EGR 342 Dynamic Systems
Lab 5 - Modeling of a Motor-Generator Set
Fall 2013
Objective
The objective of this part of the lab exercise is to construct a Simulink model for a motorgenerator set, where both machines are armature-controlled. Nominal system parameters are
given and steady-state and transient data are provided to identify better values for system
parameters. Once system parameters are identified the model will be used to predict the system
response of this motor-generator system.
Introduction:
Electric motors and generators are both machines which are used to convert one type of energy
into another usable form. In most commercial electric motors and generators, a magnetic field
acts as a coupler between a stationary member, called the stator, and a moving member called the
rotor. The stator may be made from permanent magnets, which is usually used in smaller
machines, or made from coils of wire, which is common in larger machines. The purpose of the
stator in an armature controlled motor is to set up a constant magnetic flux through the machine.
The rotor contains a number of wire coils, each identical and distributed around the periphery of
the rotor. These wire coils on the rotor are called the armature windings. The armature windings
are connected to a segmented disk called the commutator. The segmented nature of the
commutator acts as a switching mechanism as the rotor turns. At different positions of the rotor,
different windings of the armature are connected to the external leads. If electricity is supplied
to the leads, then the two magnetic fields which are set up by the stator and the armature will
effectively push off of one another. When the armature coils are powered in just the correct
sequence, forces are produced which cause the rotor to continually rotate. This is what we call
an electric motor.
If the leads are connected to a closed circuit when the rotor is manually driven, a current is
induced in the wire leads as the armature coils cut through the magnetic flux field supplied by
the stator. In this mode of operation the device is
called a generator and provides a way to convert
Tg
rotational mechanical energy back into electrical
Generator
energy.
Jg
+θg, ωg
Simply put, an electric motor and electric generator are
k
B
g
essentially the same device. The difference is that the
B
.
m
motor converts electrical energy into rotation
Motor
mechanical energy while a generator converts
Jm
rotational mechanical energy into electrical energy.
Tm
A schematic of a coupled motor-generator system is
+θm, ωm
shown in Figure 1.
Figure 1: Schematic of a motor generator system
1
Task 1 – Examination of DC Motors
A large and small DC motor have been disassembled and made available for your examination.
The smaller of the two motor has two permanent magnets which create the magnetic field of the
stator. The larger motor has wire field coils housed in the stator to create the magnetic field of
the stator.
In the space below, draw a diagram of each motor, identifying and labeling the following parts.
If you wish to turn in an electric copy, take a picture of the motor using the camera on your
laptop. If you do not have a different program installed, you can use the “Lenovo – Web
Conferencing” application. Paste the photo into PowerPoint to add labels.
-- stator
-- rotor
-- armature
-- commutator
-- brushes
-- bearings
-- permanent magnets or field winding
2
Task 2-- Motor Generator Demonstration Station.
A motor-generator system has been assembled for you to examine and run. This system has
been built from two small DC motors each with a permanent magnet generated stator field.
Figure 3a : Diagram of Motor / Generator Setup
Start by identifying important components on the motor/generator experiment board and
connecting wires to the locations indicated as shown in Figure 3. Connect the motor to the power
supply using alligator clips on the screws. (Do not turn on the power supply until ready to run.)
Connect a different set of alligator clips directly to the motor leads. These will go to the NI
myDAQ inputs AI+/-0. Connect another set of wires for the generator with alligator clips to the
screws on the other side of the board and to the AI +/-1 inputs to the NI myDAQ system. You
will not need connect to the load cell at this point.
3
Figure 3b : Motor / Generator Setup Experiment Board
Figure 3c: Power supply
Figure 3d: NI myDAQ
Once you are satisfied that all of the connections have been made, test the system. Turn on the
power supply to approximately 6 – 8 V and turn the switch on the experiment board to the “ON”
position. From this point forward, the switch on the board should be used to turn the motor on
and off. Turn the switch on the power supply off if you need to move wires around. Once it
appears that the system is working mechanically, launch the Oscilloscope in Lab view software.
As shown in Figure 4, this can be found under the NI ELVISmx Instrument Launcher. Once the
Launcher is running, select “Scope”
4
Figure 4 : Launching the NI ELVISmx Oscilloscope.
Once the oscilloscope is open, make sure both the AI 0 and the AI 1 channels are enabled. This
must be done both in the upper right and just under the display plot. In addition, change the
Volts/Div scales on both of the inputs to 2V and change the Time/Div to 20 ms. You are now
ready to collect data.
To collect data, press the “Run” button at the bottom of the screen. To stop collecting data and
save, select the “Stop” button and “Log” . This will open a dialog box asking you where you
want to save the file. Note, that this will only save the data currently on the screen.
You are to run the system and examine the following for a voltage step input to the motor.
Record the following values:
Table 1: Parameters of Demonstration Motor-Generator System.
Quantity
Steady State Motor Voltage, vam_ss
Steady State Voltage across the Load, vload
Load Resistance, Rload
Armature Resistance of Motor/Generator, Ram = Rag
5
Units
On the graph below draw or paste
a) the transient behavior of the Motor Voltage, vam as shown on the oscilloscope trace.
b) the transient signal of the Generator Voltage, vgm as shown on the oscilloscope trace.
Note, to capture the transient behavior, you must stop and save the data shortly after flipping the
switch on the experimental board. (Note, do not use the switch on the power supply.) It may take
a few tries to get the timing correct. Your data should look like the data on the oscilloscope in
Figure 4 above.
-6.60E+00
29:58.6 29:58.6 29:58.6 29:58.6 29:58.6 29:58.6 29:58.6
-6.65E+00
-6.70E+00
-6.75E+00
Voltage ( V )
-6.80E+00
-6.85E+00
-6.90E+00
-6.95E+00
-7.00E+00
-7.05E+00
Time ( s )
6
Task 3 – Develop a Simulink model
The first step in development of a motor generator Simulink model is to develop the equation of
motion for the system. The diagram shown in Figure 3 is a schematic representation for the
motor-generator set that will be modeled in the lab. The subscripts “M” and “G” refer to motor
and generator respectively. The compliance of the shaft is also accounted for on the model. The
model shows both circuits for the armature and the field coils. The circuits identified with "F"
refer to field coils used to create the magnetic field of the stator that the armature field pushes
against.
Figure 5: Model of motor generator system
The output variables we will consider in this lab will be the armature current, iam, the generator
current, iag, the angular velocity of the motor, m, the angular velocity of the generator, g, and
the voltage across the load, vg. Since the machines are armature-controlled, both field currents
are assumed to be held constant at their rated values of 0.5 A.
The equation of motion (EOM) for the motor armature circuit can be obtained using Kirchoff's
Voltage Law (KVL) and applying the definition of the back EMF to result in Eq. 1.
di am
(eq 1)
 K e m  0
dt
A Free Body Diagram (FBD) for the motor rotor is shown in Figure 6. Applying the
rotational form of Newton's 2nd Law, the EOM for the rotor of the motor is given by Eq. 2.
 v m  i am R am  L am
Jm
d 2m
d
 K t i am  B m  k  m   g 
2
dt
dt
7
(eq 2)
A FBD for the generator rotor is shown in Figure 7. Applying Newton's 2nd Law gives the
equation for the rotor of the generator as Eq. 3.
Jg
d 2 g
dt 2
 K t i ag  B
d g
dt
 k g   m 
(eq 3)
Applying KVL the EOM for the armature circuit of the generator (in the Laplace domain) is
given by Eq. 4.
diag
 K e g  iag Rag  Lag
 iag RL  0
(eq 4)
dt
8
Procedure for drawing a block diagram
Take the equations numbered (1) to (4) and develop a simulation diagram for both machines
coupled together. Each summing block corresponds to one of the numbered equations. The output
of each summing junction is labeled to help you get started. Note: you will find it helpful to
rewrite equations (1) to (4) as Laplace transforms before completing the block diagram. If you
wish to turn in an electronic copy, turn in the final Simulink diagram.
( Lam s  Ram ) I am
Σ
( Js  B) m
Σ
( Js  B) G
Σ
( Rag  RL  Lag s ) I ag
Σ
9
The Figures 8a and 8b below show the setup of a larger motor-generator pair and the
instrumentation that has been used to collect data for this experiment. .
Measurement devices
Control panel
Motor-generator
Figure 8a: Instrumentation Equipment
Generator
Motor
Flexible coupling
Figure 8b. A close-up of the motor-generator.
10
Nominal values for the system parameters (nominal means ball park, but maybe not accurate) for
the motor and generator are shown below:
Armature Resistance = 3.33 
Armature Inductance = 81.7 mH
Motor Voltage = 120 V
Load Resistance = 85.7 
Torque Constant = Voltage Constant = 0.9234 V/(rad/s) when both field currents are at 0.5A
Shaft Stiffness = 0.5 N-m/rad
Rotor Inertia = 2.33 x 10-3 kg-m2
Viscous Damping = 6 x 10-4 N-m-sec
Using the nominal values for the system parameters set up a Simulink model which implements
your simulation diagram. The input is vm which is applied as a step at t = 0. The output variables
are: iam, iag, vg, m, and g. Be sure to use integrators instead of differentiators.
Task 4 – System parameter identification
Instead of having you collect the results from this system, you be given both the steady state and
transient results. You are to use this information to make a comparison to your simulated model
behavior.
Steady State Results:
The following data represents measured results from running the motor-generator with the
specifications given above. These were generated for a power supply step function input set to
120 Volts applied to the motor input. .
Va = 112.2 V - the reason this is not 120 V is the supply has some impedance.
Ia = 1.64 A
Ig = 1.20 A
Speed = 1358 RPM
Motor Torque = 1.2 N-m
Generator Torque = 0.93 N-m
Compare these steady-state values with the predictions given by your Simulink model. There
may be some difference because not all of the parameters originally given were known with
great certainty. The parameters that are known best are probably the resistance values. The
other parameters, such as Armature Inductance, Damping Constant, Rotor Inertia, Shaft
Stiffness, and the Torque and Motor Constants have much more uncertainty associated with
them. The last page of this handout has some suggestions on how to perform this steady-state
system identification. In other words, can you slightly change the parameters with uncertain
values to better match the results which were measured.
11
Transient analysis
From the steady-state equations it is clear that not all of the system parameters appear in the
steady-state equations so in order to better identify these parameters some transient results are
required. The measured transient for the armature current of the motor is shown in Figure 9
when a step input, vm is applied to the system. Figure 9b is the same data as Figure 9a but the
scale is different. The sharp spike in Figures 9 due to a phenomenon called “armature reaction”
that is not included in your model. The transient for the generator armature current is shown in
Figure 9.
Try to match the approximate peak amplitude and low frequency transient responses by altering
parameters of your Simulink model until you have model with matches both the steady state
results and also shows some of the peak transient behavior shown. Report the final values you
have selected for your parameters.
y-axis = 2A/div
x-axis = 0.1 sec/div
y-axis = 5A/div
x-axis = 0.1 sec/div
Figure 9a: Motor armature current,
enlarged scale
Figure 9b: Motor armature current,
reduced scale
y-axis = 0.5 A/div
x-axis = 0.1 sec/div
Figure 9: Generator armature current
12
Task 5 – Use the model to explore system characteristics
After obtaining parameter values that match the transient and steady-state response use your
model to explore this motor-generator system. For example, plot the efficiency of the system as
a function of the load on the system, RL, or as a function of the steady-state angular velocity for
the system. The efficiency is defined to be:
P
  in ,motor
Pout , generator
where power, P, is defined as
P V I
It might be useful to fill out a table similar to the one shown in Table 2. This table may have as
many rows as needed to get a good plot.
Table 2: Sample table format that can be used to generate plots for the
efficiency of the motor-generator system
Load, RL
Speed
(rad/s)
V
(V)
motor
I
P
(A) (W)
generator
V
I
P
(V)
(A) (W)
overall

Reporting your results
First, your team will write a memo for this lab which will be due in 1 week.
The purpose the memo is to describe
 the simulation
 how you performed the system identification and made parameter modifications
 to present your results.
Be sure to discuss what parameters were adjusted and why. Include a comparison of the steadystate results from your model and those from the experiment. You should also have plots of all
your output variables and any others figures that help the reader understand this system. Every
figure should have a figure number and title and should be referred to by number, and discussed,
in the text. You should also discuss the transient simulation results and the actual oscilloscope
output traces.
13
Suggestions for lab 4 system identification and modeling
A steady state analysis can help us determine the necessary values of our parameters. At steady
state the four equations given
(1)
v m  R am i am  K e m
K T i am  B m  k ( m   g )
(2)
k ( m   g )  Bg  K T i ag
(3)
K e g  (R ag  R L )i ag
(4)
Parameter Values (Identical machines) from handout:
Armature Resistance = 3.33 
Armature Inductance = 81.7 mH,
Motor Voltage = 120 V,
Load Resistance = 85.7 
Torque Constant = Voltage Constant = 0.9234 V/rad/s,
Shaft Stiffness = 0.5 N-m/rad,
Rotor Inertia = 2.33 x 10-3 kg-m2,
Viscous Damping = 6 x 10-4 N-m-sec.
Steady-State values from handout (assume these values are correct):
Va = 112.2 V - the reason this is not 120 V is the supply has some impedance.
Ia = 1.64 A
Ig = 1.20 A
Speed = 1358 RPM = 142.21 rad/s
Motor Torque = 1.2 Nm
Generator Torque = 0.93 Nm
Observations/Notes:
 From Eq.1 or Eq. 4 you can solve for Ke. How do these numbers compare to the original
value given. The original number given may be wrong so use this value instead.
 Substituting Eq. 3 into Eq. 2 give you a way of estimating B.
 The motor torque is about KT iam whereas the generator torque is less because it is KT iag
(it is smaller because of losses due to damping).
 You can modify your block diagram to include the
source resistance as shown below. You can choose
the
value of Rs knowing the steady state voltage (112.2V)
and the steady state current (1.64 A)
14
TG
Generator
JG
EGR 342 Lab5: Summary of Motor-Generator system:
+θG, ωG
k
Motor: a device which transforms
electrical to mechanical energy.
Generator: a device which transforms
mechanical to electrical energy
Motor
BG
BM
JM
TM
+θM, ωM
In practice a motor may be used as a
generator and vise versa.
if
RaM
LaM
BMωM
Lf
iAM
Rf
TM
vin
vbM M
Tout
JM
Jm Bm
TM
KT
Ke
Armature Controlled Motor Model:
Electrical:
di
LaM aM  RaM iaM  K eM  vin
dt
Motor
+θM, ωM
Mechanical
d M
JM
 BM M  KT iaM  TOut
dt
LaG
TG
RaG
Generator
JG
Tin
+θG, ωG
BGωG
TG
iaG
G
vbG
Ke
JG BG
KT
Generator( using armature output circuit):
Electrical:
K eG
di
 LaG aG  RaG iaG  RLoad iaG  0
dt
so
di
LaG aG  RaG iaG  RLoad iaG  K eG  0
dt
Mechanical:
Tin  BGG  KT iaG  J G
dG
dt
so
dG
JG
 BGG  KT iaG  TIn
dt
15
Loa
d
RLoad