--- title: "G-S two iterations, lines 0.02+j0.11" subject: "Power System Analysis II" chapter: 2 key: "semester_5/power_system_analysis_ii:gs-two-iterations-2080" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-two-iterations-2080 site: BCT Engineering --- # G-S two iterations, lines 0.02+j0.11 Power System Analysis II · Chapter 2: Load Flow Analysis - Asked in: 2080 Bhadra, 12 marks - Video: [Gauss Seidal Method Solved Problem -2 | GS Method | Power flow Analysis | Power System Analysis](https://www.youtube.com/watch?v=MQBtptz5VaY) — 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-two-iterations-2080 - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2 ## Question Perform the two iteration of Gauss-Seidal load flow analysis for the data table given below to find the unknown variables of the buses. Line impedance of each line is (0.02+j0.11) pu. All data are in per unit and have usual meanings. | Bus no. | PG | QG | PL | QL | Bus voltage | Configuration | |---|---|---|---|---|---|---| | 1 | ? | ? | 1.0 | 0.5 | 1.03∠0 | 1-2 | | 2 | 1.5 | ? | 0.5 | 0 | 1.03 | 2-3 | | 3 | 0.0 | 0 | 1.2 | 0.5 | ? | 3-1 | ## Short answer After two G-S iterations: V2 = 1.030∠1.93° pu, Q2 = 0.1798 pu; V3 = 0.989∠-2.48° pu. With flat start, Q2 is computed each iteration and V2 magnitude reset to 1.03 pu. Slack bus power is determined from final voltages and Y-bus. ## Answer Bus 1 is the slack bus, bus 2 a PV (generator) bus and bus 3 a PQ (load) bus. Lines 1-2, 2-3 and 3-1 each have $z = 0.020 + j0.110$ pu (no shunt admittance given). Gauss-Seidel is used with no acceleration factor and no Q-limits (none are given). #### Bus data | Bus | Type | Specified | Unknown | |---|---|---|---| | 1 | Slack | $V_1 = 1.03\angle 0^\circ$ | $P_1$, $Q_1$ | | 2 | PV | $P_2 = 1.5 - 0.5 = 1.0$, $\lvert V_2\rvert = 1.03$ | $Q_2$, $\delta_2$ | | 3 | PQ | $P_3 = -1.2$, $Q_3 = -0.5$ | $\lvert V_3\rvert$, $\delta_3$ | Initial values: $V_2^{(0)} = 1.03\angle 0^\circ$, $V_3^{(0)} = 1.0\angle 0^\circ$. #### Y-bus $$ \begin{aligned} y_{12} &= \frac{1}{0.0200 + j0.1100} = 1.6000 - j8.8000\ \text{pu} \\ y_{23} &= \frac{1}{0.0200 + j0.1100} = 1.6000 - j8.8000\ \text{pu} \\ y_{13} &= \frac{1}{0.0200 + j0.1100} = 1.6000 - j8.8000\ \text{pu} \end{aligned} $$ $$ Y_{bus} = \begin{bmatrix} 3.2000 - j17.6000 & -1.6000 + j8.8000 & -1.6000 + j8.8000 \\ -1.6000 + j8.8000 & 3.2000 - j17.6000 & -1.6000 + j8.8000 \\ -1.6000 + j8.8000 & -1.6000 + j8.8000 & 3.2000 - j17.6000 \end{bmatrix}\ \text{pu} $$ #### Gauss-Seidel iterations For the PV bus, $Q_2$ is calculated first, then $V_2$ is found and its magnitude reset to 1.03 pu, keeping the new angle. Bus 3 uses the newest $V_2$. **Iteration 1** $$ \begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 0.0480 - j0.2640 \\ Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.2719\ \text{pu} \end{aligned} $$ $$ \begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{1.0000 - j0.2719}{1.0300} = 0.9709 - j0.2640 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -3.2480 + j17.8640 \\ V_2^{(1)} &= \frac{(0.9709 - j0.2640) - (-3.2480 + j17.8640)}{3.2000 - j17.6000} = 1.0392 + j0.0508 \\ \delta_2 &= 2.7962^\circ \Rightarrow V_2^{(1)} = 1.0300\angle 2.7962^\circ = 1.0288 + j0.0502\ \text{(magnitude reset to }1.03\text{)} \end{aligned} $$ $$ \begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.2000 + j0.5000}{1.0000} = -1.2000 + j0.5000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -3.7362 + j18.0368 \\ V_3^{(1)} &= \frac{(-1.2000 + j0.5000) - (-3.7362 + j18.0368)}{3.2000 - j17.6000} = 0.9899 - j0.0359 \\ &= 0.9905\angle -2.0757^\circ\ \text{pu} \end{aligned} $$ **Iteration 2** $$ \begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 1.2603 - j0.1132 \\ Q_2^{(2)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.1798\ \text{pu} \end{aligned} $$ $$ \begin{aligned} V_2^{(2)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{1.0000 - j0.1798}{1.0288 - j0.0502} = 0.9782 - j0.1270 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -2.9161 + j17.8324 \\ V_2^{(2)} &= \frac{(0.9782 - j0.1270) - (-2.9161 + j17.8324)}{3.2000 - j17.6000} = 1.0267 + j0.0346 \\ \delta_2 &= 1.9298^\circ \Rightarrow V_2^{(2)} = 1.0300\angle 1.9298^\circ = 1.0294 + j0.0347\ \text{(magnitude reset to }1.03\text{)} \end{aligned} $$ $$ \begin{aligned} V_3^{(2)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.2000 + j0.5000}{0.9899 + j0.0359} = -1.1924 + j0.5483 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -3.6003 + j18.0674 \\ V_3^{(2)} &= \frac{(-1.1924 + j0.5483) - (-3.6003 + j18.0674)}{3.2000 - j17.6000} = 0.9876 - j0.0428 \\ &= 0.9886\angle -2.4788^\circ\ \text{pu} \end{aligned} $$ #### Unknown variables after the 2nd iteration **Slack bus power** (with the 2nd-iteration voltages) $$ \begin{aligned} P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0300)(0.1397 - j0.3651) \\ &= 0.1439 - j0.3761 \\ \Rightarrow P_1 &= 0.1439\ \text{pu},\quad Q_1 = 0.3761\ \text{pu} \\ P_{G1} &= P_1 + P_{L1} = 0.1439 + 1.00 = 1.1439\ \text{pu} \\ Q_{G1} &= Q_1 + Q_{L1} = 0.3761 + 0.50 = 0.8761\ \text{pu} \end{aligned} $$ 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 | 0.1439 | 0.3761 | | 2 | 1.0773 | 0.2251 | | 3 | -1.2011 | -0.4906 | | Sum | 0.0201 | 0.1106 | $$ \begin{aligned} P_{loss} &= \textstyle\sum P_i = 0.0201\ \text{pu} \\ Q_{loss} &= \textstyle\sum Q_i = 0.1106\ \text{pu} \end{aligned} $$ | Unknown | Value | |---|---| | $V_2$ | $1.0300\angle 1.9298^\circ$ pu | | $Q_{G2} = Q_2 + Q_{L2}$ | $0.1798$ pu | | $V_3$ | $0.9886\angle -2.4788^\circ$ pu | | $P_{G1}$ | $1.1439$ pu | | $Q_{G1}$ | $0.8761$ pu | Note: $Q_{G2} = 0.1798$ pu is the value used in the 2nd iteration (found from the 1st-iteration voltages). Recomputing $Q_2$ from the final voltages gives 0.2251 pu (see the table above). The two differ because Gauss-Seidel has not yet converged. **Answer:** after two iterations $V_2 = 1.0300\angle 1.9298^\circ$ pu, $V_3 = 0.9886\angle -2.4788^\circ$ pu, $Q_{G2} = 0.1798$ pu, $P_{G1} = 1.1439$ pu, $Q_{G1} = 0.8761$ pu, and real power loss $\approx 0.0201$ pu. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In this problem bus 2 has PG = 1.5 and PL = 0.5 with |V| fixed at 1.03. What is it treated as, and what is its specified P? A. Slack bus, P2 = 1.0 B. PQ bus, P2 = 1.5 C. PV bus, P2 = 2.0 D. PV bus, P2 = 1.0 Answer: D. Fixed voltage magnitude with specified generation makes it a PV bus, and the net injection is PG - PL = 1.5 - 0.5 = 1.0 pu. (Key point: Bus classification and net injection) 2. Each line has z = 0.02 + j0.11 pu and there are two lines at each bus with no shunts. How are Y22 and Y21 obtained? A. Y22 = 1/(2y), Y21 = -1/y B. Y22 = 2y, Y21 = -y C. Y22 = y, Y21 = -y D. Y22 = 2y, Y21 = +y Answer: B. The diagonal is the sum of admittances connected to the bus (2y = 3.2 - j17.6) and the off-diagonal is the negative of the line admittance (-1.6 + j8.8). (Key point: Y-bus formation) 3. What is the correct order of operations at the PV bus in each Gauss-Seidel iteration? A. Compute V2 with old Q2, keep the new magnitude and angle B. Compute V2, then reset its angle to zero and keep the magnitude C. Compute Q2 from latest voltages, compute V2, then reset |V2| to 1.03 keeping the new angle D. Fix Q2 = 0 and solve only for V2 Answer: C. Q2 is first updated from the present voltages, V2 is calculated, and since |V2| is specified only the angle of the result is retained. (Key point: PV bus step in Gauss-Seidel) 4. After the second iteration, how are slack bus power and total real power loss found? A. Loss = sum of all PG B. Slack power is fixed to the specified load and loss is zero C. Slack power from P1 - jQ1 = V1* sum(Y1k Vk); loss = sum of all bus net injections P D. Loss = P1 only Answer: C. Once the voltages are known, P1 and Q1 follow from V1* times the sum Y1k Vk, and the sum of calculated bus net injections equals the line losses (0.0201 pu). 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