dynamic model of induction machine by dr. ungku anisa ungku amirulddin department of electrical...
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Dynamic Model of Induction MachineByDr. Ungku Anisa Ungku AmirulddinDepartment of Electrical Power EngineeringCollege of Engineering
Dr. Ungku Anisa, July 2008 1EEEB443 - Control & Drives
OutlineIntroductionThree-phase Dynamic ModelSpace Phasors of Motor VariablesThree-phase to Two-phase Transformation (Stationary)Two-phase (Stationary) Dynamic Model (in dsqs frame)Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq
frame)Voltage equationsTorque equation
Commonly-used Induction Motor ModelsStationary (Stator) Reference Frame ModelRotor Reference Frame ModelSynchronously Rotating Reference Frame Model
Equations in Flux LinkagesReferencesDr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 2
IntroductionPer phase equivalent circuit model only useful for analysing IM
performance in steady-stateall transients neglected during load and frequency variationsused in scalar control drives which do not require good transient
responseexample: drive systems for fans, blowers, compressors
Dynamic model used to observe dynamic (steady-state and transient) behaviour of IM since:Considers instantaneous effects of varying:
Voltages and currents Stator frequency Torque disturbances
machine is part of the feedback loop elements to control the dynamics of the drive system
High performance drive control schemes are based on dynamic model of IM
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 3
IntroductionDynamic model – complex due to magnetic coupling
between stator phases and rotor phasesCoupling coefficients vary with rotor position and
rotor position vary with timeDynamic behavior of IM can be described by
differential equations with time varying coefficientsComplexity of dynamic model can be reduced by
employing space vector equations
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 4
Three-phase Dynamic ModelIM consists of three-phase windings spaced at 120 apartModel windings using simplified equivalent stator winding located on the magnetic axis of each phase.
5Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives
a
b
b’c’
c
a’Simplified equivalent stator winding
ias
Magnetic axis of phase A
Magnetic axis of phase B
Magnetic axis of phase C
ics
ibs
Three-phase Dynamic Model
6Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives
Similar model is applied to represent the rotor ‘windings’
Rotor rotates at speed r
Rotor phase ‘a’ winding displaced from stator phase ‘a’ winding by angle r
stator, b
stator, c
stator, arotor, b
rotor, a
rotor, c
r
Three-phase Dynamic ModelVoltage equation for each stator phase:
Similarly, voltage equation for each rotor phase:
These equations can be written in a compact form.
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 7
dtdiRv
dtdiRv
dtdiRv
cscsscs
bsbssbs
asassas
dtdiRv
dtdiRv
dtdiRv
crcrrcr
brbrrbr
ararrar
'
'
'
Three-phase Dynamic ModelStator voltage equation (compact form):
Rotor voltage equation (compact form):
where:
abcs = stator flux linkage (flux linking stator windings )abcr = rotor flux linkage (flux that links rotor windings)
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 8
, , ,
cs
bs
as
cs
bs
as
cs
bs
as
i
i
i
v
v
v
abcsabcsabcs Ψiv
cr
br
ar
cr
br
ar
cr
br
ar
i
i
i
v
v
v
abcrabcrabcr Ψiv , ,
(1)
(2)
dtdRs abcsabcsabcs Ψiv
dtdRr abcrabcrabcr Ψiv '
Three-phase Dynamic ModelThe displacements between 3 phase stator (rotor) windings
are non-quadrature (i.e. not 90)Magnetic coupling exists between the 3 stator (or rotor)
phases, i.e. the flux linkage each stator (or rotor) phase is sum of:fluxes produced by the winding itselffluxes produced from the other two stator (or rotor) windingsfluxes produced by all three rotor (or stator) windings
Example: Flux linkage for stator phase ‘a’ is sum of:Fluxes produced by stator phase ‘a’ winding itselfFluxes produced by stator phase ‘b’ and stator phase ‘c’Fluxes produced by rotor phase ‘a’, ‘b’ and ‘c’
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 9
Three-phase Dynamic ModelIn general, the stator flux linkage vector:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 10
rabcs,sabcs,abcs ΨΨΨ
cs
bs
as
csbcsacs
bcsbsabs
acsabsas
i
i
i
LLL
LLL
LLL
sabcs,Ψ
cr
br
ar
crcsbrcsarcs
crbsbrbsarbs
crasbrasaras
i
i
i
LLL
LLL
LLL
,,,
,,,
,,,
rabcs,Ψ
Flux linking stator winding due to stator currents
Flux linking stator winding due to rotor currents
(3)
Three-phase Dynamic ModelIn general, the rotor flux linkage vector:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 11
Flux linking rotor winding due to rotor currents
Flux linking rotor winding due to stator currents
sabcr,rabcr,abcr ΨΨΨ
cr
br
ar
crbcracr
bcrbrabr
acrabrar
i
i
i
LLL
LLL
LLL
rabcr,Ψ
cs
bs
as
cscrbscrascr
csbrbsbrasbr
csarbsarasar
i
i
i
LLL
LLL
LLL
,,,
,,,
,,,
sabcr,Ψ
(4)
Three-phase Dynamic ModelSelf inductances in (3) and (4) consists of magnetising
inductance and leakage inductance:For stator:
For rotor:Due to symmetry in windings, mutual inductances between
stator phases in (3)(and rotor phases in (4)) can be written in terms of magnetising inductances:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 12
lsmscsbsas LLLLL Stator leakage inductances
cs
bs
as
lsmsmsms
mslsms
ms
msmslsms
s,abcs
i
i
i
LL2
L
2
L2
LLL
2
L2
L
2
LLL
Note: Subsrcipt ‘s’ is replaced with ‘r’ for rotor phase leakage inductances, currents and flux linkage
lrmscrbrar LLLLL Rotor leakage inductances
Three-phase Dynamic ModelMutual inductances between the stator and rotor windings
depends on rotor position r:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 13
cr
br
ar
rrr
rrr
rrr
mss
rr,abcs
iii
cos32cos3
2cos3
2coscos32cos
32cos3
2coscos
LNN
cs
bs
as
rrr
rrr
rrr
mss
rs,abcr
iii
cos32cos3
2cos3
2coscos32cos
32cos3
2coscos
LNN
Three-phase Dynamic ModelEquations (1) – (4):
completely describe dynamic characteristics of 3-phase IMconsists of 6 equations (3 for stator and 3 for rotor), i.e.
large number of equationsall equations are coupled to one another
Magnetic coupling complicates dynamic model in 3-phase!
Better to develop model based on space phasors:reduces number of equationseliminates magnetic coupling between phases
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 14
Space Phasors of Motor VariablesIf xa, xb, and xc are the 3-phase IM quantities, whereby:
The space phasor in the 3-phase system is obtained from the vectorial sum of the 3-phase quantities, i.e.:
(5)
x is called the space phasor or complex space vector
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives
cba
j
c
j
ba xaaxxexexx 23
4
3
2
3
2
3
2x
, where a = ej2/3
15
3
4cosˆ
32cosˆ
cosˆ
txx
txx
txx
c
b
a
Three-phase to Two-phase Transformation (Stationary)Any three-phase machine can be represented by an
equivalent two-phase machine using Park’s transformation
16
Three-phaseTwo-phase equivalent
There is magnetic coupling between phases
There is NO magnetic coupling between phases (due to 90 angle between phases)
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives
easier way to obtain dynamics of IM
r
stator, qs
rotor,
stator, ds
rotatingr
Two-phase equivalent
rotor,
Three-phase to Two-phase Transformation (Stationary)Dynamic model of IM usually obtained using the two-phase
equivalent machinestator, b
rotor, b rotor, a
rotor, cstator, c
stator, a
r
rrotating
Three-phase
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 17
Three-phase to Two-phase Transformation (Stationary)All 3-phase system quantities have to be transformed
to 2-phase system quantitiesEquivalence between the two systems is based on
the equality of mmf produced and current magnitudes, i.e.:MMF produced by 2-phase system = MMF of 3-phase
systemcurrent magnitude of 2-phase system = current of 3-phase
systemThe use of space phasors enables the transformations
from 3-phase to 2-phase system.
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 18
Three-phase to Two-phase Transformation (Stationary)In the stationary 2-phase system, the space phasor is defined
as : (6)
The space phasor in the 2-phase system must equal that in the 3-phase system.
Hence, by comparing (5) and (6):
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives
sq
sd xx jx
19
cbacba
sd xxxxaaxxx
2
1
2
1
3
2
3
2RexRe 2
cbcbasq xxxaaxxx
3
1
3
2ImxIm 2
Zero-sequence components, which may or may not be present.
Three-phase to Two-phase Transformation (Stationary)The abc dsqs transformation is given by:
(7)
Under balanced conditions, the zero-sequence component adds to zero, i.e.:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives
c
b
a
o
sq
sd
x
x
x
x
x
x
31
31
31
31
31
31
31
32
0
20
0 cbao xxxx
Three-phase to Two-phase Transformation (Stationary)Assuming balanced conditions, the abc dsqs transformation:
(8)The inverse transform (dsqs abc transformation) is given by:
(9) where:
(10)
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 21
abcabcsdq xTx
sdq
1abcabc xTx
3
13
1
00
0
1abcT
23
23
21
211
01
Tabc
Three-phase to Two-phase Transformation (Stationary)Transformation equations (8) and (9) apply to all 3-phase
quantities of the IM (i.e. voltages, current and flux linkages)Transformation matrices (Tabc and Tabc
-1) given by (10) causes the space phasor magnitude to be equal to peak value of the phase quantities, i.e.:
This is one of many abc dsqs and dsqs abc transformation matrices in literature, eg.: space phasor magnitude to be equal to 1.5 times peak value of the
phase quantities ( ) space phasor magnitude to be equal to rms value of the phase
quantities ( )
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 22
x̂x
x̂5.1x
2ˆx x
SPACE VECTORS
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 23
Space vector representation of the mmf distribution in an AC machine created by balanced positive-sequence three-phase sinusoidal currents. Each of the ABC (RGB) space vectors pulsates along its respective axis. The resultant vector (in black), of 1.5 magnitude, rotates at the excitation frequency.
Source:http://www.ece.umn.edu/users/riaz/animations/listanimations.html
SPACE VECTORSThis animation shows the motion of space vectors for the case of a balanced three-phase sinusoidal signal: fA = cos(ωt), fB = cos(ωt-α), fC = cos(ωt+α) where α = 2π/3.
The corresponding space vector is obtained from fR = (fA + γ fB + γ2 fC) = ejωt where γ = ejα.
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 24
Source:http://www.ece.umn.edu/users/riaz/animations/listanimations.html
Example 1 - Space Phasor & 3-2 phase transformationAn induction motor has the following parameters:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 25
Parameter Symbol Value
Rated power Prat 30 hp (22.4 kW)
Stator connection Delta ()
No. of poles P 6
Rated stator phase voltage (rms)
Vs,rat 230 V
Rated stator phase current (rms)
Is,rat 39.5 A
Rated frequency frat 60 Hz
Rated speed nrat 1168 rpm
Example 1 - Space Phasor & 3-2 phase transformation (contd.)Assuming that the motor is operating under rated conditions,
with stator and rotor current phasors of:
Calculate the values the following stator current values at time t = 0:
ias, ibs and ics (3-phase stator phase currents) isd
s and isqs (2-phase stator phase currents)
iar, ibr and icr (3-phase rotor phase currents) ird
s and irqs (2-phase rotor phase currents)
Show that the magnitude of isds and isq
s is equal to the peak stator phase current
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 26
A/ph 5.265.39 sI
A/ph 6.1734.36 rI
Two-phase (Stationary) Dynamic Model (in dsqs frame)From the three-phase dynamic model (eq. 1 and 2):
Applying the transformation given by (8):
Note: dsqs – stator equivalent two-phase winding - rotor equivalent two-phase winding
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 27
r
stator, qs
rotor,
stator, ds
rotatingrrotor,
dtdRs abcsabcsabcs Ψiv dtdRr abcrabcrabcr Ψiv '
dtdRsssdq
ssdq
ssdq Ψiv
dtdRr rrr Ψiv '
Two-phase (Stationary) Dynamic Model (in dsqs frame)The rotor winding rotates at
a speed r
Hence, need to transform the rotor quantities from the to the stationary dsqs frame.
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 28
r
stator, qs
rotor,
stator, ds
rotatingrrotor,
r
r
xr
qs
dsBring all rotor and stator quantities to be on the same axis!
Two-phase (Stationary) Dynamic Model - dsqs frame transformOn the frame:
On the dsqs frame:
The angle between the two frames is r
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 29
rrr xx jxxr
srq
srd
srdq xx jxxr
xrds
r
r
xr
qs
ds
xrqs
ds
r
r
xr
qs
xrxr
Two-phase (Stationary) Dynamic Model - dsqs frame transformTherefore:
More elegantly :
(full derivation dss
frm trnsform)Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 30
r
r
rr
rrsrq
srd
x
x
x
x
cossin
sincos
xrds
r
r
xr
qs
ds
xrqs
r
r
xr
qs
xrxr
r
ds
rrx cossrdx rrx sin
rrx sinsrqx rrx cos
(11)
rjs
rdqre xx (12)
Two-phase (Stationary) Dynamic Model (in dsqs frame)Two-phase dynamic model:
To transform rotor quantities from the dsqs frame, from equation (12):
Substituting these into the rotor voltage equation above..
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 31
Expressed in rotating frame
dtdRsssdq
ssdq
ssdq Ψiv
dtdRr rrr Ψiv '
,srdq
jr
re vv
,srdq
jr
re ii
srdq
jr
re ΨΨ
Hence:
Therefore, the two-phase dynamic model in the stationary dsqs frame:
s
rdqrsrdq
srdqr
srdq
srdq
jsrdq
jr
srdq
j
rrrr
jdtdR
dtedeRe
dtdR
rrr
ΨΨiv
Ψiv
Ψiv
'
'
'
Two-phase (Stationary) Dynamic Model (in dsqs frame)
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 32
Expressed in stationary frame
Expressed in rotating frame
(13)
(14)
dtdRsssdq
ssdq
ssdq Ψiv
srdq
srdq
srdq
srdq ΨΨiv rr dtdR j'
Two-phase (Stationary) Dynamic Model (in dsqs frame)The flux linkages in (13) and (14) are given by:
where and Note that equations (13)-(16) each consists of two equations.
One from equating real quantitiesOne from equating imaginary quantities
Final dynamic equations in the stationary dsqs frame is given by substituting (15) and (16) into (13) and (14) and separating the real and imaginary equations.
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 33
srdq
ssdq
ssdq iiΨ ms LL
srdq
ssdq
srdq iiΨ '
rm LL (15)
(16)
lsms LLL ''lrmr LLL
Two-phase (Stationary) Dynamic Model (in dsqs frame)Final dynamic equations in the stationary dsqs frame:
Lm = mutual inductanceLr’ = rotor self inductances referred to statorRr
’ = rotor resistance referred to statorLs = stator self inductancevrd, vrq, ird, irq are the rotor voltages and currents referred to statorS = derivative operatorDr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 34
(17)
srq
srd
ssq
ssd
rrrrmmr
rrrrmrm
mss
mss
srq
srd
ssq
ssd
i
i
i
i
SLRLSLL
LSLRLSL
SLSLR
SLSLR
v
v
v
v
''
''
'
'
00
00
Example 2 – Dynamic Model of Induction Motor in dsqs frameThe induction motor from the Example 1 has the
following additional parameters:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 35
Parameter Symbol Value
Rated torque Te,rat 183 Nm
Stator resistance Rs 0.294
Stator self inductance
Ls 0.0424 H
Referred rotor resistance
Rr’ 0.156
Referred rotor self inductance
Lr’ 0.0417 H
Mutual inductance Lm 0.041 H
Example 2 - Dynamic Model of Induction Motor in dsqs frame (contd.)Using the values of stator and rotor currents obtained in
the Example 1 , calculate the stator flux s and rotor flux r vectors at time t = 0.
Given that
Calculate the torque produced by the motor using:Stator flux s and stator current Is vector Rotor flux r and stator current Is vector Stator current Is and rotor current Ir vector
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 36
**'
* Im22
3Im
22
3Im
22
3rsmrs
r
msse iiL
Pi
L
LPi
PT
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 37
Space vectors under balanced sinusoidal conditions, appears as constant amplitude vectors rotating at the excitation frequency (2f).
In the stationary dsqs (αβ in the diagram) frame:• dsqs components are time varying sinusoidal signals at stator frequency (2f)
Source:http://www.ece.umn.edu/users/riaz/animations/listanimations.html
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 38
If we want the Induction Motor to behave like a DC motor, the two-phase components must be constant values.
This can be achieved by having a two-phase frame that rotates together with the space vector.
A rotating dq frame!Source:http://www.ece.umn.edu/users/riaz/animations/listanimations.html
Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)The dq referance frame is rotated
at an arbitrary speed g
On the dsqs frame:
On the dq frame:
The angle between the two frames is g where:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 39
tgg
g
g
d
q
x
qs
ds
xds
xqs
sq
sd
sdq xx jxx
qddq xx jxx
Two-phase Dynamic Modelin Arbitrary Rotating Frame(dsqs dq frame transform)
Therefore:
More elegantly :
(full derivation dsqs
d frm trnsform)Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 40
g
g
d
q
x
qs
ds
xds
xqs
xqxd
g
gsdx cosdx g
sqx sin
gsdx sinqx g
sqx cos
sq
sd
gg
gg
q
d
x
x
x
x
cossin
sincos
sdq
jdq xex g
(18)
(19)
Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)Hence, space vectors in the
stationary dsqs frame will have to be transformed into the rotating dq frame using:
Equation (20) will have to be employed onto equations (13) –(16) to obtain the IM dynamic model in the rotating dq frame.
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 41
dqjs
dq xex g (20) g
g
d
q
x
qs
ds
) :(Note
cossin
sincos
tggsdq
jdq
sq
sd
gg
gg
q
d
ge
x
x
x
x
xx
xds
xqs
rdqrgrdqrdqrrdq
rdqj
rrdqj
sdqj
rrdqj
jdtdR
ejdtedeRe gggg
ΨΨiv
ΨΨiv
'
'
Subst. (20) into (13), stator voltage equation:
Subst. (20) into (14), rotor voltage equation:
Expressed in stationary frame dtdR ssdq
ssdqs
ssdq Ψiv
srdqr
srdq
ssdqr
srdq jdtdR ΨΨiv '
Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 42
Expressed in arbitrary rotating frame
Expressed in stationary frame
Expressed in arbitrary rotating frame
sdqgsdqsdqssdq
sdqj
sdqj
ssdqj
jdtdR
dtedeRe ggg
ΨΨiv
Ψiv
(21)
(22)
Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)Subst. (20) into the flux linkages equations of (15) and (16):
Stator flux linkage:
Rotor flux linkage:
Airgap flux linkage:
where and
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 43
lsms LLL ''lrmr LLL
rdqsdqsdq iiΨ ms LL
rdqsdqrdq iiΨ 'rm LL
rdqsdqodq iiΨ mm LL
(23)
(24)
(25)
rq
rd
sq
sd
rrrrgmmrg
rrgrrmrgm
mmgsssg
mgmsgss
rq
rd
sq
sd
i
i
i
i
SLRLSLL
LSLRLSL
SLLSLRL
LSLLSLR
v
v
v
v
''
''
')()(
)(')(
Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)Note that equations (21)-(25) each consists of two equations.
One from equating real quantitiesOne from equating imaginary quantities
Final dynamic equations in the arbitrary rotating dq frame is given by substituting (23)-(24) into the (21)-(22) and separating the real and imaginary equations.
Final dynamic equations in the arbitrary dq frame:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 44(26)
Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)Torque equation in the arbitrary dq frame:
Product of voltage and current conjugate space vectors:
It can be shown that for ias + ibs + ics = 0,
Input power to the IM:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 45
csbs2
ascs2
bsas*ss aiiai
32
vaavv32
iv
cscsbsbsasas*ss iviviv
32
ivRe
*ssivRe
2
3 cscsbsbsasasin ivivivP
Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)Torque equation in the arbitrary dq frame:
Input power to the IM:
If and then:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 46
*ssivRe
2
3 cscsbsbsasasin ivivivP
qqddqdqd*ssin iviv
23
)jii)(jvv(Re23
ivRe23
P
q
d
v
vv
q
d
i
ii vi
23
P tin
Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)Torque equation in the arbitrary dq frame:
The IM equation given by (26) can be written as:
Note that the matrices:[R] = consists of resistive elements[L] = consists of coefficients of the derivative operator S[G] = consists of coefficients of the electrical rotor speed r
[F] = consists of coefficients of the reference frame speed g
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 47
iiiiv gr FGSLR
Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)Torque equation in the arbitrary dq frame:
Hence, the input power is given by:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 48
iiiiiiiivi gt
rtttt
in FGSLRP 2
3
2
3
PowerLosses in winding resistance
Rate of change of stored magnetic energy
Mechanical power
Power associated with g – upon expansion gives zero
Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)Torque equation in the arbitrary dq frame:
The mechanical power is most important:
By observing equation (26), [G] consists of terms associated with r :
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 49
ii rt
emmech GTP 2
3
rq
rd
sq
sd
rrrrgmmrg
rrgrrmrgm
mss
mss
rq
rd
sq
sd
i
i
i
i
SLRLSLL
LSLRLSL
SLSLR
SLSLR
v
v
v
v
''
''
')()(
)(')(
00
00
Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)Torque equation in the arbitrary dq frame:
Therefore, mechanical power:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 50
ii rt
emmech GTP 2
3
r
rdrsdm
rqrsqm
t
rq
rd
sq
sd
em
iLiL
iLiL
i
i
i
i
T
'
'
0
0
2
3
i G
Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)Torque equation in the arbitrary dq frame:
Since m = r / (P/2), hence the electromagnetic torque:
(27)
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 51
r
rdrsdm
rqrsqm
t
rq
rd
sq
sd
emmech
iLiL
iLiL
i
i
i
i
TP
'
'
0
0
2
3
rqsdrdsqme iiiiLP
T 22
3
Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame) - SummaryFinal dynamic equations in the arbitrary dq frame:
Torque equation in the arbitrary dq frame:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 52
rqsdrdsqme iiiiLP
T 22
3
rq
rd
sq
sd
rrrrgmmrg
rrgrrmrgm
mmgsssg
mgmsgss
rq
rd
sq
sd
i
i
i
i
SLRLSLL
LSLRLSL
SLLSLRL
LSLLSLR
v
v
v
v
''
''
')()(
)(')(
(26)
(27)
Commonly-used Induction Motor Models - Stationary (Stator) Reference Frame Speed of reference frame: Dynamic model in the stationary reference frame:
Torque equation in the stationary reference frame:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 53
srqssd
srd
ssqme iiiiL
PT
22
3
0g
srq
srd
ssq
ssd
rrrrmmr
rrrrmrm
mss
mss
srq
srd
ssq
ssd
i
i
i
i
SLRLSLL
LSLRLSL
SLSLR
SLSLR
v
v
v
v
''
''
'
'
00
00
(28)
(29)
(30)
Assuming balanced conditions, vsds, vsq
s, isds, isq
s can be obtained from:
(31)The inverse transform is given by:
(32) where:
and (33)
vrds, vrq
s, irds, irq
s obtained from transforming vabcr vr vrdqs
using equations (31), (32), (33) and (11)
Three-phase to Two-phase Transformation (Stationary)
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 54
abcabcsdq xTx
sdqabcabc xTx 1
3
13
1
00
0
1abcT
23
23
21
211
01
Tabc
Commonly-used Induction Motor Models - Stationary (Stator) Reference Frame This model is used when:
stator variables are required to be actual (i.e. same as in the actual machine stator)
rotor variables can be fictitiousAllows elegant simulation of stator-controlled induction motor
drivesphase-controlled and inverter-controlled IM drives (i.e. this IM
model is used for variable voltage control at constant frequency)Input variables are well defined and can be used to find vsd and
vsq easily Reduce computations leading to real-time control applications
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 55
Commonly-used Induction Motor Models – Rotor Reference Frame Speed of reference frame: Dynamic model in the rotor reference frame:
Torque equation in the rotor reference frame:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 56
rg
rrq
rrd
rsq
rsd
rrm
rrm
mmrsssr
mrmsrss
rrq
rrd
rsq
rsd
i
i
i
i
SLRSL
SLRSL
SLLSLRL
LSLLSLR
v
v
v
v
'00
0'0
(34)
(35)
(36) rrqrsd
rrd
rsqme iiiiL
PT
22
3
Two-phase Transformation (from stationary stator dsqs frame to rotor drqr frame)Assuming balanced conditions, vsd
r, vsqr, isd
r, isqr can be obtained from
vsds, vsq
s, isds, isq
s using:
(37)
The inverse transform is given by:
(38)
Note: vsds, vsq
s, isds, isq
s can be obtained from vabcs and iabcs using (31), (32) and (33)
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 57
ssq
ssd
rr
rrrsq
rsd
x
x
x
x
cossin
sincos
rsq
rsd
rr
rrssq
ssd
x
x
x
x
cossin
sincos
trr :Note
Commonly-used Induction Motor Models - Rotor Reference Frame vrd
r, vrqr, ird
r, irqr obtained from transforming vabcr vrdq
r
using equations (31), (32) and (33)This model is used when:
Switching elements and power are controlled on the rotor side
Example: for simulations of slip-energy recovery scheme
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 58
Commonly-used Induction Motor Models – Synchronously Rotating Reference Frame Speed of reference frame: Dynamic model in the synchronously rotating frame:
Torque equation in the synchronously rotating frame:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 59
sg (39)
(40)
(41) rqsdrdsqme iiiiLP
T 22
3
rq
rd
sq
sd
rrrrsmmrs
rrsrrmrsm
mmgsssg
msmssss
rq
rd
sq
sd
i
i
i
i
SLRLSLL
LSLRLSL
SLLSLRL
LSLLSLR
v
v
v
v
')()(
)(')(
Two-phase Transformation (from stationary stator dsqs frame to synchronously rotating dq frame)Assuming balanced conditions, vsd, vsq, isd, isq can be obtained from
vsds, vsq
s, isds, isq
s using:
(42)
The inverse transform is given by:
(43)
Note: vsds, vsq
s, isds, isq
s can be obtained from vabcs and iabcs using (31), (32) and (33)
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 60
ssq
ssd
ss
ss
sq
sd
x
x
x
x
cossin
sincos
sq
sd
ss
ssssq
ssd
x
x
x
x
cossin
sincos
tss :Note
Commonly-used Induction Motor Models - Synchronously Rotating Reference Framevrd, vrq, ird, irq obtained from transforming vabcr vr
vrdqs vrdq using equations (31), (32), (33), (11) and
(42)Synchronous reference frame:
transforms sinusoidal inputs into dc signalsprovides decoupled torque and flux channels
Hence, IM control similar to separately excited DC motor achieved by employing vector control schemes
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 61
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 62
Space vectors under balanced sinusoidal conditions, appears as constant amplitude vectors rotating at the excitation frequency (2f).
In the stationary αβ (dsqs) frame:• αβ (dsqs) components are time varying sinusoidal signals at stator frequency (2f)
In the rotating synchronous dq frame:• dq components are constant • values depend on the orientation of the space vectors with respect to the dq axes.
Source:http://www.ece.umn.edu/users/riaz/animations/listanimations.html
Equations in Flux LinkagesAll dynamic equations presented are consists of 4 variables,
i.e. vsdq, vrdq, isdq and irdq
Note that in squirrel-cage IM: vrdq = 0 at all timesIf the equations are required to contain flux linkages (i.e.
either sdq, rdq or odq), the dynamic model can be obtained by substituting irdq using the following equations respectively:
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 63
rdqsdqsdq iiΨ ms LL
rdqsdqrdq iiΨ 'rm LL
rdqsdqodq iiΨ mm LL
(23)
(24)
(25)
ReferencesKrishnan, R., Electric Motor Drives: Modeling, Analysis and
Control, Prentice-Hall, New Jersey, 2001.Bose, B. K., Modern Power Electronics and AC drives, Prentice-
Hall, New Jersey, 2002.
Dr. Ungku Anisa, July 2008 64EEEB443 - Control & Drives
dsqs frame transform
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 65
xrds
r
r
xr
qs
xrxr
r
ds
rrx cossrdx rrx sin
rrx sinsrqx rrx cos
rj
rsrdq
rrrr
rrrr
rrrr
srq
srd
exx
xx
xx
xx
xx
sinjcosj
cossinj
sincos
jx r
xrqs
Back
dsqs dq frame transform
Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 66
g
g
d
q
x
qs
ds
xds
xqs
te
xx
xx
xx
xx
ggjs
dqdq
ggsq
sd
gsqg
sd
gsqg
sd
qd
g
where
sinjcosj
cossinj
sincos
jx
xx
xqxd
g
gsdx cosdx g
sqx sin
gsdx sinqx g
sqx cos
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