Skip to main content

G-S first iteration with PV bus

Short answer

At end of first iteration: Q2 = 0.4375 pu, V2 = 1.05∠18.08° pu, V3 = 0.8335∠-16.52° pu. Computed by forming Ybus, finding Q2, updating V2 resetting magnitude, then V3.

For the system shown in figure, determine the voltage at the end of the first iteration by G-S method. Neglect limits on reactive power generation. Bus 1: Slack bus, V = 1∠0°; Bus 2: P-V bus, V = 1.05 pu, PG = 3 pu; Bus 3: P-Q bus, PL = 4 pu, QL = 2 pu. [Figure: line 1-2 j0.3, line 1-3 j0.2, line 2-3 j0.2; generators at buses 1 and 2, load at bus 3]

Bus 1 is the slack bus (1∠0∘1\angle 0^\circ), bus 2 a PV bus (P2=3P_2 = 3 pu, ∣V2∣=1.05|V_2| = 1.05 pu) and bus 3 a PQ bus (P3=−4P_3 = -4, Q3=−2Q_3 = -2 pu). Initial values: V2(0)=1.05∠0∘V_2^{(0)} = 1.05\angle 0^\circ, V3(0)=1∠0∘V_3^{(0)} = 1\angle 0^\circ. Q-limits are neglected.

Y-bus

y12=1/j0.3=−j3.3333y13=1/j0.2=−j5.0000y23=1/j0.2=−j5.0000\begin{aligned} y_{12} &= 1/j0.3 = -j3.3333 \\ y_{13} &= 1/j0.2 = -j5.0000 \\ y_{23} &= 1/j0.2 = -j5.0000 \end{aligned} Ybus=[−j8.3333j3.3333j5.0000j3.3333−j8.3333j5.0000j5.0000j5.0000−j10.0000] puY_{bus} = \begin{bmatrix} -j8.3333 & j3.3333 & j5.0000 \\ j3.3333 & -j8.3333 & j5.0000 \\ j5.0000 & j5.0000 & -j10.0000 \end{bmatrix}\ \text{pu}

First Gauss-Seidel iteration

Iteration 1

∑k=13Y2kVk=−j0.4167Q2(1)=−Im{V2∗∑kY2kVk}=0.4375 pu\begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= -j0.4167 \\ Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.4375\ \text{pu} \end{aligned} V2(1)=1Y22[P2−jQ2V2∗−Y21V1−Y23V3]P2−jQ2V2∗=3.0000−j0.43751.0500=2.8571−j0.4167∑k≠2Y2kVk=j8.3333V2(1)=(2.8571−j0.4167)−(j8.3333)−j8.3333=1.0500+j0.3429δ2=18.0834∘⇒V2(1)=1.0500∠18.0834∘=0.9981+j0.3259 (magnitude reset to 1.05)\begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{3.0000 - j0.4375}{1.0500} = 2.8571 - j0.4167 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= j8.3333 \\ V_2^{(1)} &= \frac{(2.8571 - j0.4167) - (j8.3333)}{-j8.3333} = 1.0500 + j0.3429 \\ \delta_2 &= 18.0834^\circ \Rightarrow V_2^{(1)} = 1.0500\angle 18.0834^\circ = 0.9981 + j0.3259\ \text{(magnitude reset to }1.05\text{)} \end{aligned} V3(1)=1Y33[P3−jQ3V3∗−Y31V1−Y32V2]P3−jQ3V3∗=−4.0000+j2.00001.0000=−4.0000+j2.0000∑k≠3Y3kVk=−1.6296+j9.9907V3(1)=(−4.0000+j2.0000)−(−1.6296+j9.9907)−j10.0000=0.7991−j0.2370=0.8335∠−16.5227∘ pu\begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-4.0000 + j2.0000}{1.0000} = -4.0000 + j2.0000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -1.6296 + j9.9907 \\ V_3^{(1)} &= \frac{(-4.0000 + j2.0000) - (-1.6296 + j9.9907)}{-j10.0000} = 0.7991 - j0.2370 \\ &= 0.8335\angle -16.5227^\circ\ \text{pu} \end{aligned}

Answer (end of first iteration): Q2=0.4375Q_2 = 0.4375 pu, V2=1.0500∠18.0834∘V_2 = 1.0500\angle 18.0834^\circ pu, V3=0.8335∠−16.5227∘V_3 = 0.8335\angle -16.5227^\circ pu.

✓ Answer checked by Claude Sonnet 5.5

All Power System Analysis II chapter 2 questions →