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- 1. ECE 422 Session 7; Page 1/19 Power Flow Solutions Power Flow Equations In the system shown at right line models to have a resistance of 0.01pu in addition to a series reactance of 0.1pu. Bus 1 Bus 3 Bus 2 Bus 4 All branches have same series impedance: pu 1 Zseries 0.01 j 0.1 pu Yseries 1 Zseries Yseries 0.99 9.901i ( ) pu All branches have same shunt admittance (note that this is the value at each end, so it has already been divided by 2). Zshunt j 10 pu Yshunt 1 Zshunt Yshunt 0.1i pu Matrix elements: y11 3 Yseries 3Yshunt y33 2 Yseries 2 Yshunt y22 2 Yseries 2Yshunt y44 3 Yseries 3Yshunt Ybus y11 Yseries Yseries Yseries Yseries y22 0 Yseries Yseries 0 y33 Yseries Yseries Yseries Yseries y44 Ybus 2.97 29.403i 0.99 9.901i 0.99 9.901i 0.99 9.901i 0.99 9.901i 1.98 19.602i 0 0.99 9.901i 0.99 9.901i 0 1.98 19.602i 0.99 9.901i 0.99 9.901i 0.99 9.901i 0.99 9.901i 2.97 29.403i pu
- 2. ECE 422 Session 7; Page 2/19 Reset origin for matrix and vector subscripts from 0 to 1. ORIGIN 1 Complex Form Equations: S1 V1 V1 Ybus 1 1 V1 V2 Ybus 1 2 V1 V3 Ybus 1 3 V1 V4 Ybus 1 4 = S2 V2 V1 Ybus 2 1 V2 V2 Ybus 2 2 V2 V4 Ybus 2 4 = S3 V3 V1 Ybus 3 1 V3 V3 Ybus 3 3 V3 V4 Ybus 3 4 = S4 V4 V1 Ybus 4 1 V4 V2 Ybus 4 2 V4 V3 Ybus 4 3 V4 V4 Ybus 4 4 = Find real and imaginary parts of the Ybus matrix: G Re Ybus G 2.97 0.99 0.99 0.99 0.99 1.98 0 0.99 0.99 0 1.98 0.99 0.99 0.99 0.99 2.97 B Im Ybus B 29.403 9.901 9.901 9.901 9.901 19.602 0 9.901 9.901 0 19.602 9.901 9.901 9.901 9.901 29.403
- 3. ECE 422 Session 7; Page 3/19 Rectangular Form P1 V1 2 G 1 1 V1 V2 G 1 2 cos θ1 θ2 ( ) B 1 2 sin θ1 θ2 ( ) V1 V3 G 1 3 cos θ1 θ3 ( ) B 1 3 sin θ1 θ3 ( ) V1 V4 G 1 4 cos θ1 θ4 ( ) B 1 4 sin θ1 θ4 ( ) = P2 V2 V1 G 2 1 cos θ2 θ1 ( ) B 2 1 sin θ2 θ1 ( ) V2 2 G 2 2 V2 V4 G 2 4 cos θ2 θ4 ( ) B 2 4 sin θ2 θ4 ( ) = P3 V3 V1 G 3 1 cos θ3 θ1 ( ) B 3 1 sin θ3 θ1 ( ) V3 2 G 3 3 V3 V4 G 3 4 cos θ3 θ4 ( ) B 3 4 sin θ3 θ4 ( ) = P4 V4 V1 G 4 1 cos θ4 θ1 ( ) B 4 1 sin θ4 θ1 ( ) V4 V2 G 4 2 cos θ4 θ2 ( ) B 4 2 sin θ4 θ2 ( ) V4 V3 G 4 3 cos θ4 θ3 ( ) B 4 3 sin θ4 θ3 ( ) V4 2 G 4 = Q1 V1 2 B 1 1 V1 V2 G 1 2 sin θ1 θ2 ( ) B 1 2 cos θ1 θ2 ( ) V1 V3 G 1 3 sin θ1 θ3 ( ) B 1 3 cos θ1 θ3 ( ) V1 V4 G 1 4 sin θ1 θ4 ( ) B 1 4 cos θ1 ( = Q2 V2 V1 G 2 1 sin θ2 θ1 ( ) B 2 1 cos θ2 θ1 ( ) V2 2 B 2 2 V2 V4 G 2 4 sin θ2 θ4 ( ) B 2 4 cos θ2 θ4 ( ) = Q3 V3 V1 G 3 1 sin θ3 θ1 ( ) B 3 1 cos θ3 θ1 ( ) V3 2 B 3 3 V3 V4 G 3 4 sin θ3 θ4 ( ) B 3 4 cos θ3 θ4 ( ) = Q4 V4 V1 G 4 1 sin θ4 θ1 ( ) B 4 1 cos θ4 θ1 ( ) V4 V2 G 4 2 sin θ4 θ2 ( ) B 4 2 cos θ4 θ2 ( ) V4 V3 G 4 3 sin θ4 θ3 ( ) B 4 3 cos θ4 θ3 ( ) V4 2 B 4 = You could plug the numbers for Gi,j and Bi,j into the equations next.
- 4. ECE 422 Session 7; Page 4/19 Newton-Raphson Example 1: Reset origin for matrix and vector subscripts from 1 back to 0. ORIGIN 0 Use a simpler Ybus Ybus j 19.98 j 10 j 10 j 10 j 19.98 j 10 j 10 j 10 j 19.98 P2 0.5 P3 1 Q2 0.5 Q3 1 Slack Bus: V1m 1.0 θ1 0 Initial Guesses: V2m0 1.0 θ20 0 V3m0 1.0 θ30 0 Power flow equations using initial guess: P20 V2m0 V1m Im Ybus 1 0 sin θ20 θ1 ( ) V2m0 V3m0 Im Ybus 1 2 sin θ20 θ30 ( ) P30 V3m0 V1m Im Ybus 2 0 sin θ30 θ1 ( ) V3m0 V2m0 Im Ybus 2 1 sin θ30 θ20 ( ) Q20 V2m0 V1m Im Ybus 1 0 cos θ20 θ1 ( ) V2m0 2 Im Ybus 1 1 V2m0 V3m0 Im Ybus 1 2 cos θ20 θ30 ( ) Q30 V3m0 V1m Im Ybus 2 0 cos θ30 θ1 ( ) V3m0 V2m0 Im Ybus 2 1 cos θ30 θ20 ( ) V3m0 2 Im Ybus 2 2 Initial Mismatch Vector: Use a "1-norm" ΔH0 P2 P20 P3 P30 Q2 Q20 Q3 Q30 ΔH0 0.5 1 0.52 0.98 norm_1 out 0 out out ΔH0 x x 0 3 for norm_1 3 well out of tolerance
- 5. ECE 422 Session 7; Page 5/19 Jacobian Terms J11 submatrix JP2θ2_0 Im Ybus 1 0 V2m0 V1m cos θ20 θ1 ( ) Im Ybus 1 2 V2m0 V3m0 cos θ20 θ30 ( ) JP2θ3_0 Im Ybus 1 2 V2m0 V3m0 cos θ20 θ30 ( ) JP3θ2_0 Im Ybus 2 1 V3m0 V2m0 cos θ30 θ20 ( ) JP3θ3_0 Im Ybus 2 0 V3m0 V1m cos θ30 θ1 ( ) Im Ybus 2 1 V3m0 V2m0 cos θ30 θ20 ( ) J12 submatrix JP2Vm2_0 Im Ybus 1 0 V1m sin θ20 θ1 ( ) Im Ybus 1 2 V3m0 sin θ20 θ30 ( ) JP2Vm3_0 Im Ybus 1 2 V2m0 sin θ20 θ30 ( ) JP3Vm2_0 Im Ybus 2 1 V3m0 sin θ30 θ20 ( ) JP3Vm3_0 Im Ybus 2 0 V1m sin θ30 θ1 ( ) Im Ybus 2 1 V2m0 sin θ30 θ20 ( ) J21 submatrix JQ2θ2_0 Im Ybus 1 0 V2m0 V1m sin θ20 θ1 ( ) Im Ybus 1 2 V2m0 V3m0 sin θ20 θ30 ( ) JQ2θ3_0 Im Ybus 1 2 V2m0 V3m0 sin θ20 θ30 ( ) JQ3θ2_0 Im Ybus 2 1 V3m0 V2m0 sin θ30 θ20 ( ) JQ3θ3_0 Im Ybus 2 0 V3m0 V1m sin θ30 θ1 ( ) Im Ybus 2 1 V3m0 V2m0 sin θ30 θ20 ( ) J22 submatrix JQ2Vm2_0 Im Ybus 1 0 V1m cos θ20 θ1 ( ) 2 V2m0 Im Ybus 1 1 Im Ybus 1 2 V3m0 cos θ20 θ30 ( ) JQ2Vm3_0 Im Ybus 1 2 V2m0 cos θ20 θ30 ( ) JQ3Vm2_0 Im Ybus 2 1 V3m0 cos θ30 θ20 ( ) JQ3Vm3_0 Im Ybus 2 0 V1m cos θ30 θ1 ( ) Im Ybus 2 1 V2m0 cos θ30 θ20 ( ) 2 V3m0 Im Ybus 2 2
- 6. ECE 422 Session 7; Page 6/19 J_0 JP2θ2_0 JP3θ2_0 JQ2θ2_0 JQ3θ2_0 JP2θ3_0 JP3θ3_0 JQ2θ3_0 JQ3θ3_0 JP2Vm2_0 JP3Vm2_0 JQ2Vm2_0 JQ3Vm2_0 JP2Vm3_0 JP3Vm3_0 JQ2Vm3_0 JQ3Vm3_0 J_0 20 10 0 0 10 20 0 0 0 0 19.96 10 0 0 10 19.96 Now solve for x (note, I'm using the built-in matrix inverse, but for a large case, one would use LU factorization Δx1 J_0 1 ΔH0 Δx1 0 0.05 1.941 10 3 0.048 θ21 θ20 Δx1 0 θ21 0 deg θ31 θ30 Δx1 1 θ31 2.865 deg V2m1 V2m0 Δx1 2 V2m1 1.002 V3m1 V3m0 Δx1 3 V3m1 0.952 Power flow equations using result of iteration 1: P21 V2m1 V1m Im Ybus 1 0 sin θ21 θ1 ( ) V2m1 V3m1 Im Ybus 1 2 sin θ21 θ31 ( ) P31 V3m1 V1m Im Ybus 2 0 sin θ31 θ1 ( ) V3m1 V2m1 Im Ybus 2 1 sin θ31 θ21 ( ) Q21 V2m1 V1m Im Ybus 1 0 cos θ21 θ1 ( ) V2m1 2 Im Ybus 1 1 V2m1 V3m1 Im Ybus 1 2 cos θ21 θ31 ( ) Q31 V3m1 V1m Im Ybus 2 0 cos θ31 θ1 ( ) V3m1 V2m1 Im Ybus 2 1 cos θ31 θ21 ( ) V3m1 2 Im Ybus 2 2
- 7. ECE 422 Session 7; Page 7/19 Updated Mismatch Vector: Use a "1-norm" ΔH1 P2 P21 P3 P31 Q2 Q21 Q3 Q31 ΔH1 0.023 0.048 0.013 0.071 norm_11_NR out 0 out out ΔH1 x x 0 3 for good improvement norm_11_NR 0.155 Jacobian Terms J11 submatrix JP2θ2_1 Im Ybus 1 0 V2m1 V1m cos θ21 θ1 ( ) Im Ybus 1 2 V2m1 V3m1 cos θ21 θ31 ( ) JP2θ3_1 Im Ybus 1 2 V2m1 V3m1 cos θ21 θ31 ( ) JP3θ2_1 Im Ybus 2 1 V3m1 V2m1 cos θ31 θ21 ( ) JP3θ3_1 Im Ybus 2 0 V3m1 V1m cos θ31 θ1 ( ) Im Ybus 2 1 V3m1 V2m1 cos θ31 θ21 ( ) J12 submatrix JP2Vm2_1 Im Ybus 1 0 V1m sin θ21 θ1 ( ) Im Ybus 1 2 V3m1 sin θ21 θ31 ( ) JP2Vm3_1 Im Ybus 1 2 V2m1 sin θ21 θ31 ( ) JP3Vm2_1 Im Ybus 2 1 V3m1 sin θ31 θ21 ( ) JP3Vm3_1 Im Ybus 2 0 V1m sin θ31 θ1 ( ) Im Ybus 2 1 V2m1 sin θ31 θ21 ( ) J21 submatrix JQ2θ2_1 Im Ybus 1 0 V2m1 V1m sin θ21 θ1 ( ) Im Ybus 1 2 V2m1 V3m1 sin θ21 θ31 ( ) JQ2θ3_1 Im Ybus 1 2 V2m1 V3m1 sin θ21 θ31 ( ) JQ3θ2_1 Im Ybus 2 1 V3m1 V2m1 sin θ31 θ21 ( ) JQ3θ3_1 Im Ybus 2 0 V3m1 V1m sin θ31 θ1 ( ) Im Ybus 2 1 V3m1 V2m1 sin θ31 θ21 ( )
- 8. ECE 422 Session 7; Page 8/19 J22 submatrix JQ2Vm2_1 Im Ybus 1 0 V1m cos θ21 θ1 ( ) 2 V2m1 Im Ybus 1 1 Im Ybus 1 2 V3m1 cos θ21 θ31 ( ) JQ2Vm3_1 Im Ybus 1 2 V2m1 cos θ21 θ31 ( ) JQ3Vm2_1 Im Ybus 2 1 V3m1 cos θ31 θ21 ( ) JQ3Vm3_1 Im Ybus 2 0 V1m cos θ31 θ1 ( ) Im Ybus 2 1 V2m1 cos θ31 θ21 ( ) 2 V3m1 Im Ybus 2 2 J_1 JP2θ2_1 JP3θ2_1 JQ2θ2_1 JQ3θ2_1 JP2θ3_1 JP3θ3_1 JQ2θ3_1 JQ3θ3_1 JP2Vm2_1 JP3Vm2_1 JQ2Vm2_1 JQ3Vm2_1 JP2Vm3_1 JP3Vm3_1 JQ2Vm3_1 JQ3Vm3_1 J_1 19.545 9.525 0.477 0.477 9.525 19.032 0.477 0.952 0.476 0.476 20.531 9.507 0.501 1.001 10.007 18.043 Now solve for x Δx2 J_1 1 ΔH1 Δx2 2.542 10 5 2.894 10 3 3.621 10 3 5.998 10 3 θ22 θ21 Δx2 0 θ22 1.457 10 3 deg θ32 θ31 Δx2 1 θ32 3.031 deg V2m2 V2m1 Δx2 2 V2m2 0.998 V3m2 V3m1 Δx2 3 V3m2 0.946
- 9. ECE 422 Session 7; Page 9/19 Power flow equations using result of iteration 2: P22 V2m2 V1m Im Ybus 1 0 sin θ22 θ1 ( ) V2m2 V3m2 Im Ybus 1 2 sin θ22 θ32 ( ) P32 V3m2 V1m Im Ybus 2 0 sin θ32 θ1 ( ) V3m2 V2m2 Im Ybus 2 1 sin θ32 θ22 ( ) Q22 V2m2 V1m Im Ybus 1 0 cos θ22 θ1 ( ) V2m2 2 Im Ybus 1 1 V2m2 V3m2 Im Ybus 1 2 cos θ22 θ32 ( ) Q32 V3m2 V1m Im Ybus 2 0 cos θ32 θ1 ( ) V3m2 V2m2 Im Ybus 2 1 cos θ32 θ22 ( ) V3m2 2 Im Ybus 2 2 Updated Mismatch Vector: Use a "1-norm" ΔH2 P2 P22 P3 P32 Q2 Q22 Q3 Q32 ΔH2 2.672 10 4 4.417 10 4 7.148 10 5 5.592 10 4 norm_12_FD out 0 out out ΔH2 x x 0 3 for norm_12_FD 1.34 10 3 close to a desirable tolerance Find P and Q at slack bus: P1 V1m V2m2 Im Ybus 0 1 sin θ1 θ22 ( ) V1m V3m2 Im Ybus 0 2 sin θ1 θ32 ( ) P1 0.5 P1 P2 1 Q1 V1m 2 Im Ybus 0 0 V2m2 Im Ybus 0 1 cos θ1 θ22 ( ) V1m V3m2 Im Ybus 0 2 cos θ1 θ32 ( ) Q1 0.551 Q1 Q22 1.051
- 10. ECE 422 Session 7; Page 10/19 Newton-Raphson Example 2: Now suppose that we were given that the voltage at bus 2 is know (voltage regulated bus): V2m 1.02 Now V2m is no longer an unknown, which reduces the size of the problem to solve Power flow equations using initial guess: P20 V2m V1m Im Ybus 1 0 sin θ20 θ1 ( ) V2m V3m0 Im Ybus 1 2 sin θ20 θ30 ( ) P30 V3m0 V1m Im Ybus 2 0 sin θ30 θ1 ( ) V3m0 V2m Im Ybus 2 1 sin θ30 θ20 ( ) Q2 equation is no longer needed here Q30 V3m0 V1m Im Ybus 2 0 cos θ30 θ1 ( ) V3m0 V2m Im Ybus 2 1 cos θ30 θ20 ( ) V3m0 2 Im Ybus 2 2 Initial Mismatch Vector: Use a "1-norm" ΔH0 P2 P20 P3 P30 Q3 Q30 ΔH0 0.5 1 0.78 norm_1 out 0 out out ΔH0 x x 0 2 for norm_1 2.28 well out of tolerance Jacobian Terms J11 submatrix JP2θ2_0 Im Ybus 1 0 V2m V1m cos θ20 θ1 ( ) Im Ybus 1 2 V2m V3m0 cos θ20 θ30 ( ) JP2θ3_0 Im Ybus 1 2 V2m V3m0 cos θ20 θ30 ( ) JP3θ2_0 Im Ybus 2 1 V3m0 V2m cos θ30 θ20 ( ) JP3θ3_0 Im Ybus 2 0 V3m0 V1m cos θ30 θ1 ( ) Im Ybus 2 1 V3m0 V2m cos θ30 θ20 ( )
- 11. ECE 422 Session 7; Page 11/19 J12 submatrix (no longer have partials with respect to V2m) JP2Vm3_0 Im Ybus 1 2 V2m sin θ20 θ30 ( ) JP3Vm3_0 Im Ybus 2 0 V1m sin θ30 θ1 ( ) Im Ybus 2 1 V2m sin θ30 θ20 ( ) J21 submatrix (no longer have Q2 terms) JQ3θ2_0 Im Ybus 2 1 V3m0 V2m sin θ30 θ20 ( ) JQ3θ3_0 Im Ybus 2 0 V3m0 V1m sin θ30 θ1 ( ) Im Ybus 2 1 V3m0 V2m sin θ30 θ20 ( ) J22 submatrix JQ2Vm3_0 Im Ybus 1 2 V2m cos θ20 θ30 ( ) JQ3Vm3_0 Im Ybus 2 0 V1m cos θ30 θ1 ( ) Im Ybus 2 1 V2m cos θ30 θ20 ( ) 2 V3m0 Im Ybus 2 2 J_0 JP2θ2_0 JP3θ2_0 JQ3θ2_0 JP2θ3_0 JP3θ3_0 JQ3θ3_0 JP2Vm3_0 JP3Vm3_0 JQ3Vm3_0 J_0 20.4 10.2 0 10.2 20.2 0 0 0 19.76 Now solve for x (note, I'm using the built-in matrix inverse, but for a large case, we would use LU factorization Δx1 J_0 1 ΔH0 Δx1 3.246 10 4 0.05 0.039 θ21 θ20 Δx1 0 θ21 0.019 deg θ31 θ30 Δx1 1 θ31 2.846 deg V3m1 V3m0 Δx1 2 V3m1 0.961
- 12. ECE 422 Session 7; Page 12/19 Power flow equations using result of iteration 1: P21 V2m V1m Im Ybus 1 0 sin θ21 θ1 ( ) V2m V3m1 Im Ybus 1 2 sin θ21 θ31 ( ) P31 V3m1 V1m Im Ybus 2 0 sin θ31 θ1 ( ) V3m1 V2m Im Ybus 2 1 sin θ31 θ21 ( ) Q31 V3m1 V1m Im Ybus 2 0 cos θ31 θ1 ( ) V3m1 V2m Im Ybus 2 1 cos θ31 θ21 ( ) V3m1 2 Im Ybus 2 2 Updated Mismatch Vector: Use a "1-norm" ΔH1 P2 P21 P3 P31 Q3 Q31 ΔH1 0.02 0.04 0.055 norm_1 out 0 out out ΔH1 x x 0 2 for norm_1 0.115 Jacobian Terms J11 submatrix JP2θ2_1 Im Ybus 1 0 V2m V1m cos θ21 θ1 ( ) Im Ybus 1 2 V2m V3m1 cos θ21 θ31 ( ) JP2θ3_1 Im Ybus 1 2 V2m V3m1 cos θ21 θ31 ( ) JP3θ2_1 Im Ybus 2 1 V3m1 V2m cos θ31 θ21 ( ) JP3θ3_1 Im Ybus 2 0 V3m1 V1m cos θ31 θ1 ( ) Im Ybus 2 1 V3m1 V2m cos θ31 θ21 ( ) J12 submatrix JP2Vm3_1 Im Ybus 1 2 V2m sin θ21 θ31 ( ) JP3Vm3_1 Im Ybus 2 0 V1m sin θ31 θ1 ( ) Im Ybus 2 1 V2m sin θ31 θ21 ( ) J21 submatrix JQ3θ2_1 Im Ybus 2 1 V3m1 V2m sin θ31 θ21 ( ) JQ3θ3_1 Im Ybus 2 0 V3m1 V1m sin θ31 θ1 ( ) Im Ybus 2 1 V3m1 V2m sin θ31 θ21 ( ) J22 submatrix JQ3Vm3_1 Im Ybus 2 0 V1m cos θ31 θ1 ( ) Im Ybus 2 1 V2m cos θ31 θ21 ( ) 2 V3m1 Im Ybus 2 2
- 13. ECE 422 Session 7; Page 13/19 J_1 JP2θ2_1 JP3θ2_1 JQ3θ2_1 JP2θ3_1 JP3θ3_1 JQ3θ3_1 JP2Vm3_1 JP3Vm3_1 JQ3Vm3_1 J_1 19.985 9.785 0.483 9.785 19.379 0.96 0.503 1 18.207 Now solve for x (note, I'm using the built-in matrix inverse, but for a large case, we would use LU factorization Δx2 J_1 1 ΔH1 Δx2 4.782 10 6 2.221 10 3 3.132 10 3 θ22 θ21 Δx2 0 θ22 0.019 deg θ32 θ31 Δx2 1 θ32 2.973 deg V3m2 V3m1 Δx2 2 V3m2 0.957 Power flow equations using result of iteration 2: P22 V2m V1m Im Ybus 1 0 sin θ22 θ1 ( ) V2m V3m2 Im Ybus 1 2 sin θ22 θ32 ( ) P32 V3m2 V1m Im Ybus 2 0 sin θ32 θ1 ( ) V3m2 V2m Im Ybus 2 1 sin θ32 θ22 ( ) Q32 V3m2 V1m Im Ybus 2 0 cos θ32 θ1 ( ) V3m2 V2m Im Ybus 2 1 cos θ32 θ22 ( ) V3m2 2 Im Ybus 2 2 Updated Mismatch Vector: Use a "1-norm" ΔH2 P2 P22 P3 P32 Q3 Q32 ΔH2 7.193 10 5 1.426 10 4 2.366 10 4 norm_1 out 0 out out ΔH2 x x 0 2 for norm_1 4.512 10 4 close to a desirable tolerance
- 14. ECE 422 Session 7; Page 14/19 Find P and Q at slack bus: P1 V1m V2m2 Im Ybus 0 1 sin θ1 θ22 ( ) V1m V3m2 Im Ybus 0 2 sin θ1 θ32 ( ) P1 0.5 P1 P2 1 Q1 V1m 2 Im Ybus 0 0 V2m2 Im Ybus 0 1 cos θ1 θ22 ( ) V1m V3m2 Im Ybus 0 2 cos θ1 θ32 ( ) Q1 0.436 and now find Q2 Q2 V2m V1m Im Ybus 1 0 cos θ22 θ1 ( ) V2m 2 Im Ybus 1 1 V2m V3m2 Im Ybus 1 2 cos θ22 θ32 ( ) Q2 0.835 Repeat Using Fast Decoupled Load Flow for Example 1 Start from the same initial guesses, the same power flow equations and the same mismatch vector as the first example above. Initial Mismatch Vector: Use a "1-norm" ΔH0 P2 P20 P3 P30 Q2 Q20 Q3 Q30 ΔH0 0.5 1 0.855 0.78 norm_1 out 0 out out ΔH0 x x 0 3 for norm_1 3.135 Well out of toleranec Or using the 2-norm norm_2sq out 0 out out ΔH0 x 2 x 0 3 for norm_2 norm_2sq norm_2 1.609 well out of tolerance
- 15. ECE 422 Session 7; Page 15/19 Or we could also evaluate using the infinity norm: norm_infinity max ΔH0 norm_infinity 1 Fast Decoupled Load Flow Set Up 1: B1 for the P equations (imaginary part of Ybus with slack bus row/column 1 removed): B1 Im submatrix Ybus 1 2 1 2 ( ) ( ) B1 19.98 10 10 19.98 B2 for Q equations (same dimension, since the same number of unknown voltages and unknown angles) B2 Im submatrix Ybus 1 2 1 2 ( ) ( ) B2 19.98 10 10 19.98 Now modify mismatch vector by dividing by voltage magnitudes.... ΔH0mod P2 P20 ( ) V2m0 P3 P30 ( ) V3m0 Q2 Q20 ( ) V2m0 Q3 Q30 ( ) V3m0 ΔH0mod 0.5 1 0.855 0.78 Now break the mismatch into P and Q equations ΔP0 submatrix ΔH0mod 0 1 0 0 ( ) ΔP0 0.5 1
- 16. ECE 422 Session 7; Page 16/19 ΔQ0 submatrix ΔH0mod 2 3 0 0 ( ) ΔQ0 0.855 0.78 Δθ1 B1 1 ΔP0 Δθ1 1.915 10 3 2.8686 deg ΔVm1 B2 1 ΔQ0 ΔVm1 0.031 0.0235 θ21 θ20 Δθ1 0 θ21 1.915 10 3 deg θ31 θ30 Δθ1 1 θ31 2.869 deg V2m1 V2m0 ΔVm1 0 V2m1 1.031 V3m1 V3m0 ΔVm1 1 V3m1 0.976 Power flow equations using result of iteration 1: P21 V2m1 V1m Im Ybus 1 0 sin θ21 θ1 ( ) V2m1 V3m1 Im Ybus 1 2 sin θ21 θ31 ( ) P31 V3m1 V1m Im Ybus 2 0 sin θ31 θ1 ( ) V3m1 V2m1 Im Ybus 2 1 sin θ31 θ21 ( ) Q21 V2m1 V1m Im Ybus 1 0 cos θ21 θ1 ( ) V2m1 2 Im Ybus 1 1 V2m1 V3m1 Im Ybus 1 2 cos θ21 θ31 ( ) Q31 V3m1 V1m Im Ybus 2 0 cos θ31 θ1 ( ) V3m1 V2m1 Im Ybus 2 1 cos θ31 θ21 ( ) V3m1 2 Im Ybus 2 2
- 17. ECE 422 Session 7; Page 17/19 Updated Mismatch Vector: Use a "1-norm" ΔH1fd P2 P21 P3 P31 Q2 Q21 Q3 Q31 ΔH1fd 3.162 10 3 7.805 10 3 0.038 0.244 norm_11_fd out 0 out out ΔH1fd x x 0 3 for norm_11_fd 0.293 good improvement For comparison, after 1 iteration on the same system Newton-Raphson had: norm_11_NR 0.155 Faster improvement Now do a second iteration ΔH1modfd P2 P21 ( ) V2m1 P3 P31 ( ) V3m1 Q2 Q21 ( ) V2m1 Q3 Q31 ( ) V3m1 ΔH1modfd 3.066 10 3 7.993 10 3 0.037 0.25 Now break the mismatch into P and Q equations ΔP1 submatrix ΔH1modfd 0 1 0 0 ΔP1 3.066 10 3 7.993 10 3 ΔQ1 submatrix ΔH1modfd 2 3 0 0 ΔQ1 0.037 0.25
- 18. ECE 422 Session 7; Page 18/19 Δθ2 B1 1 ΔP1 Δθ2 0.027 0.0365 deg ΔVm2 B2 1 ΔQ1 ΔVm2 0.0108 0.0179 θ22 θ21 Δθ2 0 θ22 0.029 deg θ32 θ31 Δθ2 1 θ32 2.905 deg V2m2 V2m0 ΔVm1 0 V2m2 1.031 V3m2 V3m0 ΔVm1 1 V3m2 0.976 Power flow equations using result of iteration 2: P22 V2m2 V1m Im Ybus 1 0 sin θ22 θ1 ( ) V2m2 V3m2 Im Ybus 1 2 sin θ22 θ32 ( ) P32 V3m2 V1m Im Ybus 2 0 sin θ32 θ1 ( ) V3m2 V2m2 Im Ybus 2 1 sin θ32 θ22 ( ) Q22 V2m2 V1m Im Ybus 1 0 cos θ22 θ1 ( ) V2m2 2 Im Ybus 1 1 V2m2 V3m2 Im Ybus 1 2 cos θ22 θ32 ( ) Q32 V3m2 V1m Im Ybus 2 0 cos θ32 θ1 ( ) V3m2 V2m2 Im Ybus 2 1 cos θ32 θ22 ( ) V3m2 2 Im Ybus 2 2 Updated Mismatch Vector: Use a "1-norm" ΔH2fd P2 P22 P3 P32 Q2 Q22 Q3 Q32 ΔH2fd 5.166 10 5 5.223 10 5 0.039 0.244 norm_12_fd out 0 out out ΔH2fd x x 0 3 for norm_12_fd 0.283 close to a desirable tolerance
- 19. ECE 422 Session 7; Page 19/19 Compare to after iteration 1: norm_11_fd 0.293 Notice that we have a small change Compare to after iteration 2 of Newton Raphson: norm_12_FD 1.34 10 3 Much faster progress, but more computational intense.