--- title: "G-S first iteration, 3-bus with PV bus" subject: "Power System Analysis II" chapter: 2 key: "semester_5/power_system_analysis_ii:gs-first-iteration-2076-asoj" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-first-iteration-2076-asoj site: BCT Engineering --- # G-S first iteration, 3-bus with PV bus Power System Analysis II · Chapter 2: Load Flow Analysis - Asked in: 2076 Asoj, 8 marks - Video: [Gauss Seidal Method Solved Problem -1 | GS Method | Power flow Analysis | Power System Analysis](https://www.youtube.com/watch?v=2NRI6gZd_7o) — Simple n Easy Learning - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-first-iteration-2076-asoj - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2 ## Question For the system in figure below, perform the load flow analysis for the first iteration by G-S method. [Figure: line 1-2 X = j0.2, line 1-3 X = j0.1, line 2-3 X = j0.1; generators at buses 1 and 2, load at bus 3] | Bus No. | Voltage | Generator P | Generator Q | Load P | Load Q | |---|---|---|---|---|---| | 1 | 1.03∠0° pu | - | - | - | - | | 2 | 1.01 pu | 0.3 pu | - | - | - | | 3 | - | - | - | 0.4 pu | 0.2 pu | ## Short answer After first Gauss-Seidel iteration: Q2 = 0 pu, V2 = 1.0100∠1.1232° pu, V3 = 1.0100∠−0.5730° pu. Slack power injection: P1 = 0.0021 pu, Q1 = 0.3110 pu. ## Answer Bus 1 is the slack bus ($1.03\angle 0^\circ$), bus 2 a PV bus ($P_2 = 0.3$ pu, $|V_2| = 1.01$ pu) and bus 3 a PQ bus ($P_3 = -0.4$, $Q_3 = -0.2$ pu). Initial values: $V_2^{(0)} = 1.01\angle 0^\circ$, $V_3^{(0)} = 1\angle 0^\circ$. No Q-limits are given. #### Y-bus $$ \begin{aligned} y_{12} &= 1/j0.2 = -j5.0000 \\ y_{13} &= 1/j0.1 = -j10.0000 \\ y_{23} &= 1/j0.1 = -j10.0000 \end{aligned} $$ $$ Y_{bus} = \begin{bmatrix} -j15.0000 & j5.0000 & j10.0000 \\ j5.0000 & -j15.0000 & j10.0000 \\ j10.0000 & j10.0000 & -j20.0000 \end{bmatrix}\ \text{pu} $$ #### First Gauss-Seidel iteration **Iteration 1** $$ \begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 0.0000 \\ Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.0000\ \text{pu} \end{aligned} $$ $$ \begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{0.3000}{1.0100} = 0.2970 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= j15.1500 \\ V_2^{(1)} &= \frac{(0.2970) - (j15.1500)}{-j15.0000} = 1.0100 + j0.0198 \\ \delta_2 &= 1.1232^\circ \Rightarrow V_2^{(1)} = 1.0100\angle 1.1232^\circ = 1.0098 + j0.0198\ \text{(magnitude reset to }1.01\text{)} \end{aligned} $$ $$ \begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-0.4000 + j0.2000}{1.0000} = -0.4000 + j0.2000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -0.1980 + j20.3981 \\ V_3^{(1)} &= \frac{(-0.4000 + j0.2000) - (-0.1980 + j20.3981)}{-j20.0000} = 1.0099 - j0.0101 \\ &= 1.0100\angle -0.5730^\circ\ \text{pu} \end{aligned} $$ At the start the voltages are such that $\sum_k Y_{2k}V_k = 0$, so $Q_2^{(1)} = 0$. #### Slack bus power after the first iteration $$ \begin{aligned} P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0300)(0.0020 - j0.3019) \\ &= 0.0021 - j0.3110 \\ \Rightarrow P_1 &= 0.0021\ \text{pu},\quad Q_1 = 0.3110\ \text{pu} \\ P_{G1} &= P_1 + P_{L1} = 0.0021 + 0.00 = 0.0021\ \text{pu} \\ Q_{G1} &= Q_1 + Q_{L1} = 0.3110 + 0.00 = 0.3110\ \text{pu} \end{aligned} $$ Total loss = sum of the calculated net injections at all buses (equal to the sum of all line losses): | Bus | $P_i$ (pu) | $Q_i$ (pu) | |---|---|---| | 1 | 0.0021 | 0.3110 | | 2 | 0.4039 | -0.0951 | | 3 | -0.4060 | -0.1979 | | Sum | 0.0000 | 0.0180 | $$ \begin{aligned} P_{loss} &= \textstyle\sum P_i = 0.0000\ \text{pu} \\ Q_{loss} &= \textstyle\sum Q_i = 0.0180\ \text{pu} \end{aligned} $$ (The lines are pure reactances, so the real loss is zero; the small value of $P_1$ is the mismatch of an unconverged first iteration.) **Answer:** $Q_2 = 0.0000$ pu, $V_2 = 1.0100\angle 1.1232^\circ$ pu, $V_3 = 1.0100\angle -0.5730^\circ$ pu, $P_1 = 0.0021$ pu, $Q_1 = 0.3110$ pu after the first iteration. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In this Gauss-Seidel problem, how are buses 1, 2 and 3 classified? A. Bus 1 PQ, bus 2 PV, bus 3 slack B. Bus 1 slack, bus 2 PQ, bus 3 PV C. Bus 1 slack, bus 2 PV, bus 3 PQ D. All three are PQ buses Answer: C. Bus 1 has fixed voltage and angle (slack), bus 2 has P and |V| specified (PV), and bus 3 has P and Q specified as a load (PQ). (Key point: Bus types) 2. Why are P3 = -0.4 and Q3 = -0.2 pu used in the V3 update? A. Bus 3 is a load, so its net injected power is negative B. Bus 3 is a generator absorbing reactive power C. Negative values are needed so the iteration converges D. Because the line reactances are inductive Answer: A. Net injection = generation minus load, so a load of 0.4 + j0.2 pu is an injection of -0.4 - j0.2 pu. (Key point: Sign of load injection) 3. After computing V2 on the PV bus 2, what is done before using it for bus 3? A. Both magnitude and angle are kept as computed B. The angle is reset to 0 and the magnitude is kept C. Q2 is set to its limit and the bus becomes PQ D. The angle is kept and the magnitude is reset to the specified 1.01 pu Answer: D. At a PV bus |V| is fixed, so only the computed angle (1.1232°) is retained and the magnitude is reset to 1.01. (Key point: PV bus voltage update) 4. Why is Q2 = 0 in the first iteration? A. Bus 2 has no generator B. The sum of Y2k*Vk is zero with the initial voltages, so no reactive power is calculated C. Reactive power is ignored at PV buses D. Q-limits force it to zero Answer: B. With V1 = 1.03, V2 = 1.01, V3 = 1 the products j5(1.03) - j15(1.01) + j10(1) sum to zero, so Q2 = 0. 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