ac machines lab manual

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


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 FACULTY-

IN-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)


• 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. Name of the apparatus Type Range Quantity 1. Voltmeter MI (0-500V) 1 2. Ammeter MI (0-10A) 1 3. Wattmeter Dynamometer

type 500V, 10A, UPF 2

4. Tachometer 1 5. TPDT switch 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 5 PROCEDURE: Make the connections as shown diagram.

Precautions: i) Keep TPST switch open

ii) Keep TPDT in position 1 (Star connection)

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, I =_______ A, W1 = _______W, W2 = ________ W,

N =_______ rpm, S1 = ________Kg, S2 =________Kg

Radius of brake drum R = 0.135 m

Synchronous speed Ns = 1500rpm

Input Power W= W1 + W2 = _________ watts

Power factor, cosΦ1= 1 1 2

1 2

3 ( )cos(tan )( )

W WW W

− × −+

=__________


Percentage slip, s = 100s

s

N NN−

= × =_________%

Torque 1 2( )T R S S g= × − × = _________ N-m

Output power 260NTπ

= = _________W

Efficiency, 100outputinput

η = × = _________%

Additional kVAR required to improve the power factor to 0.95 (cosΦ2) =

= 31 2(tan tan ) 10W −Φ − Φ × =__________

Value of capacitance required to improve the power factor = 2

1000kVARVω× =

= 2

10002kVAR

f Vπ×

× × × =________μF

MODEL GRAPHS


Speed

Torque

Speed Vs Torque Characteristic

0

RESULT: i) Brake test conducted on 3 phase squirrel cage induction motor ii) Performance characteristics plotted iii) Additional kVAR required and the value of capacitance to improve the power

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

current neglected). Starting current, 1 1

SCo o

VIR jX

=+

.

V

Xo1Ro1ISC

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? i) Direct On Line (DOL) starter ii) Stator impedance (resistance/reactance) starter


iii) Autotransformer starter iv) Star-Delta Starter v) 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 ( 3

V= ), 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

induction motor. 120s

fNP

= . Synchronous speed will be slightly greater than the

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.

10. What is the speed of the stator and rotor magnetic fields with respect to stator?

Both are rotating at synchronous speed Ns with respect to stator.


11. Compare between squirrel cage and slip ring induction motors.

SCIM SRIM 1. Its rotor consists of copper/aluminium bars permanently short circuited at both ends

1. Its rotor consists of 3 phase windings which is then connected to external resistance through slip rings and brushes

2. Starting torque is poor 2. Better starting torque can be achieved by inserting external resistance in the rotor circuit.

3. Separate starting methods are needed in order to reduce the staring current

3. Rotor resistance starting can be used

4. Its rotor can adjust to any number of stator poles. So different speeds can be obtained by different arrangements of stator winding.

4. Rotor and stator are wound for same number of poles

5. It has fewer components and hence less labor since it has no rotor winding. So it is cheaper

5. Slip rings, brushes, starting resistance etc. increases the cost

6. Better efficiency 6. Lower efficiency 7. Better cooling (larger space to provide fan blades on rotor)

7. Cooling not efficient

8. Less maintenance 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?


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.

V1

X1R1I1

Xm

X2'

R2'/s

I2'

Im

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

factor at starting. ( 2 22 2

22 2

. . R Rp fXR X

= ≈+

, R2<<X2. Hence power factor is low)

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


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 de-energised to disconnect the supply to the motor.

Motor

StartStop

Remote Stop

C a

b

M M MA

S1

S2

S3

D OLC

Fuse

L1 L2 L3

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 13 ================================================================

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

Type Range Quantity

1. Voltmeter MI 0-500V 1 2. MI 0-150V 1 3. MC 0-30V 1 4. Ammeter MI 0-5A 1 5. MI 0-10A 1 6. MC 0-10A 1 7. Wattmeter Dynamometer 500/250/125V,

5/10A,UPF 2

8. Dynamometer 150V,10A,LPF 1 9. Rheostat Wire wound 9Ω,8.5A 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,

22 2 1 22

1 1 2

' sin''( ) ( ')

V VIX XRR X X

s

= = Φ+

+ + + where

1 2

2 221 1 2

'sin'( ) ( ')

X XRR R Rs

+Φ =

+ + +

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 1 2 '

VX X+

. By changing the load RL (where 2(1 )'L

sR Rs−

= ) and Φ,

the value of the current I2’ changes. The locus of the current, however, lies on a circle

(Figure 1).


Figure 1 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.


B) BLOCKED ROTOR TEST Make the connections as shown in figure.


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

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 17 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

10A

R

A

0-10A MCS1

S3 S2

V

0-30V MC

+

-

9Ω , 8.5A

+

-+

-

28V DC

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 = 32

x Delta resistance)

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

(Y-axis) V by an angle Φo where 1cos ( )3

oco

oc oc

WV I

−Φ = .

3. From the current Isc find out the ISN (short circuit current corresponding to the

normal voltage) through the formula ( )ratedSN sc

sc

VI IV

= , draw the OB vector with the

magnitude of ISN from the origin by the same current scale, which lags the voltage

(Y-axis) V by an angle ΦSC where 1cos ( )3

scSC

sc sc

WV I

−Φ = .

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, 2

2 22

2 1

' ''

I RRotor Copper lossStator Copper loss I R

= = 2

1

'RR

where R1= stator

resistance per phase and R2= rotor resistance per phase.


Or, 2

1

'

o

RBG Rotor Copper lossBD R 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.

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 ratedV current scale× × . TABULATION

NO LOAD TEST BLOCKED ROTOR TEST

Voc Ioc W1 W2 Wsc Vsc Isc W1 W2 Wsc

Stator Resistance Measurement

S.No. V (volts) I (amps) Rdc=V/I Ω

1. 2. 3. 4.

Rdc CIRCLE DIAGRAM Voc = 400V , Ioc = ___ A , Woc = _____ W

Vsc = _____ V, Isc = 7.8A, Wsc = _____ W

Per phase values are

_____ ____3

oco oc o

IV V V I A= = = =

_____ ____3sc

s sc sIV V V I A= = = =

Rdc = _____ Ω

13 1.22 dcR R= × × = ______ Ω

1 23sc

os

WRI

= = _______ Ω

'2 1 1oR R R= − = _______ Ω


2

1

'

o

RBGBD R

= = ______

____BG BD= ×

Selection of current and power scale Current scale = 1cm = ______ A

Ioc = _______A (= ______cm)

( )ratedSN sc

VI IVsc

= = _______A(= _______cm)

1cos ( )3

oco

oc oc

WV I

−Φ = = _______ ˚

1cos ( )3

scSC

sc sc

WV I

−Φ = = _______˚

Power Scale = 3 ratedV 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 = PTOP

= ______ lag

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 602 sNπ

× N-m

= _____ x ______ x 602 750π ×

= _______N-m

Motor input at full load = PT x power scale = _____ x _______ = ______W

Efficiency at full load = 100%PQPT

× = ________%

Slip at full load s = 100%QRPR

× = _________%

Speed at full load = (1 ) ss N− × = ________ rpm

Starting torque = BG x power scale x 602 sNπ

N-m

= _____ x ______ x 602 750π ×

=______N-m


Maximum torque = I I’ = ______ x _____ x 602 750π ×

=______N-m

Maximum output = HH’ = ______ x ______ = ________W

Maximum input = JJ’ = ______ x ______ = ________W

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 21 TABULATION FROM CIRCLE DIAGRAM

MODEL GRAPH – PERFORMANCE CHARACTERISTICS FROM CIRCLE DIAGRAM

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 22 EQUIVALENT CIRCUIT PARAMETERS Voc = ______ V , Ioc = ____A , Woc = _____W

Vsc = ______V, Isc = 7.8A, Wsc = ______W

_____ ____3

oco oc o

IV V V I A= = = =

_____ ____3sc


0cos3

oc

o o

WV I

Φ = = _______

0sinΦ = _______

coso

co o

VRI

=× Φ

=_________Ω

sino

mo o

VXI

=× Φ

= ________Ω

1s

os

VZI

= = _________Ω

1 23sc

os

WRI

= = _______Ω

2 21 1 1o o oX Z R= − = ________Ω

13 *1.2*2 dcR R= =1.5 x 1.2 x ____ = ______Ω

'2 1 1oR R R= − =_______Ω

' 12 2 2

oXX X= = = ______Ω

I1 I2'

IoImIc

R1 X1 R2' X2'

400V

EXACT EQUIVALENT CIRCUIT

Rc Xm 21' '( )L

sR Rs−

=

I2'

IoImIc

Ro1=R1+R2' Xo1=X1+X2'

400V

APPROXIMATE EQUIVALENT CIRCUIT

I1

XmRc 2

1' '( )LsR R

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 = 100%s

s

N NsN−

= × = 750 710 100 5.33%750−

× =

21' ' ( )L

sR Rs−

= × =_______Ω

o o oI I−

= ∠−Φ °=________per phase

From approximate equivalent circuit,

'2

1 1

0( ')o L o

VIR R jX

∠ °=

+ +=________A per phase

1 2 'oI I I= + = 1 1I ∠−Φ ° =_______A

Line current IL = 13 I× =_______A

Power factor = 1cos( )−Φ ° =______

Output = 223 ' 'LI R× × =________W

Torque = 2( )60

OutputNπ =_______N-m

Input = 13 cosLVI Φ =________W

100%OutputEfficiencyInput

= × = ______ %

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


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.

Current during blocked rotor test, 1 1

scsc

o o

VIR jX

=+

, and Xo1 » Ro1 . Hence, the

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˚

electrical. One revolution is equal to 360˚ mechanical. 2elec mechPθ θ= ×

10. What are the methods of reducing the space harmonics? i) Distributing the winding in slots ii) Using the short-pitched winding iii) Skewing the slots iv) Using fractional-slot winding


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. 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 (0-10A) 1 5. MC (0-10A) 1 6. Rheostat Wire Wound 9Ω 8.5A 1 7. Tachometer 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

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 26 armature terminal voltage will be minimum. The ratio of minimum armature terminal

voltage per phase to maximum armature current per phase gives Zsq.

IXq

IXd

V

IRaI

δφψ

Eo

I

V

Eo

IXd

IXq

IRa

ψ

δφ

V IRa

IXq

IXd

Eo

ψ =δ

p. f. lag

p. f. lead

unity p. f.

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


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.


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.

10AA

0-10A MC

0-30V MC28V DC V

+

-

V

A9Ω,8.5A

+

-

+ -

10A

U

N

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 sdZ sqZ dX qX 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, 1.2a dcR R= × =1.2 x _____ = ____Ω

Vmax = _____V, Vmin = _____V, Imax = ____A, Imin = _____A


max

minsd

VZI

= =______Ω

min

maxsq

VZI

= =______Ω

2 2d sd aX Z R= − =_______Ω

2 2q sq aX Z R= − =________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

1 sintan

cosq

a

V IXV IR

− Φ +⎛ ⎞Ψ = ⎜ ⎟Φ +⎝ ⎠

= 1 231 0.6 11.5tan _____

231 0.8 11.5q

a

XR

− × + ×⎛ ⎞= °⎜ ⎟× + ×⎝ ⎠

δ = Ψ −Φ =_____-36.87=_____˚

cos cos sinf a dE V IR IXδ= + Ψ + Ψ =_______V

% regulation = 100fE VV−

× = ______%

a) % regulation at full load V = 231V, I = 11.5A

power factor Ф Ψ δ Ef Regulation

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

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


1 sintan

cosq

a

V IXV IR

− Φ +⎛ ⎞Ψ = ⎜ ⎟Φ +⎝ ⎠

= 1 231 0.6 11.5tan ____

231 0.8 11.5q

a

XR

− − × + ×⎛ ⎞= °⎜ ⎟× + ×⎝ ⎠

δ = Ψ −Φ = _____+36.87= _____˚

cos cos sinf a dE V IR IXδ= + Ψ + Ψ =________V


× = ________%

b) % regulation at Half full load V = 231V, I = 5.75A

power factor Ф Ψ δ Ef Regulation

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

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 31 b) To draw power angle characteristics

Assume an induced emf of 120% of terminal voltage

231 120 277.2100fE V×

= =

60oδ =

P1 = Power due to field excitation (Excitation power)

P2 = Reluctance power (power due to saliency)

13 o

d

E VSinPX

δ= = 3 277.2 231 0.866 _______

d

WX

× × ×=

2

2

3 ( ) 22d q

d q

V X X SinP

X Xδ−

= =___________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 P1 watts P2 watts P watts

-180

-150

-135

-120

-90

-60

-45

-30

0

30

45

60

90

120

135

150

180

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 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

unexcited so long as the load does not exceed 23 ( )2

d q

d q

V X XX X

−(Note: this value is

around 30% of the rated power). The power, 2

2

3 ( ) 22d q

d q

V X X SinP

X Xδ−

= is called

reluctance power.


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 ?

d

q

XX

= 1.6 to 2

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 Smooth cylindrical It consists of projected poles, laminated, made of cast iron or cast steel

It is built from solid steel forging (usually chromium – nickel steel)

Poles carry concentrated field windings Poles consists of radial slots in which field windings are placed

Air gap is not uniform Air gap is nearly uniform It has large diameter and short axial length

It is of small diameter and of very long axial length.

It is used for low and medium speed machines (water wheel drive or diesel engine drive)

It is used for high speeds (Steam driven)

Noiseless operation Less windage loss Better in dynamic balancing


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 = 100fE VV−

×

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.


LOADEf

Er

Xar Xl Ra

V

Ia

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

5A

M

30A

30A

220V DC

272Ω,1.7AZ

ZZ

A

AA

0-300V MI

+

-

A

C B

L Z A

X XX

N

5A

5A

145Ω, 2.5A

145Ω, 2.5A

220V DC

Machine details : DC motor11.7HP, 220V, 48A,1500rpm

Alternator 7.5kVA, 400V, 10.9A,50Hz, 1500rpmDPST2

DPST1 TPST

+

-

A

V

A

0-15A MI

0-2A MC

+ -

OC & SC TEST

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 38 STATOR RESISTANCE MEASUREMENT Make the connections as shown in figure.

28V DC

9Ω , 8.5A 0-10A MC

0-30V MC

10A

10A

+

-

+

-+

-

A

N

A

V

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 ZPFC

Isc (A) If (A) Ia (A) If (A)


1. 2. 3. 4.

Rdc CALCULATION EMF METHOD

Rated voltage/phase V = 230V

Short circuit current corresponding to rated voltage from SCC, Isc = ______A

Synchronous impedance, ssc

VZI

= =_______Ω


Armature resistance, 1.2a dcR R= × = _______Ω

Synchronous reactance, 2 2s s aX Z R= − = ________Ω

Ef Isc

230VOCC

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. EMF METHOD

p.f. Full load Ia = 10.9A

Ef Regulation 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

0 230 0V V−

= ∠ ° = ∠ °

10.9I I−

− −= ∠ Φ° = ∠ Φ° for lag ( 10.9I I−

+ += ∠ Φ° = ∠ Φ° for lead)


0 ( )f a s fE V I R jX E δ−

−= ∠ + ∠ Φ× + = ∠ ° =________V

(OR 2 2( cos ) ( )f a a a sE V I R Vsin I X= Φ + + Φ + =________V)


× =___________%

Ef

IRaI Vδ

IXs

Φδ

IRaV

I

IXsEf

Unity power factorLeading power factor

PHASOR DIAGRAMS - EMF METHOD

δ

I

ΦIRa

IXs

Ef

Lagging power factor

V

SAMPLE CALCULATION

MMF METHOD

SAMPLE CALCULATION


0 230 0V V−

= ∠ ° = ∠ °V

cosΦ = ____ lag

10.9I I−

− −= ∠ Φ° = ∠ Φ° for lag ( 10.9I I−

+ += ∠ Φ° = ∠ Φ° for lead)

' 0 'aE V I R E σ−= ∠ °+ ∠ Φ°× = ∠− °=________V

Refer OCC and find Ifr corresponding to E’.


( 90 )fr frI I σ= ∠ − + ° =_________A Ifa is the field current required to circulate rated current on short circuit (from SC test)

( 180 )fr faI I −= ∠ Φ + ° for lag ( ( 180 )fr faI I += ∠ Φ + ° for lead)

f fr faI I I= + =_________A= (90 )fI δ∠ + ° Hence, If = ______A, δ = ______˚ Refer OCC and find Ef corresponding to If.

f fE E δ= ∠ °=________V


× =__________%

PHASOR DIAGRAMS – MMF METHOD

δ

σ

I

ΦIRa

IXs

Ef

E'

Ifa

If


V

Ifr

Ifa

IfrIf

Ef

E'IRaI V

δ

IXs

Unity power factor

Φ σδ

If

Ifr

Ifa

IRaV

I

IXsEf

Leading power factor

E'


TABULATION – MMF method Power

factor I 'E frI faI fI fE % regulation

1

2

3

4

5

6

7

8

9

10

11

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

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.

AD B

C

air gap line

OCC

ZPFC

O F

voltage

Field Current

230V

(If fromSC test)

E(If fromZPF 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.

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 43 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.

δ σ

Ia

ΦIaRa

IaXal

IaXar

Ef

ErIfr

Ifar

If


Φσ δIf

Ifr

Ifar

IaRaV

Ia

IaXal

IaXarEf

Er

Leading power factor

Ifar

IfrIf

Ef

Er

IaXal

IaRaIaV

δ σ

IaXar

Unity power factor

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 44 SAMPLE CALCULATION


alDE I X= × =_____ V(from Potier triangle)

Xal = ______Ω

0 230 0V V−

= ∠ ° = ∠ °

10.9I I−

− −= ∠ Φ° = ∠ Φ° for lag ( 10.9I I−

+ += ∠ Φ° = ∠ Φ° for lead)

0 ( )r a alE V I R jX−

−= ∠ + ∠ Φ× + =_________V= rE σ∠ ° Refer OCC and find Ifr corresponding to Er.

( 90)f r frI I σ−

= ∠ + °=_________A

( 180)f a faI I−

−= ∠ Φ + °=________A for lag ( ( 180)f a faI I−

+= ∠ Φ + ° for lead)

fr fafI I I− − −

= + =__________A= (90 )fI δ∠ + ° Hence, If = ____A and δ = _____º Refer OCC and find Ef corresponding to If.

f fE E δ−

= ∠ °=__________V


× =_________%

TABULATION – Potier Method

Power factor I

−

rE−

f rI−

f aI−

fI−

fE−

% regulation

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

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


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.

Ef

Xs Ra

S.C.V = 0

Ia

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?

SCR= Field current required to produce rated voltage on open circuit 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


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 third-harmonic 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

Y2

B2

V1 V2

L

L

L

VR1

VR2

VY1

VY2

VB1

VB2

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.


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:


Type Range Quantity

1. Voltmeter MI 0-500V 1 2. MI 0-150V 1 3 MC 0-30V 1 4. Ammeter MI 0-5A 1 5. MI 0-10A 1 6. MC 0-10A 1 7. Wattmeter Dynamometer 500/250/125V,

5/10A,UPF 2

8. 150V,10A,LPF 1 9. Rheostat Wire wound 9Ω,8.5A 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.



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).

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 51 B) BLOCKED ROTOR TEST Make the connections as shown in the diagram.

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 52 Precautions : i) Keep the autotransformer in minimum voltage position





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

10A

R

A

0-10A MCA

B C

V0-30V MC

+

-

9Ω, 8.5A

+

-+

-

28V DC

Precautions : Keep the rheostat in maximum resistance position.


different positions of rheostat. (Note: Resistance/phase = 32

x Delta resistance)

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

(Y-axis) V by an angle Φo where 1cos ( )3

oco

o o

WV I

−Φ = .

15. From the current Isc find out the ISN (short circuit current corresponding to the

normal voltage) through the formula ( )ratedSN sc

sc

VI IV

= , draw the OB vector with the

magnitude of ISN from the origin by the same current scale, which lags the voltage

(Y-axis) V by an angle ΦSC where 1cos ( )3

scSC

sc sc

WV I

−Φ = .

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.


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

using the formula (rotor copper loss/stator copper loss) = 2

2 22

2 1

' ''

I RI R

= 2

1

'RR

where

R1= stator resistance per phase and R2= rotor resistance per phase. Or, 2

1

'

o

RBGBD R

= .

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 ratedV current scale× × . TABULATION

NO LOAD TEST BLOCKED ROTOR TEST

Voc Ioc W1 W2 Woc Vsc Isc W1 W2 Wsc


1. 2. 3. 4.

Rdc CIRCLE DIAGRAM Voc = 400V , Ioc = ___ A , Woc = _____ W

Vsc = _____ V, Isc = 7.8A, Wsc = _____ W

Per phase values are

_____ ____3

oco oc o

IV V V I A= = = =

_____ ____3sc


Rdc = _____ Ω


13 1.22 dcR R= × × = ______ Ω

1 23sc

os

WRI

= = _______ Ω

'2 1 1oR R R= − = _______ Ω

2

1

'

o

RBGBD R

= = ______

____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

( )ratedSN sc

VI IVsc

= = ____A = ____ cm

1cos ( )3

oco

o o

WV I

−Φ = = _______˚

1cos ( )3

scSC

sc sc

WV I

−Φ = = _______˚

Power Scale = 3 ratedV current scale× × = _______W = 1cm

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 55 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 = PTOP

= ______lag

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 602 sNπ

N-m

= ________N-m

Stator input at full load = PT x power scale = ________W

Efficiency at full load = 100%PQPT

× = _________%

Slip at full load, s = QRPR

= ________

Speed at full load = (1 )* ss N− = _______rpm

Starting torque developed = BG x power scale x 602 sNπ

N-m = _______N-m

Maximum torque developed = I I’ x power scale x 602 sNπ

N-m = _______N-m

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

_____ ____3

oco oc o

IV V V I A= = = =

_____ ____3sc


0cos3

oc

o o

WV I

Φ = = _______

0sinΦ = _______


coso

co o

VRI

=× Φ

=_________Ω

sino

mo o

VXI

=× Φ

= ________Ω

1s

os

VZI

= = _________Ω

1 23sc

os

WRI

= = _______Ω

2 21 1 1o o oX Z R= − = ________Ω

13 *1.2*2 dcR R= =1.5 x 1.2 x ____ = ______Ω

'2 1 1oR R R= − =_______Ω

' 12 2 2

oXX X= = = ______Ω

I1 I2'

IoImIc

R1 X1 R2' X2'

400V

EXACT EQUIVALENT CIRCUIT

Rc Xm 21' '( )L

sR Rs−

=

I2'

IoImIc

Ro1=R1+R2' Xo1=X1+X2'

400V

APPROXIMATE EQUIVALENT CIRCUIT

I1

XmRc 2

1' '( )LsR R

s−

=

PERFORMANCE AT RATED SPEED FROM EQUIVALENT CIRCUIT Synchronous speed, Ns = 1500 rpm Rated speed = N = 1400 rpm

Slip = 100%s

s

N NsN−

= × = 1500 1400 100 6.67%1500−

× =

21' '( )L

sR Rs−

= =_______Ω

oI (per phase) = o oI ∠−Φ °=_______A From approximate equivalent circuit,


'2

1 1

0( ')o L o

VIR R jX

∠ °=

+ +=________A per phase

1 2 'oI I I= + = 1 1I ∠−Φ ° = _________A

Line current IL = 13 I× = ______A Power factor = 1cos( )Φ ° = _______lag Output = 2

23 ' 'LI R× × =________W

Torque = 2( )60

OutputNπ =_________N-m (N=1400rpm)

Input = 13 cosLVI Φ = __________W

Efficiency = 100%OutputInput

× = ________%

RESULT

vi) No-load and blocked rotor tests were conducted on 3 phase squirrel cage induction motor

vii) Equivalent circuit parameters were determined viii) Circle diagram was drawn ix) Performance at full load from equivalent circuit and circle diagram were

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

2( )

120s rP N Nf +

= Hz which is higher than the supply frequency 2 120sPNf = .

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).


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.


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:


Type Range Quantity

1. Voltmeter MI (0-500V) 1 2. MC (0-500V) 1 3. MC (0-250V) 1 4. Ammeter MI (0-10A) 1 5. (10-0-10A) 2 6. Wattmeter Dynamometer 250V,10A,upf 1 7. Rheostat Wire Wound 272Ω 1.7A 1 8. Stopwatch 1 9. Tachometer 1

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 reactive-

power, 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:


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.

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 62 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 = 2 100ff× =_______%

Speed, (1 ) sN s N= − × = _______ rpm


Efficiency = 100outputinput

× =________%

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 = 2 100ff

− × = ______% (Note: Slip is negative)

Speed, (1 ) sN s N= − × = ________rpm (Note: Speed will be above synchronous speed,

1000rpm)

Efficiency = 100outputinput

× =________%

From 3Wph Vs % slip characteristics,

Hysteresis power =2

AB =______W

MODEL GRAPHS

A

B% Slip

3Wph

Effic

ienc

y %

Output in W

Generator

Motor

RESULT

a) Performance characteristics were plotted while the induction machine is operating as generator and motor.

b) hysteresis power = _______W.

================================================================


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.


5. Draw the torque-slip characteristics of an induction machine.


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:


Type Range Quantity

1. Voltmeter MI 0-500V 1 2. MI 0-150V 1 3 MC 0-30V 1 4. Ammeter MI 0-5A 1 5. MI 0-10/20A 1 6. MC 0-10A 1 7. Wattmeter Dynamometer 500/250/125V,

5/10A,UPF 2

8. Wattmeter Dynamometer 150V, 10/20A,UPF

2

9. Dynamometer 150V,10A,LPF 1 10. Rheostat Wire wound 9Ω,8.5A 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 change-

over 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.


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-S5-

S6 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.


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, 3 cosY YP VI= Φ

2. For parallel-star connection, 2 3 cosY YYP VI= Φ

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


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 S1-

S2-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, 3 cosP VIΔ Δ= Φ

2. For parallel-star connection, 2 3 cos 3.46 cosY YY YYP VI VI= Φ = Φ

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




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 BLOCKED ROTOR TEST Voc Ioc W1 W2 Woc Vsc Isc W1 W2 Wsc

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 71 B) BLOCKED ROTOR TEST ON LOW SPEED MOTOR Make the connections as shown in figure.






interchange the connections of the pressure coil (or current coil) of that wattmeter).

A) NO LOAD TEST ON HIGH SPEED MOTOR




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

NO LOAD TEST BLOCKED ROTOR TEST Voc Ioc W1 W2 Woc Vsc Isc W1 W2 Wsc

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 74 B) BLOCKED ROTOR TEST ON HIGH SPEED MOTOR Make the connections as shown in figure.






interchange the connections of the pressure coil (or current coil) of that wattmeter).

C) STATOR RESISTANCE MEASUREMENT Make the connections as shown in figure.

10A

10A

R

A

0-10A MCS1

S3 S2

V0-30V MC

+

-

9Ω, 8.5A

+

-+

-

28V DC

S4

S5 S6

Precautions : Keep the rheostat in maximum resistance position


different positions of rheostat. Switch off the supply.

(Note: For low speed connection, Resistance/phase = 32

x Delta resistance. For

high speed connection, Resistance/phase= 38

x Delta resistance)


S.No. V (volts) I (amps) Rdc=V/I Ω 1. 2. 3. 4.

Average Rdc

EQUIVALENT CIRCUIT PARAMETERS (LOW SPEED) Voc = _____V , Ioc = ____A , Woc = _____W

Vsc = ______V, Isc = _____A, Wsc = ______W

_____ ____3

oco oc o

IV V V I A= = = =


_____ ____3sc


0cos3

oc

o o

WV I

Φ = = _______

0sinΦ = _______

coso

co o

VRI

=× Φ

=_________Ω

sino

mo o

VXI

=× Φ

= ________Ω

1s

os

VZI

= = _________Ω

1 23sc

os

WRI

= = _______Ω

2 21 1 1o o oX Z R= − = ________Ω

13 *1.2*2 dcR R= =1.5 x 1.2 x ____ = ______Ω

'2 1 1oR R R= − =_______Ω

' 12 2 2

oXX X= = = ______Ω

I2'

IoImIc

Ro1=R1+R2' Xo1=X1+X2'

400V

APPROXIMATE EQUIVALENT CIRCUIT (HGH SPEED)

I1

XmRc 2

1' '( )LsR R

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)

Slip = 100%s

s

N NsN−

= × = 750 710 100 5.33%750−

× =

V = 400V

Rotor current, '2

2 ' 221 1 2

'( ) ( )

VIRR X Xs

=+ + +

= ____________A per phase

Rotor Input P2= '

2 223 ' RI

s× × = _______W


Torque developed = 2

2( )60s

PNπ = 260

2 750P

π×

× ×= _________N-m

EQUIVALENT CIRCUIT PARAMETERS (HIGH SPEED) Voc = _____ V , Ioc = ____A , Woc = _____W

Vsc = ______V, Isc = _____A, Wsc = ______W

_____ ____3

oco o oc

VV V I I A= = = =

_____ ____3sc

s s scvV V I I A= = = =

0cos3

oc

o o

WV I

Φ = = _______

0sinΦ = _______

coso

co o

VRI

=× Φ

=_________Ω

sino

mo o

VXI

=× Φ

= ________Ω

1s

os

VZI

= = _________Ω

1 23sc

os

WRI

= = _______Ω

2 21 1 1o o oX Z R= − = ________Ω

13 1.28 dcR R= × × = ________Ω

'

2 1 1oR R R= − = ________Ω ' 1

1 2 2oXX X= = = _______ Ω

I2'

IoImIc

Ro1=R1+R2' Xo1=X1+X2'

254V

APPROXIMATE EQUIVALENT CIRCUIT (HIGH SPEED)

I1

XmRc 2

1' '( )LsR R

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)


Slip = 100%s

s

N NsN−

= × = 1500 1440 100 41500−

× = %

440 2543

V V= =

Rotor current, '2

2 ' 221 1 2

'( ) ( )

VIRR X Xs

=+ + +

= _________ A per phase

Rotor Input P2 = '

2 223 ' RI

s× × = ________W

Torque developed = 2

2( )60s

PNπ = 260

2 1500P

π×

× ×= ________N-m

Note: Find the torque for different values of speed and hence plot torque Vs speed characteristics using MATLAB, C, EXCEL or any software. TABULATION

LOW SPEED HIGH SPEED Sl.No. Speed Slip Torque Speed Slip Torque

1 0 0 2 30 60 3 60 120 4 90 180 5 120 240 6 150 300 7 180 360 8 210 420 9 240 480 10 270 540 11 300 600 12 330 660 13 360 720 14 390 780 15 420 840 16 450 900 17 480 960 18 510 1020 19 540 1080 20 570 1140 21 600 1200 22 630 1260 23 660 1320 24 690 1380 25 710 1440 26 750 1500


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.


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.


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. Name of the

apparatus Type Range Quantity

1. Voltmeter MI (0-250V) 1 2. MI (0-100V) 1 3. MC (0-30V) 1 4. Ammeter MI (0-10A) 1 5. MC (0-10A) 1 6. Wattmeter Dynamometer

type 250,10A, LPF 1

7. 125V,10A, UPF 1 8. Rheostat Wire-wound 9Ω,8.5A 1

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

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 82 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


R

M L

C V

10AA

0-10A MI

0-250V MI230V1-phase50Hz

250V,10ALPF

NL

V

1 2

1'

2'C

CS

Machine Details : 230V, 7.4A, 1HP, 1425rpm

A

C

E

P

N

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


R

M L

C V

10AA

0-10A MI

0-100V MI230V1-phase50Hz

100V,10AUPF

NL

V

1 2

1'

2'C

CS

Machine Details : 230V, 7.4A, 1HP, 1425rpm

A

C

E

P

N

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.


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


10AA

0-10A MC

0-30VMC28V

DC

V

1 2+

-

V

A9Ω,8.5A

+

-

+ -

10A

Precaution: Keep the rheostat at maximum resistance position.


different positions of rheostat. Switch off the supply.

TABULATION

NO LOAD TEST BLOCKED ROTOR TEST Vo Io Wo Vsc Isc Wsc


S.No. V (volts) I (amps) Rdc=V/I Ω 1. 2. 3. 4.

Rdc


R1 X1

X2'/2

R2'/2(2-s)

X2'/2

Xm/2

Xm/2

I2f'Imf

ImbI2b'

I1

R2'/2s

Equivalent circuit during normal running condition

R1 X1

Xf

I1

Rf

Rb

Xb

Ef

Eb

V1V1

R1 X1

R2'

X2'

Equivalent circuit during blocked rotor test

Vsc

Xm/2

R1 X1

R2'/4

X2'/2

Vo

Equivalent circuit during no load test

Isc

Io

Xm/2Vo

Io

R1+R'2/4+j(X1+X'2/2) << jXm/2

CALCULATION EQUIVALENT CIRCUIT PARAMETERS BLOCKED ROTOR TEST

Vsc = _____V, Isc = _____A, Wsc = ______W


cos scsc

sc sc

WV I

Φ = = ______

1 2 2' scsc

sc

WR R RI

= + = = _______Ω

R1dc = ______ Ω R1 = 1.2 x R1dc = _______Ω

2 1scR R R= − = _______Ω

scSC

sc

VZI

= = _______ Ω

2 21 2 'sc sc scX X X Z R= + = − = __________ Ω

1 2 '2

scXX X= = = _________ Ω

NO LOAD TEST Vo = _______V , Io = _____A , Wo = _______W

Rotational loss Wrot2 2

1'( )

4o oRW I R= − + =______W

2m o

o

X VI

= = ________ Ω

Xm = ________Ω Predetermination of the steady state performance at rated speed (N = 1425 rpm)

At rated speed, slip = s

s

N NsN−

= = 1500 1425 0.051500−

=

Equivalent circuit at rated speed is given by,

Forward impedance, 2 2' '//2 2 2

mf f f

jX R jXZ R jXs

⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= _______ Ω

Backward impedance, 2 2' '//2 2(2 ) 2

mb b b

jX R jXZ R jXs

⎛ ⎞⎛ ⎞= + = +⎜ ⎟⎜ ⎟ −⎝ ⎠ ⎝ ⎠= _______ Ω

Total impedance, 1 1( ) f bZ R jX Z Z= + + + =_______ Ω

Current, 11

0VIZ∠ °

∠−Φ = = _______ Ω

Power factor = cosΦ = ______lag

Input power = 1 1 cosV I Φ =________W

Forward torque, 21f fT I R= sync.watts = _________sync.watts


Reverse torque, 21b bT I R= sync.watts = ________sync.watts

Resultant torque, f bT T T= − = ________ sync.watts

Mechanical power developed Pmech= (1 )s T− = _______watts

Output power = Pmech – Wrot = _______W

Resultant torque in N-m = 2 60

s

TNπ

= _______N-m

Efficiency = 100OutputInput

× = _______%

RESULT :

i) No load and blocked rotor tests on single phase induction motor were conducted

ii) Equivalent circuit parameters determined and iii) its performance at rated speed determined from equivalent circuit

================================================================ Do you know?

1. What is double field revolving theory ? A alternating sinusoidal flux sinm tωΦ = Φ can be represented by 2 revolving fluxes

each equal to one-half of the maximum value of alternating flux (= 2mΦ ) and each

rotating at synchronous speed ( 120s

fNP

= ) in opposite directions.

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.


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.

Speed of secondary = U = Us(1-2) and slip, s

s

U UsU−

= .

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.


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 Type Range Quantity 1. Voltmeter MI (0-500V) 2 2. MC (0-300V) 1 3 MC (0-30V) 1 4. Ammeter MI (0-15A) 1 5. MC (0-3A) 1 6. Rheostat Wire Wound 145Ω 2.5A 2 7. 272Ω 1.7A 1 8. Tachometer 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 cosaI Φ is constant; that is, the in-

phase 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.

Efsinδ

Ef3Ef1 Ef2

V

Ia1

Ia3

Ia2

Locus of Ia for constant power

-Ia1Xs

-Ia2Xs-Ia3Xs

Locus of Ia for constant power

Synchronous machine operating as motor

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 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

f a sE V jI X= − for synchronous motor and f a sE V jI X= + for synchronous generator.

The power can also be expressed as 3 sinf

s

VEP

Xδ= . Again for constant power operation,

sinfE 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 sinfE 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 over-

excited 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.


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 V-

curves (Ia Vs If) for synchronous motor on no-load. Now, for getting V curves of

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 93 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.

Sl. No.

GENERATING MODE MOTOR MODE MOTOR MODE

OUTPUT = ________W ON NO LOAD (INPUT = _______W) 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

If in amps

Ia, p.f.

Ia Vs If

p.f. Vs If

If in amps

Ia, p.f.

Ia Vs If

p.f. Vs If

laglead leadlag

SYNC.GENERATOR SYNC.MOTOR

No-load

Input=____W

output=____W

V-CURVES & INVERTED V-CURVES

SAMPLE CALCULATION (Set No. ____) Generator delivering an output of ______W.

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 94 V = 400V, Ia = ____A, If = ____A, Iamin = _____A

mincos a

a

II

Φ = = _______ lag (/lead) (Note : under-excited → lead & over-excited → lag)

Motor drawing an input of _______W. V = 400V, Ia = _____A, If = _____A, Iamin = _____A

mincos a

a

II

Φ = = _______ lead (/lag) (Note : under-excited → lag & over-excited → lead)

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?


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 inter-locking 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).

V1 V2

R1

Y1

B1

R2

Y2

B2

L1

L2L3

VR1

VR2

VY1

VY2

VB1

VB2

L1

L2

L3

VR1 VR2

VY1

VY2

VB1

VB2 L3 L2

L1

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.


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 Type Range Quantity 1. Voltmeter MI (0-500V) 1 2. Frequency meter Digital (0-60Hz) 1 3. Ammeter MC (0-3A) 1 4. MI (0-10A) 1 5. Tachometer 1 6. Rheostat Wire Wound 145Ω, 2.5A 2 7. 272Ω 1.7A 1

PRINCIPLE:

The synchronous speed Ns of an induction motor is related to supply frequency f

and number of poles P by the equation, 120s

fNP

= . Rotor speed is given by

(1 ) sN s N= − 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 4.44 wE K f T= Φ volts. The induced

emf E is nearly equal to the applied voltage V (neglecting drop in stator impedance).

Thus, we can write 4.44 wV K Tf= Φ .

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:

v) TPST in open position vi) DPST1 and DPST2 in open position vii) Motor field rheostat in minimum position viii) Potential divider in minimum voltage position ix) Autotransformer at minimum voltage position x) Keep the belt on the brake drum of induction motor in loose position

(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 ( 400 850

Vf= = ) and note down the frequency, voltage and speed

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.

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 99 TABULATION

Induction motor on no load Line voltage

in volts 352 368 384 400 400 400

Frequency in Hz

44 46 48 50 52 54

Speed of IM in rpm

Induction motor on load IL = 4A

Line voltage in volts

352 368 384 400 400 400

Frequency in Hz

44 46 48 50 52 54

Speed of IM in rpm

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.


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 101 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.

S6 EM Lab II Manual as on 1-1-2010 prepared by TGS, GEC Thrissur 102 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.

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