For the system of Fig. E7.8.1, assume bus 1 is the swing bus (δ1=0 and |V˜1|=1.0 pu). Then we can make an initial guess for bus voltage vector x^-0 = (δ2, |V˜2|, δ3, |V˜3|, δ4, |V˜4|) t =(0, 1, 0, 1, 0, 1)t . Note, the superscript 0 means initial guess. Compute the mismatch power vector for this

Question

For the system of Fig. E7.8.1, assume bus 1 is the swing bus [latex]\left(\delta_1=0 ext { and }\left|\widetilde{V}_1 ight|=1.0 \mathrm{pu} ight)[/latex]. Then we can make an initial guess for bus voltage vector [latex]\bar{x}^0=\left(\delta_2,\left|V_2 ight|, \delta_3,\left|V_3 ight|, \delta_4 ight., \left.\left|\widetilde{V}_4 ight| ight)^t=(0,1,0,1,0,1)^t[/latex]. Note, the superscript 0 means initial guess. Compute the mismatch power vector for this initial condition.

Accepted Answer

[latex]\begin{aligned} & \bar{X}^0=\left(\delta_2,\left| ilde{V}_2 ight|, \delta_3,\left| ilde{V}_3 ight|, \delta_4,\left| ilde{V}_4 ight| ight)^t=(0,1,0,1,0,1)^t \ & \Delta \bar{W}^0=\left(P_2+F_{r, 2}, Q_2+F_{i, 2}, P_3+F_{r, 3}, Q_3+F_{i, 3}, P_4+F_{r, 4}, Q_4+F_{i, 4} ight)^t \ & \Delta P_2=P_2+F_{r, 2}=P_2+\sum_{j=1}^4 y_{2 j} V_j V_2 \cos \left(- heta_{2 j}-\delta_j+\delta_2 ight) \ & =P_2+y_{21} V_1 V_2 \cos \left(- heta_{21}-\delta_1+\delta_2 ight) \ & +y_{22} V_2 V_2 \cos \left(- heta_{22}-\delta_2+\delta_2 ight) \ & +y_{23} V_3 V_2 \cos \left(- heta_{23}-\delta_3+\delta_2 ight) \ & +y_{24} V_4 V_2 \cos \left(- heta_{24}-\delta_4+\delta_2 ight) \ & \Delta P_2=P_2+F_{r, 2}=0.10+\{70.72 \cdot 1 \cdot 1 \cdot \cos (-2.356- \ & 0+0)+81.42 \cdot 1 \cdot 1 \cdot \cos (0.863-0+0)+12.127 ext {. } \ & 1 \cdot 1 \cdot \cos (-1.816-0+0)+0\}=0.10-49.996+ \ & 52.936-2.942=0.0978 \ & \Delta Q_2=Q_2+F_{i, 2}=Q_2+\sum_{j=1}^4 y_{2 j} V_j V_2 \sin \left(- heta_{2 j}-\delta_j+\delta_2 ight) \ & =Q_2+y_{21} V_1 V_2 \sin \left(- heta_{21}-\delta_1+\delta_2 ight) \ & +y_{22} V_2 V_2 \sin \left(- heta_{22}-\delta_2+\delta_2 ight) \ & +y_{23} V_3 V_2 \sin \left(- heta_{23}-\delta_3+\delta_2 ight) \ & +y_{24} V_4 V_2 \sin \left(- heta_{24}-\delta_4+\delta_2 ight) \ & \end{aligned}\ \begin{aligned} & \Delta Q_2=Q_2+F_{i, 2}=0.10+\{70.72 \cdot 1 \cdot \sin (-2.356-0 \ & +0)+81.42 \cdot 1 \cdot \sin (0.863-0+0)+12.127 . \ & 1 \cdot 1 \cdot \sin (-1.816-0+0)+0\}=0.10-50.016+ \ & 61.86-11.57=0.374 \ & \Delta P_3=P_3+F_{r, 3}=P_3+y_{31} V_1 V_3 \cos \left(- heta_{31}-\delta_1+\delta_3 ight)+ \ & y_{32} V_2 V_3 \cos \left(- heta_{32}-\delta_2+\delta_3 ight)+y_{33} V_3 V_3 \cos \ & \left(- heta_{33}-\delta_3+\delta_3 ight)+y_{34} V_4 V_3 \cos \left(- heta_{34}-\delta_4+\delta_3 ight) \ & \Delta P_3=P_3+F_{r, 3}=0.0+0.0+\{12.127 \cdot 1 \cdot 1 \cdot \cos (-1.816-0+0) \ & +56.62 \cdot 1 \cdot 1 \cdot \cos (1.154-0+0)+44.72 \cdot 1 \cdot 1 \cdot \cos (-2.034-0+0)\} \ & =0.0+0.0-2.944+22.92-19.98=-0.0056 \ & \Delta Q_3=Q_3+F_{i, 3}=Q_3+y_{31} V_1 V_3 \sin \left(- heta_{31}-\delta_1+\delta_3 ight)+ \ & y_{32} V_2 V_3 \sin \left(- heta_{32}-\delta_2+\delta_3 ight)+y_{33} V_3 V_3 \sin \left(- heta_{33}-\delta_3+\delta_3 ight) \ & +y_{34} V_4 V_3 \sin \left(- heta_{34}-\delta_4+\delta_3 ight) \ & \Delta Q_3=Q_3+F_{i, 3}=0.0+0.0+\{12.127 \cdot 1 \cdot 1 . \ & \sin (-1.816-0+0)+56.62 \cdot 1 \cdot 1 \cdot \sin (1.154-0+0) \ & +44.72 \cdot 1 \cdot 1 \cdot \sin (-2.034-0+0)\}=0.0+0.0- \ & 11.764+51.77-40.008=-0.00166 \ & \Delta P_4=P_4+F_{r, 4}=P_4+y_{41} V_1 V_4 \cos \left(- heta_{41}-\delta_1+\delta_4 ight)+ \ & y_{42} V_2 V_4 \cos \left(- heta_{42}-\delta_2+\delta_4 ight)+y_{43} V_3 V_4 \cos \left(- heta_{43}-\delta_3+\delta_4 ight) \ & +y_{44} V_4 V_4 \cos \left(- heta_{44}-\delta_4+\delta_4 ight) \ & \Delta P_4=P_4+F_{\mathrm{r}, 4}=0.25+\{44.72 \cdot 1 \cdot 1 \cdot \cos (-2.034- \ & 0+0)+0+44.72 \cdot 1 \cdot 1 \cdot \cos (-2.034-0+0)+ \ & 89.44 \cdot 1 \cdot 1 \cdot \cos (1.1076-0+0)\}=0.25- \ & 19.98+0-19.98+39.96=0.2527 \ & \Delta Q_4=Q_4+F_{i, 4}=Q_4+y_{41} V_1 V_4 \sin \left(- heta_{41}-\delta_1+\delta_4 ight)+ \ & y_{42} V_2 V_4 \sin \left(- heta_{42}-\delta_2+\delta_4 ight)+y_{43} V_3 V_4 \sin \left(- heta_{43}- ight. \ & \left.\delta_3+\delta_4 ight)+y_{44} V_4 V_4 \sin \left(- heta_{44}-\delta_4+\delta_4 ight) \ & \Delta Q_4=Q_4+F_{i, 4}=0.1+\{44.72 \cdot 1 \cdot 1 \cdot \sin (-2.034- \ & 0+0)+0+44.72 \cdot 1 \cdot 1 \cdot \sin (-2.034-0+0)+ \ & 89.44 \cdot 1 \cdot 1 \cdot \sin (1.1076-0+0)\}=0.1- \ & \end{aligned}[/latex]

40.0076 + 0-40.0076 + 80.016 = 0.1004.

The mismatch power vector is now

[latex]\begin{gathered} \Delta \bar{W}^0=(0.0978,0.374,-0.0056 \ -0.00166,0.2527,0.1004)^t \end{gathered}[/latex]

Question 7.AE.9: For the system of Fig. E7.8.1, assume bus 1 is the swing bus......

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