Machine Dynamic Equations

Machine Dynamic Equations - Synchronous Machine Dynamic...

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Unformatted text preview: Synchronous Machine Dynamic Equations March 30, 2011 DYNAMIC EQUATIONS 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 Voltage transducer with simple thyristor exciter d dt V1 (t) = 1 TR [Et − V1 (t)] Ef d (t) = KA (Vref − V1 (t)) 1 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 Tw V2 (t) ￿ ￿ d V2 (t) + T1 dt V2 (t) − Vs (t) 1 T2 1 TA Kstab d ωs dt ω (t) [−Ef d + KA (Vref − Et + Vs (t))] EXAMPLE LINEARIZATION Single machine, infinite bus, neglecting transient saliency (i.e., model 2) calculation of K constants Again define 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 m1 = EB (XT q sin δ −RE cos δ ) D n1 = EB (RE sin δ +XT d cos δ ) D m2 = XT q D n2 = RE D Now the K constants ￿ K1 = n1 (Vq + Xq Id ) + m1 (Vd − Xd Iq ) ￿ K2 = n2 (Vq + Xq Id ) + m2 (Vd − Xd Iq ) + Iq K3 = 1 ￿ 1+m2 (Xd −Xd ) ￿ K4 = m1 (Xd − Xd ) K5 = K6 = Vd Et Xq n1 Vd Et Xq n2 + − Vq ￿ Et Xd m1 Vq Et (1 ￿ − Xd m2 ) Now the linearized dynamics matrix 2 x = Ax + Bu ˙ x= ￿ ∆ωr D − KH 2 ωo 0 0 D −Kstab KH 2 stab D − T1 K2 KH T 2 ∆δ ￿ ∆Eq K1 − 2H 0 K − T ￿4 d0 AK − KTA 5 K1 −Kstab 2H ∆Ef d stab K1 − T1 K2 2H T ∆ V2 A= K2 − 2H 0 − K31 ￿ Td0 AK − KTA 6 K2 −Kstab 2H stab K2 − T1 K2 2H T 1 2H and B = 3 0 0 0 0 0 ∆Vs ￿T and u = ∆Pm 0 0 0 0 0 0 1 ￿ Td0 − T1 A 0 0 1 T2 0 0 0 1 ￿− TW 1− T1 TW KA TA ￿ 0 1 − T2 ...
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This note was uploaded on 07/14/2011 for the course ECE 522 taught by Professor Tomsovic during the Summer '10 term at University of Florida.

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