To discover the factors affecting the direction of rotation and speed of three-phase motors.

EXPERIMENT 12 Direction of Rotation of Three-Phase Motor PURPOSE: To discover the factors affecting the direction of rotation and speed of three-phase motors. BRIEFING: The stators of three-phase motors are made up of steel laminations. These laminations are stacked into what looks like solid core. The core is slotted for the stator windings. There are no protruding pole pieces (salient poles) on the stator. The conductors of the stator winding, however, are wound in coil groups. Each group produces a pair of electromagnetic poles when current passes through it. The coil groups are spaced evenly around the stator. In a two-pole motor there is one coil group per phase. Since poles always come in pairs (North and South), two poles is the minimum number possible. Each coil group is connected to one phase of a three-phase supply. The Phase B sine wave is 120 electrical degrees behind the sine wave of Phase A. Similarly, Phase C lags Phase B by 120 degrees. This causes the magnetic characteristics of coil group #1 to be passed along to coil group #2; then passed along to coil group #3. If coil group #1 is connected to Phase A; group #2 to Phase B; and #3 to Phase C, this pass along will be in a particular direction, say clockwise. However, what if group #3 is connected to Phase B? The magnetic characteristics of coil group #1 will go to coil group #3. The magnetic field will revolve counterclockwise. In fact, interchanging any two leads will reverse the direction of a three-phase motor. In a two-pole motor the revolving magnetic field of the stator goes around once during each AC cycle. If you had four poles per phase (a four-pole motor) the field would go only half a revolution during each cycle. You can compute the speed of the revolving field (called synchronous speed) from the frequency of the applied voltage and from the number of pairs of poles. The equation is: Synchronous Speed = Frequency no. of pairs of poles *60 12-1

For a two-pole motor (60 Hertz): For a four-pole motor (60 Hertz): S = 60 x 60/1 = 3600 rpm S = 60 x 60/2 = 1800 rpm 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 * 100 PERFORMANCE OBJECTIVES: Upon successful completion of this experiment, the student will be able to: 1. Explain the generation of revolving magnetic field in the stator of a three-phase motor. 2. Successfully connect a three-phase motor to produce the desired direction of rotation. MACHINE REQUIRED IM-100 Three-Phase Induction Motor POWER REQUIRED Fixed 3 AC Supply ADDITIONAL MATERIAL REQUIRED MGB-100-DG Bedplate HT-100 Series Tachometer 12-2

PROGRAM PLAN: Step 1. Clamp the motor on the bedplate. Install coupling guards. Step 2. Connect the motor as shown in Figure 12-1. Note that Phase A of the supply is connected to terminal L1 the motor; B to L2; and C to L3. Note also that terminal T1 is connected to terminal A; T2 to B; T3 to C. Figure 12-1 Step 3. Turn ON the Main AC power. Step 4. Turn ON the motor circuit breaker switch. Noting the direction of rotation as that viewed from the right-hand end, indicate the direction on Figure A of TEST RE- SULTS. Step 5. Turn OFF the motor. Step 6. Reconnect the stator as follows: Leaving T3 connected to C, interchange the other two leads so that T1 is connected to B; T2 to A. Step 7. Repeat Step 4 for Figure B of TEST RESULTS, then turn OFF the motor. Step 8. Reconnect the stator as follows: Leaving T1 connected to B, interchange the other two leads so that T2 is connected to C; T3 to A. Step 9. Repeat Step 4 for Figure C of TEST RESULTS, then turn OFF the motor. Step 10. Reconnect the stator as follows: Leaving T3 connected to A, interchange the other two leads so that T1 is connected to C; T2 to B. Step 11. Repeat Step 4 for Figure D of TEST RESULTS, then turn OFF the motor. 12-3

Step 12. Reconnect the stator as follows: Leaving T1 connected to C, interchange the other two leads so that T2 is connected to A; T3 to B. Step 13. Repeat Step 4 for Figure E of TEST RESULTS, then turn OFF the motor. Step 14. Reconnect the stator as follows: Leaving T3 connected to B, interchange the other two leads so that T1 is connected to A; T2 to C. Step 15. Repeat Step 4 for Figure F of TEST RESULTS, then turn OFF the motor. Step 16. Leaving T1 connected to A, interchange the other two leads so that the connection is the same as shown in Figure 12-1. Step 17. Turn ON the motor circuit breaker switch. Measure the speed of the unloaded motor. Record in TEST RESULTS. Step 18. Turn OFF all circuit breaker switches. Disconnect all leads. TEST RESULTS: Figure A Figure B Figure C Figure D Figure E Figure F MOTOR SPEED (Step 17) RPM. 12-4

DE-BRIEFING: 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 (Note: The IM-100 is a 4-pole motor). Then compute Slip Speed. Synchronous Speed = Synchronous Speed = Frequency no. of pairs of poles *60 Slip speed = Synchronous Speed - Rotor Speed Slip Speed = 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? 12-5

QUICK QUIZ: 1. 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. 2. 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. 3. 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. 4. Three-phase Induction Motors must: (a) Run faster than synchronous speed. (b) Run slower than synchronous speed. (c) Run at synchronous speed. 5. Synchronous speed is determined by: (a) Frequency and number of poles. (b) Torque and speed of driven load. (c) Field strength and armature current. 12-6

EXPERIMENT 16 Losses and Efficiency of Induction Motors PURPOSE: To discover the sources of losses in a three-pole squirrel-cage induction motor and to compute efficiency. BRIEFING: There are two major classifications of losses in induction motors. The first is the copper losses. 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 is the equivalent resistance of the motor. It is this equivalent resistance that must be multiplied times the current squared to determine the I 2 R 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, there is such a slight change in motor speed, you can consider them constant. PERFORMANCE OBJECTIVES: Upon successful completion of this experiment, the student will be able to: 1. Perform locked rotor tests and determine the equivalent resistance of induction motors. 2. Explain the source of losses and compute efficiency induction motors. 16-1

MACHINES REQUIRED IM-100 Induction Motor DYN-100A-DM Dynamometer POWER REQUIRED Variable 3 AC Supply 0-150 volt Variable DC, 1 amp MACHINES REQUIRED 0-2.5/4 amp AC Ammeter 0-300 volt AC Voltmeter (2) 0-300/600 Watt AC Wattmeters 0-150 volt DC Voltmeter ADDITIONAL MATERIAL REQUIRED MGB-100-DG Bedplate RL-100A Resistance Load PROGRAM PLAN: Step 1. Place the IM-100 induction motor on the left side of the bedplate. Clamp securely and install guards. Step 2. Connect the motor to the variable 240V AC supply as shown in Figure 16-1. Be sure to observe polarities when connecting the wattmeter. The two-wattmeter method of measuring three-phase power is being used. Figure 16-1 16-2

Step 3. Have someone check your connections to be sure they are correct. Step 4. Turn ON the main AC and the 0-240V supply. Adjust the output to 208 volts. Step 5. Put temporary jumpers across the terminals of the ammeter and the current coils of the wattmeter to protect the movements from the large inrush of current. NOTE: The power factor of an unloaded induction motor is likely to be less than 0.5. It may therefore be necessary in the next step to reverse the leads of the voltage terminals of one wattmeter and to subtract its reading from the other one to get total watts. Step 6. Turn ON the motor, remove the jumpers and read the motor current and voltage. Record these values in TABLE 16-1 of TEST RESULTS. Step 7. Turn OFF all circuit breakers. Step 8. Place the dynamometer on the right side of the bedplate. Couple it to the induction motor and clamp securely. Install guards. Step 9. Install the rotor locking device securely on the dynamometer. Step 10. Have someone check your connections to be sure they are correct. Then turn the voltage control knob fully counterclockwise to its zero output position. Turn ON the main AC, the 0-240V AC, and the motor circuit breakers. Step 11. Slowly increase the output of the 0-240V AC supply until the ammeter reads the rated current of 2.4 amperes. Read the applied voltage and the wattage. Record these values in TABLE 16-2 of TEST RESULTS. Step 12. Turn OFF all circuit breakers. Step 13. Connect the dynamometer as shown in Figure 16-2. Note that this a separately excited shunt generator connection. Be sure the rotor locking device has been removed from the dynamometer shaft. Adjust its field rheostat to its maximum resistance position, fully clockwise. Step 14. With the motor switch OFF, turn on the main AC, the 0-240V AC, and the 0-150V DC supplies. Adjust the AC supply to 208 volts. Adjust the excitation supply to maximum. Be sure all of the switches on the RL-100A resistance load banks are in the downward (OFF) position. 16-3

Figure 16-2 Step 15. Put temporary jumpers across the terminals of the ammeter and the current coils of the wattmeters. Step 16. Turn the motor ON. Remove the jumpers from the ammeter and the current coils of the wattmeters. Step 17. Use the dynamometer field rheostat to adjust its terminal voltage to 120 volts DC. Step 18. Read the motor current and input watts and record these readings in TABLE 16-3. Step 19. Turn ON load steps 1,2, and 3, on the RL-100A Load Bank. Step 20. Readjust the dynamometer s field rheostat or the excitation supply, as required, to maintain a terminal voltage of 120 volts. Then repeat Step 18. Step 21. Turn ON loads steps 4, 5, and 6. Repeat Step 20. Step 22. Turn ON loads steps 7, 8, and 9. Repeat Step 20. Step 23. Turn ON loads steps 10, 11, and 12. Repeat Step 20. Step 24. Turn OFF all circuit breaker switches. Disconnect all leads. 16-4

TEST RESULTS: V A I NL W #1 W #2 TOTAL WATTS COMP P RL 208 TABLE 16-1 - NO-LOAD TEST V A I LR W #1 W #2 TOTAL WATTS COMP R EQ 2.4 TABLE 16-2 - LOCKED ROTOR TEST 0 1-3 1-6 1-9 1-12 MOTOR CURRENT WATTS #1 WATTS #2 TOTAL WATTS IN P RL P CL TOTAL LOSSES P L % EFFICIENCY TABLE 16-3 - LOAD TEST DE-BRIEFING: 1. During the Locked Rotor Test, all of the input power was lost in the motor s equivalent resistance. The equation is: W=3x(I LR ) 2 xr EQ Solving for R EQ the equation becomes R EQ =W/(3 x I 2 ). From the data you recorded in TABLE 16-2, compute the total power in (TOT. W) and R EQ. Record these values in TABLE 16-2. 16-5

2. During the No-Load-Test, the input power was lost in both the equivalent resistance (I 2 R EQ ) and in the rotational losses (P RL ). To find the value of P RL, first compute the total power in. Then compute the total copper losses (P CL =3x(I NL ) 2 xr EQ ) using the value of R EQ = TOT. W - P CL. Record this value of P RL in TABLE 16-1 and for every load listed in TABLE 16-3. (P RL assumed constant). 3. For each of the motor current values in TABLE 16-3, combine the two wattmeter readings to provide the total power in. Record these values in TABLE 16-3. 4. For each of the motor current values in TABLE 16-3 compute the total copper loss from the equation P CL =3xI 2 xr EQ using the value of R EQ from TABLE 16-2. Record the copper loss values in TABLE 16-3. Add each of these P CL values to the P RL value computed in No. 2 to produce the total loss (P L ) value for each of the loads. Record these values in TA- BLE 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 = List these efficiencies in TABLE 16-3. Total Watts IN \ PL Total Watts IN * 100 QUICK QUIZ: 1. As load increases, rotational losses: (a) Increase. (b) Decrease. (c) Remain the same. 2. As load increases, copper losses: (a) Increase. (b) Decrease. (c) Remain the same. 3. 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. 16-6

4. As load increase, the motor: (a) Operates more efficiently. (b) Operates less efficiently. (c) Operates with the same efficiency. 5. Power Out equals: (a) The power in. (b) The total of copper and rotational losses. (c) The power in minus the total losses. 16-7