--- title: "G-S first iteration with line charging" subject: "Power System Analysis II" chapter: 2 key: "semester_5/power_system_analysis_ii:gs-first-iteration-line-charging" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/gs-first-iteration-line-charging site: BCT Engineering --- # G-S first iteration with line charging Power System Analysis II · Chapter 2: Load Flow Analysis - Asked in: 2068 Chaitra, 4×4 marks - Video: [Numerical on Gauss seidel method of load flow](https://www.youtube.com/watch?v=4nhmkQaw6n8) — Fun N Facts - 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-line-charging - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2 ## Question A 3-bus power system is shown in figure. The series impedance and shunt admittance of each line are 0.026+j0.11 pu and j0.04 pu respectively. The bus specification, power and bus voltage are as under. (PG, QG, PL, QL and bus voltages are in pu). [Figure: three buses with lines 1-2, 1-3 and 2-3] | Bus No. | PG | QG | PL | QL | Bus voltage | Bus type | |---|---|---|---|---|---|---| | 1 | ? | ? | 1.0 | 0.5 | 1.03∠0° | Slack | | 2 | 1.5 | ? | 0.5 | 0 | 1.03 | PV | | 3 | 0 | 0 | 1.2 | 0.5 | ? | PQ | i) Form YBUS ii) Generator reactive power at bus 2 using Gauss-Seidel method (up to first iteration) iii) Voltage at bus 2 and 3 using Gauss-Seidel method (up to first iteration) iv) Total network real power loss (after first iteration) ## Short answer Y‑bus: diagonal 4.0701 − j17.1798 pu, off‑diagonal −2.0351 + j8.6099 pu. Generator reactive power Q2 = 0.2236 pu. V2 = 1.0300∠2.76° pu, V3 = 0.9891∠−2.05° pu. Total real power loss = 0.0290 pu. ## Answer Bus 1 is the slack bus, bus 2 a PV bus and bus 3 a PQ bus. The line shunt admittance j0.04 pu is taken as the **total** line-charging admittance of each line, so $j0.02$ pu is placed at each end (nominal-π model). Gauss-Seidel is used with no acceleration factor. Net injections: $P_2 = 1.5 - 0.5 = 1.0$ pu; $P_3 = -1.2$ pu, $Q_3 = -0.5$ pu. Initial values: $V_2^{(0)} = 1.03\angle 0^\circ$, $V_3^{(0)} = 1\angle 0^\circ$. #### i) Y-bus $$ \begin{aligned} y_{series} &= \frac{1}{0.026 + j0.110} = 2.0351 - j8.6099\ \text{pu} \\ Y_{ii} &= 2y_{series} + 2(j0.02) = 4.0701 - j17.1798 \\ Y_{ij} &= -y_{series} = -2.0351 + j8.6099 \end{aligned} $$ $$ Y_{bus} = \begin{bmatrix} 4.0701 - j17.1798 & -2.0351 + j8.6099 & -2.0351 + j8.6099 \\ -2.0351 + j8.6099 & 4.0701 - j17.1798 & -2.0351 + j8.6099 \\ -2.0351 + j8.6099 & -2.0351 + j8.6099 & 4.0701 - j17.1798 \end{bmatrix}\ \text{pu} $$ #### ii) Generator reactive power at bus 2 (first iteration) $$ \begin{aligned} \textstyle\sum_k Y_{2k}V_k^{(0)} &= 0.0611 - j0.2171 \\ Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.2236\ \text{pu} \\ Q_{G2} &= Q_2 + Q_{L2} = 0.2236 + 0 = 0.2236\ \text{pu} \end{aligned} $$ No Q-limits are given, so bus 2 remains a PV bus. #### iii) Voltages at buses 2 and 3 (first iteration) **Iteration 1** $$ \begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 0.0611 - j0.2171 \\ Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.2236\ \text{pu} \end{aligned} $$ $$ \begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{1.0000 - j0.2236}{1.0300} = 0.9709 - j0.2171 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -4.1312 + j17.4781 \\ V_2^{(1)} &= \frac{(0.9709 - j0.2171) - (-4.1312 + j17.4781)}{4.0701 - j17.1798} = 1.0419 + j0.0501 \\ \delta_2 &= 2.7554^\circ \Rightarrow V_2^{(1)} = 1.0300\angle 2.7554^\circ = 1.0288 + j0.0495\ \text{(magnitude reset to }1.03\text{)} \end{aligned} $$ $$ \begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.2000 + j0.5000}{1.0000} = -1.2000 + j0.5000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -4.6161 + j17.6254 \\ V_3^{(1)} &= \frac{(-1.2000 + j0.5000) - (-4.6161 + j17.6254)}{4.0701 - j17.1798} = 0.9885 - j0.0353 \\ &= 0.9891\angle -2.0473^\circ\ \text{pu} \end{aligned} $$ So $V_2^{(1)} = 1.0300\angle 2.7554^\circ$ pu and $V_3^{(1)} = 0.9891\angle -2.0473^\circ$ pu. #### iv) Total network real power loss (after first iteration) Slack bus power with the first-iteration voltages: $$ \begin{aligned} P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0300)(-0.0351 - j0.3556) \\ &= -0.0362 - j0.3662 \\ \Rightarrow P_1 &= -0.0362\ \text{pu},\quad Q_1 = 0.3662\ \text{pu} \\ P_{G1} &= P_1 + P_{L1} = -0.0362 + 1.00 = 0.9638\ \text{pu} \\ Q_{G1} &= Q_1 + Q_{L1} = 0.3662 + 0.50 = 0.8662\ \text{pu} \end{aligned} $$ 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 | -0.0362 | 0.3662 | | 2 | 1.2690 | 0.0843 | | 3 | -1.2038 | -0.4518 | | Sum | 0.0290 | -0.0013 | $$ \begin{aligned} P_{loss} &= \textstyle\sum P_i = 0.0290\ \text{pu} \\ Q_{loss} &= \textstyle\sum Q_i = -0.0013\ \text{pu} \end{aligned} $$ **Answer:** $Q_{G2} = 0.2236$ pu, $V_2 = 1.0300\angle 2.7554^\circ$ pu, $V_3 = 0.9891\angle -2.0473^\circ$ pu, total real power loss $= 0.0290$ pu (about 2.90 MW on a 100 MVA base). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The line shunt admittance j0.04 pu is the total charging of each line. How does it enter Y11 if two lines meet at bus 1? A. j0.02 per line at each end, so 2(j0.02) = j0.04 is added to the diagonal B. j0.04 is added to every off-diagonal element C. It is subtracted from the diagonal D. j0.04 is added once to Y11 Answer: A. Nominal-pi puts half (j0.02) at each end, so the diagonal is 2y + 2(j0.02) = 4.0701 - j17.1798; off-diagonals are unchanged. (Key point: Line charging in Y-bus) 2. Which statement about the off-diagonal Yij is right? A. Yij = -(y_series + j0.02) B. Yij = -y_series, not affected by line charging C. Yij = y_series + j0.04 D. Yij = 0 for lines with shunt Answer: B. Shunt charging is connected to ground, so it only affects the diagonal terms; Yij = -2.0351 + j8.6099. (Key point: Off-diagonal elements) 3. How is the generator reactive power QG2 found in the first iteration? A. Q2 = P2 tan(angle) B. Q2 = -Im{V2* sum Y2k Vk} with initial voltages, then QG2 = Q2 + QL2 C. QG2 = QL2 = 0 D. QG2 from the slack bus power Answer: B. Q2 is computed from the initial voltages (0.2236 pu); QL2 = 0, so QG2 = 0.2236 pu. (Key point: Reactive power at bus 2) 4. How is the total real power loss found after the first iteration? A. Sum of the calculated net P injections at all buses (0.0290 pu) B. Sum of PG C. Difference of V magnitudes D. P1 alone Answer: A. Since the sum of injected power equals the sum of line losses, adding calculated Pi over all buses gives the loss. 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