Chapter 9 unbalanced operation and single phase induction machines

336
9.1. INTRODUCTION
The method of symmetrical components, as developed by Fortescue [1] , has been used
to analyze unbalanced operation of symmetrical induction machines since the early
1900s. This technique, which has been presented by numerous authors [2–5] , is
described in its traditional form in the fi rst sections of this chapter. The extension of
symmetrical components to analyze unbalanced conditions, such as an open-circuited
stator phase, is generally achieved by revolving fi eld theory. This approach is not used
here; instead, reference-frame theory is used to establish the method of symmetrical
components and to apply it to various types of unbalanced conditions [6] . In particular,
it is shown that unbalanced phase variables can be expressed as a series of balanced
sets in the arbitrary reference frame with coeffi cients that may be constant or time
varying. This feature of the transformation to the arbitrary reference frame permits the
theory of symmetrical components to be established analytically and it provides a
straightforward means of applying the concept of symmetrical components to various
types of unbalanced conditions.
In this chapter, unbalanced applied stator voltages, unbalanced stator impedances,
open-circuited stator phase, and unbalanced rotor resistors of the three-phase induction
Analysis of Electric Machinery and Drive Systems, Third Edition. Paul Krause, Oleg Wasynczuk,
Scott Sudhoff, and Steven Pekarek.
© 2013 Institute of Electrical and Electronics Engineers, Inc. Published 2013 by John Wiley & Sons, Inc.
UNBALANCED OPERATION
AND SINGLE-PHASE
INDUCTION MACHINES
9
SYMMETRICAL COMPONENT THEORY 337
machine are considered. Single-phase induction motors are analyzed, and several unbal-
anced and fault modes of synchronous machine operation are illustrated.
9.2. SYMMETRICAL COMPONENT THEORY
In 1918, C.L. Fortescue [1] set forth the method of symmetrical components for the
purpose of analyzing the steady-state behavior of power system apparatus during unbal-

anced operation. Fortescue showed that the phasors representing an unbalanced set
of steady-state multiphase variables could be expressed in terms of balanced sets of
phasors. For example, the phasors representing an unbalanced three-phase set can be
expressed in terms of (1) a balanced set of phasors with an abc sequence (the positive
sequence), (2) a balanced set of phasors with an acb sequence (the negative sequence),
and (3) a set of three equal phasors (the zero sequence). Although the method of sym-
metrical components is widely used in the analysis of unbalanced static networks [2] ,
it is perhaps most appropriate for the analysis of symmetrical induction machines during
unbalanced operations.
Fortescue ’ s change of variables is a complex transformation that may be written
for three-phase stationary circuits in phasor form as


FSF
+−
=
0s abcs
(9.2-1)
where the symmetrical components are

()

F
+− + −
=
⎡
⎣
⎤
⎦
00s

T
sss
FFF
(9.2-2)
The unbalanced phasors are

()

F
abcs
T
as bs cs
FFF=
⎡
⎣
⎤
⎦

(9.2-3)
and the transformation is expressed

S =
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥

⎥
⎥
1
3
1
1
11 1
2
2
aa
aa
(9.2-4)
The quantity a is complex, denoting a counterclockwise rotation of 2 π /3 rad. That is

ae j
j
==−+
(/)23
1
2
3
2
π
(9.2-5)

ae j
j243
1
2
3

2
==−−
(/)
π
(9.2-6)
338 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES
The inverse transformation is

()S
−
=
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
12
2
111
1
1
aa
aa
(9.2-7)
The + , − , and 0 subscripts denote the positive, negative, and zero sequences, respec-

tively. Hence, for the induction machine shown in Figure 4.2-1,

F
s+
,
aF
s
2

+
, and
aF
s

+

make up the positive sequence set of phasors, which we will denote as,

F
as+
,

F
bs+
, and


F
cs+
, respectively. Similarly,


F
s−
,
aF
s

−
, and
aF
s
2

−
are the negative sequence set, here
denoted as

F
as−
,

F
bs−
, and

F
cs−
. The zero sequence is the steady-state version of the zero
quantities introduced in Chapter 3 . Although we will assume that only one frequency
is present, the method of symmetrical components can be applied to each frequency

present in the system [6] .
In order to compute the total instantaneous, steady-state power, it is fi rst necessary
to convert the phasors to the variables in sinusoidal form and then multiply phase
voltage times phase current. Thus

PVIVIVI
VVVI I I
abc as as bs bs cs cs
as as s as as s
=++
=++ +++
+− +−
()()(
00
VVVVI I I
VVVI I I
bs bs s bs bs s
cs cs s cs cs
+− +−
+− +−
++ ++
+++ ++
00
0
)( )
()(
00s
)

(9.2-8)

Uppercase letters are used to denote steady-state sinusoidal quantities. We will show that
the instantaneous power consists of an average value and time-varying components.
9.3. SYMMETRICAL COMPONENT ANALYSIS OF
INDUCTION MACHINES
Although the method of symmetrical components can be used to analyze unbalanced
conditions other than unbalanced stator voltages, this application of symmetrical com-
ponents is perhaps the most common. Once the unbalanced applied stator voltages are
known, (9.2-1) can be used to determine

V
as+
,

V
as−
, and

V
s0
. It is clear that the currents
due to the positive-sequence balanced set can be determined from the voltage equations
given by (6.9-11)–(6.9-13) . These equations are rewritten here with the + added to the
subscript to denote positive-sequence phasors.


VrjXIjXII
as s
e
b

ls as
e
b
Mas ar++++
=+
⎛
⎝
⎜
⎞
⎠
⎟
++
′
ω
ω
ω
ω
()
(9.3-1)



′
=
′
+
′
⎛
⎝
⎜

⎞
⎠
⎟
′
++
′
+
+++
V
s
r
s
jXI jXI I
ar r e
b
lr ar
e
b
Mas ar
ω
ω
ω
ω
()
(9.3-2)
where ω
b
is the base electrical angular velocity generally selected as rated and
UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES 339

s
er
e
=
−
ωω
ω
(9.3-3)
As mentioned earlier, the method of symmetrical components can be applied to all
frequencies present in the system. For example, the unbalanced applied voltages could
be of any arbitrary periodic waveform whereupon the frequencies that exist in the
Fourier series expression of these voltages could each be broken up into positive, nega-
tive, and zero sequences.
It is customary to obtain the voltage equations for the negative-sequence quantities
by reasoning. In particular, slip is the normalized difference between the speed of the
rotating air-gap MMF and the rotor speed. The negative-sequence currents establish
an air-gap MMF that rotates a ω
e
in the clockwise direction. Hence, the normalized
difference between the air-gap MMF and rotor speed is ( ω
e
+ ω
r
)/ ω
e
which can be
written as (2 − s ), where s is defi ned by (9.3-3) . Therefore, the voltage equations for
the negative-sequence quantities can be obtained by replacing s by (2 − s ) in (9.3-1)
and (9.3-2) . In particular,


VrjXIjXII
as s
e
b
ls as
e
b
Mas ar−−−−
=+
⎛
⎝
⎜
⎞
⎠
⎟
++
′
ω
ω
ω
ω
()
(9.3-4)



′
−
=

′
−
+
′
⎛
⎝
⎜
⎞
⎠
⎟
′
++
′
−
−−
V
s
r
s
jXI jXI I
ar r e
b
lr ar
e
b
Mas ar
22
ω
ω
ω

ω
(
−−
)
(9.3-5)
If the zero-sequence quantities exist in a three-phase induction machine, the steady-state
variables may be determined from the phasor equivalent of (6.5-24) and (6.5-27) . The
electromagnetic torque may be calculated using sequence quantities; however, the deri-
vation of the torque expression is deferred until later.
9.4. UNBALANCED STATOR CONDITIONS OF INDUCTION
MACHINES: REFERENCE-FRAME ANALYSIS
The theory of symmetrical components set forth in the previous sections can be used
to analyze most unbalanced steady-state operating conditions. However, one tends to
look for a more rigorous development of this theory and straightforward procedures
for applying it to unbalanced conditions, such as an open stator phase or unbalanced
rotor resistors. Reference-frame theory can be useful in achieving these goals [6] .
Although simultaneous stator and rotor unbalanced conditions can be analyzed, the
notation necessary to formulate such a generalized method of analysis becomes quite
involved. Therefore, we will consider stator and rotor unbalances separately.
In Reference 6 , the stator variables are expressed as a series of sinusoidal functions
with time-varying coeffi cients. Such an analysis is notationally involved and somewhat
diffi cult to follow. We will not become that involved because the concept can be estab-
lished by assuming a single-stator (rotor) frequency for stator (rotor) unbalances with
constant coeffi cients. We will discuss the restrictions imposed by these assumptions as
we go along.
340 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES
Unbalanced Machine Variables in the Arbitrary Reference Frame
If we assume that the rotor is a three-wire symmetrical system and the stator applied
voltages are single frequency, then unbalanced steady-state stator variables may be
expressed as

F
F
F
FF
FF
FF
as
bs
cs
as as
bs bs
cs cs
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
=
⎡
⎣
⎢
⎢
⎢
⎤

⎦
⎥
⎥
⎥
αβ
αβ
αβ

cos
ωω
ω
e
e
t
tsin
⎡
⎣
⎢
⎤
⎦
⎥
(9.4-1)
Transforming the stator variables as expressed by (9.4-1) to the arbitrary reference
frame by (3.3-1) reveals an interesting property. In particular, we have

F
F
tttt
qs
ds

eee e
⎡
⎣
⎢
⎤
⎦
⎥
=
−−++
−
cos( ) sin( ) cos( ) sin( )
sin(
ωθ ωθ ωθ ωθ
ωωθ ωθ ωθ ωθ
eee e
qsA
qsB
qsC
q
ttt t
F
F
F
F
−−+−+
⎡
⎣
⎢
⎤
⎦

⎥
) cos( ) sin( ) cos( )
ssD
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
(9.4-2)
and

FF tF t
s abcs e abcs e0
=+
αβ
ωω
cos sin
(9.4-3)
where

FFFF FF
F
qsA asbscs bscs

dsB
=−−+ −
⎡
⎣
⎢
⎤
⎦
⎥
=−
1
3
1
2
1
2
3
2
ααα ββ
()

(9.4-4)


FFFF FF
F
qsB asbscs bscs
dsA
=−−− −
⎡
⎣

⎢
⎤
⎦
⎥
=
1
3
1
2
1
2
3
2
βββ αα
()

(9.4-5)

FFFF FF
F
qsC asbscs bscs
dsD
=−−− −
⎡
⎣
⎢
⎤
⎦
⎥
=

1
3
1
2
1
2
3
2
ααα ββ
()

(9.4-6)

FFFF FF
F
qsD asbscs bscs
dsC
=−−+ −
⎡
⎣
⎢
⎤
⎦
⎥
=−
1
3
1
2
1

2
3
2
βββ αα
()

(9.4-7)

FFFF
abcs asbscs
αααα
=++
(
)
1
3
(9.4-8)
UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES 341

FFFF
abcs asbscs
ββββ
=++
()
1
3
(9.4-9)
It is recalled that

d

dt
θ
ω
=
(9.4-10)
where ω is the electrical angular velocity of the arbitrary reference frame.
It is interesting that the qs and ds variables form two, two-phase balanced sets in
the arbitrary reference frame. In order to emphasize this, (9.4-2) is written with sinu-
soidal functions of ( ω
e
t − θ ) separated from those with the argument of ( ω
e
t + θ ). It is
possible to relate these balanced sets to the positive- and negative-sequence variables.
For this purpose, let us consider the induction machine in Figure 6.2-1. A balanced
three-phase set of currents of abc sequence will produce an air-gap MMF that rotates
counterclockwise at an angular velocity of ω
e
. By defi nition, the positive sequence is
the abc sequence, and it is often referred to as the positively rotating balanced set. The
negative sequence, which has the time sequence of acb , produces an air-gap MMF that
rotates in the clockwise direction. It is commonly referred to as the negatively rotating
balanced set. Let us think of the qs and ds variables as being associated with windings
with their magnetic axes positioned relative to the magnetic axis of the stator windings
as shown in Figure 6.2-1. Now consider the series of balanced sets formed by the qs
and ds variables in (9.4-2) . The two balanced sets of currents formed by the fi rst and
second column of the 2 × 4 matrix produce an air-gap MMF that rotates counterclock-
wise relative to the arbitrary reference frame whenever ω < ω
e
and always counter-

clockwise relative to the actual stator winding. Hence, the balanced sets with the
argument ( ω
e
t − θ ) may be considered as positive sequence or positively rotating sets
because they produce counterclockwise rotating air-gap MMFs relative to the stator
windings. It follows that the balanced sets formed by the third and fourth columns with
the argument ( ω
e
t + θ ) can be thought of as negative sequence or negatively rotating
sets. Therefore, we can express (9.4-2) as

FF F
qs qs qs
=+
+−
(9.4-11)

FF F
ds ds ds
=+
+−
(9.4-12)
The zero-sequence variables are expressed by (9.4-3) . It is important to note that F
qs

+

( F
ds

+
) is the positive-sequence terms of F
qs
( F
ds
), and that together I
qs

+
with I
ds

+
produce
the positive-sequence rotating air-gap MMF. Similarly, I
qs

−
with I
ds

−
, produce the
negative-sequence rotating air-gap MMF.
It is understood that the qs and ds variables may be expressed in any reference
frame by setting ω , in (9.4-2) , equal to the angular velocity of the reference frame of
interest. For example, ω = 0 for the stationary reference frame and ω = ω
r
for the rotor
reference frame. It is also clear that the instantaneous sequence sets may be identifi ed

in these reference frames. In particular, when ω is set equal to zero, the frequency
of the variables is ω
e
; however, the positive and negative sequence are immediately
342 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES
identifi able. When ω = ω
r
, we see that there are two electrical angular frequencies
present; ω
e
− ω
r
and ω
e
+ ω
r
. The latter occurs due to the air-gap MMF established by
negative-sequence stator variables. Because we have assumed that the rotor circuits
form a three-wire symmetrical system only one positive and one negative sequence set
will be present in the rotor. However, the balanced negative-sequence set of rotor vari-
ables will appear in the arbitrary reference frame with the same frequency ( ω
e
t + θ ) as
the negative sequence set established by the stator unbalance.
It follows that for stator unbalances and the assumption that the rotor is a three-wire
symmetrical system, (9.4-2)–(9.4-9) can be used to identify the positive- and negative-
sequence rotor variables. We only need to (1) replace all s subscripts with r in (9.4-2)–
(9.4-7) , (2) add a prime to all quantities associated with the rotor variables and (3) set
θ in (9.4-2) equal to ω
r

t . Recall that the rotor variables are transformed to the arbitrary
reference frame by ( 6.4-1 ), wherein β is defi ned by ( 6.4-5 ). Please don ’ t confuse the β
given by ( 6.4-5 ) and the β used in the subscripts starting with (9.4-1)–(9.4-9) .
The instantaneous electromagnetic torque may be expressed in arbitrary reference-
frame variables as

T
PX
II II
e
M
b
qs dr ds qr
=
⎛
⎝
⎜
⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟
′
−
′
()

3
22
ω
(9.4-13)
If we use (9.4-2) to express I
qs
and I
ds
and the equivalent expression for
′
I
qr
and
′
I
dr
, the
torque, for a stator unbalance, may be expressed as

T
PX
II II II
e
M
b
qsA qrB qsB qrA qsC q
=
⎛
⎝
⎜

⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟
′
−
′
(
−
′
3
22
ω

rrD qsD qrC
qsA qrD qsC qrB qsB qrC qsD qrA
II
II II II II
+

′
)
+−
′
+
′
−
′
+
′
()
coos
sin
2
2
ω
ω
e
qsA qrC qsB qrD qsC qrA qsD qrB e
t
II II II II t+
′
−
′
−
′
+
′
()

(9.4-14)

Torque may be expressed in per unit by expressing X
M
in per unit and eliminating the
factor (3/2)( P /2)(1/ ω
b
).
Phasor Relationships
The phasors representing the instantaneous sequence quantities F
qs
and F
ds
for a stator
unbalance given in (9.4-2)–(9.4-7) may be written as

2

FFjF
qs qsA qsB+
=−
(9.4-15)

2
2


FFjF
jF
ds qsB qsA

qs
+
+
=+
=

(9.4-16)

UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES 343

2

FFjF
qs qsC qsD−
=−
(9.4-17)

2
2


FFjF
jF
ds qsD qsC
qs
−
−
=− −
=−

(9.4-18)
If (9.4-15)–(9.4-18) are used to express (9.4-11) and (9.4-12) in phasor form, we have





F
F
jj
F
F
qs
ds
qs
qs
⎡
⎣
⎢
⎤
⎦
⎥
=
−
⎡
⎣
⎢
⎤
⎦
⎥

⎡
⎣
⎢
⎤
⎦
⎥
+
−
11

(9.4-19)
Solving (9.4-19) for

F
qs+
and

F
qs−
yields





F
F
j
j
F

F
qs
qs
qs
ds
+
−
⎡
⎣
⎢
⎤
⎦
⎥
=
−
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
1
2
1

1

(9.4-20)
Equation (9.4-20) is the symmetrical component transformation for a two-phase system
and (9.4-19) is the inverse. The zero sequence may be expressed in phasor form from
(9.4-3) as

2
0

FF jF
s abcs abcs
=−
αβ
(9.4-21)
If we add the zero sequence to (9.4-20) , we can write







F
F
F
j
j
F
F

F
qs
qs
s
qs
ds
+
−
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
=
−
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥

0
1
2
10
10
002

00s
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
(9.4-22)
Recall from our work in Chapter 3 that for balanced sets, the phasors are the same in
all asynchronously rotating reference frames. The frequency of the balance set in a
specifi c reference frame need to be considered only when expressing the instantaneous
balanced set. Therefore, except for

F
s0
, all phasors in (9.4-22) are valid in any asyn-
chronous reference frame, and it is not necessary to use a superscript, although we will
for clarity.
Let us write (9.4-22) as





F
F
F
j
j
F
qs
s
qs
s
s
s
s
as+
−
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥

=
−
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥


0
1
2
10
10
002
K


F
F
bs
cs
⎡
⎣
⎢

⎢
⎢
⎤
⎦
⎥
⎥
⎥
(9.4-23)
What have we done? Well, fi rst we added a superscript to the variables given in
(9.4-22) to give us the sense of being in the stationary reference frame. Next, we trans-
formed

F
qs
s
,

F
ds
s
, and

F
s0
to

F
as
,


F
bs
, and

F
cs
by
K
s
s
. With θ = 0,
K
s
s
becomes
344 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES

K
s
s
=
−−
−
⎡
⎣
⎢
⎢
⎢
⎢
⎢

⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
2
3
1
1
2
1
2
0
3
2
3
2
1
2
1
2
1
2
(9.4-24)

Because
K
s
s
is not a function of time, it is permissible to use it to transform phasors. It
can be shown that

SK=
−
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
1
2
10
10
002
j
j
s
s
(9.4-25)
Thus (9.4-23) may be written as


FSF
qs s
s
abcs±
=
0
(9.4-26)
where S is the symmetrical component transformation given by (9.2-4) . Therefore,
(9.4-26) is (9.2-1) , and we have


FF
qs
s
s++
=
(9.4-27)


FF
qs
s
s−−
=
(9.4-28)
Now that we are working in the stationary reference frame, let us rewrite (9.4-20) as






F
F
j
j
F
F
qs
s
qs
s
qs
s
ds
s
+
−
⎡
⎣
⎢
⎤
⎦
⎥
=
−
⎡
⎣
⎢

⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
1
2
1
1

(9.4-29)
For stator unbalances with symmetrical three-wire rotor circuits, (9.4-29) also applies
to rotor phasors in the stationary reference frame. That is





′
′
⎡
⎣
⎢
⎤
⎦
⎥

=
−
⎡
⎣
⎢
⎤
⎦
⎥
′
′
⎡
⎣
⎢
⎤
⎦
⎥
+
−
F
F
j
j
F
F
qr
s
qr
s
qr
s

dr
s
1
2
1
1

(9.4-30)
The steady-state voltage equations in the stationary reference frame may be obtained
from (6.5-34) by setting ω = 0 and p = j ω
e
. Thus





V
V
V
V
rj X j X
r
qs
s
ds
s
qr
s
dr

s
s
e
b
ss
e
b
M
s
′
′
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
=
+
ω
ω
ω
ω
00

0
++
−+
′
−
′
jX jX
jX X rjX X
X
e
b
ss
e
b
M
e
b
M
r
b
Mr
e
b
rr
r
b
rr
r
b
ω

ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
0
MM
e
b
M
r
b
rr r
e
b
rr
qs
jX X rjX
I
ω
ω
ω

ω
ω
ω
′′
+
′
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥


ss
ds
s
qr
s
dr
s
I
I
I



′
′
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
(9.4-31)
UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES 345

If we now incorporate (9.4-29) and (9.4-30) into (9.4-31) , the sequence voltage equa-
tions become





V
V
s
V
V
s
rj X
qs
s
qr
s
qs
s
qr
s
s
e
b
+
+
−
−
′

′
−
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
=
+
2
ω
ω
sss
e
b
M
e

b
M
re
b
rr
s
e
b
ss
e
b
M
jX
jX
r
s
jX
rX jX
j
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
ω

00
00
00
00
′
+
′
+
ee
b
M
re
b
rr
qs
s
qr
X
r
s
jX
I
I
ω
ω
ω
′
−
+
′

⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
′
+
+
2



ss
qs
s
qr
s
I
I


−
−
′
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥

(9.4-32)
where

s
er

e
=
−
ωω
ω
(9.4-33)
The steady-state electromagnet torque, for a stator unbalance, may be expressed

T
PX
jI I I I j
e
M
b
qs qr
s
qs qr
s
=
⎛
⎝
⎜
⎞
⎠
⎟
′
−
′
⎡
⎣

⎤
⎦
+
++ −−
3
2
ω
Re ( ) Re
 
*s *s
−−
′
+
′
()
⎡
⎣
⎤
⎦
+
′
+− −+
+−
 

II II t
II
qs
s
qr

s
qs
s
qr
s
e
qs
s
qr
s
cos
Re
2
ω
−−
′
⎡
⎣
⎤
⎦
−+

II t
qs
s
qr
s
e
sin2
ω

(9.4-34)
where the asterisk denotes the conjugate.
Steady-state instantaneous stator variables, F
as
, F
bs
, and F
qs
, may be obtained from
the phasors

F
as
,

F
bs
, and

F
cs
determined from (9.4-23) . With the assumption of a sym-
metrical rotor, the frequency of the stator variables is ω
e
; however, as has been men-
tioned the rotor variables
′
F
ar

,
′
F
br
, and
′
F
cr
each contain two frequencies ( ω
e
− ω
r
) and
( ω
e
+ ω
r
). The instantaneous rotor variables can be obtained by using the rotor equiva-
lent of (9.4-15)–(9.4-18) to identify
′
F
qrA
through
′
F
qrD
, which can be substituted into the
rotor equivalent of (9.4-2) . To obtain
′
F

qr
r
and
′
F
dr
r
, θ in (9.4-2) must be set to ω
r
t , where-
upon
K
r
r
may be used to obtain
′
F
ar
,
′
F
br
, and
′
F
cr
. Since we have assumed symmetrical
three-wire stator and rotor circuits, neither F
0

s
nor
′
F
r0
exist; however, we have included
the zero-sequence notation in order to show the equivalence to the symmetrical com-
ponent transformation. If a zero sequence were present in the rotor circuits, it would
consist of two frequencies, unlike the stator zero sequence, which would contain
only ω
e
.
346 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES
9.5. TYPICAL UNBALANCED STATOR CONDITIONS OF
INDUCTION MACHINES
Although it is not practical to consider all stator unbalanced conditions that might occur,
the information given in this section should serve as a guide to the solution of a large
class of problems. Unbalanced source voltages, unbalanced phase impedances, and an
open-circuited stator phase are considered, and the method of calculating the steady-
state performance is set forth in each case.
Unbalanced Source Voltages
Perhaps the most common unbalanced stator condition is unbalanced source voltages.
This can occur in a power system due to a fault or a switching malfunction that may
cause unbalanced conditions to exist for a considerable period of time. The stator circuit
of an induction machine for the purpose of analysis is given in Figure 9.5-1 . From
Figure 9.5-1 , we can write

evv
ga as ng
=+

(9.5-1)

evv
gb bs ng
=+
(9.5-2)

evv
gc cs ng
=+
(9.5-3)
In a symmetrical three-wire system, the zero sequence is not present, and it can be
shown that in the steady state, we have


VEEE
qs
s
ga gb gc
=−−
⎛
⎝
⎜
⎞
⎠
⎟
2
3
1
2

1
2
(9.5-4)


VEE
ds
s
gb gc
=−+
()
1
3
(9.5-5)
Figure 9.5-1. Stator circuit of induction machine.
e
gc
e
gb
e
ga
g
i
bs
i
cs
i
as
r
s

r
s
r
s
v
cs
v
as
v
bs
n
+
+
+
TYPICAL UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES 347
Substituting (9.5-4) and (9.5-5) into (9.4-29) yields

 
VEEEjEE
qs
s
ga gb gc gb gc+
=−−
⎛
⎝
⎜
⎞
⎠
⎟
−−+

()
1
3
1
2
1
2
1
23
(9.5-6)

 
VEEEjEE
qs
s
ga gb gc gb gc−
=−−
⎛
⎝
⎜
⎞
⎠
⎟
+−+
()
1
3
1
2
1

2
1
23
(9.5-7)
The steady-state phasor currents can be calculated from (9.4-32) . The torque may then
be calculated using (9.4-34) .
Unbalanced Stator Impedances
For this unbalanced condition, let us consider the stator circuit shown in Figure 9.5-2 ,
wherein an impedance z(p) is placed in series with the as winding. The following equa-
tions may be written as

eizpvv
ga as as ng
=++()
(9.5-8)

evv
gb bs ng
=+
(9.5-9)

evv
gc cs ng
=+
(9.5-10)
In a three-wire system, i
0

s
is zero, therefore v

0

s
is zero because the stator circuits are
symmetrical. Let us now assume that the source voltages are balanced; hence, we can
add (9.5-8)–(9.5-10) , and we have

vizp
ng as
=−
1
3
()
(9.5-11)
Figure 9.5-2. Stator circuit with impedance in series with as winding.
e
gc
e
ga
e
gb
i
cs
i
as
i
bs
r
s
r

s
r
s
v
cs
v
as
v
bs
g
z(p)
+
+
+
n
348 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES
Equation (9.5-11) is valid as long as e
ga
+ e
gb
+ e
gc
is zero. Substituting (9.5-11) into
(9.5-8)–(9.5-10) and solving for the phase voltages yields

ve izp
as ga as
=−
2
3

()
(9.5-12)

ve izp
bs gb as
=+
1
3
()
(9.5-13)

ve izp
cs gc as
=+
1
3
()
(9.5-14)
In a problem at the end of the chapter, you are asked to write ν
as
, ν
bs
, and ν
cs
assuming
e
ga
, e
ga
, and e

gc
are not balanced. Substituting the steady-state phasor equivalents of
(9.5-12)–(9.5-14) into (9.4-29) yields

 
VE IZ
qs
s
ga as+
=−
1
3
(9.5-15)


VIZ
qs
s
as−
=−
1
3
(9.5-16)
From the inverse of (9.4-29) and because

I
s0
is zero, we have

 

II I
as qs
s
qs
s
=+
+−
(9.5-17)
If we now substitute (9.5-17) into (9.5-15) and (9.5-16) and then substitute the results
into (9.4-32) , we obtain



E
V
s
Zr j X j
ga
qr
s
s
e
b
ss
e
′
⎡
⎣
⎢
⎢

⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
=
++
+
0
0
1
3
ω
ω
ω

ω
bb
M
e
b
M
re
b
rr
s
e
b
ss
e
b
M
XZ
jX
r
s
jX
ZZrjXjX
1
3
0
00
1
3
0
1

3
0
ω
ω
ω
ω
ω
ω
ω
ω
′
+
′
++
00
2
jX
r
s
jX
I
I
e
b
M
re
b
rr
qs
s

ω
ω
ω
ω
′
−
+
′
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥

⎥
⎥
′
+


qqr
s
qs
s
qr
s
I
I
+
−
−
′
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥





(9.5-18)
The above equation, which can be reduced to a 3 × 3 matrix, may be used to solve for
the currents in phasor form and the torque may then be determined from (9.4-34) . It is
important to note that unbalanced external stator impedances cause the positive- and
negative-sequence voltage equations to be coupled.
TYPICAL UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES 349
Open-Circuited Stator Phase
For the purpose of analyzing an open-circuited stator phase, which is equivalent to
single-phase operation, let us consider the stator circuits of a three-phase, wye-connected
induction machine as shown in Figure 9.5-3 , where phase a is open-circuited. Because
the stator circuit is a three-wire symmetrical system, i
0

s
and v
0

s
are zero. Therefore, with
θ = 0,
ff
qs
s
as
=
and

vv
p
as qs
s
b
qs
s
==
ω
ψ
(9.5-19)
From (6.5-28) with
i
qs
s
= 0

ψ
qs
s
Mqr
s
Xi=
′
(9.5-20)
Thus

v
p

Xi
as
b
Mqr
s
=
′
ω
(9.5-21)
Therefore, if i
as
is zero and the voltage
(/ )pXi
bMqr
s
ω
′
is applied to phase a, the current
i
as
will be forced to remain at zero [7] . From Figure 9.5-3

vev
bs gb ng
=−
(9.5-22)

vev
cs gc ng
=−

(9.5-23)
In this system, ν
0

s
is zero; therefore, adding (9.5-21)–(9.5-23) and solving the result for
ν
ng
yields
Figure 9.5-3. Stator circuit with provisions for switching i
as
.
e
gc
e
ga
e
gb
i
cs
i
as
i
bs
r
s
r
s
r
s

v
cs
v
as
v
sa
v
bs
g
+
+
+
+
n
–
350 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES

veev
ng gb gc as
=++
1
2
1
2
()
(9.5-24)
Substituting (9.5-24) into (9.5-22) and (9.5-23) gives

veev
bs gb gc as

=−−
1
2
1
2
1
2
(9.5-25)

veev
cs gb gc as
=− + −
1
2
1
2
1
2
(9.5-26)
The above relationships are valid for transient and steady-state conditions. It is assumed
that the source voltages contain only one frequency, therefore, if we substitute the
steady-state phasor equivalent of (9.5-21) , (9.5-25) , and (9.5-26) into (9.4-26) , we
obtain


Vj XIE
qs
s
e
b

Mqr
s
+
=
′
+
1
2
ω
ω
(9.5-27)


Vj XIE
qs
s
e
b
Mqr
s
−
=
′
−
1
2
ω
ω
(9.5-28)
where


Ej E E
gb gc
=−
1
23
()
(9.5-29)
We can write

 
′
=
′
+
′
+−
II I
qr
s
qr
s
qr
s
(9.5-30)
Also,

I
qs

s
may also be expressed in terms of its symmetrical components as

 
II I
qs
s
qs
s
qs
s
=+
+−
(9.5-31)
However,

I
as
is zero, and since θ and

I
s0
are both zero, then

I
qs
s
, which is

I

as
, is zero.
Thus


II
qs
s
qs
s
−+
=−
(9.5-32)
If we substitute (9.5-30) into (9.5-27) and (9.5-28) , and then substitute the results into
(9.4-32) , and if we incorporate (9.5-32) , we can write
UNBALANCED ROTOR CONDITIONS OF INDUCTION MACHINES 351



E
V
s
rj X j X j
qr
s
s
e
b
ss
e

b
M
e
b
′
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
=
+−
+
0
1
2
1

2
ω
ω
ω
ω
ω
ω
XX
jX
r
s
jX
jX
r
s
jX
M
e
b
M
re
b
rr
e
b
M
re
b
rr
ω

ω
ω
ω
ω
ω
ω
ω
′
+
′
−
′
−
+
′
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
0
0
2
⎢⎢
⎤
⎦
⎥

⎥
⎥
⎥
⎥
⎥
⎥
′
′
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
+
+
−



I
I
I
qs
s
qr

s
qr
s
(9.5-33)
9.6. UNBALANCED ROTOR CONDITIONS OF INDUCTION MACHINES
In the analysis of unbalanced rotor conditions, it will be assumed that the stator circuits
are symmetrical and the stator applied voltages are balanced and have only one fre-
quency. Since the analysis of steady-state operation during unbalanced rotor conditions
is similar in many respects to the analysis for unbalanced stator conditions, the relation-
ships will be given without a lengthy discussion. The principal difference is the refer-
ence frame, in which the analysis is carried out. It is convenient, in the case of rotor
unbalanced conditions with symmetrical stator circuits, to conduct the analysis in the
rotor reference frame since therein the variables are of one frequency.
Unbalanced Machine Variables in the Arbitrary Reference Frame
Assuming only one rotor frequency is present, the rotor variables may be expressed as

′
′
′
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
=

′′
′′
′′
⎡
⎣
⎢
⎢
⎢
F
F
F
FF
FF
FF
ar
br
cr
ar ar
br br
cr cr
αβ
αβ
αβ
⎤⎤
⎦
⎥
⎥
⎥
−
(

)
−
(
)
⎡
⎣
⎢
⎤
⎦
⎥

cos
sin
ωω
ωω
er
er
t
t
(9.6-1)
Transforming the rotor variables as expressed by (9.6-1) to the arbitrary reference fame
by (6.4-1) yields

′
′
⎡
⎣
⎢
⎤
⎦

⎥
−−−+−F
F
tt t
qr
dr
ee er e
cos( ) sin( ) cos[( ) ] sin[(
ωθ ωθ ω ω θ ω
222
22
ωθ
ωθ ωθ ω ω θ ω ω
r
e e er er
t
tt t
)]
sin( ) cos( ) sin[( ) ] cos[(
+
−− − −+− −))]t
F
F
F
F
qrA
qrB
qrC
qrD
+

⎡
⎣
⎢
⎤
⎦
⎥
′
′
′
′
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
θ

(9.6-2)
and

′
=
′

−+
′
−FF tF t
r abcr e r abcr e r0
αβ
ωω ωω
cos( ) sin( )
(9.6-3)
352 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES
where

′
=
′
−
′
−
′
+
′
−
′
()
⎡
⎣
⎢
⎤
⎦
⎥
=−

′
FFFF FF
F
qrA arbrcr brcr
drB
1
3
1
2
1
2
3
2
ααα ββ

(9.6-4)

′
=
′
−
′
−
′
−
′
−
′
(
)

⎡
⎣
⎢
⎤
⎦
⎥
=
′
FFFF FF
F
qrB arbrcr brcr
drA
1
3
1
2
1
2
3
2
βββ αα

(9.6-5)

′
=
′
−
′
−

′
−
′
−
′
()
⎡
⎣
⎢
⎤
⎦
⎥
=
′
FFFF FF
F
qrC arbrcr brcr
drD
1
3
1
2
1
2
3
2
ααα ββ

(9.6-6)

′
=
′
−
′
−
′
+
′
−
′
(
)
⎡
⎣
⎢
⎤
⎦
⎥
=−
′
F FFF FF
F
qrD qrbrcr brcr
drC
1
3
1
2
1

2
3
2
βββ αα

(9.6-7)

′
=
′
+
′
+
′
(
)
FFFF
abcr arbrcr
αααα
1
3
(9.6-8)

′
=
′
+
′
+
′

()
FFFF
abcr arbrcr
ββββ
1
3
(9.6-9)
Also, in (6.4-1) , we have substituted for β , where

d
dt
r
β
ωω
=−
(
)
(9.6-10)
Note that in (9.6-2) , the positive sequence variables again have the argument of ( ω
e
t − θ );
however, the argument of the negative sequence is [( ω
e
− 2 ω
r
) t + θ ]. It follows that

′
=
′

+
′
+−
FF F
qr qr qr
(9.6-11)

′
=
′
+
′
+−
FF F
dr dr dr
(9.6-12)
We have assumed that the stator is a symmetrical, three-wire system and the applied
voltages are balanced, containing only one frequency, ω
e
. Therefore, we can use (9.6-2)–
(9.6-7) to express the stator variables by (1) replacing all r subscripts with s , (2) remov-
ing the primes, and (3) setting θ = 0 in (9.6-2) . This gives rise to two stator frequencies;
ω
e
and ( ω
e
− 2 ω
r
). It is assumed that the source voltages are of the frequency ω
e

, and we
will also assume a zero impedance source. Thus, the stator currents and the phase volt-
ages of the stator windings will contain ω
e
and ( ω
e
− 2 ω
r
). The ( ω
e
− 2 ω
r
) frequency,
which is the stator negative sequence, is induced into the stator windings due to the
UNBALANCED ROTOR CONDITIONS OF INDUCTION MACHINES 353
rotor unbalance. It is interesting to note that the stator negative sequence currents are
not present when ω
r
= (1/2) ω
e
. Therefore, because

I
qs−
r
is zero at ω
r
= (1/2) ω
e
, we would

expect the negative sequence torque to also be zero.
The instantaneous steady-state electromagnetic torque for a rotor unbalance may
be expressed as

T
PX
II II II
e
M
b
qsA qrB qsB qrA qsC qrD
=
⎛
⎝
⎜
⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟

′
−
′
−
′
3
22
ω
((
+
′
)
+−
′
+
′
−
′
+
′
()
II
II II II II
qsD qrC
qsA qrD qsC qrB qsB qrC qsD qrA
coss( )
sin (
2
2
ωω

er
qsA qrC qsB qrD qsC qrA qsD qrB
t
II II II II
−
+
′
−
′
−
′
+
′
()
ωωω
er
t− )

(9.6-13)

The pulsing component in (9.6-13) is commonly referred to as the “twice slip-frequency”
torque.
Phasor Relationships
It is convenient to conduct the steady-state analysis of unbalanced rotor conditions
much the same as unbalanced stator conditions. The phasor expressions from
(9.6-2) are

2

′

=
′
−
′
+
FFjF
qr qrA qrB
(9.6-14)

2
2


′
=
′
+
′
=
′
+
+
FFjF
jF
dr qrB qrA
qr

(9.6-15)

2

′
=
′
−
′
−
FFjF
qr qrC qrD
(9.6-16)

2
2


′
=−
′
−
′
=−
′
−
−
FFjF
jF
dr qrD qrC
qr

(9.6-17)

Following a procedure identical to that in the case of stator unbalance, we can write


′
=
′
±
FSF
qr r
r
abcr0
(9.6-18)
and


FSF
qs s
s
abcs±
=
0
(9.6-19)
We can write (6.5-34) in the rotor reference frame by setting ω = ω
r
, and then, by setting
p = j( ω
e
− ω

r
), we can obtain the voltage equations for steady-state conditions.
If we then substitute (9.4-29) and (9.4-30) for stator variables and similar relationships
for rotor variables from (9.6-15) and (9.6-16) into the steady-state equations, we will
obtain
354 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES





V
V
s
V
V
s
rj X
qs
r
qr
r
qs
r
qr
r
s
e
e
ss

+
+
−
−
′
′
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
=
+
ω
ω
jjX
jX

r
s
jX
rj X j
e
b
M
e
b
M
re
b
rr
s
er
b
ss
ω
ω
ω
ω
ω
ω
ωω
ω
ω
00
00
00
2

′
+
′
+
−
⎛
⎝
⎜
⎞
⎠
⎟
eer
b
M
e
b
M
re
b
rr
X
jX
r
s
jX
−
⎛
⎝
⎜
⎞

⎠
⎟
′
+
′
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
2
00
ω
ω
ω
ω

ω
ω
⎥⎥
⎥
⎥
⎥
⎥
′
′
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
+
+
−
−




I

I
I
I
qs
r
qr
r
qs
r
qr
r


(9.6-20)
where

s
er
e
=
−
ωω
ω
(9.6-21)
Since we are dealing with symmetrical three-wire systems, F
0

s
and
′

F
r0
are zero.
The steady-state electromagnetic torque for a rotor unbalance may be expressed

T
PX
jI I I I j
e
M
b
qs
r
qr
r
qs
r
qr
r
=
⎛
⎝
⎜
⎞
⎠
⎟
′
−
′
()

⎡
⎣
⎤
⎦
+
++ −−
3
2
ω
Re Re
**
 
−−
′
+
′
()
⎡
⎣
⎤
⎦
−
+
′
+− −+
+
 

II II t
I

qs
r
qr
r
qs
r
qr
r
er
qs
r
cos( )
Re
ωω
2
IIII t
qr
r
qs
r
qr
r
er−−+
−
′
()
−

sin( )
ωω

2

(9.6-22)
where the asterisk denotes the conjugate. As we have mentioned, the negative sequence
torque is zero when ω
r
= (1/2) ω
e
, since

I
qs
r
−
becomes zero; however, the pulsating torque
component is still present.
9.7. UNBALANCED ROTOR RESISTORS
In some applications where it is necessary to accelerate a large-inertia mechanical load,
a wound-rotor induction machine equipped with variable external rotor resistors is often
used. As the speed of the machine increases, the value of the external rotor resistors
is decreased proportionally so as to maintain nearly maximum electromagnetic torque
during most of the acceleration period. Care must be taken, however, in order not to
unbalance the external rotor resistors during this process, otherwise a torque pulsation
of twice slip frequency occurs as noted in (9.6-13) and (9.6-22) , which may cause low-
frequency oscillations in the connected mechanical system. For the purpose of analyz-
ing unbalanced rotor resistors, we will consider the rotor circuit given in Figure 9.7-1 .
The rotor phase voltages may be written as

′
=−

′′
vviR
ar pm ar ar
(9.7-1)
UNBALANCED ROTOR RESISTORS 355

′
=−
′′
vviR
br pm br br
(9.7-2)

′
=−
′′
vviR
cr pm cr cr
(9.7-3)
Since the rotor is assumed to be a three-wire system,
′
=i
r0
0
, and hence
′
=v
r0
0
, there-

fore, if we add (9.7-1)–(9.7-3) and solve for ν
pm
, we obtain

viRiRiR
pm ar ar br br cr cr
=
′′
+
′′
+
′′
(
)
1
3
(9.7-4)
Substituting (9.7-4) into (9.7-1)–(9.7-3) yields

′
′
′
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥

⎥
⎥
=
−
′′ ′
′
−
′′
′′
v
v
v
RR R
RRR
R
ar
br
cr
ar br cr
ar br cr
ar
1
3
2
2
RRR
i
i
i
br cr

ar
br
cr
−
′
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
′
′
′
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
2
(9.7-5)

For the analysis of steady-state operation, it is convenient to express (9.7-5) as


′
=
′
VRI
abcr abcr
(9.7-6)
where the terms of (9.7-6) can be determined by comparison with (9.7-5) . If we sub-
stitute (9.7-6) into (9.6-18) , we obtain


′
=
′
±
VSRI
qr r
r
abcr0
(9.7-7)
If we now substitute the inverse of (9.6-18) for

′
I
abcr
, we can write



′
=
′
±
−
±
VSRSI
qr r
r
qr r
r
0
1
0
()
(9.7-8)
It is clear that

′
V
r0
is zero, therefore, if we solve (9.7-8) , we obtain

 
′
=−
′
+
′
+

′′
++ −
VRIRjRI
qr
r
abcr qr
r
ABCr bcr qr
r
()
(9.7-9)
Figure 9.7-1. Rotor circuits with unbalanced external resistors.
r
r
¢
r
r
¢
r
r
¢
V
cr
¢
R
cr
¢
R
ar
¢

R
br
¢
V
ar
¢
V
br
¢
i
cr
¢
i
ar
¢
i
br
¢
m
+
+
+
p
356 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES


′
=
′
−

′′
−
′′
−+−
VR jRIRI
qr
r
ABCr bcr qr
r
abcr qr
r
()
(9.7-10)
where

′
=
′
+
′
+
′
RRRR
abcr arbrcr
1
3
()
(9.7-11)

′

=−
′
+
′
+
′
RRRR
ABCr arbrcr
1
3
1
2
1
2
()
(9.7-12)

′
=
′
−
′
RRR
bcr br cr
1
23
()
(9.7-13)
Substituting (9.7-9) and (9.7-10) into (9.6-20) yields



V
V
rj X j X
jX
r
qs
r
qs
r
s
e
b
ss
e
b
M
e
b
M
r
+
−
⎡
⎣
⎢
⎢
⎢
⎢

⎤
⎦
⎥
⎥
⎥
⎥
=
+
′
+
0
0
00
ω
ω
ω
ω
ω
ω
′′
+
′
−
′
−
′
+
−
⎛
⎝

⎜
⎞
⎠
⎟
R
s
jX
RjR
s
rj X j
abcr e
b
rr
ABCr bcr
s
er
b
ss
ω
ω
ωω
ω
0
00
2
ωωω
ω
ω
ω
ω

ω
er
b
M
ABCr bcr e
b
M
r abcr e
b
X
RjR
s
jX
rR
s
j
−
⎛
⎝
⎜
⎞
⎠
⎟
−
′
+
′′
+
′
+

′
2
0 XX
I
I
I
I
rr
qs
r
qr
r
qs
r
qr
r
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦

⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
′
′
⎡
⎣
⎢
⎢
+
+
−
−




⎢⎢
⎢
⎤
⎦
⎥
⎥

⎥
⎥

(9.7-14)
Because the stator is assumed to be balanced,

V
qs
r
−
in the above equation is zero and


V
qs
r
+
is

V
as
. Because

V
qs
r
−
is zero, then at ω
r
= (1/2) ω

e
,

I
qs
r
−
must be zero.
The acceleration characteristics of an induction machine with unbalanced rotor
resistors and a constant load torque are illustrated in Figure 9.7-2 . The values of the
external rotor resistors are
′
=R
ar
02.pu
,
′
=R
br
01.pu
, and
′
=R
cr
005.pu
. The machine
is a six-pole, three-phase, 220-V (line-to-line), 10-hp, 60-Hz machine. The parameters
expressed in per unit are

rX r

XHsX
sMr
ls lr
==
′
=
==
′
=
0 0453 2 042 0 0222
0 0775 0 05 0 0322



UNBALANCED ROTOR RESISTORS 357
Figure 9.7-2. Starting characteristics of a 10-hp induction motor with unbalanced rotor
resistors and 1.0 pu torque load; R’
ar
= 0.2 pu, R’
br
= 0.1 pu, and R’
cr
= 0.05 pu.
i
as
i
bs
i
cs
4.0

–4.0
0
4.0
–4.0
0
4.0
–4.0
0
4.0
–4.0
0
4.0
–4.0
0
4.0
–4.0
0
i
ar
¢
i
br
¢
i
cr
T
e
¢
w
r

w
b
4
2
0
1.0
0.5
0
0.1second
Balanced, rated stator voltages are applied, and the torque load is maintained constant
at 1.0 pu.
The steady-state torque versus speed characteristics are shown in Figure 9.7-3 for
balanced, rated stator voltages with the same rotor unbalance. The average torque, T
e

( + )
,
T
e

( − )
, and the zero to peak amplitude of the instantaneous, steady-state torque are plotted.
The “dip” in the average torque, which occurs soon after ω
r
= (1/2) ω
e
and T
e

( − )

has
become negative, is commonly referred to as “Goerges phenomenon,” named after
Hans Goerges who, in the late 1800s, explained the reason why loaded induction motors
358 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES
with unbalanced rotor impedances often accelerated only to approximately half syn-
chronous speed. This phenomenon is described in Reference 8 .
9.8. SINGLE-PHASE INDUCTION MACHINES
The single-phase induction machine is used widely in household applications: refrigera-
tor, washing machine, clothes dryer, furnace fan, garbage disposal, air conditioner, and
so on. Although the brushless dc machine is starting to be used in some of these appli-
cations, the single-phase induction motor remains in wide use.
As we have mentioned, the single-phase induction machine cannot develop a start-
ing torque. This is overcome by causing the currents in the stator windings of a two-
phase induction machine to be out of phase. To accomplish this the impedances of the
stator windings are made unequal. For example, the split-phase machine is a two-phase
induction motor with the impedance of the start winding larger than that of the run
winding. The asymmetry of the stator windings causes the winding currents to be
Figure 9.7-3. Steady-state torque-speed characteristics of a 10-hp induction motor with
unbalanced rotor resistors as given in Figure 9.7-2 .
3.0
2.5
2.0
1.5
1.0
0.5
0
–0.5
–1.0
Torque, per unit
0.2 0.4 0.6 0.8 1.0

T
e(+)
T
e(–)
T
e(avg)
T
e(pul)
w
r
w
b
SINGLE-PHASE INDUCTION MACHINES 359
out of phase when the phases are connected to the same single-phase source. This
provides a starting torque, and the start winding is generally mechanically disconnected
once the rotor speed has reached 60–80% of rated speed. The starting torque may
be increased by placing a capacitor in series with the start winding. This is called a
capacitor-start machine. The capacitor with the start winding can be switched-out at
60–80% of rated speed or the start winding may remain connected to the source and
the value of the series capacitance changed by switching out one branch of a parallel
arrangement of capacitors, thus increasing the average torque at rated speed. This is
referred to as a capacitor-start capacitor-run machine.
Although we may have made it appear that the split-phase machine is the type of
two-phase machine that is used in single-phase applications, this is not the case. In fact,
symmetrical two-phase induction machines are often used, whereupon the sizing of the
series capacitor becomes the means by which starting torque is produced.
It is appropriate that we consider the analysis of the single-phase machines at this
time because it is always operated in an unbalanced mode. In particular, the work we
have done involving unbalanced stator voltages, unbalanced stator impedances, and
open-circuited stator phase for the three-phase induction machine can, with minor

modifi cations, be used to analyze the operating modes of single-phase machines. We
will follow closely to the development for three-phase unbalanced case; however, we
will limit our work to the symmetrical, two-phase induction machine. The unsymmetri-
cal or split-phase induction machine will not be considered. It is covered in extensive
detail in Reference 7 .
Voltage Equations in Arbitrary Reference Frame
Before considering single-phase operation, it is necessary to set forth the voltage
equations in the arbitrary reference frame for the symmetrical two-phase induction
machine. Most readers will fi nd this a review; therefore, the equations are given with
a minimum of explanation. The two-phase symmetrical induction machine is shown in
Figure 9.8-1 .
The change of variables that formulates a transformation of a two-phase set of
variables associated with a stationary circuit to the arbitrary reference frame may be
expressed as

fKf
qds s abs
=
(9.8-1)
where

()f
qds
T
qs ds
ff=
[]

(9.8-2)

()f
abs
T
as bs
ff=
[]

(9.8-3)

K
s
=
−
⎡
⎣
⎢
⎤
⎦
⎥
cos sin
sin cos
θθ
θθ
(9.8-4)
360 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES
where K
s
= ( K
s
)

− 1
and the superscript T denotes the transpose of a matrix, and the
angular position may be expressed as

d
dt
θ
ω
=
(9.8-5)
The change of variables that transforms the rotor variables to the arbitrary reference
frame is

′
=
′
fKf
qdr r abr
(9.8-6)
where

()
′
=
′′
[]
f
qdr
T
qr dr

ff
(9.8-7)

()
′
=
′′
[]
f
abr
T
ar br
ff
(9.8-8)

K
r
=
−
⎡
⎣
⎢
⎤
⎦
⎥
cos sin
sin cos
ββ
ββ
(9.8-9)

Figure 9.8-1. A two-pole two-phase symmetrical induction machine.