Modelling Of Im Using Dq Transformation

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 λas vas = ias rs + dt d λbs vbs = ibs rs + dt d λcs vcs = ics rs + dt

Voltage Equations (cont’d) Rotor Voltage Equations: d λar var = iar rr + dt d λbr vbr = ibr rr + dt d λcr vcr = icr rr + dt

Flux Linkage Equations

Model of Induction Motor To build up our simulation equation, we could just differentiate each expression for λ, e.g.

d λas 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 π/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   f0   fc  where f can be i, v, or λ.

Park’s Transformation (cont’d)

  cos θ q  2 Tqd 0 (θ q )  =  − sin θ q 3  1   2

2π  2π     cos θ q −  cos θ q +  3 3     2π  2π     − sin θ q −  − sin θ q +  3 3      1 1  2 2 

Park’s Transformation (cont’d) The inverse transform is given by:   cos θ q  −1   2π Tqd 0 (θ q )  =  cos θ q − 3     2π cos θ +   q 3  

− sin θ q 2π    − sin θ q − 3   2π   − sin θ +   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(θ)], and

λ dq 0 r = [Tr ]λ abcr

(θ to be defined)

on the rotor,

Induction Motor Model in qd0 (cont’d) STATOR: “abc”:

λabcs = Ls iabcs + Lsr iabcr

“qd0”: λqd0s= Ts λabcs= Ts Ls Ts-1 iqd0s +Ts Lsr Ts-1 iqd0r

ROTOR: λqd0r= Tr λabcr= 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

0  0  where Lar= Lr-Lab Lar 

0 Lar 0

−1 s

Ts LsT and similarly for

.

But what about the cross terms? They

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 β and θr must be to make the transformation work but we still have not determined what to set θ to. We’ll come back to this but let us first look at our new qd0 constitutive law and work out simulation equations. d vqd 0 s = T s vabcs = T s Riabcs + T s λabcs dt d −1 −1 = T s RT s iqd 0 s + T s T s λqd 0 s dt d −1 = Riqd 0 s + T s T s λqd 0 s dt

(

)

Induction Motor Model in qd0 (cont’d) Using the differentiation product rule: vqd 0 s

(

)

d  d −1  = Riqd 0 s + λqd 0 s + T s T s  λqd 0 s dt  dt 

(

d = Riqd 0 s + λqd 0 s dt

)

  0  dθ  +  dt   0 

−

dθ dt 0 0

 0  0  λqd 0 s   0 

Induction Motor Model in qd0 (cont’d) For the stator this matrix is: 0 ω   0

−ω 0 0

0 0  0 

For the rotor the terminal equation is essentially identical but the matrix is: 0 −(ω − ω r ) 0   (ω − ω )  0 0 r    0 0 0 

Induction Motor Model in qd0 (cont’d) Simulation model; Stator Equations: vqs = iqs rs + ωλds +

d λqs dt

d λds vds = ids rs − ωλqs + dt d λ0 s v0 s = i0 s rs + dt

Induction Motor Model in qd0 (cont’d) Simulation model; Rotor Equations: vqr = iqr rr + (ω − ω r )λdr +

d λqr dt

d λdr vdr = idr rr − (ω − ω r )λqr + dt d λ0 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:  λqs   Las λ   0  ds    λ0 s   0  =  λqr  3 2 Lsr  λdr   0    λ0 r   0

0

0

3 2 Lsr

0

Las 0

0 Las 0

0 0

3 2 Lsr 0

0

0

Lar

0

3 2 Lsr 0

0 0

0 0

Lar 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 θ? One good choice is:

dθ = ωe dt where ωe is synchronous frequency. Remember that this choice makes a balanced 3Φ 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 τ m =   ( λqr idr − λdr iqr ) 2 2  where P= # of poles F=ma, so:

dω r J = (τ m − τ l ) dt

where

τl

= load torque

Induction Motor Model in qd0 (cont’d) Example: The equations for a balanced 3Φ, squirrel cage, 2-pole rotor induction motor: Constitutive Laws:

3 τ m = ( λqr idr − λdr iqr ) 2  λqs   Las λ   0  ds  =   λqr  3 2 Lsr     λdr   0

0 Las 0 3 2 Lsr

3 2 Lsr 0 Lar 0

0  iqs  3 2 Lsr  ids  0  iqr    Lar  idr 

Induction Motor Model in qd0 (cont’d) State equations: d − λqs = rs iqs + ωλds − vqs dt d − λds = rs ids − ωλqs − vds dt d − λqr = rr iqr + (ω − ω r )λdr dt d − λdr = rr idr − (ω − ω r )λqr dt dω r (τ m − τ l ) dt

=

J

ωr= rotor speed ω= frame speed J= shaft inertia τl = 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 ω=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 + λqs dt d = rs ids + λds dt d = rr iqr + λqr − ω r λdr dt d = rr idr + λdr + ω r λqr dt

d v0 s = rs i0 s + λ0 s dt d λ0 r v0 r = rr i0 r + dt

qd0 Induction Motor Model in Stationary Reference Frame (cont’d) Flux Linkage Equations:  λqs   xls + xm λ   0  ds    λ0 s   0  =  λqr   xm  λdr   0    λ0 r   0

0

0

xm

0

xls + xm 0

0 xls

0 0

xm 0

0

0

xlr + xm

0

xm 0

0 0

0 0

xlr + xm 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 = (λqr idr − λdr iqr ) 22 3P = (λds iqs − λqs 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 ω= ωe . 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 λqs vqs = iqs rs + ω e λds + dt d λds vds = ids rs − ω e λqs + dt d λ0 s v0 s = i0 s rs + dt

vqr = iqr rr + (ω e − ω r )λdr +

d λqr

dt d λdr vdr = idr rr − (ω e − ω r )λqr + dt d λ0 r v0 r = i0 r rr + dt

qd0 Induction Motor Model in Synchronous Reference Frame (cont’d) Flux Linkage Equations:  λqs   xls + xm λ   0  ds    λ0 s   0  =  λqr   xm  λdr   0    λ0 r   0

0

0

xm

0

xls + xm 0

0 xls

0 0

xm 0

0

0

xlr + xm

0

xm 0

0 0

0 0

xlr + xm 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 = (λqr idr − λdr iqr ) 22 3P = (λds iqs − λqs 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 3Φ phase operation are given by:

vas = Vms cos(ω et )

ias = I ms cos(ω et − φ s )

2π vbs = Vms cos(ω et − ) 3 4π vcs = Vms cos(ω et − ) 3

2π ibs = I ms cos(ω et − − φs ) 3 4π ics = I ms cos(ω et − − φs ) 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( sω et − θ r (0) − δ )

iar = I mr cos( sω et − θ r (0) − δ − φr )

2π vbr = Vmr cos( sω et − − θ r (0) − δ ) ibr = I mr cos( sω et − 2π − θ r (0) − δ − φr ) 3 3 4π 4π i = I cos( s ω t − − θ r (0) − δ − φr ) vcr = Vmr cos( sω et − − θ r (0) − δ ) 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 jωe t s s v s = vqs − jvds = Vms e v s s − jφ jωe t i s = iqs − jids = I ms e e

v v r = (vqrr − jvdrr )e jθ r ( t ) = (Vmr e j ( sωet −θ r (0) −δ ) )e jθ r ( t ) v j ( sωe t −θ r (0) −δ ) jθ r ( t ) jθ r ( t ) r r i r = (iqr − jidr )e = ( I mr e )e 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 ωe(1-s),

θ r (t ) = ω e (1 − s)t + θ r (0) This equation can be used to simplify the rotor voltage and current space vectors which become:

v v r = vqrs − jvdrs = Vmr e − jδ e jωet v s − j (δ +φr ) jω e t s i r = iqr − jidr = I mr e e

Steady State Model of Induction Motor (cont’d) Use phasors to perform steady state analysis. Notation: A - rms values of space vectors ° - rms time phasors B Thus,

°V as = Vms e j 0 2

I ms − jφs % I as = e 2

°V ar = Vmr e jδ 2

I mr − j (δ +φr ) % I ar = e 2

Steady State Model of Induction Motor (cont’d) and

ur s ur s vqss − jvdss ° as e jωet V qs − j V ds = =V 2

rs r s iqss − jidss jω et I qs − j I ds = = I% as e 2 ur s ur s vqrs − jvdrs ° ar e jωet V qr − j V dr = =V 2

rs r s iqrs − jidrs jωe t I qr − j I dr = = I% ar e 2

Steady State Model of Induction Motor (cont’d) Referring the rotor voltages and currents to the stator side gives: ur ' s ur ' s  N V qr − j V dr =  s  Nr

r' s r' s  N I qr − j I dr =  r  Ns

' ° jω e t jω e t ° V e = V e ar ar  

'  % jωet % jω e t 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: ur s ur s rs rs rs rs V qs − j V ds = [rs + jω e ( Lls + Lm )](I qs − j I ds ) + jω e Lm (I 'qr − j I 'dr ) uur s ur s rs rs V 'qr − j V 'dr = j (ω e − ω r ) Lm (I qs − j I ds )

rs rs +[r + j (ω e − ω r )( L + Lm )(I 'qr − j I 'dr ) ' r

' lr

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 (ωe-ωr) by s ωe, and dropping the common ejωt term, we get:

° as = (r + jω L )I% % %' ) V + j ω L ( I + I as as s e ls e m ar

± 'ar = (r ' + jsω L ' )I%' + jsω L (I% %' ) V + I as r e lr ar e m ar ± 'ar r 'r V %' + jω L (I% %' ) = ( + j ω L ' ) I + I as e lr ar e m ar ÷s => s s

Steady State Model of Induction Motor (cont’d) The relations on the previous slide can be rewritten as:

ωe °V as = (r + j ω e x )I% % %' ) + j x ( I + I as as s ls m ar ωb ωb ± 'ar ωe ωe r 'r V % % =( + j x 'lr )I 'ar + j xm (I% as + I 'ar ) s s ωb ωb where ωb is the base or rated angular freq. given by

ω b = 2π 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 I% m = I as + I 'ar is shown below 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: ωe

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:

Pmech 3I r (1 − s ) '2 ' Tem = = 3I ar rr = ω rm sω sm (1 − s ) sω sm

'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:

° th = V

jxm ° as V 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 = 2ω 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: ° as V I% as = Zin

* ° % Sin = Pin + jQin = 3VΙas as

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 3Φ induction motor with the following equivalent circuit parameters: rs = 0.1062Ω

xls = 0.2145 Ω

rr’ = 0.0764Ω

xlr’ = 0.2145 Ω

xm = 5.834 Ω

Jrotor= 2.8 kgm2

A comparison of the performance under constant voltage and constant current is shown in the accompanying handout.