Machine Circuit Equations

# Machine Circuit Equations - Synchronous Machine Circuit...

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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 d-q axis models but makes simpliﬁcations that ignore certain aspects of the d-q axis diﬀerences. Notation follows Kundur except for using caps on some phasors. Deﬁne the terminal quantities in reference to the d-q axes: It = id + jiq (assumes d-q axis as reference) Et = ed + jeq (assumes d-q axis as reference) Now make some subsitutions to turn ﬂux equations into voltage-current 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 - steady-state 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 d-q 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 d-axis voltage ed ∠ (δ − 90◦ ) = [Vt sin(δ − δ1 )] ∠ (δ − 90◦ ) q-axis voltage eq ∠δ = [Vt cos(δ − δ1 )] ∠δ 4 d-axis current id ∠ (δ − 90◦ ) = [It sin(δ − δ1 + φ)] ∠ (δ − 90◦ ) q-axis current iq ∠δ = [It cos(δ − δ1 + φ)] ∠δ Generated power ∗ Pe =real{Et It }= ed id + eq iq q-axis 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 d-axis current id = ( ￿ Eq −EB cos δ )XT q −RE EB sin δ D Id = id ∠ (δ − 90◦ ) (inﬁnite bus reference) q-axis 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 q-axis currents −1 I q = [I − Mq ] (Pq E ￿ − Pd E ￿ ) q d and d-axis 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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