Final Exam Equations

Final Exam Equations - 1 Final Exam Equations Power Flow...

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Unformatted text preview: 1 Final Exam Equations Power Flow ∂ gi = −Vi Vj (Gij cos(δi − δj ) + Bij sin(δi − δj )) ∂δj FORMULATION Vj Admittance Matrix ∂ gi = Vi Vj (Gij sin(δi − δj ) − Bij cos(δi − δj )) ∂ Vj ∀i = j ∂ fi = −gi (δ , V ) − Bii Vi2 ∂δi Ybus = G + jB = Yij Power flow equations Pi = fi (δ , V ) = Vi n ￿ Vj (Gij cos(δi − δj ) + Bij sin(δi − δj )) n ￿ Vj (Gij sin(δi − δj ) − Bij cos(δi − δj )) j =1 Qi = gi (δ , V ) = Vi Vi j =1 ∂ fi = fi (δ , V ) + Gii Vi2 ∂ Vi ∂ gi = fi (δ , V ) − Gii Vi2 ∂δi Vi NUMERICAL ITERATION ∂ gi = gi (δ , V ) − Bii Vi2 ∂ Vi Jacobian dimensions General form update = old value + iteration matrix*error ∂ fi : (n − 1) × (n − 1) ∂δj x(k+1) = x(k) + A−1 [y − f (x(k) )] ∂ fi : (n − 1) × nPQ ∂ Vj Gauss ∂ gi : nPQ × (n − 1) ∂δj A = diag {ai } ∂ gi : nPQ × nPQ ∂ Vj Newton-Raphson A = J = jacobian = [ ∂ fi | (k) ∂ xj x Updates are of the form GAUSS-SEIDEL ITERATION Vi (k+1) i−1 ∗ ￿ (k+1) Si − + ... Yij Vj (k ) Vi j =1 n ￿ −1 (k ) . . . − Yii Yij Vj = Vi (k ) + Yii −1 j =i ￿ δ (k+1) V (k+1) ￿ ￿ = ￿ δ (k ) V (k ) ∆P ∆Q(k) (k ) ￿ ￿ = + ￿ ∂ fi ∂δj ∂ gi ∂δj with −1 ∂ fi (k ) ∂ Vj ∂ gi ∂ Vj (k ) Pi − fi (δ (k) , V (k) ) (k ) Qi − gi (δ (k) , V (k) ) Fast Decoupled Newton-Raphson Notes: 1) At slack bus simply calculate S after solution found. 2) At PV bus, calculate Q from most recent voltage and angle updates; and update only the voltage angle. Jacobian approximation NEWTON-RAPHSON LOAD FLOW ∂ fi ≈ −Vi Vj Bij ≈ −Vi Bij ∂δj Jacobian ∀i ￿= j ∂ fi = Vi Vj (Gij sin(δi − δj ) − Bij cos(δi − δj )) ∂δj Vj ∂ fi = Vi Vj (Gij cos(δi − δj ) + Bij sin(δi − δj )) ∂ Vj ∀i ￿= j Vj ∂ gi ≈ −Vi Vj Bij ≈ −Vi Bij ∂ Vj ∀i = j n ￿ ∂ fi ≈ Vi Vj Bij ≈ −Vi Bii ∂δi j =1,j ￿=i ￿ ￿ ∆P (k ) ∆Q(k) ￿ 2 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 K ω ,2 = Vi ∂ gi ≈ Vi ∂ Vi n ￿ j =1,j ￿=i Vj Bij ≈ −Vi Bii then define B ￿ = −B with all shunts ignored and off-nominal transformers set at a = 1 B ￿￿ = −B with all shunts include and phase-shifting transformers set to ∠a = 0◦ So in vector notation δ (k+1) = δ (k) + (B ￿ )−1 ∆P (k) /V (k) V (k+1) = V (k) + (B ￿￿ )−1 ∆Q(k) /V (k) DC load flow equation - can be solved directly 1 (Kω,1 + 2Kω,2 + 2Kω,3 + Kω,4 ) 6 1 Kδ = (Kδ,1 + 2Kδ,2 + 2Kδ,3 + Kδ,4 ) 6 ω (t + ∆t) = ω (t) + Kω ∆t Kω = δ (t + ∆t) = δ (t) + Kδ ∆t δ ∼ (B ￿ )−1 P = Small signal analysis Generation Control ∆pref = n ￿ ∆prefi π fs max P cos(δ0 )∆δ (t) He ￿ π fs max ω0 = P cos(δ0 ) He ¨ ∆δ (t) = − i=1 ∆ pm = n ￿ ∆pmi i=1 β= n ￿1 Ri i=1 ∆pm = ∆pref − β ∆f ACE = (ptie − ptiescheduled ) + Bf ∆f ￿ ∆prefi = −Ki ACEi dt Power System Dynamics 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) ¨ ∆δ (t) = − 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 ∆ωd˙(t) = − ∆δ (t) − ∆ωd (t) 2H 2H ˙ ∆δ (t) = ωs ∆ωd (t) KD K1 ω s + )∆δ (s) = 0 2H 2H ￿ ￿ ￿ 1 2 − 8K H ω s= −KD ± KD =0 1 s 4H (s2 + 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 V = − dΨ dt Electrical to mechanical conversions . p = # of poles 3 pωm 2 ωs = Ψ0 = −L0 i0 Ψf d = Lf f d if d + Lf 1d i1d − 3 Laf d id 2 for constant speed Ψ1d = L11d i1d + Lf 1d if d − 3 La1d id 2 Pm = ω m T m Ψ1q = L11q i1q − 3 La1q iq 2 Voltage Equations - abc axes in Fundamental Parameters ea = d dt Ψa eb = d dt Ψb ec = d dt Ψc Voltage Equations - 0dq axes in Fundamental Parameters − Ra ia Note: − Ra ib d dt Ψabc − Ra ic d dt Ψf d + Rf d if d 0= d dt Ψ1d d dt Ψ1q edq0 + R1d i1d 0= + R1q i1q (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 ￿ ￿ π ic Lab0 + Laa2 cos(2θ − 3 ) + i1d La1d cos θ − i1q La1q sin θ (and the rotor equations) Ψf d = Lf f d if d + L￿ kd ikd − ￿ f ￿ ￿ π Laf d ia cos θ + ib cos θ − 23 + ic cos θ + Ψ1d = Lf 1d if d + L11d i1d − ￿ ￿ ￿ ￿ π La1d ia cos θ + ib cos θ − 23 + ic cos θ + ￿ 2π 3 2π 3 2π 3 ￿ ￿￿ ￿￿ + ic sin θ + P −1 1 2 cos (θ) ￿ ￿ π = cos ￿θ − 23 ￿ 2π cos θ + 3 1 2 and − sin (θ) ￿ ￿ π − sin ￿θ − 23 ￿ 2π − sin θ + 3 i dq0 = P iabc and iabc = P −1 1 2 d −1 dt P ￿￿ 2π 3 1 1 1 assuming near synchronous speed (i.e., θ(t) = ωs t + θ0 ) − sin (θ) ￿ − cos (θ) ￿ 0 ￿ ￿ π π = ωs − sin ￿θ − 23 ￿ − cos ￿θ − 23 ￿ 0 2π 2π − sin θ + 3 − cos θ + 3 0 d P −1 ￿ P dt2π ￿ = ￿ ￿ π − sin (θ) ￿ cos (θ) cos ￿ − 3 ￿ cos ￿ + 23 ￿ θ θ ￿ π π π 2 − sin θ + 23 − sin ￿θ − 23 ￿ ωs − sin (θ) − sin θ − 23 3 2π 1 1 1 − sin θ + 3 2 2 2 0 −1 0 d P dt P −1 = ωs 1 0 0 000 ed = d dt Ψd eq = d dt Ψq e0 = ef d = L0 = Laa0 − 2Labo then Ψd = −Ld id + Laf d if d + La1d i1d Ψq = −Lq iq + La1q i1q d dt Ψ0 d dt Ψf d − Ra i0 + Rf d if d + R1d i1d 0= d dt Ψ1q + R1q i1q Synchronous Machine Dynamic Modeling Time constants from fundamental parameters ￿ Td0 ≈ ￿ Td ≈ Ψdq0 = P Ψabc and Ψabc = P −1 Ψdq0 Lq = Laa0 + Labo − 3 Laa2 2 + Ψd ωs − Ra iq d dt Ψ1d edq0 = P eabc and eabc = P −1 edq0 Ld = Laa0 + Labo + 3 Laa2 2 d dt Ψabc − Ψq ωs − Ra id 0= idq0 Flux Equations 0dq axes in Fundamental Parameters d dt θ and substituting for Ψabc and 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 Ψd d d = −Ridq0 + dt Ψdq0 + P dt P −1 Ψq Ψ0 now if ωs = Flux Equations abc axes in Fundamental Parameters Ψ1q = ￿ ￿ L11q i1q + La1q ia sin θ + ib sin θ − d −1 Ψdq0 dt P and with edq0 = P eabc (single damper winding) ef d = d = P −1 dt Ψdq0 + 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 4 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 ￿ ￿ Ed = − notice the simple relationship ￿ ￿ Eq = eq + Ra iq + Xd id 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 ￿ ￿ ￿ Ed = ed + Ra id − Xq iq or equivalently q0 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 )] ￿ ￿ ￿ ￿ ￿ 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 − = − ￿ ￿ d 1 d dt Vs (t) = T2 V2 (t) + T1 dt V2 (t) − Vs (t) d dt Ef d (t) = 1 TA Kstab d ωs dt ω (t) ￿ Eq = Eq + Xd id − 1 Tw V2 (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) 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 ￿ ￿ ￿ E = Et + Ra It + jXd id − Xq iq or equivalently Ψd = −Ld id + Laf d if d ￿ Ed 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 Now define some common relationships (later some quantities will go to zero in the different models we’ll develop): ωs Ψ Xaf if d = Xf ffdd + 3 Xf f d id since i1d 2 d ωs Ψ Xq i1q = X111q + 3 Xa1q iq so 2 11 q q ed = −Ra id + Xq iq − 3 Xa1q Xa1q iq − 2 11 − Xd id + = 0 and ωs Ψ1q Xa1q X11q Xaf d 3 2 Xaf d Xf f d id 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 + ￿ jEq ￿ ￿ = (ed + jeq ) + Ra (id + jiq ) + jXd id − Xq iq 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 ￿ then Xq = Xq d = 0; dt Ψf d ￿= 0 ￿ ￿ jEq = Et + Ra It + jXd id − Xq iq eq = −Ra iq − Xd id + Xaf d if d eq = −Ra iq + ￿= 0 Then is just what we have above eq = −Ra iq + ωs Ψd ωs Ψ Xaf d Xf ffdd − Xd id ￿ ￿ Eq = Eq + (Xd − Xd )id = Ef d ed = −Ra id − ωs Ψq X 3 Xaf d 2 Xf f d id rearrange and we have Et = ed + jeq (assumes d-q axis as reference) Notice 3 Xaf d 2 Xf f d id We can add and subtract an Xd id term and let Eq = Xaf d if d V (or equivalently in some textsEf d = Xaf d Rf d ) fd Single stage power system stabilizer with washout filter d dt V2 (t) ωs Ψ1q Xa1q X11q ￿ 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◦ 5 ￿ Iq , Eq , Eq , Ef d aligned along δ axis Id , Ed aligned along δ − 90◦ axis angle references if ∠Id = 0 ◦ id = ￿ (Eq −EB cos δ)XT q −RE EB sin δ D Id = id ∠ (δ − 90 ) (infinite bus reference) ◦ q-axis current ￿ Iq , Eq , Eq , Ef d aligned along 90 ◦ iq = EB sin δ +RE id XT q Id , Ed aligned along 0◦ Iq = iq ∠δ (infinite bus reference) ∠Et = 90◦ − δ Terminal current Example Analysis - Single Machine - Find Circuit Quantities from State Variables It = It ∠ (δ1 − φ) = id ∠ (δ − 90◦ ) + iq ∠δ (infinite bus reference) Given power output, terminal voltage, using infinite bus as reference then one can find d-q axes machine quantities (needed for initial conditions) Terminal voltage E t = Et ∠δ1 = EB + It RE + j It XE (infinite bus reference) Generated power ￿ 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 ∗ Pe =real{Et It }= ed id + eq iq Fault Current Calculations Positive, negative and zero sequence transformations E = E ∠δ = E t + jXq It a = 1∠120◦ d-axis voltage V q-axis voltage d-axis current id ∠ (δ − 90 ) = [It sin(δ − δ1 + φ)] ∠ (δ − 90◦ ) ◦ 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+ ￿ (X d − 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 RE = R E + R a = AI = AV 2 D = RE + X T q X T d d-axis current 012 A −1 111 A = 1 a2 a 1 a a2 111 1 1∗ 1 a a2 = A = 3 3 1 a2 a Fault Impedances Fault current - single line to ground fault 0 I a = 3I a = 3E a z + z + z 2 + 3z f 0 1 Fault current - line to line fault √ 1 I b = −I c = (a2 − a)I a = −j 3 Ea z + z2 + zf 1 Fault current - double line to ground fault 1 0 I b + I c = 3I a = −3 1 Ia = z+ 1 XT q = XE + Xq ￿ XT d = XE + Xd 012 eq ∠δ = [Vt cos(δ − δ1 )] ∠δ q-axis current abc abc I ed ∠ (δ − 90◦ ) = [Vt sin(δ − δ1 )] ∠ (δ − 90◦ ) 2 Ia = − E a − z1I a z 0 + 3z f Ea z 2 (z 0 +3z f ) z 2 +z 0 +3z f 1 E a − z1I a z2 ...
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