Transcript Microelectromechanical Devices

ECE 8830 - Electric Drives
Topic 4: Modeling of Induction Motor using
qd0 Transformations
Spring 2004
Introduction
Steady state model developed in previous
topic 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 (cont’d)
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.
Circuit Model of a Three-Phase Induction
Machine (State-Space Approach)
Voltage Equations
Stator Voltage Equations:
d as
vas  ias rs 
dt
d bs
vbs  ibs rs 
dt
d cs
vcs  ics rs 
dt
Voltage Equations (cont’d)
Rotor Voltage Equations:
d ar
var  iar rr 
dt
d br
vbr  ibr rr 
dt
d cr
vcr  icr rr 
dt
Flux Linkage Equations
Model of Induction Motor
To build up our simulation equation, we
could just differentiate each expression
for , e.g.
d as d [first row of matrix]
vas 

dt
dt
But since Lsr depends on position,
which will generally be a function of
time, the trig. terms will lead to a mess!
Park’s transform to the rescue!
Park’s Transformation
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 q-axis is aligned with the magnetic
axis of the field winding and the +ve d-axis
is defined as leading the +ve q-axis by /2.
(see Fig. 5.16c Ong on next slide).
Park’s Transformation (cont’d)
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.
Park’s Transformation (cont’d)
The Park’s transformation equation is of
the form:
fq 
f a 
f    T   f 
 d   qd 0   b 
 f 0 
 f c 
where f can be i, v, or .
Park’s Transformation (cont’d)

 cos  q

2
 Tqd 0 ( q )     sin  q
3
 1

 2
2 
2  


cos  q 
 cos  q 

3
3




2 
2  


 sin  q 
  sin  q 

3
3





1
1

2
2

Park’s Transformation (cont’d)
The inverse transform is given by:


cos  q

1
 
2
Tqd 0 ( q )    cos  q 
3
 
 
2
cos  q 
3
 
 sin  q
2


  sin  q 
3


2


  sin  q 
3


Of course, [T][T]-1=[I]







1


1


1

Park’s Transformation (cont’d)
Thus,
 vq 
 va 
 v   T   v 
 d   qd 0   b 
 v0 
 vc 
and
 iq 
ia 
i   T  i 
 d   qd 0   b 
 i0 
 ic 
Induction Motor Model in qd0
Acknowledgement:
The following notes covering the induction
motor modeling in qd0 space are mostly
courtesy of Dr. Steven Leeb of MIT.
Induction Motor Model in qd0 (cont’d)
This transform lets us define new “qd0”
variables.
Our induction motor has two subsystems the rotor and the stator - to transform to
our orthogonal coordinates:
So,
λ qd 0  Ts  λ abc
on the stator,
where [Ts]= [T()],
and
λ dq 0r  [Tr ]λ abcr
( to be defined)
on the rotor,
where [Tr]= [T()], ( to be defined)
Induction Motor Model in qd0 (cont’d)
STATOR:
“abc”:
abcs = Ls iabcs + Lsr iabcr
“qd0”: qd0s= Ts abcs= Ts Ls Ts-1 iqd0s +Ts Lsr Ts-1 iqd0r
ROTOR:
qd0r= Tr abcr= Tr LsrT Ts-1 iqd0s +Tr Lr Tr-1 iqd0r
Induction Motor Model in qd0 (cont’d)
After some algebra, we find:
 Lar
Tr LrTr1   0
 0
and similarly for
0
Lar
0
0 
0  where Lar= Lr-Lab
Lar 
1
s s .
Ts L T
But what about the cross terms? They
depend on the choice of  and .
Let  =  - r , where r is the rotor position.
Induction Motor Model in qd0 (cont’d)
Now:
 3 Lm
 2
Tr LTsrTs1  Ts LsrTr1   0

 0

0
3 L
2 m
0
0

0

0

Just constants!!
Our double reference frame transformation
eliminates the trig. terms found in our
original equations.
Induction Motor Model in qd0 (cont’d)
We know what  and r must be to make
the transformation work but we still have
not determined what to set  to. We’ll
come back to this but let us first look at
our new qd0 constitutive law and work
out simulation equations.
vqd 0 s

d
 T s vabcs  T s Riabcs  T s
abcs
dt
d 1
1
 T s RT s iqd 0 s  T s T s qd 0 s
dt
d 1
 Riqd 0 s  T s T s qd 0 s
dt

Induction Motor Model in qd0 (cont’d)
Using the differentiation product rule:
vqd 0 s


d
 d 1 
 Riqd 0 s 
qd 0 s  T s T s  qd 0 s
dt
 dt


d
 Riqd 0 s 
qd 0 s
dt


 0

d


 dt

 0


d
dt
0
0

0

0  qd 0 s


0

Induction Motor Model in qd0 (cont’d)
For the stator this matrix is:
0


 0

0
0
0
0 
0 
For the rotor the terminal equation is
essentially identical but the matrix is:
0
(   r ) 0 

(   )

0
0
r



0
0
0 
Induction Motor Model in qd0 (cont’d)
Simulation model; Stator Equations:
vqs  iqs rs  ds 
d qs
dt
d ds
vds  ids rs  qs 
dt
d 0 s
v0 s  i0 s rs 
dt
Induction Motor Model in qd0 (cont’d)
Simulation model; Rotor Equations:
vqr  iqr rr  (   r )dr 
d qr
dt
d dr
vdr  idr rr  (   r )qr 
dt
d 0 r
v0 r  i0 r rr 
dt
Induction Motor Model in qd0 (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 qd0 (cont’d)
We can also write down the flux linkages:
 qs   Las
   0
 ds  
 0 s   0
 
 qr  3 2 Lsr
 dr   0
  
0 r   0
0
Las
0
0
3 2 Lsr
0
0
3 2 Lsr
0
Las 0
0
0
0
3 2 Lsr
0
0
Lar
0
0
Lar
0
0
0
0
0   iqs 
0   ids 
0  i0 s 
 
0  iqr 
0  idr 
 
Lar 0  i0 r 
Induction Motor Model in qd0 (cont’d)
How do we pick ?
One good choice is:
d
 e
dt
where e is synchronous frequency.
Remember that this choice makes a
balanced 3 voltage set applied to the
stator look like a constant.
Induction Motor Model in qd0 (cont’d)
The torque of the motor in qd0 space is
given by:
3 P
 m     qr idr  dr iqr 
2 2 
where P= # of poles
F=ma, so:
d r
J
 ( m   l )
dt
where
l
= load torque
Induction Motor Model in qd0 (cont’d)
Example: The equations for a balanced 3,
squirrel cage, 2-pole rotor induction motor:
Constitutive Laws:
3
 m  qr idr  dr iqr
2

 qs   Las
   0
 ds   
qr  3 2 Lsr
  
dr   0
0
3 2 Lsr
Las
0
0
Lar
3 2 Lsr
0

 iqs 
3 2 Lsr  ids 
0  iqr 
 
Lar  idr 
0
Induction Motor Model in qd0 (cont’d)
State equations:
d
 qs  rs iqs  ds  vqs
dt
d
 ds  rs ids  qs  vds
dt
d
 qr  rr iqr  (   r )dr
dt
d
 dr  rr idr  (   r )qr
dt
d r ( m   l )
dt

J
r= rotor speed
= frame speed
J= shaft inertia
l = load torque
qd0 Induction Motor Model in
Stationary Reference Frame
The qd0 induction motor model in the
stationary reference frame can be obtained
by setting =0. This model is known as the
Stanley model and the equivalent circuits
are given on the next slide.
qd0 Induction Motor Model in
Stationary Reference Frame (cont’d)
qd0 Induction Motor Model in
Stationary Reference Frame (cont’d)
Stator and Rotor Voltage Equations:
vqs
vds
vqr
vdr
d
 rs iqs  qs
dt
d
 rs ids  ds
dt
d
 rr iqr  qr   r dr
dt
d
 rr idr  dr   r qr
dt
d
0 s
dt
d
v0 r  rr i0 r  0 r
dt
v0 s  rs i0 s 
qd0 Induction Motor Model in
Stationary Reference Frame (cont’d)
Flux Linkage Equations:
 qs   xls  xm
   0
 ds  
 0 s   0
 
 qr   xm
 dr   0
  
0 r   0
0
xls  xm
0
0
xm
0
0
xm
0
xls
0
0
0
xm
0
0
xlr  xm
0
0
xlr  xm
0
0
0
0
0   iqs 
0   ids 
0  i0 s 
 
0  iqr 
0  idr 
 
xlr  i0 r 
qd0 Induction Motor Model in
Stationary Reference Frame (cont’d)
Torque Equation:
3P
Tem 
(qr idr  dr iqr )
22
3P

(ds iqs  qs ids )
22
3P

xm (idr iqs  iqr ids )
22
Induction Motor Model in qd0 Example
Example 5.3 Krishnan
qd0 Induction Motor Model in
Synchronous Reference Frame
The qd0 induction motor model in the
synchronous reference frame can be
obtained by setting = e . This model
is known as the Kron model and the
equivalent circuits are given on the
next slide.
qd0 Induction Motor Model in
Synchronous Reference Frame (cont’d)
qd0 Induction Motor Model in
Synchronous Reference Frame (cont’d)
Stator and Rotor Voltage Equations:
d qs
vqs  iqs rs  e ds 
dt
d ds
vds  ids rs   e qs 
dt
d 0 s
v0 s  i0 s rs 
dt
vqr  iqr rr  (e   r )dr 
d qr
dt
d dr
vdr  idr rr  ( e   r )qr 
dt
d 0 r
v0 r  i0 r rr 
dt
qd0 Induction Motor Model in
Synchronous Reference Frame (cont’d)
Flux Linkage Equations:
 qs   xls  xm
   0
 ds  
 0 s   0
 
 qr   xm
 dr   0
  
0 r   0
0
xls  xm
0
0
xm
0
0
xm
0
xls
0
0
0
xm
0
0
xlr  xm
0
0
xlr  xm
0
0
0
0
0   iqs 
0   ids 
0  i0 s 
 
0  iqr 
0  idr 
 
xlr  i0 r 
qd0 Induction Motor Model in
Synchronous Reference Frame (cont’d)
Torque Equation:
3P
Tem 
(qr idr  dr iqr )
22
3P

(ds iqs  qs ids )
22
3P

xm (idr iqs  iqr ids )
22
Induction Motor Model in Synchronous
Reference Frame Example
Example 5.5 Krishnan
Steady State Model of Induction
Motor
The stator voltages and currents for an
induction machine at steady state with
balanced 3 phase operation are given by:
vas  Vms cos(et )
ias  I ms cos(et  s )
2
vbs  Vms cos( et  )
3
4
vcs  Vms cos( et  )
3
2
ibs  I ms cos( et 
 s )
3
4
ics  I ms cos( et 
 s )
3
Steady State Model of Induction
Motor (cont’d)
Similarly, the rotor voltages and currents with
the rotor rotating at a slip s are given by:
var  Vmr cos(set  r (0)   )
vbr  Vmr cos( s et 
iar  I mr cos(set  r (0)    r )
2
  r (0)   ) ibr  I mr cos( s et  2   r (0)    r )
3
3
4
4
i

I
cos(
s

t

  r (0)    r )
vcr  Vmr cos( s et 
  r (0)   ) cr mr
e
3
3
Steady State Model of Induction
Motor (cont’d)
Transforming these stator and rotor abc
variables to the qd0 reference with the q-axis
aligned with the a-axis of the stator gives:
v s  v  jv  Vms e
s
qs
s
ds
jet
 j
i s  i  ji  I ms e e
s
qs
v r  (v  jv )e
r
qr
r
dr
i r  (i  ji )e
r
qr
r
dr
s
ds
j r ( t )
j r ( t )
 (Vmr e
 ( I mr e
jet
j ( set  r (0)  )
j ( set  r (0)  )
)e
j r ( t )
)e jr (t )
where s and r= qd0 components in stationary
frame and rotating ref. frames, respectively.
Steady State Model of Induction
Motor (cont’d)
In steady state operation with the rotor
rotating at a constant speed of e(1-s),
 r (t )  e (1  s)t   r (0)
This equation can be used to simplify the
rotor voltage and current space vectors
which become:
vr  v  jv  Vmr e
s
qr
s
dr
i r  i  ji  I mr e
s
qr
s
dr
 j
e
jet
 j ( r )
e
jet
Steady State Model of Induction
Motor (cont’d)
Use phasors to perform steady state analysis.
Notation:
A - rms values of space vectors
B - rms time phasors
Thus,
V as 
Vms
V ar 
Vmr
2
2
e
j0
e j
I ms  js
I as 
e
2
I mr  j ( r )
I ar 
e
2
Steady State Model of Induction
Motor (cont’d)
and
s
qs
s
ds
V  jV 
s
qs
s
ds
I  jI 
s
qr
s
dr
s
dr
I  jI 
2
iqss  jidss
2
V  jV 
s
qr
vqss  jvdss
 I as e jet
vqrs  jvdrs
2
iqrs  jidrs
2
 V as e jet
 V ar e jet
 I ar e jet
Steady State Model of Induction
Motor (cont’d)
Referring the rotor voltages and currents
to the stator side gives:
 Ns
V  jV  
 Nr
's
qr
's
dr
 Nr
I  jI  
 Ns
's
qr
's
dr
'

jet
jet
V
e

V
e
ar
ar


'

jet
jet
I
e

I
e
ar
ar


where the primed quantities indicate rotor
quantities referred to the stator side.
Steady State Model of Induction
Motor (cont’d)
In the stationary reference frame, the
qd0 voltage and flux linkage equations
can be rewritten in terms of the complex
rms space voltage vectors as follows:
s
qs
s
ds
s
qs
s
ds
s
s
V  j V  [rs  j e ( Lls  Lm )](I  j I )  je Lm (I 'qr
 j I 'dr
)
s
qr
s
qs
s
ds
V '  j V '  j (e   r ) Lm (I  j I )
s
dr
s
s
[rr'  j ( e   r )( L'lr  Lm )(I 'qr
 j I 'dr
)
Steady State Model of Induction
Motor (cont’d)
Using the relationships between the rms
space vectors and rms time phasors
provided earlier, and re-writing (e-r) by
se, and dropping the common ejt term,
we get:
Vas  (rs  je Lls )Ias  je Lm (Ias  I 'ar )
V 'ar  (r 'r  jse L 'lr )I 'ar  jse Lm (Ias  I 'ar )
r 'r
V 'ar
 (  je L 'lr )I 'ar  je Lm (I as  I 'ar )
s =>
s
s
Steady State Model of Induction
Motor (cont’d)
The relations on the previous slide can be
rewritten as:
e
e
V as  (rs  j
xls )I as  j
xm (I as  I 'ar )
b
b
e
e
r 'r
V 'ar
(  j
x 'lr )I 'ar  j
xm (I as  I 'ar )
s
s
b
b
where b is the base or rated angular freq.
given by b  2 f rated where frated =rated
frequency in Hz of the machine.
Steady State Model of Induction
Motor (cont’d)
A phasor diagram of the stator and rotor
variables with Im  Ias  I ' is shown below
ar
together with an equivalent circuit diagram.
Steady State Model of Induction
Motor (cont’d)
By adding and subtracting rr’ and
regrouping terms, we get the alternative
equivalent circuit representation shown
below:
e
Steady State Model of Induction
Motor (cont’d)
The rr’ (1-s)/s resistance term is
associated with the mechanical power
developed.
The rr’/s resistance term is associated
with the power through the air gap.
Steady State Model of Induction
Motor (cont’d)
If our main interest is on the torque
developed, the stator side can be
replaced by the Thevenin equivalent
circuit shown below:
Steady State Model of Induction
Motor (cont’d)
In steady state:
The average power developed is given by:
 1 s  '
Pem  3I 
 rr
 s 
'2
ar
The average torque developed is given by:
Tem 
Pmech
rm
3I r
(1  s)
 3I r

ssm (1  s) ssm
'2 '
ar r
'2 '
ar r
Steady State Model of Induction
Motor (cont’d)
The operating characteristics are quite
different if the induction motor is
operated at constant voltage or constant
current.
Constant voltage -> stator series
impedance drop is small => airgap
voltage close to supply voltage over wide
range of loading.
Constant current -> terminal and airgap
voltage could vary significantly.
Steady State Model of Induction
Motor- Constant Voltage Supply
Shorting the rotor windings and
operating the stator windings with a
constant voltage supply leads to the
below Thevenin equivalent circuit.
Steady State Model of Induction
Motor- Constant Voltage Supply
The Thevenin circuit parameters are:
jxm
V th 
V as
rs  j ( xls  xm )
jxm (rs  jxls )
Zth  rth  jxth 
rs  j ( xls  xm )
Steady State Model of Induction
Motor- Constant Voltage Supply
The average torque developed for a P-pole
machine with constant voltage supply is
given by:
Vth2 (rr' / s)
3P
Tem 
2e (rth  rr' / s)2  ( xth  xlr' )2
We can use this equation to generate the
torque-slip characteristics of an induction
motor driven by constant voltage supply.
Steady State Model of Induction
Motor- Stator Input Impedance
The stator input impedance is given by:
jxm (rr' / s  jxlr' )
Zin  rs  jxls  '
rr / s  j ( xlr'  xm )
The stator input current and complex
power are given by:
V as
I as 
Zin
*
as as
S in  Pin  jQin  3V Ι
Steady State Model of Induction
Motor- Constant Current Supply
With a constant current supply, the stator
current is held fixed and the stator voltage
varies with the input impedance given on
the previous slide.
The rotor current Iar’ can be used to
determine the torque and is given by:
2 2
m as
'
lr
x I
I  ' 2
2
(rr / s)  ( x  xm )
'2
ar
Comparison of Constant Voltage
vs. Constant Current Operation
Consider a 20 hp, 60Hz, 220V 3 induction
motor with the following equivalent circuit
parameters:
rs = 0.1062
rr’ = 0.0764
xm = 5.834 
xls = 0.2145 
xlr’ = 0.2145 
Jrotor= 2.8 kgm2
A comparison of the performance under
constant voltage and constant current is
shown in the accompanying handout.