(1/k)ωrλ
M
LlM
L
dr
lr
+
+
-
iqs
iqr
Vqs
rr
R
L
qfe
mq
iqfe
-
r
kωrλ
A
LlA
L
qr
lR
+
ids
idr
- +
r
V
R
ds
R
L
dfe
md
idfe
-
Fig. 1. The d-q axes two-phase induction motor Equivalent circuit with iron losses.
The machine model may be expressed by the following voltage and flux equations:
Voltage Equations (1):
=
+
q
v s
m
r i qs pλ qs (1)
v =
+
ds
a
r i ds pλ ds (2)
0 = r i
− (1 / k) * ω λ + λ
r qr
r
dr
p qr (3)
0 = r i + k * ω λ + λ
R ds
r qr
p dr (4)
0 = − i
R
+ L ( p i + p i − p i
)
qfe
qfe
mq
qs
qr
qfe
(5)
0 = − i
R
+ L
( p i + p i − p i
)
dfe
dfe
md
ds
dr
dfe (6)
Flux Equations:
λ = L i
+ L
( i + i
− i
)
qs
lm qs
mq
qs
qr
qfe
(7)
λ
= L i
+ L
( i + i
− i
)
ds
la ds
md
ds
dr
dfe
(8)
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Electric Machines and Drives
λ
= L i
+ L
( i + i
− i
)
qr
lr qr
mq
qs
qr
qfe
(9)
λ
= L i
+ L
( i + i
− i
)
dr
lR dr
md
ds
dr
dfe
(10)
Electrical torque equation is expressed as:
P
1
Te =
( k L i ( i + i − i ) − L
i ( i + i − i ) (11)
2
mq dr qs
qr
qfe
md qr
ds
dr
qfe
k
Te − T =
ω +
ω
l
j m p r B m r (12)
2.2 Field-Oriented Controller [FOC]
The stator windings of the motor are unbalanced. The machine parameters differ from the d
axis to the q axis. The waveform of the electromagnetic torque demonstrates the unbalance
of the system. The torque in equation (11) contains an AC term; it can be observed that two
values are presented for the referred magnetizing inductance. It is possible to eliminate the
AC term of electro-magnetic torque by an appropriate control of the stator currents.
However, these relations are valid only in linear conditions. Furthermore, the model is
implemented using a non-referred equivalent circuit, which presumes some complicated
measurement of the magnetizing mutual inductance of the stator and the rotor circuits.
The indirect field-oriented control scheme is the most popular scheme for field-oriented
controllers. It provides decoupling between the torque and the flux currents. The electric
torque must be a function of the stator currents and rotor flux in synchronous reference
frame [Popescu & Navrapescu, 2000]. Assuming that the stator currents can be imposed as:
s
s
i =
(13)
ds
ids 1
s
s
i
=
(14)
qs
k i qs 1
Where: k= Msrd / Msrq
P
=
⎡
s
s
s
s
λ
−
λ ⎤ (15)
e
T
M i
M i
2
sqr qs
⎣
dr
sdr ds
qr ⎦
r
L
By substituting the variables ids, and iqs by auxiliary variables ids1, and iqs1 into (15) the torque can be expressed by
P M
sdr
=
⎡ s
s
s
s
λ
−
λ ⎤
e
T
i
i
⎣
1
1
2
qs
dr
ds
qr ⎦ (16)
r
L
In synchronous reference frame, the electromagnetic torque is expressed as:
P M
sdr
=
⎡ e
e
e
e
λ
−
λ ⎤
e
T
i
i
⎣
1
1
2
qs
dr
ds
qr ⎦ (17)
r
L
P M
sdr
=
⎡ e
e
λ ⎤
e
T
i
⎣
1
2
qs
r ⎦ (18)
r
L
Swarm Intelligence Based Controller for
Electric Machines and Hybrid Electric Vehicles Applications
165
e
λ
e
r
i
=
ds 1
(19)
Msdr
M
sdr
e
ω −ω =
e
r
i
(20)
1
τ *
qs
λ
r
r
2.3 Synchronous reference frame for losses model
It is very complex to find the losses expression for the two asymmetrical windings induction
motor with losses model. In this section, a simplified induction motor model with iron
losses will be developed. For this purpose, it is necessary to transform all machine variables
to the synchronous reference frame. The voltage equations are written in expanded form as
follows [Hegazy, 2006; Amin et al., 2006; Amin et al., 2009]:
e
e
di
di
e
e
qs
qm
v = r i + L
+ L
+ ω (
e
e
L i + L i )
qs
m qs
lm
mq
e
la ds
md dm (21)
dt
dt
e
e
di
di
e
e
ds
dm
v = r i + L
+ L
− ω (
e
e
L i + L i )
ds
a ds
la
md
e
lm qs
mq qm (22)
dt
dt
e
e
di
di
ω
0
e
qr
qm
sl
= r i + L
+ L
+
(
e
e
L i + L i )
r qr
lr
mq
lR dr
md dm (23)
dt
dt
k
e
e
di
di
0
e
dr
dm
= r i + L
+ L
− k * ω (
e
e
L i + L i )
R dr
lR
md
sl
lr qr
mq qm (24)
dt
dt
e
e
e
e
+
=
+
q
i s iqr iqfe iqm (25)
e
e
e
e
i +
=
+
ds
idr idfe idm (26)
ω L L
e
e
lr mqs e
v
= −
dm
iqs (27)
r
L
e
e
= ω
q
v m
e
m
L dsids (28)
Where:
e
e
v
e
qm
e
vdm
i
=
=
dfe
; idfe
Rqfe
Rdfe
The losses in the motor are mainly:
a. Stator copper losses,
b. Rotor copper losses,
c. Core losses, and
166
Electric Machines and Drives
d. Friction losses
The total electrical losses can be expressed as follows
Plosses = Pcu1 + Pcu2 +Pcore
(29)
Where:
Pcu1 : Stator copper losses
Pcu2 : Rotor copper losses
Pcore : Core losses
The stator copper losses of the two asymmetrical windings induction motor are caused by
electric currents flowing through the stator windings. The core losses of the motor are
produced from the hysteresis and eddy currents in the stator. The total electrical losses of
motor can be rewritten as:
2
2
2
2
2
2
e
e
v
v
e
e
e
e
qm
dm
=
+
+
+
+
+
l
Posses
m
r qs
i
a
r ids
r
r iqr
R
r idr
(30)
qf
R e
d
R fe
The total electrical losses are obtained as follows:
⎛
⎞
2
2 2 2
⎜
⎟
⎡
⎤
2 2
⎛
2 2
⎞ 2
r L
ω L L
⎜
T L
⎟
ω L
λ
r mqs
e
lr mqs
e r
e mds
r
= ⎢ +
+
⎥
+ ⎜ +
⎟
lo
P sses
m
r
r
(31)
2
2
⎜
2
a
⎟ ⎜
⎟ 2
⎢
L
L R
⎥
⎛ L
⎞
R
⎣
r
r dfe
⎦
2
L
mds
2
⎜
⎟ ⎝
qfe
⎠ mds
P
λ
⎜
⎜
⎟ r
⎝
K
⎟
⎝
⎠
⎠
2 T * r
e
r
ω =
sl
(32)
2
P * λ r
Where:
ωe = ωr+ ωsl, and ωsl is the slip speed (rad/sec).
Equation (31) is the electrical losses formula, which depends on rotor flux (λr) according to
operating point (speed and load torque). The total losses of the motor ( TP losses
) are given
as follows:
TP
=
+
=
losses
l
P osses F
P ric i
P n - o
P ut (33)
P
Efficiency (η
)
out
=
(34)
P + TP
out
losses
Where:
Friction power losses = F ∗ωr2, and
Output power ( Pout) = TL∗ωr.
2.4 Losses minimization control scheme
The equation (33) is the cost function, the total losses, which depends on rotor flux (λr)
according to the operating point. Figure 2 presents the distribution of losses in motor and its
variation with