ac electric machines lab manul

7/27/2019 AC Electric Machines Lab manul

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ELECTRICAL MACHINES LAB.

LAB SAFETY RULES:

Read ALL of the following rules carefully and remember them while working in the

laboratory.

1. Some of the experiments involve voltages that could conceivably lead to

serious injury or death. Therefore strict adherence to the following rules

will greatly decrease the probability that accidents will occur.

2. Never hurry. Haste causes many accidents.

3. Use one hand to make connections

4. Always see that power is connected to your equipment through a circuit

breaker or load switch.

5. Connect the power source last. Disconnect the power source first.

6. Never make wiring changes on live circuits. Work deliberately and carefully

and check your work as you proceed.

7. Before connecting the power, check the wiring carefully for agreement with

the wiring diagram for an accidental short-circuit and for loose connections.

8. Check out the supply voltage to make sure that is what you expect. For

example: AC or DC, 120V, 208V or 240V.

9. Do not cause short-circuits or high currents arcs. Burn from arcs may be

very severe even at a distance of a few meters. Report all electrical burns to

your instructor.

10. Be careful to keep metallic accessories of apparel or jewelry out of contact

with live circuit parts and loose articles of clothing out of moving

machinery.

11. When using a multiple range meter always use the high range first, to

determine the feasibility of using a lower range.

12. Check the current rating of all rheostats before use. Make sure that nocurrent overload will occur as the rheostat setting is changed.

13. Never overload any electrical machinery by more than 25% of the rated

voltage or current for more than a few seconds.

14. Select ratings of a current coil (CC) and potential coil (PC) in a wattmeter

properly before connecting in a test circuit.

15. Do not permit a hot leg of a three phase 208V supply, or of a 240V or 120V

supply to come in contact with any grounded objects, as a dangerous short-

circuits will result.16. If you know or suspect that an accident is about to occur, take

immediate steps to prevent it but do not jeopardize your own safety indoing so.



LIST OF EXPERIMENTS:

Sr.

No.Name of Experiment

Page

No:

1- Open Circuit Test of Transformer. 01

2- Short Circuit Test of Transformer. 06

3 Transformer Efficiency. 09

4 Direction of Rotation of 3-Phase Induction Motor. 15

5 Starting Characteristics of Squirrel Cage Induction Motor. 20

6 Running Characteristics of Squirrel Cage Induction Motor. 24

7 Starting Characteristics of Wound Rotor Induction Motors. 31

8 Speed Control of Wound Rotor Induction Motors. 35

9 Losses & Efficiency of Induction Motors. 40

10 Saturation Curve of an Alternator. 46

11 Effect of Speed on an Alternator. 52

12 Load Characteristics of an Alternator. 56

13 Losses & Efficiency of Alternators. 62

14 Paralleling Alternators. 70

15 Starting & Synchronizing, Synchronous Machines. 80

16 Synchronous Moto V-Curves. 86

17 Power Factor Correction Using Synchronous Motors. 91



Lab Manual—EE-314—AC-Machines

University College of Engineering & Technology, IUB υ

EXPERIMENT # 1:

OPEN CIRCUIT TEST OF TRANSFORMER.

PERFORMANCE OBJECTIVE:

After completion of this laboratory experiment, the student will be able to perform an open circuit

transformer test,

• Measure the exciting current and determine the core losses in a transformer.

• Specifically, the student will be able to determine the Equivalent Circuit parameters (shunt branch) of

the transformer.

DISCUSSION:

The current input to the primary winding, without a load connected to the secondary winding

is usually from 1 to 5 percent of the full load current rating. This no load primary current is called the

exciting current and consist of the following:

1. The magnetizing current that supplies the alternating flux in the core, which produces the

primary and secondary induced voltages. This current component lags the applied voltage by

90°.

2. Due to eddy currents and hysteresis, the core will lose power. The changing flux induces

voltage and current in the iron core causing an I²R loss. This eddy current loss is minimized by

laminating the core and insulating each lamination with a varnish or oxide coating. The

inability of the magnetic domains of the core material to instantly follow the changing flux

(due to inter-domain friction) incurs a power loss as heat. This hysteresis loop loss is

minimized by the use of special steel and various core configurations. The lost core current is

in phase with the applied voltage.

The vectorial sum of the in-phase core loss and lagging magnetizing currents produces the

exciting current of a transformer.

The open circuit test is used to determine the values of parameters of the shunt branch of the

equivalent circuit; Rp and Xp. We can see from Figure-1 that with the secondary winding left open,

the only part of the equivalent circuit that affects our measurement is the parallel branch.

FIGURE-1




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EQUIPMENT:

• Single Phase Transformer [MV-1911]

• AC Power Meter [Goodwill Instek]

• Power Pack [MV-1300]

• Connecting Wires

CONNECTION DIAGRAM:

FIGURE-2

PROCEDURE:

Step-1 -Make sure that all the equipment stated above is available.

-All connections are to be made when the equipment is not connected to supply.

Step-2 Transformer [MV-1911] has four set of windings on the L.V side. Create a 1:2 ratio step-up

transformer by connecting the two windings on the L.V side, in series (i.e. 115V) and keep

the high voltage winding (i.e. 230V) as the secondary side.

Step-3 Connect the circuit as shown in figure-2.

Step-4 After connecting the circuit, let it be checked by your Lab Instructor.

Step-5 Make sure that the voltage regulator knob of the power supply is at its zero mark, and

then switch on the variable AC-Voltage supply.

Step-6 Vary the input voltage staring at 0V, in 20V increments to go upto full rated voltage of the

primary side (i.e. 115V)

-At each step change, record IP, W0 & V1 in table-1.

Step-7 Turn OFF all circuit breakers. Disconnect all leads.

PRECAUTIONS:-This transformer is rated at 1.0KVA. The rated current is 1000VA/230V = 4.34A on the 230V

side & 1000VA/115V = 8.69A on the 115V side. In any case, do not increase the current beyond the

rated current limit on either side.

-The impedance of the parallel branch is usually very high but appears lower when referred to

the low voltage side. We have selected low voltage side as primary for our test. The rated voltage on

the primary side is lower than secondary side and therefore more manageable.




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TEST RESULTS:

Sr.

No

V1

(volts)

Ip

(Amp)

W0

(Watts)

Ic =

W0 / V1

(Amp)

Im =

(Ip2

- Ic2)

1/2

(Amp)

cosφ =

W0 / V1 Ip

Rp Xp

1

2

3

4

5

6

TABLE-1

CALCULATIONS:

1. Compute the parameters Rp and Xp at the rated voltage by using

a. Rp = W0 / (Ic)2

= V12

/ W0

&

b. Xp = V1 / Im

-These parameters are referred to the low voltage side.

2. Find the values of Rp and Xp as referred to the high voltage side.

3. Plot the no-load current Ip , magnetizing current Im and core loss W0 and no-load power factor

cosφ, against the applied voltage V1(on x-axis) on the graph paper.




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GRAPH:




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REVIEW QUESTIONS:

1. The current that produces the primary and secondary induced voltage in the core.

a. Lags the applied voltage by 90°.

b. Leads the applied voltage by 90°.

c. Is in phase with applied voltage.

2. Why does the core lose power?

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

3. How can you minimize the losses in the core?

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

FINAL CHECKLIST:

1. Clean your equipment/materials and work benches before you leave.

2. Return all equipment and materials to their proper storage area.

3. Submit your answers to the questions, together with your data, calculations, and results

before the next laboratory session.

____________

Signature:

Lab Instructor




University College of Engineering & Technology, IUB ϊ

EXPERIMENT # 2:

SHORT CIRCUIT TEST OF TRANSFORMER

PERFORMANCE OBJECTIVES:

Upon successful completion of this laboratory experiment, the student will be able to perform

a short circuit transformer test. Specifically, he will be able to determine the equivalent circuit

parameters (series branch) of a transformer by the short circuit method.

DISCUSSION:

It is possible to represent a transformer as an ordinary series electric circuit (neglecting thesmall, no-load exciting current) that has three elements: 1) the equivalent resistance, 2) the

equivalent leakage reactance, and 3) the load [see Figure-1].

Figure-1

Note that the transformer, as an electrical circuit, merely acts like an impedance voltage drop,

which depends not only upon the actual load current but also upon the power factor of the load.

The short-circuit test is an experimental method of determining the equivalent series

resistance and reactance (implied as impedance) of a transformer. In this test the windings are made

to carry the rated currents without requiring the transformer to deliver a load. This is done by

shorting the secondary winding and increasing the primary voltage from zero to that value which

causes the rated current to flow. In this way, it is possible to simulate the pattern of flux leakage in

the primary and secondary because the later depend upon the load currents in the two windings.

From the data of watts, amperes, and volts obtained from this test, the values of equivalent

resistance impedance, and reactance can then be calculated using the following equations:

R01 =ୱୡ

²ୱୡ

Z01 = VSC / ISC

X01 = [Z012-R01

2]

½




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EQUIPMENT:

• Single Phase Transformer

• AC Power Meter

• Variable Power Supply

• Connecting Wires

CONNECTION DIAGRAM:

Figure-2

PROCEDURE:



Step-2 Transformer [MV-1911] has four set of windings on the L.V side. Create a 2:1 ratio step-

down transformer by connecting the two windings on the L.V side, in series (i.e. 115V)

and keep the high voltage winding (i.e. 230V) as the primary side.

Step-3 Connect the circuit as shown in figure-2.




Step-6 Vary the input voltage very slowly and carefully, starting at 0V in very small increments to

go up to full rated current of the primary side (i.e. 4.34A).

-Record VSC, ISC and WSC in Table 1.

Step-7 Compute the values R01, Z01 & X01

PRECAUTIONS:

• This transformer is rated at 1.0KVA. The rated current is 1000VA/230V = 4.34A on the

230V side and 1000VA/115V = 8.69A on the 115V side. In any case, do not increase the

current beyond the rated current limit on either side.

• The current on the low voltage side will be comparatively high than the current on the

high voltage side. This test can be performed on either side of transformer. However, we

will perform this test on the high voltage side of the transformer to keep the current

passing through the measuring instruments within a measureable range (i.e without

exceeding/overloading the measuring instruments).

CAUTION!!! Be sure the variable supply knob is turned to full counter-clockwise position (i.e zero

position) before energizing circuit.




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TEST RESULTS:

VSC ISC WSC R01 Z01 X01

Table-1

REVIEW QUESTIONS:

1. What is the equivalent circuit found in the short-circuit test?

2. Write the formula for calculating the value of equivalent series resistance, impedance, and

reactance of a transformer. ____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

3. Refer these calculations to the secondary.

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

FINAL CHECKLIST:





_________________

Signature:

Lab Instructor




University College of Engineering & Technology, IUB ύ

EXPERIMENT # 3:

TRANSFORMER EFFICIENCY


Upon successful completion of this experiment the student will be able to:

1. Investigate the methods of determining percent of Efficiency of a Transformer, at unity

power factor loads.

2. The effect of load on Efficiency of Transformer.

DISCUSSION:

Referring to the information obtained from the short circuit test and open circuit transformer

tests, it is easily seen that losses are confined to two basic categories:

1. Core losses- (Eddy currents and hysteresis losses), which are proportional to applied voltage

and are measured in terms of true power. These losses are essentially constant for all values

of transformer loading provided that the applied voltages remain same. This is true because

flux is essentially constant for these values.

2. Copper losses- Vary as the square of the load current. This loss is not constant but must be

computed for all values of load current.

The percent efficiency can be computed from the following formula for each load value from

no-load to full load:

Efficiency = ˟ 100 =ሺ ା ሻ

˟ 100

Since the losses can be reduced to a very small percentage by good design, it is not uncommon to

have efficiencies as high as 98 – 99 % in large transformers. Because the relationship to power-in and

power-out are so close, it may prove difficult to achieve good results from the measuring procedure.

EQUIPMENT:

• Single Phase Transformer

• 2 - AC Power Meter


• Connecting Leads.




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CONNECTION DIAGRAM:

Figure-1

PROCEDURE:

Step-1 -Make sure that all the equipment stated above is available.-All connections are to be made when the equipment is not connected to supply.

Step-2 Note the value of core losses, from OC test data & the value of REQ = R01, from SC test

data.

Step-3 Connect the circuit as shown in connection diagram (Figure-1).


Step-5 Make sure that the voltage regulator knob of the power supply is at its zero mark, andthen switch on the variable AC-Voltage supply.

Step-6 Increase the supply voltage to the rated voltage of the transformer.

Step-7 Then change the load in order to increase load current from 1.5 to 2.4 Amp in 10 steps at

unity power factor. Record the load current I1 & value of PIN [W1] for each step in Table 1.

Step-8 Calculate copper losses [I12R] at each load current and note it down besides the

respective load current.

Step-9 Calculate the %Efficiency for each level of load and note it down in Table 1.

Step-10 Now, decrease the load current in ten steps from 2.4 to 1.5 Amp and record all the meter

readings for each step in Table 2

Step-11 Calculate %Efficiency using PIN & POUT.

Step-12 Plot the values of load current versus %Efficiency for increasing & decreasing load current

on the same Graph.

Step-13 Turn OFF all circuit breaker switches. Disconnect all leads.




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TEST RESULTS:

R01 = REQ = ________ (from SC test data)

Core Losses = ________ (from OC test data)

Load

Current [I1]

Copper

Losses[I12R1]

PIN [W1]

% Efficiency

Table-1

PR

I

Volts

Amperes

Watts

S

E

C

Volts

Amperes

Watts

% EFF

Measured

Table-2




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GRAPH:




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REVIEW QUESTIONS:

1. How the change in load current does effects the efficiency of transformer?

__________________________________________________________________________________

__________________________________________________________________________________

2. What is the condition for the maximum efficiency of the transformer?

__________________________________________________________________________________

__________________________________________________________________________________

1. During the Locked Rotor Test, all of the input power was lost in the motor’s equivalent

resistance. The equation is:

W = 3 x (ILR)2

x REQ

Solving for REQ the equation becomes REQ = W/(3 x I2). From the data you recorded in Table-2

compute the total power in (TOT. W) and REQ. Record these values in TABLE-2.

2. During the No-Load-Test, the input power was lost in both the equivalent resistance and in

the rotational losses (PRL) To find the value of PRL first compute the total power in. Then

compute the total copper losses:

PCL = 3 x (INL)2

x REQ

Using the value of REQ = Total. W. PCL. Record this value of PRL Table-1 and for every load

listed in Table-3. (PRL assumed constant).

3. For each of the motor current values in TABLE-3, add the wattmeter readings to provide the

total power in. Record these values in TABLE-3.

4. For each of the motor current values in TABLE 16-3 compute the total copper loss from the

equation PCL = 3 x x REQ using the value of REQ from TABLE-2. Record the copper loss values

in TABLE-3. Add each of these values to the value computed in No. 2 to produce the

total loss value for each of the loads. Record these values in TABLE 16-3.

5. Compute efficiency, the ratio of output power to input power. For output power subtract

the losses from the output power. The equation is:

% Efficiency = Total Watts in - PL x 100

Total Watts in

List these efficiencies in TABLE-3.

6. As load increases, rotational losses:

a. Increase.b. Decrease.

c. Remain the same.




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7. As load increases, copper losses:

a. Increase.

b. Decrease.

c. Remain the same.

8. As load increases, the total losses become:

a. A larger share of the total power in.

b. A smaller share of the total power in.

c. The same share of the total power in.

9. As load increase, the motor:

a. Operates more efficiently.

b. Operates less efficiently.c. Operates with the same efficiency.

10. Power Out equals:

a. The power in.

b. The total of copper and rotational losses.

c. The power in minus the total losses.

FINAL CHECKLIST:





____________Signature:

Lab Instructor




University College of Engineering & Technology, IUB υϊ

The rotors of induction motors (squirrel-cage or wound rotor) can never run at synchronous

speed. There must be relative motion between the field and the rotor so that induction may take

place. The difference between synchronous speed and rotor speed is called slip speed, or simply slip.

The percent slip can be computed from the following equation:

%Slip = Slip Speed____

Synchronous Speed

EQUIPMENT:

• Three-Phase Induction Motor

• Torque-Speed measuring unit



CONNECTION DIAGRAM:

Figure-1

PROCEDURE:



Step-2 Clamp the motor on the machine bed & Install coupling guards.

Step-3 Connect the circuit as shown in connection diagram (Figure-1). Note that Phase-L1 of the

supply is connected to terminal U1 of the motor, L2 to V1, and L3 to W1. Note also that

terminals U2, V2, W2 are connected in star connection.




-Then increase the supply voltage to the motor up-to 220VAC.




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Step-6 Note the direction of rotation as that viewed from the right-hand end, indicate the

direction on Figure. A of TEST RESULTS.

Step-7 Turn OFF the motor.

Step-8 Reconnect the stator as follows: Leaving L3 connected to W1, interchange the other two

leads so that L1 is connected to V1; L2 to U1.

Step-9 Repeat Step 5 & 6 for Figure. B of TEST RESULTS, then turn OFF the motor.

Step-10 Reconnect the stator as follows: Leaving L1 connected to V1, interchange the other two

leads so that L2 is connected to W1; L3 to U1.

Step-11 Repeat Step 5 & 6 for Figure. C of TEST RESULTS, then turn OFF the motor.

Step-12 Reconnect the stator as follows: Leaving L3 connected to U1, interchange the other two

leads so that L1 is connected toW1; L2 to V1.

Step-13 Repeat Step 5 & 6 for Figure. D of TEST RESULTS, then turn OFF the motor.

Step-14 Reconnect the stator as follows: Leaving L1 connected to W1, interchange the other two

leads so that L2 is connected to U1; L3 to V1.

Step-15 Repeat Step 5 & 6 for Figure. E of TEST RESULTS, then turn OFF the motor.

Step-16 Reconnect the stator as follows: Leaving L3 connected to V1, interchange the other two

leads so that L1 is connected to U1; L2 to W1.

Step-17 Repeat Step 5 & 6 for Figure. F of TEST RESULTS, then turn OFF the motor.

Step-18 Leaving L1 connected to U1, interchange the other two leads so that the connection is

the same as shown in Figure 1.

Step-19 Turn ON the motor. Measure the speed of the unloaded motor. Record it in Test Results.





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TEST RESULTS:

Motor Speed (Step 19): ____________ RPM.

REVIEW QUESTIONS:

1. Each time you reconnected the stator, you interchanged two of the three leads. What was the

effect on the motor’s direction?

____________________________________________________________________________

____________________________________________________________________________

2. There are no electrical connections made to the rotor of a squirrel-cage induction motor. Yet

there is current in the squirrel-cage bars. Explain what causes the current.

____________________________________________________________________________

____________________________________________________________________________

3. If the rotor of an induction motor turned at the same speed as the revolving magnetic field

(synchronous speed) what would happen to rotor current?

____________________________________________________________________________

____________________________________________________________________________

4. Slip speed is the difference between synchronous speed and rotor speed.

First compute synchronous speed;

Synchronous Speed = (Frequency x 120) / no. of pairs of poles [rpm]

Synchronous Speed = ________________

Then compute Slip Speed;

Slip speed = Synchronous Speed - Rotor Speed

Slip Speed = ___________________




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5. Percent slip is the ratio of slip speed to synchronous speed [%Slip = (Slip Speed/Synchronous

Speed) x 100]. What is the percent slip of the test motor running unloaded?

____________________________________________________________________________

____________________________________________________________________________

6. The main (stator) field of a three-phase motor revolves because:

a) DC is applied to the rotor coils.

b) DC is applied to the stator coils.

c) Three out-of-phase voltages are applied to the stator coils.

7. If the coil that is connected to Phase B line is to the left of the coil connected to Phase A line,

the magnetic field of the stator will:

a) Revolve to the right.

b) Revolve to the left.

c) Go straight ahead.

8. To reverse the direction of a three-phase motor, you must:

a) Reverse the motor connections.

b) Change all three stator connections.

c) Interchange any two stator connections.

9. 3-phase Induction Motors must:

a) Run faster than synchronous speed.

b) Run slower than synchronous speed.

c) Run at synchronous speed.

10. Synchronous speed is determined by:

a) Frequency and number of poles.

b) Torque and speed of driven load.

c) Field strength and armature current.

FINAL CHECKLIST:





____________

Signature:

Lab Instructor




University College of Engineering & Technology, IUB φυ

Knowing this allows us to lock the rotor and measure starting torque at a reduced voltage; then

compute what it would be at full voltage. If full voltage were applied, the high starting current would

trip the circuit breakers before we could get a reading.

EQUIPMENT:

• Three Phase Squirrel-Cage Induction Motor

• Torque-Speed measuring unit

• AC power meters



CONNECTION DIAGRAM:

Figure-1

PROCEDURE:



Step-2 Clamp the motor on the machine bed & Install coupling guards.



Step-5 Install the rotor locking device securely on the torque measuring unit.

Step-6 This step is to be performed as quickly as possible. Make sure that the voltage regu;ator

knob is turned fully counter-clockwise. With the motor load switch off, turn the voltage

control knob clockwise until the voltmeter reads 55 volts. (NOTE: when the motor is

turned on, the voltage will drop. This is the voltage at which readings are to be taken.)

Turn the motor ON and quickly read line amps, torque & wattmeter readings. Turn themotor OFF and record these readings in TABLE 1 of TEST RESULTS

Step-7 Repeat Step-6 two additional times. Allow two minutes before tests.




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Step-8 Average the three tests and record the average values of current, torque, and power in

TABLE 1.


TEST RESULTS:

Line Volts Line Amps Torque W1 Total Watts

3 X W1

Test 1

Test 2

Test 3

Average

Table-1

Volts Amps Torque

Full Voltage Starting

Full Load Running

Starting / Running %

Table-2

REVIEW QUESTIONS:

1. Full voltage is 220 volts. Your tests were run at one-quarter full voltage (55 volts). Starting

current would therefore be four times the value you measured. Compute full voltage starting

current and record in TABLE 2.

2. Torque is proportional to applied voltage squared. Therefore, full voltage starting torque

would be four (16) times (4)2

the torque you measured at one-quarter voltage. Compute full

voltage starting torque and record in TABLE 2.

3. Compute the ratio of full load starting current (which you computed in #1) to the rated full

load running current 1.38( amps).

4. The total apparent power is computed from the equation:

PS = 1.73 E x I volt-amperes

Compute the reduced voltage starting apparent power using the current read in step 5.PS = ___________________




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5. The True power (P) at reduced voltage is the sum of the two-wattmeter readings. The power

factor is the ratio of true power to apparent power-P.F. = P/PS. Compute the power factor

(cosɵ).

P.F = _________________

6. Starting current is:

a. Greater than full load current.

b. Less than full load current

c. The same as full load current.

7. Full load running torque is:

a. Greater than starting torque.

b. Less than starting torque.

c. The same as starting torque.

8. Each amp of starting current provides:

a. The same torque as each amp of running current.

b. More torque than each amp of runnings current.

c. Less torque than each amp of running current.

9. You would get more in-oz of starting torque from each amp of starting current if:

a. There was more inductive reactance in the rotor.

b. There was more resistance in the rotor.

c. The rotor bars did not form a complete circuit.

10. At the instant of start, an induction motor has:

a. A leading phase angle

b.

A good power factor, small lagging phase angle.c. A poor power factor, large lagging phase angle.

FINAL CHECKLIST:





____________

Signature:

Lab Instructor




University College of Engineering & Technology, IUB φψ

EXPERIMENT # 6:

RUNNINIG CHARACTERISTICS OF SQUIRREL CAGE INDUCTION MOTOR



1. Explain and predict the changes in speed and current that will occur as a squirrel- cage

induction motor is loaded.

2. Perform load tests on three-phase motors.

DISCUSSION:

The speed at which the stator’s field revolves is the synchronous speed. As this field is cut by

the bars of the squirrel-cage rotor winding, current is induced into the bars. The rotor’s magnetic

field (caused by this current) interacts with the stator’s field to produce torque on the rotor.

The torque is directly proportional to rotor current, and the cosine of the phase angle

between the rotor and stator fields (cos ɵ). Another way of expressing this relationship is that

torque is directly proportional to the in-phase component of rotor current, IR cos ɵ.

At the instant of start, IR is high but the in-phase component is low because of the poor

power factor (cos ɵ). As the rotor picks up speed, both the induced rotor voltage and the inductivereactance decrease. Basically IR is going down while cos ɵ is going up.

Look at a trig table, or even at a cosine curve like this one:

You can see that there is not very much difference in the value of cosɵ when ɵ is 0o

(cos ɵ = 1)

and when ɵ is 20o

(cosɵ = 0.94).

Therefore, over the operating range of the motor, the rotor power factor does not play an

important part in the torque output. More important is rotor current. Rotor current falls off sharply

as the rotor approaches synchronous speed (i.e., slip approaches zero). Speed doesn’t have to drop

back very much to increase the rotor current, stator power factor, and torque.

When you are running an induction motor without load, it draws almost as much current as it

does fully loaded. This no-load current, however, is made up of two components. The in-phase

component supplies electrical and mechanical losses. The quadrature (90 degrees out of phase)




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component is the magnetizing current. It is quite large in comparison with the in- phase part. As the

motor is loaded, it is like putting a resistive load on the secondary of a transformer. The in-phase

component gets larger. The stator’s power factor improves accordingly. The increased rotor current

does not necessarily add to the total current being drawn by the motor. It simply uses more of that

current for useful work.

In this experiment we will be using the two-wattmeter method of measuring power input. At

no-load, power factor is less than 0.5. That means that one wattmeter must be connected with its

voltage coils reversed and its reading subtracted from the other one. As the motor is loaded, the

power factor improves. When it reaches 0.5, the potential coil connections must be connected

normally, its reading added to the other one.

EQUIPMENT:

• Induction Motor (Squirrel-Cage)

• DC Machine (operating as DC-Generator.)

• Torque Speed Measuring Unit.

• Variable Power Supply.

• AC-Power Meters.

• Digital Multi-Meters

• Resistive load.


CONNECTION DIAGRAM:

Figure-1




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PROCEDURE:



Step-2 Place the DC-Machine (operating as DC-Generator) on the machine bed on the left side of

torque speed measuring unit & Induction Motor on the right side. Couple and clamp the

machines securely & Install guards.

Step-3 Connect the circuit as shown in connection diagram (Figure-1). Note that the DC-Machine

is connected as a separately-excited shunt generator. Make sure that the switch for the

load connected to the generator is in OFF position.



Step-6 Increase the supply voltage to motor’s stator winding, up-to the rated voltage.

Step-7 Turn on the DC-Voltage supply and adjust the excitation of DC-Machine until the

generator terminal voltages reach 220 V. Note that the DC-Machine is connected as

shunt excited generator.

Step-8 Record the values of line & load amps, W, torque, and speed in TABLE-1 of TEST RESULTS,

under “NO LOAD”.

Step-9 Turn ON the resistive load and adjust its value until load current reaches 0.25A.

Step-10 Readjust the generator’s field rheostat or the excitation supply, as required, to maintain a

terminal voltage of 220 volts. Then repeat Step 8.

Step-11 Now vary the resistive load and adjust its value until load current reaches 0.5A.

Step-12 Repeat step-10.

Step-13 Now vary the resistive load and adjust its value until load current reaches 1.0A.

Step-14 Repeat step-10





University College of Engineering & Technology, IUB φϋ

TEST RESULTS:

Load current NO LOAD 0.25 A 0.5 A 1.0 A

Line

voltage

V1

V2 V3

Line

current

I1

I2

I3

W1

W2

W3

TOTAL W

VA

TORQUE

SPEED

P.F = W / VA

TORQUE / A

Table-1

CALCULATIONS & REVIEW QUESTIOINS:

1. Add the wattmeter readings and record under TOTAL WATtS.

2. The equation for computing total apparent power input in voltamperes is:

VA = Line Volt x Line Amps x 1.73

Compute the volt amperes for each of the load steps and record under VA in TABLE-1

3. The motor power factor is the ratio of the true power (watts) to the apparent power (volt-

amperes). Perform this division for each of the load steps and record in TABLE-1.

4. Compute the torque per unit of the source current for each load step and record in Table-1.

Discuss how does the change in load affect it?

5. Using the data you have compiled in TABLE-1, plot three curves on the graphs provided.

a. Show how motor current changes as the torque output of the motor increases.

b. Show how speed changes as the torque output of the motor increases.

c. Show how the power factor changes as the torque of the motor increases.

6. Someone suggests that you buy a motor rated for twice the torque you need so as to be sure

you are not working the motor too hard. Discuss why you think this is or is not a good idea.

_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________




University College of Engineering & Technology, IUB φό

7. Over the operating range of an induction motor (no load to full load):

a. There is a large variation in speed.

b. There is absolutely no variation in speed.

c. There is a small variation in speed.

8. From no load to full load, there is:

a. Considerable change in power factor.

b. A small change in power factor.

c. Absolutely no change in power factor.

9. At no load, poor motor power factor is due to:

a. The high frequency of induced rotor voltage.

b. The rotor power factor.

c. The quadrature stator magnetizing current.

10. The mechanical power of the rotor is supplied by:

a. Active Power input to the stator.

b. Reactive Power input to the stator.

c. Apparent Power input to the stator.

11. The frequency and value of induced rotor voltage depends on:

a. The rotor speed only.

b. The difference between rotor speed and the speed of the revolving stator field.

c. Synchronous speed only.




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GRAPH:

No. 1

No. 2




University College of Engineering & Technology, IUB χτ

No. 3

FINAL CHECKLIST:





____________

Signature:

Lab Instructor




University College of Engineering & Technology, IUB χυ

EXPERIMENT # 7:

STARTING CHARACTERISTICS OF WOUND-ROTOR INDUCTION MOTORS


After successful completion of this experiment the student will be able to:

• Explain the addition of resistance to the rotor circuit of wound-rotor motors.

• Perform locked-rotor tests on wound-rotor motors.

• Determine the effect that rotor circuit resistance has on the starting torque of a wound-rotor

induction motor.

DISCUSSION:

The reason you don’t get more starting torque from a squirrel-cage induction motor is that it

has such a poor power factor. Its inductive reactance is high in comparison to its resistance. Of

course, squirrel-cage rotors could be made with greater resistance, but that would hurt the running

characteristics.

The rotor of a wound-rotor motor has coils. It is wound to have the same number of poles as

the stator. The rotor coils are Y-connected internally with the other end of each coil terminating

outside of the motor housing. An external variable resistance box is connected to the rotor terminals.This box is designed to add and remove resistance from the three phases simultaneously. Generally

speaking, the more resistance you have in the rotor circuit at start, the more starting torque you will

get for each ampere of starting current. The disadvantage is that power is being dissipated (lost)

outside of the motor. While it is improving starting current phase angle, the resistance also has the

effect of reducing rotor current. If you put in too much resistance, the lower current will wipe out the

advantage you had from improved power factor.

EQUIPMENT:

• Three Phase Wound-Rotor Induction Motor.


• Variable AC Supply.


• Induction Motor Starter.





University College of Engineering & Technology, IUB χφ

CONNECTION DIAGRAM:

FIGURE-1

PROCEDURE:



Step-2 Place the wound rotor induction meter on the machine bed on the left of torque speedmeasuring unit. Couple and clamp the machine securely & Install guards.

Step-3 Connect the circuit as shown in connection diagram.

Step-4 Install the rotor locking device securely on the torque-speed measuring unit.




Step-7 Set the Induction Motor Starter knob at R=0 Ω.

Step-8 Increase the supply voltage to motor’s stator winding slowly so that the rated current

flows through the winding.

-This step must be performed quickly. Voltage is to be applied to the motor for no longer

than 5-10 seconds. Turn the motor on and quickly read current and torque & record

these values in Table-1.

Step-9 Turn OFF the variable AC supply. Set the Induction Motor Starter knob to the 2

nd

position.

Step-10 Repeat Step-6.




University College of Engineering & Technology, IUB χχ

Step-11 Turn OFF the variable AC supply. Set the Induction Motor Starter knob to the 3rd

position.


Step-13 Turn OFF the variable AC supply. Set the Induction Motor Starter knob to the 4th

position.



TEST RESULTS:

STEP-1 (R= 0 Ω) STEP-2 STEP-3 STEP-4

VOLTS

AMPSTORQUE

TABLE-1

STEP-1 (R= 0 Ω) STEP-2 STEP-3 STEP-4

VOLTS

AMPS

TORQUE

Nm/Amp

TABLE-2 (FULL RATED VOLTAGE VALUES)

CALCULATIONS:

• Full voltage is 220 volts. Your tests were run at reduced voltage. Starting current would

therefore be different from the value you measured. Compute full voltage starting current

for each of the three resistance positions and record in TABLE-2.

• Torque is proportional to applied voltage squared. Therefore, full voltage starting torquewould also be different than the value you measured at reduced voltage. Compute full

voltage starting torque for each of the three resistance positions and record in TABLE-2.

• The reason for adding resistance during start was to get more torque for each ampere of

starting current. Compute the Nm per amp by dividing each of the full voltage starting

torques by their respective full voltage starting currents. Record these values in TABLE-2.




University College of Engineering & Technology, IUB χψ

REVIEW QUESTIONS:

1. Addition of rotor resistance tends to reduce starting torque. Why, then, is it used?

____________________________________________________________________________

____________________________________________________________________________

________________________________________________________________________

2. With maximum rotor resistance connected, the starting current is:

a. Greater than without the resistance.

b. Less than without the resistance.

c. The same as without the resistance.

3. As rotor resistance was increased, the starting torque;

a. Decreased.

b. Increased.

c. Remained the same.

4. up to a point, adding resistance to the rotor resistance to the rotor circuit provides:

a. More torque per ampere of starting current.

b. Less torque per ampere of starting current.

c. The same amount of torque per ampere of starting current.

5. The reason you add resistance is to:

a. Increase torque by increasing current.

b. Increase torque by improving power factor.

c. Increase torque per ampere by improving power factor.

6. Resistance in the rotor circuit:

a. Has no effect on current inrush.

b. Increases current inrush.

c. Decreases current inrush.

____________

Signature:

Lab Instructor




University College of Engineering & Technology, IUB χω

EXPERIMENT # 8:

SPEED CONTROL OF WOUND-ROTOR INDUCTION MOTORS.



1. Explain control systems which cut out starting rotor resistance as the motor gains speed.

2. Explain how rotor resistance controls speed of a wound rotor motor.

DISCUSSION:

At the instant of start, resistance in the rotor circuit prevents a large current inrush. The price

you pay for that is a reduced starting torque. This is compensated for by the fact that you get more

torque per ampere of starting current. Resistance accomplishes this by making the rotor field closer

to being in-phase with the stator field. In other words, it improves the power factor of the rotor.

The reason that the rotor has such a poor power factor at start is that the induced rotor frequency is

at a maximum i.e. equal to the frequency of the incoming power. Once the rotor starts turning,

however, this rotor frequency starts going down. With a wound rotor motor running without load,

the induced rotor frequency may be only 5 hertz or so. At that frequency, the rotor windings have

practically no inductive reactance. If you had resistance in series with the rotor windings, it would

not improve rotor power factor. All it would do is increase the losses in the rotor circuit.

Here is what happens: The motor, itself, automatically picks out the amount of slip it needs to

produce the rotor current that will drive the connected load at that speed. When rotor resistance is

added, the rotor starts losing some of its power to the resistance, it needs more slip so it can produce

more current from a higher induced rotor voltage.

That makes the rotor slow down. But the load hasn’t changed. Therefore, the rotor draws

enough current so it can produce extra torque. The load, remember, is proportional to torque times

speed. If the speed goes down, the torque has to go up since the load is constant. Rotor resistance,

then, can provide speed control of a wound rotor motor. It is not absolutely accurate, however,

because speed changes with load. There are bigger changes in speed with load if there is resistancein the rotor circuit.

EQUIPMENT:

• Three Phase Wound-Rotor Induction Motor.

• DC-Machine.




• Variable Resistor.





University College of Engineering & Technology, IUB χϊ

CONNECTION DIAGRAM:

Figure-1

PROCEDURE:



Step-2 Place the wound rotor induction motor on the machine bed on the left side of torque

speed measuring unit & DC-Machine on the right side. Couple and clamp the machines

securely & Install guards.

Step-3 Connect the circuit as shown in connection diagram.




Step-6 Set the resistance (connected to rotor) to its max. resistance position.

Step-7 Increase the supply voltage to motor’s stator winding, up-to the rated voltage & then

gradually reduce the resistance (connected to the rotor), to its minimum value.

Step-8 Turn on the DC-Voltage supply and adjust the excitation of DC-Machine to the rated

voltage. Note that the DC-Machine is connected as shunt excited generator.

Step-9 Load the generator by adjusting the variable resistor connected to it until the load

current reaches 1.5 A.

Step-10 Read torque, stator current, speed & rotor current and record these values in Table-1.




University College of Engineering & Technology, IUB χϋ

Step-11 Now increase the resistance (connected to rotor), to 10 Ω & by keeping the load

constant, repeat step-11.

Step-12 Increase rotor resistance in steps to 15 Ω & 20 Ω and for each step, record the

corresponding values of torque, stator current, speed and rotor current in Table-1.

Step-13 With rotor resistance at its max position (20 Ω), record the speed in Table-2.

Step-14 Reduce the load on the generator in steps & record motor speed in Table-2 as per given

load current.

Step-15 Repeat step-14 with rotor resistance at its minimum position.


TEST RESULTS:

ROTOR RESISTANCE:

0 Ω 10 Ω 15 Ω 20 Ω

SPEED

TORQUE

ROTOR CURRENT

STATOR CURRENT

Table-1

SPEED (rpm)

LOAD CURRENT: 1.5 A 0.5A CHANGE

MAX. ROTOR RESISTANCE:

(20 Ω)

MIN. ROTOR RESISTANCE:

(0 Ω)




University College of Engineering & Technology, IUB χό

REVIEW QUESTIONS:

1. As you added resistance externally to the rotor circuit, while the load remained constant,

what happened to the rotor speed and why?

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

2. As rotor speed decreased, what happened to the current being inducted into the rotor &

why?

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

3. As rotor speed decreased, what happened to the current being inducted into the rotor &

why?

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

4. Compute the change in speed (in Table-2) due to change in load step. Do this for both max.

resistance & zero resistance.

____________________________________________________________________________ ____________________________________________________________________________

____________________________________________________________________________

5. An increase in the external rotor resistance:

a. Increase wound-rotor motor speed.

b. Decrease wound-rotor motor speed.

c. Has no effect on wound-rotor motor speed.

6. When the rotor slows down, the rotor frequency:

a. Goes up.

b. Goes down.

c. Remains the same.

7. With resistance in the rotor circuit, as load changes:

a. There is no change in speed at all.b. There is a slight change in speed.

c. There is a noticeable change in speed.




University College of Engineering & Technology, IUB χύ

8. The larger the rotor resistance:

a. The less the rotor current.

b. The greater the rotor current.

c. No effect on rotor current.

9. Wound rotor motors are usually:

a. Started without external resistance in the rotor circuit, then cut in as the motor speeds up.

b. Started with some external resistance in the rotor circuit, then cut inemor as the motor

speeds up.

c. Started with maximum external resistance in the rotor circuit, then cut out as the motor

speeds up.

FINAL CHECKLIST:





____________

Signature:

Lab Instructor




University College of Engineering & Technology, IUB ψτ

EXPERIMENT # 9:

LOSSES AND EFFICIENCY OF INDUCTION MOTORS



1. Perform locked rotor tests and determine the equivalent resistance of induction motors.

2. Explain the source of losses and compute efficiency induction motors.

DISCUSSION:

There are two major classifications of losses in induction motors. The first is the copperlosses. These losses are electrical in nature and are due not only to the stator resistance, but the

referred resistance of the rotor as well the total equivalent resistance of the motor. It is this

equivalent resistance that must be multiplied times the current squared to determine the I2R copper

losses.

To find the equivalent resistance of a motor you must perform a “locked rotor” test. In this

test, the rotor of the induction motor is locked so that it cannot move. In this condition, there

cannot be any rotational losses. All of the electrical power must therefore be lost electrically. The

voltage is slowly increased until rated current flows. The power measurement at that point is used

to compute the equivalent resistance.

The second classification of losses is the rotational losses. Although you could use torque and

speed measurements to compute these losses, it is easier to measure the power input to an

unloaded motor. This power is made up of (1) the no-load copper losses plus (2) the rotational

losses. You can use the no-load current and the equivalent resistance to compute the no-load

copper losses. By subtracting this from the total power in, you have rotational losses.

Rotational losses tend to change with speed. However, speed you can consider them constant.

EQUIPMENT:

• Induction Motor (Squirrel-Cage)






• Resistive load.• Connecting Leads.




University College of Engineering & Technology, IUB ψυ

CONNECTION DIAGRAM:

Figure-1

Figure-2

PROCEDURE:



Step-2 Place the Induction Motor on the right side of the torque speed measuring unit. Couple

and clamp the machine securely & Install guards.






Step-7 Read and record the values of motor current and voltage in TABLE-1 of TEST RESULTS.

Step-8 Turn OFF all circuit breakers

Step-9 Install the rotor locking device securely on the torque-speed measuring unit.




University College of Engineering & Technology, IUB ψφ



Step-11 Slowly increase the output of the 0 - 220V AC supply until the the rated current of the

motor flows through it. Read the applied voltage, current and the wattage. Record these

values in TABLE-2 of TEST RESULTS.

Step-12 Turn OFF all circuit breakers.

Step-13 Now place the DC-Machine on the left side of the Torque-Speed Measuring Unit. Couple

it to the induction motor and clamp securely. Install guards..

Step-14 Make the connections as shown in Figure-2. Note that this is a separately excited shunt

generator connection. Adjust its field rheostat to its maximum resistance position,

fully clockwise.


Step-16 Turn on the DC-Voltage supply and adjust the excitation of DC-Machine until the terminal

voltage of generator become 220 Volts.

Step-17 Read the motor current and input watts and record these readings in TABLE-3.

Step-18 Turn ON the resistive load and adjust its value until 0.25A current flows through it.

Step-19 Readjust the generator’s field rheostat or the excitation supply, as required, to maintain

a terminal voltage of 220 volts. Then repeat Step 17.

Step-20 Now adjust the load resistor value until 0.5A current flows through it. Repeat Step-19.




TEST RESULTS:




University College of Engineering & Technology, IUB ψχ

VA INL W1 W2 W3 Total W PRL

Table-1

VA IFL W1 W2 W3 Total W REQ

Table-2

Load Current: NO-LOAD

Motor

Current:

W1 W2

W3

TOTAL W

PRL

PCL

TOTAL LOSSES

PL

% EFFICIENCY

Table-3

CALCULATIONS & REVIEW QUESTIOINS:

1. During the Locked Rotor Test, all of the input power was lost in the motor’s equivalent

resistance. The equation is:

W = 3 x (ILR)2

x REQ

Solving for REQ the equation becomes REQ = W/(3 x I2). From the data you recorded in Table-2

compute the total power in (TOT. W) and REQ. Record these values in TABLE-2.

2. During the No-Load-Test, the input power was lost in both the equivalent resistance and in

the rotational losses (PRL) To find the value of PRL first compute the total power in. Then

compute the total copper losses:

PCL = 3 x (INL)2

x REQ

Using the value of REQ = Total. W. PCL. Record this value of PRL Table-1 and for every load

listed in Table-3. (PRL assumed constant).

3. For each of the motor current values in TABLE-3, add the wattmeter readings to provide the




University College of Engineering & Technology, IUB ψψ

total power in. Record these values in TABLE-3.

4. For each of the motor current values in TABLE 16-3 compute the total copper loss from the

equation PCL = 3 x x REQ using the value of REQ from TABLE-2. Record the copper loss values

in TABLE-3. Add each of these values to the value computed in No. 2 to produce the

total loss value for each of the loads. Record these values in TABLE 16-3.

5. Compute efficiency, the ratio of output power to input power. For output power subtract

the losses from the output power. The equation is:

% Efficiency = Total Watts in - PL x 100

Total Watts in

List these efficiencies in TABLE-3.

6.

As load increases, rotational losses:

a. Increase.

b. Decrease.

c. Remain the same.

7. As load increases, copper losses:

a. Increase.

b. Decrease.

c. Remain the same.

8. As load increases, the total losses become:

a. A larger share of the total power in.

b. A smaller share of the total power in.

c. The same share of the total power in.

9. As load increase, the motor:

a. Operates more efficiently.b. Operates less efficiently.

c. Operates with the same efficiency.

10. Power Out equals:

a. The power in.

b. The total of copper and rotational losses.

c. The power in minus the total losses.




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FINAL CHECKLIST:





____________

Signature:

Lab Instructor




University College of Engineering & Technology, IUB ψϊ

EXPERIMENT # 10:

SATURATION CURVE OF AN ALTERNATOR.



1. Describe how voltage is generated in an armature and the effect of field current.

2. Perform saturation tests on alternators & to discover the effect of saturation on the terminal

voltage of an alternator.

DISCUSSION:

Alternators are designed to run at a specific speed (known as synchronous speed) to produce

voltage at a specific frequency. That’s why they are referred to as “synchronous alternators”.

Alternators are driven at synchronous speed by a prime mover. Typical prime movers are diesel

engines, jet engines, steam turbines, hydro-turbines, and DC motors.

The field coil of a synchronous alternator is wound on a rotor, which has salient poles. The

laminated iron core is called the “spider”. The armature coils are imbedded in slots on the stator. As

the rotor (field) is driven, its magnetic field sweeps around inside the housing. This moving field is

cut by the turns of the armature coil of the stator. This induces a voltage into the armature coils.

The amount of voltage induced depends on two things: (1) the speed of the rotor and (2) the

strength of the magnetic field. Magnetic field strength, in turn, depends on the amount of current

passing through the field coil. As the current increases, so does the field strength up to a point. That

point we call saturation. When the spider becomes magnetically saturated, further increase of field

current produce little or no further increase of field strength. Thus, the alternator’s terminal voltage

levels off.

EQUIPMENT:

• DC Machine (operating as a motor.)

• Synchronous Machine (operating as an alternator.)








University College of Engineering & Technology, IUB ψϋ

CONNECTION DIAGRAM:

Figure-1

PROCEDURE:



Step-2 Place the DC-Machine (operating as motor) on the machine bed on the left side of torquespeed measuring unit & Synchronous Machine on the right side. Couple and clamp the


Step-3 Connect the circuit as shown in connection diagram. Note that the DC-Machine is

connected as a self-excited shunt motor and a DC excitation is supplied to the field coil of

the alternator.



Step-6 Turn ON the main AC circuit breaker; the DC circuit breaker; and the motor.

Step-7 Slowly increase the output of the DC supply to 220 volts to start the motor.

Step-8 Then turn ON the DC excitation supply to the Alternator.

Step-9 Use the motor's field rheostat to adjust motor speed to 1500 RPM.

Step-10 Adjust the output of the alternator’s DC-excitation supply until the alternator's field

current is approximately 0.2 ampere.




University College of Engineering & Technology, IUB ψό

Step-11 Record the exact value of field current and terminal voltage in TABLE-1. Note that the

voltage across each of the three armature coils is the same, since they are being

produced by the same field.

Step-12 Repeat steps 10 and 11 for the following values of alternator field current: 0.4, 0.6, 0.8,

and 1.0 amperes.

Step-13 Slowly decrease the excitation voltage until the ammeter reads approximately 0.8 amps.

Record the exact value of field current and terminal voltage in TABLE-2.

Step-14 Repeat step 13 for the following approximate values of field current: 0.6, 0.4, 0.2, and 0

amperes.


TEST RESULTS:

INCREASING FIELD CURRENT:

FIELD CURRENT:

TERMINAL VOLTAGE:

Table-1

DECREASING FIELD CURRENT

FIELD CURRENT:TERMINAL VOLTAGE:

Table-2

REVIEW QUESTIONS:

1. Using the data you recorded in TABLE-1, plot a curve of Terminal Volts versus Field Amperes

on the graph provided at the end. Label this curve INCREASING.

2. Using the data you recorded in TABLE-2, plot a curve of Terminal Volts versus Field Amperes

on the same graph. Label this curve DECREASING.

3. What effect did saturation have on terminal voltage?

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________




University College of Engineering & Technology, IUB ψύ

4. From your observations would you regard residual magnetism in the rotor core an important

or unimportant factor? What led you to this conclusion?

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

5. Why is it good to use low hysteresis material for the rotor core?

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

6. Induced (generated) voltage in an alternator results from:

a. A stationary field and moving conductors.

b. A moving field and stationary conductors.

c. Both moving field and moving conductors.

7. You cannot have a self-excited alternator without rectifying the AC output, because:

a. The field poles must be constant North-South.

b. There is not enough residual magnetism to begin the generating process.

c. Armature coils must have DC applied to them.

8. Below the saturation point, as field current increases, terminal voltage:

a. Increases in direct proportion.

b. Decreases in direct proportion.

c. Stays the same.

9. Above saturation, increases in field current produce:

a. Larger increase in terminal voltage.

b. Smaller increases in terminal voltage.

c. No difference in terminal voltage.

10. When running at a constant speed, the terminal voltage of an alternator can be changed by:

a. Reversing the polarity of the field

b. Changing the direction of rotation.

c. Changing field current.




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GRAPH:




University College of Engineering & Technology, IUB ωυ

FINAL CHECKLIST:





____________

Signature:

Lab Instructor




University College of Engineering & Technology, IUB ωφ

EXPERIMENT # 11:

EFFECT OF SPEED ON ALTERNATOR.



1. Discover the effect of speed on the terminal voltage and frequency of generated voltage &

vary the output voltage and frequency of an alternator by changing speed.

2. Compute alternator frequency, given the number of poles and speed.

DISCUSSION:

The armature coils of an alternator are wound on the stator. There are no salient poles on

the stator; the conductors are imbedded into slots. The coils are arranged into distinct coil groups.

These coil groups are arranged on the stator to produce three-phase power. The coils are known as

A, B, and C.

Assume the magnetic field from the rotor is sweeping around. As it passes under Phase A

coil, Phase A voltage is at peak. 120 electrical degrees later, it passed under Phase B coil. Then, 120

electrical degrees later, it passes under Phase C coil. If there is only one group of three coils on the

stator, a full cycle is completed during one revolution of the rotor.

The rotor has one pair of salient poles (north and south) for each coil group on the stator. An

alternator is known by the number of rotor poles. Thus, the lowest number is 2. We call that a two-

pole alternator. The induction motor in our lab is a four pole machine. That means that two cycles of

AC are produced each time the rotor completes one revolution.

The frequency of the output voltage depends on the number of poles and the speed at

which the rotor is being driven.

Frequency (Hertz) = Speed (RPM) x (1minute/60seconds) x no. of pairs of poles

For example, a 2-pole (1 pair of poles) alternator driven at 3600 RPM has a frequency of

60Hz.

Speed affects terminal voltage also. The faster the rotor is driven, the faster the magnetic flux

lines get cut by the stationary armature conductors. Therefore, the larger the terminal voltage for

the same field current.




University College of Engineering & Technology, IUB ωχ

EQUIPMENT:







CONNECTION DIAGRAM:

Figure-1

PROCEDURE:







the alternator.








University College of Engineering & Technology, IUB ωψ




Step-10 Adjust the output of the alternator’s DC-excitation supply until the alternator is

generating 220 volts.

Step-11 Adjust the motor’s field rheostat until the motor is running at 1400 RPM.

Step-12 Record the terminal voltage in Table-1

Step-13 Repeat steps 11 & 12 for speed 1600 rpm.


TEST RESULTS:

SPEED:

TERMINAL VOLTAGE:

Table-1

REVIEW QUESTIONS:

1. Using the equation f = S x P where f is the frequency in hertz; S is speed in revolutions per

second; and P is the number of pair of poles, compute the frequency of the generated voltage

at 1800 RPM.

____________________________________________________________________________

2. Compute the frequency at 2000 rpm.

____________________________________________________________________________

3. Compute the frequency at 1600 rpm.

____________________________________________________________________________

4. What relationship did you observe between terminal voltage and speed?

____________________________________________________________________________

____________________________________________________________________________ ____________________________________________________________________________




University College of Engineering & Technology, IUB ωω

5. Is it better to energize the alternator’s field before or after it reaches full speed? What led

you to this conclusion?

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

6. When speed increases, the frequency:

a. Increases

b. Decreases

c. Remains the same

7. When Speed increases, the terminal voltage:

a. Increases

b. Decreases

c. Remains the same

8. To obtain a higher terminal voltage at the same frequency, you should increase:

a. Speed.

b. Armature Resistance

c. Field Current.

9. The terminal voltage of an alternator depends on:

a. Direction of Rotation.

b. Direction of Field Current.

c. Speed and Field Current.

10. In three-phase power, each phase voltage lags behind the one in front of it by:

a. 30 Electrical Degrees

b. 120 Electrical Degrees.

c. 360 Electrical Degrees.

FINAL CHECKLIST:





____________

Signature:

Lab Instructor




University College of Engineering & Technology, IUB ωϊ

EXPERIMENT # 12:

LOAD CHARACTERISTICS OF AN ALTERNATOR.



1. Explain why loading has an effect on terminal voltage. 2. Differentiate between unity, lagging, and leading power factor loads.

DISCUSSION:

If there is no load on an alternator, its terminal voltage depends solely on speed and field

current. However, load current flows through the armature coils making terminal voltage depend on

the nature of the load. In this experiment we will maintain a constant speed and a constant field

current while loading the alternator with a unity, leading, and lagging loads.

There are three reasons why terminal voltage is different from that generated. First is

armature resistance. Second is armature reaction. Third is armature reactance.

With a resistive (unity power factor) load, there is the voltage drop due to the armature coil’s

resistance. This IR drop increases as the load increases. Also there is the inductance of the armature

coil. This increases as the load increases. Armature reaction is the effect that the magnetic field of

the armature has on the main rotor field. It weakens the main field reducing the generated voltage.

It, too, acts like a voltage drop, increasing as the load increases.

When the load is inductive (lagging power factor) all three elements are still present.

However, the load current is already lagging terminal voltage. This doesn’t change the IR drop but it

does increase the effects of armature reactance and armature reaction.

With a capacitive (leading power factor) load, there is a completely different situation. You

still have the IR drop due to resistance, but the adds to the generated voltage instead of

subtracting from it. From Lenz’s law we know that inductive reactance tends to oppose whatever

causes it. Its cause in the alternator is the load current. The load current, however, leads the

generated voltage. If the angle of lead is great enough, the coil’s back-voltage (which is where

inductive reactance comes from) makes the terminal voltage larger than the generated voltage.

Armature reaction helps too. Instead of weakening the main field, it strengthens it. Therefore, with a

leading power factor load, terminal voltage increases as the load increases. Voltage regulation (V.R.)

is the ratio between the total drop in voltage and the full load voltage. The equation is:

% V.R. =No-load volts - Full load volts x 100

Full load volts




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EQUIPMENT:




• Variable DC Supply.


• Ammeters

• Volt meters

• Variable Resistive Load.

• Variable Inductive Load.

• Variable Capacitive Load.


CONNECTION DIAGRAM:

Figure-1

PROCEDURE:



Step-2 Place the DC-Machine (operating as motor) on the machine bed on the left side of torque

speed measuring unit & Synchronous Machine on the right side. Couple and clamp the





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the alternator. Connect the RLC load to the alternator’s terminals.



then switch on the variable DC-Voltage supply.

Step-6 Turn ON the main DC circuit breaker; the DC circuit breaker; and the motor.




Step-10 Adjust the output of the alternator’s DC-excitation supply until the alternator is

generating 220 volts.

Step-11 Turn on the resistive load & increase it gradually until the rated current flows through the

alternator stator windings. Re-adjust the alternator’s excitation supply until the

alternator’s terminal voltages are exactly 220 Volts.

Step-12 Repeat Steps 9 and 10 as necessary until the full load terminal volts of the alternator is

exactly 220 volts at 1500 RPM. (Note that the load should also be varied accordingly in

order to keep the load current equal to the alternator’s rated current.)

Step-13 Record the full load terminals as 220 Volts in Table-1.

Step-14 Remove the load from alternator terminals and repeat Step-9 to re-adjust motor speed

to 1500 RPM.

Step-15 Read the no-load terminal voltages at the alternator terminals and record these values in

Table-1.

Step-16 Connect the inductive load (2-steps) to alternator terminals in parallel with alreadyconnected resistive load, to produce lagging power factor and vary the resistive load

accordingly until the rated current flows through the alternator terminals.





Step-19 Remove the load from alternator terminals and repeat Step-9 to re-adjust motor speedto 1500 RPM.




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Table-1.

Step-21 Disconnect the inductive load & connect the capacitive load (2-steps) to alternator

terminals in parallel with already connected resistive load, to produce leading power

factor and vary the resistive load accordingly until the rated current flows through the

alternator terminals.





Step-24 Remove the load from alternator terminals and repeat Step-9 to re-adjust motor speed

to 1500 RPM.


Table-1.


TEST RESULTS:

TYPE OF LOAD:UNITY LAGGING LEADING

NO LOAD VOLTS

FULL LOAD VOLTS

VOLTAGE REGULATION

Table-1

CALCULATIONS:

1. From the data recorded in TABLE-1 compute the voltage regulation of this alternator when

connected to a unity power factor load.

2. From the data recorded in TABLE-l compute the voltage regulation of this alternator when

connected to a lagging power factor load.

3. From the data recorded in TABLE-1 compute the voltage regulation of this alternator when

connected to a leading power factor load.




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REVIEW QUESTIONS:

1. From your observations, was the percentage voltage regulation better or poorer when a reactive

load was connected? What led you to this conclusion?

_______________________________________________________________________________ _______________________________________________________________________________

_______________________________________________________________________________

2. From your observations, was the percentage voltage regulation better or poorer when a reactive

load was connected? Where does the additional voltage come from?

_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

3. As resistive load is added, terminal voltage:

a. Increases

b. Decreases

c. Remains the same

4. The power factor of the connected load:

a. Affects armature resistance and inductive reactance

b. Affects armature reactance and reaction.

c. Affects armature resistance and reactance.

5. As inductive load increases, terminal voltage:

a. Increases

b. Decreases

c. Remains the same

6. A zero percent voltage regulation means:

a. No-load voltage is greater than full-load voltage

b. No-load voltage is less than full-load voltage

c. No-load voltage is the same as full-load voltage




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7. As capacitive load increases, terminal voltage:

a. Increases

b. Decreases

c. Remains the same

FINAL CHECKLIST:





____________

Signature:

Lab Instructor




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EXPERIMENT # 13:

LOSSES & EFFICIENCY OF AN ALTERNATOR.



1. Identify the sources of losses in alternators.

2. Compute efficiency given the power out and the magnitude of the losses.

DISCUSSION:

Whenever we convert one form of energy into another there are bound to be losses. No

machine is perfect. Power is supplied to an alternator both in the form of electrical energy and in

the form of mechanical energy. The electrical energy is supplied to the field coil. This energy is used

to set up the main magnetic field. This field is constant. There is no energy taken from it in the

generation of electricity. Therefore, since none of the power out comes from this energy, the power

by the field must be counted as a loss.

Most of the power comes from the prime mover. Some of this mechanical power is lost to

the windings and friction of the alternator. The mechanical losses do not depend on the alternator’s

load. To find these losses, it is necessary to determine the overall mechanical losses then subtract

the losses of the prime mover.

Another class of losses that does not vary with load is the core losses. We are speaking here

about the armature’s core. Since there is an alternating voltage generated, the core is continually

becoming magnetized with one polarity, de-magnetized, then magnetized with the other polarity

each cycle. All of this magnetic activity in the core causes eddy current and hysteresis losses. These

core losses depend on the alternator’s voltage, not on load.

As load current flows through the armature coils the resistance of the wire causes a power

loss. This copper loss is proportional to the square of the current, P=I2R. Copper losses, therefore,

increase rapidly with load.

Percent efficiency is the ratio between the power out and the power in.

% Eff. = x100

If the load has unity power factor,

= V x I x 1.73.

Regardless of load,

= +Losses.




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EQUIPMENT:




• Variable DC Supply.


• Ammeters

• Volt meters

• Variable Resistive Load.

• Variable Inductive Load.

• Variable Capacitive Load.


CONNECTION DIAGRAM:

Figure-1

Figure-2




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Figure-3

PROCEDURE:



Step-2 Place the DC-Machine (operating as motor) on the machine bed on the left side of torque

speed measuring unit & Synchronous Machine on the right side. Clamp the machines

securely but do not couple them & Install guards.

A. ROTATIONAL LOSSES

Step-3 Connect the DC machine as a shunt motor as shown in Figure-1.




Step-6 Turn ON the main DC circuit breaker; the DC circuit breaker; and the motor.



Step-9 Read the motor’s voltage and current and record these readings in TABLE-1

Step-10 Turn OFF the main DC-Supply and motor circuit breaker switches.

Step-11 Multiply voltage and current read in Step-9 to compute rotational losses in motor.

Step-12 Couple the alternator to the motor. Clamp securely. Install guards.

Step-13 With no connections made to the alternator, repeat Steps 6, 7, 8, 9 and 10.

Step-14 Multiply voltage and current read in Step 13 to compute total rotational losses in the




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motor and alternator, PMAL. Record this value in TABLE-1.

Step-15 Compute the alternator’s rotational losses, PAL by subtracting PML from PMAL

Record in TABLE-1 and TABLE-5.

B. DETERMINE THE ARMATURE RESISTANCE

Step-16 Connect the DC-Supply to one of the alternator’s coils as shown in Figure-2.



Step-18 Slowly increase the voltage until the ammeter reads 1.0 amperes. Read the voltage and

record in TABLE-2.

Step-19 Turn off the circuit breaker switches and disconnect the leads from the alternator only.

Step-20 Compute armature resistance as follows:

a. RDC = V / I. Record in TABLE-2.

b. Multiply RDC times 1.5 to find the AC resistance of one coil. Record in TABLE-2.

c. Multiply the result of (b) times 3 to find the total armature resistance of the 3

coils. Record in TABLE-2.

C. DETERMINE THE FIELD LOSS.

Step-21 Connect the alternator field to the DC-Supply as shown in Figure-3. Be sure all of the

resistance toggle switches are in the downward (OFF) position.

Note that the alternator is wye-connected.

Step-22 Repeat steps-6, 7, and 8.

Step-23 Slowly increase the excitation voltage until the terminal voltage of the alternator is

220 volts.

Step-24 Read the field voltage and amps and record in TABLE-3.

Step-25 Multiply field volts and amps to compute field loss PFL. Record in TABLE -3 and TABLE-5.

D. CORE LOSSES

Step-26 Read the motor’s voltage and current and record these readings in TABLE-4.

Step-27 Multiply the voltage and current read in Step26 to compute total no-load losses,

PNLL. Record in TABLE-4.

Step-28 Compute the alternator’s core losses, PCL

by subtracting the total rotational losses

PMAL from the total no-load losses, PNLL. Record in TABLE-4 and TABLE-5.




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E. POWER OUT

Step-29 Add resistive load at the alternator terminals.


Step-31 Read the terminal voltage of the alternator and the load current. Record these values inTABLE -6.





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TEST RESULTS:

ROTATIONAL LOSSES:

V I V x I

UNCOUPLED MOTOR PML

=

COUPLED MOTOR PMAL=

PAL=PMAL-PML

Table-1

ARMATURE RESISTANCE:

I V RDC=V/I RDC x 1.5 ARMATURE RES.

Table-2

FIELD LOSSES:

CURRENT VOLTAGE PFL=V x I

Table-3

CORE LOSSES:

CURRENT VOLTAGE PNLL PCL=(PNLL-PMAL)

Table-4

TOTAL LOSSES:

ROTATIONAL (PAL)

FIELD (PFL)

CORE (PCL)

IALT

2

x R (PLOAD)TOTAL

Table-5

EFFICIENCY:

V I POUT PIN=POUT+LOSSES

EFFICIENCY:

Table-6




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CALCULATIONS:

1. From the current recorded in Step 29 and the armature resistance computed in Part B, compute

the full load armature loss Record this value in TABLE-5.

2. Add the rotational, field, core, and armature losses and record the value in TABLE-5.

3. Compute the power output from the equation:

= E x I x 1.73 Record in TABLE-6.

4. Add the losses computed in #2 to the power output computed in #3. Record the value in TABLE-6.

5. Compute the alternator’s efficiency from the equation:

Efficiency = Power Out x100

Power Out + Losses

Record your answer in TABLE-6.

REVIEW QUESTIONS:

1. Of the following types of losses, which one varies with load:

a. core loss

b. Copper loss in armature

c. Field loss

2. Of the following types of losses, which one is a mechanical loss:

a. Rotational loss

b. Copper loss in armature

c. Field loss

3. At low loads, the efficiency of an alternator is:

a. Greater than at rated load.

b. Less than at rated load.

c. The same as rated load.

4. Most of the electrical power delivered to the load is supplied to the alternator as:

a. Electrical energy to the field.

b. Electrical energy to the armature.

c. Mechanical energy in the form of torque on the rotor shaft.




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5. The sum of electrical power supplied to the load and alternator losses equals:

a. The total power lost in the motor generator set.

b. The power input to the alternator.

c. The power supplied to the prime mover.

FINAL CHECKLIST:





____________

Signature:

Lab Instructor




University College of Engineering & Technology, IUB ϋτ

EXPERIMENT # 14:

PARALLELING ALTERNATORS



1. Explain the conditions necessary to parallel alternators

2. Demonstrate the proper procedure for bringing an additional alternator on line. To learn the

technique of bringing an alternator on-line and having it assume a share of the load.

DISCUSSION:

Power companies usually have two or more alternators at each generating station. If

both alternators are “online”, their voltages, frequencies and phases are identical. Additionally, there

are usually a number of generating stations in any power system. The stations are also interlocked.

Then, a number of systems are tied together in a network, or power grid. This provides what is called

the “infinite bus”. When any alternator is brought on line, its voltage, frequency, and phase must

match those of the infinite bus. Once on-line, it is locked-in to the bus and can pick up its share of the

load. Just how much load is handled by each of the alternators is controlled by its set points.

To bring an alternator on line and parallel it with those already on-line, you first must have itspinning at the proper speed. This produces the proper frequency. Second, you must provide the

proper excitation. That produces the proper voltage level. Third, you must be sure the phase

sequence of the new alternator matches that of the on-line alternators.

Fortunately, there is a simple device to help with phasing and frequency. It consists of three

lamps, one in each phase, which operates from the difference in voltage. When the two voltages have

the same phase sequence and frequency, there is no difference between the voltages at any point in

any of the cycles. At that point, all lamps are dark.

If one frequency is greater, the lamps flash together at a rate equal to the difference infrequency. If either phase sequence is reversed, the lamps do not go bright and dark together, but

flash alternately.

In the first part of this experiment you will parallel the alternator with the voltage distribution

system in your lab. While in the second part, you will parallel two alternators




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PART. 1 – PARALLELING WITH DISTRIBUTION SYSTEM

EQUIPMENT:





• Synchronizing Device.




CONNECTION DIAGRAM:

Figure-1




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PROCEDURE:







the alternator.


Step-5 Turn on the main AC circuit breaker and measure the line voltage and frequency. Write itdown for future reference. Then turn power OFF.


then switch on the “variable” AC-Voltage supply.





Step-11 Adjust the output of the alternator’s DC-excitation supply until the terminal voltages of

alternator read exactly the same as the line voltages measured in step-5.

Step-12 At this point the lights on the SYNCHRONIZING DEVICE should be flashing OFF and ON

together. If they are flashing alternately turn the excitation supply to zero, interchange

any two leads from the alternator to the SYNCHRONIZING DEVICE. Then repeat Step-10,

11 & 12.

Step-13 When the lights are bright, the two voltages are 180” out of phase. When they are out,

the two voltages are exactly in phase. This turning ON & OFF alternately is due to the

difference in generated frequency and the line frequency. Adjust the speed of the motor

until the lights of the SYNCHRONIZING DEVICE are out (generated frequency matches

the line frequency).

Step-14 Repeat step-10 & 12 until the generated voltages and frequency matches exactly with

that of line voltages and frequency.

Step-15 Push the toggle switch of the SYNCHRONIZING DEVICE to CLOSE. Your alternator is now in

parallel with the supply and can supply power to it. Its speed and terminal voltage are

locked in and cannot be changed independently.




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Step-16 Read the current and power & record the readings in TABLE-1.

Step-17 Normally, adjusting the DC motor’s field rheostat would change motor speed. Attempt

to increase motor speed by turning the field rheostat knob clockwise about 45o

Read

and record in TABLE-1, the alternator’s terminal voltage, speed, current, and power.Return the knob to its original position.

Step-18 Repeat Step-17 for a counterclockwise rotation of about 45o. Use a minus sign if power is

being supplied to the alternator.

Step-19 Normally, adjusting the alternator’s field excitation would change terminal voltage.

Attempt to increase the voltage by increasing excitation current by 0.1 ampere. Read and

record in TABLE-1 the terminal voltage, speed, current, and power. Return to the original

excitation current value.

Step-20 Repeat Step-19 for a reduction of 0.1 amps in excitation current. Use a minus (-) sign if

power is being supplied to the alternator.

Step-21 Cut off the supply to the DC-motor i.e. there is no voltage being applied to the motor.

Make a note of what happens for the REVIEW QUESTIONS.


TEST RESULTS:

LINE VOLTAGE: ______________

FREQUENCY: ______________

TERMINAL

VOLTAGESPEED CURRENT POWER

STEP-16

STEP-17

STEP-18STEP-19

STEP-20

Table-1




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PART. 2 – PARALLELING TWO ALTERNATORS

EQUIPMENT:

• 2-DC Machines (operating as motors.)

• 2-Synchronous Machines (operating as alternators.)



• Synchronizing Device.


• Digital Multi-Meters.

• Resistive Load.

• Inductive Load.





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PROCEDURE:



Step-2 On each of the two machine beds, place the DC-Machine (operating as motor) on the left

side of torque speed measuring unit & Synchronous Machine on the right side. Couple

and clamp the machines securely & Install guards.

Step-3 Connect the circuit as shown in connection diagram. Note that the DC-Machines are

connected as self-excited shunt motors and a DC excitation is supplied to the field coil of

the alternators.


Step-5 Turn the field rheostat knob on each motor fully counterclockwise to their minimum

resistance position. All of the resistance toggle switches on the Load Banks should be

downward (OFF). Be sure the switch on the SYNCHRONIZING DEVICE is in the OPEN

position.

Step-6 Call the two Motor-Generator sets No. 1 and No. 2.

Start motor #l as follows:

Step-7 Make sure that the voltage regulator knob of the power supply is at its zero mark, andthen switch on the “variable” AC-Voltage supply.





Step-12 Adjust the output of the alternator’s DC-excitation supply until the terminal voltages of

alternator. 1 read 220 volts.

Step-13 Turn on the circuit breaker switch of Alternator No. 1 and apply a resistive load until the

ammeter reads approximately 0.25 amps.

Step-14 Readjust speed to maintain 1500 RPM by using the motor’s field rheostat. Also readjust

the excitation to maintain 220 volts output.

Step-15 Repeat Steps-6 to step-11 for Motor No. 2 and Alternator No. 2. Turn on the circuit

breaker switch on Alternator No. 2.




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Step-16 The phasing lamps on the SYNCHRONIZING DEVICE should now be flashing on and off

together. If they flash alternately, the phase rotation of the two alternators are not

identical. To correct this condition, simply interchange any two leads from the output of

Alternator No. 2.

Step-17 With the lamps flashing together, adjust the speed of Alternator No. 2 until the flashing

stops. If all lamps are lit, the voltages are 180° out of phase. Speed should be adjusted

until all lamps are out.

Step-18 Check to be sure the terminal voltage of both alternators is 220 volts. Then parallel the

alternators by closing the switch on the SYNCHRONIZING DEVICE. Alternator No. 2 is

now floating on the line.

Step-19 Adjust the field rheostat of Motor No. 2 until Alternator No. 2 is carrying half of the load.

Step-20 Reduce the speed of Alternator No. 1 with the Motor’s field rheostat until it is floating onthe line. Then turn OFF its circuit breaker switch.

Step-21 Reduce the load to zero. Turn OFF all circuit breaker switches. Disconnect all leads.

REVIEW QUESTIONS:

1. When the alternator was parallel with power lines, what changes in terminal voltage and

speed did you observe?

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

Explain the change or lack of change?

____________________________________________________________________________ ____________________________________________________________________________

____________________________________________________________________________

2. What happened when you removed power from the DC-motor driving the alternator?

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________




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3. Explain how you were able to cause the paralleled alternators to pick-up a greater share of

load?

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

4. What four conditions must match in order to bring an alternator on line?

a. ________________________________________________________________________

b. ________________________________________________________________________ c. ________________________________________________________________________

d. ________________________________________________________________________

5. When an alternator is “floating on the line, it is:

a. Supplying a portion of the load current.

b. Receiving AC power from the lines.

c. Neither supplying nor receiving AC power.

6. The speed of prime mover determines:

a. Frequency

b. Phase rotationc. Phase relationship with bus

7. The field excitation of the alternator determines:

a. Frequency

b. Voltage

c. Phase Rotation

8. The synchronizing (phasing) lamps operate from the difference between two voltages. When

they remain dark:

a. Voltages are equal and 180’out of phase.

b. Voltages are equal and in-phase.

c. Voltages are unequal and out of phase.

9. When an alternator is paralleled with the infinite bus, you cannot change:

a. The load current it supplies.

b. The power it supplies.

c. The frequency of its output.




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FINAL CHECKLIST:





____________Signature:

Lab Instructor




University College of Engineering & Technology, IUB ότ

EXPERIMENT # 15:

STARTING AND SYNCHRONIZING SYNCHRONOUS MACHINES



1. Explain the principle of synchronous motors.

2. Start and synchronize a synchronous motor.

DISCUSSION:

When three-phase is applied to the stator of a three-phase motor, a revolving stator

magnetic field is created. This field revolves at synchronous speed, which is a speed determined by

the number of poles per phase of the motor and the frequency of the incoming power. The rotor of

a synchronous motor becomes “locked in” on the revolving stator field. It then rotates at

synchronous speed. To accomplish this, the rotor contains a DC field winding.

The problem is in starting. If you have DC applied to the field coil, while the rotor is standing

still, the revolving field passes the stationary field much too fast to be locked onto. First the DC field

coils on the rotor must be made to rotate almost as fast as the revolving stator field. Then, when

you apply DC to it, the rotor is pulled into synchronism. That means that the rotor turns at

synchronous speed.

To get the rotor turning in the first place, a squirrel-cage winding is used. The bars are

embedded in the rotor core. When power is applied to the stator, the revolving field induces voltage

into these windings. In other words, a synchronous motor starts as an induction motor. When the

rotor reaches 95% of synchronous speed, DC is switched into the rotor field winding.

Now, during the start process, there will also be voltage induced into the DC field winding as

the rotor turns. Rather than have a charged-up field coil, its terminals are shorted through a resistor

while the squirrel-cage winding is getting the rotor started. Because synchronous motors must

achieve 95% synchronous speed before being synchronized, they are rarely started under load. Load

is applied after it is running as a synchronous motor. The rotor, however, continues to turn at

synchronous speed.

It is possible to load a synchronous motor beyond its ability to stay in synchronism. The

counter-torque of a load overcomes the torque (pull) on the rotor from the revolving stator field.

When that happens, the motor pulls out of step with the stator field. It will not simply fall back to




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running smoothly as an induction motor, however. Induced currents, added to the excitation

current in the DC field winding, make the rotor pulsate. Therefore, field excitation should be

removed as soon as possible after the rotor pulls out of synchronism. Then, if you want to re-

synchronize the motor, you must first remove the overload.

EQUIPMENT:


• Synchronous Machine (operating as motor.)





• Resistive load.


CONNECTION DIAGRAM:

Figure-1




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PROCEDURE:




torque speed measuring unit & Synchronous Machine on the right side. Couple and

clamp the machines securely & Install guards.


connected as a separately-excited shunt generator and a DC excitation is supplied to the

rotor of the synchronous machine. Make sure that the switch for the load connected to

the generator is in OFF position.


Step-5 With the motor switch OFF, turn ON the main AC and the 0-220 V DC-excitation supply.

Step-6 Adjust the excitation supply to the motor’s rotor until rated current flows in the field coil.

Step-7 Turn ON the motor circuit breaker and increase the supply voltage to 220 VAC.

Step-8 The synchronous motor is now running as an induction motor. Measure the speed, and

stator current. Record these values in Table-1.

Step-9 Then turn ON the DC excitation supply to the Synchronous Motor.

Step-10 Turn ON the DC supply to the generator field windings and adjust its output to rated

volts. Adjust the DC-generator’s field rheostat until the terminal voltage is 220 volts.

Step-11 Adjust the DC excitation current to the synchronous motor until the stator current is at its

lowest point.

Step-12 Measure speed and stator current. Record these values in Table-1.

Step-13 Turn ON the resistive load and increase it in steps. For each step, measure speed, statorcurrent and torque. Record these values in TABLE-2.

Step-14 While you are increasing the load in steps, there will be a value of load when the load

torque on the motor’s shaft increases the max. pull-out torque of the motor and the

motor gets pulled out of synchronism.

-immediately remove the DC-Excitation from the synchronous motor rotor.

Step-15 Now make an attempt to re-synchronize the motor by removing the load in steps.

Step-16 Make a note of load resistance when you get able to re-synchronize the motor.





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TEST RESULTS:

SPEED: STATOR CURRENT:

BEFORE SYNCHRONIZING:

AFTER SYNCHRONIZING:

Table-1

LOAD STEPS: 1 2 3 4 5 6 7 8

SPEED:

TORQUE:

STATOR AMPS:

Table-2




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REVIEW QUESTIONS:

1. The damper winding (squirrel-cage) starts the rotor and gets it up to 95% synchronous speed.

After DC is applied to the rotor field coil and the rotor pulls up to synchronous speed, what is

the job of the damper winding? Explain.

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

2. Why isn’t DC applied to the rotor field coil right away instead of waiting until the rotor is up to

speed?

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

3. What happened to rotor speed as you synchronized the rotor?

____________________________________________________________________________

____________________________________________________________________________

____________________________________________________________________________

What happened to stator current?

____________________________________________________________________________Explain why rotor speed changed the way it did.

____________________________________________________________________________

4. What was the “pull-out torque” for this motor? (Pull-out torque is the maximum torque the

motor will produce before dropping out of synchronism).

____________________________________________________________________________

5. When a synchronous motor is loaded, its speed:

a. Increases.

b. Decreases.


6. Normally, the stator current of a synchronous motor is:

a. Higher than that of an induction motor.

b. Lower than that of an induction motor.

c. The same as that of an induction motor.




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7. A synchronous motor will not “synchronize” if:

a. The stator has a revolving field.

b. The load on the rotor is too great.

c. DC is applied to the field coil on the rotor.

8. As the motor was loaded, stator current:

a. Increased.

b. Decreased.


9. You can easily tell when a synchronous motor drops out of synchronism, because the rotor

begins to:

a. Speed up.b. Pulsate.

c. Reverse direction.

FINAL CHECKLIST:



3. Submit your answers to the questions, together with your data, calculations, and resultsbefore the next laboratory session.

____________

Signature:

Lab Instructor




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EXPERIMENT # 16:

SYNCHRONOUS MOTOR V-CURVES



1. Demonstrate change in synchronous motor power factor.

2. Explain the effect of excitation current in terms of V-curves.

DISCUSSION:

Synchronous motors have the unique ability to run at different power factors.

Motors actually require electric power for two reasons. The first is to supply power current

that gets converted to mechanical power for the load and rotational losses. The second kind of

electrical input to a motor is the excitation current. Excitation current stores energy in the magnetic

field and releases it back to the source. Excitation current, which does no actual work, is ninety

degrees out of phase with power current.

Induction motors must draw both the power current and the excitation current from the AC

lines. That’s ’why typical induction motors operate with 0.8 lagging power factor.

Synchronous motors, on the other hand, have a separate source of excitation current. If you

wanted to supply less than normal excitation current to the DC field coil on the rotor, a synchronous

motor would run at 0.8 lagging power factor, the same as induction motors. This is seldom done,

however.

Instead the excitation current is increased to point where it magnetizes the rotor, stator, and

air gap so that no excitation current is taken from the AC lines at all. The entire stator current,

therefore, is converted to mechanical power. The synchronous motor has a unity power factor.

Excitation current can, however, be increased above normal. Now, not only does the motor

not take any excitation current from the AC lines, it actually supplied excitation current to the AC

lines. Typical synchronous motors can run at 0.8 P.F., leading.




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PROCEDURE:




torque speed measuring unit & Synchronous Machine on the right side. Couple and

clamp the machines securely & Install guards.


connected as a separately-excited shunt generator and a DC excitation is supplied to the

rotor of the synchronous machine. Make sure that the switch for the load connected to

the generator is in OFF position.



Step-6 Adjust the excitation supply to the motor’s rotor until rated current flows in the field coil.

Then, turn OFF the supply to the motor’s field.

Step-7 Turn ON the motor circuit breaker and increase the supply voltage to 220 VAC. The

synchronous motor is now running as an induction motor.

Step-8 Then turn ON the DC excitation supply to the Synchronous Motor. The motor should nowbe synchronized with the stator’s revolving field.

Step-9 With rated amperes flowing in the DC field, record stator current in TABLE-1.

Step-10 Reduce the value of rotor current by 0.1 amps. Read and record stator current in TABLE-l.

Step-11 Continue to repeat Step 10 until the motor pulls out of synchronism and for each step,

record stator current in TABLE-l.

-When the motor pulls out of synchronism, immediately remove the DC-Excitation from

the rotor.

Step-12 Turn ON the 220 volt DC supply to the field coil of the DC-Machine.

Step-13 Adjust the field rheostat of DC-Machine until its terminal voltage is 220 volts.

Step-14 Turn ON load to the DC-Generator and vary the load resistor value until 0.5 A current

flows through the load.

Step-15 Repeat step 13.

Step-16 Repeat Steps 9, 10, 11, 12, and 13 for TABLE-2.





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TEST RESULTS:

NO-LOAD

FIELD AMPS

STATOR AMPS

Table-1

0.5 A LOAD CURRENT

FIELD AMPS

STATOR AMPS

Table-2

REVIEW QUESTIONS:

1. From the data you recorded in TABLE-1, plot a curve on the graph provided showing how

stator current changes as the field excitation current changes with no load on the motor.

Label this curve NO LOAD.

2. From the data you recorded in TABLE-2, plot a curve on the same graph showing how stator

current changes with field excitation at 0.5A load current. Label this curve 0.5A LOAD

CURRENT.

3. Connect with a dotted line the lowest point of the three curves. Label this line UNITY POWER

FACTOR.

4. On the left side of the unity power factor line, label the area LAGGING POWER FACTOR. On

the right side of the line, label the area LEADING POWER FACTOR

5. As excitation current was decreased from 1.0 amps to pull-out, the stator current:

a. Went up then down.

b. Went down then up.

c. Remained the same.

6. At the lowest point on the stator current curve:

a. Current is leading voltage.

b. Current is lagging voltage.c. Current is in-phase with voltage.




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7. Normal excitation is when the synchronous motor has:

a. Unity power factor.

b. Leading power factor.

c. Lagging power factor.

8. Under-excitation produces a:




9. Over-excitation produces a:




GRAPH:

FINAL CHECKLIST:





____________

Signature:

Lab Instructor




University College of Engineering & Technology, IUB ύυ

EXPERIMENT # 17:

POWER FACTOR CORRECTION USING SYNCHRONOUS MOTORS.



1. Explain how synchronous motors improve power factor.

2. Connect a synchronous motor in a way that improves system power factor.

DISCUSSION:

Power factor, quite simply, is the ratio of power current to the total current.

An induction motor operates at less than unity power factor because it draws both power

current and excitation voltage from the AC power line. Factories that have a lot of induction motors

in use may have a low, lagging power factor.

Three main detrimental effects of a low, lagging power factor are:

1. Low power factor cuts down system loadability. That is, it reduces the capacity of the power

system to carry kilowatts. The capacity of all apparatus is determined by the KVA it can carry.

Hence, larger generators, transmission lines, transformers, feeders and switches must be

provided for each kilowatt of load when power factor is low than when it is high. Thus, capitalinvestment per kilowatt of load is higher.

2. Low power factor means more current per kilowatt. Hence, each kilowatt must carry a higher

burden of line losses, making it cost more to transport each kilowatt of power.

3. Low power factor may depress the voltage, reducing the output of practically all electrical

apparatus.

At low, lagging power factor also affects the following:

1. GENERATORS: Reduces generator capacity and efficiency.

2. TRANSFORMERS: Increases the voltage drop across transformers so that voltage regulationof the transformer is impaired.

3. DISTRIBUTION LOSSES: Makes larger distribution lines necessary and causes a greater

voltage drop in these lines.

4. POWER COST: A majority of power companies have penalties in their rates for low, lagging

power factor and incentives for high power factor. The customer thus ben- efits when he

keeps the power factor of his plant high.

Power factor can be corrected with a capacitor. However, if a synchronous motor is run with

a leading power factor, it can perform useful work and correct power factor at the same time. What

this means is that the exciting current, instead of flowing back and forth from induction motor to

power company, flows back and forth between the induction and synchronous motors.




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EQUIPMENT:

• 2-Induction Motors

• 2-DC Machine (operating as DC-Generator.)

• 1-Synchronous Machine (operating as motor.)





• 2-Resistive loads.


CONNECTION DIAGRAM:

Figure-1




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Figure-2

PROCEDURE:



Step-2 Place the DC-Machines (operating as DC-Generator) on the machine bed on the left side

of torque speed measuring unit & Induction Motors on the right side. Couple and clamp

the machines securely & Install guards. Name the two motor-generator sets as SET-1 &

SET-2.

Step-3 Connect the circuit as shown in connection diagram (Figure-1). Note that the DC-Machine

is connected as a separately-excited shunt generator. Make sure that the switch for the

load connected to the generator is in OFF position.



Step-6 Adjust the excitation supply to the DC-Machine’s (generator) field until rated current




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flows in the field coil. Then, turn OFF the supply to field.

Step-7 Turn ON the motor circuit breaker and increase the supply voltage to 220 VAC.

Step-8 Then turn ON the DC excitation supply to the Generator.

Step-9 Adjust the field rheostat until the terminal voltages of generator equals 220V.

-(Repeat these steps for both the motor-generator sets)

-Note that both that sets are fed from the same source.

Step-10 Now turn on the loads connected to the generator terminals for both the sets one after

another. Adjust the load resistor value until rated current flows through the induction

motor. Fix the load resistor value, i.e. load is not to be changed during the next

procedure.

Step-11 Read and record the value of both the motor’s voltages, currents and power & the value

of power, voltage and current drawn from the main source in Table-1.

Step-12 Turn OFF the Main AC & DC supply.

Step-13 Now replace the induction motor in the SET-2 with an equal power rated Synchronousmotor. Connect the circuit diagram as shown in Figure-2.


Step-15 Repeat steps 5-9 for the motor generator SET-1

Step-16 For starting and running synchronous motor in motor-generator SET-2, adapt the

procedure as demonstrated in the previous experiment.

Step-17 Now turn on the loads for both motor-generator sets.

-the load value must be the same as that of the previous procedure.

Step-18 Read and record the value of both the motor’s voltages, currents and power & the value

of power, voltage and current drawn from the main source in Table-2.

-mention the field current of the synchronous motor on the top of the respective table.





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TEST RESULTS:

WITH INDUCTON MOTORS

SOURCE

VOLTAGE

SOURCE

CURRENT

SOURCE

POWER

MOTOR-1

VOLTAGE

MOTOR-1

CURRENT

MOTOR-1

POWER

MOTOR-2

VOLTAGE

MOTOR-2

CURRENT

MOTOR-2

POWER

Table-1

WITH SYNCHRONOUS MOTOR

____ A FIELD CURRENT

SOURCEVOLTAGE

SOURCECURRENT

SOURCEPOWER

MOTOR-1VOLTAGE

MOTOR-1CURRENT

MOTOR-1POWER

MOTOR-2VOLTAGE

MOTOR-2CURRENT

MOTOR-2POWER

Table-2

REVIEW QUESTIONS:

1. From the data you recorded in TABLE-1, Compute the system power factor.

P.F = WATTS/VOLT-AMPERES

__________________________________________________________________________________



__________________________________________________________________________________



__________________________________________________________________________________

4. Was there much improvement in power factor when the synchronous motor was put

into the circuit? Explain the change.

__________________________________________________________________________________

__________________________________________________________________________________

__________________________________________________________________________________




5. Induction motors typically have:

a. A high leading power factor.

b. A low leading power factor.

c. c) A low lagging power factor.

6. Excitation current furnishes:

a. Total apparent power.

b. Real (true) active power.

c. Reactive (wattless) power.

7. When the excitation current for an induction motor comes from the AC lines:

a. Excitation current makes up a large share of the total current.

b. Excitation current makes up a small share of the total current.

c. None of the total current is made up of excitation current.

8. When some of the excitation current for an induction motor comes from a synchronous

motor:

a. Excitation current makes up a large share of the total current.

b. Excitation current makes up a small share of the total current.

c. None of the total current is made up of excitation current.

9. Synchronous motors usually run at:

a. Unity or lagging power factor.

b. Unity or leading power factor.

c. Lagging or leading power factor.

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