Torque Equation
(See section 4.9)
Our goal is to combine the state-space voltage equations with the
state-space torque equations.
To achieve this, we need to do the following three things to the
torque equation:
1. Address the difference in power bases.
2. Address the difference in speed (time) bases.
3. Express the electromagnetic torque in terms of id and iq
quantities instead of a-b-c quantities.
Let’s take them in that order.
1. Power Base (see first part of Section 4.9):
Consider the electrical torque in MKS units (ntn-meters); denote it
as Te.
The electrical torque that is computed from the voltage equations
will be on a per-phase base, because, as we have seen, all
quantities in the voltage equations are per-unitized on a per-phase
base. Let’s denote this torque as Teφu (A&F call it Teφ). Therefore,
Teφu =
Te
(S B / ω B )
However, the swing equation is usually written on a 3-phase power
base, e.g.,
2H
ω Re
ω& = Tmu − Teu in per-unit.
Here, we have that:
Teu =
Te
(3S B / ω B )
1
We will continue to write it (and use it) like this, because the
network equations are typically given on a three-phase base. In
addition, this is the convention in the literature.
So, from the above relations for per unit torque, it is clear that we
must divide the torque obtained from the voltage equations by 3
before using it in the swing equation, i.e.,
Teu =
Teφu
3
2. Speed (time) Base (see Section 4.9.1)
In the voltage equations, both speed and time were per-unitized, so
we also need to do this in the swing equation.
2 H dω
= Tmu − Teu
ω Re dt
Let’s substitute for speed and time according to ω=ωuωB and
t=tutB, resulting in
2 H d (ω uω B )
2 H d (ω uω B )
= Tmu − Teu
= Tmu − Teu Î
ω Re d (t u / ω B )
ω Re d (t u t B )
2H
2 d (ω u )
ω
= Tmu − Teu
B
Îω
d
(
t
)
Re
u
With ωRe =ωB, we have that
dω u
2 Hω B
= Tmu − Teu
dt u
Now define τj=2HωB, and the swing equation becomes
dω u
= Tmu − Teu
τj
Mechanical starting time:
total time required to
accelerate the unit from
standstill to rated speed
ωR if rated torque
(Tau=1.0) is applied as a
step function at t=0.
dt u
Aside: Recall (eq. (46) of “Swing equation” notes, and pg. 450
A&F) that the mechanical starting time is T4=2H, therefore we see
τj=T4ωB.
2
3. The electromagnetic torque (see Section 4.10)
Our basic approach is to obtain an expression for the electric
power in terms of the 0dq quantities and then use that to obtain an
expression for the electric torque in terms of the 0dq quantities.
In what follows, assume that all quantities are in per-unit on a perphase base.
The instantaneous 3-phase power is given in terms of a-b-c
quantities as
T
p out = v a ia + vb ib + vc ic = v abc i abc
We want it in terms of the 0dq quantities.
Recall that vodq=Pvabc and iodq=Piabc, implying that vabc=P-1v0dq and
iabc=P-1i0dq.
But we need vabcT, which will be vabcT =[ P-1v0dq ] T.
How do we deal with the transpose of a vector product? Consider:
⎡ x ⎤ ⎡1 2⎤ ⎡5⎤ ⎡17 ⎤
⎢ y ⎥ = ⎢3 4⎥ ⎢6⎥ = ⎢39⎥
⎣ ⎦ ⎣
⎦⎣ ⎦ ⎣ ⎦
T
⎡ x⎤
⎢ y ⎥ = [x
⎣ ⎦
⎡1 3 ⎤
= [17 39]
y ] = [5 6]⎢
⎥
⎣2 4⎦
From the above illustration, we may infer that
vabcT =[ P-1v0dq ]T= v0dqT[P-1] T
But here, we recall that P is orthogonal. Then [P-1] T= P.
3
So finally, we have that vabcT= v0dqTP. Therefore, we can substitute
into the instantaneous power expression to obtain:
T
−1
T
T
p out = v abc i abc = [v 0 dq P][ P i 0 dq ] = v 0 dq i 0 dq
The above proves that our version of Park’s transformation is
power invariant, i.e., the instantaneous power is obtained from
either the a-b-c quantities or the 0-d-q quantities using the same
form of expression, according to:
p out = v a i a + vb ib + v c ic = v 0 i0 + v d i d + v q iq
Aside: Observe that power invariance depends on the
orthogonality of P. Without an orthogonal P, then [P-1]T≠P, and
T
−1
T
−1
T
p out = v abc i abc = [v 0 dq ( P )T ][ P i 0 dq ] ≠ v 0 dq i 0 dq
We will again consider only balanced conditions so that zerosequence quantities are zero, and
p out = v d i d + v q i q
Returning to the voltage equations, we can extract the expressions
for vd and vq as (see last page of notes in “perunitization”):
v d = −rid − Ld i&d − kM F i&F − kM D i&D − ωλq
v q = −riq − Lq i&q − kM Q i&Q − kM G i&G + ωλd
Now substitute this into the expression for pout to obtain:
p out = −rid − Ld i&d id − kM F i&F id − kM D i&D id − ωλq id +
2
− riq − Lq i&q iq − kM Q i&Q iq − kM G i&G iq + ωλd iq
2
Gathering together
• The derivative terms
• The ω terms
• The resistive terms
4
we obtain:
λ&
&
λd
644447
44448
644447q 4444
8
p out = −id ( Ld i&d + kM F i&F + kM D i&D ) − iq ( Lq i&q − kM Q i&Q − kM G i&G )
+ ω (λ d i q − λ q i d )
2
2
− r (id + iq )
Note the expressions in brackets of the first line are flux linkage
derivatives according to the notes in “macheqts” (see eq. 4.20’).
Making the substitution indicated by the brackets above the first
line of the expression,
2
2
p out = − [id λ&d + iq λ&q ] + ω (λd iq − λq id ) − r (id + iq )
eq. 4.94’
144244
3 144244
3 14243
TERM 1
TERM 2
TERM 3
Note that this is identical to eq. 4.94 in the text except for the
minus sign in front of the term 1. I believe that this is an error in
the text. But it does not matter, because we will not use this term
anyway.
The text, on page 106, and Charles Concordia in his book on
synchronous machines (see page 28 of “Synchronous Machines:
Theory and Performance,” 1951) indicate that the three terms may
be understood to represent:
• Term 1: rate of change in the stator magnetic field energy
(recognizing flux linkage derivatives as voltages, id λ&d is d-axis
winding power and iq λ&q is q-axis winding power, where “power”
here is of course reactive).
• Term 2: Power crossing the air gap (the speed-voltage terms)
• Term 3: Stator ohmic losses due to the armature resistance
Therefore, terms 1 and 3 represent power that is entirely on the
stator side. But we need the power transferred from the rotor to the
stator, which represents the electromagnetic torque. Therefore, we
are only interested in term 2.
5
From any text on electromechanics (see, for example, pg. 104 of
Fitzgerald, Kingsley, and Kusko), we know that a body
experiencing a force f over a distance ∂x undergoes a change in
energy according to
∂W = f∂x
Analogously, a body experiencing a torque T over an angle ∂θ
undergoes a change in energy according to
∂W = T∂θ
For magnetically coupled coils for which at least one of them may
experience rotation, the exerted electromagnetic torque is related to
the variation in field energy with angular motion according to
T fld =
∂W fld
∂θ m
But we may write this in terms of time derivatives according to:
∂W fld ∂t
∂W fld / ∂t
=
T fld =
∂θ m / ∂t
∂t ∂θ m
Note that the numerator is the power and the denominator is the
speed, therefore:
∂W fld / ∂t Pfld
T fld =
=
∂θ m / ∂t
ωm
Note that
T fld =
Pfld
ω m is expressed in MKS units. In per-unit, we
have:
T fldu =
Pfld / S B
ω m / ω mB
=
Pfldu
ωe /ω B
=
Pfldu
ωu
Here ωu is the same as ω in our voltage equation, eq. 4.94’ above.
Therefore,
6
T fldu =
Pfldu
ωu
=
ω (λ d i q − λ q i d )
= λ d iq − λ q id
ω
This is Teφu, as discussed on page 1 above.
Now, from the first set of notes (see page 20), eqt. 4.20’, we have:
λd = Ld id + kM F i F + kM D i D
λq = Lq iq + kM Q iQ + kM G iG
Substitution of the above flux linkage relations into our torque
expression yields:
Teφu = λ d iq − λq id = ( Ld iq )id + (kM F iq )i F + (kM D iq )i D
− ( Lq id )iq − (kM Q id )iQ − (kM G id )iG
The above can be written as the product of 2 vectors, according to
[
Teφu = Ld iq
kM F iq
kM D iq
Recall the swing equation:
τj
− Lq i d
− kM Q id
⎡id ⎤
⎢i ⎥
⎢ F⎥
⎢i ⎥
− kM G id ⎢ D ⎥
⎢ iq ⎥ (4.98’)
⎢i Q ⎥
⎢ ⎥
⎣⎢iG ⎦⎥
]
dω u
= Tmu − Teu
dt u
where τj=2HωB and the torque is given on a three-phase base.
Three issues:
1. As discussed before, we must divide Teφu in (4.98’) by 3 to
account for power base difference before using it in the above.
2. We will bring in a damping term.
3. Drop the per-unit notation, and realize that per-unit is implied
throughout.
7
So the swing equation becomes:
Teφ
dω
τj
= Tm − Te = Tm −
− Td
dt
3
Here, the damping term is Td. Typically, it is written as a linear
function of speed with the constant of proportionality D; thus,
Td=Dω, and we have:
Teφ
dω
= Tm − Te = Tm −
− Dω
dt
3
We want a state-space equation so as to combine with our statespace “current-form” of the voltage equations (given by eq. 4.75),
which is
τj
v = −( R + ω N )i − Li&
(eq. 4.75)
where each term is defined on pg. 26 of the “per-unitization” notes.
So let’s divide both sides of the swing equation above by τj.
ω& =
Tm
τj
+
⎡− D⎤
1
[−Teφ ] + ⎢
⎥ω
3τ j
⎢⎣ τ j ⎥⎦
Substituting into the last expression eqt. 4.98’ for Teφ, we have
ω& =
⎡ − Ld i q
+⎢
τ j ⎣⎢ 3τ j
Tm
− kM F iq
− kM D iq
Lq i d
kM Q id
3τ j
3τ j
3τ j
3τ j
⎡ id ⎤
⎢i ⎥
⎢ F⎥
kM G id ⎤ ⎢i D ⎥ ⎡ − D ⎤
⎥⎢ ⎥ + ⎢
⎥ω
3τ j ⎦⎥ ⎢ iq ⎥ ⎣⎢ τ j ⎦⎥
⎢i Q ⎥
⎢ ⎥
⎣⎢iG ⎦⎥
(4.101’)
Now let’s bring in ω into the state vector….
8
ω& =
⎡ − Ld i q
+⎢
τ j ⎢⎣ 3τ j
Tm
− kM F i q
− kM D i q
Lq i d
kM Q i d
3τ j
3τ j
3τ j
3τ j
kM G i d
3τ j
⎡id ⎤
⎢i ⎥
⎢ F⎥
⎢i D ⎥
− D ⎤⎢ ⎥
⎥ i
τ j ⎥⎦ ⎢ q ⎥
⎢i Q ⎥
⎢ ⎥
⎢iG ⎥
⎢⎣ω ⎥⎦
Finally, we recall that there are two states for each machine: speed
and angle, yet above, we only have angle. But we must be careful
here, and use per-unit.
Recall that:
θ = ω Re t + δ +
π
2 Î
θ& = ω = ω Re + δ&
Dividing through by ωB=ωRe, we obtain that
ω u = 1 + δ&u
& = ω −1
u
Î δu
Dropping the per-unit notation, we have
δ& = ω − 1
(4.102)
Now we have three different sets of state equations, summarized as
follows:
−1
i& = − L ( R + ω N )i − L−1 v
9
(eq. 4.75)
ω& =
⎡ − Ld i q
+⎢
τ j ⎣⎢ 3τ j
Tm
− kM F i q
− kM D i q
Lq i d
kM Q i d
3τ j
3τ j
3τ j
3τ j
δ& = ω − 1
kM G i d
3τ j
⎡id ⎤
⎢i ⎥
⎢ F⎥
⎢i D ⎥
− D ⎤⎢ ⎥
⎥ i
τ j ⎦⎥ ⎢ q ⎥
⎢i Q ⎥
⎢ ⎥
⎢iG ⎥
⎢⎣ω ⎥⎦
(4.102)
And we can put all of this together into a single state equation that
looks like the following:
⎡ i&d ⎤
⎢i& ⎥
⎢ F⎥
⎢i&D ⎥ ⎡
⎢ i& ⎥ ⎢ ⎡ − L i
⎢ q ⎥ = ⎢⎢ d q
⎢i&Q ⎥ ⎢ ⎢ 3τ j
⎢ & ⎥ ⎢⎣ ⎣ 0
⎢iG ⎥
⎢ω& ⎥
⎢⎣ δ& ⎥⎦
−1
− kM F iq
3τ j
0
− L (R + ω N )
− kM D iq Lq id
3τ j
3τ j
0
0
kM Q id
3τ j
0
⎡ id ⎤
⎢i ⎥
⎢ F⎥
⎤ ⎢iD ⎥ ⎡[− L−1 v ]⎤
0
kM G id ⎤ ⎡ − D
⎤ ⎥ ⎢ iq ⎥ ⎢ ⎡Tm ⎤ ⎥
0
⎥ ⎥ ⎢i ⎥ + ⎢ ⎢ τ j ⎥ ⎥
3τ j ⎥ ⎢ τ j
⎥ ⎢
⎥ ⎥ ⎢⎢ Q ⎥⎥ ⎢ ⎢ ⎥ ⎥
0 ⎦ ⎣ 1 0⎦ ⎥⎦ i
⎣⎢ ⎣ − 1⎦ ⎦⎥
⎢ G⎥
⎢ω ⎥
⎢⎣ δ ⎥⎦
The above relation is called the “current state-space model.” We
will derive a “flux-linkage state-space model” in the next set of
notes.
10