the induction machine handbook chuong (7)

Author: Ion Boldea, S.A.Nasar………… ………
Chapter 7
STEADY STATE EQUIVALENT CIRCUIT AND
PERFORMANCE
7.1 BASIC STEADY-STATE EQUIVALENT CIRCUIT
When the IM is fed in the stator from a three-phase balanced power source,
the three-phase currents produce a traveling field in the airgap. This field
produces emfs both in the stator and in the rotor windings: E
1
and E
2s
. A
symmetrical cage in the rotor may be reduced to an equivalent three-phase
winding.
The frequency of E
2s
is f
2
.

111
1
1
1
1
1
112
Sfpn
n
nn
pn

p
f
npff =
−
=








−=−=
(7.1)
This is so because the stator mmf travels around the airgap with a speed n
1

= f
1
/p
1
while the rotor travels with a speed of n. Consequently, the relative speed
of the mmf wave with respect to rotor conductors is (n
1
– n) and thus the rotor
frequency f
2
in (7.1) is obtained.
Now the emf in the short-circuited rotor “acts upon” the rotor resistance R

r

and leakage inductance L
rl
:

()
r
rl1r
2s2
ILjSRESE ω+==
(7.2)

r
rl1
r
2
s2
ILj
S
R
E
S
E







ω+==
(7.3)
If Equation (7.2) includes variables at rotor frequency Sω
1
, with the rotor in
motion, Equation (7.3) refers to a circuit at stator frequency ω
1
, that is with a
“fictious” rotor at standstill.
Now after reducing E
2
, I
r
, R
r
,and L
rl
to the stator by the procedure shown in
Chapter 6, Equation (7.3) yields

'I'Lj1
S
1
'R'RE'E
r
rl1rr
12







ω+






−+==
(7.4)

rotors cagefor 2/1 W,1K ;K
WK
WK
E
E
2w2E
11w
22w
1
2
====

© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………

rotors cagefor Nm ,2/1 W;K
WKm
WKm
'I
I
r22I
22w2
11w1
r
r
====
(7.5)

I
E
rl
rl
r
r
K
K
'L
L
'R
R
==
(7.6)

skew2y2q2w1y1q1w
KKKK ;KKK ⋅⋅=⋅=
(7.7)
W
1
, W
2
are turns per phase (or per current path)
K
w1
, K
w2
are winding factors for the fundamental mmf waves
m
1
, m
2
are the numbers of stator and rotor phases, N
r
is the number of rotor slots
The stator phase equation is easily written:

()
sl1s
s
s
1
LjRIVE ω+−=−
(7.8)
because in addition to the emf, there is only the stator resistance and leakage

inductance voltage drop.
Finally, as there is current (mmf) in the rotor, the emf E
1
is produced
concurrently by the two mmfs (I
s
, I
r
′).

()
'IILj
dt
d
E
rs
m11
m1
1
+ω−=
ψ
−=
(7.9)
If the rotor is not short-circuited, Equation (7.4) becomes

'I'Lj1
S
1
'R'R
S

'V
E
r
rl1rr
r
1






ω+






−+=−
(7.10)
The division of V
r
(rotor applied voltage) by slip (S) comes into place as the
derivation of (7.10) starts in (7.2) where

()
r
rl1r
r

2
ILjSRVES ω+=−
(7.11)
The rotor circuit is considered as a source, while the stator circuit is a sink.
Now Equations (7.8) – (7.11) constitute the IM equations per phase reduced to
the stator for the rotor circuit.
Notice that in these equations there is only one frequency, the stator
frequency ω
1
, which means that they refer to an equivalent rotor at standstill,
but with an additional “virtual” rotor resistance per phase R
r
(1/S−1) dependent
on slip (speed).
It is now evident that the active power in this additional resistance is in fact
the electro-mechanical power of the actual motor

()
2
r
rem
'I1
S
1
'R3n2TP







−=π⋅=
(7.12)
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




with
()
S1
p
f
n
1
1
−=
(7.13)

()
elm
2
r
r
e
1
1
P
S

'I'R3
T
p
==
ω
(7.14)
P
elm
is called the electromagnetic power, the active power which crosses the
airgap, from stator to rotor for motoring and vice versa for generating.
Equation (7.14) provides a definition of slip which is very useful for design
purposes:

()
elm
Cor
elm
2
r
r
P
P
P
'I'R3
S
==
(7.15)
Equation (7.15) signifies that, for a given electromagnetic power P
elm
(or

torque, for given frequency), the slip is proportional to rotor winding losses.
I
s
V
s
E
1
R
s
j L
ω
1sl
E
1
R’
r
j L’
ω
1rl
R’
r
(1-S)
S
V
S
r
I
s
V
s

R
s
j L =jX
ω
1sl sl
R
1m
I
0a
j L =jX
ω
11m 1m
I
m
R’
r
R’
r
(1-S)
S
V
S
r
+
-
j L’ =jX’
ω
1rl rl
E =+j L ( + ’)=j L
ωω

II I
111msr1m1m

a.)
b.)

Figure 7.1 The equivalent circuit
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




Equations (7.8) – (7.11) lead progressively to the ideal equivalent circuit in
Figure 7.1.
Still missing in Figure 7.1a are the parameters to account for core losses,
additional losses (in the cores and windings due to harmonics), and the
mechanical losses.
The additional losses P
ad
will be left out and considered separately in
Chapter 11 as they amount, in general, to up to 3% of rated power in well-
designed IM.
The mechanical and fundamental core losses may be combined in a
resistance R
1m
in parallel with X
1m
in Figure 7.1b, as at least core losses are
produced by the main path flux (and magnetization current I

m
). R
1m
may also be
combined as a resistance in series with X
1m
, for convenience in constant
frequency IMs. For variable frequency IMs, however, the parallel resistance R
1m

varies only slightly with frequency as the power in it (mainly core losses) is
proportional to E
1
2
= ω
1
2
K
w1
2
Φ
1
2
, which is consistent to eddy current core loss
variation with frequency and flux squared.
R
1m
may be calculated in the design stage or may be found through standard
measurements.

moa
iron
2
m
2
m1
iron
2
1
m1
II ;
P
IX3
P
E3
R <<==
(7.16)
7.2 CLASSIFICATION OF OPERATION MODES
The electromagnetic (active) power crossing the airgap P
elm
(7.14) is
positive for S > 0 and negative for S < 0.
That is, for S < 0, the electromagnetic power flows from the rotor to the
stator. After covering the stator losses, the rest of it is sent back to the power
source. For ω
1
> 0 (7.14) S < 0 means negative torque T
e
. Also, S < 0 means n >
n

1
= f
1
/p
1
. For S > 1 from the slip definition, S = (n
1
– n)/n
1
, it means that either
n < 0 and n
1
(f
1
) > 0 or n > 0 and n
1
(f
1
) < 0.
In both cases, as S > 1 (S > 0), the electromagnetic power P
elm
> 0 and thus
flows from the power source into the machine.
On the other hand, with n > 0, n
1
(ω
1
) < 0, the torque T
e
is negative; it is

opposite to motion direction. That is braking. The same is true for n < 0 and
n
1
(ω
1
) > 0. In this case, the machine absorbs electric power through the stator
and mechanical power from the shaft and transforms them into heat in the rotor
circuit total resistances.
Now for 0 < S < 1, T
e
> 0, 0 < n < n
1
, ω
1
> 0, the IM is motoring as the
torque acts along the direction of motion.
The above reasoning is summarized in Table 7.1.
Positive ω
1
(f
1
) means positive sequence-forward mmf traveling wave. For
negative ω
1
(f
1
), a companion table for reverse motion may be obtained.


© 2002 by CRC Press LLC

Author: Ion Boldea, S.A.Nasar………… ………




Table 7.1 Operation modes (f
1
/p
1
> 0)
S - - - - 0 + + + + 1 + + + +
n + + + + f
1
/p
1
+ + + + 0
T
e

P
elm

0 - - - - 0
0
+ + + +
+ + + +
+ + + +
+ + + +
0
0

Operation
mode
Generator Motor Braking
7.3 IDEAL NO-LOAD OPERATION
The ideal no-load operation mode corresponds to zero rotor current. From
(7.11), for I
r0
= 0 we obtain

2
R
0
R
2
0
E
V
S ;0VES ==−
(7.17)
The slip S
0
for ideal no-load depends on the value and phase of the rotor
applied voltage V
R
. For V
R
in phase with E
2
: S
0

> 0 and, with them in opposite
phase, S
0
< 0.
The conventional ideal no – load-synchronism, for the short-circuited rotor
(V
R
= 0) corresponds to S
0
= 0, n
0
= f
1
/p
1
. If the rotor windings (in a wound
rotor) are supplied with a forward voltage sequence of adequate frequency f
2
=
Sf
1
(f
1
> 0, f
2
> 0), subsynchronous operation (motoring and generating) may be
obtained. If the rotor voltage sequence is changed, f
2
= sf
1

< 0 (f
1
> 0),
supersynchronous operation may be obtained. This is the case of the doubly fed
induction machine. For the time being we will deal, however, with the
conventional ideal no-load (conventional synchronism) for which S
0
= 0.
The equivalent circuit degenerates into the one in Figure 7.2a (rotor circuit
is open).
Building the phasor diagram (Figure 7.2b) starts with I
m
, continues with
jX
1m
I
m
, then I
0a


m1
m
m1
oa
R
IjX
I =
(7.18)
and

moas0
III +=
(7.19)
Finally, the stator phase voltage V
s
(Figure 7.2b) is

0s
sl
s0
s
m
m1
s
IjXIRIjXV ++=
(7.20)
The input (active) power P
s0
is dissipated into electromagnetic loss,
fundamental and harmonics stator core losses, and stator windings and space
harmonics caused rotor core and cage losses. The driving motor covers the
mechanical losses and whatever losses would occur in the rotor core and
squirrel cage due to space harmonics fields and hysteresis.
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




I

s0
V
s
R
s
jX
sl
R
1m
I
0a
I
m
jX
1m
a.)
jX
1m
I
m
V
s
jX
sl
I
s0
R
s
I
s0

I
0a
I
m
I
s0
ϕ
oi
b.)
P
s0
p
iron Co
p
c.)
synchronous
motor
2p poles
3~
f
1
Power
analyser
VA R I A C
3~
f
1
d.)
n=f /p
11

2p poles
1
1

Figure 7.2 Ideal no-load operation (V
R
= 0):
a.) equivalent circuit b.) phasor diagram c.) power balance d.) test rig
For the time being, when doing the measurements, we will consider only
stator core and winding losses.

2
0ssiron
2
0ss
2
m1
2
m1
2
m1
m1
2
0ss
2
a0m10s
IR3p
IR3
RX
X

R3IR3IR3P
+=
=








+
+
=+≈
(7.21)
From d.c. measurements, we may determine the stator resistance R
s
. Then, from
(7.21), with P
s0
, I
s0
measured with a power analyzer, we may calculate the iron
losses p
iron
for given stator voltage V
s
and frequency f
1
.

We may repeat the testing for variable f
1
and V
1
(or variable V
1
/f
1
) to
determine the core loss dependence on frequency and voltage.
The input reactive power is

2
0ssl
2
m1
2
m1
2
m1
m10s
IX3
RX
R
X3Q









+
+
=
(7.22)
From (7.21)-(7.22), with R
s
known, Q
s0
, I
s0
, P
s0
measured, we may calculate
only two out of the three unknowns (parameters): X
1m
, R
1m
, X
sl
.
We know that R
1m
>> X
1m
>> X
sl
. However, X

sl
may be taken by the design
value, or the value measured with the rotor out or half the stall rotor (S = 1)
reactance X
sc
, as shown later in this chapter.
Consequently, X
1m
and R
1m
may be found with good precision from the
ideal no-load test (Figure 7.2d). Unfortunately, a synchronous motor with the
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




same number of poles is needed to provide driving at synchronism. This is why
the no-load motoring operation mode has become standard for industrial use,
while the ideal no-load test is mainly used for prototyping.

Example 7.1 Ideal no-load parameters
An induction motor driven at synchronism (n = n
1
= 1800rpm, f
1
= 60Hz, p
1
=

2) is fed at rated voltage V
1
= 240 V (phase RMS) and draws a phase current I
s0

= 3 A, the power analyzer measures P
s0
= 36 W, Q
s0
= 700 VAR, the stator
resistance R
s
= 0.1 Ω, X
sl
= 0.3 Ω. Let us calculate the core loss p
iron
, X
1m
, R
1m
.
Solution
From (7.21), the core loss p
iron
is

W3.3331.0336IR3Pp
2
2
0ss0siron

=⋅⋅−=−=
(7.23)
Now, from (7.21) and (7.22), we get

Ω==
⋅
⋅⋅−
=
−
=
+
233.1
27
3.33
33
31.0336
I3
IR3P
RX
XR
2
2
2
0s
2
0ss0s
2
m1
2
m1

2
m1m1
(7.24)

Ω=
⋅
⋅⋅−
=
−
=
+
626.25
33
33.03700
I3
IX3Q
RX
XR
2
2
2
0s
2
0ssl0s
2
m1
2
m1
m1
2

m1
(7.25)
Dividing (7.25) by (7.26) we get

78.20
233.1
626.25
X
R
m1
m1
==
(7.26)
From (7.25),

()
685.25X626.25
178.20
X
;626.25
1
R
X
X
m1
2
m1
2
m1
m1

m1
=⇒=
+
=
+








(7.27)
R
1m
is calculated from (7.26),

Ω=⋅=⋅= 74.53378.20626.2578.20XR
m1m1
(7.28)
By doing experiments for various frequencies f
1
(and V
s
/f
1
ratios), the slight
dependence of R
1m

on frequency f1 and on the magnetization current I
m
may be
proved.
As a bonus, the power factor cos
ϕ
oi
may be obtained as

05136.0
P
Q
tancoscos
0s
0s
1
i0
=

















=ϕ
−
(7.29)
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




The lower the power factor at ideal no-load, the lower the core loss in the
machine (the winding losses are low in this case).
In general, when the machine is driven under load, the value of emf (E
1
=
X
1m
I
m
) does not vary notably up to rated load and thus the core loss found from
ideal no-load testing may be used for assessing performance during loading,
through the loss segregation procedure. Note however that, on load, additional
losses, produced by space field harmonics,occur. For a precise efficiency
computation, these “stray load losses” have to be added to the core loss
measured under ideal no-load or even for no-load motoring operation.
7.4 SHORT-CIRCUIT (ZERO SPEED) OPERATION
At start, the IM speed is zero (S = 1), but the electromagnetic torque is

positive (Table 7.1), so when three-phase fed, the IM tends to start (rotate); to
prevent this, the rotor has to be stalled.
First, we adapt the equivalent circuit by letting S = 1 and R
r
′ and X
sl
, X
rl
′ be
replaced by their values as affected by skin effect and magnetic saturation
(mainly leakage saturation as shown in Chapter 6): X
slstart
, R′
rstart
, X′
rlstart
(Figure
7.3).
I
sc
V
ssc
R
s
jX
slstart
R
1m
1m
I

m
R’
rstart
jX
I’
rsc
jX’
rlstart
V
ssc
R =R +R’
sc
sc
jX
srstart
sc
X =X +X’
slstart rlstart
a.)
b.)
I
sc
mandatory at low frequency

© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………

P
sc
p
iron
(very
small)
p =3(R +R’ )I
Cosc
c.)
R I
sc sc
sc
j
X
I
sc
V
ssc
ϕ
sc
I
sc
d.
)
srstart
sc
2

Three
phase

power
analyser
VA R I A C
3~
f
1
e.)
a
b
c
n=0
S=1
Z - impedance per phase
sc

Single
phase
power
analyser
VA R I A C
1~
f
1
f.)
a
b
c
n=0
S=1
Z - impedance per phase

sc
3
2

Figure 7.3 Short-circuit (zero speed) operation:
a.) complete equivalent circuit at S = 1, b.) simplified equivalent circuit S = 1, c.) power balance, d.)
phasor diagram, e.) three-phase zero speed testing, f.) single-phase supply zero speed testing
For standard frequencies of 50(60) Hz, and above, X
1m
>> R′
rstart
. Also, X
1m

>> X′
rlstart
, so it is acceptable to neglect it in the classical short-circuit equivalent
circuit (Figure 7.3b).
For low frequencies, however, this is not so; that is, X
1m
<> R
′
rstart
, so the
complete equivalent circuit (Figure 7.3a) is mandatory.
The power balance and the phasor diagram (for the simplified circuit of
Figure 7.3b) are shown in Figure 7.3c and d. The test rigs for three-phase and
single-phase supply testing are presented in Figure 7.3e and f.
It is evident that for single-phase supply, there is no starting torque as we
have a non-traveling stator mmf aligned to phase a. Consequently, no rotor

stalling is required.
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




The equivalent impedance is now (3/2)Z
sc
because phase a is in series with
phases b and c in parallel. The simplified equivalent circuit (Figure 7.3b) may
be used consistently for supply frequencies above 50(60) Hz. For lower
frequencies, the complete equivalent circuit has to be known. Still, the core loss
may be neglected (R
1m

≈

∞
) but, from ideal no-load test at various voltages, we
have to have L
1m
(I
m
) function. A rather cumbersome iterative (optimization)
procedure is then required to find R′
rstart
, X′
rlstart
, and X

slstart
with only two
equations from measurements of R
sc
, V
sc
/I
sc
.

2
rscrstart
2
scssc
'I'R3IR3P +=
(7.30)

()
()
rlstartm1rlstart
rlstartrlstartm1
slstarts
sc
'XXj'R
'jX'RjX
jXRZ
++
+
++=
(7.31)

This particular case, so typical with variable frequency IM supplies, will not
be pursued further. For frequencies above 50(60) Hz the short-circuit impedance
is simply

rlstartslstartscrlstartsscscsc
sc
'XXX ;'RRR ;jXRZ +=+=+≈
(7.32)
and with P
sc
, V
ssc
, I
sc
measured, we may calculate

2
sc
2
sc
ssc
sc
2
sc
sc
sc
R
I
V
X ;

I3
P
R −








== (7.33)
for three phase zero speed testing and

2
sc
2
~sc
~ssc
sc
2
~sc
~sc
sc
R
2
3
I
V
3

2
X ;
I
P
3
2
R






−








==
(7.34)
for single phase zero speed testing.
If the test is done at rated voltage, the starting current I
start
(I
sc
)

Vsn
is much
larger than the rated current,

0.85.4
I
I
n
start
÷≈
(7.35)
for cage rotors, larger for high efficiency motors, and

1210
I
I
n
start
÷≈
(7.36)
for short-circuited rotor windings.
The starting torque T
es
is:
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………

1
1
2
startrstart
es
p
I'R3
T
ω
=
(7.37)
with

()
enes
T3.27.0T ÷=
(7.38)
for cage rotors and

()
enes
T3.01.0T ÷=
(7.39)
for short-circuited wound rotors.
I
sat
E
s
V

ssc
I
sc
x
x
x
x
x
x
h
or
closed
rotor
slots
semiclosed
rotor
slots
R
s
jX
sl
R
1m
E
s
R’
r
S
jX
1m

j L’
ω
rl1

Figure 7.4 Stator voltage versus short-circuit current
Thorough testing of IM at zero speed generally completed up to rated
current. Consequently, lower voltages are required, thus avoiding machine
overheating. A quick test at rated voltage is done on prototypes or random IMs
from, production line to measure the peak starting current and the starting
torque. This is a must for heavy starting applications (nuclear power plant
cooling pump-motors, for example) as both skin effect and leakage saturation
notably modify the motor starting parameters: X
slstart
, X′
rlstart
, R′
rstart
.
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




Also, closed slot rotors have their slot leakage flux path saturated at rotor
currents notably less than rated current, so a low voltage test at zero speed
should produce results as in Figure 7.4.
Intercepting the I
sc
/V

sc
curve with abscissa, we obtain, for the closed slot
rotor, a non-zero emf E
s
.[1] E
s
is in the order of 6 to 12V for 220 V phase RMS,
50(60) Hz motors. This additional emf is sometimes introduced in the
equivalent circuit together with a constant rotor leakage inductance to account
for rotor slot−bridge saturation. E
s
is 90
0
ahead of rotor current I
r
′ and is equal
to

()
eorsbridgebridge1wbridge11s
LhB ;KW2f2
4
E ⋅⋅=ΦΦπ
π
≈
(7.40)
B
sbridge
is the saturation flux density in the rotor slot bridges (B
sbridge

= (2 –
2.2)T). The bridge height is h
or
= 0.3 to 1 mm depending on rotor peripheral
speed. The smaller, the better.
A more complete investigation of combined skin and saturation effects on
leakage inductances is to be found in Chapter 9 for both semiclosed and closed
rotor slots.

Example 7.2
Parameters from zero speed testing
An induction motor with a cage semiclosed slot rotor has been tested at zero
speed for V
ssc
= 30 V (phase RMS, 60 Hz). The input power and phase current
are: P
sc
= 810 kW, I
sc
= 30 A. The a.c. stator resistance R
s
= 0.1Ω. The rotor
resistance, without skin effect, is good for running conditions, R
r
= 0.1Ω. Let us
determine the short-circuit (and rotor) resistance and leakage reactance at zero
speed, and the start-to-load rotor resistance ratio due to skin effect.
Solution
From (7.33) we may determine directly the values of short-circuit resistance
and reactance, R

sc
and X
sc
,

Ω=−






=−








=
Ω=
⋅
==
954.03.0
30
30
R
I

V
X
;3.0
303
810
I3
P
R
2
2
2
sc
2
sc
ssc
sc
22
sc
sc
sc
(7.41)
The rotor resistance at start R′
rstart
is

Ω=−=−= 2.01.03.0RR'R
sscrstart
(7.42)
So, the rotor resistance at start is two times that of full load conditions.

0.2
1.0
2.0
'R
'R
r
rstart
==
(7.43)
The skin effect is responsible for this increase.
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




Separating the two leakage reactances X
slstart
and X′
rlstart
from X
sc
is hardly
possible. In general, X
slstart
is affected at start by leakage saturation, if the stator
slots are not open, while X′
rlstart
is affected by both leakage saturation and skin
effect. However, at low voltage (at about rated current), the stator leakage and

rotor leakage reactances are not affected by leakage saturation; only skin effect
affects both R′
rstart
and X′
rlstart
. In this case it is common practice to consider

rlstartsc
'XX
2
1
≈
(7.44)
7.5 NO-LOAD MOTOR OPERATION
When no mechanical load is applied to the shaft, the IM works on no-load.
In terms of energy conversion, the IM input power has to cover the core,
winding, and mechanical losses. The IM has to develop some torque to cover
the mechanical losses. So there are some currents in the rotor. However, they
tend to be small and, consequently, they are usually neglected.
The equivalent circuit for this case is similar to the case of ideal no-load,
but now the core loss resistance R
1m
may be paralleled by a fictitious resistance
R
mec
which includes the effect of mechanical losses.
The measured values are P
0
, I
0

, and V
s
. Voltages were tested varying from,
in general, 1/3V
sn
to 1.2V
sn
through a Variac.
R
s
jX
sl
R
1m
R’
r
S
jX
1m
jX’
rl
I
0
V
s
I’
r0
0m
S
S

=(0.02 - 0.1)
0m
n
a.)

x
x
x
x
x
x
x
x
x
x
x
p
iron
p
mec
1/9
1.0
V /V
ssn
b.)
P -3R I
0
s0
2
( )

2
p
cos
3R I
s
0
2
p
core
3R I
1m
0a
2
p
cor
3R’ I’
r
r0
2
p
mec
P
0
c.)

© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………

Power
analyser
VA R I A C
3~
f
1
d.)
n =n (1-S )
10
0m

Figure 7.5 No-load motor operation
a.) equivalent circuit, b.) no-load loss segregation, c.) power balance, d.) test arrangement
As the core loss (eddy current loss, in fact) varies with (ω
1
Ψ
1
)
2
, that is
approximately with V
s
2
, we may end up with a straight line if we represent the
function

(
)
2

smeciron
2
0s0
VfppIR3P =+=−
(7.44)
The intersection of this line with the vertical axis represents the mechanical
losses p
mec
which are independent of voltage. But for all voltages the rotor speed
has to remain constant and very close to synchronous speed.
Subsequently the core losses p
iron
are found. We may compare the core
losses from the ideal no-load and the no-load motoring operation modes. Now
that we have p
mec
and the rotor circuit is resistive (R
r
′/S
0n
>>X
rl
′), we may
calculate approximately the actual rotor current I
r0
.

sn0r
n0
r

V'I
S
'R
≈
(7.45)

()
'IV3
S
'I'R3
S1
S
'I'R3
p
0rsn
n0
2
0rr
n0
n0
2
0rr
mec
≈≈−≈
(7.46)
Now with R
r
′ known, S
0n
may be determined. After a few such calculation

cycles, convergence toward more precise values of I
r0
′ and S
0n
is obtained.

Example 7.3. No-load motoring
An induction motor has been tested at no-load at two voltage levels: V
sn
=
220V, P
0
= 300W, I
0
= 5A and, respectively, V
s
′ = 65V, P
0
′ = 100W, I
0
′ = 4A.
With R
s
= 0.1Ω, let us calculate the core and mechanical losses at rated voltage
p
iron
, p
mec
. It is assumed that the core losses are proportional to voltage squared.
Solution

The power balance for the two cases (7.44) yields

()
mec
n
iron
2
0s0
ppIR3P +=−

© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………





()
mec
2
sn
s
n
iron
2
0s0
p
V
'V
p'IR3'P +









=−
(7.47)

()
()
2.95p
220
65
p41.03100
5.292pp51.03300
mec
2
n
iron
2
mec
n
iron
2
=+







=⋅⋅−
=+=⋅⋅−

From (7.47),

W328.7617.2165.292p
W17.216
0873.01
25.955.292
p
mec
iron
=−=
=
−
−
=
(7.48)
Now, as a bonus, from (7.46) we may calculate approximately the no-load
current I
r0
′.

A1156.0
2203
328.76

V3
p
'I
sn
mec
0r
=
⋅
=≈
(7.49)
It should be noted the rotor current is much smaller than the no-load stator
current I
0
= 5A! During the no-load motoring tests, especially for rated voltage
and above, due to teeth or/and back core saturation, there is a notable third flux
and emf harmonic for the star connection. However, in this case, the third
harmonic in current does not occur. The 5th, 7th saturation harmonics occur in
the current for the star connection.
For the delta connection, the emf (and flux) third saturation harmonic does
not occur. It occurs only in the phase currents (Figure 7.6).
As expected, the no-load current includes various harmonics (due to mmf
and slot openings).
1
1.2
Y
V
V
s
sn
I (t)

t
0
0.8
1.2
∆
V
V
s
sn
I (t)
t
0

Figure 7.6 No-load currents, a.) star connection, b.) delta connection
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




They are, in general, smaller for larger q slot/pole/phase chorded coils and
skewing of rotor slots. More details in Chapter 10. In general the current
harmonics content decreases with increasing load.
7.6 THE MOTOR MODE OF OPERATION
The motor mode of operation occurs when the motor drives a mechanical
load (a pump, compressor, drive-train, machine tool, electrical generator, etc.).
For motoring, T
e
> 0 for 0 < n < f
1

/p
1
and T
e
< 0 for 0 > n > –f
1
/p
1
. So the
electromagnetic torque acts along the direction of motion. In general, the slip is
0 < S < 1 (see paragraph 7.2). This time the complete equivalent circuit is used
(Figure 7.1).
The power balance for motoring is shown in Figure 7.7.
p
cus
p
iron
p
s
p
cor
p
mec
stator
copper
losses
core
losses
stray
load

losses
rotor
winding
losses
mechanical
losses
P
m
P
elm
P
1e

Figure 7.7 Power balance for motoring operation
The motor efficiency η is

mecsCorironCosm
m
e1
m
pppppP
P
P
P
power electricinput
powershaft
+++++
===η
(7.50)
The rated power speed n is

()
S1
P
f
n
1
1
−=
(7.51)
The rated slip is S
n
= (0.08 – 0.006), larger for lower power motors.
The stray load losses p
s
refer to additional core and winding losses due to
space (and eventual time) field and voltage time harmonics. They occur both in
the stator and in the rotor. In general, due to difficulties in computation, the
stray load losses are still assigned a constant value in some standards (0.5 or 1%
of rated power). More on stray losses in Chapter 11.
The slip definition (7.15) is a bit confusing as P
elm
is defined as active
power crossing the airgap. As the stray load losses occur both in the stator and
rotor, part of them should be counted in the stator. Consequently, a more
realistic definition of slip S (from 7.15) is

sironCose1
cor
selm

cor
pppP
p
pP
p
S
−−−
=
−
=
(7.52)
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




As slip frequency (rotor current fundamental frequency) f
2n
= Sf
1n
it means
that in general for f
1
= 60(50) Hz, f
2n
= 4.8(4) to 0.36(0.3) Hz.
For high speed (frequency) IMs, the value of f
2
is much higher. For

example, for f
1n
= 300 Hz (18,000 rpm, 2p
1
= 2) f
2n
= 4 – 8 Hz, while for f
1n
=
1,200 Hz it may reach values in the interval of 16 – 32 Hz. So, for high
frequency (speed) IMs, some skin effect is present even at rated speed (slip).
Not so, in general, in 60(50) Hz low power motors.
7.7 GENERATING TO POWER GRID
As shown in paragraph 7.2, with S < 0 the electromagnetic power travels
from rotor to stator (P
eml
< 0) and thus, after covering the stator losses, the rest
of it is sent back to the power grid. As expected, the machine has to be driven at
the shaft at a speed n > f
1
/p
1
as the electromagnetic torque T
e
(and P
elm
) is
negative (Figure 7.8).
Driving
motor

Induction
generator
n>f /p
11
f
3~
power
grid
1

Figure 7.8 Induction generator at power grid
The driving motor could be a wind turbine, a Diesel motor, a hydraulic
turbine etc. or an electric motor (in laboratory tests).
The power grid is considered stiff (constant voltage and frequency) but,
sometimes, in remote areas, it may also be rather weak.
To calculate the performance, we may use again the complete equivalent
circuit (Figure 7.1) with S < 0. It may be easily proved that the equivalent
resistance R
e
and reactance X
e
, as seen from the power grid, vary with slip as
shown on Figure 7.9.
actual
generating
ideal generating motoring braking
0
S
-
8

S
og2
S
og1
1
R (S)
X (S)
e
e

Figure 7.9 Equivalent IM resistance R
e
and reactance X
e
versus slip S
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




It should be noted that the equivalent reactance remains positive at all slips.
Consequently, the IM draws reactive power in any conditions. This is necessary
for a short-circuited rotor. If the IM is doubly fed as shown in Chapter 19, the
situation changes as reactive power may be infused in the machine through the
rotor slip rings by proper rotor voltage phasing, at f
2
= Sf
1
.

Between S
0g1
and S
0g2
(both negative), Figure 7.9, the equivalent resistance
R
e
of IM is negative. This means it delivers active power to the power grid.
The power balance is, in a way, opposite to that for motoring (Figure 7.10).
p
Cos
p
iron
p
s
p
Cor
p
mec
P
shaft
P
2electric

Figure 7.10. Power balance for generating
The efficiency η
g
is now

mecsCorironCoselectric2

electric2
shaft1
electric2
pppppP
P
P
P
inputpower shaft
outputpower electric
+++++
===η
(7.53)
Above the speed
()
0S ;S1
P
f
n
2g02g0
1
1
max
<−=
, (7.54)
as evident in Figure 7.9, the IM remains in the generator mode but all the
electric power produced is dissipated as loss in the machine itself.
Induction generators are used more and more for industrial generation to
produce part of the plant energy at convenient timing and costs. However, as it
still draws reactive power, “sources” of reactive power (synchronous generators,
or synchronous capacitors or capacitors) are also required to regulate the voltage

in the power grid.

Example 7.4
Generator at power grid
A squirrel cage IM with the parameters R
s
= R
r
′ = 0.6 Ω, X
sl
= X
rl
′ = 2 Ω, X
1ms
=
60 Ω, and R
1ms
= 3 Ω (the equivalent series resistance to cover the core losses,
instead of a parallel one)–Figure 7.11 – works as a generator at the power grid.
Let us find the two slip values S
0g1
and S
0g2
between which it delivers power to
the grid.
Solution
The switch from parallel to series connection in the magnetization branch,
used here for convenience, is widely used.
The condition to find the values of S
0g1

and S
0g2
for which, in fact, (Figure
7.9) the delivered power is zero is

(
)
0.0SR
g0e
=
(7.55)
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




I
s
V
s
R
s
jX
sl
R
1ms
1ms
R’
S

r
jX
jX’
rl
A
B
A
B
parallel
series
R
1m
1m
jX

Figure 7.11 Equivalent circuit with series magnetization branch
Equation (7.55) translates into

()
()
()
0'XXRRR'XR
RR2XR
S
'R
S
'R
RR
2
rlms1ss

2
ms1
2
rlms1
sms1
2
ms1
2
ms1
g0
r
2
g0
r
sms1
=++++
++++








+
(7.56)
In numbers:

()

()
02606.06.0323
S
1
6.0326036.0
S
1
6.06.3
2
22
g0
22
2
g0
2
=++⋅+⋅+
+⋅⋅+++








⋅
(7.57)
with the solutions S
og1
= −0.33⋅10

-3
and S
0g2
= −0.3877.
Now with f
1
= 60 Hz and with 2p
1
= 4 poles, the corresponding speeds (in
rpm) are

()
()()
()
()()
rpm86.2497603877.01
2
60
60S1
P
f
n
rpm594.1800601033.01
2
60
60S1
P
f
n
2g0

1
1
2og
3
1g0
1
1
1og
=⋅−−=⋅−=
=⋅⋅−−=⋅−=
−
(7.58)
7.8 AUTONOMOUS INDUCTION GENERATOR MODE
As shown in paragraph 7.7, to become a generator, the IM needs to be
driven above no-load ideal speed n
1
(n
1
= f
1
/p
1
with short-circuited rotor) and to
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




be provided with reactive power to produce and maintain the magnetic field in

the machine.
As known, this reactive power may be “produced” with synchronous
condensers (or capacitors)–Figure 7.12.
The capacitors are
∆
connected to reduce their capacitance as they are
supplied by line voltage. Now the voltage V
s
and frequency f
1
of the IG on no-
load and on load depend essentially on machine parameters, capacitors C
∆
, and
speed n. Still n > f
1
/p
1
.
Let us explore the principle of IG capacitor excitation on no-load. The
machine is driven at a speed n.
Driving
motor
3~
f
1
3
p
hase
load

n>f /p
11
2
p

p
oles
1
C
∆

Figure 7.12 Autonomous induction generator (IG) with capacitor magnetization
The d.c. remanent magnetization in the rotor, if any (if none, d.c.
magnetization may be provided as a few d.c. current surges through the stator
with one phase in series with the other two in parallel), produces an a.c. emf in
the stator phases. Then three-phase emfs of frequency f
1
= p
1
⋅
n cause currents to
flow in the stator phases and capacitors. Their phase angle is such that they are
producing an airgap field that always increases the remanent field; then again,
this field produces a higher stator emf and so on until the machine settles at a
certain voltage V
s
and frequency f
1
≈ p
1

n. Changing the speed will change both
the frequency f
1
and the no-load voltage V
s0
. The same effect is produced when
the capacitors C
∆
are changed.
I
m
I
m
E
rem
f =n
p
11
jX
1m
V
C =C /3
Y
∆
s0

Figure 7.13 Ideal no-load IG per phase equivalent circuit with capacitor excitation
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………

A quasiquantitative analysis of this selfexcitation process may be produced
by neglecting the resistances and leakage reactances in the machine. The
equivalent circuit degenerates into that in Figure 7.13.
The presence of rotor remanent flux density (from prior action) is depicted
by the resultant small emf E
rem
(E
rem
= 2 to 4 V) whose frequency is f
1
= np
1
.
The frequency f
1
is essentially imposed by speed.
The machine equation becomes simply

()
m0s
m
Y1
remm
m10s
IVI
C
1

jEIjXV =
ω
−=+=
(7.59)
As we know, the magnetization characteristic (curve)–V
s0
(I
m
)–is, in general,
nonlinear due to magnetic saturation (Chapter 5) and may be obtained through
the ideal no-load test at f
1
= p
1
n. On the other hand, the capacitor voltage
depends linearly on capacitor current. The capacitor current on no-load is,
however, equal to the motor stator current. Graphically, Equation (7.59) is
depicted on Figure 7.14.
E
rem
L (I )
1m m
V (I )
s0 m
A
A’
I
m
f =p n
11

f’ =p n’
11
n’< n
I’
1
C
ω
m
1Y
V
V’
s0
s0

Figure 7.14 Capacitor selfexcitation of IG on no-load
Point A represents the no-load voltage V
s0
for given speed, n, and capacitor
C
Y
. If the selfexcitation process is performed at a lower speed n′ (n′ < n), a
lower no-load voltage (point A′), V
s0
′, at a lower frequency f
1
′ ≈ p
1
n′ is
obtained.
Changing (reducing) the capacitor C

Y
produces similar effects. The
selfexcitation process requires, as seen in Figure 7.14, the presence of remanent
magnetization (E
rem
≠ 0) and of magnetic saturation to be successful, that is, to
produce a clear intersection of the two curves.
When the magnetization curve V
1m
(I
m
) is available, we may use the
complete equivalent circuit (Figure 7.1) with a parallel capacitor C
Y
over the
terminals to explore the load characteristics. For a given capacitor bank and
speed n, the output voltage V
s
versus load current I
s
depends on the load power
factor (Figure 7.15).
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




I
s

R
s
j L
ω
1sl
R
1m
I
0a
j L (I )
ω
11mm
I
m
R’
S
r
j L’
ω
1rl
( )
ω
1
I’
r
a.c. load
excitation
capacitor
R
l

j L
ω
1l
1
j C
ω
1
l
C
Y
I
l
I
c
I
L
(load current)
S (slip)
capacitor load
resistive load
inductive load
f (frequency)
1
V
S
C =ct.
n=ct.
Y
(load voltage)
S=1-

np
f
<0
l
l

Figure 7.15 Autonomous induction generator on load
a.) equivalent circuit b.) load curves
The load curves in Figure 7.15 may be obtained directly by solving the
equivalent circuit in Figure 7.15a for load current, for given load impedance,
speed n, and slip S. However, the necessary nonlinearity of magnetization curve
L
1m
(I
m
), Figure 7.14, imposes an iterative procedure to solve the equivalent
circuit. This is now at hand with existing application software such Matlab, etc.
Above a certain load, the machine voltage drops gradually to zero as there
will be a deficit of capacitor energy to produce machine magnetization. Point A
on the magnetization curve will drop gradually to zero.
As the load increases, the slip (negative) increases and, for given speed n,
the frequency decreases. So the IG can produce power above a certain level of
magnetic saturation and above a certain speed for given capacitors. The voltage
and frequency decrease notably with load.
A variable capacitor would keep the voltage constant with load. Still, the
frequency, by principle, at constant speed, will decrease with load.
Only simultaneous capacitor and speed control may produce constant
voltage and frequency for variable load. More on autonomous IGs in Chapter
19.
7.9 THE ELECTROMAGNETIC TORQUE

By electromagnetic torque, T
e
, we mean, the torque produced by the
fundamental airgap flux density in interaction with the fundamental rotor
current.
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




In paragraph 7.1, we have already derived the expression of T
e
(7.14) for
the singly fed IM. By singly fed IM, we understand the short-circuited rotor or
the wound rotor with a passive impedance at its terminals. For the general case,
an R
l
L
l
C
l
impedance could be connected to the slip rings of a wound rotor
(Figure 7.16).
Even for this case, the electromagnetic torque T
e
may be calculated (7.14)
where instead of rotor resistance R
r
′, the total series rotor resistance R

r
′ + R
l
′ is
introduced.
Notice that the rotor circuit and thus the R
l
, C
l
, L
l
have been reduced to
primary, becoming R
l
′, C
l
′, L
l
′.
Both rotor and stator circuit blocks in Figure 7.16 are characterized by the
stator frequency ω
1
.
I
s
R
s
j L
ω
1sl

R
1m
I
0a
j L
ω
11m
I
m
R’
S
r
j L’
ω
1rl
I’
r
E
1
V’
S
r
I’
r
R’
S
l
1
S C’
ω

1l
2
j L’
ω
1l

Figure 7.16 Singly – fed IM with additional rotor impedance
Again, Figure 7.16 refers to a fictitious IM at standstill which behaves as
the real machine but delivers active power in a resistance (R
r
′ + R
l
′)(1 – S)/S
dependent on slip, instead of producing mechanical (shaft) power.

'R'R'R ;
p
'I
S
'R3
T
lrre
1
1
2
r
re
e
+=
ω

=
(7.60)
In industry the wound rotor slip rings may be connected through brushes to
a three-phase variable resistance or a diode rectifier, or a d.c. – d.c. static
converter and a single constant resistance, or a semi (or fully) controlled
rectifier and a constant single resistance for starting and (or) limited range speed
control as shown later in this paragraph. In essence, however, such devices have
models that can be reduced to the situation in Figure 7.16. To further explore
the torque, we will consider the rotor with a total resistance R
re
′ but without any
additional inductance and capacitance connected at the brushes (for the wound
rotor).
From Figure 7.16, with V
r
′ = 0 and R
r
′ replaced by R
re
′, we can easily
calculate the rotor and stator currents I
r
′ and I
s
as

m1
rl
re
m11

r
Z'jX
S
'R
ZI
'I
++
⋅
−= (7.61)
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………





m1
rl
re
rl
re
m1
sls
s
s
Z'jX
S
'R
'jX
S

'R
Z
jXR
V
I
++






+
++
= (7.62)

s0rsma0
I'IIII =+=+
(7.63)

ml1ml1
m1m1
m1m1
m1
jXR
jXR
jXR
Z +=
+
=

(7.64)
With constant parameters, we may approximate

()
08.102.1C
X
XX
Z
jXZ
1
ml1
slml1
m1
sl
m1
−≈=
+
≈
+
(7.65)
In this case
()
'XCXj
S
'RC
R
V
I
rl1sl
re1

s
s
s
+++
≈ (7.66)
Substituting I
s
from (7.66) into (7.61) and then I
r
′ from (7.61) into (7.60), T
e

becomes

()
2
rl1sl
2
re1
s
rl
1
1
2
s
e
'XCX
S
'RC
R

S
'R
PV3
T
++






+
ω
=
(7.67)
As we can see, T
e
is a function of slip S. Its peak values are obtained for

()
2
rl1sl
2
s
re1
k
e
'XCXR
'RC
S0

S
T
++
±
=→=
∂
∂
(7.68)
Finally, the peak (breakdown) torque T
ek
is obtained from the critical slip S
k
in
(7.67)

()
()






++±
ω
==
'XCXRRC2
V
P3
TT

rl1sl
2
ss1
2
s
1
1
sk
eek
(7.69)
The whole torque/slip (or speed) curve is shown in Figure 7.17.
Concerning the T
e
versus S curve, a few remarks are in order
• The peak (breakdown) torque T
ek
value is independent of rotor equivalent
(cumulated) resistance, but the critical slip S
k
at which it occurs is
proportional to it.
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar………… ………




• When the stator resistance R
s
is notable (low power, subkW motors), the

generator breakdown torque is slightly larger than the motor breakdown
torque. With R
s
neglected, the two are equal to each other.
T /T
een
T /T
ekm
en
A
A
-S
S
S
n
1
0
+
-
88
n
S
S
+
-
88
nk
1
-1
m

g
generator
motor
brake
rated
torque
0
1
T /T
ekg
en
T /T
es en
f >0
1
k
n=
f
P
(1-S)
1
1

Figure 7.17 The electromagnetic torque T
e
(in relative units) versus slip (speed)
• With R
s
neglected (large power machines above 10 kW) S
k

, from (7.68) and
T
ek
from (7.69), become

()()
sc1
re
rl1sl1
re1
rl1sl
re1
k
L
'R
'LCL
'RC
'XCX
'RC
S
ω
±
≈
+ω
±
≈
+
±
≈
(7.70)

()
sc
2
1
s
1
rl1sl1
2
1
s
1ek
L2
1V
p3
'LCLC2
1V
p3T








ω
≈
+









ω
±≈
(7.71)
• In general, as long as I
s
R
s
/V
s
< 0.05, we may safely approximate the
breakdown torque to Equation (7.71).
• The critical slip speed in (7.70) S
k
ω
1
= ± R
re
′/L
sc
is dependent on rotor
resistance (acumulated) and on the total leakage (short-circuit) inductance.
Care must be exercised to calculate L
sl

and L
rl
′ for the actual conditions at
breakdown torque, where skin and leakage saturation effects are notable.
• The breakdown torque in (7.71) is proportional to voltage per frequency
squared and inversely proportional to equivalent leakage inductance L
se
.
Notice that (7.70) and (7.71) are not valid for low values of voltage and
frequency when V
s
≤ 0.05I
s
R
s
.
• When designing an IM for high breakdown torque, a low short-circuit
inductance configuration is needed.
• In (7.70) and (7.71), the final form, C
1
= 1, which, in fact, means that L
1m
≈
∞ or the iron is not saturated. For deeply saturated IMs, C
1
≠ 1 and the first
form of T
ek
in (7.71) is to be used.
© 2002 by CRC Press LLC