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Question 7.AE.13: For the system of Fig. E7.8.1, compute the correction voltag......

For the system of Fig. E7.8.1, compute the correction voltage vector \Delta \bar{x}^0, updating the mismatch power vector, and check the convergence of the load flow algorithm (using a convergence tolerance of 0.0001).

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Based on the results of Application Examples 7.9 to 7.11, the correction voltage vector \Delta \bar{x}^0 can be computed from

\Delta \vec{x}^0=\left(J^0\right)^{-1} \Delta W\left(\bar{x}^0\right)=(2.68 \mathrm{E}-4,2.3 \mathrm{E}-3,2.79 \mathrm{E}-3,3.57 \mathrm{E}-3,3.39 \mathrm{E}-3,4.03 \mathrm{E}-3)^t

The updated mismatch power vector is

\Delta \bar{W}^1=(1.41 \mathrm{E}-4,3.01 \mathrm{E}-4,-9.63 \mathrm{E}-6,7.62 \mathrm{E}-5,1.20 \mathrm{E}-3,5.96 \mathrm{E}-4)^t

Since the components of \Delta \bar{W}^1 are larger than the specified convergence tolerance (e.g.,0.0001), the Newton–Raphson solution has not converged. However, it will converge in two iterations:
Fundamental power flow iteration summary (Fig. E7.8.1):

Fundamental power flow output solution (Fig. E7.8.1), where δ angles are rounded (approximate) values:

“Active and reactive powers in the line between “from bus” and “to bus.”

Absolute real power mismatch Absolute reactive power mismatch
Iteration Average (%) Worst (%) Worst bus Average (%) Worst (%) Worst bus
0 11.67 25 4 6.67 10.00 4
1 0.05 0.12 4 0.03 0.06 4
2 0.00 0.00 4 0.00 0.00 4
Bus Voltage Bus generation (G) and load (L) powers Line power ^a
From bus |v| (%) \delta\ (^{\circ}) P_G (%) Q_G (%) P_L (%) Q_L (%) To bus P_{line} (%)
Q_{line} (%)
1 100.0 0.00 35.09 20.15 0.00 0.00 2 13.35 10.69
4 21.73 9.47
2 99.76 -0.02 0.00 0.00 10.00 10.00 1 -13.32 -10.66
3 3.32 0.66
3 99.64 -0.16 0.00 0.00 0.00 0.00 2 -3.32 -0.65
4 3.32 0.65
4 99.59 -0.20 0.00 0.00 25.00 10.00 3 -3.32 -0.65
1 -21.68 -9.35

^a Active and reactive powers in the line between “from bus” and “to bus.”

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