Transcript DC Motors II

Lesson 29:
DC Motors II
1
Learning Objectives
• Analyze the circuit equivalent of a permanent magnetic DC
motor that accounts for armature resistance, induced
electromotive force (back EMF), developed electromagnetic
torque, and applied (input) voltage.
• Define the power output of a permanent magnetic DC motor
in terms of developed electromagnetic torque and angular
velocity. Relate power output in terms of horse power.
• Determine the efficiency of a permanent magnetic DC motor
using the given or calculated power in and power out.
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Basic DC Motor Operation
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Parts of a DC Motor
SOURCE: Gears Educational Systems.
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Rotary DC Motor
• Torque developed:
Td = Kv Ia
• Power Developed:
Pd = Td ω = Kv Ia ω
• Back EMF:
Ea = Kv ω
• Angular velocity (ω): 2 (RPM/60)
 , TLOAD
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DC Motor Power Flow
Pelecloss
Pmechloss
Pout
Pin
Pd = Pout + Pmechloss
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Electrical Power Losses
• Electrical loss (Pelecloss) occurs due to the armature
resistance and is expressed as:
Pelecloss = Ia2 Ra
Pelecloss
Ia
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Ra
Torque Losses (Tloss)
• Mechanical loss (Pmechloss) represents losses due to the
friction of mechanical parts, magnetic
inefficiencies of the material, and losses coupling
brushes and commutator and is expressed as:
Pmechloss = Tloss 
Pmechloss
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Output Power
• Pd is the power developed by the motor which includes power
out and mechanical losses. It is expressed:
Pd = Pmechloss + Pout = Td  = Kv IA 
• Power out is the power that ultimately gets to the load and is
expressed:
Pout = TLOADm
Pout
Pd = Pout + Pmechloss
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Power Conversion Diagram
Electrical
Mechanical
Pd = Ea I a = Td 
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Motor Efficiency
• Developed power is:
Pd = Ea I a = Pout  Pmechloss
• If we Ignore rotational losses, Pd = Pout, and
machine efficiency can be calculated as:
Pout
Ea I a
PD
=
100 =
100 =
100
Pin
Pin
VDC I a
Ea
=
(Tloss = 0)
VDC
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Magnetic field
• Instead of permanent magnet, we could raise the field strength
B with an electromagnet.
• The wires wrapped around a ferromagnetic core are known as
field windings.
• The field windings are stationary and are part of the stator.
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Magnetic Field
2 poles
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Magnetic Poles
• Increasing the number of poles will increase and
smooth the output torque.
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1 2
2
3
4
8
7
6
4
5
3
eight-pole dc machine
four-pole dc machine
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Example Problem 1
We wish to design a 1/4 hp, 28 V DC motor with an efficiency of 96%.
a) What current can we expect to draw?
b) If the machine constant is Kv= 0.2139 ν·s, determine Tout if we ignore mechanical losses.
c) Calculate rated speed in rpm.
a) 1hp = 746W =Pout = 0.25hp * 746W = 186.5W
=
Pout
P
186.5W
= Pin = out =
194.27W
Pin

0.96
Pi n = VDC * I a
Ia =
Pi n 194.27W
=
= 6.94 A
VDC
28V
c)  = 2
0
b) Pd = Pout  Pmechloss
Pd = Pout = K v I a 
=
Pd
186.5W
=
= 125.6 rad/s
K v I a (0.2139vs )(6.94 A)
TLOAD =
Pout

=
186.5W
= 1.485 N  m
125.6 rad/s
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RPM
125.6rad / s
= RPM =
60 = 1199.4 RPM
60
2
QUESTIONS?
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