw line PK parallel to output line meeting the circle at point P.
12. Draw line PT parallel to Y-axis meeting output line at Q, torque line at R, constant
loss line at S and X-axis at T.
Or,
Note: Choose the current scale such that the circle diagram will be as large as possible.
The larger the circle diagram more will be the accuracy. Select power scale =
3 × Vrated × current scale .
TABULATION
NO LOAD TEST
Voc
Ioc
W1
BLOCKED ROTOR TEST
W2
Wsc
Vsc
Isc
W1
Stator Resistance Measurement
S.No.
V (volts)
I (amps)
1.
2.
3.
4.
Rdc
CIRCLE DIAGRAM
Voc = 400V , Ioc = ___ A ,
Vsc = _____ V,
Woc = _____ W
Isc = 7.8A, Wsc = _____ W
Per phase values are
Vo = Voc = _____ V
Io =
I oc
= ____ A
3
Vs = Vsc = _____ V
Is =
I sc
= ____ A
3
Rdc = _____ Ω
3
R1 = ×1.2 × Rdc = ______ Ω
2
W
Ro1 = sc2 = _______ Ω
3I s
R2 ' = Ro1 − R1 = _______ Ω
Rdc=V/I Ω
W2
Wsc
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
BG R2 '
= ______
=
BD Ro1
BG = ____× BD
Selection of current and power scale
Current scale = 1cm = ______ A
Ioc = _______A (= ______cm)
Vrated
) I sc = _______A(= _______cm)
Vsc
Woc
Φ o = cos −1 (
) = _______ ˚
3Voc I oc
Wsc
Φ SC = cos −1 (
) = _______˚
3Vsc I sc
I SN = (
Power Scale =
3 × Vrated × current scale = _______ W = 1cm
PERFORMANCE AT FULL LOAD FROM CIRCLE DIAGRAM
Full load output = 3000W = PQ = _____cm
Full load current = OP x current scale = ____ x _____ = ______A
Full load power factor =
PT
= ______ lag
OP
Rotor copper loss at full load = QR x power scale = ____ x _______ = _______W
Stator copper loss at full load = RS x power scale = _____ x ______ = _______W
Constant loss = ST x power scale = ___ x ______ = ________W
Rotor input at full load = PR x power scale = _____ x ______ = _______W
Torque at full load = PR x power scale (sync. watts) = PR x power scale ×
60
N-m
2π N s
60
= _______N-m
2π × 750
Motor input at full load = PT x power scale = _____ x _______ = ______W
PQ
Efficiency at full load =
× 100% = ________%
PT
QR
Slip at full load s =
×100% = _________%
PR
Speed at full load = (1 − s ) × N s = ________ rpm
60
Starting torque = BG x power scale x
N-m
2π N s
60
=______N-m
= _____ x ______ x
2π × 750
= _____ x ______ x
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
60
=______N-m
2π × 750
Maximum output = HH’ = ______ x ______ = ________W
Maximum torque = I I’ = ______ x _____ x
Maximum input = JJ’ = ______ x ______ = ________W
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21
TABULATION FROM CIRCLE DIAGRAM
MODEL GRAPH – PERFORMANCE CHARACTERISTICS FROM CIRCLE
DIAGRAM
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
EQUIVALENT CIRCUIT PARAMETERS
Voc = ______ V , Ioc = ____A , Woc = _____W
Vsc = ______V, Isc = 7.8A, Wsc = ______W
Vo = Voc = _____ V
Io =
I oc
= ____ A
3
Vs = Vsc = _____ V
Is =
I sc
= ____ A
3
Woc
= _______
3Vo I o
sin Φ 0 = _______
Vo
=_________Ω
Rc =
I o × cos Φ o
Vo
Xm =
= ________Ω
I o × sin Φ o
V
Z o1 = s = _________Ω
Is
W
Ro1 = sc2 = _______Ω
3I s
cos Φ 0 =
X o1 = Z o12 − Ro12 = ________Ω
3
R1 = *1.2* Rdc =1.5 x 1.2 x ____ = ______Ω
2
'
R2 = Ro1 − R1 =_______Ω
X
X 2 = X 2 ' = o1 = ______Ω
2
EXACT EQUIVALENT CIRCUIT
R1
I1
X1
R2'
I2'
X2'
Io
Ic
400V
Rc
Im
Xm
1− s
)
RL ' = R2 '(
s
APPROXIMATE EQUIVALENT CIRCUIT
Ro1=R1+R2'
Xo1=X1+X2'
I
I'
1
2
Io
Ic
400V
Rc
Im
Xm
RL ' = R2 '(
1− s
)
s
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
23
PERFORMANCE AT RATED SPEED FROM EQUIVALENT CIRCUIT
Synchronous speed, Ns = 750 rpm
Rated speed = N = 710 rpm
Slip = s =
Ns − N
750 − 710
× 100% =
×100 = 5.33%
Ns
750
1− s
) =_______Ω
RL ' = R2 '× (
s
−
I o = I o ∠ − Φ o ° =________per phase
From approximate equivalent circuit,
I 2' =
V ∠0°
=________A per phase
( Ro1 + RL ') + jX o1
I1 = I o + I 2 ' = I1∠ − Φ1 ° =_______A
Line current IL =
3 × I1 =_______A
Power factor = cos(−Φ1 °) =______
Output = 3 × I 2 '2 × RL ' =________W
Torque =
Output
=_______N-m
(2π N )
60
Input =
3VI L cos Φ1 =________W
Efficiency =
Output
× 100% = ______ %
Input
RESULT
i)
No-load and blocked rotor tests were conducted on 3 phase squirrel cage
induction motor
ii)
Equivalent circuit parameters were determined
iii)
Circle diagram was drawn
iv)
Performance at full load from equivalent circuit and circle diagram were
determined
v)
Performance characteristics were plotted from circle diagram.
================================================================
Do you know?
1. Why is no load current in a 3 phase induction motor more than that in a
transformer?
Because of the air gap, the magnetizing current is far larger in an induction motor
than in a transformer for the same VA rating. Also, induction motor has friction and
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
24
windage loss in addition to core loss. In induction motor, no load current is about 30
to 40% of full load current.
2. Why is the power factor on no load less than 0.5?
The no load current is mainly used for magnetizing the core which is inductive in
nature. Hence the power factor is less.
3. On blocked rotor test, the power factor may be less than 0.5. Give reason.
Vsc
, and Xo1 » Ro1 . Hence, the
Current during blocked rotor test, I sc =
Ro1 + jX o1
power factor is less.
4. The rotor core loss of an induction motor under running condition is
usually neglected. Why ?
During running condition, rotor frequency = slip x supply frequency. Since slip is
very small (2 to 8% during running condition), rotor frequency is small; the rotor
core loss is less and neglected.
5. What happens if one of the supply phases is dead at the instant of
starting?
Motor will not start because the resulting supply system is single phase.
6. During running condition, if one of the fuses blows, what happens?
If the motor is on no load or light load, the motor will continue to run.
7. No load test should always be conducted at rated voltage but the blocked
rotor test may be conducted at any current (rated current not necessary).
Why?
In blocked rotor test, wattmeter reading gives copper loss which is proportional to
square of the current. By knowing copper loss at any load, we can calculate the
copper loss at rated current. But, in no load test, wattmeter reads core loss which has
two components, hysteresis loss (proportional to V1.6) and eddy current loss
(proportional to V2).Hence, Core loss is equal to K1 V1.6 + K2 V2 . Two
proportionality constants, make it not possible to convert core loss at one voltage to
another voltage.
8. What you mean by plugging?
It is a type of electric braking used in induction motors where the braking torque
is produced by interchanging any two supply terminals. Here, the direction of rotation
of the rotating magnetic field is reversed with respect to the rotation of the motor. The
electromagnetic torque developed provides the braking action and brings the rotor to
a quick stop.
9. What is the difference between electrical degree & mechanical degree?
Electrical angle is a measure of one cycle of emf or current wave. 1 cycle=360˚
P
electrical. One revolution is equal to 360˚ mechanical. θ elec = × θ mech
2
10. What are the methods of reducing the space harmonics?
i)
Distributing the winding in slots
Using the short-pitched winding
ii)
Skewing the slots
iii)
Using fractional-slot winding
iv)
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
================================================================
Experiment No. 3
SLIP TEST ON 3-PHASE SALIENT POLE SYNCHRONOUS
MACHINE
================================================================
AIM: i) To conduct the slip test on 3-phase salient pole synchronous machine
ii) To determine the direct axis and quadrature axis synchronous reactances
iii) To predetermine the voltage regulation at different loads and power factors and
iv) To draw the power Vs torque angle characteristics for a specified induced emf.
APPARATUS:
S.No.
1.
2.
3
4.
5.
6.
7.
Name of the apparatus
Voltmeter
Ammeter
Rheostat
Tachometer
Type
MI
MI
MC
MI
MC
Wire Wound
Range
(0-500V)
(0-300V)
(0-30V)
(0-10A)
(0-10A)
9Ω 8.5A
Quantity
2
1
1
1
1
1
1
PRINCIPLE:
The direct and quadrature axis reactances can be measured by slip test. The
machine is driven by a dc motor at a speed slightly less or slightly more than synchronous
speed. The field winding is kept open circuited and a low voltage 3 phase supply (about
25% of the rated voltage) is applied to the armature terminals. The direction of rotation
should be same as the direction of rotating field. If this condition is fulfilled, a small ac
voltage would be indicated by the voltmeter across the field winding.
The relative velocity between armature mmf and field poles is equal to slip speed
i.e. difference between synchronous speed and rotor speed. The stator mmf moves slowly
past the field poles at slip speed. This would cause the armature current to vary cyclically
at twice the slip frequency. When the peak of the armature mmf is in line with the field
poles, the reluctance offered by the magnetic circuit is minimum, the armature current,
required for the establishment of constant air-gap flux, will be minimum. Constant
applied voltage minus the minimum impedance voltage drop (armature current being
minimum) in the leads and 3-phase variac gives maximum armature-terminal voltage.
The ratio of maximum armature terminal voltage per phase to minimum armature current
per phase gives Zsd. After one quarter of slip cycle, the peak of armature mmf is in line
with q-axis and the reluctance offered by the magnetic circuit is maximum. The armature
current, required for the establishment of constant air-gap flux, will be maximum and the
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
armature terminal voltage will be minimum. The ratio of minimum armature terminal
voltage per phase to maximum armature current per phase gives Zsq.
IXd
Eo
V
δ
IXq
φ
ψ
I
p. f. lag
I
φ
Eo
IRa
ψ =δ
IXd
ψ
IXd
IXq
IXq
unity p. f.
V
IRa
Eo
δ
V
IRa
p. f. lead
PHASOR DIAGRAMS
When the armature mmf is in line with field poles, the armature flux linkage with
field winding is maximum and rate of change of this flux linkage is zero, so that induced
voltage across the field winding is zero. On the other hand, when armature mmf is in line
with q-axis, the flux linkage with field winding is minimum and rate of change of this
flux linkage is maximum, so that induced voltage across the field winding is maximum.
PROCEDURE:
SLIP TEST
Make the connections as shown in figure.
Precautions : i) Keep the autotransformer at minimum voltage position
ii) Keep DPST, TPST and SPST switches open
iii) Keep dc motor field rheostat at minimum resistance position
Switch on the d.c. supply by closing the DPST switch. Using the three point
starter, start the motor. Run the motor at synchronous speed by varying the motor field
rheostat. Close the TPST switch. By adjusting the autotransformer, apply 20% to 30% of
the rated voltage to the armature of the synchronous machine. Make sure that the
direction of rotation of the prime mover and the direction of rotation of the magnetic field
produced in the armature are the same by closing the SPST switch. If the voltmeter
reading across the alternator field winding is very small, both the directions are correct. If
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
27
the voltmeter reading is high, interchange the two lines of 3 phase supply after switching
off the 3 phase supply. SPST switch is kept open.
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
The speed is slightly reduced/increased from synchronous speed, so that slip is
increased and the voltmeter and ammeter readings are oscillating. The maximum and
minimum readings of voltmeter and ammeter are noted. The above said procedure can be
repeated with two more different autotransformer settings.
(During slip test, it would be observed that swing of the ammeter pointer is very wide,
whereas the voltmeter has only small swing because of the low impedance voltage drop in
the leads and 3-phase autotransformer).
STATOR RESISTANCE MEASUREMENT
Make the connections as shown in the diagram.
+
10A
+
9Ω,8.5A
A
A
-
0-10A MC
+
U
V
V
0-30V MC -
28V DC
N
10A
-
Precaution: Keep the rheostat at maximum resistance position.
Switch on 28V dc supply. Adjusting the rheostat for different values of current,
note down the ammeter and voltmeter readings.
TABULATION – SLIP TEST
Sl.No.
Vmax
Vmin
Imax
Imin
Z sd
Z sq
1.
2.
3.
TABULATION – Stator resistance measurement
Stator Resistance Measurement
S.No.
V (volts) I (amps)
Rdc=V/I Ω
1.
2.
3.
4.
Rdc
SAMPLE CALCULATION (SET No. ___ )
Armature resistance, Ra = 1.2 × Rdc =1.2 x _____ = ____Ω
Vmax = _____V, Vmin = _____V, Imax = ____A, Imin = _____A
Xd
Xq
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
Z sd =
Vmax
=______Ω
I min
Z sq =
Vmin
=______Ω
I max
X d = Z sd 2 − Ra 2 =_______Ω
X q = Z sq 2 − Ra 2 =________Ω
a) To find Percentage regulation at full load and 0.8 p.f. lag
V=231V, I=11.5A, Ф = +36.87˚, cosФ=0.8, sinФ=0.6
231× 0.6 + 11.5 × X q ⎞
⎛ V sin Φ + IX q ⎞
−1 ⎛
Ψ = tan −1 ⎜
⎟ = _____ °
⎟ = tan ⎜
⎝ 231× 0.8 + 11.5 × Ra ⎠
⎝ V cos Φ + IRa ⎠
δ = Ψ − Φ =_____-36.87=_____˚
E f = V cos δ + IRa cos Ψ + IX d sin Ψ =_______V
% regulation =
E f −V
V
× 100 = ______%
a) % regulation at full load V = 231V, I = 11.5A
power factor
Ф
0
lag
90
0.2
lag
78.46
0.4
lag
66.42
0.6
lag
53.13
0.8
lag
36.87
1
Ψ
δ
Ef
0
0.8
lead
-36.87
0.6
lead
-53.13
0.4
lead
-66.42
0.2
lead
-78.46
0
lead
-90
b) To find Percentage regulation at full load and 0.8 p.f. lead
V=231V, I=11.5A, Ф = -36.87˚, cosФ=0.8, sinФ= -0.6
Regulation
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
−231× 0.6 + 11.5 × X q ⎞
⎛ V sin Φ + IX q ⎞
−1 ⎛
Ψ = tan −1 ⎜
⎟ = ____ °
⎟ = tan ⎜
⎝ 231× 0.8 + 11.5 × Ra ⎠
⎝ V cos Φ + IRa ⎠
δ = Ψ − Φ = _____+36.87= _____˚
E f = V cos δ + IRa cos Ψ + IX d sin Ψ =________V
% regulation =
E f −V
V
× 100 = ________%
b) % regulation at Half full load V = 231V, I = 5.75A
power factor
Ф
0
lag
90
0.2
lag
78.46
0.4
lag
66.42
0.6
lag
53.13
0.8
lag
36.87
1
0
0.8
lead
-36.87
0.6
lead
-53.13
0.4
lead
-66.42
0.2
lead
-78.46
0
lead
-90
MODEL GRAPH
Ψ
δ
Ef
Regulation
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
b) To draw power angle characteristics
Assume an induced emf of 120% of terminal voltage
Ef =
231× 120
= 277.2V
100
δ = 60o
P1 = Power due to field excitation (Excitation power)
P2 = Reluctance power (power due to saliency)
P1 =
P2 =
3EoVSinδ 3 × 277.2 × 231× 0.866
= _______ W
=
Xd
Xd
3V 2 ( X d − X q ) Sin 2δ
2Xd Xq
=___________W
Resultant Power P = P1 + P2 = _________W
TABULATION – Power Angle Characteristics
V = 231V, Ef = 1.2 * V = 277.2volts , Xd = ______Ω, Xq = _______Ω
Load
Angle δ
-180
-150
-135
-120
-90
-60
-45
-30
0
30
45
60
90
120
135
150
180
P1 watts
P2 watts
P watts
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32
MODEL GRAPH
RESULT
Slip Test was conducted, direct axis and quadrature axis synchronous reactances were
determined and % regulation at various power factors was found out. Also regulation
curves and power-angle characteristics were plotted.
================================================================
Do you know?
1. During slip test, what does happen if rated voltage is applied to the armature?
Since the excitation emf is zero, heavy currents would be drawn by the armature if
connected to the rated voltage supply. Also, if the voltage is large, the reluctance
torque due to saliency may bring the rotor in synchronism with the rotating flux (ie.
the synchronous machine will run as a reluctance motor).
2. What you mean by the reluctance power?
A salient pole synchronous machine can stay synchronized to mains with its field
3V 2 ( X d − X q )
unexcited so long as the load does not exceed
(Note: this value is
2Xd Xq
around 30% of the rated power). The power, P2 =
reluctance power.
3V 2 ( X d − X q ) Sin 2δ
2Xd Xq
is called
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
33
3. What is meant by a reluctance motor?
A synchronous motor with salient poles but with no field winding is known as
reluctance motor.
4. What is the normal value of the ratio Xd/Xq ?
Xd
= 1.6 to 2
Xq
5. What is two reaction theory?
The armature mmf Fa (and hence the armature current) can be resolved into two
components – one acting along the d-axis, Fd, and the other acting along the q-axis,
Fq. The component mmf’s (Fd, Fq) or current (Id, Iq) produce fluxes (фad, фaq) along
the respective axes. These fluxes can be represented by the following reactances: Xad
= d-axis armature reactance to account for the flux фad produced by the d-axis
current Id . Xaq = q-axis armature reactance to account for the flux фaq produced by
the d-axis current Iq. If the leakage inductance Xal is included to account for the
leakage flux produced by the armature current, then Xd = Xad + Xal, d-axis
synchronous reactance and Xq = Xaq + Xal, q-axis synchronous reactance.
6.
Can you find the values of Xd and Xq by conducting OC and SC tests on
salient-pole synchronous machine?
Value of Xs obtained from OC and SC tests of salient pole machine corresponds to
Xd.
7. Compare between salient-pole type and smooth cylindrical type synchronous
machines.
Salient Pole
It consists of projected poles,
laminated, made of cast iron or cast
steel
Poles carry concentrated field windings
Air gap is not uniform
It has large diameter and short axial
length
It is used for low and medium speed
machines (water wheel drive or diesel
engine drive)
Smooth cylindrical
It is built from solid steel forging
(usually chromium – nickel steel)
Poles consists of radial slots in
which field windings are placed
Air gap is nearly uniform
It is of small diameter and of very
long axial length.
It is used for high speeds (Steam
driven)
Noiseless operation
Less windage loss
Better in dynamic balancing
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
34
================================================================
Experiment No. 4
VOLTAGE REGULATION OF ALTERNATOR
================================================================
AIM: To predetermine the voltage regulation of the given 3 phase alternator by i) emf
method ii) mmf method and iii) Zero power factor (Potier) method.
APPARATUS:
S.No. Name of the apparatus
Type
Range
Quantity
1.
Voltmeter
MI
(0-500V)
2
2.
MI
(0-300V)
1
3.
MC
(0-30V)
1
4.
Ammeter
MI
(0-15A)
1
5.
MC
(0-10A)
1
6.
MC
(0-5A)
1
7.
MC
(0-2A)
1
8.
Rheostat
Wire Wound
9Ω 8.5A
1
9.
272Ω 1.7A
1
10.
145Ω, 2.5A
2
11.
Tachometer
1
PRINCIPLE:
The terminal voltage of an alternator under load conditions is different from the
open circuit voltage due to the effects of armature resistance, leakage reactance and
armature reaction. Voltage regulation is defined as the rise in voltage, expressed as per
cent of rated voltage, when the load current is reduced to zero, the field excitation and
frequency being maintained constant. Thus,
Voltage regulation =
E f −V
V
× 100
The term rise in voltage used in the above definition pre-supposes a resistive or
inductive load. If the load is capacitive, the magnetizing effect of armature reaction, due
to the leading current, may cause V to be higher than Ef, thus causing a drop in voltage,
when the load current is reduced to zero. In that case, the regulation is negative.
The regulation of a synchronous generator can be predetermined by the following
methods: a) synchronous impedance or emf method, b) mmf or ampere-turn method c)
zero power factor or potier method.
Open circuit characteristic (OCC) : The open circuit characteristic of an alternator is a
curve of the armature terminal voltage on open circuit as a function of field excitation
when the machine is running at synchronous speed.
35
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
Xar
Xl
Ra
Ia
L
O
A V
D
Er
Ef
Short circuit characteristic (SCC) : It is the plot of short circuit armature current as a
function of field current when the machine is running at synchronous speed.
Zero power factor curve (ZPFC) : Zero power factor characteristic of an alternator
gives the variation of terminal voltage with field current, when the alternator is delivering
its full load current to a zero power factor (lagging) load.
PROCEDURE:
i) OPEN CIRCUIT & SHORT CIRCUIT CHARACTERISTICS (OCC & SCC)
Make the connections as shown in diagram.
Precautions/Initial settings:
i)
TPST in open position
ii)
DPST1 and DPST2 in open position
iii)
Motor field rheostat in minimum position
iv)
Potential divider in minimum voltage position
30A
+
OC & SC TEST
0-15A MI
L Z A
0-300V
MI
A
A
220V
DC
272Ω,1.7A
30A
Z
+
DPST1
+
B
C
AA
ZZ
A
V
N
M
5A
-
A
X
-
XX
TPST
0-2A MC
5A
145Ω, 2.5A
220V
DC
5A
-
DPST2
145Ω, 2.5A
Machine details : DC motor
11.7HP, 220V, 48A,1500rpm
Alternator 7.5kVA, 400V, 10.9A,
50Hz, 1500rpm
Switch on the DC supply to the DC motor by closing the switch DPST1. Start the DC
shunt motor using 3-point starter. Increase the resistance of dc motor field rheostat and
drive the alternator at rated speed. Now, dc supply is given to the alternator field winding
and for different values of field current, note down the open circuit voltage across the
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
36
armature terminals. Take care to keep the speed constant (rated value) through out the
experiment. The above procedure is repeated till the open circuit voltage reaches 120% of
rated value. Open circuit voltage/phase Ef Vs field current If gives OCC.
For SCC, reduce the armature voltage to zero by bringing the potential divider to
minimum voltage position. Now, close the TPST switch. By varying the potential divider,
increase the current through the short circuited armature up to rated value. Note both the
ammeter readings. Isc Vs If gives SCC.
ii) ZERO POWER FACTOR CHARACTERISTICS (ZPFC)
Make the connections as shown in diagram.
Precautions/Initial settings:
i)
TPST in open position
ii)
DPST1 and DPST2 in open position
iii)
Motor field rheostat in minimum position
iv)
Potential divider in minimum voltage position
Switch on the DC supply to the DC motor by closing the switch DPST1. Start the
DC shunt motor using 3-point starter. Increase the resistance of dc motor field rheostat
and drive the alternator at rated speed. Now, switch on the dc supply to the alternator field
by closing the switch DPST2 and vary the potential divider so that the generated voltage
is nearly equal to rated value. Close the switch TPST1 and note down the 3 phase supply
voltage. Adjust the potential divider and make the generator terminal voltage equal to the
3 phase supply voltage. Now, the lamps will flicker uniformly (All the lamps become dim
or bright at a time). If the lamps are not flickering uniformly (phase sequence is wrong),
then interchange the two terminals of the 3 phase supply voltage after opening the switch
TPST1. If the flickering is so fast, the motor field rheostat is adjusted very slightly so that
the frequency of flickering is convenient and the synchronization switch is closed at the
middle of the dark period. Synchronization is over. After synchronization, adjust the
motor field rheostat and make the wattmeter reading equal to zero. Now, increase the
synchronous generator field current by varying the potential divider so that the armature
current reaches rated value (10.9A). Note down the readings.
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
37
38
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
STATOR RESISTANCE MEASUREMENT
Make the connections as shown in figure.
9Ω, 8.5A
0-10A MC
+
10A
+
A
-
28V DC
+
A
-
N
V
0-30V MC
10A
-
Precaution: Keep the rheostat at maximum position.
Switch on 28V d.c. supply. Note down the voltmeter and ammeter readings for
different positions of rheostat (If possible, take readings for rated armature current).
OCC
Field Current
If
O.C. Volt Ef
SCC
Isc (A)
ZPFC
If (A)
S.No.
1.
2.
3.
4.
Ia (A)
If (A)
Stator Resistance Measurement
V (volts) I (amps)
Rdc=V/I Ω
Rdc
CALCULATION
EMF METHOD
Rated voltage/phase V = 230V
Short circuit current corresponding to rated voltage from SCC, Isc = ______A
Synchronous impedance, Z s =
V
=_______Ω
I sc
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
Armature resistance, Ra = 1.2 × Rdc = _______Ω
Synchronous reactance, X s = Z s 2 − Ra 2 = ________Ω
Ef
Isc
OCC
230V
SCC
0
If
SAMPLE CALCULATION
Regulation at full load and ____ pf. Lag
Full load current = 10.9A, V = 230V, Xs = _____ Ω, Ra = ______Ω, cosΦ = ____ lag
Sl. No.
p.f.
1
0
lag
2
0.2
lag
3
0.4
lag
4
0.6
lag
5
0.8
lag
6
1
7
0.8
lead
8
0.6
lead
9
0.4
lead
10
0.2
lead
11
0
lead
EMF METHOD
Full load Ia = 10.9A
Ef
Regulation
−
V = V ∠0° = 230∠0°
−
−
I = I ∠ − Φ° = 10.9∠ − Φ° for lag ( I = I ∠ + Φ° = 10.9∠ + Φ° for lead)
40
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
−
E f = V ∠0 + I ∠ − Φ × ( Ra + jX s ) = E f ∠δ ° =________V
(OR E f = (V cos Φ + I a Ra ) 2 + (VsinΦ + I a X s ) 2 =________V)
% regulation =
E f −V
V
× 100 =___________%
PHASOR DIAGRAMS - EMF METHOD
Ef
IXs
IXs
Ef
I
δ
δ
IRa
V
I
Φ
Unity power factor
IRa
V
Leading power factor
Ef
IXs
δ
V
Φ
IRa
Lagging power factor
I
SAMPLE CALCULATION
MMF METHOD
SAMPLE CALCULATION
Regulation at full load and ____ pf. Lag
−
V = V ∠0° = 230∠0° V
cosΦ = ____ lag
−
−
I = I ∠ − Φ° = 10.9∠ − Φ° for lag ( I = I ∠ + Φ° = 10.9∠ + Φ° for lead)
E ' = V ∠0° + I ∠ − Φ°× Ra = E ' ∠ − σ ° =________V
Refer OCC and find Ifr corresponding to E’.
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
I fr = I fr ∠(−σ + 90°) =_________A
Ifa is the field current required to circulate rated current on short circuit (from SC test)
I fr = I fa ∠( − Φ + 180°) for lag ( I fr = I fa ∠( + Φ + 180°) for lead)
I f = I fr + I fa =_________A= I f ∠(90 + δ °)
Hence, If = ______A, δ = ______˚
Refer OCC and find Ef corresponding to If.
E f = E f ∠δ ° =________V
% regulation =
E f −V
V
× 100 =__________%
PHASOR DIAGRAMS – MMF METHOD
Ef
IXs
If
Ifr
Ifa
δ
V IR
a
σ
Φ
E'
I
Lagging power factor
Ef
If
IXs
Ifr
δ
Ifa
V
I
IRa
E'
Unity power factor
Ef
IXs
Ifr
I
If
E'
δ
Φ
Ifa
σ
V
Leading power factor
IRa
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
TABULATION – MMF method
1
Power
factor
0 lag
2
0.2 lag
3
0.4 lag
4
0.6 lag
5
0.8 lag
6
1
7
0.8 lead
8
0.6 lead
9
0.4 lead
10
0.2 lead
11
0 lead
I
E'
I fr
I fa
If
Ef
%
regulation
POTIER METHOD
The zero power factor curve can be used to determine leakage reactance Xl and armature
reaction mmf Ifa . It is not necessary to plot full curve. Only two points F and A are
sufficient. Point F on ZPF characteristics corresponds to field current to circulate full-load
short circuit current during SC test. The point A corresponds to rated terminal voltage and
rated armature current condition when the load is zero power factor lagging.
voltage
air gap line
OCC
C
230V
O
D
F
(If from
SC test)
A
B
ZPFC
E
Field Current
(If from
ZPF test)
Draw AD = OF (parallel to X-axis)
Draw line DC from D parallel to air gap line meeting OCC at C.
Drop a vertical CB from C meeting the line AD at B.
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
Now, ΔABC is the potier triangle.
Drop more potier triangles to complete ZPFC.
BC = IaXal = voltage drop due to armature leakage reactance
AB = Ifar = Field current necessary to overcome the demagnetizing effect of armature
reaction at full load.
BD = Field current necessary to induce an emf required for balancing leakage reactance
drop AB.
Ef
IaXar
If
Er
Ifr
Ifar
δ
IaXal
σ
Φ
IaRa
Ia
Lagging power factor
Ef
IaXar
Er
Ia
Ifr
σ δ
If
IaXal
Φ
Ifar
IaRa
V
Leading power factor
Ef
IaXar
Er
If
IaXal
Ifr
δ σ
Ifar
Ia
V
IaRa
Unity power factor
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
SAMPLE CALCULATION
Regulation at full load and ____ pf. Lag
DE = I × X al =_____ V(from Potier triangle)
Xal = ______Ω
−
V = V ∠0° = 230∠0°
−
−
I = I ∠ − Φ° = 10.9∠ − Φ° for lag ( I = I ∠ + Φ° = 10.9∠ + Φ° for lead)
−
E r = V ∠0 + I ∠ − Φ × ( Ra + jX al ) =_________V= Er ∠σ °
Refer OCC and find Ifr corresponding to Er.
−
I f r = I fr ∠(σ + 90)° =_________A
−
−
I f a = I fa ∠( − Φ + 180)° =________A for lag
−
−
( I f a = I fa ∠( + Φ + 180)° for lead)
−
I f = I fr + I fa =__________A= I f ∠(90 + δ )°
Hence, If = ____A and δ = _____º
Refer OCC and find Ef corresponding to If.
−
E f = E f ∠δ ° =__________V
% regulation =
E f −V
V
× 100 =_________%
TABULATION – Potier Method
Power
factor
0 lag
0.2 lag
0.4 lag
0.6 lag
0.8 lag
1
0.8 lead
0.6 lead
0.4 lead
0.2 lead
0 lead
−
I
−
Er
−
Ifr
−
Ifa
−
If
−
Ef
%
regulation
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
45
MODEL GRAPHS
RESULT
Voltage regulation of the given 3 phase alternator at various power factors was
predetermined i) by emf method ii) by mmf method and iii) by potier method.
================================================================
Do you know?
1. What is armature reaction of a synchronous machine?
The effect of armature flux on the main field flux is known as armature reaction.
Armature reaction has distorting effect on unity power factor, wholly demagnetizing
at zero power factor lagging and wholly magnetizing at zero power factor leading.
2. When the load on an alternator is varied, it terminal voltage is also found
to vary. Why?
The terminal voltage under load conditions is different from that under no-load
conditions due to following three factors: i) effect of armature resistance ii) effect of
armature leakage reactance and iii) effect of armature reaction.
3. Why voltage regulation on alternator is negative for leading power factor?
When the power factor is leading, the effect of armature flux is to assist the main
flux, hence to generate more emf and so to increase the terminal voltage when the
alternator is loaded. Thus the terminal voltage of an alternator decreases when the
load of leading p.f. (ie capacitive load) is thrown off and voltage regulation is
negative.
4. Why does synchronous impedance method give a poorer voltage
regulation?
In synchronous impedance method of determination of voltage regulation
synchronous reactance is assumed to be constant while actually it varies with the
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
46
saturation (at low saturation its value is higher because the effect of armature
reaction is larger than that at high saturation). Now under short-circuit conditions,
saturation is very low and therefore, the value of synchronous impedance measured is
higher than that in actual operating conditions and the regulation determined is
higher than actual one.
5. What is the power factor of an alternator under short-circuit condition?
Since Ra<<Xs, power factor of the machine is about zero lagging.
Xs
Ra
Ia
Ef
S.C.
V= 0
6. Why is the short circuit characteristic of an alternator linear?
Air gap flux under short circuit condition is only about 15% of that under rated
voltage condition since the field current is very small. Also, the power factor under
short circuit condition is about zero lagging and hence the armature reaction is
demagnetizing in nature. Therefore there is no saturation of the magnetic circuit and
hence SCC is a straight line.
7. Define short circuit ratio (SCR). What is its improtance?
Field current required to produce rated voltage on open circuit
SCR=
Field current required to produce rated current on short circuit
SCR is the reciprocal of the per unit value of saturated synchronous reactance.
The SCR has an important effect of the performance of the machine and the cost. A
lower value of SCR means a greater change in field current to maintain constant
terminal voltage and a lower value of steady state stability limit. Lesser the SCR,
lesser is the size, weight and cost of the machine. Evidently, the short circuit ratio has
an important role to play in determining the current through the armature under fault
conditions. Modern alternators are built with short circuit ratio between 0.5 and 1.5.
8. Why the star-connection is preferred for the stator winding of a
generator?
In case of star-connection, the voltage per phase is only 1/√3 or 58% pf the
voltage between the lines. This reduces the amount of insulation required in the slots
which, in turn, enables to increase the cross section of the conductors. A larger
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
conductor permits to increase the current and hence, the power output of the
machine.
When a synchronous generator is under load, the voltage induced in each phase
becomes distorted, and the waveform is no longer sinusoidal. The distortion is mainly
due to an undesired third harmonic voltage. With a star-connection, the distorting
line-to-neutral harmonics do not appear between the lines because they affectively
cancel each other. Consequently, the line voltages remain sinusoidal under all load
conditions. Unfortunately, when a delta connection is used, the harmonic voltages do
not cancel, but add up. Because the delta is closed on itself, they produce a thirdharmonic circulating current, which increases the I2R losses.
9. What is infinite busbar?
A supply system with large number of synchronous generators in parallel and
operating at constant voltage and frequency is called infinite busbar. (It has zero
synchronous impedance Zs and infinite rotational inertia).
10. What are the conditions for paralleling an alternator with the infinite bus?
Before the alternator can be connected to the infinite bus, the incoming alternator
and the infinite bus must have the same i) voltage ii) frequency iii) phase sequence
and iv) phase.
11. What is the effect of increase in excitation of a synchronous generator
connected to an infinite busbar?
An under-excited generator operates at leading power factor, a normal excited
generator at unity power factor and overexcited generator at lagging power factor.
12. What is meant by synchronizing?
The process of connecting an alternator in parallel with another alternator or
with infinite bus bar is called synchronizing.
13. Explain synchronization by dark-lamp method?
In this method, three lamps are connected as shown in figure. By running the
alternator at synchronous speed and by adjusting the field excitation, the armature
voltage is increased near to rated value. If all the three lamps become bright and dim
simultaneously, the phase sequences of both the incoming generator and bus-bar are
the same. If they become bright and dim in sequence, the phase sequence of the
incoming alternator should be reversed (by interchanging any two leads of incoming
alternator).
R1
Y1
B1
R2
V1
V2
VR1
VR2
Y2
B2
L
VB2
VY1
L
VB1
VY2
L
Now, the field excitation is adjusted such that voltages of the incoming alternator and
the bus-bar are equal. The speed of the prime-mover of the incoming machine is
further adjusted slowly until the lamps flicker at a very low rate. The paralleling
switch is closed at the instant all the three lamps are dark. The incoming alternator
thus gets connected in parallel with the bus-bar.
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
================================================================
Experiment No. 5
NO-LOAD AND BLOCKED-ROTOR TESTS ON A 3-PHASE
SLIP RING INDUCTION MOTOR
================================================================
AIM: i) To conduct no load and blocked rotor tests on 3 phase slip ring induction motor
ii)To determine the equivalent circuit parameters and hence predetermine the
performance at full load from the equivalent circuit and
iii)To draw the circle diagram and hence predetermine the performance at full load
from circle diagram.
APPARATUS:
Name of the
S.No.
apparatus
1.
Voltmeter
2.
3
4.
Ammeter
5.
6.
7.
Wattmeter
8.
9.
Rheostat
Type
Range
Quantity
MI
MI
MC
MI
MI
MC
Dynamometer
0-500V
0-150V
0-30V
0-5A
0-10A
0-10A
500/250/125V,
5/10A,UPF
150V,10A,LPF
9Ω,8.5A
1
1
1
1
1
1
2
Wire wound
1
1
PRINCIPLE:
Depending upon the construction of rotor, there are two types of 3-phase
induction motors - a) squirrel cage and b) wound rotor or slip ring type.
A squirrel cage rotor has a number of conducting bars (made of copper or
aluminum) laid in the slots of the rotor core. These bars are short-circuited at both ends
by conducting end rings. The cage winding is adaptable to any number of poles. The cage
rotor motor is cheap and robust. However it’s starting torque is low.
A wound rotor has a laminated core with slots on its outer surface. These slots
carry 3-phase rotor winding, which is similar to the stator winding. Both the stator and
rotor windings are designed for the same number of poles. The 3-phase rotor winding is
usually star connected. The ends of three phases are tied to slip rings mounted on the
motor shaft. The rotor windings are shorted through brushes, which ride on the slip rings.
Thus the rotor currents are accessible at these brushes. Extra resistance can be connected
to the slip rings. This extra resistance is usually necessary to give a high starting torque.
The simplest and cheapest method of starting wound-rotor induction motors is by
means of added rotor resistance, with full-line voltage across the stator terminals. At the
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
49
time of start, the addition of external resistance in the rotor circuit of a wound-rotor
induction motor i) decreases its starting current ii) increases its starting torque and iii)
improves its starting power factor.
A wound-rotor induction motor is used for loads requiring large starting torque or
for loads requiring speed control. A wound-rotor induction motor is more expensive than
a squirrel-cage motor and also it requires more maintenance because of the brushes and
slip rings. A wound-rotor motor may be used for hoists, cranes, elevators, compressors
etc.
PROCEDURE:
A) NO LOAD TEST
Make the connections as shown in the diagram.
Precautions : i) Keep the autotransformer in minimum voltage position
ii) Keep belt on brake drum in loose position (motor on no load)
iii) Starter handle should be in maximum anticlockwise position (External
rotor resistance is maximum)
Switch on the 3 phase supply. Adjust the autotransformer and apply rated voltage
to the stator. By pressing the push button provided on the starter, rotate the handle of the
starter slowly in clockwise direction so that rotor resistance is gradually cut off. Now, the
motor runs on no-load. Since the power factor on no-load is quite low, less than 0.5, (the
no load current is mainly used for magnetizing the core which is largely inductive in
nature), one of the wattmeter will read negative. Then switch off the supply and
interchange the connections of the pressure coil (or current coil) of that wattmeter and
again start the motor by the above procedure. Note down the ammeter, voltmeter and
wattmeter readings. The sum of the wattmeter readings shows the rotational losses
(rotational losses = core loss + mechanical losses).
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
B) BLOCKED ROTOR TEST
Make the connections as shown in the diagram.
51
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Precautions : i) Keep the autotransformer in minimum voltage position
ii) Rotor is blocked by tightening the belt on the brake drum.
Switch on the 3 phase supply. Adjust the autotransformer so that rated current (to
get full load copper loss) flows in the ammeter. Note down voltmeter, ammeter and
wattmeter readings. (If any of the wattmeter reads negative, switch off the supply and
interchange the connections of the pressure coil (or current coil) of that wattmeter and
again take readings.)
C) STATOR RESISTANCE MEASUREMENT
Make the connections as shown in the diagram.
+
10A
28V DC
-
9Ω, 8.5A
+
0-10A MC
A
0-30V MC
A
+
V
-
R
10A
C
B
Precautions : Keep the rheostat in maximum resistance position.
Switch on 28V d.c. supply. Note down voltmeter and ammeter readings for
different positions of rheostat. (Note: Resistance/phase =
3
x Delta resistance)
2
Procedure to draw the circle diagram: (Do not write in the fair record)
13. Draw the lines by taking the current (I) in X-axis, voltage (V) in Y-axis. (V & I
are line values)
14. From the No-load test find out the current Io and draw the OA vector with the
magnitude of Io from the origin by suitable current scale, which lags the voltage
Woc
(Y-axis) V by an angle Φo where Φ o = cos −1 (
).
3Vo I o
15. From the current Isc find out the ISN (short circuit current corresponding to the
V
normal voltage) through the formula I SN = I sc ( rated ) , draw the OB vector with the
Vsc
magnitude of ISN from the origin by the same current scale, which lags the voltage
Wsc
(Y-axis) V by an angle ΦSC where Φ SC = cos −1 (
).
3Vsc I sc
16. Join the points B and A to get the output line.
17. Draw the parallel line for the X-axis from point A and for the Y-axis from point B
upto the X-axis (point E), let both the lines intersects at point D.
18. Then draw the bisector for the output line and extend it to the line AD let the point
of intersection be C.
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
19. By keeping the point C as center draw a semi circle with radius CA.
20. Let EB be the line of total loss [EB = ED + DB where ED = constant loss and DB
= variable loss]
21. In the line DB locate the point G to separate the stator and rotor copper losses by
I '2 R ' R '
using the formula (rotor copper loss/stator copper loss) = 2 2 2 = 2 where
I 2 ' R1
R1
BG R2 '
R1= stator resistance per phase and R2= rotor resistance per phase. Or,
=
.
BD Ro1
22. To get the torque line, join the points A and G.
23. To find the full load quantities, draw line BK (=Full load output/power scale).
Now, draw line PK parallel to output line meeting the circle at point P.
24. Draw line PT parallel to Y-axis meeting output line at Q, torque line at R, constant
loss line at S and X-axis at T.
Note: Choose the current scale such that the circle diagram will be as large as possible.
The larger the circle diagram more will be the accuracy. Select power scale =
3 × Vrated × current scale .
TABULATION
NO LOAD TEST
Voc
Ioc
W1
BLOCKED ROTOR TEST
W2
Woc
Vsc
Isc
W1
Stator Resistance Measurement
S.No.
V (volts) I (amps)
Rdc=V/I Ω
1.
2.
3.
4.
Rdc
CIRCLE DIAGRAM
Voc = 400V , Ioc = ___ A ,
Vsc = _____ V,
Woc = _____ W
Isc = 7.8A, Wsc = _____ W
Per phase values are
Vo = Voc = _____ V
Io =
I oc
= ____ A
3
Vs = Vsc = _____ V
Is =
I sc
= ____ A
3
Rdc = _____ Ω
W2
Wsc
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
54
3
R1 = × 1.2 × Rdc = ______ Ω
2
W
Ro1 = sc2 = _______ Ω
3I s
R2 ' = Ro1 − R1 = _______ Ω
BG R2 '
=
= ______
BD Ro1
BG = ____× BD
Selection of current and power scale
Current scale = 1cm = _____A (Take the current as large as possible, 1cm = 1 or 1.5A)
Io = ______A =_____cm
Vrated
) I sc = ____A = ____ cm
Vsc
Woc
Φ o = cos −1 (
) = _______˚
3Vo I o
Wsc
Φ SC = cos −1 (
) = _______˚
3Vsc I sc
I SN = (
Power Scale =
3 × Vrated × current scale = _______W = 1cm
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
PERFORMANCE AT FULL LOAD FROM CIRCLE DIAGRAM
Full load output = 3700W = PQ = _______cm
Full load current = OP x current scale = ______A
Full load power factor =
PT
= ______lag
OP
Rotor copper loss at full load = QR x power scale = _______W
Stator copper loss at full load = RS x power scale = _______W
Constant loss = ST x power scale = _______W
Rotor input at full load = PR x power scale = ________ W
Torque developed at full load = PR x power scale (sync.watts)
= PR x power scale x
60
N-m
2π N s
= ________N-m
Stator input at full load = PT x power scale = ________W
PQ
Efficiency at full load =
× 100% = _________%
PT
QR
Slip at full load, s =
= ________
PR
Speed at full load = (1 − s) * N s = _______rpm
60
Starting torque developed = BG x power scale x
N-m = _______N-m
2π N s
60
Maximum torque developed = I I’ x power scale x
N-m = _______N-m
2π N s
Maximum output = HH’ x power scale = ________W
Maximum input = JJ’ x power scale = _________W
EQUIVALENT CIRCUIT PARAMETERS
Voc = ____V , Ioc = ____A , Woc = _____W
Vsc = ______V, Isc = ____A, Wsc = ______W
Vo = Voc = _____ V
Io =
I oc
= ____ A
3
Vs = Vsc = _____ V
Is =
I sc
= ____ A
3
Woc
= _______
3Vo I o
sin Φ 0 = _______
cos Φ 0 =
55
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
Vo
=_________Ω
I o × cos Φ o
Vo
= ________Ω
Xm =
I o × sin Φ o
V
Z o1 = s = _________Ω
Is
W
Ro1 = sc2 = _______Ω
3I s
Rc =
X o1 = Z o12 − Ro12 = ________Ω
3
R1 = *1.2* Rdc =1.5 x 1.2 x ____ = ______Ω
2
R2 ' = Ro1 − R1 =_______Ω
X
X 2 = X 2 ' = o1 = ______Ω
2
EXACT EQUIVALENT CIRCUIT
R1
I1
X1
R2'
I2'
X2'
Io
Ic
400V
Rc
Im
Xm
1− s
)
RL ' = R2 '(
s
APPROXIMATE EQUIVALENT CIRCUIT
Ro1=R1+R2'
Xo1=X1+X2'
I
I'
1
2
Io
Ic
400V
Rc
Im
Xm
RL ' = R2 '(
1− s
)
s
PERFORMANCE AT RATED SPEED FROM EQUIVALENT CIRCUIT
Synchronous speed, Ns = 1500 rpm
Rated speed = N = 1400 rpm
N −N
1500 − 1400
× 100% =
× 100 = 6.67%
Slip = s = s
Ns
1500
1− s
RL ' = R2 '(
) =_______Ω
s
I o (per phase) = I o ∠ − Φ o ° =_______A
From approximate equivalent circuit,
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
57
V ∠0°
=________A per phase
( Ro1 + RL ') + jX o1
I1 = I o + I 2 ' = I1∠ − Φ1 ° = _________A
I 2' =
Line current IL = 3 × I1 = ______A
Power factor = cos(Φ1 °) = _______lag
Output = 3 × I 2 '2 × RL ' =________W
Output
Torque =
=_________N-m
(N=1400rpm)
2
( πN )
60
Input = 3VI L cos Φ1 = __________W
Output
Efficiency =
× 100% = ________%
Input
RESULT
vi)
No-load and blocked rotor tests were conducted on 3 phase squirrel cage
induction motor
Equivalent circuit parameters were determined
vii)
viii) Circle diagram was drawn
Performance at full load from equivalent circuit and circle diagram were
ix)
determined
================================================================
Do you know?
1.
What happens to a slip ring induction motor if 3 phase supply is given to
rotor windings keeping the stator terminals shorted?
The three phase rotor current will produce a rotating field in the air gap, which
will rotate at the synchronous speed with respect to the rotor. Voltage and current
will be induced in the stator windings. According to Lenz’s law, the rotor will
rotate opposite to the direction of the rotating field so that the induced voltage in
the stator winding is decreased.
2.
How can frequencies greater than the supply frequency can be obtained
with the use of a 3 phase slip ring induction motor?
By running the rotor against the direction of rotating magnetic field by means of a
prime-mover. If rotor speed is Nr rpm and rotating magnetic field speed Ns, then
relative speed between rotor conductors and rotating magnetic field would be
(Ns+Nr) rpm. This gives rotor frequency f2 of the voltage at slip rings as
P( N s + N r )
PN s
f2 =
Hz which is higher than the supply frequency f 2 =
.
120
120
3.
What is the advantage of slip ring induction motor?
The slip ring induction motor gives high starting torque with low starting current.
These motors are well suited for high inertia loads which take a long time to
accelerate (lifts, cranes etc).
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
58
4.
What are the speed control methods used in slip ring induction motors?
a) stator voltage variation, b) Rotor resistance variation, c) slip power recovery
5.
What are the advantages of addition of external rotor resistance at
starting?
a) It decreases the starting current, b) increases the starting torque and c)
improves the starting power factor.
6.
What are the differences between a transformer and induction motor?
i. A transformer is a static device where as an induction motor is
a rotating device. Therefore an air gap exists in an induction
motor. Due to the presence of air gap, the magnetizing current
is pretty high in an induction motor.
ii. Because of the presence of air gap, the leakage reactances in
an induction motor are higher than in a transformer.
iii. The losses are higher and the efficiency is lower in an
induction motor than in transformer.
iv. The transformer winding consists of concentrated coils. In
induction motor, the windings are distributed in slots.
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
================================================================
Experiment No. 6
INDUCTION MACHINE AS GENERATOR AND MOTOR
================================================================
AIM: i) To operate the given 3 phase induction machine as a) induction motor and b)
induction generator
ii) To conduct load test in both generating and motor modes
iii) To plot the performance characteristics and
iv) To plot W Vs slip and hence determine the hysteresis power.
APPARATUS:
S.No.
Name of the
apparatus
1.
Voltmeter
2.
3.
4.
Ammeter
5.
6.
Wattmeter
7.
Rheostat
8.
Stopwatch
9.
Tachometer
Type
Range
Quantity
MI
MC
MC
MI
(0-500V)
(0-500V)
(0-250V)
(0-10A)
(10-0-10A)
250V,10A,upf
272Ω 1.7A
1
1
1
1
2
1
1
1
1
Dynamometer
Wire Wound
PRINCIPLE:
An induction generator is asynchronous in nature because of which it is
commonly used as windmill generator since a windmill runs at non-fixed speed. These
are used in remote areas to supplement power received from weak transmission links.
If the induction machine is driven at a speed greater than synchronous speed by a
prime mover, the direction of induced torque reverses and it acts as an induction
generator. The rotating magnetic field is set up by the magnetizing current drawn from
the mains. Based on the way with which the generator gets the required lagging reactivepower, they are classified into i) Line excited induction generator and ii) Self-excited
induction generator.
Line excited induction generator : Induction machine connected to supply mains and
driven at super-synchronous speed by its prime mover is called a line excited induction
generator. The generator draws the required lagging reactive power from the mains.
Self excited induction generator : In this generators, the necessary lagging reactive
power for its excitation is obtained by a capacitor bank connected across the generator
terminals. If the rotor of the machine is driven by its prime-mover, the presence of
residual flux (present in rotor core) causes a small emf to get induced in the stator
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
60
windings at a frequency proportional to the rotor speed. This voltage impresses over the
3-phase capacitor gives rise to leading current drawn by the capacitor which is equivalent
to lagging current supplied to the generator. The flux set up by this current assists the
initial residual flux. Hence the net flux will increase causing the voltage to build up
further. The steady state voltage induced on no load is given by the intersection of
magnetization characteristic of the machine and capacitance V-I characteristic.
PROCEDURE:
Make the connections as shown in figure.
Precaution: i) Keep the dc machine (separately excited) field rheostat at maximum
resistance position (Reason: Initially dc machine will act as dc generator)
ii) Keep DPST, SPST and TPST switches open
iii) Keep rotor rheostat at maximum resistance position
Switch on the three phase supply by closing the TPST switch. Induction machine
will start as induction motor. As the motor gathers speed, gradually cut off the rotor
resistance. Now, switch on the DC supply by closing the DPST switch. Decrease the
resistance of the dc motor field rheostat (i.e., excitation is increased) and note the
voltmeter reading across the SPST switch. If it is increasing, switch off the dc supply and
interchange the armature terminals A & AA. Again switch on the dc supply keeping the
motor field rheostat at maximum position. Decrease the resistance of the field rheostat
gradually and make the voltmeter reading across the SPST switch zero. Now, the SPST
switch is closed.
Again increase the excitation till the dc ammeter shows rated current (10A).
Please note that during this time, the connection of the reversing switch across the
wattmeter should be such that the wattmeter reading is positive. Note down all the meter
readings and time for 10 (or 5) oscillations. Now, decrease the excitation in steps for
different values of dc ammeter and note all the readings each time. This procedure is
continued till the wattmeter reads zero. Note the speed using tachometer.
If the excitation is again decreased, induction machine will be in generator mode.
Now, interchange the connections of the pressure coil of the wattmeter using the
reversing switch. Decrease the excitation and take readings for different values of dc
ammeter. This procedure is continued till the dc ammeter reading reaches rated value
(10A). Before switching off the dc supply, increase the excitation till the dc ammeter
reads zero. Now switch off the ac supply also.
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TABULATION
Note :i) Wattmeter reading negative indicates induction generator mode.
ii) Since the supply frequency is usually slightly less than 50Hz, induction machine
acts as generator even if the speed is less than 1000rpm.
iii) Here, the dc generator is separately excited type; not self-excited.
SAMPLE CALCULATION
INDUCTION MACHINE WORKING AS MOTOR (Set No. __)
Vdc = _____V, Idc = _____A, Vac = ______V, Iac = ______A Wph = _____W
Time for 10 oscillations = ____ sec
Ns = 1000rpm
Rotor frequency, f2 = 10/T = ____Hz
Input = 3 Wph = 3 x ____ = ______W
Output = Vdc x Idc = ______W
% Slip =
f2
× 100 =_______%
f
Speed, N = (1 − s ) × N s = _______ rpm
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Efficiency =
output
×100 =________%
input
INDUCTION MACHINE WORKING AS GENERATOR (Set No. ___)
Vdc = ______V, Idc = ____A, Vac = ______V, Iac = _____A, Wph = _____W
Time for 10 oscillations = _____ sec
Ns = 1000rpm
Rotor frequency, f2 = 10/T = ______Hz
Input = Vdc x Idc = ______W
Output = 3 Wph = _______W
% Slip = −
f2
× 100 = ______% (Note: Slip is negative)
f
Speed, N = (1 − s) × N s = ________rpm (Note: Speed will be above synchronous speed,
1000rpm)
Efficiency =
output
×100 =________%
input
From 3Wph Vs % slip characteristics,
Hysteresis power =
AB
=______W
2
MODEL GRAPHS
Efficiency %
3Wph
Generator
Motor
A
B
% Slip
Output in W
RESULT
a) Performance characteristics were plotted while the induction machine is
operating as generator and motor.
b) hysteresis power = _______W.
================================================================
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64
Do you know?
1. What is an induction generator? What are its limitations?
When a 3 phase induction motor is made to run at a speed higher than its
synchronous speed by means of a prime-mover coupled with it, the 3 phase induction
motor becomes a 3 phase induction generator.
Its limitations are : i) it can not run in isolation, it must take reactive power from an
existing supply network,
ii) its output voltage and frequency can not be controlled and
iii) it always operates at a leading power factor.
2. Explain the working principle of self excited induction generator.
If the rotor of the machine is driven by its prime-mover, the presence of residual
flux causes a small emf to get induced in the stator windings at a frequency proportional
to the rotor speed. This voltage impresses over the 3-phase capacitor gives rise to leading
current drawn by the capacitor which is equivalent to lagging current supplied to the
generator. The flux set up by this current assists the initial residual flux. Hence the net
flux will increase causing the voltage to build up further. The steady state voltage induced
on no load is given by the intersection of magnetization characteristic of the machine and
capacitance V-A characteristic.
The frequency generated is slightly less than that corresponding to the speed of
rotation.
The terminal voltage of the generator increases with the capacitance, but its
magnitude is limited by saturation in the iron. If the capacitance is insufficient, the
generator voltage will not build up. The capacitor bank must be able to supply at least as
much reactive power as the machine normally absorbs when operating as a motor.
3. In case of a line-excited induction generator, how the slip is affecting the
active power delivered?
The active power delivered to the line is directly proportional to the slip above
synchronous speed. Thus, a higher prime-mover speed produces a greater electrical
output.
4. What you mean by single phasing?
Single phasing is a fault condition in which a 3 phase motor is operating with one
line open (due to blowing of a fuse in one phase). Although the 3 phase motor will not
start with one line open, if the motor is running when single phasing occurs, it will
continue to run as long as the shaft load is less than 80% rated load and the remaining
single phase voltage is normal; rotation of the rotor produces a quadrature field that
maintains the rotation.
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
5. Draw the torque-slip characteristics of an induction machine.
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================================================================
Experiment No. 7
NO LOAD & BLOCKED ROTOR TESTS ON POLE
CHANGING INDUCTION MOTOR
================================================================
AIM: i) To study the different modes of operation of a 3 phase pole changing induction
motor
ii) To perform no load and blocked rotor tests on pole-changing induction motor,
determine the equivalent circuit parameters and plot the torque-speed
characteristics for both low speed and high speed connections.
APPARATUS:
S.No.
Type
Range
Quantity
Wattmeter
MI
MI
MC
MI
MI
MC
Dynamometer
1
1
1
1
1
1
2
8.
Wattmeter
Dynamometer
9.
10.
Rheostat
Dynamometer
Wire wound
0-500V
0-150V
0-30V
0-5A
0-10/20A
0-10A
500/250/125V,
5/10A,UPF
150V,
10/20A,UPF
150V,10A,LPF
9Ω,8.5A
1.
2.
3
4.
5.
6.
7.
Name of the
apparatus
Voltmeter
Ammeter
2
1
1
PRINCIPLE:
If an induction motor is to run at different speeds, one way is to have different
windings for the motor so that it will have different synchronous speeds and the running
speeds. Another method is to use one winding but with suitable connections for a changeover to double the number of poles.
In pole changing induction motors, the stator winding of each phase is divided
into two equal groups of coils. These coil groups are connected in series and parallel with
the current direction being reversed only in one group, to create two different numbers of
poles (even) in the ratio 2:1 respectively. When the connection is changed from series to
parallel or vice versa, the current in one group of coils is also reversed at the same time.
This technique, termed the consequent pole method, is applied to all three windings
(phases). This type of induction motor has always the squirrel cage rotor, which can adapt
to any number of stator poles.
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67
Figure 1 (a) shows schematically only four coils of one phase of the windings
connected in series, along with the direction of current in them, producing four poles in
the stator. If the current in coils 2 and 4 is reversed and the connection is changed to
parallel with two coils (1 and 3, and 2 and 4) connected in series for each path, eight poles
are formed in the stator (Figure 1 (b)). It may be noted that the direction of current in
coils 1 and 3 remains the same. Only one type of connection is shown.
Constant torque and constant power operations
The choice of winding arrangements of 3 phase, two speed motors depends on the
operating characteristics required at the two speeds. For constant torque operation, the
change of stator winding is made from series-star to parallel-star, while for constant
power operation the change is made from series-delta to parallel-delta.
a) Constant torque type
If it is desired to have constant torque at both the speeds, the arrangement is as
shown in figure 2. For low speed, the voltage is applied to terminals S1-S2-S3 while S4-S5S6 are left open and for high speed operation of the motor, the voltage is applied to
terminals S4-S5-S6 and the terminals S1-S2-S3 are shorted. The flux densities for the two
speeds are approximately equal; the motor can be considered to have a constant torque.
The output of the motor is approximately proportional to the speed.
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
68
Let V = Line voltage, I = Maximum current that the winding can carry.
Then, the power drawn from the supply is given by,
1. For series-star connection, PY = 3VI cos ΦY
2. For parallel-star connection, PY = 2 3VI cos ΦYY
It is assumed that the power factor remains unchanged and the motor losses are
negligible. With the changeover of stator winding from series-star to parallel-star, the
power drawn from the supply is doubled. Simultaneously, the speed is also doubled. So,
the motor torque remains constant.
b) Constant power type
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If constant power output is to be obtained from the motor at both the speeds, i.e.
the torque is inversely proportional to the speed, the connection of the winding may be
done as shown in figure 3. For low speed operation, the voltage is applied to terminals S1S2-S3 and terminals S4-S5-S6 are left open, while for high speed operation, the voltage is
applied to terminals S4-S5-S6 and S1-S2-S3 are shorted.
1. For series-delta connection, PΔ = 3VI cos Φ Δ
2. For parallel-star connection, PY = 2 3VI cos ΦYY = 3.46VI cos ΦYY
After changeover from series-delta to parallel-star, the power increases slightly
(about 15%), if power factor is assumed to remain constant. The constant power
connection is the most expensive, because in this case the motor size becomes the largest.
PROCEDURE:
A) NO LOAD TEST ON LOW SPEED MOTOR
Make the connections as shown in figure.
Precautions : i) Keep the autotransformer in minimum voltage position
ii) Keep belt on brake drum in loose position (motor on no load)
Switch on the 3 phase supply by closing TPST switch. Adjust the autotransformer
and apply rated voltage to the stator. Note down the ammeter, voltmeter and wattmeter
readings. (If any of the wattmeter reads negative, switch off the supply and interchange
the connections of the pressure coil (or current coil) of that wattmeter).
TABULATION
LOW SPEED
NO LOAD TEST
Voc
Ioc
W1
W2
BLOCKED ROTOR TEST
Woc
Vsc
Isc
W1
W2
Wsc
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B) BLOCKED ROTOR TEST ON LOW SPEED MOTOR
Make the connections as shown in figure.
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72
Precautions : i) Keep the autotransformer in minimum voltage position
ii) Rotor is blocked by tightening the belt on the brake drum.
Switch on the 3 phase supply. Adjust the autotransformer so that rated current (to
get full load copper loss) flows in the ammeter. Note down voltmeter, ammeter and
wattmeter readings. (If any of the wattmeter reads negative, switch off the supply and
interchange the connections of the pressure coil (or current coil) of that wattmeter).
A) NO LOAD TEST ON HIGH SPEED MOTOR
Make the connections as shown in figure.
Precautions : i) Keep the autotransformer in minimum voltage position
ii) Keep belt on brake drum in loose position (motor on no load)
Switch on the 3 phase supply by closing TPST switch. Adjust the autotransformer
and apply rated voltage to the stator. Note down the ammeter, voltmeter and wattmeter
readings. (If any of the wattmeter reads negative, switch off the supply and interchange
the connections of the pressure coil (or current coil) of that wattmeter).
TABULATION
HIGH SPEED
Voc
NO LOAD TEST
Ioc
W1
W2
Woc
Vsc
BLOCKED ROTOR TEST
Isc
W1
W2
Wsc
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B) BLOCKED ROTOR TEST ON HIGH SPEED MOTOR
Make the connections as shown in figure.
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Precautions : i) Keep the autotransformer in minimum voltage position
ii) Rotor is blocked by tightening the belt on the brake drum.
Switch on the 3 phase supply. Adjust the autotransformer so that rated current (to
get full load copper loss) flows in the ammeter. Note down voltmeter, ammeter and
wattmeter readings. (If any of the wattmeter reads negative, switch off the supply and
interchange the connections of the pressure coil (or current coil) of that wattmeter).
C) STATOR RESISTANCE MEASUREMENT
Make the connections as shown in figure.
+
9Ω, 8.5A
10A
28V
DC
-
+
0-10A MC
A
0-30V MC
S1
S5
+
V
-
S6
R
10A
S3
S4
S2
Precautions : Keep the rheostat in maximum resistance position
Switch on 28V d.c. supply. Note down voltmeter and ammeter readings for
different positions of rheostat. Switch off the supply.
(Note: For low speed connection, Resistance/phase =
3
x Delta resistance. For
2
3
high speed connection, Resistance/phase= x Delta resistance)
8
S.No.
1.
2.
3.
4.
Stator Resistance Measurement
V (volts) I (amps)
Rdc=V/I Ω
Average
Rdc
EQUIVALENT CIRCUIT PARAMETERS (LOW SPEED)
Voc = _____V , Ioc = ____A , Woc = _____W
Vsc = ______V, Isc = _____A, Wsc = ______W
Vo = Voc = _____ V
Io =
I oc
= ____ A
3
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S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
Vs = Vsc = _____ V
Is =
I sc
= ____ A
3
Woc
= _______
3Vo I o
sin Φ 0 = _______
Vo
Rc =
=_________Ω
I o × cos Φ o
Vo
Xm =
= ________Ω
I o × sin Φ o
V
Z o1 = s = _________Ω
Is
Wsc
Ro1 = 2 = _______Ω
3I s
cos Φ 0 =
X o1 = Z o12 − Ro12 = ________Ω
3
R1 = *1.2* Rdc =1.5 x 1.2 x ____ = ______Ω
2
'
R2 = Ro1 − R1 =_______Ω
X
X 2 = X 2 ' = o1 = ______Ω
2
APPROXIMATE EQUIVALENT CIRCUIT (HGH SPEED)
I1
Ro1=R1+R2'
I2'
Xo1=X1+X2'
Io
Ic
400V
Rc
Im
Xm
RL ' = R2 '(
1− s
)
s
TORQUE SPEED CHARACTERISTICS FROM EQUIVALENT CIRCUIT – LOW
SPEED
Synchronous speed, Ns = 750 rpm
Assume speed = N = 710 rpm (Take N = 0 to 750rpm in steps of 30rpm)
N −N
750 − 710
× 100% =
×100 = 5.33%
Slip = s = s
750
Ns
V = 400V
V
= ____________A per phase
Rotor current, I 2 ' =
R2 ' 2
' 2
( R1 +
) + ( X1 + X 2 )
s
R'
Rotor Input P2= 3 × I 2 '2 × 2 = _______W
s
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Torque developed =
(
P2
2π N s
=
)
60
60 × P2
= _________N-m
2 × π × 750
EQUIVALENT CIRCUIT PARAMETERS (HIGH SPEED)
Voc = _____ V , Ioc = ____A , Woc = _____W
Vsc = ______V, Isc = _____A, Wsc = ______W
Vo =
Voc
= _____ V
3
I o = I oc = ____ A
Vs =
vsc
= _____ V
3
I s = I sc = ____ A
Woc
= _______
3Vo I o
sin Φ 0 = _______
Vo
Rc =
=_________Ω
I o × cos Φ o
Vo
Xm =
= ________Ω
I o × sin Φ o
V
Z o1 = s = _________Ω
Is
W
Ro1 = sc2 = _______Ω
3I s
cos Φ 0 =
X o1 = Z o12 − Ro12 = ________Ω
3
R1 = × 1.2 × Rdc = ________Ω
8
R2 ' = Ro1 − R1 = ________Ω
X
X 1 = X 2 ' = o1 = _______ Ω
2
APPROXIMATE EQUIVALENT CIRCUIT (HIGH SPEED)
I1
Ro1=R1+R2'
I2'
Xo1=X1+X2'
Io
Ic
254V
Rc
Im
Xm
RL ' = R2 '(
1− s
)
s
TORQUE SPEED CHARACTERISTICS FROM EQUIVALENT CIRCUIT –
HIGH SPEED
Synchronous speed, Ns = 1500 rpm
Assume speed = N = 1440 rpm (Take N = 0 to 1500rpm in steps of 60rpm)
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
Slip = s =
V=
78
Ns − N
1500 − 1440
× 100% =
×100 = 4 %
Ns
1500
440
= 254V
3
Rotor current, I 2 ' =
V
R '
( R1 + 2 ) 2 + ( X 1 + X 2' ) 2
s
= _________ A per phase
R2'
= ________W
s
P2
60 × P2
Torque developed =
=
= ________N-m
2π N s
2
1500
×
π
×
(
)
60
Rotor Input P2 = 3 × I 2 '2 ×
Note: Find the torque for different values of speed and hence plot torque Vs speed
characteristics using MATLAB, C, EXCEL or any software.
TABULATION
Sl.No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
LOW SPEED
Speed
Slip
0
30
60
90
120
150
180
210
240
270
300
330
360
390
420
450
480
510
540
570
600
630
660
690
710
750
Torque
Speed
0
60
120
180
240
300
360
420
480
540
600
660
720
780
840
900
960
1020
1080
1140
1200
1260
1320
1380
1440
1500
HIGH SPEED
Slip
Torque
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
79
RESULT
i) Different modes of operation of a 3-phase pole-changing induction motor
studied.
ii) No-load and blocked rotor tests conducted with low and high speed
connections
iii) Equivalent circuit parameters determined for both low and high speed.
iv) Torque-speed characteristics plotted for both low and high speed.
================================================================
Do you know?
1. Explain the reasons for lower power factor of low speed, 3 phase induction
motors as compared to that of high speed motors.
Because of larger number of poles in low speed 3 phase induction motors as
compared to high speed motors, magnetizing current is more due to increase in
leakages, which increases with the increase in number of poles. So the power
factor of low speed induction motors is poor in comparison to that for high speed
induction motors.
2. If the number of poles on a motor is increased to lower the speed, how will
the power factor be affected?
Power factor will be reduced.
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3. Speed control by pole changing method is applicable only to squirrel cage
induction motors. Why?
The rotor of a squirrel cage induction motor can adjust to any number of stator
poles. For wound rotor induction motor, rotor is wound for same number of poles
as the stator. When the stator winding is changed for different poles, rotor also
should be changed for the same number of poles which is not practical.
4. What you mean by plugging?
While an induction motor is rotating in one direction, if the phase sequence is
changed suddenly, the stator rotating magnetic field will rotate opposite to the
rotation of the rotor. The motor will come to zero speed rapidly. This type of
braking of an induction motor is called plugging. At zero speed, unless the supply
is disconnected, the motor will accelerate in the opposite direction.
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================================================================
Experiment No. 8
NO-LOAD AND BLOCKED-ROTOR TESTS ON SINGLE
PHASE INDUCTION MOTOR
================================================================
AIM: i) To conduct the no load and blocked rotor tests on single phase induction motor
ii) To find the equivalent circuit parameters
iii) To predetermine its performance at rated speed.
APPARATUS:
S.No.
1.
2.
3.
4.
5.
6.
Name of the
apparatus
Voltmeter
Ammeter
Wattmeter
7.
8.
Rheostat
Type
Range
Quantity
MI
MI
MC
MI
MC
Dynamometer
type
(0-250V)
(0-100V)
(0-30V)
(0-10A)
(0-10A)
250,10A, LPF
1
1
1
1
1
1
125V,10A, UPF
9Ω,8.5A
1
1
Wire-wound
PRINCIPLE:
Since a single phase induction motor does not have a starting torque, it needs
special methods of starting. The stator is provided with two windings, called main and
auxiliary windings, whose axes are space displaced by 90 electrical degrees. The rotor is
of squirrel cage construction. The auxiliary winding is excited by a current which is out of
phase with the current in the main winding, both current derived from the same supply
mains. The auxiliary winding is disconnected by a centrifugal switch after the motor has
achieved about 75% speed. Depending upon the starting methods, single phase induction
motors are classified into i) split phase motor, ii) capacitor start motor, iii) Capacitor start
and run motor, iv) shaded pole motor and v) repulsion start induction run motor. In all
these methods, it will act as a two-phase motor at the time of starting.
Here, no load and blocked rotor tests on capacitor start induction motor are done.
In a capacitor start motor, the main (or running) and auxiliary (or starting) windings are
space displaced by 90˚. The time displacement between the currents in the main and
auxiliary winding is achieved by connecting a capacitor in series with auxiliary winding.
By using a capacitor of proper value, the current Ia in the auxiliary winding can be made
to lead the current Im in the main winding by 90˚ at standstill. Thus the motor develops a
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starting torque. Both the capacitor and auxiliary winding are designed for short time duty
and are disconnected by centrifugal switch when the motor has reached 75% speed.
PROCEDURE:
NO LOAD TEST
Make the connections as shown in the diagram.
P
250V,10A
LPF
1
L
0-10A MI
10A
M
A
C
V
1'
A
R
E
230V
1-phase
50Hz
2
0-250V MI
CS
2'
V
C
C
N
Machine Details : 230V, 7.4A, 1HP, 1425rpm
NL
Precaution: Keep the autotransformer at minimum voltage position.
Switch on the supply. Adjust the autotransformer and apply rated voltage. Note
down the ammeter, voltmeter and wattmeter readings. Switch off the supply.
BLOCKED ROTOR TEST
Make the connections as shown in the diagram.
0-10A MI
10A
A
P
M
C
100V,10A
UPF
1
L
V
1'
A
R
E
230V
1-phase
50Hz
C
2
0-100V MI
V
CS
2'
C
N
NL
Machine Details : 230V, 7.4A, 1HP, 1425rpm
Note: For blocked rotor test, the auxiliary winding is disconnected and only the main
winding is connected to ac supply.
Precaution: Keep the autotransformer at minimum voltage position.
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Switch on the supply. Adjust the autotransformer and make the current equal to
rated value. Note down the ammeter, voltmeter and wattmeter readings. Switch off the
supply.
STATOR RESISTANCE MEASUREMENT
Make the connections as shown in the diagram.
+
0-10A MC
10A
+
9Ω,8.5A
A
A
1
2
+
0-30V
MC
28V
DC
V
V
-
10A
-
Precaution: Keep the rheostat at maximum resistance position.
Switch on 28V d.c. supply. Note down voltmeter and ammeter readings for
different positions of rheostat. Switch off the supply.
TABULATION
NO LOAD TEST
Vo
Io
Wo
S.No.
1.
2.
3.
4.
BLOCKED ROTOR TEST
Vsc
Isc
Wsc
Stator Resistance Measurement
V (volts) I (amps)
Rdc=V/I Ω
Rdc
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I1
R1
X1
I1
I2f'
Imf
R1
X1
R2'/2s
Rf
Xm/2
Ef
X2'/2
Xf
V1
V1
I2b'
Imb
Rb
R2'/2(2-s)
Eb
Xm/2
Xb
X2'/2
Equivalent circuit during normal
running condition
R1
R1
Isc
X1
Io
X1
Xm/2
Io
R2'
Vsc
Vo
R2'/4
Vo
Xm/2
X2'
X2'/2
Equivalent circuit during
blocked rotor test
R1+R'2/4+j(X1+X'2/2) << jXm/2
Equivalent circuit during
no load test
CALCULATION
EQUIVALENT CIRCUIT PARAMETERS
BLOCKED ROTOR TEST
Vsc = _____V, Isc = _____A, Wsc = ______W
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
Wsc
= ______
Vsc I sc
W
Rsc = R1 + R2 ' = sc2 = _______Ω
I sc
R1dc = ______ Ω
R1 = 1.2 x R1dc = _______Ω
R2 = Rsc − R1 = _______Ω
V
Z SC = sc = _______ Ω
I sc
cos Φ sc =
X sc = X 1 + X 2 ' = Z sc 2 − Rsc 2 = __________ Ω
X sc
= _________ Ω
2
NO LOAD TEST
Vo = _______V , Io = _____A , Wo = _______W
X1 = X 2 ' =
Rotational loss Wrot = Wo − I o 2 ( R1 +
R2 '
) =______W
4
X m Vo
= = ________ Ω
2
Io
Xm = ________Ω
Predetermination of the steady state performance at rated speed (N = 1425 rpm)
At rated speed, slip = s =
N s − N 1500 − 1425
=
= 0.05
Ns
1500
Equivalent circuit at rated speed is given by,
⎛ jX ⎞ ⎛ R ' jX ' ⎞
Forward impedance, Z f = R f + jX f = ⎜ m ⎟ // ⎜ 2 + 2 ⎟ = _______ Ω
2 ⎠
⎝ 2 ⎠ ⎝ 2s
jX ' ⎞
⎛ jX ⎞ ⎛ R2 '
Backward impedance, Z b = Rb + jX b = ⎜ m ⎟ // ⎜
+ 2 ⎟ = _______ Ω
2 ⎠
⎝ 2 ⎠ ⎝ 2(2 − s )
Total impedance, Z = ( R1 + jX 1 ) + Z f + Z b =_______ Ω
Current, I1∠ − Φ =
V1∠0°
= _______ Ω
Z
Power factor = cos Φ = ______lag
Input power = V1 I1 cos Φ =________W
Forward torque, T f = I12 R f sync.watts = _________sync.watts
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86
Reverse torque, Tb = I12 Rb sync.watts = ________sync.watts
Resultant torque, T = T f − Tb = ________ sync.watts
Mechanical power developed Pmech= (1 − s)T = _______watts
Output power = Pmech – Wrot = _______W
Resultant torque in N-m =
Efficiency =
2π
T
Ns
= _______N-m
60
Output
×100 = _______%
Input
RESULT :
i)
No load and blocked rotor tests on single phase induction motor were
conducted
ii)
Equivalent circuit parameters determined and
its performance at rated speed determined from equivalent circuit
iii)
================================================================
Do you know?
1. What is double field revolving theory ?
A alternating sinusoidal flux Φ = Φ m sin ωt can be represented by 2 revolving fluxes
Φ
each equal to one-half of the maximum value of alternating flux (= m ) and each
2
120 f
) in opposite directions.
rotating at synchronous speed ( N s =
P
If the rotor is stationary and the stator winding is connected to a single phase supply,
a pulsating flux (not rotating) is produced. This pulsating flux induces current by
transformer action in the rotor circuit, which in turn produces a pulsating rotor flux
along the same axis as the stator flux. By Lenz’s law, these two fluxes tend to oppose
each other. As the angle between these fluxes is zero, no starting torque is developed.
2. How will you reverse the direction of rotation of a single phase induction
motor ?
Interchange the connections of either main winding or auxiliary winding (not both).
3. Why are the single phase induction motors with one stator winding not self
starting?
If the rotor is stationary and the stator winding is connected to a single phase supply,
a pulsating flux (not rotating) is produced. This pulsating flux induces current by
transformer action in the rotor circuit, which in turn produces a pulsating rotor flux
along the same axis as the stator flux. By Lenz’s law, these two fluxes tend to oppose
each other. As the angle between these fluxes is zero, no starting torque is developed.
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87
4. How will you start a single phase induction motor?
Single phase motor can be started either by spinning the rotor or by using auxiliary
winding.
A pulsating field is equivalent to two rotating fields of half the magnitude but rotating
at the same synchronous speed in opposite directions. Both these rotating fluxes
produce torque, although in opposite directions. At standstill, these two torques,
forward and backward, are equal in magnitude and therefore the resultant starting
torque is zero. At any other speed, the two torques are unequal and the resultant
torque keeps the motor rotating in the direction rotation.
5. What is the typical capacitor value for a capacitor run induction motor?
300 μF for 0.5 hp motor
6. What is a universal motor?
Single phase series motor can be used with either a dc source or a single phase ac
source, hence there are called universal motors. They are used in domestic appliances
such as portable tools, drills, mixers and vacuum cleaners and usually are light in
weight and operate at high speeds (1500 to 10,000rpm).
7. What you mean by synchronous speed in a linear induction motor?
Synchronous speed is speed of the primary flux. It is a function of the frequency of the
applied voltage and the span of the primary coils (pole pitch). In one cycle of applied
voltage, the magnetic field travels a linear distance equal to two pole pitches.
If Us = synchronous speed (m/s), τ = pole pitch (m) and f = supply frequency (Hz)
Us = 2τf
Synchronous speed is not dependent on the number of poles. Any number of poles may
be used, odd or even.
U −U
.
Speed of secondary = U = Us(1-2) and slip, s = s
Us
Reversal of direction of speed linear induction motor is accomplished by reversing the
phase sequence of the primary voltage.
Primary of linear induction motor is wound similar to the stator of a squirrel cage
motor, except that the windings are laid in a straight line. The secondary is a
conducting sheet or rail of copper or aluminum. Although the conducting rail has no
squirrel cage bars, the induced eddy currents in the rail develop a force in a direction
to oppose the relative motion.
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================================================================
Experiment No. 9
V CURVES OF SYNCHRONOUS MACHINE
================================================================
AIM: i) To synchronize a 3 phase alternator to the supply mains using Dark lamp
method
ii) Plot the V curves and inverted V curves when synchronous machine is acting as
generator and motor at no load and constant power.
APPARATUS:
S.No. Name of the apparatus
1.
Voltmeter
2.
3
4.
Ammeter
5.
6.
Rheostat
7.
8.
Tachometer
Type
MI
MC
MC
MI
MC
Wire Wound
Range
(0-500V)
(0-300V)
(0-30V)
(0-15A)
(0-3A)
145Ω 2.5A
272Ω 1.7A
Quantity
2
1
1
1
1
2
1
1
PRINCIPLE:
Assume constant power operation of a synchronous machine connected to an
infinite bus. The equivalent circuit, neglecting the stator resistance, and the phasor
diagram are shown in figures. For a 3 phase machine, the power transfer is P = 3VIacosф.
Because V is constant, for constant power operation I a cos Φ is constant; that is, the inphase component of the stator current on the axis of the phase V is constant. The locus of
the stator current is therefore the vertical line passing through the current phasor for unity
power factor.
Locus of Ia for constant power
Ia3
Ia2
-Ia1Xs
-Ia2Xs
V
-Ia3Xs
Efsinδ
Ef1
Ef2
Ef3
Ia1
Locus of Ia for constant power
Synchronous machine operating as motor
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89
In figure, phasor diagrams are drawn for three stator currents:
Ia = Ia1 , lagging V
= Ia2 , in phase with V
=Ia3
, leading
V
For these stator currents the excitation voltages Ef1, Ef2 and Ef3 (representing the field
currents If1, If2
and If3, respectively) are drawn to satisfy the phasor relationship
E f = V − jI a X s for synchronous motor and E f = V + jI a X s for synchronous generator.
The power can also be expressed as P = 3
VE f
Xs
sin δ . Again for constant power operation,
E f sin δ is constant. Thus the locus of Ef (or If) is also a straight line parallel to the
phasor V such that the vertical difference between the locus of Ef and the phasor V is
constant and equals E f sin δ .
The excitation voltage Ef changes linearly with the field current If. Therefore, as If
is changed, Ef will change along the locus of Ef and Ia will change along the locus of Ia,
signifying a change in the power factor angle ф of the stator current.
When the machine is working as synchronous motor, for low field current If1,
underexcitation (Ef = Ef1), the stator current (Ia = Ia1) is large and lagging. The stator
current is minimum (Ia = Ia2) and at unity power factor for the field current If2 (Ef = Ef2)
which is called normal excitation. For larger field current If2, overexcitation (Ef = Ef3) the
stator current (Ia = Ia3) is large and leading. The variation of the stator current with the
field current for constant-power operation is shown in figure. This is known as the V-
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
90
curve because of the characteristic shape. The variation of the power factor with the field
current is the inverted V-curve.
When the machine is working as synchronous generator, for low field current If1,
underexcitation (Ef = Ef1), the stator current (Ia = Ia1) is large and leading. The stator
current is minimum (Ia = Ia2) and at unity power factor for the field current If2 (Ef = Ef2)
which is called normal excitation. For larger field current If2, overexcitation (Ef = Ef3) the
stator current (Ia = Ia3) is large and lagging.
When the machines is working as generator, it supplies a lagging power factor
current when over-excited and leading power factor current when under-excited. When
the machine is working as motor, it draws a leading power factor current when overexcited and draws a lagging power factor current when under excited.
If the synchronous machine is not transferring any power but is simply floating on
the infinite bus, the power factor is zero, that is, the stator current either leads or lags the
stator voltage by 90˚. The magnitude of the stator current changes as the field current is
changed, but the stator current is always reactive. Looking for the machine terminals, the
machine behaves as a variable inductor or capacitor as the field current is changed. An
unloaded synchronous machine is called a synchronous condenser and may be used to
regulate the receiving end voltage of a long power transmission line.
PROCEDURE:
Make the connections as per the diagram.
Precautions : i) Keep dc motor field rheostat in minimum position
ii) Keep alternator field potential divider in the minimum voltage position
iii) Keep DPDT, DPST, TPST1, TPST2 switches open
iv) Keep the load on DC side in off position.
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92
Switch on the supply to the dc motor (DPDT switch in position 1-1’). Start the
motor using the 3 point starter and increase the speed to the synchronous speed by
varying the motor field rheostat. Now, switch on the supply to the alternator field and
vary the potential divider so that the generated voltage is nearly equal to rated value.
Close the TPST switch and note down the 3 phase supply voltage. Adjust the potential
divider and make the generator terminal voltage equal to the 3 phase supply voltage.
Now, the lamps will flicker in sequence. If the lamps are flickering uniformly ie. all the
lamps become dim or bright simultaneously (phase sequence is wrong), then interchange
the two terminals of the 3 phase supply voltage after switching off the TPST and DPST
switches. If the flickering is so fast, the motor field rheostat is adjusted very slightly so
that the frequency of flickering is convenient and the synchronization switch is closed
when two lamps show maximum brightness and the third dark. Synchronization is over.
Now, the wattmeter shows zero reading.
Synchronous machine working as generator
Now, decrease the field current of the dc motor (i.e. make synchronous machine in
generator mode) so that wattmeter reads a specified output (say, 600W per phase) (Please
take care of the multiplication factor of wattmeter). Keeping the wattmeter reading at
600W/phase, increase the synchronous machine field current If by adjusting the potential
divider so that ammeter reading Ia shows rated current. Note If and Ia. Now decrease If
gradually and note down Ia each time. We can see that Ia decreases from rated current,
reaches minimum, again increases and reaches rated value. Wattmeter reading should be
same while taking each reading (Otherwise, vary motor field rheostat). Ia Vs If curve (V
curve) is plotted for the generator for a constant power output of 1800W. Now increase If
so that armature current is minimum. Increase the motor field current, so that wattmeter
reads zero.
Synchronous machine working as motor
For getting V curves of synchronous motor on no load, switch off the supply to
the dc machine and connect the dc machine to the lamp load by using DPDT switch when
the wattmeter reading is zero and ammeter reading minimum. Now the synchronous
machine will act as motor and dc machine as generator. Note down the wattmeter reading
which shows the no-load losses of both dc machine and synchronous motor. Increase the
field current If of synchronous machine till the armature current Ia becomes rated value
and note down the readings. Decrease If in steps and note down Ia each time. Plot the Vcurves (Ia Vs If) for synchronous motor on no-load. Now, for getting V curves of
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synchronous motor drawing constant power input (say, 800W/phase), vary the lamp/water
load and make the wattmeter reading equal to 800W/phase. Increase the field current of
the synchronous machine to rated current and the above procedure is repeated. Plot the V
curves of the synchronous motor drawing constant power input.
GENERATING MODE
Sl.
No.
OUTPUT = ________W
Ia
If
cosф
(lag/lead)
MOTOR MODE
ON NO LOAD
(INPUT = _______W)
Ia
If
cosф
(lag/lead)
MOTOR MODE
INPUT = _________W
Ia
If
cosф
(lag/lead)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Note: a) Over-excited synchronous generator operates at lagging p.f. and under-excited
synchronous generator at leading p.f. and b) over-excited synchronous motor operates at
leading p.f. and under-excited synchronous motor at lagging p.f.
MODEL GRAPH
Ia, p.f.
V-CURVES & INVERTED V-CURVES
SYNC.MOTOR
I , p.f.
SYNC.GENERATOR
p.f. Vs If
a
p.f. Vs If
output=____W
Ia Vs If
Input=____W
Ia Vs If
No-load
lead
lag
If in amps
SAMPLE CALCULATION (Set No. ____)
Generator delivering an output of ______W.
lag
lead
If in amps
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
94
V = 400V, Ia = ____A, If = ____A, Iamin = _____A
I
cos Φ = a min = _______ lag (/lead) (Note : under-excited → lead & over-excited → lag)
Ia
Motor drawing an input of _______W.
V = 400V, Ia = _____A, If = _____A, Iamin = _____A
I
cos Φ = a min = _______ lead (/lag) (Note : under-excited → lag & over-excited → lead)
Ia
RESULT
3 phase synchronous machine was synchronized to the supply mains using bright
lamp method and V-curves and inverted V-curves were plotted while the synchronous
machine is working as generator and motor.
================================================================
Do you know?
1. What is infinite busbar?
A supply system with large number of synchronous generators in parallel and
operating at constant voltage and frequency is called infinite busbar. (It has zero
synchronous impedance Zs and infinite rotational inertia).
2. A synchronous motor works at a p.f. 0.8. A slight decrease of field current
worsens the power factor. Was the p.f. leading or lagging?
Lagging.
3. What are the conditions for paralleling an alternator with the infinite bus?
Before the alternator can be connected to the infinite bus, the incoming alternator
and the infinite bus must have the same i) voltage ii) frequency iii) phase sequence
and iv) phase.
4. What is the possible effect of wrong synchronization?
Wrong synchronization means connecting two sources at an instant when the
phase difference between two voltages is not zero. This may result into sudden flow of
power, excessive current circulation, and mechanical shocks resulting due to heavy
torque. This is highly undesirable.
5. What is the effect of increase in excitation of a synchronous motor?
When the excitation of a synchronous motor is increased, first power factor
improves until it becomes unity and with the further increase in excitation the power
factor becomes leading one and decreases.
6. A synchronous motor develops some mechanical power, even if the field is
unexcited. Is it cylindrical or salient pole machine?
Salient pole
7. What is meant by synchronous condenser?
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Synchronous condenser is overexcited synchronous motor on no load. It is used
for power factor improvement since it draws leading current.
8. What is meant by hunting of synchronous motor?
The oscillation of synchronous motor rotor about its equilibrium position is called
the hunting. The causes of hunting are i) sudden change in load, ii) fault in supply
system and iii) sudden change in field current. It can be reduced by using damper
bars, using flywheel or designing the synchronous machine with suitable
synchronizing-power coefficient.
9. How is the speed of a synchronous motor varied?
By varying the frequency of supply voltage.
10. Why a 3 phase synchronous motor will always run at synchronous speed?
The synchronous motor always runs at synchronous speed because there is an interlocking action between stator and rotor fields of this motor.
11. What is meant by synchronizing?
The process of connecting an alternator in parallel with another alternator or with
infinite bus bar is called synchronizing.
12. Explain synchronization by bright-lamp method?
In this method, three lamps are connected as shown in figure. By running the
alternator at synchronous speed and by adjusting the field excitation, the armature
voltage is increased near to rated value. If all the three lamps become bright and dim
in sequence, the phase sequences of both the incoming generator and bus-bar are the
same. If they become bright and dim simultaneously, the phase sequence of the
incoming alternator should be reversed (by interchanging any two leads of incoming
alternator).
L1
R1
Y1
B1
V1
L2
L3
R2
V2
Y2
VR1
L1
VR1 VR2
L1
VR2
VB2
VY1
B2
VB1
L3
L2
VY2
VB1
VB2
VY1
L3
L2
VY2
Now, the field excitation is adjusted such that voltages of the incoming alternator
and the bus-bar are equal. The speed of the prime-mover of the incoming machine is
further adjusted slowly until the lamps flicker at a very low rate. The paralleling
switch is closed at the instant when one lamp is dark and other two are maximum
bright. The incoming alternator thus gets connected in parallel with the bus-bar.
13. NOTE: 44MW, 10kV, 60Hz, 50pole, 144rpm synchronous motor is used to drive the
Queen Elizabeth II passenger ship (Second largest luxury ship after Queen Mary II).
A solid state V/f drive circuit provides speed control though frequency adjustment.
After 40 years of service, in 2009, this ship will become a seven star hotel in Dubai.
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================================================================
Experiment No. 10
SPEED CONTROL OF INDUCTION MOTOR BY
VARIABLE FREQUENCY METHOD
================================================================
AIM: To control the speed of the 3 phase induction motor by changing the supply
frequency and to plot the speed Vs frequency curve.
APPARATUS:
S.No. Name of the apparatus
1.
Voltmeter
2.
Frequency meter
3.
Ammeter
4.
5.
Tachometer
6.
Rheostat
7.
Type
MI
Digital
MC
MI
Range
(0-500V)
(0-60Hz)
(0-3A)
(0-10A)
Wire Wound
145Ω, 2.5A
272Ω 1.7A
Quantity
1
1
1
1
1
2
1
PRINCIPLE:
The synchronous speed Ns of an induction motor is related to supply frequency f
and number of poles P by the equation, N s =
120 f
. Rotor speed is given by
P
N = (1 − s ) N s where s is the slip. The basic methods of speed control of an induction
motor are a) by changing the number of poles and b) by varying the line frequency.
The emf per phase of an induction motor is given by E = 4.44 K w f ΦT volts. The induced
emf E is nearly equal to the applied voltage V (neglecting drop in stator impedance).
Thus, we can write
V
= 4.44 K wΦT .
f
When the frequency is reduced, the applied voltage also must be reduced
proportionally so as to maintain constant flux, otherwise the core will get saturated
resulting in excessive core loss and magnetizing current. The maximum torque also
remains constant under this condition. However, the voltage is not varied proportionately
in the low frequency range to account for the voltage drop in the winding resistance. This
type of control (constant V/f) is used for speed control below base frequency (line
frequency of 50Hz).
As the voltage increases above rated value, when the input frequency goes above
base frequency, only constant (rated) voltage with variable frequency (frequency control)
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
97
is used for speed control. Under this condition, both flux and maximum torque decrease
as the frequency is increased.
PROCEDURE:
Make the connections as shown in diagram.
Precautions:
TPST in open position
v)
vi)
DPST1 and DPST2 in open position
Motor field rheostat in minimum position
vii)
viii) Potential divider in minimum voltage position
ix)
Autotransformer at minimum voltage position
Keep the belt on the brake drum of induction motor in loose position
x)
(induction motor on no load).
Switch on the DC supply to the DC motor by closing the switch DPST1. Start the DC
shunt motor using 3-point starter. Increase the resistance of dc motor field rheostat and
drive the alternator at rated speed (1500rpm). Now, dc supply is given to the alternator
field winding and adjust the potential divider so that the generated voltage is rated value
(400V). Close the TPST switch.
Increase the autotransformer. Induction motor starts running on no load. Apply rated
voltage by adjusting autotransformer. Note down the frequency, voltage and speed of the
induction motor. Now, increase the frequency keeping the voltage constant (=400V).
Again, note down frequency, voltage and speed each time. Repeat the procedure till
frequency reaches 54Hz.
Now, decrease the frequency till it becomes 50Hz. Decrease the voltage and
frequency in proportion (
V 400
=
= 8 ) and note down the frequency, voltage and speed
50
f
of the induction motor each time. This procedure is continued till frequency decreases to
44Hz.
Repeat the above procedure for another load (say IL = 4A). Switch off the supply after
bringing the motor to no-load.
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98
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TABULATION
Induction motor on no load
Line voltage 352
368 384
400
in volts
Frequency
44
46
48
50
in Hz
Speed of IM
in rpm
400
400
52
54
Induction motor on load IL = 4A
Line voltage 352
368 384
400 400
in volts
Frequency
44
46
48
50
52
in Hz
Speed of IM
in rpm
400
54
RESULT
Speed of the 3 phase induction motor was controlled by variable frequency
method and speed Vs frequency characteristics were plotted.
================================================================
Do you know?
1. What does happen to the induction motor if supply frequency is reduced
keeping the supply voltage constant ?
If supply frequency f is decreased keeping supply voltage constant, speed will
decrease but flux will increase (Φ α V/f) hysteresis loss increases but eddy current loss
remains constant. Since core loss increases, it causes overheating and decreased
efficiency. Magnetizing current drawn from supply will be large.
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
100
MODEL QUESTIONS
INSTRUCTIONS
Phase 1:- Copy the question to the answer sheet. Draw a neat CONNECTION DIAGRAM
and TABULAR COLUMN with PEN, write the RELEVANT EQUATIONS, and draw the
expected shape of graph (if any) and get approval from the examiner (No need to write
procedure/principle).
Phase 2:- Make the connections and GET VERIFIED by the examiner before switching
on. Conduct the experiment. Show one set of readings to the examiner before switching
off.
Phase 3:- Complete the calculations, draw the graphs (if any), write down the RESULTs
and submit the answer sheet.
Qn.A1:- By conducting suitable test on the given slip ring induction motor (______KW,
_______V, _____A, ∆ connected, _______rpm) obtain the equivalent circuit & hence
predetermine efficiency & torque developed at a speed of ________rpm. Assume stator
resistance/phase (ac) R1 = _______Ω.
Qn.A2:- By conducting suitable test on the given slip ring induction motor (______KW,
_______V, _____A, ∆ connected, _______rpm) draw the circle diagram and hence obtain
the efficiency & torque developed at an output of _____kW. Assume stator
resistance/phase (ac) R1 = _______Ω.
Qn.A3:- By conducting suitable test on the given slip ring induction motor (______KW,
_______V, _____A, ∆ connected, _______rpm) obtain the equivalent circuit i) during no
load test ii) during blocked rotor test & iii) at a speed of ________rpm.
Assume stator resistance/phase (ac) R1 = _____Ω.
Qn.A4:- By conducting suitable test on the given slip ring induction motor (______KW,
_______V, _____A, ∆ connected, _______rpm) draw the circle diagram & hence obtain
the following i) maximum torque ii) maximum power output iii) maximum power input & iv)
starting torque. Assume stator resistance/phase (ac) R1 = _____Ω.
Qn.A5:- By conducting suitable test on the given slip ring induction motor (______KW,
_______V, _____A, ∆ connected, _______rpm), draw the circle diagram and hence obtain
the line current, power factor, slip, torque and efficiency at an output of _____kW. Assume
stator resistance/phase (ac) R1 = _____Ω.
Qn.A6:- By conducting suitable test on the given slip ring induction motor (______KW,
_______V, _____A, ∆ connected, _______rpm) draw the circle diagram & hence obtain
the slip Vs output characteristics. Assume stator resistance/phase (ac) R1 = _____Ω.
Qn.B1:- By conducting suitable test on the given 3phase squirrel cage induction motor
(______KW, _______V, _____A, ∆ connected, _______rpm), determine the torque &
efficiency at a slip of 3%.
Qn.B2:- By conducting suitable test on the given 3phase squirrel cage induction motor
(______KW, _______V, _____A, ∆ connected, _______rpm), determine the value of
capacitance required to improve the power factor to unity while the induction motor is
running at a slip of 3%.
Qn.B3:- By conducting suitable test on the given 3phase squirrel cage induction motor
(______KW, _______V, _____A, ∆ connected, _______rpm), determine the torque &
efficiency at 3/4th full load.
Qn.B4:- By conducting suitable test on the given 3phase squirrel cage induction motor
(______KW, _______V, _____A, ∆ connected, _______rpm), determine the torque, slip &
efficiency at 0.5 p.f.
Qn.B5:- By conducting suitable test on the given 3phase squirrel cage induction motor
(______KW, _______V, _____A, ∆ connected, _______rpm), determine i) the slip at no
load , ii) slip at 0.5 p.f. & iii) slip at ½ full load.
S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur
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Qn.E1:- By conducting suitable test on the given 3 phase pole changing induction motor
(______KW, _______V, _____A, _____ connected, _______rpm) while running at low
speed, obtain the equivalent circuit & hence find the efficiency at a slip of 5%. Assume
stator resistance/phase (ac) R1 = _____Ω.
Qn.E2:- By conducting suitable test on the given 3 phase pole changing induction motor
(______KW, _______V, _____A, ____ connected, _______rpm) while running at high
speed, obtain the equivalent circuit & hence find the efficiency at a slip of 5%. Assume
stator resistance/phase (ac) R1 = _____Ω.
Qn.E3:- By conducting suitable test on the given 3 phase pole changing induction motor
(______KW, _______V, _____A, _____ connected, _______rpm) while running at low
speed, draw the circle diagram & hence find the torque, slip & efficiency when the power
factor is maximum. Assume stator resistance/phase (ac) R1 = 8Ω.
Qn.C1:- By conducting suitable test on the given 3 phase alternator (______KW,
_______V, _____A, ∆ connected, _______rpm), predetermine the power factor at which
the full load regulation is zero. Use synchronous impedance method. Neglect Ra.
Qn.C2:- By conducting suitable test on the given 3 phase alternator (______KW,
_______V, _____A, ∆ connected, _______rpm), predetermine the percentage regulation
when the given 3 phase alternator is delivering full load at a) unity p.f. & b) zero p.f. (lag &
lead). Use pessimistic method. Draw the relevant phasor diagrams. Neglect Ra.
Qn.C3:- By conducting suitable test on the given 3 phase alternator (______KW,
_______V, _____A, ∆ connected, _______rpm), predetermine the full load regulation at
unity p.f. by a) pessimistic method and b) optimistic method. Draw the relevant phasor
diagrams. Neglect Ra.
Qn.C4:- By conducting suitable test on the given 3 phase alternator (______KW,
_______V, _____A, Y connected, _______rpm), predetermine the full load regulation at
0.5 p.f. lag by a) pessimistic method and b) optimistic method. Compare the results. Draw
the relevant phasor diagrams. Neglect Ra.
Qn.D1:- By conducting suitable test on the given 3 phase salient pole type synchronous
machine (______KW, _______V, _____A, ∆ connected, _______rpm), predetermine the
full load regulation at zero p.f. (lag & lead). Draw the relevant phasor diagrams. Neglect Ra.
Qn.D2:- By conducting suitable test on the given 3 phase salient pole type synchronous
machine (______KW, _______V, _____A, ∆ connected, _______rpm), predetermine the
full load regulation at zero p.f. lag & lead. Draw the relevant phasor diagrams. Neglect Ra.
Qn.D3:- By conducting suitable test on the given 3 phase salient pole type synchronous
machine (______KW, _______V, _____A, ∆ connected, _______rpm), predetermine the
excitation power & reluctance power at a load (torque) angle of 45°. Assume excitation emf
= 120% of rated voltage. Neglect Ra.
Qn.D4:- By conducting suitable test on the given 3 phase salient pole type synchronous
machine ((______KVA, _______V, _____A, ∆ connected, _______rpm), predetermine the
excitation power & reluctance power at a load (torque) angle of 60°. Assume excitation emf
= 80% of rated voltage. Neglect Ra.
Qn.E4:- By conducting suitable test on the given 3 phase pole changing induction motor
(______KVA, _______V, _____A, _____ connected, _______rpm) while running at low
speed, draw the circle diagram & hence find the efficiency when the input is maximum.
Assume stator resistance/phase (ac) R1 = _____Ω.
Qn.E5:- By conducting suitable test on the given 3 phase pole changing induction motor
(______KW, _______V, _____A, ____ connected, _______rpm) while running at low
speed, draw the circle diagram & hence find the efficiency when the output is maximum.
Assume stator resistance/phase (ac) R1 = _____Ω.
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Qn.E6:- By conducting suitable test on the given 3 phase pole changing induction motor
(______KW, _______V, _____A, ____ connected, _______rpm) while running at low
speed, draw the circle diagram & hence find the slip when the torque is maximum. Assume
stator resistance/phase (ac) R1 = ______Ω.
Qn.H1:- By conducting suitable test on the given 3 phase alternator (______KVA,
_______V, _____A, Y connected, _______rpm), predetermine the regulation at full load &
unity p.f. by POTIER method. Neglect Ra.
Qn.H2:- By conducting suitable test on the given 3 phase alternator (______KVA,
_______V, _____A, Y connected, _______rpm), predetermine the regulation at full load &
zero p.f. lag by POTIER method. Neglect Ra.
Qn.G1:- By conducting suitable test on the given 1 phase induction motor (______KW/HP,
_______V, _____A, _______rpm) obtain the equivalent circuit during running condition &
hence find the efficiency at a slip of 4%. Assume stator resistance (ac) = ____Ω.
Qn.G2:- By conducting suitable test on the given 1 phase induction motor (______KW/HP,
_______V, _____A, _______rpm) obtain the equivalent circuit during running condition &
hence find the efficiency at a slip of 6%. Assume stator resistance (ac) = _____Ω.
Qn.J1:- By conducting suitable test on the given 3 phase synchronous machine (______KVA,
_______V, _____A, Y connected, _______rpm), obtain the V curves & inverted V curves
(for Ia ≤ 6A) while working as a motor drawing a power input of _______W.
Qn.J2:- By conducting suitable test on the given 3 phase synchronous machine (______KVA,
_______V, _____A, Y connected, _______rpm), obtain the V curves & inverted V curves
(for Ia ≤ 6A) while working as a generator delivering a power output of ________W.
Qn.F1:- By conducting suitable test on the given 3 phase induction machine (______KW,
_______V, _____A, ∆ connected, _______rpm) coupled with a DC machine (______kW,
____V, _____A), determine the efficiency when the induction machine is working as motor
drawing an input of _____W.
Qn.F2:- By conducting suitable test on the given 3 phase induction machine (______KW,
_______V, _____A, ∆ connected, _______rpm) coupled with a DC machine (______kW,
____V, _____A), determine the efficiency when the induction machine is working as
generator delivering an output of _______W.
Qn.F4:- By conducting suitable test on the given 3 phase induction machine (______KW,
_______V, _____A, ∆ connected, _______rpm) coupled with a DC machine (______kW,
______V, ______A), determine the efficiency when the dc machine is working as motor
drawing an input current of _____A.
Qn.F3:- By conducting suitable test on the given 3 phase induction machine (______KW,
_______V, _____A, ∆ connected, _______rpm) coupled with a DC machine (_____kW,
_____A, ______V), determine the efficiency when the dc machine is working as generator
delivering an output current of _____A.
Qn.G1:- By conducting suitable test on the given 1 phase induction motor (______KW/HP,
_______V, _____A, _______rpm) obtain a) the equivalent circuit during running condition
at slip of 5% b) equivalent circuit during no-load test c) the equivalent circuit during blocked
rotor test. Assume stator resistance (ac) = ____Ω.
Qn.G2:- By conducting suitable test on the given 1 phase induction motor (______KW/HP,
_______V, _____A, _______rpm) obtain the efficiency at a slip of 4%. Assume stator
resistance (ac) = ____Ω.