ELEC0047 - Power system dynamics, control and stability
Dynamics of the induction machine
Thierry Van Cutsem
t.vancutsem@ulg.ac.be
www.montefiore.ulg.ac.be/~vct
November 2015
1 / 20
Brief recall
Brief recall
Induction or asynchronous machine
motor widely used in industry, tertiary sector, etc.
sometimes also as small generator
2 / 20
Brief recall
Principle of operation
Stator:
three-phrase windings carrying three-phase currents of angular frequency ωs
produces a magnetic field rotating at angular speed ωs
a single pair of poles is assumed for simplicity.
Rotor:
rotates at a speed ωr 6= ωs characterized by the motor slip :
s=
ωs − ωr
ωs
can be modeled with a set of three-phase windings
currents induced in these windings have angular frequency ωs − ωr = sωs
and produce a magnetic field rotating at angular speed g ωs with respect to
the rotor, i.e. sωs + ωr = ωs with respect to the stator.
Both rotating magnetic fields are fixed with respect to each other.
Their interaction creates the electromagnetic torque.
3 / 20
Brief recall
Two types of machines: squirrel-cage and wound rotors
Squirrel-cage rotor
non insulated aluminum or copper bars inserted in slots, connected at their
ends to allow the currents to flow
simple construction, easy maintenance, reliable operation
possible presence of a second cage aimed at providing a larger starting torque
(non considered here).
4 / 20
Brief recall
Wound rotor
the rotor carries insulated three-phase windings, which are accessed through
sliprings and brushes
used when the rotor circuits have to be accessed, e.g. to control
the starting torque (external resistance)
the starting current
the rotor speed
construction and maintenance are more expensive.
5 / 20
Brief recall
Modelling the induction machine
Modelling the induction machine
Motor sign convention at both stator and rotor.
dψa
dψb
dψc
vb = Rs ib +
vc = Rs ic +
dt
dt
dt
dψA
dψB
dψC
0 = Rr iB +
0 = Rr iC +
0 = Rr iA +
dt
dt
dt
Rs : resistance of one stator circuit
Rr : resistance of one rotor circuit
va
=
Rs ia +
single pair of poles assumed for simplicity of notation
6 / 20
Brief recall
Park transformation, equations and inductance matrix
Park transformation, equations and inductance matrix
Several reference frames can be used, depending on the application
we use a very general one where the d and q axes rotate at some speed ωref
both stator and rotor windings are transformed into this reference frame
this yields new, equivalent windings which are all fixed wrt each other.
ωr : rotor speed
stator
rotor
original
windings
a, b, c
A, B, C
transformed
into
ds, qs, os
dr , qr , or
relative
speed
ωref
ωref − ωr
7 / 20
Brief recall
Park transformation, equations and inductance matrix
By similarity with the derivations of the synchronous machine :
vds
vqs
vos








ψds
ψqs
ψos
ψdr
ψqr
ψor

dψds
dt
dψqs
= Rs iqs − ωref ψds +
dt
dψos
= Rs ios +
dt
= Rs ids + ωref ψqs +
dψdr
dt
dψqr
− ωr )ψdr +
dt
0
= Rr idr + (ωref − ωr )ψqr +
0
= Rr iqr − (ωref
0
= Rr ior +

dψor
dt
Lss
 
 
 
=
  Lsr
 
 

Lsr
Lss
Lsr







Los
Lrr
Lsr
Lrr
Lor
ids
iqs
ios
idr
iqr
ior








8 / 20
Brief recall
Park transformation, equations and inductance matrix
Typical values
Rs
Lss − Lsr
Lsr
0.01 - 0.12 pu
0.07 - 0.15 pu
1.8 - 3.8 pu
Rr
Lrr − Lsr
0.01 - 0.13 pu
0.06 - 0.18 pu
per unit values on the machine base
9 / 20
Brief recall
Energy, power and torque
Energy, power and torque
Stator power balance
Instantaneous power entering the stator =
Joule losses in stator pJs
+ d/dt magnetic energy in stator windings Wms
+ power passing from stator to rotor ps→r
pT (t)
(what type of power is it ?)
= va ia + vb ib + vc ic = vds ids + vqs iqs + vos ios
dψds
dψqs
dψos
2
2
2
= (Rs ids
+ Rs iqs
+ Rs ios
) + (ids
+ iqs
+ ios
)
dt
dt
dt
+ωref (ψqs ids − ψds iqs )
Hence:
ps→r = ωref (ψqs ids − ψds iqs )
(1)
10 / 20
Brief recall
Energy, power and torque
Rotor power balance
Power passing from stator to rotor ps→r =
Joule losses in rotor pJr
+ d/dt kinetic energy Wc
+
+
d/dt magnetic energy in rotor windings Wmr
power transferred to the mechanical load Pm .
From the Park equations :
vdr idr + vqr iqr + vor ior = 0
dψdr
dψqr
dψor
+iqr
+ior
)+(ωref −ωr )(ψqr idr −ψdr iqr ) = 0
dt
dt
dt
dWmr
pJr +
= −(ωref − ωr )(ψqr idr − ψdr iqr )
dt
2
2
2
(Rr idr
+Rr iqr
+Rr ior
)+(idr
Hence, the above rotor power balance equation can be rewritten as :
dWc
ps→r = −(ωref − ωr )(ψqr idr − ψdr iqr ) +
+ Pm
dt
Replacing ps→r by (1) and using the rotor motion equation :
(ωref − ωr )(ψqr idr − ψdr iqr ) + ωref (ψqs ids − ψds iqs ) = ωr Te
11 / 20
Brief recall
Energy, power and torque
Expressions of torque
ψqs ids − ψds iqs = (Lss iqs + Lsr iqr )ids − (Lss ids + Lsr idr )iqs = Lsr (iqr ids − idr iqs )
ψqr idr − ψdr iqr
=
(Lrr iqr + Lsr iqs )idr − (Lrr idr + Lsr ids )iqr = Lsr (iqs idr − ids iqr )
=
−(ψqs ids − ψds iqs )
Hence :
Te = ψqs ids − ψds iqs = ψdr iqr − ψqr idr = Lsr (iqr ids − idr iqs ).
Remarks
The above derivation shows that ps→r = ωref Te
ps→r is both of electromagnetic and mechanical nature
the expression of Te looks very similar to that of the synchronous machine
but both machines behave quite differently
in particular, in the synchronous machine, ps→r is of mechanical nature only.
12 / 20
Brief recall
Rotor motion equation
Rotor motion equation
Following the same derivation as for the synchronous machine yields :
2H
d
ωr = Te − Tm
dt
where H is the inertia constant, in second
ωr , Te and Tm are in per unit
t is in second.
Mechanical torque Tm : varies with the rotor speed ωr
A common model is :
Tm = Tmo Aωr2 + Bωr + C
with A + B + C = 1
where :
Tmo is the torque value at synchronous speed, i.e. when ωr = 1
A, B et C depend on the driven mechanical load.
13 / 20
Brief recall
Rotor motion equation
Typical inertia and torque parameters
component
heat pump, air conditioning
refrigerator, freezer
dishwasher
clothes washer
clothes dryer
pumps, fans, other motors
small industrial motor
large industrial motor
power plant auxiliaries
agricultural water pump
A
0.2
0.2
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
B
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
C
0.8
0.8
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
H (s)
0.28
0.28
0.28
1.50
1.30
0.70
0.70
1.50
1.50
0.4
14 / 20
Brief recall
Model under the phasor approximation
Model under the phasor approximation
Neglecting transformer voltages and dropping the “os” winding:
vds
=
Rs ids + ωref ψqs
(2)
vqs
=
Rs iqs − ωref ψds
(3)
The other equations are unchanged. Dropping the “or ” winding:
dψdr
dt
dψqr
dt
ψds
= −Rr idr − (ωref − ωr )ψqr
(4)
= −Rr iqr + (ωref − ωr )ψdr
(5)
= Lss ids + Lsr idr
(6)
ψqs
= Lss iqs + Lsr iqr
(7)
ψdr
= Lsr ids + Lrr idr
(8)
ψqr
d
2H ωr
dt
= Lsr iqs + Lrr iqr
(9)
= ψdr iqr − ψqr idr − Tmo Aωr2 + Bωr + C
(10)
15 / 20
Brief recall
Exercice
Exercise
Bring the model of the previous slide in the form:
d
x = f(x, y, u)
dt
0 = g(x, y, u)
with the vector of differential states:
x = [ψdr ψqr ωr ]T ,
the vector of algebraic states:
y = [ids iqs ]T ,
and the vector of “inputs”:
u = [vds vqs ]T .
This model is referred to as third-order model of the (single-cage) induction
machine.
16 / 20
Brief recall
Machine behaviour just after a change of network voltage
Machine behaviour just after a change of network voltage
For instance, immediately after a short-circuit
fluxes ψdr and ψqr in rotor windings do not change
rotor currents vary significantly (to keep fluxes constant !)
From Eqs. (8, 9) :
ψqr − Lsr iqs
ψdr − Lsr ids
iqr =
Lrr
Lrr
and the stator equations can be rewritten as :
idr =
vds
= Rs ids + ωref (Lss iqs + Lsr iqr )
(11)
L2sr
Lsr
) iqs + ωref
ψqr
= Rs ids + ωref (Lss −
Lrr
L
| {z }
| {zrr }
00
L00
s
vqs
eds
= Rs iqs − ωref (Lss ids + Lsr idr )
00
Lsr
= Rs iqs − ωref Ls ids − ωref
ψdr
Lrr
| {z }
(12)
00
eqs
17 / 20
Brief recall
Machine behaviour just after a change of network voltage
Following the same derivation as for the synchronous machine, the above two
equations can be combined into the single complex equation:
00
00
V a = Rs I a + jωref Ls I a + E
in the very first moment after a voltage change, the machine behaves as an
00
00
e.m.f E behind the stator resistance Rs and an equivalent inductance Ls
hence, in case of a fault, motors somewhat contribute to the fault current
00
note the analogy between Ls and the transient inductance of the synchronous
machine (compare formulae)
however, the rotor windings of the induction machine behave as the damper
00
windings of the synchronous machine (hence, the notation which refers to
subtransient dynamics)
00
00
after 1 or 2 cycles, eds , eqs have changed significantly.
18 / 20
Brief recall
Simplified (first-order) model
Simplified (first-order) model
Rotor windings contribute with fast transients
approximation: assume their dynamics infinitely fast, and set dψr /dt = 0
this yields a first-order model, with rotor motion as the only dynamics.
0
=
Rr idr + (ωref − ωr )ψqr = Rr idr + (ωref − ωr )Lrr iqr + (ωref − ωr )Lsr iqs
0
=
Rr iqr − (ωref − ωr )ψdr = Rr idr − (ωref − ωr )Lrr idr − (ωref − ωr )Lsr ids
Dividing by
ωref − ωr
:
ωref
0
=
0
=
ωref Rr
idr + ωref Lrr iqr + ωref Lsr iqs
ωref − ωr
ωref Rr
iqr − ωref Lrr idr − ωref Lsr ids
ωref − ωr
(13)
(14)
ωref − ωr
is the machine slip with respect to ωref .
ωref
19 / 20
Brief recall
Simplified (first-order) model
By analogy with the synchronous machine, one can interpret :
ids and iqs as projections on the (d, q) axes of a rotating vector representing
the current in phase a, with corresponding phasor I¯;
idr and iqr as projections on (d, q) axes of a rotating vector representing the
current in one rotor winding, seen from stator, with corresponding phasor I¯r .
Combining (11, 12) and (13, 14), respectively, into complex equations:
V = Rs I + jωref Lss I + jωref Lsr I r
0=
ωref Rr
I r + jωref Lrr I r + jωref Lsr I
ωref − ωr
which corresponds to the well-known (steady-state) equivalent circuit :
where ωr varies according to
the rotor motion equation (10)
20 / 20