Exam 2 - Midterm 2 Equations April 5, 2011 Swing Equation -...

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Unformatted text preview: Midterm 2 Equations April 5, 2011 Swing Equation - Equal Area Criterion max Pe (δ, t) = Pe sin δ (t) max Pe = Ea V Xeq π fs ¨ δ (t) = (Pm − Pe (δ, t)) H ￿ δcc ￿ Aa = (Pm − Pe (δ, t))dδ δ0 Ad = ￿ δ2 δcc ￿￿ (Pe (δ, t) − Pm )dδ Aa = Ad Runge-Kutta Integration K ω ,1 = πf ￿ (Pm − Pe max sin(δ (t))) H Kδ,1 = ω (t) πf Kδ,1 ∆t ￿ (Pm − Pe max sin(δ (t) + )) H 2 K ω ,1 ∆ t Kδ,2 = ω (t) + 2 πf Kδ,2 ∆t ￿ K ω ,3 = (Pm − Pe max sin(δ (t) + )) H 2 K ω ,2 ∆ t Kδ,3 = ω (t) + 2 πf ￿ K ω ,4 = (Pm − Pe max sin(δ (t) + Kδ,3 ∆t)) H Kδ,4 = ω (t) + Kω,3 ∆t 1 Kω = (Kω,1 + 2Kω,2 + 2Kω,3 + Kω,4 ) 6 K ω ,2 = 1 Kδ = 1 (Kδ,1 + 2Kδ,2 + 2Kδ,3 + Kδ,4 ) 6 ω (t + ∆t) = ω (t) + Kω ∆t δ (t + ∆t) = δ (t) + Kδ ∆t Small signal analysis ¨ ∆δ (t) = − ω0 = ¨ ∆δ (t) = − π fs max P cos(δ0 )∆δ (t) He ￿ π fs max P cos(δ0 ) He π fs max π fs ˙ P cos(δ0 )∆δ (t) − PD ∆δ (t) He H max K1 = Pe cos(δ0 ) K D = PD ω s K1 KD ∆δ (t) − ∆ωr (t) ∆ωr˙(t) = − 2H 2H ˙ ∆δ (t) = ωs ∆ωr (t) (s2 + s 1 s= 4H ￿ KD K1 ω s + )∆δ (s) = 0 2H 2H −KD ± ￿ 2 KD − 8K 1 H ω s ￿ Synchronous Machine Fundamental Circuit Models Magnetic Circuit for High Permeability Core Rc = Lc = 1 Pc N2 Rc F = NI = = = lc µAc N 2 µAc lc lc µAc φ = Rc φ Faraday’s Law 2 V = − dΨ dt Electrical to mechanical conversions . p = # of poles pωm 2 ωs = for constant speed Pm = ω m T m Voltage Equations - abc axes in Fundamental Parameters ea = d dt Ψa − Ra ia eb = d dt Ψb − Ra ib ec = d dt Ψc − Ra ic (single damper winding) ef d = d dt Ψf d + Rf d if d 0= d dt Ψ1d + R1d i1d 0= d dt Ψ1q + R1q i1q Flux Equations abc axes in Fundamental Parameters (just one armature phase shown for illustration, showing self and mutual inductances) ￿ ￿ Ψa = −ia [Laa0 + Laa2 cos 2θ] +ib Lab0 + Laa2 cos(2θ + π ) + . . . 3 ￿ ￿ ic Lab0 + Laa2 cos(2θ − π ) + i1d La1d cos θ − i1q La1q sin θ 3 (and the rotor equations) ￿ ￿ ￿ ￿ ￿￿ π π Ψf d = Lf f d if d + Lf kd ikd − Laf d ia cos θ + ib cos θ − 23 + ic cos θ + 23 ￿ ￿ ￿ ￿ ￿￿ π π Ψ1d = Lf 1d if d + L11d i1d − La1d ia cos θ + ib cos θ − 23 + ic cos θ + 23 ￿ ￿ ￿ ￿ ￿￿ π π Ψ1q = L11q i1q + La1q ia sin θ + ib sin θ − 23 + ic sin θ + 23 Park’s Transformation ￿ ￿ ￿ ￿ π π cos (θ) cos ￿ − 23 ￿ cos ￿ + 23 ￿ θ θ π π − sin θ + 23 P = 2 − sin (θ) − sin θ − 23 3 P −1 1 2 cos (θ) ￿ ￿ π = cos ￿θ − 23 ￿ 2π cos θ + 3 1 2 3 − sin (θ) ￿ ￿ π − sin ￿θ − 23 ￿ 2π − sin θ + 3 1 2 1 1 1 and i dq0 = P iabc and iabc = P −1 idq0 edq0 = P eabc and eabc = P −1 edq0 Ψdq0 = P Ψabc and Ψabc = P −1 Ψdq0 Flux Equations 0dq axes in Fundamental Parameters Ld = Laa0 + Labo + 3 Laa2 2 Lq = Laa0 + Labo − 3 Laa2 2 L0 = Laa0 − 2Labo then Ψd = −Ld id + Laf d if d + La1d i1d Ψq = −Lq iq + La1q i1q Ψ0 = −L0 i0 Ψf d = Lf f d if d + Lf 1d i1d − 3 Laf d id 2 Ψ1d = L11d i1d + Lf 1d if d − 3 La1d id 2 Ψ1q = L11q i1q − 3 La1q iq 2 Voltage Equations - 0dq axes in Fundamental Parameters Note: d dt Ψabc d = P −1 dt Ψdq0 + d −1 Ψdq0 dt P and with edq0 = P eabc edq0 now if ωs = Ψd d d = −Ridq0 + dt Ψdq0 + P dt P −1 Ψq Ψ0 assuming near synchronous speed (i.e., θ(t) = ωs t + θ0 ) − sin (θ) ￿ − cos (θ) ￿ 0 ￿ ￿ π π d −1 = ωs − sin ￿θ − 23 ￿ − cos ￿θ − 23 ￿ 0 dt P 2π 2π − sin θ + 3 − cos θ + 3 0 d dt θ ￿ ￿ π cos (θ) cos ￿ − 23 ￿ θ 2 − sin (θ) − sin θ − 2π 3 3 ωs 1 2 1 2 d P dt ￿ −1 = ￿ P π − sin (θ) ￿ cos ￿ + 23 ￿ θ ￿ π π − sin θ + 23 − sin ￿θ − 23 ￿ 2π 1 − sin θ + 3 2 4 − cos (θ) ￿ ￿ π − cos ￿θ − 23 ￿ 2π − cos θ + 3 0 0 0 0 = ωs 1 0 d P dt P −1 −1 0 0 0 0 0 and substituting for Ψabc and d dt Ψabc ed = d dt Ψd − Ψq ωs − Ra id eq = d dt Ψq + Ψd ωs − Ra iq e0 = ef d = d dt Ψ0 − Ra i0 d dt Ψf d + Rf d if d 0= d dt Ψ1d + R1d i1d 0= d dt Ψ1q + R1q i1q Synchronous Machine Dynamic Modeling Time constants from fundamental parameters ￿ Td0 ≈ ￿ Td ≈ Xf f d ωs Rf d ￿ Xd Xf f d ωs Rf d Xd ￿ Tq 0 ≈ X11q ωs R1q Classical machine model d dt ω (t) = π fs H (Pm d dt δ (t) − Pe − PD )) = ω (t) − ωs (t) 2 ∗ Pe (t) =Real{Et It } + Ra It PD (t) = KD (ω (t) − ωs (t)) To include field flux variations also have (flux decay model) and have d dt Ψf d ￿= 0 (equation traditionally written in terms of voltage not flux). Applies to both model 1 and model 2. ￿ ￿ d 1 ￿ ￿ ￿ Ef d (t) − Eq (t) − (Xd − Xd )Id (t) dt Eq (t) = T ￿ d0 To include q-axis damper winding (i.e., model 1) for cylindrical rotor also have ￿ ￿ d 1 ￿ ￿ ￿ −Ed (t) + (Xq − Xq )Iq (t) dt Ed (t) = T ￿ q0 5 Voltage transducer with simple thyristor exciter d dt V1 (t) = 1 TR [Et − V1 (t)] Ef d (t) = KA (Vref − V1 (t)) Or can model as a time constant in the exciter d dt Ef d (t) = 1 TA [−Ef d + KA (Vref − Et )] Single stage power system stabilizer with washout filter d dt V2 (t) d dt Vs (t) d dt Ef d (t) = = = 1 T2 1 TA Kstab d ωs dt ω (t) − 1 Tw V2 (t) ￿ ￿ d V2 (t) + T1 dt V2 (t) − Vs (t) [−Ef d + KA (Vref − Et + Vs (t))] Simplified Circuit Equations for Stability Studies - Transient Period It = id + jiq (assumes d-q axis as reference) Et = ed + jeq (assumes d-q axis as reference) ed = −Ra id − ωs Ψq eq = −Ra iq + ωs Ψd 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 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 6 ωs Ψ1q X11q + ωs Ψ1q Xa1q X11q X af − Xd id + 3 Xaf d Xf f d id 2 d 3 Xa1q 2 X11q iq so 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 Model 1: Considering q axis damper winding current but no d axis damper current flux constant except for field - transient period 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 7 Model 2: Neglecting both damper winding currents and variation in damper flux - transient period 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 ￿ ￿ Xd ≈ Xq ￿ ￿ jEq = Et + (Ra + jXd )It ￿ Ed = 0 Model 4: Neglecting damper windings, saliency, and variations in field flux - steady-state equation 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 - Find Circuit Quantities from State Variables Given power output, terminal voltage, using infinite bus as reference then one can find d-q axes machine quantities (needed for initial conditions) 8 ￿ 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 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 ￿ 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 9 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) 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 10 ...
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