Bird J. Electrical Circuit Theory and Technology
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406 Electrical Circuit Theory and Technology
Figure 22.14
Typical applications are with machine tools, pumps and mill motors. The
squirrel cage rotor type is the most widely used of all a.c. motors.
22.18 Further problems
on three-phase induction
motors
Synchronous speed
1 The synchronous speed of a 3-phase, 4-pole induction motor is
60 rev/s. Determine the frequency of the supply to the stator
windings. [120 Hz]
2 The synchronous speed of a 3-phase induction motor is 25 rev/s and
the frequency of the supply to the stator is 50 Hz. Calculate the
equivalent number of pairs of poles of the motor. [2]
3 A 6-pole, 3-phase induction motor is connected to a 300 Hz supply.
Determine the speed of rotation of the magnetic field produced by
the stator. [100 rev/s]
Slip
4 A 6-pole, 3-phase induction motor runs at 970 rev/min at a certain
load. If the stator is connected to a 50 Hz supply, find the percentage
slip at this load. [3%]
5 A 3-phase, 50 Hz induction motor has 8 poles. If the full load slip is
2.5%, determine (a) the synchronous speed, (b) the rotor speed, and
(c) the frequency of the rotor e.m.f.’s.
[(a) 750 rev/min (b) 731 rev/min (c) 1.25 Hz]
6 A three-phase induction motor is supplied from a 60 Hz supply and
runs at 1710 rev/min when the slip is 5%. Determine the synchronous
speed. [1800 rev/min]
Three-phase induction motors 407
7 A 4-pole, 3-phase, 50 Hz induction motor runs at 1440 rev/min at
full load. Calculate (a) the synchronous speed, (b) the slip and (c) the
frequency of the rotor induced e.m.f.’s.
[(a) 1500 rev/min (b) 4% (c) 2 Hz]
Rotor frequency
8 A 12-pole, 3-phase, 50 Hz induction motor runs at 475 rev/min.
Determine (a) the slip speed, (b) the percentage slip and (c) the
frequency of rotor currents. [(a) 25 rev/min (b) 5% (c) 2.5 Hz]
9 The frequency of the supply to the stator of a 6-pole induction motor
is 50 Hz and the rotor frequency is 2 Hz. Determine (a) the slip, and
(b) the rotor speed in rev/min. [(a) 0.04 or 4% (b) 960 rev/min]
Losses and efficiency
10 The power supplied to a three-phase induction motor is 50 kW and
the stator losses are 2 kW. If the slip is 4%, determine (a) the rotor
copper loss, (b) the total mechanical power developed by the rotor,
(c) the output power of the motor if friction and windage losses
are 1 kW, and (d) the efficiency of the motor, neglecting rotor iron
losses. [(a) 1.92 kW (b) 46.08 kW (c) 45.08 kW (d) 90.16%]
11 By using external rotor resistance, the speed of the induction motor in
problem 15 is reduced to 40% of its synchronous speed. If the torque
and stator losses are unchanged, calculate (a) the rotor copper loss,
and (b) the efficiency of the motor. [(a) 28.80 kW (b) 36.40%]
Torque equation
12 A 400 V, three-phase, 50 Hz, 2-pole, star-connected induction motor
runs at 48.5 rev/s on full load. The rotor resistance and reactance per
phase are 0.4 " and 4.0 " respectively, and the effective rotor-stator
turns ratio is 0.8:1. Calculate (a) the synchronous speed, (b) the slip,
(c) the full load torque, (d) the power output if mechanical losses
amount to 500 W, (e) the maximum torque, (f) the speed at which
maximum torque occurs, and (g) the starting torque.
[(a) 50 rev/s or 3000 rev/min (b) 0.03 or 3%
(c) 22.43 Nm (d) 6.34 kW (e) 40.74 Nm
(f) 45 rev/s or 2700 rev/min (g) 8.07 Nm]
13 For the induction motor in problem 12, calculate at full load (a) the
rotor current, (b) the rotor copper loss, and (c) the starting current.
[(a) 10.62 A (b) 135.3 W (c) 45.96 A]
14 If the stator losses for the induction motor in problem 12 are 525 W,
calculate at full load (a) the power input, (b) the efficiency of the
motor and (c) the current taken from the supply if the motor runs at
a power factor of 0.84. [(a) 7.49 kW (b) 84.65% (c) 12.87 A]
15 For the induction motor in problem 12, determine the resistance of
the rotor winding required for maximum starting torque. [4.0 "]
Assignment 7
This assignment covers the material contained in chapters 21
and 22.
The marks for each question are shown in brackets at the end of
each question.
1 A 6-pole armature has 1000 conductors and a flux per pole
of 40 mWb. Determine the e.m.f. generated when running at
600 rev/min when (a) lap wound (b) wave wound. (6)
2 The armature of a d.c. machine has a resistance of 0.3 and is
connected to a 200 V supply. Calculate the e.m.f. generated when it
is running (a) as a generator giving 80 A (b) as a motor taking 80 A
(4)
3 A 15 kW shunt generator having an armature circuit resistance of 1
and a field resistance of 160 generates a terminal voltage of 240 V
at full-load. Determine the efficiency of the generator at full-load
assuming the iron, friction and windage losses amount to 544 W.
(6)
4 A 4-pole d.c. motor has a wave-wound armature with 1000 conduc-
tors. The useful flux per pole is 40 mWb. Calculate the torque exerted
when a current of 25 A flows in each armature conductor. (4)
5 A 400 V shunt motor runs at it’s normal speed of 20 rev/s when the
armature current is 100 A. The armature resistance is 0.25 . Calcu-
late the speed, in rev/min when the current is 50 A and a resistance
of 0.40 is connected in series with the armature, the shunt field
remaining constant. (7)
6 The stator of a three-phase, 6-pole induction motor is connected to a
60 Hz supply. The rotor runs at 1155 rev/min at full load. Determine
(a) the synchronous speed, and (b) the slip at full load. (6)
7 The power supplied to a three-phase induction motor is 40 kW and
the stator losses are 2 kW. If the slip is 4% determine (a) the rotor
copper loss, (b) the total mechanical power developed by the rotor,
(c) the output power of the motor if frictional and windage losses are
1.48 kW, and (d) the efficiency of the motor, neglecting rotor iron
loss. (9)
8 A 400 V, three-phase, 100 Hz, 8-pole induction motor runs at
24.25 rev/s on full load. The rotor resistance and reactance per
phase are 0.2 and 2 respectively and the effective rotor-stator
turns ratio is 0.80:1. Calculate (a) the synchronous speed, (b) the
percentage slip, and (c) the full load torque. (8)
Main formulae for Part 2
A.c. theory:
T D
1
f
or f D
1
T
I D
i
2
1
C i
2
2
C i
2
3
CÐÐÐCi
n
n
For a sine wave: I
AV
D
2
I
m
or 0.637I
m
I D
1
p
2
I
m
or 0.707I
m
Form factor D
rms
average
Peak factor D
maximum
rms
General sinusoidal voltage:
v D V
m
sinωt š
Single-phase circuits:
X
L
D 2fL X
C
D
1
2fC
Z D
V
I
D
R
2
C X
2
Series resonance: f
r
D
1
2
p
LC
Q D
V
L
V
or
V
C
V
D
2f
r
L
R
D
1
2f
r
CR
D
1
R
L
C
Q D
f
r
f
2
f
1
or f
2
f
1
D
f
r
Q
Parallel resonance (LR-C circuit): f
r
D
1
2
1
LC
R
2
L
2
I
r
D
VRC
L
R
D
D
L
CR
Q D
2f
r
L
R
D
I
C
I
r
P D VI cos or I
2
RSD VI
Q D VI sin power factor D cos D
R
Z
D.c. transients:
C–R circuit D CR
Charging:
v
C
D V
1 e
t/CR
v
r
D Ve
t/CR
i D Ie
t/CR
Discharging: v
C
D v
R
D Ve
t/CR
i D Ie
t/CR
L-R circuit D
L
R
Current growth:
v
L
D Ve
Rt/L
v
R
D V
1 e
Rt/L
i D I
1 e
Rt/L
Current decay: v
L
D v
R
D Ve
Rt/L
i D Ie
Rt/L
410 Electrical Circuit Theory and Technology
Operational amplifiers:
CMRR D 20log
10
differential voltage gain
common mode gain
dB
Inverter: A D
V
o
V
i
D
R
f
R
i
Non-inverter: A D
V
o
V
i
D 1 C
R
f
R
i
Summing: V
o
DR
f
V
1
R
1
C
V
2
R
2
C
V
3
R
3
Integrator: V
o
D
1
CR
V
i
dt
Differential: If V
1
>V
2
: V
o
D V
1
V
2
R
f
R
1
If V
2
>V
1
: V
o
D V
2
V
1
R
3
R
2
C R
3
1 C
R
f
R
1
Three-phase systems: Star I
L
D I
p
V
L
D
p
3V
p
Delta V
L
D V
p
I
L
D
p
3I
p
P D
p
3V
L
I
L
cos or P D 3I
2
p
R
p
Two-wattmeter method P D P
1
C P
2
tan D
p
3
P
1
P
2
P
1
C P
2
Transformers:
V
1
V
2
D
N
1
N
2
D
I
2
I
1
I
0
D
I
2
M
C I
2
C
I
M
D I
0
sin
0
I
C
D I
0
cos
0
E D 4.44f
m
N Regulation D
E
2
E
1
E
2
ð 100%
Equivalent circuit: R
e
D R
1
C R
2
V
1
V
2
2
X
e
D X
1
C X
2
V
1
V
2
2
Z
e
D
R
2
e
C X
2
e
Efficiency, & D 1
losses
input power
Output power D V
2
I
2
cos
2
Total loss D copper loss Ciron loss
Input power D output power Closses
Resistance matching: R
1
D
N
1
N
2
2
R
L
D.c. machines:
Generated e.m.f. E D
2pnZ
c
/ ω
(c D 2 for wave winding, c D 2p for lap winding)
Generator: E D V C I
a
R
a
Efficiency, & D
VI
VI C I
2
a
R
a
C I
f
V C C
ð 100%
Main formulae for Part 2 411
Motor: E D V I
a
R
a
Efficiency, & D
VI I
2
a
R
a
I
f
V C
VI
ð 100%
Torque D
EI
a
2n
D
pZI
a
c
/ I
a
Three-phase induction
motors:
n
S
D
f
p
s D
n
S
n
r
n
S
ð 100 f
r
D sf X
r
D sX
2
I
r
D
E
r
Z
r
D
s
N
2
N
1
E
1
[R
2
2
C sX
2
2
]
s D
I
2
r
R
2
P
2
Efficiency, & D
P
m
P
1
D
input stator loss rotor copper loss friction & windage loss
input power
Torque, T D
mN
2
/N
1
2
2n
S
sE
2
1
R
2
R
2
2
C sX
2
2
/
sE
2
1
R
2
R
2
2
C sX
2
2
Part3AdvancedCircuit
Theoryand
Technology
23 Revision of complex
numbers
At the end of this chapter you should be able to:
ž define a complex number
ž understand the Argand diagram
ž perform calculations on addition, subtraction, multiplication,
and division in Cartesian and polar forms
ž use De Moivres theorem for powers and roots of complex
numbers
23.1 Introduction
A complex number is of the form (a C jb) where a is a real number and
jb is an imaginary number. Hence (1 C j2) and (5 j7) are examples
of complex numbers.
By definition,
j D
p
1
and
j
2
D1
Complex numbers are widely used in the analysis of series,
parallel and series-parallel electrical networks supplied by alternating
voltages (see Chapters 24 to 26), in deriving balance equations
with a.c. bridges (see Chapter 27), in analysing a.c. circuits using
Kirchhoff’s laws (Chapter 30), mesh and nodal analysis (Chapter 31),
the superposition theorem (Chapter 32), with Th
´
evenin’s and Norton’s
theorems (Chapter 33) and with delta-star and star-delta transforms
(Chapter 34) and in many other aspects of higher electrical engineering.
The advantage of the use of complex numbers is that the manipulative
processes become simply algebraic processes.
A complex number can be represented pictorially on an Argand
diagram. In Figure 23.1, the line OA represents the complex number
(2 C j3), OB represents (3 j), OC represents (2 C j2) and OD
represents (4 j3).
A complex number of the form a C jb is called a Cartesian or rect-
angular complex number. The significance of the j operator is shown in
Figure 23.2. In Figure 23.2(a) the number 4 (i.e., 4 C j0) is shown drawn
as a phasor horizontally to the right of the origin on the real axis. (Such
a phasor could represent, for example, an alternating current, i D 4sinωt
amperes, when time t is zero.)
The number j4(i.e.,0C j4) is shown in Figure 23.2(b) drawn verti-
cally upwards from the origin on the imaginary axis. Hence multiplying
416 Electrical Circuit Theory and Technology
Figure 23.1 The Argand diagram
the number 4 by the operator j results in an anticlockwise phase-shift of
90
°
without altering its magnitude.
Multiplying j4byj gives j
2
4, i.e. 4, and is shown in Figure 23.2(c)
as a phasor four units long on the horizontal real axis to the left of the
origin—an anticlockwise phase-shift of 90
°
compared with the position
shown in Figure 23.2(b). Thus multiplying by j
2
reverses the original
direction of a phasor.
Multiplying j
2
4byj gives j
3
4, i.e. j4, and is shown in Figure 23.2(d)
as a phasor four units long on the vertical, imaginary axis downward
from the origin — an anticlockwise phase-shift of 90
°
compared with the
position shown in Figure 23.2(c).
Multiplying j
3
4byj gives j
4
4, i.e. 4, which is the original position of
the phasor shown in Figure 23.2(a).
Summarizing, application of the operator j to any number rotates it
90
°
anticlockwise on the Argand diagram, multiplying a number by j
2
rotates it 180
°
anticlockwise, multiplying a number by j
3
rotates it 270
°
anticlockwise and multiplication by j
4
rotates it 360
°
anticlockwise, i.e.,
back to its original position. In each case the phasor is unchanged in its
magnitude.
By similar reasoning, if a phasor is operated on by j then a phase
shift of 90
°
(i.e., clockwise direction) occurs, again without change of
magnitude.
In electrical circuits, 90
°
phase shifts occur between voltage and current
with pure capacitors and inductors; this is the key as to why j notation
is used so much in the analysis of electrical networks. This is explained
in Chapter 24 following.
Figure 23.2
Revision of complex numbers 417
23.2 Operations
involving Cartesian
complex numbers
(a) Addition and subtraction
a C jb C c C jd D a C c C jb C d
and a C jb c C jd D a c C jb d
Thus, 3 C j2 C 2 j4 D 3 C j2 C 2 j4 D 5
− j2
and 3 C j2 2 j4 D 3 C j2 2 C j4 D 1 Y j6
(b) Multiplication
a C jbc C jd D ac C ajd C jbc C jbjd
D ac C jad C jbc C j
2
bd
But j
2
D1, thus
a C jbc C jd D ac bd C jad C bc
For example, 3 C j22 j4 D 6 j12 C j4 j
2
8
D 6 18 C j12 C4
D 14 C j8 D 14
− j8
(c) Complex conjugate
The complex conjugate of (a C jb)is(a jb). For example, the conju-
gate of (3 j2) is (3 C j2).
The product of a complex number and its complex conjugate is always
a real number, and this is an important property used when dividing
complex numbers. Thus
a C jba jb D a
2
jab C jab j
2
b
2
D a
2
b
2
D a
2
C b
2
(i.e. a real number)
For example, 1 C j21 j2 D 1
2
C 2
2
D 5
and 3 j43 C j4 D 3
2
C 4
2
D 25
(d) Division
The expression of one complex number divided by another, in the form
a C jb, is accomplished by multiplying the numerator and denominator by
the complex conjugate of the denominator. This has the effect of making
the denominator a real number. Hence, for example,
2 C j4
3 j4
D
2 C j4
3 j4
ð
3 C j4
3 C j4
D
6 C j8 C j12 C j
2
16
3
2
C 4
2
D
6 C j8 C j12 16
25