G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Chapter 18 Instructor Notes
The content of Chapter 18 is somewhat unusual for a textbook of this nature. The intent of this chapter is to provide a reasonably quantitative overview of the operation of small electric machines (mostly motors). In many practical industrial applications, ranging from servos for robots, to drug delivery systems, to actuation devices for control systems, to manufacturing equipment, to fluid power systems, small motors find widespread application.
The first section discusses the brushless DC motor, including the basics of the electronic circuits that make its operation possible. The second section introduces the stepper motor and its drive. In section
18.3, the switched reluctance machine is introduced; this is a new section in the third edition, motivated by the increasing interest in this family of machines for industrial applications. Next, single phase AC motors are discussed in Section 18.4, starting with the universal motor, and continuing with a classification of single phase induction motors, which includes split-phase, capacitor-type and shaded-pole motors. The presentation detail is sufficient to permit quantitative analysis of these motors using circuit models. The final section, 18.5, on motor selection and application, introduces some of the basic ideas behind motor selection and performance calculations. This section, which describes calculations related to reflected load inertias in the presence of mechanical gear reductions, and calculations of acceleration, torque, efficiency, and thermal loading, could be covered at any point in Chapter 17 or 18, even if the material in Sections
18.1-18.4 is not covered.
The examples given in the chapter are supplemented by over thirty homework problems, some of which are extensions of the examples presented in the text. Problems 18.5 and 18.9 require some background in digital logic circuits (Chapter 13); problems 18.7, 18.36 and 18.37 require some background in system dynamics; all remaining problems can be solved strictly on the basis of the material covered in the chapter.
Learning Objectives
1. Understand the basic principles of operation of brushless DC motors, and the trade-offs between these and brush-type DC motors. Section 1.
2. Understand the operation and basic configurations of step motors, and understand step sequences for the different classes of step motors. Section 2.
3. Understand the operating principles of switched-reluctance machines. Section 3.
4. Classify and analyze single-phase AC motors, including the universal motor and various types of single-phase induction motors, using simple circuit models. Section 4.
5. Outline the selection process for an electric machine given an application; perform calculations related to load inertia, acceleration, efficiency, and thermal characteristics.
Section 5.
18.1

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Section 18.1: Brushless DC Motors
Problem 18.1
Solution:
Known quantities:
A permanent magnet six-pole two-phase synchronous machine.
λ m
=
0 .
1 V
⋅ s .
Find:
The amplitude of the open-circuit phase voltage measured when the rotor is turned at n
=
60 rev sec .
Assumptions:
None.
Analysis:
We know that
λ m p
=
=
6
0 .
1 V m
=
2
⋅ s
ω m
=
60 rev
Let flux linkage
λ = s
λ m
=
2
π sin
ω
× t
60 rad
, where s
ω =
= p
2
2
π
ω
× m
=
180
3
× rad
60 s
=
180 rev
=
360
π s rad s
Then, the generated voltage:
V
= e
= d
λ dt
= d (
λ m sin dt
ω
V m
=
=
V m
ωλ cos m
ω
= t
360
π ×
0 .
1
= t )
= ωλ
113 .
1 V m cos
ω t
______________________________________________________________________________________
Problem 18.2
Solution:
Known quantities:
A four-pole two-phase brushless dc motor. n
=
3600 rev min . The open-circuit voltage across one of the phases is 50 V .
Find: a)
λ
. b) The no-load rotor speed
ω in rad/s when the mechanical source is removed and
V a
=
2 25 cos
θ
, V b
=
2 25 sin
θ
, where
θ = ω e t .
Assumptions:
None.
Analysis:
We know that
18.2

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 a) let e
=
V
= p
=
4 m
ω m
V n
=
=
2
=
3600
50 V rev min
2 V sin
θ =
2 V sin
ω t
ω = p
ω m
2
⋅
2
π rad min
=
2
×
60
×
2
π rad s
λ =
³ t
0 edt
=
³ t
0
2 V sin
ω tdt
=
ω
2
V cos
ω t
=
2
240
π
50 cos 240
π t
=
0 .
094 cos
ω t b)
Symmetric voltages in symmetric windings produce a rotational field in voltage with frequency
Let f s
=
3600 rev min . Then, the rotor speed is
ω m
= p
ω
2
=
3600
2
=
1800 rev min
=
2
π
60
×
1800 rad s
=
60
π rad s f s
.
______________________________________________________________________________________
Problem 18.3
Solution:
Known quantities:
T
1
= T
3
= 1 s (see Figure 18.7); maximum motor rpm, n max
= 1,800 rev/min.
Find:
T
2
.
Assumptions:
The motor covers 0.5 m in 100 revolutions.
Analysis:
The maximum rotational velocity of the motor is: v = n max
/60 = 1,800/60 = 30 rev/s. Using the expression derived in Example 18.2, we know that the maximum motor rotational velocity is: d v
1
2
T
1
T
2
1
2
T
3 and we can calculate T
T
2 d v
1
2
T
1
1
2
T
3
2
as follows:
100 rev
30 rev/s
1 s 2.33 s
Thus, the total trapezoidal profile has been shortened by 2/3 s.
______________________________________________________________________________________
Problem 18.4
Solution:
Known quantities:
Desired load motion profile (Figure P18.4). The motor covers 0.5 m (100 revolutions) in 3 s.
Find:
Maximum motor speed, acceleration and deceleration times.
Assumptions:
Assume a triangular speed profile..
18.3

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Analysis:
To simplify the analysis, choose a symmetrical speed profile; thus, the motor will accelerate for 1.5 s and decelerate for 1.5 s, or T
1
= T
2
= 1.5 s.
Using the results of Example 18.2, if we set the flat portion of the speed profile ( T
2
in Example 18.2) to zero, we can write an expression for the total motor travel. d v
1
2
T
1
1
2
T
2 and calculate the maximum motor speed to be: v
1
2
T
1 d
1
2
T
2
1
2
100 rev
1.5
which corresponds to n max
1
1.5
s
66.67 rev/s
2
= 66.67
×
60 = 4,000 rev/min.
______________________________________________________________________________________
18.4

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Section 18.2: Stepping Motors
Problem 18.5
Solution:
Known quantities:
Variable-reluctance step motor of Example 18.4 (Figure 18.11)
Find:
Design a logic circuit to achieve the step sequence given in Table 18.4 (see below)
Assumptions:
Hint: Use a counter and logic gates
Analysis:
Table 18.4: Current Excitation Sequence for VR Step Motor
S
A
S
B
S
C
S
D
Rotor
Position
1 0 0 0 0°
1 1 0 0 45°
0 1 0 0 90°
There are eight possible configurations for the motor. Hence, a 3-bit binary counter was chosen that pulses every 45°. The table below lists the corresponding logic for the binary counter.
Binary Counter Step Motor Response b
0 b
1 b
2
S
A
S
B
S
C
S
D
Rotor
Position
0 0 0 1 0 0 0 0°
1 0 0 1 1 0 0 45°
0 1 0 0 1 0 0 90°
Next, convert the truth table to a logical expression for each of the outputs:
S
A
= b
0
⋅ b
1
⋅ b
2
+ b
0
⋅ b
1
⋅ b
2
+ b
0
⋅ b
1
⋅ b
2
= b
1
⋅ b
2
( b
0
+ b
0
)
+ b
0
⋅ b
1
⋅ b
2
S
A
S
B
S
C
=
=
= b
1 b
0 b
0
⋅
⋅
⋅ b
2 b
1 b
1
⋅
⋅
+ b
0 b
2 b
2
+
+
⋅ b
1 b
0 b
0
⋅
⋅
⋅ b
2 b
1 b
1
⋅
⋅ b
2 b
2
+
+ b
0 b
0
⋅
⋅ b
1 b
1
⋅ b
2
⋅ b
2
=
= b
2 b
1
⋅
( b
0 b
2
⋅
( b
1 b
0
+
+ b
0 b
0
⋅
) b
1
+
+ b
0
⋅ b
0 b
1
⋅
⋅ b
1 b
2
)
S
C
S
D
=
= b
1 b
0
⋅ b
2
⋅ b
1
⋅
+ b
0 b
2
+
⋅ b
1 b
0
⋅ b
2
⋅ b
1
⋅ b
2
+ b
0
⋅ b
1
⋅ b
2
= b
2
( b
0
⋅ b
1
+ b
0
⋅ b
1
+ b
0
⋅ b
1
)
18.5

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
From these four expressions, a logic circuit diagram can be made for each individual output, and tied together to achieve the desired step response.
3-B it B inary
Cou nter b0 b1 b2
NO T
A ND
NO T
O R
S a
A ND
A ND
A ND
O R
NO T
A ND
O R
NO T
S b
A ND
A ND
NO T
NO T
A ND
NO T O R
S c
A ND
A ND
A ND
NO T
O R
A ND
O R
NO T
S d
A ND
A ND
______________________________________________________________________________________
18.6

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Problem 18.6
Solution:
Known quantities:
PM stepper motor with 6 poles, bipolar supply,
Find:
Smallest achievable step size.
Assumptions:
None
Analysis:
With reference to Example 18.3, we see that the half-step sequence for the 2-phase 4-pole motor leads to
45-degree steps. The addition of two poles will reduce the step size by 50%, resulting in 30-degree steps.
______________________________________________________________________________________
Problem 18.7
Solution:
Known quantities:
J m
, J
L
, D , T f
.
Find:
The dynamic equation for a stepping motor coupled to a load.
Assumptions:
None.
Analysis:
The equation will have the following form:
V
=
Ri
+
L di
+
K
E
ω dt
T
=
K
T i
=
( J m
+
J
L
) d
ω
+
D
ω +
T
F
+
T
L dt
______________________________________________________________________________________
Problem 18.8
Solution:
Known quantities:
A hybrid stepper motor capable of
Find:
Sketch the rotor-stator configuration of the motor.
Assumptions:
None.
Analysis:
The rotor and stator configuration is shown below:
18.7

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
The motor has 5 rotor teeth and 4 stator teeth (two phases).
______________________________________________________________________________________
Problem 18.9
Solution:
Known quantities:
Shown in Check Your understanding on page 928.
Find:
A binary counter and logic gates to implement the stepping motor binary sequence.
Assumptions:
None.
Analysis:
______________________________________________________________________________________
Problem 18.10
Solution:
Known quantities:
A two-phase permanent magnet stepper motor has 50 rotor teeth. When driven at
ω measured open circuit phase peak-to-peak voltage is 25 V .
=
100 rad s , the
Find: a) Calculate
λ b) Express the developed torque when i a
=
1 A and i b
=
0 .
Assumptions:
The winding resistance is 0 .
1
Ω
.
Analysis: a)
We know that p
=
50 m
=
2
ω m
=
100 rad s
18.8

G. Rizzoni, Principles and Applications of Electrical Engineering
Strikes per rev.:
N
= p
⋅ m
=
100
Problem solutions, Chapter 18
∆ θ
λ =
=
=
³ t
0
V
0
ω m
2
π
N edt
⋅
=
=
2
π
=
100
³
0
π
50
V
0
ω m
2
π
50 d
θ
π
50
=
12 .
5
100
⋅
π
50
=
0 .
00785 V
⋅ s b)
Let the winding resistance be represented by R , then w
V
= k a
ω m
+
R w
I where
Then T k a
=
T
= k
T
I
= k
T k a
I
=
=
V
V
−
ω
−
R w
I
ω m
R w
I m
N
⋅ m
______________________________________________________________________________________
Problem 18.11
Solution:
Known quantities:
The schematic diagram of a four-phase, two-pole PM stepper motor is shown in Figure P18.11. The phase coils are excited in sequence by means of a logic circuit.
Find:
The no-load voltage of the generator and terminal voltage at half load.
Assumptions: a) The logic schedule for full-stepping of this motor. b) The displacement angle of the full step .
Analysis: a)
For full step clockwise rotation is:
Phase 1
→ phase 4
→ phase 3
→ phase 2
→ phase 1 b)
18.9

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
The displacement angle of the full step sequence is 90
,
.
______________________________________________________________________________________
Problem 18.12
Solution:
Known quantities:
A PM stepper motor provides a full-step angle of 15
,
.
Find:
The number of stator and rotor poles.
Assumptions:
None.
Analysis:
The motor will require 24 stator teeth and 2 rotor teeth.
______________________________________________________________________________________
Problem 18.13
Solution:
Known quantities:
A bridge driver scheme for a two-phase stepping motor is as shown in Figure P18.13.
Find:
The excitation sequences of the bridge operation.
Assumptions:
None.
Analysis:
CK R 1 2 3 4 5 6 7 8
S1 1 1 0 0 0 0 0 1 1
S2 0 0 0 1 1 1 0 0 0
S3 0 0 0 1 1 1 0 0 0
S4 1 1 0 0 0 0 0 1 1
S5 0 1 1 1 0 0 0 0 0
S6 0 0 0 0 0 1 1 1 0
S7 0 0 0 0 0 1 1 1 0
S8 0 1 1 1 0 0 0 0 0
Where "1" means switch is closed.
______________________________________________________________________________________
18.10

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Problem 18.14
Solution:
Known quantities:
A PM stepper motor provides a full-step angle of 15
,
. It is used to directly drive a 0 .
100 in .
lead screw.
Find: a) The resolution of the stepper motor in steps/revolution. b) The distance the screw travels in inches for each step c) The number of full 15
, steps required to move the lead screw and the stepper motor shaft through
17 .
5 revolution. d) The shaft speed (in rev/min) when the stepping frequency is 220 pps .
Assumptions:
None.
Analysis: a) steps revolution
=
360
=
24
15 b) d
=
0 .
1
'' ×
15
,
360
,
=
0 .
0042
'' c) steps
=
175 rev
×
24 steps rev
=
420 steps d)
# pulses
=
# steps
Ÿ n
SH
=
220 steps
× s
1 rev
24 steps
×
60 s min
=
550 rpm
______________________________________________________________________________________
18.11

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Section 18.4: Single-Phase AC Motors
Problem 18.15
Solution:
Known quantities:
The motor data are the following: a) b)
3
4 hp , 900 rev min
1
1
2 hp , 3600 rev min c) d)
3
4 hp , 1800 rev min
1
1
2 hp , 6000 rev min
Find:
Whether the following motors are integral- or fractional-horse power motors.
Assumptions:
None.
Analysis: a)
The power is 0 .
75
1800
900
Integral.
=
1 .
5 hp . b)
The power is 1 .
5
1800
3600
Fractional.
=
0 .
75 hp . c)
The power is 0 .
75
1800
1800
Fractional.
=
0 .
75 hp . d)
The power is 1 .
5
1800
6000
=
0 .
45 hp .
Fractional.
______________________________________________________________________________________
Problem 18.16
Solution:
Known quantities:
F
1
=
F
1 ( peak ) cos
θ
, F
1 ( peak )
=
F
1 (max) cos
θ
, .
Find:
The expression for F and verify that for a single-phase winding, both forward and backward components
1 are present.
18.12

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Assumptions:
None.
Analysis:
The stator mmf F
1
can be expressed as:
F
1
=
F
1 max cos(
ω t ) cos
θ
=
1
2
F
1 max cos
θ
=
F
CW
+
1
+
2
F
1 max
F
CCW cos(
ω t ) cos
θ
−
1
2 cos(
ω
F
1 max cos
θ cos(
ω t ) t )
+
1
2
F
1 max cos
θ cos(
ω t ) where:
F
CW
is a clockwise-rotating mmf.
F
CCW
is a counter clockwise-rotating magnetic mmf.
______________________________________________________________________________________
Problem 18.17
Solution:
Known quantities:
A 200 V , 60 Hz , 10 hp single-phase induction motor operates at an efficiency of 0 .
86 a power factor of
0 .
9 .
Find:
The capacitor that should be placed in parallel with the motor so that the feeder supplying the motor will operate at unity power factor.
Assumptions:
None.
Analysis:
We have:
P out
=
746
×
10
=
7460 W
From P in
P in
I
S
=
P out =
8674 .
4 W
=
V
S
= eff .
I
S cos
θ
S
=
8674 .
4 W
48 .
2 A ,
θ
, we have:
=
25 .
84
, lagging
Therefore,
I
S
=
48 .
2
∠ −
25 .
84
,
To get unit power factor, I
C
=
=
43 .
38 j 21 .
01
−
From j 21 .
01
C
=
=
−
200
−
1 j
278 .
6
ω
C
µ
F
,
ω =
377 rad j 21 .
01 A s , we have:
______________________________________________________________________________________
18.13

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Problem 18.18
Solution:
Known quantities:
A 230 V , 50 Hz single-phase two-pole induction motor operates at 3 percent slip.
Find:
The slip in the opposite direction of rotation. Find the speed of the motor in the normal direction of rotation.
Assumptions:
None.
Analysis:
For a 2-pole machine, the synchronous speed is 3000 rev min for an excitation frequency of 50 Hz . From s
=
0
2910
.
03 rev
, the slip in the opposite direction of rotation is 0 .
97 , the motor speed is min
=
304 .
7 rad s .
______________________________________________________________________________________
Problem 18.19
Solution:
Known quantities:
A stepper motor with a 15
,
step angle operates in one-phase excitation mode.
Find:
The amount of time to take for the motor to rotate through 28 rev when the pulse rate is 180 pps .
Assumptions:
None.
Analysis:
ω =
180
×
15
=
2700
,
/ s
=
7 .
5 rev s t
=
28 rev
7 .
5 rev s
=
3 .
73 s
______________________________________________________________________________________
Problem 18.20
Solution:
Known quantities:
A
1
4 hp , 110 V , 60 Hz , four-pole capacitor-start motor has the following parameters:
R
S
=
2 .
02
Ω
R
R
X m
=
=
4 .
12
Ω
66 .
8
Ω
X
S
=
2 .
8
Ω s
X
=
R
=
2
0 .
05
.
12
Ω
Find: a) The stator current. b) The mechanical power. c) The rotor speed.
Assumptions:
None.
18.14

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Analysis: a)
0
Z
.
5 in
Z
0 .
5 Z
= b f
=
=
0 .
991
+
15 .
93
+ j 1 .
057 j 20 .
07
=
=
0 .
5 ( R b
0 .
5 ( R f
+
+ jX b jX
) f
)
18 .
94
+ j 23 .
93
=
30 .
52
∠
51 .
64
,
∴
I
1
=
V
1
Z in
=
3 .
60
∠ −
51 .
64
,
A b)
P f
=
I
2
1
( 0 .
5 R f
)
=
207 .
0 W
P b
=
P mech
I
1
2
=
( 0 .
5 R b
( 1
− s
)
)( P
= f
12 .
84 W
−
P b
)
=
184 .
45 W c)
For a 4-pole machine,
ω s
=
188 .
5 rad s ,
Thus, the rotor speed is:
179 .
1 rad s
=
1710 rev
ω min m
=
179 .
1 rad s
______________________________________________________________________________________
Problem 18.21
Solution:
Known quantities:
A
1
4 hp , 110 V , 60 Hz , four-pole, single-phase induction motor has the following parameters:
R
S
=
1 .
86
Ω
R
R
X m
=
=
3 .
56
53 .
5
Ω
Ω
X
S
=
2 .
56 s
X
=
R
=
2
0 .
05
.
56
Ω
Ω
Find:
The mechanical power output.
Assumptions:
None.
Analysis:
0 .
5 Z b
0 .
5 Z f
=
=
0 .
830
+
12 .
41
+
Z in
=
15 .
1
+ j j
1 .
16
248
.
98 j 20 .
79
=
25 .
7
∠
54 .
0
,
I
1
=
P f
V
1
Z in
=
4 .
28
∠ −
=
I
1
2
( 0 .
5 R f
)
54 .
0
,
=
227 .
5 W
A
P b
=
P mech
I
2
1
=
( 0 .
5 R b
( 1
− s )(
)
=
P f
15 .
2 W
−
P b
)
=
201 .
68 W
______________________________________________________________________________________
18.15

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Problem 18.22
Solution:
Known quantities:
A 115 V , 60 Hz , four-pole, one-phase induction motor has the following parameters:
R
S
=
0 .
5
Ω
R
R
X m
=
=
0 .
25
Ω
35
Ω
X
X
S
R
=
=
0 .
4
0 .
4
Ω
Ω
Find:
The input current and developed torque when the motor speed is 1 , 730 rev min .
Assumptions:
None.
Analysis:
The synchronous speed is 1 , 800 rev
( 1
− s )
=
0 .
961 , therefore the slip s
0 .
5
0 .
5
Z
Z b f
=
=
0 .
064
+
3 .
034
+ j 0 .
2 j 0 .
747
I
Z in
P
S f
=
3 .
842
∠
20 .
52
,
=
=
29 .
9
∠ −
2717 .
5 W
20 .
52
,
P b
=
57 .
22 W
P mech
ω m
=
=
2660 .
3
181 .
2 rad
W s
A min for 1 , 730 rev min
=
0 .
039 . We have:
.
The torque developed is:
T dev
=
2551 .
6
=
14 .
68 N
⋅ m
181 .
2
______________________________________________________________________________________
Problem 18.23
Solution:
Known quantities:
No-load test of a single-phase induction motor at rated voltage and rated frequency.
Find:
The equivalent circuit of a single-phase induction motor for the no-load test.
Assumptions:
None.
Analysis:
At no load, s
≈
0 . The circuit model is shown below:
18.16

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
R
1
X
1
+
I
1
0.5
X m
0.5X'
2
V
1
0.5
X m
R'
2
4
-
______________________________________________________________________________________
Problem 18.24
Solution:
Known quantities:
The locked-rotor test of the single-phase induction motor.
Find:
The equivalent circuit.
Assumptions:
Neglect the magnetizing current.
Analysis:
For locked rotor,
ω m
=
0 , s
=
1 . The circuit is shown below:
R
1
X
1
X'
2
+
V
1
I
1
R'
2
-
______________________________________________________________________________________
Problem 18.25
Solution:
Known quantities:
A
4
1
8
Ω hp , 115
and 6
V
Ω
two-pole universal motor has the effective resistances of the armature and series field as
. The output torque is 0 .
17 N
⋅ m when the motor is drawing rated current of 1 .
5 A at a power factor of 0 .
88 at rated speed.
Find: a) The full-load efficiency. b) The rated speed. c) The full-load copper losses. d) The combined windage, friction, and iron losses. e) The motor speed when the rms current is 0 .
5 A .
Assumptions:
Phase differences and saturation.
18.17

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Analysis:
From P f
=
0 .
88 , we have
θ = −
28 .
4
,
and I
S
The rated speed is:
1 8
×
746
0 .
17
=
548 .
53 rad s a)
P in
=
151 .
8 W and the efficiency is: eff
=
1 .
5
∠ −
28 .
4
,
A
=
93 .
25
151 .
8
=
61 .
43 %
. b)
The speed is 5238 .
1 rev min . c) The copper loss is:
1 .
5
2
Other loss will be: 151 .
8
×
10
−
=
22 .
5
22 .
5 W
−
93 .
25
=
36 .
05 W d) e)
P in
=
( 115 )( 0
=
12 .
05 W
.
5
+
( 0 .
5
2 ×
10 )
+
P out
)( 0 .
88 )
Assume T is proportional to
2
I .
T new
=
0
(
1
.
5
.
5
)
2
( 0 .
17 )
=
0 .
019 N
⋅ m
ω n
=
12 .
05
=
637 .
9 rad
0 .
019
=
6091 .
9 rpm s
______________________________________________________________________________________
Problem 18.26
Solution:
Known quantities:
A 240 V , 60 Hz , two-pole universal motor operates at 12 , 000 rev min on full load and draws a current of
4 .
55
+ j 3
6 .
5 A at a power factor of 0 .
94 lagging. The series field-winding impedance is
.
2
Ω
and the armature circuit impedance is 6 .
15
+ j 9 .
4
Ω
Find: a) The back emf of the motor. b) The mechanical power developed by the motor. c) The power output if the rotational loss is
65 W . d) The efficiency of the motor.
Assumptions:
None.
Analysis:
, we have
θ = −
19 .
95
,
and I
S
=
6 .
5
∠ −
19 .
95
,
A . a) b)
E b
=
146 .
68
− j 53 .
25
=
156 .
05
∠ −
19 .
95
,
V
P dev
=
I
S
E b cos 0
, =
1014 .
3 W c) d)
P out
=
1014 .
3
−
65
=
949 .
3 W
18.18

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
P in eff
=
=
1466 .
4 W
64 .
7 %
______________________________________________________________________________________
Problem 18.27
Solution:
Known quantities:
A single-phase motor is drawing 20 A from a 400 V , 50 Hz supply. The power factor is 0 .
8 lagging.
Find:
The value of capacitor connected across the circuit to raise the power factor to unity.
Assumptions:
None.
Analysis:
V
S
=
400
∠
0
,
,
θ = −
36 .
9
,
, therefore,
For a unity power factor,
We have 12
=
400
ω
C ,
ω =
I
C
=
314 j 12
.
16
. rad
Therefore, C
=
95 .
5
µ
F s .
______________________________________________________________________________________
Problem 18.28
Solution:
Known quantities:
A way of operating a three-phase induction motor as a single-phase source is shows in Figure P18.28.
Find:
Whether the motor will work. Explain why or why not.
Assumptions:
None.
Analysis:
It will work. The b and c windings will produce a magnetic field similar to a single phase machine, that is, two components rotating in opposite directions and the a winding would act as a starting winding. The phase shift provided by the capacitor is needed to provide a nonzero starting torque.
______________________________________________________________________________________
Problem 18.29
Solution:
Known quantities:
A
1
4 hp capacitor-start motor with its output adjusted to rated value.
E
=
115 volts ; I
=
3 .
8 A ; P
=
310 W ; rev min
=
1725 .
Find: a) Efficiency. c) Torque in pound-inches.
18.19

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Assumptions:
None.
Analysis: a) b)
P out eff
=
=
P
1
4
×
746
=
186 .
5 W out
P in
=
0 .
602
=
60 .
2 % pf
=
P
VI
=
310
( 115 )( 3 .
8 )
=
0 .
709 lagging c)
T
=
=
7 .
04
9 .
13
P out n
R lb
⋅ in
=
7 .
04
186 .
5
1725
=
0 .
761 lb
⋅ ft
×
12 in ft
______________________________________________________________________________________
18.20

G. Rizzoni, Principles and Applications of Electrical Engineering
Section 18.5: Motor Selection and Applications
Problem 18.30
Solution:
Known quantities:
The tasks are the following: b) Refrigerator. c) Air conditioner compressor. d) e)
Air conditioner fan.
Variable-speed sewing machine.
f) Clock.
Problem solutions, Chapter 18
Find:
The type of motor that can perform the above tasks.
Assumptions:
None.
Analysis:
The universal motor speed is easily controlled and thus it would be used for variable speed, that is, (e) and
(g). The vacuum cleaner motors are often universal motors. This motor could also be used for the fan motors.
A single-phase induction motor is used for (b) and (c).
The clock should use a single-phase synchronous motor.
The tape drive would be a single-phase synchronous motor also.
An X-Y plotter uses a stepper motor.
______________________________________________________________________________________
Problem 18.31
Solution:
Known quantities:
A 5 hp , 1150 rev min shunt motor. The speed control by means of a tapped field resistor is shown Figure
P18.31.
Find:
The speed of the motor and the torque available at the maximum permissible load with the tap at position 3.
Assumptions:
None.
Analysis: n
T
=
=
230 %
40 %
×
1150
×
T rated
=
2645 rpm
=
( 0 .
4 )
( 33 , 000 )( 5 )
( 2
π
)( 1150 )
=
9 .
13 lb
⋅ ft
______________________________________________________________________________________
18.21

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Problem 18.32
Solution:
Known quantities:
The applications are the following: a) Inexpensive analog electric clock. c) Escalator which must start under all load conditions. e) f)
Table model circular saw operating at about 3 , 500 rev min .
Hand-held circular saw operating at
15 , 000 rev min
.
Find:
The single-phase motor that can apply to the above cases.
Assumptions:
None.
Analysis: a) reluctance b) shaded-pole c) capacitor-start d) universal e) permanent split capacitor f) universal g) permanent split capacitor
______________________________________________________________________________________
Problem 18.33
Solution:
Known quantities:
The power required to drive a fan varies as a cube of the speed. The motor driving a shaft-mounted fan is loaded to 100 percent of its horsepower rating on the top speed connection.
Find:
The horsepower output in percent of rating at the following speed reduction. a) 20 percent. b) 30 percent. c) 50 percent.
Assumptions:
None.
Analysis: a)
HP
=
( 0 .
8 )
3 b)
HP
=
( 0 .
7 )
3
=
0 .
512
=
51 .
2 % of
=
0 .
343
=
34 .
3 % of rated rated c)
HP
=
( 0 .
5 ) 3 =
0 .
125
=
12 .
5 % of rated
______________________________________________________________________________________
18.22

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Problem 18.34
Solution:
Known quantities:
An industrial plant has a load of 800 kW at a power factor of 0 .
8 lagging. A synchronous motor will be used to change them to 200 kW and 0 .
92 .
Find:
The KVA input rating and the power factor at which the motor operates.
Assumptions:
The motor has an efficiency of 91 percent.
Analysis:
P m
P new
=
=
200 K
0 .
91
800 K
=
+
219 .
8 KW
219 .
8 K
=
1019 .
8 KW
Q new
=
P new tan(cos
−
1
0 .
92 )
=
434 .
4 KVAR
Q old
Q m
=
800
=
Q old
K
− tan(cos
Q new
=
−
1
0 .
8 )
165 .
6
=
600
KVAR
KVAR
S m
=
P
2 m
+ 2
Q m
=
274 .
8 KVA pf m
= cos tan
−
1
(
Q m
P m
)
=
0 .
8 leading
______________________________________________________________________________________
Problem 18.35
Solution:
Known quantities:
An electric machine is controlled so that its torque-speed characteristics exhibit a constant-torque region and a constant-power region as shown in Figure P18.35.
Average efficiency of the electric drive is 87%.
Machine torque is constant at 150 N-m from 0 to 2500 rpm.
The constant power region is from 2500 to 6000 rpm.
Machine drives a constant torque load requiring 75 N-m.
Find: a) Operating speed of the machine b) Electric power needed to operate the machine
Assumptions:
None
Analysis: a) Since the load requires less than the starting torque, the motor should be operating in the constant power regime. The constant power is determined from the starting torque and the base speed:
P
= ϖ base
T start
=
(
2500 r / min
)(
(
2
π rad
60 s /
/ rev min
)
) ( 150 N
− m
)
=
39 .
3 kW
The operating speed is determined from the load torque:
18.23

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
ϖ n
=
=
P
T
L
=
(
523 .
6
(
(
39300 W
75 N
− m
)
) rad /
=
523 .
6 rad s
) ( 60 s /
2
π rad / min rev
)
)
=
/ s
5000 r / min b) The electric power needed to operate the machine:
P e
=
P
η
=
39
0
.
.
3 kW
87
=
45 .
1 kW
______________________________________________________________________________________
Problem 18.36
Solution:
Known quantities:
The PM synchronous (brushless DC) motor in Figure P18.36
Electrical subsystem parameters: R
S
, L
S
, k . (motor constant), V
S
( t ), I
S
( t ).
Mechanical subsystem parameters: inertia and damping coefficient, J , b .
Thermal subsystem parameters: thermal resistance, specific heat, mass, R t
, c , m .
Find:
Write the differential equations describing the electrothermomechanical dynamics of the systems
Assumptions:
All heat is generated in the stator by the stator current (i.e. the heat generated in the rotor is negligible)
The rotor and stator are at the same temperature and specific heat c
The stator is highly thermally conductive
The dominant heat-transfer term is convection
Overall thermal resistance
The motor generates torque according to the equation T m
= kI
S
The back emf is equal to E
R from stator to air t b
= kw
Analysis:
Mechanical subsystem
J d
ω dt
( t )
=
T m
− b
ω
( t )
= kI
S
( t )
− b
ω
( t )
Electrical subsystem
V
S
( t )
=
L
S dI
S
( t ) dt
+
R
S
I
S
( t )
+
E b
=
L
S dI
S
( t ) dt
+
R
S
I
S
( t )
+ k
ω
( t )
Thermal subsystem
R
S
I
S
2
( t )
S
( t )
R t air mc d
S
( t ) dt
______________________________________________________________________________________
Problem 18.37
Solution:
Known quantities:
The wound separately excited motor in Figure P18.37
18.24

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18
Electrical subsystem parameters: R f
, L f
, R a
, L a
, (motor field and armature electrical parameters), k f
. k a , k
T
(motor field and armature constants), V
S
( t ), V f
( t ), I a
( t ), I f
( t ).
Mechanical subsystem parameters: load inertia, damping coefficient, load torque, J , b , T
L
.
Thermal subsystem parameters: C t-rotor
, h t-rotor
, A rotor
(rotor thermal capacitance, film coefficient of heat transfer from rotor surface to air and from air to stator inner surface, rotor and inner stator surface area (assumed equal).
C t-stator
, h t-stator
, A stator
(stator thermal capacitance, film coefficient of heat transfer from stator outer surface to air, stator outer surface area.
Find:
Write the differential equations describing the electrothermomechanical dynamics of the systems
Assumptions:
Heat is generated in the stator and the rotor by the respective currents
The stator and rotor are highly thermally conductive
The dominant heat-transfer term is convection through the air gap and to ambient
Heat storage in the air gap is negligible, and the air gap is infinitely thin
The motor generates torque according to the equation T m
= k
T
I a
The back emf is equal to E b
= k a w
The stator and rotor each act as a lumped thermal mass
Analysis:
Mechanical subsystem
J k
T d
ω
= dt k
( t a
)
φ
,
=
φ
T m
=
− k f b
I
ω f
( t
( )
,
)
J d
ω
( t ) dt
= k a k f
I f
( )
I a
− k
T
T
L
( t )
=
−
= k
T k a k f
I a
I b
ω
( t ) f
( t )
−
T
L
− b
ω
( t )
Electrical subsystem
Armature Circuit:
−
T
L
V a
( t )
=
L a
Field Circuit: dI a
( t )
+ dt
R a
I a
( t )
+
E b
=
L a dI a
( t )
+
R a
I a
( t )
+ k a
ω
( t ) dt
V f
( t )
=
L f dI f
( dt
Thermal subsystem
Rotor: t )
R a
I a
2
( t )
− h rotor
+
R f
I f
( t )
A rotor
( θ
R
( t )
− θ
S
( ) )
= mc rotor d
θ
R
( t ) dt
Stator:
R f
I
2 f
( t )
− h stator
A stator
( θ
S
( t )
− θ air
)
+ h rotor
A rotor
( θ
R
( t )
− θ
S
( ) )
= mc stator d
θ
S
( t ) dt
______________________________________________________________________________________
18.25