March 2024 Directors Meeting, Division of Student Affairs and Academic Support
268320253 hmwk5-solutions
1. ECE 320
HW #5
Due: Wed, March 2
by 11:30am
Short answer questions:
1. What is commutation? How can a commutator convert ac voltages on a machine's
armature to dc voltages on it's terminals?
Commutation is the process of coverting AC voltage internal to the rotor of a DC machine to a
constant DC output voltage. To do this, the commutator bars are strategically aligned with rotor
windings and stator poles so that when the internal voltage reverses polarity, the output voltage
will remain constant.
2. Why does curving the pole faces in a dc machine contribute to a smoother dc output
voltage from it?
Curving the pole faces to match with the curvature of the rotor is done to create a constant air gap
between the rotor and the stator. Therefore, the voltage induced on a winding should theoretically
be constant the entire time it's under a pole face. (eind=2vBl for a 2 pole machine)
3. Explain the concept of electrical degrees. How is the electrical angle of the voltage in
a rotor conductor related to the machanical angle of the machine's shaft?
360 electrical degrees is the time-interval for an alternating voltage or current to complete one full
alternating cycle. In a full-pitched coil, a coil spans 180 electrical degrees. This means that if
one side of the coil is under a magnetic pole face, the opposite side of the coil will be under the
magnetic pole of opposite polarity.
The relationship of electrical to mechanical angles is shown below.
θelec
Poles
2
θmech⋅=
4. What types of losses are present in a dc machine?
Losses in a DC Machine include:
1. Electrical or Copper Losses (I2R losses) in the armature and field windings.
2. Brush losses due to a resistance in the carbon contact brush
3. Core losses from hysteresis and eddy currents
4. Mechanical losses due to windage and friction
5. Stray load losses incorporate all other miscellaneous losses
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2. 5. Given the simple rotating loop below and the defined parameters,
System parameters:
B5 0.75T:= VB 27V:=
l5 0.5m:= R5 0.5Ω:=
r5 0.15m:= ω5 250
rad
sec
:=
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3. (a) Is this machine operating as a motor or a generator? Explain.
eind_a 2 r5⋅ l5⋅ B5⋅ ω5⋅:= eind_a 28.125 V=
Since the battery voltage is only 27V, the machine is in generator mode. In this scenario, the
machine is charging the battery.
(b) What is the current flowing? Is it flowing into or out of the machine?
ib
eind_a VB−
R5
:= ib 2.25 A= Flowing out of the machine (again, this is
generator mode.)
(c) What is the power? Is it flowing into or out of the machine?
Ap π r5⋅ l5⋅:= Ap 0.236m
2
=
φc Ap B5⋅:=
τind_c
2
π
φc⋅ ib⋅:= τind_c 0.253 N m⋅⋅=
Pind τind_c ω5⋅:= Pind 63.281 W= Power at the shaft
as a check: Pind_c eind_a ib⋅ 63.281 W=:=
PB VB ib⋅:= PB 60.75W= Power at the battery
Again, the power at the shaft is larger than the power at the battery so the power flows from the
machine to the battery. This is expected because the machine is acting as a generator.
(d) If the speed of the rotor were changed to 280 rad/s, what would happen to the current?
ω5d 280
rad
sec
:=
eind_d 2 r5⋅ l5⋅ B5⋅ ω5d⋅:= eind_d 31.5 V=
id
eind_d VB−
R5
:= id 9 A= Armature current increases
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4. (e) If the speed of the rotor were changed to 220 rad/s, what would happen to the
curent? Caculate the power for this scenario.
ω5e 220
rad
sec
:=
eind_e 2 r5⋅ l5⋅ B5⋅ ω5e⋅:= eind_e 24.75V=
ie
eind_e VB−
R5
:= ie 4.5− A= Armature current decreases and now is
flowing in the opposite direction. The
machine is now operating as a motor.
In motor convention:
ie
VB eind_e−
R5
:= ie 4.5A=
τind_e
2
π
φc⋅ ie⋅:= τind_e 0.506 N m⋅⋅=
Pind_e τind_e ω5e⋅:= Pind_e 111.375 W= Power at the shaft
as a check: Pind_e eind_e ie⋅ 111.375 W=:=
PB_e VB ie⋅:= PB_e 121.5W= Power at the battery
Power at the shaft is smaller than the power at the battery so the power flows from the battery
to run the the machine. This is expected because the machine is a motor for this speed
scenario.
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