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Lý thuyết về quá trình Ổn định quá độ (Transient Stability) bao gồm các nội dung cơ bản sau: • The Swing Equation. • Application to Synchronous Machines. • StepbyStep Solution Method.
eBook for You TRANSMISSION & DISTRIBUTION A Division of Global Power POWER SYSTEM STABILITY CALCULATION TRAINING Day D 2-T Transient Stability i t St bilit Prepared by: Peter Anderson July 5, 2013 OUTLINE • Application to Synchronous Machines • Step-by-Step Solution Method eBook for You • The Swing Equation THE SWING EQUATION d2θ = dω = ω0.(Pm – Pe) dt2 dt 2H H = Inertia Constant = J.ω2/2S in MW-s/MVA or seconds Metric: GD2 in kg-m2 Imperial: WR2 in lb.ft2 H= H= 5.48163 × (GD ) × (RPM) kVA.10 0.231 × (WR ) × (RPM) kVA.10 eBook for You Power-Angle Relationship ANALYSIS OF THE SWING EQUATION dω = ω0.(Pm – Pe) dt 2H Pm is constant Rotor acceleration is a function of the Electrical Power eBook for You In terms of short-term transient stability studies studies, APPLICATION TO SYNCHRONOUS MACHINES Increase in Mechanical Power Input 1.4 1.2 eBook for You Pm increased Rotor accelerates from 25 deg to new operating point at 40 deg g Rotor overshoots to 60 deg, where area above Pm equals the area below Pm Now Pe >Pm and rotor Pm decelerates towards θ=40 deg Rotor will oscillate around θ 40 θ=40 deg If the overshoot reaches θ=140 deg The rotor will not be able to return to the new operating point and will slip to the next pole position 0.8 0.6 0.4 0.2 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 STEP-BY-STEP STEP BY STEP SOLUTION METHOD Pm Increased (2) θ (3) Pm (4) Pe ‐0.0 +0.0 0.5 1.5 θ0 θ0 θ1 θ2 θ3 θ4 Pm0 Pm1 Pm1 Pm1 Pm1 Pm1 (5) (6) Pa Acceleration (3)‐(4) (5)*k Pm0‐Pe0 Pm1‐Pe0 α0 Pm1‐Pe1 α1 Pm1‐Pe2 α2 Pm1‐Pe3 α3 Pm1‐Pe4 α4 Pe0 Pe0 Pe1 Pe2 Pe3 Pe4 (7) ∆t1 ‐ ∆t/2 ∆t ∆t ∆t ∆t (8) ∆ω (6)*(7) ∆ω1 ∆ω2 ∆ω3 ∆ω4 ∆ω5 (9) ω deg/s ω0=0 ω0+∆ω1=ω1 ω1+∆ω2=ω2 ω2+∆ω3=ω3 ω3+∆ω4=ω4 ω4+∆ω5=ω5 1.4 1.2 0.8 08 0.6 0.4 0.2 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 (10) ∆t2 ‐ ∆t ∆t ∆t ∆t ∆t (11) (12) ∆θ θ (9)*(10) (2)+(11) ∆θ1 ∆θ2 ∆θ3 ∆θ4 ∆θ5 θ1 θ2 θ3 θ4 θ5 eBook for You (1) T STEP-BY-STEP STEP BY STEP SOLUTION METHOD Pm Increased from 0.42 to 0.8 pu 1.20 eBook for You 1.00 0.80 0.60 0.40 0.20 0.00 30 60 90 120 150 180 STEP-BY-STEP STEP BY STEP SOLUTION METHOD θ 25.0 25 25.0 26.4 30.3 36.3 43.9 52.2 60.6 68.5 75.4 81.1 85.5 88.5 90.0 90.1 88.7 85.9 85 81.7 76.1 69.3 61.5 53.2 Pm 0.42 42 0.80 0.80 0.80 0.80 0.80 0.80 0.80 0.80 0.80 0.80 0.80 0.80 0.80 0.80 0.80 0.80 80 0.80 0.80 0.80 0.80 0.80 Pe 0.42 42 0.42 0.44 0.50 0.59 0.69 0.79 0.87 0.93 0.97 0.99 1.00 1.00 1.00 1.00 1.00 1.00 00 0.99 0.97 0.94 0.88 0.80 Pa 0.38 0.36 0.30 0.21 0.11 0.01 ‐0.07 ‐0.13 ‐0.17 ‐0.19 ‐0.20 ‐0.20 ‐0.20 ‐0.20 ‐0.20 ‐0.20 20 ‐0.19 ‐0.17 ‐0.14 ‐0.08 0.00 α 10.87 10.25 8.52 5.98 3.08 0.29 ‐2.05 ‐3.75 ‐4.83 ‐5.42 ‐5.67 ‐5.75 ‐5.76 ‐5.76 ‐5.75 ‐5.69 69 ‐5.46 ‐4.92 ‐3.90 ‐2.27 ‐0.01 ∆t1 0.25 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 05 0.5 0.5 0.5 0.5 0.5 ∆ω 2.72 5.13 4.26 2.99 1.54 0.14 ‐1.02 ‐1.87 ‐2.41 ‐2.71 ‐2.84 ‐2.87 ‐2.88 ‐2.88 ‐2.88 ‐2.84 84 ‐2.73 ‐2.46 ‐1.95 ‐1.14 0.00 ω 0.0 00 2.7 7.8 12.1 15.1 16.6 16.8 15.8 13.9 11.5 8.8 5.9 3.0 0.2 ‐2.7 ‐5.6 ‐8.4 84 ‐11.2 ‐13.6 ‐15.6 ‐16.7 ‐16.7 ∆t2 ∆θ θ 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 05 0.5 0.5 0.5 0.5 0.5 1.4 3.9 6.1 7.5 8.3 8.4 7.9 6.9 5.7 4.4 3.0 1.5 0.1 ‐1.4 ‐2.8 ‐4.2 42 ‐5.6 ‐6.8 ‐7.8 ‐8.4 ‐8.4 26.4 30.3 36.3 43.9 52.2 60.6 68.5 75.4 81.1 85.5 88.5 90.0 90.1 88.7 85.9 81.7 81 76.1 69.3 61.5 53.2 44.8 eBook for You T ‐0.0 00 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 75 8.5 9.5 10 STEP-BY-STEP STEP BY STEP SOLUTION METHOD Pm Increased from 0.42 to 0.8 pu 100.0 90.0 70.0 eBook for You Machine Angle (deg) 80.0 80 60.0 50.0 40.0 30.0 20.0 10.0 10 0.0 2.5 7.5 Time (s) 10 12.5 15 eBook for You TRANSMISSION & DISTRIBUTION A Division of Global Power POWER SYSTEM STABILITY CALCULATION TRAINING Day D 3-T Transient Stability i t St bilit Prepared by: Peter Anderson July 8, 2013 BASIC MODELS IN STABILITY STUDIES Exciter pE fd = [K E ( Vs ‐ Vt ‐ Vfb )‐ E fd ] TE pVfb = (K f ‐ pE fd ‐ Vfb ) Tf Vs + Vt ‐ Exciter KE 1 + TEp Limiter Emax, , Emin ‐ Vfb pKf 1 + Tfp eBook for You E E Vs = fd0 + Vt0 ; E ≤VE ≤ max ; pE fd ≤D max KE Rate Limiter Dmax Efd BASIC MODELS IN STABILITY STUDIES Governor pC on = pPm = [Ps ‐ (ω ‐ ω )/(ω 0R)‐ C on ] Tc [C on ‐ Pm ] Ts ω0 ‐ ω + eBook for You PM ≤Glim Regulator Limiter ω0.R Glim ‐ Ps Control Valve 1 + Tcp Turbine Con 1 + Tsp Pm SOLUTION OF DIFFERENTIAL EQUATIONS 90 Method of Integration: Implicit Trapezoidal Rule h Y(t+h) = Y(t) + pY(t+h) + pY(t) ( ) 80 70 60 50 Y 40 30 Y = Integrable Variable g 20 h = Integration Step Length 10 X Iterative Transformation of Differential Equations to I i T f i f Diff i lE i Algebraic Equations: Y(t+h) = YC + YX X (t+h) h YC = Cons tan t = Y(t) + F Y(t) , F X (t) [ ( ) ( )] h YX.X (t+h) = F X (t+h) [( )] 10 eBook for You X = Non ‐ Integrable Variable SOLUTION OF DIFFERENTIAL EQUATIONS 1 3 6 Establish Initial Conditions by Solving for Y and X at t=0 Solve for Y(t+h) using values of X(t) by means of Trapezoidal Rule Solve for X(t+h) using values of Y(t+h) from above Resolve for Y(t+h) using values of X(t+h) Repeat Steps 2, & until convergence is achieved Proceed to next time step Numerically Stable g Allows Larger Time Steps to be used than other methods In Step in practice, an extrapolated value of X(t) is used except immediately after a discontinuity using X(t+h) = 2X(t) – X(t-h) eBook for You Solution Algorithm: Calculate Integrable Variables Transform from Machine Reference Frame to System T f f M hi R f F t S t Reference Frame Calculate Norton equivalent Nodal Injected Current Solve for V using I = Y.V where Y = Nodal Admittance Matrix Use New values for V to move to next time step Network Solution carried out simultaneously with Integration Process eBook for You NETWORK SOLUTION 10 SAMPLE CASES 4x500MW‐ST 250MW‐H 4x590MVA,21/400kV 1x295MVA,21/400kV GBUS‐1 GBUS‐2 600MVA, 0.975pf 600MVA 975pf 400km eBook for You 100km HVBUS‐1 100km 50km GBUS‐3 HVBUS‐3 1x590MVA,21/400kV 1200MVA, 0.975pf 600MVA, 0.975pf 500MW‐ST 11 FAULT POSITION-1 POSITION Line Fault close to Generator HV Bus Base‐Fault at GBus1 Base‐Fault at GBus2 60 30 ‐30 ‐60 ‐90 60 30 ‐30 ‐60 ‐90 50 100 150 200 G2 Time (cycles) 50 G3 Time (cycles) Base‐Fault at GBus3 90 60 30 ‐30 ‐60 60 ‐90 50 100 Time (cycles) 100 150 200 G2 G3 150 200 G2 G3 eBook for You Machine Angle e (degrees wrt G1) 90 Mac chine Angle (degrees w wrt G1) Machine Angle e (degrees wrt G1) 90 12 FAULT POSITION-2 POSITION Line Fault close to System HV Bus Base‐Fault at HVBus3 Base‐Fault at HVBus1 90 30 ‐30 ‐60 60 30 ‐30 ‐60 ‐90 ‐90 50 100 150 200 G2 Time (cycles) 50 Time (cycles) Time (cycles) G3 Base‐All Faults 20 10 ‐10 ‐20 ‐30 ‐40 ‐50 HVBus1 10 20 30 Time (cycles) HVBus3 100 GBus1 40 GBus2 50 GBus3 150 200 G2 G3 eBook for You Machine An ngle (degrees wrt G1) 60 Ma achine Angle (degrees s wrt G1) Machine Ang gle (degrees wrt G1) 90 13 GENRATOR INERTIA Line Fault close to Generator HV Bus (GBUS-3_ 20 10 eBook for You Machine Ang gle (degrees w wrt G1) G3‐H increased ‐10 ‐20 20 ‐30 ‐40 50 100 150 200 Time (cycles) G3 G30 14 GENERATOR EXCITATION CONTROL Line Fault close to Generator HV Bus (GBUS-3_ 20 10 eBook for You Machine Angl le (degrees wr rt G1) Exciters added ‐10 ‐20 ‐30 ‐40 ‐50 ‐60 60 50 100 Time (cycles) 150 G30 200 G3 15 GENERATOR REACTANCE Line Fault close to Generator HV Bus (GBUS-3_ G3‐Xd reduced 60 eBook for You Machine Angle e (degrees wrt t G1) 90 30 ‐30 ‐60 ‐90 50 100 Time (cycles) 150 200 G3 G30 16 GENERATOR MODEL TYPE Line Fault close to Generator HV Bus (GBUS-3_ 90 60 30 eBook for You Machine Ang gle (degrees wr rt G1) G2‐Fixed V/Xd' ‐30 ‐60 ‐90 50 100 Time (cycles) G2 150 G3 G20 200 G30 17 FAULT CLEARING TIME Line Fault close to Generator HV Bus (GBUS-3) 180 150 120 90 eBook for You Machine Angle (degrees w wrt G1) Base‐All Faults 60 30 ‐30 ‐60 ‐90 5‐cycles 10 20 30 Time (cycles) 7‐cycles 10‐cycles 40 50 12‐cycles 18 LOAD REPRESENTATION Line Fault close to Generator HV Bus (GBUS-3_ 90 60 eBook for You M Machine Angle e (degrees wr rt G1) Constant Power‐All Faults 30 ‐30 ‐60 10 cycles 3‐cycles 20 30 Time (cycles) cycles 4‐cycles 40 cycles 5‐cycles 50 19 SUMMARY OF CASES Parameter Stability Increase Fault Position Decrease System Dependent Y ‐ Generator Excitation Control Y ‐ Decreased Synchronous Reactance Y ‐ Simplified Generator Model Si lifi d G t M d l Y ‐ Increased Fault Clearing Time ‐ Y Decreased Voltage Dependency of Load ‐ Y Fault Clearing Time & Voltage Dependency of Load have the most significant impacts on stability eBook for You Increased Inertia ... 1.4 3.9 6. 1 7.5 8.3 8.4 7.9 6. 9 5.7 4.4 3.0 1.5 0.1 ‐1.4 ‐2.8 ‐4.2 42 ‐5 .6 ? ?6. 8 ‐7.8 ‐8.4 ‐8.4 26. 4 30.3 36. 3 43.9 52.2 60 .6 68.5 75.4 81.1 85.5 88.5 90.0 90.1 88.7 85.9 81.7 81 76. 1 69 .3 61 .5 53.2... 0 .60 0.40 0.20 0.00 30 60 90 120 150 180 STEP-BY-STEP STEP BY STEP SOLUTION METHOD θ 25.0 25 25.0 26. 4 30.3 36. 3 43.9 52.2 60 .6 68.5 75.4 81.1 85.5 88.5 90.0 90.1 88.7 85.9 85 81.7 76. 1 69 .3 61 .5... ‐539 ‐592 ? ?63 4 63 4 ? ?66 5 ? ?68 7 0.01 0.02 0.02 0.02 0.02 0.01 0.01 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 s ∆ω ω (6) *(7) (9)N‐1 +(8)N deg/s deg/s 10 ∆t2 28.80 57 .60 57 .60 57 .60 57 .60 28.80