Energy Conversion Lab Experiment

The Single-Phase Transformer

2010-01-25T11:19:00.001-08:00

Objective:

To study the voltage and current ratios of a transformer.
To learn about transformer-exciting currents, volt-ampere capacity and short-circuit currents.

Discussion:

Transformers are probably the most universally-used pieces of equipment in the electrical industry. They range in size from miniature units in transistor radios to huge units, weighing tons, used in central power distributing stations. However, all transformers have the same basic properties which you are about to examine.

When mutual induction exists between two coils or windings, a change in current through one induces a voltage in the other. Every transformer has a primary winding and one or more secondary windings. The primary winding receives electrical energy from a power source and couples this energy to the secondary winding by means of a changing magnetic field. The energy appears as an electromotive force across the secondary winding, and if a load is connected to the secondary, the energy is transferred to the load. Thus, electrical energy can be transferred from one circuit to another, with no physical connection between the two. Transformers are indispensable in AC power distribution, since they can convert electrical power at a given current and voltage into an equivalent power at some other current and voltage.

When a transformer is in operation, AC currents flow in its windings and an alternating magnetic field is set-up in the iron core. As a result, copper and iron losses are produced which represents active power (watts) and causes the transformer to heat up. Establishing a magnetic field requires reactive power (var) which is drawn from the power line. For these reasons the total power delivered to the primary winding is always slightly larger than the total power delivered by the secondary winding. However, we can say, to a good approximation, that in most transformers:

a) Primary Power (watts) = Secondary Power (watts)

b) Primary Volt-amperes (VA) = Secondary Volt-amperes (VA)

c) Primary vars = Secondary vars.

When the primary voltage is raised beyond its rated value, the iron core (laminations) begins to saturate, and the magnetizing (exciting) current increases rapidly.

Transformers are subject to accidental short-circuits caused by natural and man-made disasters. The short-circuit currents can be very large and, unless interrupted, will quickly burn out a transformer. It is the purpose of this Experiment to show these major points.

Equipment Required:

A Single Phase Transformer
Power Supply
AC Ammeter
AC Voltmeter
Wires

Procedure:

CAUTION!!!

High voltages are Present In the Experiment! Do not make any connections with the power on! The power should be turned off after completing each individual measurement!!!

o 1. Examine the construction of the Single-Phase Transformer paying particular attention to the transformer, connection terminals and the wiring.

The transformer core is made up of thin sheets (lamentations) of steel. Identify it.

Note that the transformer windings are brought out to terminals mounted on the transformer coil.
Note that these windings are then wired to the connection terminals mounted on the module face.

o 2. Identify the three separate transformer windings maked on the module face:

List the rated voltage for each of the three windings:

Terminals 1 to 2 = 220 V ac

Terminals 3 to 4 = 380 V ac

Terminals 5 to 6 = 220 V ac

List the rated voltage between the following connection terminals:

Terminals 3 to 7 = 190 V ac

Terminals 7 to 8 = 139 V ac

Terminals 8 to 4 = 51 V ac

Terminals 3 to 8 = 329 V ac

Terminals 7 to 4 = 190 V ac

Terminals 5 to 9 = 110 V ac

Terminals 9 to 6 = 110 V ac

List the rated current for each of the following connections:

Terminals 1 to 2 = 0.25 A ac

Terminals 3 to 4 = 0.15 A ac

Terminals 5 to 6 = 0.25 A ac

Terminals 3 to 7 = 0.15 A ac

Terminals 8 to 4 = 0.15 A ac

o 3. Using the lowest range of our ohmmeter, measure and record the DC resistance of each winding:

Terminals 1 to 2 = 880 Ω

Terminals 3 to 4 = 2533.333 Ω

Terminals 3 to 7 = 1266.667 Ω

Terminals 7 to 8 = 926.667 Ω

Terminals 8 to 4 = 340 Ω

Terminals 5 to 6 = 880 Ω

Terminals 5 to 9 = 440 Ω

Terminals 9 to 6 = 440 Ω

o 4. Now we measure the unloaded secondary voltages with 220 V ac applied to the primary winding.

a. Connect the circuit shown in Figure-1.

b. Turn on the power supply and adjust for 220 V ac as indicated by the voltmeter across the power supply terminals 4 to N.

c. Measure and record the output voltage E₂.

d. Return the voltage to zero and turn off the power supply.

e. Repeat ( b, c and d) measuring the output voltage E₂ for each of the listed windings.

f. Winding 1 to 2 = 220 V ac

Winding 3 to 4 = 380 V ac

Winding 5 to 6 = 220 V ac

Winding 3 to 7 = 185 V ac

Winding 7 to 8 = 138 V ac

Winding 8 to 4 = 50 V ac

Winding 5 to 9 = 108 V ac

Winding 9 to 6 = 106 V ac

o 5. a. We measured voltage correspond well with the rated values. But we have very something difference between the rated values and the measured values due to different kind of losses which is negligible.

b. The current for which the magnetic flux produced is called magnetizing current. The current I₁ is associated with the magnetic flux so here I₁ is magnetizing current. We measure it.

o 6. Windings 1 to 2 and 5 to 6 each have 1100 turns of wire. Winding 3 to 4 has 1900 turns. Calculate the following turn ratios:

o 7. a. Connect the circuit shown in Figure-2. Notice that current meter I₂ short-circuits winding 5 to 6.

b. Turn on the power supply and gradually increase the voltage until the short-circuit current I₂ is 0.2 A ac.

c. Measure and record E₁, I₁, and I₂.

I₁ = 0.4 A ac

E₁ = 30 V ac

I₂ = 0.2 A ac

Return the voltage to zero and turn off the power supply.
Calculate the current ratio:

For a transformer the turn and current ratio are inversely equal. So theoretically the voltage and current ratio is not equal. Because-

Figure - 3

o 8. a. Connect the circuit shown in Figure-3. Notice that winding 3 to 4 is short-circuited by the current meter I₃.

b. Turn on the power supply and gradually increase the voltage until the current through the primary winding I₁, is 0.2 A ac.

c. Measure and record I₃ and E₁.

I₃ = 0.12 A ac

E₁ = 18 V ac

d. Return the voltage to zero and turn off the power supply.

e. Calculate the current ratio:

o 9. Now we determine the effect of core saturation upon the exciting current of a transformer.

a. Connect the circuit shown in Figure-4. Notice that power supply terminals 4 and 5 are now being used. These terminals will furnish variable 0 to 380 V ac.

b. Turn on the power supply and adjust for 50 V ac as indicated by the voltmeter across power supply terminals 4 to 5.

c. Measure and record the exciting-current I₁ and the output voltage E₂ for each of the input voltages listed in Table-1.

d. Return the voltage to zero and turn off the power supply.

E₁ V ac	I₁ mA ac	E₂ V ac
50	0	45
100	0	95
150	0	145
200	0	198
250	0.1	240
300	0.2	298
350	0.3	340
400	0.5	378

o 10. a. Plot our recorded current values on the graph of Figure-5. Draw a smooth curve through our plotted points.

b. Note that the magnetizing current increase rapidly after a certain input voltage has been reached.

c. When the voltages increase regularly; the current is increase. One time they are in saturation point.

Transformer Regulation

2010-01-25T10:02:00.000-08:00

Objective:

To study the voltage regulation of the transformer with varying loads.
To study transformer regulation with inductive and capacitive loading.

Discussion:

The load on a large power transformer in a sub-station will vary from a very small value in the early hours of the morning to a very high value during the heavy peaks of maximum industrial and commercial activity. The transformer secondary voltage will vary somewhat with the load and, because motors and incandescent lamps and heating devices are all quite sensitive to voltage changes, transformer regulation is of considerable importance. The secondary voltage is also dependent upon whether the power factor of the load is leading, lagging or unity. Therefore, it should be known how the transformer will behave when it is loaded with a capacitive, an inductive or a resistive load.

If a transformer were perfect (ideal) its windings would have no resistance. Furthermore, it would require no reactive power (vars) to set up the magnetic field within it. Such a transformer would have perfect regulation under all load conditions and the secondary voltage would remain absolutely constant. But, practical transformers do have winding resistance and they do require reactive power to produce their magnetic fields. The primary and secondary winding possess, therefore, an overall resistance R and an overall reactance X. The equivalent circuit of a power transformer having a turn ratio of 1 to 1, can be approximated by the circuit shown in Figure-1. The actual transformer terminals are P₁ P₂ on the primary side and S₁ S₂ on the secondary.

In between these terminals we have shown the transformer as being composed of a perfect (ideal) transformer in series with impedance consisting of R and X, which represents its imperfections. It is clear that if the primary voltage is held constant, then the secondary voltage will vary with loading because of R and X.

An interesting feature arises with a capacitive load, because partial resonance is set up between the capacitance and the reactance X so that the secondary voltage E₂ may actually tend to rise as the capacitive load value increases.

Equipment Required:

A Single Phase Transformer
Power Supply
Resistive, Capacitive and Inductive Load
AC Ammeter
AC Voltmeter
Wires

Procedure:

CAUTION!!!

High voltages are Present In the Experiment! Do not make any connections with the power on! The power should be turned off after completing each individual measurement!!!

1. Using Single-Phase Transformer, Power Supply, Resistive Load, AC Ammeter and AC Voltmeter, connect the circuit shown in Figure-2.

2. a. Place all of the Resistive Load switches in their open position for zero load current.

b. Turn on the power supply and adjust for exactly 220 V ac as indicated by voltmeter E₁.

c. Measure and record in Table-1 the input current I₁, the output current I₂ and the output voltage E₂.

d. Adjust the load resistance Z_L to 4400 Ω. Make sure that the input voltage remains at exactly 220 V ac. Measure and record I₁, I₂ and E₂.

e. Repeat (d) for each of the listed values in Table-1.

f. Return the voltage to zero and turn off the power supply.

Z_L (ohms)	I₂ (mA ac)	E₂ (V ac)	I₁ (mA ac)
Infinitive	0	210	0
4400	0.05	215	0.06
2200	0.095	210	0.11
1467	0.14	200	0.15
1100	0.18	198	0.19
880	0.225	195	0.23

3. a. Calculate the transformer regulation using the no-load and full-load output voltage from Table-1.

b. For different value of load resistance, we can see the primary and secondary winding VA is not equal. For fixed value of E₁ is 220 V ac and for decreasing load resistance, the primary winding current I₁ increasing, the secondary winding current I₂ increasing and voltage E₂ decrease. For every value of load resistance, we can measure the primary winding VA always greater then the secondary winding VA.

4. a. Repeat procedure 2 using the Inductive Load in place of the Resistive Load.

b. Record our measurements in Table-2.

Z_L (ohms)	I₂ (mA ac)	E₂ (V ac)	I₁ (mA ac)
Infinitive	0	210	0
4400	0.05	210	0.06
2200	0.10	200	0.11
1467	0.14	198	0.15
1100	0.17	193	0.19
880	0.22	185	0.23

5. a. Repeat procedure 2 using the Capacitive Load in Place of the Resistive Load.

b. Record our measurements in Table-3.

Z_L (ohms)	I₂ (mA ac)	E₂ (V ac)	I₁ (mA ac)
Infinitive	0	210	0
4400	0.055	215	0.05
2200	0.11	223	0.12
1467	0.155	230	0.16
1100	0.23	235	0.21
880	0.25	240	0.25

6. Now we construct an output voltage E₂ vs output current I₂ regulation curve for each type of transformer load.

a. Plot recorded values of E₂ (at each value of I₂ listed in Table-1) on the graph of Figure-1.

b. Draw a smooth curve through plotted points. Label this curve “Resistive Load”.

c. Repeat (a) for the Inductive (Table-2) and Capacitive (Table-3) Loads. Label these curves “Inductive Load” and “Capacitive Load”.

The AutoTransformer

2010-01-25T08:59:00.001-08:00

Objective:

To study the voltage and current relationship of an autotransformer.
To learn how to connect a standard transformer as an autotransformer.

Discussion:

There is a special type of transformer which physically has only one winding. Functionally, though, the one winding serves as both the primary and secondary. This type of transformer is called an autotransformer. When an autotransformer is used to step up the voltage, part of the single winding acts as the primary, and the entire winding acts as the secondary. When an autotransformer is used to step down the voltage, the entire winding acts as the primary, and part of the winding acts as the secondary.

Figure-1 and Figure-2 show autotransformers connected for both step-up and step-down operation.

The action of the autotransformer is basically the same as the standard two-winding transformer. Power is transferred from the primary to the secondary by the changing magnetic field, and the secondary in turn, regulates the current in the primary to set up the required condition of equal primary and secondary power. The amount of step-up or step-down in voltage depends on the turn’s ratio between the primary and secondary, with each winding considered as separate, even though some turns are common to both the primary and secondary.

Voltages and currents in the various windings can be found by two simple rules:

a) Primary apparent power (VA) equals Secondary apparent power (VA).

(VA)_P = (VA)_S---------------- (1)

E_PI_P = E_SI_S------------------ (2)

b) The primary (source) voltage and the secondary (load) voltage are directly proportional to the number of turns N.

These equations depend upon one important fact, that voltage E_{A to B}and E_{B to C} add in the same direction and do not oppose each other. We have assumed that the voltages are in phase.

The load current, of course, cannot exceed the current carrying capacity of the winding. Once this is known it is relatively easy to calculate the VA load which a particular autotransformer can supply.

A disadvantage of the autotransformer is the lack of isolation between the primary and secondary circuits, because the primary and secondary both use some of the same turns.

Equipment Required:

A Single Phase Transformer
Power Supply
Resistive Load
AC Ammeter
AC Voltmeter
Wires

Procedure:

CAUTION!!!

High voltages are Present In the Experiment! Do not make any connections with the power on! The power should be turned off after completing each individual measurement!!!

1. Using a Single-Phase Transformer, Power Supply, Resistive Load, AC Ammeter and AC Voltmeter, connect the circuit shown in Figure-3. Note that winding 5 to 6 is connected as the primary winding across the 220 V ac source. The centre tap of the winding, terminal-9 is connected to one side of the load and the 6 to 9 portion of the primary winding is connected as the secondary winding.

2. a. Place all of the Resistive Load switches in their open positions for zero loads current.

b. Turn on the power supply and adjust for exactly 220 V ac as indicated

by voltmeter E₁. (This is the rated voltage for winding 5 to 6).

c. Adjust the load resistance R_L to 4400 Ω.

d. Measure and record currents I₁, I₂ and the output voltage E₂.

I₁ = 0.01 A ac

I₂ = 0.02 A ac

E₂ = 105 V ac

e. Return the voltage to zero and turn off the power supply.

3. a. Calculate the apparent power in the primary and secondary circuits.

E₁ (220) × I₁ (0.01) = 2.2 (VA)_P

E₂ (105) × I₂ (0.02) = 2.1(VA)_S

b. The primary and secondary apparent powers are not equal. Because we can see from Figur-1; the primary winding turns is greater than the secondary winding turns. So the secondary winding apparent power is less than the primary winding apparent power. This is a step-down Autotransformer.

4. Connect the circuit shown in Figure-3. Notice that winding 6 to 9 is now connected as the primary winding across the 110 V ac source. The 5 and 6 winding is now connected as the secondary winding.

5. a. Place all of the Resistive Load switches in their open positions for zero load current.

b. Turn on the power supply and adjust for exactly 110 V ac as indicated by voltmeter E₁.(This is the rated voltage for winding 6 to 9.

c. Adjust the load resistance R_L to 2200 Ω.

d. Measure and record currents I₁, I₂ and the output voltage E₂.

I₁ = 0.01 A ac

I₂ = 0.10 A ac

E₂ = 210 V ac

e. Return the voltage to zero and turn off the power supply.

6. a. Calculate the apparent power in the primary and secondary circuits.

E₁ (110) × I₁ (0.01) = 1.1 (VA) _P

E₂ (210) × I₂ (0.10) = 21(VA)_S

b. The primary and secondary apparent powers are not equal. Because we can see from Figur-2; the primary winding turns is less than the secondary winding turns. So the secondary winding apparent power is greater than the primary winding apparent power. This is a step-up Autotransformer.

Transformers in Parallel

2010-01-25T07:14:00.000-08:00

Objective:

· To learn how to connect transformers in parallel.

· To determine the efficiency of parallel-connected transformers.

Discussion:

Transformers may be connected in parallel to furnish load currents greater than the rated current of each transformer. There are two precautions to be observed when connecting transformers in parallel.

1. The windings to be paralleled must have identical output voltage ratings.

2. The windings to be paralleled must have identical polarities.

Very large short-circuit currents can be developed if these rules are not followed. In fact, transformers, circuit breakers and associated circuitry can be severely damaged, or may even explode, if these short-circuit currents are large enough.

The efficiency of any machine or electrical device is given by the ratio of output power to input power. (Apparent power and reactive power are not used in calculating transformer efficiency). The equation for percent efficiency is:

Equipment Required:

A Single Phase Transformer
Resistive Load
Power Supply
AC Ammeter
AC Voltmeter
Wires

Procedure:

CAUTION!!!

High voltages are Present In the Experiment! Do not make any connections with the power on! The power should be turned off after completing each individual measurement!!!

1. Using your Single-Phase Transformer, Power Supply, Single-Phase Wattmeter, Resistive Load, AC Ammeter and AC Voltmeter, connect the circuit shown in Figure – 1. Note that the two transformers are connected in parallel. The primary windings (1 and 2) are connected together to the 220 V ac power source. The wattmeter will indicate the input power. Each secondary winding (3 to 4) is connected in parallel to the load R_L. Ammeters are inserted to measure load current I_L and transformer secondary currents I₁ and I₂.

Connect two resistance sections of the Variable resistance module in series to implement the resistive load R_L shown in Figure – 1. This is required to dissipate the large amount of power involved in this exercise.

2. Place all the resistance switches in their open positions for zero load current. Note that the windings are connected for voltage step-up operation (220 V primary to 380 V secondary).

3. Please your circuit wiring approved by the instructor before proceeding.

4. a. Turn on the power supply and slowly advance the voltage output control knob while noting the transformer secondary current meters I₁, if the windings are properly phased, no load or secondary currents should be flowing.

b. Adjust the power supply voltage to 220 V ac as indicated by the voltmeter connected across the wattmeter.

5. a. Gradually increase the load R_L until the load current I_L equals 250mA ac. Check to see that the input voltage is exactly 220 v ac.

b. Measure and record the load voltage, load current, transformer secondary currents and the input power.

E_L = 340 V ac

I_L = 0.24 A ac

I₁ = 0.12 A ac

I₂ = 0.12 A ac

Pin = 95 W

c. Return the voltage to zero and turn off the power supply.

6. a. Calculate the load power,

E₁ (340) × I_L (0.24) = 81.6 W

b. Calculate the circuit efficiency,

c. Calculate the transformer losses

Pin (95) – Pout (81.6) = 13.4 W

d. Calculate the power delivered by transformer-1.

I₁ (0.12) × E_L (340) = 40.8 W

e. Calculate the power delivered by transformer – 2.

I₂ (0.12) × E_L (340) = 40.8 W

7. Is the load reasonably distributed between the two transformers? Yes.

Review Questions:

1. Show how you would parallel connect the transformers to the source and the load in Figure – 2. Windings 1 to 2 and 3 to 4 are rated for 2.4 KV ac and windings 5 to 6 and 7 to 8 are rated for 400 V ac.

Figure – 3

In a parallel connection of two transformers, should be properly connected with regard to polarity. The percentage impedance should be equal in magnitude. The regulation must be same.

2. The efficiency of transformers which supplies a pure capacitive load is zero. Explain.

Ans: We know the efficiency of a transformer,

For capacitive load, Pout = VI cosθ = VI cos 90^o = 0

3. Name the losses which cause a transformer to heat up.

Ans: The I²R losses that means copper losses and core losses are cause a transformer to heat up.

4. How does the efficiency of our Single-Phase Transformer compare to the efficiency of DC Motor/Generator? Explain.

Ans: In DC generator or motor; there is a friction loss which plays an important role on efficiency calculation. But in Transformers there is no friction loss, so transformer is more efficiency than a dc motor or generator of same rating.

Three-Phase to Two-Phase Conversion (Transformers)

2010-01-25T04:45:00.000-08:00

Objective:

· To connect two transformers in such a way as to convert a three-phase system to a two-phase system.

· To show that a balanced two-phase system gives rise to a balanced three-phase system.

Discussion:

Two-Phase power was wide-spread in industry some time ago, but has since been replaced by three-phase power. However, there still exist requirements for two-phase power, principally in the small motor control field. It is useful, therefore, to know how to convert a three-phase, or from three-phase to two-phase, with or without a change in voltage, by means of special transformers which have the required ratio of transformers. In this regard, the Scott connection is probably the best known, and it is the one we will examine here.

In a balanced, three-phase system, the voltages and currents are all equal and displaced by 120^o from each other. In a balanced two-phase system, the voltages and currents are also equal but are displaced only 90^o from each other. The voltage is, therefore, maximum in one phase when it is zero in the other. Similarly, the current is maximum in one phase when is zero in the other.

Equipment Required:

A Single Phase Transformer
Resistive Load
Power Supply
AC Ammeter
AC Voltmeter
Wires

Procedure:

CAUTION!!!

High voltages are Present In the Experiment! Do not make any connections with the power on! The power should be turned off after completing each individual measurement!!!

Figure – 1

1. Using your Single-Phase Transformer, Resistive Load, Power Supply, AC Ammeter and AC Voltmeter, connect the circuit shown in Figure – 1. Note the connections to taps 3 and 6 on T₁ and taps 3, 7 and 4 on T₂. This particular configuration is known as the Scott connection. The winding ratios have been carefully chosen for a balanced two-phase output. The transformer inputs are connected to the variable 380 V 3Ф out put of the power supply, terminals 4, 5 and 6.

2. a. Adjust the resistance module switches for R₁ = R₂ = 880 Ω.

b. Turn on the power supply and adjust for a line-to-line voltage of 380 V ac.

c. Measure and record the three input line currents I₄, I₅ and I₆.

I₄ = 0.11 A ac; I₅ = 0.11 A ac; I₆ = 0.11 A ac;

d. Return the voltage to zero and turn off the power supply.

e. Are the three line currents reasonably well balanced? Yes

3. a. Remove the two current meters A₅ and A₆ from the input circuit and place one in series with each of the load resistors R₁ and R₂ as shown by the dashed lines in Figure – 1.

b. Turn on the power supply and adjust for a line-to-line voltage of 380 V ac.

c. Measure and record the voltage across and the currents through, the two load resistances R₁ and R₂.

E₁ = 140 V ac; I₁ = 0.155 A ac

E₂ = 135 V ac; I₂ = 0.16 A ac

d. Return the voltage to zero and turn off the power supply.

e. Are the load currents and voltages reasonably well balanced? No

4. a. Calculate the power in each load resistance.

P₁ = E₁ (140) × I₁ (0.155) = 21.7 W

P₂ = E₂ (135) × I₂ (0.16) = 21.6 W

2Ф total power = P₁ + P₂ = (21.7 + 21.6) W = 43.3 W

b. Calculate the apparent power delivered to the transformers from the three-phase source.

380V × 1.73 × I₄(0.11) = 72.4 VA

3Ф total power = 72.4 W

Figure – 2

5. Remove the two resistance loads from the circuit and connect the transformer secondaries as shown in Figure – 2. Do not disturb the wiring to the primary windings.

6. a. Turn on the power supply and adjust for a line-to-line voltage of 380 V ac.

b. Measure and record the voltage across each winding and the voltage across both windings in series.

E₁ = 140 Vac; E₂ = 135 Vac; E₃ = 195 Vac

c. Return the voltage to zero and turn off the power supply.

Review Question:

1. How much larger is the apparent power delivered by three-phase power source in procedure 3 than the apparent two-phase power delivered to the load?

Ans: 3Ф total power = 72.4 W and 2Ф total power = 43.3 W. The power difference between source and load = (72.4 – 43.3) W = 29.1 W. We can observe that about 40% larger the apparent power delivered by three-phase power source than the apparent two-phase power delivered to the load.

2. In the space provided, show how you would connect two Single-Phase Transformers to obtain three-phase output from a two-phase power source. Identify all windings and taps.

Ans: Shown in Figure – 3.

Figure – 3

The Wound-Rotor Induction Motor Part (1)

2010-01-25T00:43:00.000-08:00

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Objective:

· To examine the construction of the three-phase wound-rotor induction motor.

· To understand exciting current, synchronous speed and slip in a three-phase induction motor.

· To observe the effect of the revolving field and rotor speed upon the voltage induced in the rotor.

Discussion:

You have, sp far, been introduced to rotating stator fields produced by single-phase power; electric power companies normally generate and transmit three-phase power. Single-phase power for the individual home is obtained from one phase of the three-phase power lines. Three-phase (polyphase) motors are commonly used in industry and electric power companies normally supply three-phase power to industrial users.

The creation of a rotating stator field using three-phase power is similar to the principle of the split-phase or two-phase (capacitor-run) system. In the three-phase system, a rotating magnetic field is generated in three phases instead of two. When the stator of a three-phase motor is connected to a three-phase power source, currents flow in the three stator windings and a revolving magnetic field is established. These three exciting currents supply the reactive power to establish the rotating magnetic field. They also supply the power consumed by the copper and iron losses in the motor.

The speed of the rotating magnetic field is entirely determined by the frequency of the three-phase power source, and is known as the synchronous speed. The frequency of electric power systems is accurately maintained by the electric power companies; therefore, the synchronous speed of the stator field (in rpm) remains constant. (It is, in fact, used to operate electric clocks).

The wound-rotor consists of a rotor core with the three windings in place of the conducting bars of the squirrel-cage rotor, in this case, currents are induced in the windings just as they wound be in shorted turns. However, the advantage of using windings is that the wires can be brought out through slip rings so that resistance, and, therefore, the current through the windings, can be controlled.

The rotating stator field induces an alternating voltage in each winding of the rotor. When the rotor is at standstill the frequency of the induced rotor voltage is equal to that of the power source. If the rotor is now rotated by some external means, in the same direction as the rotating stator field, the rate at which the magnetic flux cuts the rotor windings will diminish. The induced voltage and its frequency will drop. When the rotor revolves at the same speed and in the same direction as the rotating stator field, the induced voltage, as well as its frequency, will drop to zero. (The rotor is now at synchronous speed.) Conversely, if the rotor is driven at synchronous speed, but in the opposite direction to the rotating stator field, the induced voltage, as well as its frequency, will be twice the value as when the rotor was stand still.

Although the rotor will be driven by an external motor is this Experiment, it should be noted that for a given rotor speed the induced voltage value and its frequency will be the same even if the rotor were turning by itself.

Equipment Required:

Three-Phase Wound-Rotor Induction Motor
DC Motor/ Generator
Three-Phase Wattmeter
AC Ammeter
AC Voltmeter
Wires

Procedure:

CAUTION!!!

High voltages are Present In the Experiment! Do not make any connections with the power on! The power should be turned off after completing each individual measurement!!!

1. Examine the construction of the Three-Phase Wound-Rotor Induction Motor, paying particular attention to the motor, slip rings, connection terminals and the wiring.

2. Viewing the motor from the rear of the module:

a. Identify the three rotor slip rings and brushes.

b. Can the brushes be moved?

c. Note that the three rotor windings are brought out to the three slip rings via a slot in the rotor shaft.

d. Identify the stator windings. Note that they consist of many turns of small diameter wire evenly spaced around the stator.

e. Identify the rotor windings. Note that they consist of many turns of slightly larger diameter wire evenly spaced around the rotor.

f. Note the spacing of the air gap between the rotor and the stator.

3. Viewing the front face of the module:

a. The three separate stator windings are connected to terminals 1 and 4, 2 and 5, 3 and 6.

b. What is the rated current of the stator windings? 0.7 A

c. What is the rated voltage of the stator windings? 220 V

d. The three rotor windings are (wye/ delta) wye connected.

e. They are connected to terminals 7, 8 and 9.

f. What is the rated voltage of the rotor windings? 110 V

g. What is the rated current of the rotor windings? 1 A

h. What is the rated speed and mechanical output power of the rotor?

Speed = 1450 rpm

Power = 175 W

Figure – 1

4. Using your DC Motor/ Generator, Three-Phase Wound-Rotor Induction Motor, Three-Phase Wattmeter, Power Supply, AC Ammeter and AC Voltmeter, connect the circuit shown in Figure – 1.

5. a. Note that the DC motor/generator is connected with fixed shunt field excitation to power supply terminals 8 and N, (220 Vdc). The field rheostat should be turned to its full CW position (for minimum resistance)
b. Note that the armature is connected to the variable DC output of the power supply, terminals 7 and N, (0-220 Vdc).
c. Note that the stator of the wound-rotor motor is wye connected, in series with three ammeters and the wattmeter to the fixed 380 V, 3Ф output of the power supply, terminals 1, 2 and 3.
d. Note that the 3Ω input voltage is measured by V₁ and that the 3Ф rotor output voltage is measured by V₂.

6. a. Couple the DC motor/generator to the wound-rotor motor with the timing belt.

b. Turn on the power supply. Keep the variable output voltage control at zero (the DC motor should not be turning).

c. Measure and record the following:

E₁ = 380 V ac; W₁ = 110 W; W₂ = 110 W;

I₁ = 0.4 A ac; I₂ = 0.4 A ac; I₃ = 0.4 A ac;

E₂= 198 V ac;

d. Turn off the power supply.

7. Calculate the following:

a. Apparent power = VI = 380 × 0.4 = 152 VA

b. Active Power = VI cosθ = 380 × 0.4 × 0.4 = 60.8 W

c. Power Factor = cosθ = 0.4

d. Reactive Power = VI sinθ = 380 × 0.4 × 0.92 = 139.3 var

8. a. Turn on the power supply and adjust the variable DC output voltage for a motor speed of exactly 750 rpm.

b. Measure and record the following:

Note: If the value of E₂ is less than in procedure 6, turn off the power supply and interchange any two of the three stator leads

E₁ = 380 V ac; W₁ = 110 W; W₂ = 110 W;

I₁ = 0.4 A ac; I₂ = 0.4 A ac; I₃ = 0.4 A ac;

E₂= 92 V ac;

9. a. Increase the variable DC output voltage to 220 V dc and adjust the field rheostat for a motor speed of exactly 1500 rpm.

b. Measure and record the following:

E₁ = 380 V ac; W₁ = 110 W; W₂ = 110 W;

I₁ = 0.4 A ac; I₂ = 0.4 A ac; I₃ = 0.4 A ac;

E₂= 0 V ac;

c. Return the voltage to zero and turn off the power supply.

d. In procedure 8 and 9 is the rotor being turned with against the rotating stator field? Explain.

Ans: When we consider a motor speed is less than the rated speed; then we observed the motor can run and we found a terminal voltage E₂. And when the motor speed is greater than the rated speed; the rotor cannot run and we found a terminal voltage is zero. Finally we can say, if the magnetic flux speed is less than the rotor speed; the motor can run with rotating stator field. In equally or greater than rotor speed; the motor cannot run.

10. a. Interchange your DC armature connections in order to reverse the motor direction. Turn the field rheostat to its full CW position.

b. Turn on the power supply and adjust the DC output voltage for a motor speed of 750 rpm.

c. Measure and record the following:

E₁ = 380 V ac; W₁ = 110 W; W₂ = 110 W;

I₁ = 0.4 A ac; I₂ = 0.4 A ac; I₃ = 0.4 A ac;

E₂= 295 V ac;

11. a. Increase the variable DC output voltage to 220 V dc and adjust the field rheostat for a motor speed of 1500 rpm.

b. Measure and record the following:

E₁ = 380 V ac; W₁ = 110 W; W₂ = 110 W;

I₁ = 0.4 A ac; I₂ = 0.4 A ac; I₃ = 0.4 A ac;

E₂= 395 V ac;

c. Return the voltage to zero and turn off the power supply.

The Wound-Rotor Induction Motor Part (2)

2010-01-24T03:02:00.000-08:00

Objective:

· To observe the characteristics of the wound-rotor induction motor at no-load and full-load.

· To observe speed control using an external variable resistance.

Discussion:

The three ends of the three-phase rotor windings are brought out to three slip rings mounted on the rotor shaft. The brushes bearing on the slip rings play an important role in realizing maximum advantage from the wound-rotor motor. By connecting the brushes through rheostats, it becomes possible to develop a higher starting torque than is possible with a squirrel-cage motor. On starting, the full resistance of the rheostats is maintained in the rotor circuit, thus providing the very maximum starting torque.

As the motor approaches normal operating speed, the rheostat resistance is gradually reduced until it is out of the circuit entirely at full speed. Although the starting torque of the wound-rotor motor is higher, it is not as efficient as the squirrel-cage motor at full speed, because the resistance of the rotor windings is always more than that of a squirrel-cage motor.

A special feature of the wound-rotor motor is its variable speed capability. By varying the rheostat resistance, it is possible to vary the percentage of slip and thus, vary the motor speed. In such cases, below full speed operation means the motor is running at reduced efficiency and mechanical output power. In addition, because of a high rotor resistance, the motor is made more susceptible to variation in speed as the load changes.

Equipment Required:

Three-Phase Wound-Rotor Induction Motor
Electrodynamometer
Single-Phase Power Factor Meter
Three-Phase Rheostat
Power Supply
AC Ammeter
AC Voltmeter
Wires

Procedure:

CAUTION!!!

High voltages are Present In the Experiment! Do not make any connections with the power on! The power should be turned off after completing each individual measurement!!!

1. a. Examine the construction of the Three-Phase Rheostat, paying particular attention to the circuit schematic diagramed on the face of the module.

b. Note that the arms of the three rheostats are separately brought out to terminals 1, 2 and 3. The remaining ends of the rheostats are wired together internally and brought out to the N terminal.

c. Note that the three rheostats are ganged together and that their individual resistances can be varied simultaneously by turning the single control knob.

d. When the control Knob is fully CCW the resistance of each rheostat is 75 Ω.

2. Using your Three-Phase Wound-Rotor Induction Motor, Electrodynamometer, Single-Phase Power Factor Meter, Three-Phase Rheostat, Power Supply, AC Ammeter and AC Voltmeter, connect the circuit shown in Figure – 1. Do not couple the motor to the electrodynamometer at this time!

3. a. Set the speed control rheostat knob at its full CCW position for zero resistance.

b. Turn on the power supply and adjust E1 to 380 Vac. The motor should be running.

c. Measure and record in Table – 1, the three line currents, the two power factor meter indications (remember, to observe the polarities) and the motor speed.

d. Return the voltage to zero and turn off the power supply.

4. a. Couple the motor to the electrodynamometer with the timing belt.

b. Set the dynamometer control knob at its full CCW position.

c. Repeat procedures 3 for each of the torques listed in Table – 1, maintaining the input voltage at 380 Vac.

d. Return the voltage to zero and turn off the power supply.

5. a. Set the speed control rheostat knob at its full CCW position for maximum resistance.

b. Uncouple the motor from the electrodynamometer.

6. a. Turn on the power supply and adjust E1 to 380 V ac. The motor should be running.

b. Measure and record in Table – 2, the three line currents, the two power factor meter indications and the motor speed.

c. Return the voltage to zero and turn off the power supply.

7. a. Couple the motor to the electrodynamometer with the timing belt.

b. Set dynamometer control knob at its full CCW position.

c. Repeat procedure 6 for each of the torques listed in Table – 2, maintaining the input voltage at 380 V ac.

d. with a developed torque of 0.9 N-m, rotate the speed control rheostat knob from full CW to full CCW.

e. Does the motor speed change? Yes

f. Does the developed torque change? Yes

g. Return the voltage to zero and turn off the power supply.

The Three-Phase Alternator

2010-01-23T22:59:00.000-08:00

Objective:

To obtain the no-load saturation curve of the alternator.
To obtain the short-circuit characteristics of the alternator.

Discussion:

The terms alternating current generator, synchronous generator, synchronous alternator and alternator are commonly used interchangeably in engineering literature. Because synchronous generators are so much more commonly used than induction generators, the terms alternator, as often used, and as used here, applies only, to synchronous generators.

Alternators are, by far, the most important source of electric energy. Alternators generate an AC voltage whose frequency depends entirely upon the speed of rotation. The generated voltage value depends upon the speed, the DC field excitation and the power factor of the load.

As the DC field excitation of an alternator is increased, its speed being held constant, the magnetic flux and hence, the output voltage will also increase in direct proportion to the current. However, with progressive increase in DC field current, the flux will eventually reach a high enough value to saturate the iron in the alternator.

There will be a smaller increase in flux for a given increase in DC field current. Because the generated voltage is directly related to the magnetic flux intensity, it can be used as a measure of the degree of saturation.

The three phase of the alternator are mechanically spaced at equal intervals from each other, and therefore, the respective generated voltage are not in phase, but are displaced from each other by 120 electrical degrees.

When an alternator delivering full rated output voltage is suddenly subjected to a short-circuit, very large currents will initially flow. However, these large short-circuit currents drop off rapidly to save values if the short-circuit is maintained.

Equipment Required:

Three-Phase Synchronous Motor/Generator
Four-Pole Squirrel-Cage Induction Motor
Power Supply
DC Voltmeter/Ammeter
AC Voltmeter
Wires

Procedure:

CAUTION!!!

High voltages are Present In the Experiment! Do not make any connections with the power on! The power should be turned off after completing each individual measurement!!!

Three-Phase Synchronous Motor/Generator

1. Using your Three-Phase Synchronous Motor/Generator, Four-Pole Squirrel-Cage Induction Motor, Power Supply, AC Voltmeter and DC Voltmeter/Ammeter, connect the circuit shown in Figure-1. The squirrel-cage motor will be used to drive the synchronous motor/generator as an alternator. Its speed will be assumed constant during this Experiment. Note that the squirrel-cage motor is connected to the fixed 380 V 3Ф output of the power supply, terminal 1, 2 and 3. The rotor of the alternator is connected to the variable 0-220 output of the power supply, terminal 7 and N.

2. a. Couple the squirrel-cage motor to the alternator with the timing belt.

b. Set the alternator field rheostat at its full CW position (for zero resistance). Open switch S.

c. Set the power supply voltage control at its full CCW position (for zero DC voltage).

3. a. Turn on the power supply. The motor should be running.

b. With zero DC excitation measure and record E₁, E₂ and E₃ (use the lowest ranges of the voltmeters.

E₁ = 10 Vac, E₂ = 10 Vac, E₃ = 10 Vac

c. Explain why there is an AC voltage generated in the absence of DC excitation.

Ans: In a Synchronous Generator Rotor; there is some residual magnetism. For this reason we have found an AC voltage generated without DC excitation.

4. a. Close the switch S.

b. Gradually increase the DC excitation from zero to 0.05 Adc.

c. Measure and record in Table – 1 the three generated voltages E₁, E₂, E₃.

d. Repeat (b) for each of the DC current listed in Table – 1

e. Return the voltage to zero and turn off the power supply.

Table – 1

5. Calculate and record in Table – 1 the average output of the alternator for each of the listed DC currents.

6. a. Turn on the power supply and adjust the DC excitation until E₁ = 380 Vac. Measure and Record E₂ and E₃.

E₁ = 380 Vac, E₂ = 380 Vac, E₃ = 380 Vac

b. Turn on the power supply without touching the voltage adjusts control.

c. Reconnect the three AC voltmeters so they will measure the voltages across each of the three stator windings.

d. Turn on the power supply. Measure and record the generated voltages across each of the wye connected stator windings

E_{1 to 4} = 220 Vac, E_{2 to 5} = 220 Vac, E_{3 to 8} = 220 Vac

e. Return the voltage to zero and turn off the power supply.

Figure – 2

7. Using your Synchronizing Module, connect the circuit shown in Figure – 2. Note that the switch is wired to present a dead shot across the alternator windings when it is closed.

8. a. Set the synchronizing switch to its open position.

b. Turn on the power supply and adjust the DC excitation until E₁ = 380 Vac. The motor should be running and the three lamps on the synchronizing module should be illuminated.

c. measure and record the DC exciting current I₁.

I₁ = 0.25 Adc

d. Apply a short-circuit to your alternator by closing the synchronizing switch and note the behavior of the AC current I₂.

e. To what approximate peak value did I₂ increase?

I₂ = 0.11 Aac

f. What is the final steady-state value of I₂ and I₁?

I₁ = 0.25 Adc and I₂ = 0.11 Aac

g. Return the voltage to zero and turn off the power supply.

Review Questions:

1. Plot your recorded average voltage values vs DC current values from Table – 1 on the graph of Figure – 3.

2. Draw a smooth curve through your plotted points.

3. Up to what voltage is the curve a reasonably straight line?
E = 220 Vac

4. Where would you say is the knee of the saturation curve?
E = 300 Vac

5. Explain why the voltage increases less rapidly as the DC current increases.

Figure – 3

Ans: Alternators generate an AC voltage whose frequency depends upon the speed of rotation, DC field excitation and power factor of the load. If the DC field excitation is increased; its speed constant, the magnetic flux increased and the output voltage will also increased in direct proportion to the current. So there are a smaller increased in flux for a given increase in DC field current; the output voltage increases.

The Synchronous Motor

2010-01-23T01:38:00.000-08:00

Objective:

To examine the construction of the 3Ф synchronous motor.
To obtain the starting characteristics of the 3Ф synchronous motor.

Discussion:

The synchronous motor gets its name from the term synchronous speed, which is the natural speed of the rotating magnetic field of the stator. As you have learned, this natural speed of rotation is controlled strictly by the number of pole pairs and the frequency of the applied power.

Like the induction motor, the synchronous motor makes use of the rotating magnetic field. Unlike the induction motor, however, the torque developed does not depend on the induction currents in the rotor. Briefly, the principle of operation of the synchronous motor is as follows. A multiphase source of AC is applied to the stator windings and a rotating magnetic field is produced. A direct current is applied to the rotor windings and a fixed magnetic field is produced. The motor is so constructed that these two magnetic fields react upon each other causing the rotor to rotate at the same speed as the rotating magnetic field. If a load is applied to the rotor shaft, the rotor will momentarily fall behind the rotating field but will continue to rotate at the same synchronous speed.

The falling behind is analogous to the being tied to the rotating field with a rubber band. Heavier loads will cause stretching of the band so the rotor position lags the stator field but the rotor continues at the same speed. If the load is made too large, the rotor will pull out of synchronism with the rotating field and, as a result, will no longer rotate at the same speed. The motor is then said to be overloaded.

The synchronous motor is not a self-starting motor. The rotor is heavy and, from a dead stop, it is not possible to bring the rotor into magnetic lock with the rotating magnetic field. For this reason, all synchronous motors have some kind of starting device. A simple starter is another motor which brings the rotor up to approximately 90% of its synchronous speed. The starting motor is then disconnected and the rotor locks in step with the rotating field. The more commonly used starting method is to have the rotor include a squirrel cage induction winding. The induction winding brings the rotor almost to synchronous speed as an induction motor. The squirrel-cage is also useful even after the motor has attained synchronous speed, because it tends to dampen rotor oscillations caused by sudden changes in loading. Your Three-Phase Synchronous Motor/Generator contains a squirrel-cage-type rotor.

Equipment Required:

Three-Phase Synchronous Motor/Generator
Electro Dynamometer
AC Ammeter
AC Voltmeter
Wires

Procedure:

CAUTION!!!

High voltages are Present In the Experiment! Do not make any connections with the power on! The power should be turned off after completing each individual measurement!!!

1. Examine the construction of the Three-Phase Synchronous Motor/Generator, paying particular attention to the motor, slip rings, rheostat, connection terminals and the wiring.

2. Viewing the motor from the rear of the module:

a. Identify the two slip rings and brushes.

b. Can the brushes be moved?

c. Note that the two rotor windings are brought out to the two slip rings via a slot in the rotor shaft.

d. Identify the DC damper windings on the rotor.(Although there are only two windings, they are connected so that their magnetomotive forces act in opposition, thus, creating four poles.

e. Identify the four salient poles just beneath the damper windings.

f. Identify the stator winding and note that it is identical to that of the three-phase squirrel cage and wound rotor motors.

3. Viewing the front face of the module:

a. The three separate stator windings are connected to terminals 1 and 4, 2 and 5, 3 and 6.

b. What is the rated voltage of the stator windings? 220 V

c. What is the rated current of the stator windings? 0.5 A

d. The rotor winding is connected through the 380 Ω rheostat (and a toggle switch S) to terminals 7 and 8.

e. What is the rated voltage of the rotor winding?

f. What is the rated speed and mechanical output power of the motor? Speed=1450 rpm and Power = 175W.

Starting Characteristics:

Figure – 1

4. Using your Three-Phase Synchronous Motor/Generator, Power Supply and AC Ammeter, connect the circuit shown in Figure – 1. Note that the three stator windings are wye-connected to the fixed 380 V 3 Ф output of the power supply, terminals 1, 2 and 3.

5. a. Turn on the power supply. Note that the motor starts smoothly and continues to run as an ordinary induction motor.

b. Note the direction of rotation

Rotation = CCW, I₁ = 0.5 Aac

c. Turn off the power supply and interchange any two of the leads from the power supply.

d. turn on the power supply and note the direction of rotation.

Rotation = CW, I₁ = 0.5 Aac

e. Turn off the power supply.

6. a. Connect the circuit shown in Figure – 2. Note that the synchronous motor is wired in its normal starting configuration (as a three-phase squirrel-cage induction motor).

b. Set the dynamometer control knob at its full CW position (to provide a maximum starting load for the synchronous motor).

c. Close the switch S.

Figure – 2

7. a. Turn on the power supply and quickly measure E₁, E₂, I₁ and the developed starting torque. Turn off the power supply.

E₁ = 398 Vac, E₂ = 410 Vac, I₁ = 1.5 Aac.

Starting torque = 3.0 N-m.

b. Calculate the full-load torque corresponding to 175W at 1500 rpm

Full-load Torque = Pout ×60/2πN = 175×60 / 2π×1500 = 1.11N-m

c. Calculate the ratio of starting torque to full-load torque.

Ratio = 1.11 × 100/ 3.0 = 37%

d. Why a large AC voltage E2 was induced in the rotor windings?

Because of a higher starting torque a large AC voltage is induced as both are proportional.