lecture6

lecture6 - Example Using the Newton-Raphson PF, find the...

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Unformatted text preview: Example Using the Newton-Raphson PF, find the power flow solution 1 0.01 + j0.03 y12 = 10 − j 20 pu y13 = 10 − j 30 pu y23 = 16 − j 32 pu Power Systems I 0.0125 + j0.025 0.02 + j0.04 2 3 |V3| = 1.04 200 MW 400 + j 250 S =− = −4.0 − j 2.5 pu 100 200 P3sch = = 2.0 pu 100 sch 2 Slack Bus V1 = 1.05∠0° 400 MW 250 MVAR Example Ybus Ⱥ 20 − j 50 − 10 + j 20 − 10 + j 30Ⱥ = Ⱥ− 10 + j 20 26 − j52 − 16 + j 32Ⱥ Ⱥ Ⱥ Ⱥ− 10 + j 30 − 16 + j 32 26 − j 62 Ⱥ Ⱥ Ⱥ 31.6∠1.89 Ⱥ Ⱥ53.9∠ − 1.90 22.4∠2.03 = Ⱥ 22.4∠2.03 58.1∠ − 1.11 35.8∠2.03 Ⱥ angles are in radians Ⱥ Ⱥ Ⱥ 31.6∠1.89 35.8∠2.03 67.2∠ − 1.17Ⱥ Ⱥ Ⱥ 2 P2 = V2 V1 Y21 cos(θ 21 − δ 2 + δ 1 ) + V2 Y22 cos(θ 22 ) + V2 V3 Y23 cos(θ 23 − δ 2 + δ 3 ) 2 P3 = V3 V1 Y31 cos(θ 31 − δ 3 + δ 1 ) + V3 V2 Y32 cos(θ 32 − δ 3 + δ 2 ) + V3 Y33 cos(θ 33 ) 2 Q2 = − V2 V1 Y21 sin (θ 21 − δ 2 + δ 1 ) − V2 Y22 sin (θ 22 ) − V2 V3 Y23 sin (θ 23 − δ 2 + δ 3 ) Power Systems I Example Ⱥδ 2 Ⱥ Ⱥ Ⱥ x = Ⱥδ 3 Ⱥ ȺV2 Ⱥ Ⱥ Ⱥ Ⱥ P2 (δ 2 , δ 3 ,V2 ) Ⱥ Ⱥ Ⱥ f ( x ) = Ⱥ P3 (δ 2 , δ 3 ,V2 ) Ⱥ ȺQ2 (δ 2 , δ 3 ,V2 )Ⱥ Ⱥ Ⱥ Ⱥ V 1.05 22.3 cos(2.03 − δ ) + V 2 58.1 cos(− 1.11) + V 1.04 35.8 cos(2.03 − δ + δ ) Ⱥ 21 2 2 2 3 Ⱥ 2 Ⱥ 2 = Ⱥ V3 1.05 31.6 cos(1.89 − δ 3 ) + 1.04 V2 35.8 cos(2.03 − δ 3 + δ 2 ) + 1.04 67.2 cos(− 1.17 )Ⱥ Ⱥ 2 Ⱥ − V2 1.05 22.3 sin (2.03 − δ 2 ) − V2 58.1 sin (− 1.11) − V2 1.04 35.8 sin (2.03 − δ 2 + δ 3 ) Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ P2sch Ⱥ Ⱥ P2 (δ 2 , δ 3 ,V2 ) Ⱥ Ⱥ− 4.0Ⱥ Ⱥ P2 (δ 2 , δ 3 ,V2 ) Ⱥ Ⱥ ΔP2 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Δc = Ⱥ ΔP3 Ⱥ = c − f ( x ) = Ⱥ P3sch Ⱥ − Ⱥ P3 (δ 2 , δ 3 ,V2 ) Ⱥ = Ⱥ 2.0 Ⱥ − Ⱥ P3 (δ 2 , δ 3 ,V2 ) Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ sch ȺQ2 Ⱥ ȺQ2 (δ 2 , δ 3 ,V2 )Ⱥ Ⱥ − 2.5Ⱥ ȺQ2 (δ 2 , δ 3 ,V2 )Ⱥ ȺΔQ2 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Power Systems I Example ∂P2 = ∂δ 2 3 ∑V 2 V j Y2 j sin (θ 2 j − δ 2 + δ j ) j =1, j ≠ 2 = V2 V1 Y21 sin(θ 21 − δ 2 ) + V2 V3 Y23 sin (θ 23 − δ 2 + δ 3 ) = V2 1.05 22.4 sin(2.03 − δ 2 ) + V2 1.04 35.8 sin (2.03 − δ 2 + δ 3 ) ∂P2 = − V2 V3 Y23 sin(θ 23 − δ 2 + δ 3 ) = − V2 1.04 35.8 sin(2.03 − δ 2 + δ 3 ) ∂δ 3 ∂P2 = 2 V2 Y22 cos(θ 22 ) + ∂ V2 3 ∑V j Y2 j cos(θ 2 j − δ 2 + δ j ) j =1, j ≠ 2 = 2 V2 Y22 cos(θ 22 ) + V2 Y21 cos(θ 21 − δ 2 + δ 1 ) + V2 Y23 cos(θ 23 − δ 2 + δ 3 ) = 2 V2 58.1 cos(2.03) + 1.05 22.4 cos(2.03 − δ 2 ) + 1.04 35.8 cos(2.03 − δ 2 + δ 3 ) Power Systems I Example ∂P3 = − V3 V2 Y32 sin (θ 32 − δ 3 + δ 2 ) = − 1.04 V2 35.8 sin (2.03 − δ 2 + δ 3 ) ∂δ 2 ∂P3 = ∂δ 3 3 ∑V 3 V j Y3 j sin (θ 3 j − δ 3 + δ j ) j =1, j ≠3 = V3 V1 Y31 sin (θ 31 − δ 3 + δ 1 ) + V3 V2 Y32 sin (θ 32 − δ 3 + δ 2 ) = 1.04 1.05 31.6 sin (1.89 − δ 3 ) + 1.04 V2 35.8 sin (2.03 − δ 3 + δ 2 ) ∂P3 = V3 Y32 cos(θ 32 − δ 3 + δ 2 ) = 1.04 35.8 cos(2.03 − δ 2 + δ 3 ) ∂ V2 Power Systems I Example ∂Q2 = ∂δ 2 3 ∑V 2 V j Y2 j cos(θ 2 j − δ 2 + δ j ) j =1, j ≠2 = V2 V1 Y21 cos(θ 21 − δ 2 + δ 1 ) + V2 V3 Y23 cos(θ 23 − δ 2 + δ 3 ) = V2 1.05 22.4 cos(2.03 − δ 2 ) + V2 1.04 35.8 cos(2.03 − δ 2 + δ 3 ) ∂Q2 = − V2 V3 Y23 cos(θ 23 − δ 2 + δ 3 ) = − V2 1.04 35.8 cos(2.03 − δ 2 + δ 3 ) ∂δ 3 ∂Q2 = −2 V2 Y22 sin (θ 22 ) − ∂ V2 3 ∑V j Y2 j sin (θ 2 j − δ 2 + δ j ) j =1, j ≠2 = −2 V2 Y22 sin (θ 22 ) − V1 Y21 sin (θ 21 − δ 2 + δ 1 ) − V3 Y23 sin (θ 23 − δ 2 + δ 3 ) = −2 V2 58.1 sin (− 1.11) − 1.05 22.4 sin (2.03 − δ 2 ) − 1.04 35.8 sin (2.03 − δ 2 + δ 3 ) Power Systems I Example x [ k +1] = x [ k ] + J −1 ⋅ Δc[ k ] Ⱥδ 2 Ⱥ Ⱥ Ⱥ = Ⱥδ 3 Ⱥ ȺV2 Ⱥ Ⱥ Ⱥ [ k +1] Ⱥδ 2 Ⱥ Ⱥ Ⱥ = Ⱥδ 3 Ⱥ ȺV2 Ⱥ Ⱥ Ⱥ Power Systems I [k ] Ⱥ ∂P2 ∂δ 2 + Ⱥ ∂P3 ∂δ 2 Ⱥ Ⱥ∂Q2 ∂δ 2 Ⱥ ∂P2 ∂δ 3 ∂P3 ∂δ 3 ∂Q2 ∂δ 3 −1 ∂P2 ∂V2 Ⱥ Ⱥ ΔP2 Ⱥ ∂P3 ∂V2 Ⱥ ⋅ Ⱥ ΔP3 Ⱥ Ⱥ Ⱥ Ⱥ ∂Q2 ∂V2 Ⱥ Ⱥ ΔQ2 Ⱥ Ⱥ Ⱥ Ⱥ [k ] Newton-Raphson PF Example Ⱥ P2sch Ⱥ Ⱥ P2[ 0 ] Ⱥ Ⱥ − 4.0Ⱥ Ⱥ − 1.14 Ⱥ Ⱥ − 2.86Ⱥ Ⱥ0.0Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ x [ 0 ] = Ⱥ0.0Ⱥ Δc[ 0 ] = Ⱥ P3sch Ⱥ − Ⱥ P3[ 0 ] Ⱥ = Ⱥ 2.0 Ⱥ − Ⱥ 0.562 Ⱥ = Ⱥ 1.438 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ sch [0] ȺQ2 Ⱥ ȺQ2 Ⱥ Ⱥ − 2.5Ⱥ Ⱥ− 2.28Ⱥ Ⱥ − 0.22Ⱥ Ⱥ1.0 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Δx[ 0 ] = J −1Δc[ 0 ] Δx [ 0 ] x [1] [ Ⱥ Δδ 2 0 ] Ⱥ Ⱥ 54.28 − 33.28 24.86 Ⱥ −1 Ⱥ− 2.86Ⱥ Ⱥ − 0.04526Ⱥ Ⱥ [ 0 ] Ⱥ Ⱥ = Ⱥ Δδ 3 Ⱥ = − 33.28 66.04 − 16.64Ⱥ Ⱥ 1.438 Ⱥ = Ⱥ − 0.00772Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ [ 0 ] Ⱥ Ⱥ Δ V2 49.72 Ⱥ Ⱥ− 0.22Ⱥ Ⱥ − 0.02655Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ− 27.14 16.64 [ Ⱥ δ 21] Ⱥ Ⱥ0.0 + (− 0.04526 )Ⱥ Ⱥ − 0.04526Ⱥ Ⱥ Ⱥ = Ⱥ δ 3[1] Ⱥ = Ⱥ0.0 + (− 0.00772 )Ⱥ = Ⱥ − 0.00772Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ [1] Ⱥ Ⱥ V2 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ1.0 + (− 0.02655)Ⱥ Ⱥ 0.9734 Ⱥ Power Systems I Newton-Raphson PF Example x [1] Ⱥ P2sch Ⱥ Ⱥ P2[1] Ⱥ Ⱥ − 4.0Ⱥ Ⱥ − 3.901Ⱥ Ⱥ− 0.099Ⱥ Ⱥ − 0.04526Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ = Ⱥ− 0.00772Ⱥ Δc[1] = Ⱥ P3sch Ⱥ − Ⱥ P3[1] Ⱥ = Ⱥ 2.0 Ⱥ − Ⱥ 1.978 Ⱥ = Ⱥ 0.0217 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ sch [1] ȺQ2 Ⱥ ȺQ2 Ⱥ Ⱥ − 2.5Ⱥ Ⱥ− 2.449Ⱥ Ⱥ − 0.051Ⱥ Ⱥ 0.9734 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Δx [1] Ⱥ 51.72 − 31.77 21.30 Ⱥ = Ⱥ − 32.98 65.66 − 15.38Ⱥ Ⱥ Ⱥ 48.10 Ⱥ Ⱥ− 28.54 17.40 Ⱥ Ⱥ −1 Ⱥ− 0.099Ⱥ Ⱥ − 0.001795Ⱥ Ⱥ 0.0217 Ⱥ = Ⱥ − 0.000985Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ − 0.051Ⱥ Ⱥ − 0.001767Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ [ Ⱥ δ 2 2 ] Ⱥ Ⱥ − 0.04526 + ( −0.001795)Ⱥ Ⱥ− 0.04706Ⱥ Ⱥ [ 2 ] Ⱥ Ⱥ [2] x = Ⱥ δ 3 Ⱥ = − 0.00772 + ( −0.000985)Ⱥ = Ⱥ− 0.00870Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ [ 2 ] Ⱥ Ⱥ V2 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ 0.9734 + ( −0.001767) Ⱥ Ⱥ 0.9717 Ⱥ Power Systems I Newton-Raphson PF Example x [2] Ⱥ P2sch Ⱥ Ⱥ P2[1] Ⱥ Ⱥ − 4.0Ⱥ Ⱥ − 3.999Ⱥ Ⱥ − 0.0002Ⱥ Ⱥ− 0.04706Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ = Ⱥ− 0.00870Ⱥ Δc[ 2 ] = Ⱥ P3sch Ⱥ − Ⱥ P3[1] Ⱥ = Ⱥ 2.0 Ⱥ − Ⱥ 1.999 Ⱥ = Ⱥ 0.00004 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ sch [1] ȺQ2 Ⱥ ȺQ2 Ⱥ Ⱥ − 2.5Ⱥ Ⱥ− 2.499Ⱥ Ⱥ − 0.0001Ⱥ Ⱥ 0.9717 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Δx [ 2 ] x [ 3] Ⱥ 51.60 − 31.69 21.14 Ⱥ = Ⱥ − 32.93 65.60 − 15.35Ⱥ Ⱥ Ⱥ Ⱥ− 28.55 17.40 47.95 Ⱥ Ⱥ Ⱥ −1 Ⱥ− 0.000216Ⱥ Ⱥ − 0.000038Ⱥ Ⱥ 0.000038 Ⱥ = Ⱥ − 0.000002Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ − 0.000143Ⱥ Ⱥ − 0.000004Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ [ Ⱥ δ 23] Ⱥ Ⱥ − 0.04706 + ( −0.000038) Ⱥ Ⱥ − 0.04706 Ⱥ Ⱥ Ⱥ = Ⱥ δ 3[ 3] Ⱥ = Ⱥ− 0.00870 + ( −0.000002)Ⱥ = Ⱥ − 0.008705Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ [ 3] Ⱥ Ⱥ V2 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ 0.9717 + ( −0.000004) Ⱥ Ⱥ 0.97168 Ⱥ Power Systems I Newton-Raphson PF Example Ⱥ P2sch Ⱥ Ⱥ P2[1] Ⱥ Ⱥ− 4.0Ⱥ Ⱥ− 4.0Ⱥ Ⱥ0.0000Ⱥ Ⱥ − 0.04706 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ x [ 3] = Ⱥ− 0.008705Ⱥ Δc[ 2 ] = Ⱥ P3sch Ⱥ − Ⱥ P3[1] Ⱥ = Ⱥ 2.0 Ⱥ − Ⱥ 2.0 Ⱥ = Ⱥ0.0000Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ sch [1] ȺQ2 Ⱥ ȺQ2 Ⱥ Ⱥ − 2.5Ⱥ Ⱥ − 2.5Ⱥ Ⱥ0.0000Ⱥ Ⱥ 0.97168 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ ε max = 2.5 × 10 −4 2 P = V1 Y11 cos(θ11 ) + V1 V2 Y12 cos(θ12 − δ 1 + δ 2 ) + V1 V3 Y13 cos(θ13 − δ 1 + δ 3 ) 1 2 Q1 = − V1 Y11 sin (θ11 ) − V1 V2 Y12 sin (θ12 − δ 1 + δ 2 ) − V1 V3 Y13 sin (θ13 − δ 1 + δ 3 ) 2 Q3 = − V3 V1 Y31 sin (θ 31 − δ 3 + δ 1 ) − V3 V2 Y32 sin (θ 32 − δ 3 + δ 2 ) − V3 Y33 sin (θ 33 ) P = 2.1842 pu 1 Q1 = 1.4085 pu Q3 = 1.4617 pu Power Systems I Fast Decoupled Power Flow Transmission lines and transformers have high X/R ratios Real power change, ΔP Reactive power changes, ΔQ is less sensitive to changes in the voltage magnitude, Δ|V| is more sensitive to changes in the phase angle, Δδ is less sensitive to changes in the phase angle , Δδ is more sensitive to changes in the voltage magnitude, Δ|V| Jacobian submatrices JQd and JPV tend to be much smaller in magnitude compared to JPd and JQV Jacobian submatrices JQd and JPV can be set to zero ΔP = J Pδ ⋅ Δδ = ∂P ∂δ Δδ ΔP Ⱥ = Ⱥ J Pδ 0 Ⱥ Ⱥ Δδ Ⱥ Ⱥ Ⱥ ΔQ Ⱥ Ⱥ 0 J QV Ⱥ ȺΔ V Ⱥ ΔQ = J QV ⋅ Δ V = ∂Q ∂ V Δ V Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Power Systems I Fast Decoupled Power Flow JPV elements ∂Pi = Vi Yij cos(θ ij − δ i + δ j ) ∂Vj θ ij ≈ 90° δ i ≈ δ j ∂Pi ≈ Vi Yij cos(90°) = 0.0 ∂Vj JQd elements ∂Qi = − Vi V j Yij cos(θ ij − δ i + δ j ) ∂δ j θ ij ≈ 90° δ i ≈ δ j ∂Qi ≈ − Vi V j Yij cos(90° ) = 0.0 ∂δ j Power Systems I Fast Decoupled Power Flow The matrix equation is separated into two decoupled equations requires considerably less time to solve compared to the full Newton-Raphson method JPd and JQV submatrices can be further simplified to eliminate the need for recomputing of the submatrices during each iteration some terms in each element are relatively small and can be eliminated the remaining equations consist of constant terms and one variable term the one variable term can be moved and coupled with the change in power variable the result is a Jacobian matrix with constant term elements Power Systems I Jacobian JPd Diagonal Terms n ∂Pi = ∑ Vi V j Yij sin(θ ij − δ i + δ j ) ∂δ i j =1 j ≠i n = − Vi Yii sin(θ ii ) + ∑ Vi V j Yij sin (θ ij − δ i + δ j ) 2 j =1 ∂Pi 2 = − Vi Yii sin(θ ii ) − Qi ∂δ i Yii sin(θ ii ) = Bii 2 Vi ≈ Vi Power Systems I ⇒ Bii >> Qi n Qi = − ∑ Vi V j Yij sin(θ ij − δ i + δ j ) j =1 ∂Pi 2 = − Vi Bii ∂δ i ∂Pi = − Vi Bii ∂δ i Jacobian JPd Off-diagonal Terms ∂Pi = − Vi V j Yij sin(θ ij − δ i + δ j ) ∂δ j δ j − δi ≈ 0 ∂Pi = − Vi V j Yij sin(θ ij ) ∂δ i Yij sin(θ ij ) = Bij ∂Pi = − Vi Bij ∂δ i Power Systems I Vj ≈ 1 Jacobian JQV Diagonal Terms n ∂Qi = −2 Vi Yii sin(θii ) − ∑ V j Yij sin (θ ij − δ i + δ j ) ∂ Vi j =1 j ≠i n ∂Qi −1 = − Vi Yii sin(θ ii ) − Vi ∑ Vi V j Yij sin (θ ij − δ i + δ j ) ∂ Vi j =1 ∂Qi −1 = − Vi Yii sin(θ ii ) + Vi Qi ∂ Vi Yii sin(θ ii ) = Bii Power Systems I Bii >> Qi ⇒ n Qi = − ∑ Vi V j Yij sin (θij − δ i + δ j ) j =1 ∂Qi = − Vi Bii ∂ Vi Jacobian JQV Off-diagonal Terms ∂Qi = − Vi Yij sin(θ ij − δ i + δ j ) ∂Vj δ j − δi ≈ 0 ∂Qi = − Vi Yij sin(θ ij ) ∂Vj Yij sin(θ ij ) = Bij ∂Qi = − Vi Bij ∂Vj Power Systems I Fast Decoupled Power Flow Individual power change equations in JPd and JQV n ΔPi = ∑ − Vi Bij Δδ j ⇒ n ΔPi = ∑ − Bij Δδ j Vi j =1 ⇒ n ΔQi = ∑ − Bij Δ V j Vi j =1 j =1 n ΔQi = ∑ − Vi Bij Δ V j j =1 Matrix equation for JPd and JQV ΔP = − B+ Δδ Vi ΔQ = − B+ + Δ V Vi Power Systems I −1 ⇒ Δδ = −[B+ ΔP V −1 ⇒ Δ V = −[B+ + ΔQ V Example Using the fast decoupled PF, find the power flow solution 1 0.01 + j0.03 y12 = 10 − j 20 pu y13 = 10 − j 30 pu y23 = 16 − j 32 pu Power Systems I 0.0125 + j0.025 0.02 + j0.04 2 3 |V3| = 1.04 200 MW 400 + j 250 S =− = −4.0 − j 2.5 pu 100 200 P3sch = = 2.0 pu 100 sch 2 Slack Bus V1 = 1.05∠0° 400 MW 250 MVAR Example Ⱥ− 52 32 Ⱥ B+ = Ⱥ 32 − 62Ⱥ Ⱥ Ⱥ Ⱥ− 0.028182 − 0.014545Ⱥ −1 [B+ = Ⱥ − 0.014545 − 0.023636Ⱥ Ⱥ Ⱥ B+ + = [− 52] [B+ +−1 = [− 0.019231] Power Systems I Example Initial values: V [ 0] First iteration: Ⱥ1.05∠0°Ⱥ = Ⱥ1.00∠0°Ⱥ Ⱥ Ⱥ Ⱥ1.00∠0°Ⱥ Ⱥ Ⱥ Ⱥ P2sch Ⱥ Ⱥ− 4.0Ⱥ Ⱥ Ⱥ y = Ⱥ P3sch Ⱥ = Ⱥ 2.0 Ⱥ Ⱥ Ⱥ ȺQ2sch Ⱥ Ⱥ− 2.5Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ x [k ] [ Ⱥδ 2k ] Ⱥ Ⱥ [ Ⱥ = Ⱥδ 3 k ] Ⱥ ȺV2[ k ] Ⱥ Ⱥ Ⱥ n x [0] Ⱥ0.0Ⱥ = Ⱥ0.0Ⱥ Ⱥ Ⱥ Ⱥ1.0 Ⱥ Ⱥ Ⱥ Ⱥ Pinj 2 (x ) Ⱥ Pinj i = ∑ Vi V j Yij cos(θ ij − δ i + δ j ) Ⱥ Ⱥ j =1 f ( x ) = Ⱥ Pinj 3 ( x ) Ⱥ n ȺQinj 2 ( x )Ⱥ Qinj i = −∑ Vi V j Yij sin (θ ij − δ i + δ j ) Ⱥ Ⱥ j =1 Power Systems I Example Ⱥ V2 2 Y22 cos(θ 22 ) + V2 V1 Y21 cos(θ 21 − δ 2 + δ 1 ) + V2 V3 Y23 cos(θ 23 − δ 2 + δ 3 ) Ⱥ Ⱥ 2 Ⱥ V Y cos(θ 33 ) + V3 V1 Y31 cos(θ 31 − δ 3 + δ 1 ) + V3 V2 Y32 cos(θ 32 − δ 3 + δ 2 ) Ⱥ f ( x ) = Ⱥ 3 2 33 Ⱥ− V Y sin (θ ) − V V Y sin (θ − δ + δ ) − V V Y sin (θ − δ + δ )Ⱥ 22 2 1 21 21 2 1 2 3 23 23 2 3 Ⱥ 2 22 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ V2 2 58.1 cos(− 1.11) + V2 1.05 22.4 cos(2.03 − δ 2 ) + V2 1.04 35.8 cos(2.03 − δ 2 + δ 3 ) Ⱥ Ⱥ Ⱥ 2 1.04 67.2 cos(− 1.17 ) + 1.04 1.05 31.6 cos(1.89 − δ 3 ) + 1.04 V2 35.8 cos(2.03 − δ 3 + δ 2 )Ⱥ = Ⱥ Ⱥ − V 2 58.1 sin (− 1.11) − V 1.05 22.4 sin (2.03 − δ ) − V 1.04 35.8 sin (2.03 − δ + δ ) Ⱥ 2 2 2 2 2 3 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Power Systems I Example Ⱥ P2sch Ⱥ Ⱥ P2[ 0] Ⱥ Ⱥ − 4.0Ⱥ Ⱥ − 1.14 Ⱥ Ⱥ− 2.86Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Δy [ 0 ] = Ⱥ P3sch Ⱥ − Ⱥ P3[ 0] Ⱥ = Ⱥ 2.0 Ⱥ − Ⱥ 0.562 Ⱥ = Ⱥ 1.438 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ sch [ 0] ȺQ2 Ⱥ ȺQ2 Ⱥ Ⱥ − 2.5Ⱥ Ⱥ− 2.28Ⱥ Ⱥ− 0.22Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ [ ȺΔδ 20 ] Ⱥ Ⱥ0.028182 0.014545Ⱥ Ⱥ − 2.86 1.0 Ⱥ Ⱥ− 0.06048Ⱥ Ⱥ [ 0 ] Ⱥ = Ⱥ Ⱥ Ⱥ1.438 1.04Ⱥ = Ⱥ − 0.00891Ⱥ Ⱥ Ⱥ Ⱥ ȺΔδ 3 Ⱥ Ⱥ0.014545 0.023636Ⱥ Ⱥ [Δ V ] = [0.019231] [− 0.22 1.0] = [− 0.004231] [0] 2 [ δ 21] = 0.0 + (− 0.06048) = −0.06048 [ δ 31] = 0.0 + (− 0.00891) = −0.00891 V2[1] = 1.0 + (− 0.004231) = 0.995769 Power Systems I Example Remaining iterations: Iter 1 2 3 4 5 6 7 8 9 10 δ2 -0.060482 -0.056496 -0.044194 -0.044802 -0.047665 -0.047614 -0.046936 -0.046928 -0.047087 -0.047094 Power Systems I δ3 -0.008909 -0.007952 -0.008690 -0.008986 -0.008713 -0.008645 -0.008702 -0.008720 -0.008707 -0.008702 |V2| 0.995769 0.965274 0.965711 0.972985 0.973116 0.971414 0.971333 0.971732 0.971762 0.971669 Δ P2 -2.860000 0.175895 0.640309 -0.021395 -0.153368 0.000520 0.035980 0.000948 -0.008442 -0.000470 Δ P3 1.438400 -0.070951 -0.457039 0.001195 0.112899 0.002610 -0.026190 -0.001411 0.006133 0.000510 Δ Q2 -0.220000 -1.579042 0.021948 0.365249 0.006657 -0.086136 -0.004067 0.020119 0.001558 -0.004688 ...
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This note was uploaded on 02/01/2012 for the course EEL 4213 taught by Professor Thomasbaldwin during the Fall '11 term at FSU.

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