modeling of induction motor using dq0 transformations first semester 1431/1432
TRANSCRIPT
Modeling of Induction Motor using dq0 Transformations
First Semester 1431/1432
Steady state model developed in previous studies of induction motor neglects electrical transients due to load changes and stator frequency variations. Such variations arise in applications involving variable-speed drives.
Variable-speed drives are converter-fed from finite sources, which unlike the utility supply, are limited by switch ratings and filter sizes, i.e. they cannot supply large transient power.
Introduction
Thus, we need to evaluate dynamics of converter-fed variable-speed drives to assess the adequacy of the converter switches and the converters for a given motor and their interaction to determine the excursions of currents and torque in the converter and motor. Thus, the dynamic model considers the instantaneous effects of varying voltages/currents, stator frequency and torque disturbance.
Introduction (cont’d)
Circuit Model of a Three-Phase IM
1. Space mmf and flux waves are considered to be sinusoidally distributed, thereby neglecting the effect of teeth and slots.
2. The machine is regarded as group of linear coupled circuits, permitting superposition to be applied, while neglecting saturation, hysteresis, and eddy currents.
3. Ls : self inductance per phase of the stator windings.4. Ms: mutual inductance per phase of the stator windings.5. rs: resistance per phase of the stator windings.6. Lr : self inductance per phase of the rotor windings.7. Mr: mutual inductance per phase of the rotor windings8. rr: resistance per phase of the rotor windings.9. Msr: maximum value of mutual inductance between any
stator phase and any rotor phase.
Assumptions and Definitions:
ias
i bs
ics
iar
i br
icrvas+
+v
bs
+v cs
v ar+
+vbr
vcr +
axis of stator phase c
axis of stator phase a
axis of stator phase b
axis of rotor phase c
axis of rotor phase b
axis of rotor phase a
r
r
r
Circuit Model of a Three-Phase IM
r
Stator Voltage Equations:
asas as s
dv i r
dt
bsbs bs s
dv i r
dt
cscs cs s
dv i r
dt
Voltage Equations
ias
i bs
ics
iar
i br
icrvas+
+v
bs
+v cs
v ar+
+vbr
vcr +
axis of stator phase c
axis of stator phase a
axis of stator phase b
axis of rotor phase c
axis of rotor phase b
axis of rotor phase a
r
r
r
r
Rotor Voltage Equations:
a ra r a r r
dv i r
dt
brbr br r
dv i r
dt
crcr cr r
dv i r
dt
Voltage Equations (cont’d)
ias
i bs
ics
iar
i br
icrvas+
+v
bs
+v cs
v ar+
+vbr
vcr +
axis of stator phase c
axis of stator phase a
axis of stator phase b
axis of rotor phase c
axis of rotor phase b
axis of rotor phase a
r
r
r
r
cos( ) cos( 120 ) cos( 120 )sr sr sar r br r c
as s as s bs s c
r r r
s
M M
L i M i M i
i i M i
cos( 120 ) cos( ) cos( 120 )ar r
bs s as s bs s
br r csr s r rr s
cs
rM M
M i L i M i
Mi i i
Flux Linkage Equations
ias
i bs
ics
iar
i br
icrvas+
+v
bs
+v cs
v ar+
+vbr
vcr +
axis of stator phase c
axis of stator phase a
axis of stator phase b
axis of rotor phase c
axis of rotor phase b
axis of rotor phase a
r
r
r
r
(
cos( ) cos( 120 ) cos 20 )
)
( 1sr sar r br r cr
as s as s bs cs
rr srM M
L i M
i i M
i i
i
cos( 120 ) cos( ) c
( )
os( 120 )
bs s as cs s b
sr sr sar r br r cr r
s
r
M i
M M Mi i
i L i
i
Flux Linkage Equations
0as bs csi i i
In general, we can assume:ias
i bs
ics
iar
i br
icrvas+
+v
bs
+v cs
v ar+
+vbr
vcr +
axis of stator phase c
axis of stator phase a
axis of stator phase b
axis of rotor phase c
axis of rotor phase b
axis of rotor phase a
r
r
r
r
ss s sL L M
Let:
cos( ) cos( 120 ) cos( 120 )
( )ar r r r
r ar r br c
as bs cs
r
sr sr sr
L i M
i i M
i i
M M i
Flux Linkage Equations
0ar br cri i i
In general, we can assume:
ias
i bs
ics
iar
i br
icrvas+
+v
bs
+v cs
v ar+
+vbr
vcr +
axis of stator phase c
axis of stator phase a
axis of stator phase b
axis of rotor phase c
axis of rotor phase b
axis of rotor phase a
r
r
r
r
cos( 120 ) cos( ) cos( 120 )
( )assr sr sbr r r r
r b
bs cs
r c
r
r ar r
M M M
L i M i
i i i
i
rr r rL L M
Let:
Flux Linkage Equations
cos( 120 ) cos( 120 ) cos( )cs ar r bss c r r crs sr ri iM iL i
cos( ) cos( 120 ) cos( 120 )as ss sr ar r br r ra crsL i i i iM
cos( 120 ) cos( ) cos( 120 )bs ss bs ar r br r r rsr ci iL M ii
Stator:
Rotor:
cos( ) cos( 120 ) cos( 120 )as bs cssar r r r rrr ariM i ii L
cos( 120 ) cos( ) cos( 120 )as bs cbr r r rs rr brsr i i i L iM
cos( 120 ) cos( 120 ) cos( )ascr r r r rrbs s cr rcs i i i L iM
0 0
0
0 0
0
0
0
0
0
0 0
ar arrr
ar brrr
ar crr
sr
Ts
asssas
bsssb
r
s
ccs
r
sss
iL
iL
i
iL
L
L
L
iL
iL
cos( ) cos( 120 ) cos( 120 )
cos( 120 ) cos( ) cos( 120 )
cos( 120 ) cos( 120 ) cos( )
r r r
r rsr r
r
s
r r
rL M
Flux Linkage Equations
asas as s
dv i r
dt
To build up our simulation equations, we could just differentiate each expression for , e.g.
But since Lsr depends on position ,
which will generally be a function of time, the trigonometric terms will lead to a mess!
First raw of the Matrixasas as s
d dv i r
dt dt
Model of Induction Motor
The Park’s transformation is a three-phase to two-phase transformation for synchronous machine analysis. It is used to transform the stator variables of a synchronous machine onto a dq reference frame that is fixed to the rotor.
The +ve d-axis is aligned with the magnetic axis of the field winding and the +ve q-axis is defined as leading the +ve d-axis by /2.
Park’s Transformation
Park’s Transformation (cont’d)
ias
i bs
ics
iar
i br
icrvas+
+v
bs
+v cs
v ar+
+vbr
vcr +
axis of stator phase c
axis of stator phase a
axis of stator phase b
axis of rotor phase c
axis of rotor phase b
axis of rotor phase a
r
r
r
r
d-axis
q-axis
The result of this transformation is that all time-varying inductances in the voltage equations of an induction machine due to electric circuits in relative motion can be eliminated.
In induction machine, the d-axis is assumed to align on a-axis at t = 0 and rotate with synchronous speed ()
The Park’s transformation equation is of the form:
where f can be i, v, or .
0
0
f f
f T f
f f
d a
q dq b
c
Park’s Transformation (cont’d)
0
2 2cos cos cos
3 3
2 2( ) sin sin sin
3 3
1 1 1
2 2 2
T
d d d
dq d d d dK
Park’s Transformation (cont’d)
where K is a convenient constant. The current id and iq are proportional to the components of mmf in the direct and quadrature axes, respectively, produced by the resultant of all three armature currents, ia, ib, and ic. For balanced phase currents of a given maximum magnitude, the maximum value of id and iq can be of the same magnitude. Under balanced conditions, the maximum magnitude of any one of the phase currents is given by . To achieve this relationship, a value of 2/3 is assigned to the constant K.
2 2, , ,a peak b peak c peak d qi i i i i
The inverse transform is given by:
Of course, [T][T]-1=[I]
1
0
cos sin 1
2 2( ) cos sin 1
3 3
2 2cos sin 1
3 3
T
d d
dq d d d
d d
Park’s Transformation (cont’d)
Thus,
and
0 0
0
d a
q dq b dq abc
c
v v
v T v T v
v v
0 0
0
d a
q dq b dq abc
c
i i
i T i T i
i i
Park’s Transformation (cont’d)
ids
vds+
iqr +vqr
idr
vdr+
iqs +vqs
Induction Motor Model in dq0
d-axis
q-axis
Lets us define new “dq0” variables.
Our induction motor has two subsystems - the rotor and the stator - to transform to our orthogonal coordinates:
So, on the stator,
where [Ts]= [T()], ( = t)
and on the rotor,
where [Tr]= [T()], ( = - r = ( r) t )
0dq s s abcsT
0 [ ]dq r r abcrT
Induction Motor Model in dq0 (cont’d)
Induction Motor Model in dq0 (cont’d)
0
0 0
1
10
"abc": [ ]
"dq0": [ ]
[ ]
STATOR:
[ ]
[ ]
[ ]
abcs ss abcs
dq s s abcs ss s abcs s
dq s ss dq s s
sr
sr
sr
abcr
r r abcr
r dq r
L i
T L T i
L
L
L
T
L
i
T i
i TT
T
i
1
1
0
0 00
"abc": [ ]
"dq0": [ ]
ROTOR:
[ ]
[ ]
[ ][ ]
Tsr
T
abcs
s s abcs
s dq
abcr rr abcr
dq r r abcr r rr r abcr
dq s r rr dq rs
sr
Tsr
L iL
L
i
T T i
TL
T T L T i
T i L i
1 0 0
0 1 0
0 0 1ss ssL L
1 0 0
0 1 0
0 0 1rr rrL L
Now:
Just constants!!
Our double reference frame transformation eliminates the trigonometric terms found in our original equations.
1 1
30 0
23
0 02
0 0 0
sr
Tsr sr ss s rr rT
M
T L MT T T
Induction Motor Model in dq0 (cont’d)
Let us look at our new dq0 constitutive law and work out simulation equations.
0dq s s abcs s abcs s abcs
dv T v T R i T
dt
1 10 0s s dq s s s dq s
dT RT i T T
dt
Induction Motor Model in dq0 (cont’d)
10 0dq s s s dq s
dR i T T
dt
Using the differentiation product rule:
0 0 0
0 0
0 0
0 0 0
dq s dq s dq s
d
dtd d
R idt dt
Induction Motor Model in dq0 (cont’d)
10 0 0 0dq s dq s dq s s s dq s
d dv R i T T
dt dt
For the stator this matrix is:
For the rotor the terminal equation is
essentially identical but the matrix is:
0 0
0 0
0 0 0
0 ( ) 0
( ) 0 0
0 0 0
r
r
Induction Motor Model in dq0 (cont’d)
Simulation model; Stator Equations:
dsds ds s qs
dv i r
dt
qsqs qs s ds
dv i r
dt
00 0
ss s s
dv i r
dt
Induction Motor Model in dq0 (cont’d)
Simulation model; Rotor Equations:
( ) drdr dr r r qr
dv i r
dt
( ) qrqr qr r r dr
dv i r
dt
00 0
rr r r
dv i r
dt
Induction Motor Model in dq0 (cont’d)
Zero-sequence equations (v0s and v0r) may be ignored for balanced operation.
For a squirrel cage rotor machine,
vdr= vqr= 0.
Induction Motor Model in dq0 (cont’d)
We can also write down the flux linkages:
0
0
0
0
0
0
0 0 3 2 0 0
0 0 0 3 2 0
0 0 0 0 0
3 2 0 0 0 0
0 3 2 0 0 0
0 0 0 0 0
dr rr dr
qr rr qr
r r
ds ss ds
qs ss q
sr
sr
sr
sr
s
r r
s
s s s
L i
L
ML i
M
M
M i
i
L
L i
L
i
Induction Motor Model in dq0 (cont’d)
The torque of the motor in qd0 frame is given by:
where P= # of poles
F=ma, so:
where = load torque
3
2 2 qr dr d re r qi iP
( )re l
dJ
dt
l
Induction Motor Model in dq0 (cont’d)