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Unformatted text preview: Synchronous Machine Circuit Equations
March 30, 2011
SIMPLIFIED CIRCUIT EQUATIONS FOR STABILITY STUDIES
 TRANSIENT PERIOD
The following considers dq axis models but makes simpliﬁcations
that ignore certain aspects of the dq axis diﬀerences. Notation
follows Kundur except for using caps on some phasors.
Deﬁne the terminal quantities in reference to the dq axes:
It = id + jiq (assumes dq axis as reference)
Et = ed + jeq (assumes dq axis as reference)
Now make some subsitutions to turn ﬂux equations into
voltagecurrent relationships assuming close to synchronous and
stator transients are zero:
d
d
dt Ψq , dt Ψd = 0; ed = −Ra id − ωs Ψq
eq = −Ra iq + ωs Ψd
Now also assume i1d = 0 (since small relative to ﬁeld current):
Ψd = −Ld id + Laf d if d
Ψq = −Lq iq + La1q i1q
and deﬁne
Xq = ωs Lq ; Xa1q = ωs La1q and ωs Ψq = −Xq iq + Xa1q i1q
Xd = ωs Ld ; Xaf d = ωs Laf d so ωs Ψd = −Xd id + Xaf d if d
ed = −Ra id + Xq iq − Xa1q i1q
eq = −Ra iq − Xd id + Xaf d if d
1 Now deﬁne some common relationships (later some quantities will
go to zero in the diﬀerent models we’ll develop):
Notice if d = ωs Ψf d
Xf f d + 3 Xaf d
2 Xf f d id since i1d = 0 and i1q =
X q
ed = −Ra id + Xq iq − 3 Xa1q Xa1q iq −
2
11 eq = −Ra iq + Xaf d ωs Ψf d
Xf f d ωs Ψ1q
X11q + 3 Xa1q
2 X11q iq so ωs Ψ1q Xa1q
X11q
X af
− Xd id + 3 Xaf d Xf f d id
2
d Now deﬁne the following to simplify the equations
Xd = Xd −
Xq = Xq −
Eq = 2
3 Xaf d
2 Xf f d
2
3 Xa1q
2 X11q ωs Ψf d Xaf d
Xf f d
Ed = − ωs Ψ1q Xa1q
X11q notice the simple relationship
Eq = eq + Ra iq + Xd id
Ed = ed + Ra id − Xq iq or equivalently
E = Ed + jEq = Et + Ra It + jXd id − Xq iq Also will need to look at the ﬁeld voltage and the related circuit equations so
substitute in Eq the expression for Ψf d
Eq = Xaf d if d − 3 Xaf d
2 Xf f d id We can add and subtract an Xd id term and let Eq = Xaf d if d (or equivalently
V
in some textsEf d = Xaf d Rf d )
fd
Eq = Eq + Xd id − 3 Xaf d
2 Xf f d id − Xd id rearrange and we have
Eq = Eq + (Xd − Xd )id = Ef d CIRCUIT EQUATIONS  TRANSIENT PERIOD 2 Diﬀerent models are numbered here as: 1) Cylindrical rotors neglects d axis damper winding currents; 2) Salient pole  neglect
both damper winding currents; 3) Neglect salient transiency; 4)
Neglect all but electromechanical transient. Model three and four
nearly identical.
In all models, we have from above:
d
d
d
dt Ψq , dt Ψ1d , dt Ψd , i1d = 0; Model 1: Considering q axis damper winding current but no d axis
damper current ﬂux constant except for ﬁeld  transient period
(This is used for cylindrical rotors since for salient pole the q axis damper
winding current decays much more quickly. Also note that d axis damper
winding current eﬀect can be included with the ﬁeld current eﬀects)
d
d
dt Ψ1q , dt Ψf d , ikq = 0 Then is just what we have above
E = Et + Ra It + jXd id − Xq iq or equivalently
Ed + jEq = (ed + jeq ) + Ra (id + jiq ) + jXd id − Xq iq Model 2: Neglecting both damper winding currents and variation in
damper ﬂux  transient period
(typically used for salient pole machines)
d
d
d
d
dt Ψd , dt Ψ1d , dt Ψq , dt Ψ1q ,i1d , i1q d
= 0; dt Ψf d = 0
then Xq = Xq
jEq = Et + Ra It + jXd id − Xq iq
Ed = 0 Model 3: Neglect transient saliency
(typically used for simple text book problems)
Xd ≈ Xq
jEq = Et + (Ra + jXd )It
Ed = 0 3 Model 4: Neglecting damper windings, saliency, and variations in
ﬁeld ﬂux  steadystate equation
(typically used for simple text book problems and classical models neglecting
armature resistance)
d
d
d
dt Ψd , dt Ψq , ikd , ikq , dt Ψf d =0 Xd ≈ Xq (= Xs )
E = Et + (Ra + jXs )It
Note on phasor representations
angle references if ∠Et = 0◦
Iq , Eq , Eq , Ef d aligned along δ axis Id , Ed aligned along δ − 90◦ axis
angle references if ∠Id = 0◦
Iq , Eq , Eq , Ef d aligned along 90◦ Id , Ed aligned along 0◦
∠Et = 90◦ − δ
EXAMPLE ANALYSIS  SINGLE MACHINE SYSTEM
CIRCUIT EQUATIONS
Given power output, terminal voltage, using inﬁnite bus as reference then one
can ﬁnd dq axes machine quantities (needed for initial conditions)
Assume Xq ≈ Xq (i.e., model 2) and armature resistance outside terminal
voltage (lumped with RE ) E t = Et ∠δ1 and It = It ∠ (δ1 − φ)
For convenience deﬁne
E = E ∠δ = E t + jXq It
daxis voltage
ed ∠ (δ − 90◦ ) = [Vt sin(δ − δ1 )] ∠ (δ − 90◦ )
qaxis voltage
eq ∠δ = [Vt cos(δ − δ1 )] ∠δ
4 daxis current
id ∠ (δ − 90◦ ) = [It sin(δ − δ1 + φ)] ∠ (δ − 90◦ )
qaxis current
iq ∠δ = [It cos(δ − δ1 + φ)] ∠δ
Generated power
∗
Pe =real{Et It }= ed id + eq iq qaxis voltage behind the reactance
Eq = E + (Xd − Xq ) id Field voltage
Ef d = Eq + (Xd − Xd )id Reference voltage
Vref = Ef d
KA + Et CALCULATING CIRCUIT QUANTITIES FROM STATE
VARIABLES
Single machine, inﬁnite bus and Xq ≈ Xq (i.e., again model 2) then one can
ﬁnd network quantities (needed at each step of integration) Some intermediary variables
R E = R E + Ra
XT q = XE + Xq
XT d = XE + Xd
2
D = RE + X T q X T d daxis current
id = (
Eq −EB cos δ )XT q −RE EB sin δ
D Id = id ∠ (δ − 90◦ ) (inﬁnite bus reference)
qaxis current
iq = EB sin δ +RE id
XT q Iq = iq ∠δ (inﬁnite bus reference)
5 Terminal current
It = It ∠ (δ1 − φ) = id ∠ (δ − 90◦ ) + iq ∠δ (inﬁnite bus reference)
Terminal voltage
E t = Et ∠δ1 = EB + It RE + j It XE (inﬁnite bus reference)
Generated power
∗
Pe =real{Et It }= ed id + eq iq CALCULATING CIRCUIT QUANTITIES FROM STATE
VARIABLES FOR MULTIMACHINES ASSUMING IMPEDANCE
LOADS
Did not cover this but following is for model 1 with modiﬁcations for model 2
shown at end.
reduced system admittance matrix
−1
−1
Y gen = Y nn − Y nr Y rr Y rn
deﬁne intermediary variables −1
−1
Y = Y gen + [Ra ] + X (1∠90◦ )
d
Yδ = [1∠ (90◦ − δ )] Y [1∠δ ]
Y = Y · (X − X )(1∠90◦ )
X δ d q Yδ = Pd + jPq
YX = Md + jMq then the circuit equation is
E ∠δ + E ∠ (δ − 90◦ ) = E t + [Ra ] I t + X (1∠90◦ ) I d ∠ (δ − 90◦ ) . . .
q
d
d
+ X (1∠90◦ ) I q ∠δ
q
then rearrange
−1
E ∠δ + E ∠ (δ − 90◦ ) = Y gen + [Ra ] + X (1∠90◦ ) I t . . .
q
d
d
− (X − X )(1∠90◦ ) I q ∠δ
d
q and interchange order of diagonal matrices
6 [1∠δ ] (E + E ∠ (−90◦ )) = Y
q
d −1
I t − [1∠δ ] · (X − X )(1∠90◦ ) I q
d
q then multiply by Y and rotate by [1∠ (90◦ − δ )] so
Yδ (E + E ∠ (−90◦ )) = [1∠ (90◦ − δ )] I t − Yδ · (X − X )(1∠90◦ ) I q
q
d
d
q
ﬁnally substituting the next two equations
[1∠ (90◦ − δ )] I t = I d + jI q
Yδ · (X − X )(1∠90◦ ) I q = YX I q
d
q
into
Yδ (E + E ∠ (−90◦ )) = I d + jI q − YX I q
q
d and switching to all rectangular coordinates
(Pd + jPq ) (E − jE ) = I d + jI q − (Md + jMq ) I q
q
d
which gives qaxis currents
−1 I q = [I − Mq ] (Pq E − Pd E )
q
d and daxis currents
−1
−1
I d = Pd + Md [I − Mq ] Pq E + (Pq − Md [I − Mq ] Pd E )E
q
d
d
or more simply I d = Pd E + Pq E + M d I q
q
d
Terminal current
I t = I d ∠ (δ − 90◦ ) + I q ∠δ (inﬁnite bus reference)
Terminal voltage
E t = Y gen I t
Generated power
∗
Pei =real{Eti Iti } If assuming E = 0 and X q = X (model 2) then the above changes to
d
q
YX = Yδ · j X − X q
d
E ∠δ = E t + [Ra ] I t + X (1∠90◦ ) I d ∠ (δ − 90◦ ) + X q (1∠90◦ ) I q ∠δ
q
d
7 −1
E ∠δ = Y gen + [Ra ] + X (1∠90◦ ) I t − (X − X q )(1∠90◦ ) I q ∠δ
q
d
d
[1∠δ ] E = Y
q −1
I t − [1∠δ ] · (X − X q )(1∠90◦ ) I q
d
Yδ E = [1∠ (90◦ − δ )] I t − Yδ · (X − X q )(1∠90◦ ) I q
q
d
(Pd + jPq ) E = I d + jI q − (Md + jMq ) I q
q
−1 I q = [I − Mq ] Pq E
q
−1
I d = Pd + Md [I − Mq ] Pq E
q 8 ...
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 Summer '10
 Tomsovic
 Electrical Engineering

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