Monday, January 25, 2010

The Wound-Rotor Induction Motor Part (1)

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

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.

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