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0. DC MOTORS
1. DC SHUNT MOTOR
1.1.. OBJECTIVE
To study the torque vs. speed characteristic of a shunt wound DC motor and calculate its efficiency.
1.2.. (^) DISCUSSION
The speed of any DC motor depends directly on its armature voltage and the strength of its magnetic field. The field winding in a shunt motor is in parallel with the armature winding and the DC supply. If the DC line voltage is constant, the armature voltage will be constant and thus the magnetic field strength will be constant. This consistency leads to a reasonably constant speed of operation.
The speed does tend to drop with increasing load on the motor. This drop in speed is a result of resistive losses in the armature winding. Shunt motors with low armature winding resistance tend to have nearly constant speed of operation.
As with any energy conversion device, the DC shunt motor is not 100% efficient. Not all of the electric energy supplied to the motor is converted into useful work (mechanical power). The difference between electrical power supplied and mechanical power available at the shaft is lost in the form of heat inside the motor. Losses occur in the DC resistance of the field and armature windings, in the magnetic circuit that couples field and armature windings, in the friction and windage of the rotating armature and in the resistance of the brush contacts on the commutator. Losses increase as the load on the motor increases, resulting in significant heating of the motor at full load.
1.3.. INSTRUMENTS AND COMPONENTS
Power Supply Module EMS 8821 DC Metering Module EMS 8412 DC Motor/Generator Module EMS 8211 Electrodynamometer Module EMS 8911 Hand Tachometer EMS 8920
1.4.. PROCEDURE
CAUTION! – High voltages are present in this Experiment. DO NOT make any connections with the power supply ON. Get in the habit of turning OFF the power supply after every measurement.
- Connect the following circuit. DO NOT APPLY POWER AT THIS TIME.
Manual Rev 1.
Set the shunt field rheostat control knob at its full clockwise position, for maximum field excitation. Make sure the brushes are in their neutral position (90 F 0 B 0and 0 F 0 B 0).
Set the Electrodynamometer control knob (or Prime Mover/Dynamometer control knob) at its full counterclockwise position (minimum load). Note that the Dynamometer will require a power source.
Turn on the power supply and adjust the voltage control to 120 V DC. Note the direction of rotation. If it is not clockwise, turn OFF the power supply and swap the connections across the shunt field. Then turn on the power supply.
Adjust the field rheostat counterclockwise for a no load motor speed of 1800 rpm as indicated by the tachometer. Double check the voltmeter to ensure the source voltage is 120 V DC. Once the source voltage and no load speed are set, DO NOT change the field rheostat setting for the remainder of this experiment section.
Measure and record the line current as indicated by the ammeter for the no load condition at 1800 rpm.
Source
(Volts)
Line Current
(Amps)
Speed
(RPM)
Torque (N-M, Lbf-
In)
120 1800 0 n-m, 0 lbf-in
120 0.35 n-m, 3 lbf-in
120 0.7 n-m, 6 lbf-in
120 1.05 n-m, 9 lbf-in
120 1.4 n-m, 12 lbf-in
Apply a load to the motor by turning the dynamometer control clockwise until the torque reading is 3 pound – inches (lbf-in) or 0.35 Newton-meters. Adjust the DC voltage control to maintain 120 V as necessary.
Measure and record the line current and motor speed for the 3 lbf-in load condition.
Increase the load to 6, 9 and 12 lbf-in, taking speed and current measurements at each point. Record them in the table provided above.
Return the voltage control to zero percent and turn OFF the power supply.
Plot the recorded points on the graph below and connect them with a smooth curve.
The completed graph represents a “speed – torque curve” for the DC shunt motor.
- Calculate the speed regulation of the motor using the following equation:
RPM (No Load) - RPM (Full Load) Speed Reg. = -------------------------------------------- x 100 % RPM (Full Load)
Note: Full load for this motor is 9 Lbf – in.
Manual Rev 1.
1.8.. Using the input and output power in watts, calculate the efficiency of the motor at full load.
Efficiency, F 0 6 8= 100% (P-out) / (P-in)
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1.9.. What are the losses, in watts, for the motor at full load? List some of the types of losses which occur in DC motors.
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1.10.. Will losses decrease if a cooling fan is mounted on the shaft of the motor? Explain:
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1.11.. How much larger is the starting current than the full load current?
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Manual Rev 1.
- DC SERIES MOTOR
2.1.. OBJECTIVE
To study the torque vs. speed characteristic of a series wound DC motor and calculate its efficiency.
2.2.. DISCUSSION
The Series DC Motor behaves differently than the Shunt DC Motor. As demonstrated in the previous lab, the shunt winding produces an almost constant speed of operation (low speed regulation). The series winding produces a machine with very high speed regulation.
The operating speed of the series motor is a function of its load current (as discussed below). Under heavy load, the motor operates at very low speeds, while at no load, the motor speed can be excessively high. An unloaded series motor can overspeed and literally spin itself apart. NEVER ALLOW A SERIES DC MOTOR TO OVERSPEED.
In the series motor, the armature and field windings both carry the same current. When the motor is lightly loaded, the magnetic field in the armature is weak as the motor is drawing a minimum current. When the motor is heavily loaded the motor draws a maximum current and the armature field is strong. Since the torque produced by the motor is proportional to the product of the armature current and the magnetic field in the armature, the series motor under heavy load at low speed will produce a very large amount of torque. Thus the series motor is very useful for starting large, high-inertia loads. Applications include locomotive drives (electric trains and buses), and traction motors.
2.3.. INSTRUMENTS AND COMPONENTS
Power Supply Module EMS 8821 DC Metering Module EMS 8412 DC Motor/Generator Module EMS 8211 Electrodynamometer Module EMS 8911 Hand Tachometer EMS 8920
2.4.. PROCEDURE
CAUTION! – High voltages are present in this Experiment. DO NOT make any connections with the power supply ON. Get in the habit of turning OFF the power supply after every measurement.
- Connect the following circuit. DO NOT APPLY POWER AT THIS TIME.
Manual Rev 1.
Turn on the DC power supply and increase the voltage control until the motor draws 3 amps of starting current. The motor will turn very slowly or not at all.
Measure and record the DC voltage and the torque developed.
V = __________ Volts DC Torque = __________ Lbf – in or N-m.
Return the voltage control to zero percent and turn off the power supply. (KEEP THIS CIRCUIT CONNECTED FOR THE NEXT SECTION)
The line current drawn by the motor in step 12) above is limited only by the equivalent DC resistance of the armature winding and field windings. Calculate the value of the starting current drawn by the motor if the full line voltage, 120 V DC, were applied:
Starting Current at 120 V DC = __________ Amps
2.. CONCLUSIONS
1.12.. Calculate the horsepower, HP, developed by the series wound DC motor when the load torque is 9 Lbf – in (1.05 N-m):
HP = (RPM)(Lbf – in)(1.59) / 100,000 OR HP = (RPM)(N-m)(14.07) / 100,
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1.13.. Based on the result above and knowing that 1 HP equals 746 Watts, what is the power developed by the motor in watts?
1.14.. What is the input power to the motor, in watts, as calculated using the voltage and current from the table for 9 Lbf – in of load?
1.15.. Using the input and output power in watts, calculate the efficiency of the motor at full load.
Efficiency, F 0 6 8= 100% (P-out) / (P-in)
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Manual Rev 1.
1.16.. What are the losses, in watts, for the motor at full load?
1.17.. How much larger is the starting current than the full load current?
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- DC COMPOUND MOTOR
2.5.. OBJECTIVE
To study the torque vs. speed characteristic of a cumulative compound wound DC motor and calculate its efficiency.
2.6.. DISCUSSION
A Compound DC Motor has both a series and a shunt winding. As such, its performance is a blend of the two types. When one considers the construction of a compound motor, the shunt winding could be placed before the series winding ( long shunt, in parallel with the DC source) or after the series winding (short shunt, in parallel with the armature). Also, the shunt winding can be energized so that its magnetic field “aides” the field produced by the series winding (cumulative compounding) or energized so that it opposes the series field (differential compounding). Each nuance in design changes the operating characteristics of the motor.
Most compound motors are cumulative compound, with a long shunt. The goal is usually to either have a ‘shunt’ motor with improved starting torque or a ‘series’ motor with improved speed regulation.
2.7.. INSTRUMENTS AND COMPONENTS
Power Supply Module EMS 8821 DC Metering Module EMS 8412 DC Motor/Generator Module EMS 8211 Electrodynamometer Module EMS 8911 Hand Tachometer EMS 8920
2.8.. PROCEDURE
CAUTION! – High voltages are present in this Experiment. DO NOT make any connections with the power supply ON. Get in the habit of turning OFF the power supply
Manual Rev 1.
Speed Regulation = _____________ %
Now set the dynamometer control to its full clockwise position to maximize the starting load for the motor. Do NOT adjust the field rheostat.
Turn on the DC power supply and increase the voltage control until the motor draws 3 amps of starting current. The motor will turn very slowly or not at all.
Measure and record the DC voltage and the torque developed.
V = __________ Volts DC Torque = __________ Lbf – in or N-m.
Return the voltage control to zero percent and turn off the power supply.
The line current drawn by the motor in step 11) above is limited only by the equivalent DC resistance of the armature winding and field windings. Calculate the value of the starting current drawn by the motor if full line voltage, 120 V DC, were applied:
Starting Current at 120 V DC = __________ Amps
3.. CONCLUSIONS
1.18.. Calculate the horsepower, HP, developed by the cumulative compound DC motor when the load torque is 9 Lbf – in (1.05 N-m):
HP = (RPM)(Lbf – in)(1.59) / 100,000 OR HP = (RPM)(N-m)(14.07) / 100,
1.19.. Based on the result above and knowing that 1 HP equals 746 Watts, what is the power developed by the motor in watts?
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1.20.. What is the input power to the motor, in watts, as calculated using the voltage and current from the table for 9 Lbf – in of load?
Manual Rev 1.
1.21.. Using the input and output power in watts, calculate the efficiency of the motor at full load.
Efficiency, F 0 6 8= 100% (P-out) / (P-in)
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1.22.. What are the losses, in watts, for the motor at full load?
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1.23.. How much larger is the starting current than the full load current?
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Manual Rev 1.
