ECE 4501
Power Systems Laboratory Manual
Rev 1.0
5.0 DC MOTORS
5.1
DC SHUNT MOTOR
5.1.1
OBJECTIVE
To study the torque vs. speed characteristic of a shunt wound DC motor and calculate its
efficiency.
5.1.2
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.
5.1.3
INSTRUMENTS AND COMPONENTS
Power Supply Module
DC Metering Module
DC Motor/Generator Module
Electrodynamometer Module
Hand Tachometer
5.1.4
EMS 8821
EMS 8412
EMS 8211
EMS 8911
EMS 8920
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.
1) Connect the following circuit. DO NOT APPLY POWER AT THIS TIME.
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
2) 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 and 0).
3) 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.
4) 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.
5) 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.
6) 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
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
7) 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.
8) Measure and record the line current and motor speed for the 3 lbf-in load condition.
9) 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.
10) Return the voltage control to zero percent and turn OFF the power supply.
11) 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.
12) Calculate the speed regulation of the motor using the following equation:
Speed Reg. =
RPM (No Load) - RPM (Full Load)
-------------------------------------------- x 100 %
RPM (Full Load)
Note: Full load for this motor is 9 Lbf – in.
Speed Regulation = _____________ %
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
13) Now set the dynamometer control to its full clockwise position to maximize the starting load
for the motor. Do NOT adjust the field rheostat.
14) 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.
15) Measure and record the DC voltage and the torque developed.
V = __________ Volts DC
Torque = __________ Lbf – in or N-m.
16) Return the voltage control to zero percent and turn off the power supply.
17) The line current drawn by the motor in step 14) 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
5.1.5
CONCLUSIONS
5.1.5.1
Calculate the horsepower, HP, developed by the shunt 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) / 100,000
_______________________________________________________________________
_______________________________________________________________________
5.1.5.2
Based on the result above and knowing that 1 HP equals 746 Watts, what is the
power developed by the motor in watts?
_______________________________________________________________________
_______________________________________________________________________
5.1.5.3
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?
_______________________________________________________________________
_______________________________________________________________________
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ECE 4501
Power Systems Laboratory Manual
5.1.5.4
Rev 1.0
Using the input and output power in watts, calculate the efficiency of the motor at
full load.
Efficiency,  = 100% (P-out) / (P-in)
_______________________________________________________________________
_______________________________________________________________________
5.1.5.5
What are the losses, in watts, for the motor at full load? List some of the types of
losses which occur in DC motors.
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
5.1.5.6
Will losses decrease if a cooling fan is mounted on the shaft of the motor? Explain:
_______________________________________________________________________
_______________________________________________________________________
5.1.5.7
How much larger is the starting current than the full load current?
_______________________________________________________________________
_______________________________________________________________________
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ECE 4501
5.2
5.2.1
Power Systems Laboratory Manual
Rev 1.0
DC SERIES MOTOR
OBJECTIVE
To study the torque vs. speed characteristic of a series wound DC motor and calculate its
efficiency.
5.2.2
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.
5.2.3
INSTRUMENTS AND COMPONENTS
Power Supply Module
DC Metering Module
DC Motor/Generator Module
Electrodynamometer Module
Hand Tachometer
5.2.4
EMS 8821
EMS 8412
EMS 8211
EMS 8911
EMS 8920
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.
1) Connect the following circuit. DO NOT APPLY POWER AT THIS TIME.
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
2) Set the Dynamometer (Or Electrodynamometer) control knob in the middle of its range (to
give the DC motor a load to start against).
3) Turn on the power supply (also the 24 Vac supply if the PM/Dynamometer is used) and
slowly increase the voltage control until the motor starts to turn. If the motor turns
counterclockwise, turn OFF the power supply and swap the leads on the series field. Adjust
the voltage for 120 V dc operation.
4) Increase the load on the series motor by adjusting the Dynamometer control knob
counterclockwise. Adjust for a load of 12 lbf-in or 1.4 Newton-meters.
5) Measure the line current and motor speed and record them in the table below.
6) Decrease the load on the motor and measure the current and speed for the other values of load
torque in the table.
7) Return the voltage control to zero percent and turn OFF the power supply.
Source (Volts) Line Current (Amps) Speed (RPM) Torque (N-M, Lbf-In)
120
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
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
8) Plot the recorded values on the graph below and connect them with a smooth curve. The
completed graph represents a “speed – torque curve” for the DC Series Motor.
9)
Calculate the speed regulation of the motor using the following equation:
Speed Reg. =
RPM (No Load) - RPM (Full Load)
-------------------------------------------- x 100 %
RPM (Full Load)
Note: Full load for this motor is 9 Lbf – in.
Speed Regulation = _____________ %
10) Now set the dynamometer control to its full clockwise position to maximize the starting load
for the motor. Do NOT adjust the field rheostat.
11) 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.
12) Measure and record the DC voltage and the torque developed.
V = __________ Volts DC
Torque = __________ Lbf – in or N-m.
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
13) Return the voltage control to zero percent and turn off the power supply. (KEEP THIS
CIRCUIT CONNECTED FOR THE NEXT SECTION)
14) 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
5.2.5
CONCLUSIONS
5.2.5.1
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) / 100,000
_______________________________________________________________________
_______________________________________________________________________
5.2.5.2
Based on the result above and knowing that 1 HP equals 746 Watts, what is the
power developed by the motor in watts?
_______________________________________________________________________
_______________________________________________________________________
5.2.5.3
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?
_______________________________________________________________________
_______________________________________________________________________
5.2.5.4
Using the input and output power in watts, calculate the efficiency of the motor
at full load.
Efficiency,  = 100% (P-out) / (P-in)
_______________________________________________________________________
_______________________________________________________________________
5.2.5.5
What are the losses, in watts, for the motor at full load?
_______________________________________________________________________
_______________________________________________________________________
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ECE 4501
Power Systems Laboratory Manual
5.2.5.6
Rev 1.0
How much larger is the starting current than the full load current?
_______________________________________________________________________
_______________________________________________________________________
5.3
DC COMPOUND MOTOR
5.3.1
OBJECTIVE
To study the torque vs. speed characteristic of a cumulative compound wound DC motor and
calculate its efficiency.
5.3.2
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.
5.3.3
INSTRUMENTS AND COMPONENTS
Power Supply Module
DC Metering Module
DC Motor/Generator Module
Electrodynamometer Module
Hand Tachometer
5.3.4
EMS 8821
EMS 8412
EMS 8211
EMS 8911
EMS 8920
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.
1) Start with the DC Series Motor as connected in 5.2.4 in the previous section, making sure
that it will turn in a clockwise direction.
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ECE 4501
Power Systems Laboratory Manual
2)
Rev 1.0
Add the shunt winding by connecting terminal 1 to 5, 6 to 7 and 8 to 4, as shown in the
diagram below:
3) With the Dynamometer control set for zero load (fully counter clockwise), turn on the
power supply and adjust the voltage control for 120 V dc. If the motor runs at
excessively high speed (> 2000 RPM) the shunt winding is wired for differentialcompound operation. If this is the case, return the voltage to zero and turn OFF the
power supply. Interchange the leads to terminals 5 and 8 of the shunt winding and repeat
step 3.
4) Adjust the shunt field rheostat until the no-load motor speed is 1800 RPM. DO NOT
CHANGE THE RHEOSTAT SETTING FOR THE REMAINDER OF THIS
EXPERIMENT.
5) Measure and record line current and motor speed for each value of load torque indicated
in the table below.
Source (Volts) Line Current (Amps) Speed (RPM) Torque (N-M, Lbf-In)
120
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
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
6) Return the voltage control to zero percent and turn OFF the power supply.
7) Plot the recorded values on the graph below and connect them with a smooth curve. The
completed graph represents a speed – torque curve for the DC Compound Motor.
8) Calculate the speed regulation of the motor using the following equation:
Speed Reg. =
RPM (No Load) - RPM (Full Load)
-------------------------------------------- x 100 %
RPM (Full Load)
Note: Full load for this motor is 9 Lbf – in.
Speed Regulation = _____________ %
9) Now set the dynamometer control to its full clockwise position to maximize the starting
load for the motor. Do NOT adjust the field rheostat.
10) 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.
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ECE 4501
Power Systems Laboratory Manual
Rev 1.0
11) Measure and record the DC voltage and the torque developed.
Torque = __________ Lbf – in or N-m.
V = __________ Volts DC
12) Return the voltage control to zero percent and turn off the power supply.
13) 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
5.3.5
CONCLUSIONS
5.3.5.1
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) / 100,000
_______________________________________________________________________
_______________________________________________________________________
5.3.5.2
Based on the result above and knowing that 1 HP equals 746 Watts, what is the
power developed by the motor in watts?
_______________________________________________________________________
_______________________________________________________________________
5.3.5.3
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?
_______________________________________________________________________
_______________________________________________________________________
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ECE 4501
Power Systems Laboratory Manual
5.3.5.4
Using the input and output power in watts, calculate the efficiency of the motor at
full load.
Efficiency,  = 100% (P-out) / (P-in)
_______________________________________________________________________
_______________________________________________________________________
5.3.5.5
What are the losses, in watts, for the motor at full load?
_______________________________________________________________________
_______________________________________________________________________
5.3.5.6
Rev 1.0
How much larger is the starting current than the full load current?
_______________________________________________________________________
_______________________________________________________________________
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