--- title: "Find |V3|, θ3 and QG2 with π lines" subject: "Power System Analysis II" chapter: 2 key: "semester_5/power_system_analysis_ii:pi-line-v3-qg2" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/pi-line-v3-qg2 site: BCT Engineering --- # Find |V3|, θ3 and QG2 with π lines Power System Analysis II · Chapter 2: Load Flow Analysis - Asked in: 2075 Chaitra, 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/pi-line-v3-qg2 - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2 ## Question The bus 1 is assumed as slack bus for the 3-bus system shown below. Find |V3|, θ3, and QG2. The transmission line is represented as nominal π equivalent network with series impedance of zL = (0.0 + j0.10) and half line charging admittance yc = j0.02. [Figure: lines 1-2, 1-3 and 2-3; bus 1 generator SG1 with V1 = 1.0 + j0; bus 2 generator PG2 = 0.6661, |V2| = 1.04; bus 3 load SD3 = 2.5 + j1.0] ## Short answer Using Newton-Raphson method with nominal-π Y-bus, converged load flow gives |V3| = 0.972 pu, θ3 = −8.15° and QG2 = 1.166 pu, which is within the 0–1.5 pu limit. ## Answer Bus 1 is the slack bus ($1.0\angle 0^\circ$), bus 2 a PV bus ($P_2 = 0.6661$ pu, $|V_2| = 1.04$ pu) and bus 3 a PQ bus ($P_3 = -2.5$, $Q_3 = -1.0$ pu). Each line is a nominal-π with $z_L = j0.1$ pu and $j0.02$ pu at each end. The unknowns are found from the converged load flow (Newton-Raphson, polar form, flat start $V_2^0 = 1.04\angle 0^\circ$, $V_3^0 = 1\angle 0^\circ$). #### Y-bus $$ \begin{aligned} y_L &= 1/j0.1 = -j10\ \text{pu} \\ Y_{ii} &= 2(-j10) + 2(j0.02) = -j19.96 \\ Y_{ij} &= +j10 \end{aligned} $$ $$ Y_{bus} = \begin{bmatrix} -j19.9600 & j10.0000 & j10.0000 \\ j10.0000 & -j19.9600 & j10.0000 \\ j10.0000 & j10.0000 & -j19.9600 \end{bmatrix}\ \text{pu} $$ #### First iteration (shown in full) With $Y_{ij} = |Y_{ij}|\angle\theta_{ij}$, $V_i = |V_i|\angle\delta_i$ and $\phi_{ij} = \theta_{ij} - \delta_i + \delta_j$: $$ \begin{aligned} P_i &= \sum_{j=1}^{n} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ Q_i &= -\sum_{j=1}^{n} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \end{aligned} $$ Jacobian elements (polar form, $j \ne i$): $$ \begin{aligned} \frac{\partial P_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial |V_i|} &= 2|V_i||Y_{ii}|\cos\theta_{ii} + \sum_{j\ne i} |V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial P_i}{\partial |V_j|} &= |V_i||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_i|} &= -2|V_i||Y_{ii}|\sin\theta_{ii} - \sum_{j\ne i} |V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_j|} &= -|V_i||Y_{ij}|\sin\phi_{ij} \end{aligned} $$ At the flat start the $\cos$ terms vanish ($\theta = \pm 90^\circ$), so $P_2^0 = P_3^0 = 0$, and $$ \begin{aligned} Q_2^{0} &= 0.7887, \quad Q_3^{0} = -0.4400 \\ \Delta P_2 &= 0.6661, \quad \Delta P_3 = -2.5000, \quad \Delta Q_3 = -1.0 - (-0.4400) = -0.5600 \end{aligned} $$ $$ \begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (1.0400)(1.0000)(10.0000)\sin(90.00^\circ) + (1.0400)(1.0000)(10.0000)\sin(90.00^\circ) = 20.8000 \\ \frac{\partial P_2}{\partial \delta_3} &= -(1.0400)(1.0000)(10.0000)\sin(90.00^\circ) = -10.4000 \\ \frac{\partial P_2}{\partial |V_3|} &= (1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial P_3}{\partial \delta_2} &= -(1.0000)(1.0400)(10.0000)\sin(90.00^\circ) = -10.4000 \\ \frac{\partial P_3}{\partial \delta_3} &= (1.0000)(1.0000)(10.0000)\sin(90.00^\circ) + (1.0000)(1.0400)(10.0000)\sin(90.00^\circ) = 20.4000 \\ \frac{\partial P_3}{\partial |V_3|} &= 2(1.0000)(19.9600)\cos(-90.00^\circ) + (1.0000)(10.0000)\cos(90.00^\circ) + (1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_2} &= -(1.0000)(1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_3} &= (1.0000)(1.0000)(10.0000)\cos(90.00^\circ) + (1.0000)(1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial |V_3|} &= -2(1.0000)(19.9600)\sin(-90.00^\circ) - (1.0000)(10.0000)\sin(90.00^\circ) - (1.0400)(10.0000)\sin(90.00^\circ) = 19.5200 \end{aligned} $$ $$ \begin{bmatrix} 0.6661 \\ -2.5000 \\ -0.5600 \end{bmatrix} = \begin{bmatrix} 20.8000 & -10.4000 & 0.0000 \\ -10.4000 & 20.4000 & 0.0000 \\ 0.0000 & 0.0000 & 19.5200 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix} \Rightarrow \begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix} = \begin{bmatrix} -0.0393 \\ -0.1426 \\ -0.0287 \end{bmatrix} $$ #### Iterations to convergence The same steps are repeated (Jacobian recalculated each time) until all mismatches are below $10^{-5}$ pu: | Iteration | $\Delta P_2$ | $\Delta P_3$ | $\Delta Q_3$ | $\delta_2$ (deg) | $\delta_3$ (deg) | $\lvert V_3\rvert$ (pu) | |---|---|---|---|---|---|---| | 1 | 0.6661 | -2.5000 | -0.5600 | -2.2493 | -8.1682 | 0.9713 | | 2 | 0.0326 | -0.0783 | -0.1688 | -2.2844 | -8.4855 | 0.9615 | | 3 | 0.0005 | -0.0011 | -0.0021 | -2.2847 | -8.4897 | 0.9614 | | 4 | 0.0000 | 0.0000 | 0.0000 | -2.2847 | -8.4897 | 0.9614 | The mismatches in each row are those at the start of that iteration; the angles and $|V_3|$ are the values at its end. #### Results Converged voltages: $V_2 = 1.0400\angle -2.285^\circ$ pu, $V_3 = 0.9614\angle -8.490^\circ$ pu. Reactive power at bus 2: $$ \begin{aligned} Q_2 &= -\sum_{j}|V_2||V_j||Y_{2j}|\sin(\theta_{2j} - \delta_2 + \delta_j) \\ &= 1.2571\ \text{pu} \end{aligned} $$ Since there is no load at bus 2, $Q_{G2} = Q_2$. **Answer:** $|V_3| = 0.9614$ pu, $\theta_3 = -8.49^\circ$, $Q_{G2} = 1.2571$ pu. (Slack bus: $P_{G1} = 1.8339$ pu, $Q_{G1} = 0.0597$ pu.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is Y11 = -j19.96 rather than -j20? A. Line resistance reduces it B. The two half-line charging admittances (+j0.02 each) are added to the diagonal C. Bus 1 is the slack bus D. The Y-bus is rounded to 4 digits Answer: B. Y_ii = 2(-j10) + 2(j0.02) = -j19.96; shunt charging is capacitive and reduces the diagonal magnitude. (Key point: Y-bus with line charging) 2. At the flat start Q3^0 = -0.44 pu and the load Q is 1.0 pu. What is DeltaQ3? A. +0.56 B. -1.44 C. -0.56 D. +1.44 Answer: C. DeltaQ3 = Q_specified - Q_calculated = -1.0 - (-0.44) = -0.56 pu. (Key point: Mismatch at bus 3) 3. How does the solution proceed beyond the first iteration? A. The first Jacobian is reused for all iterations B. One iteration is enough for a PQ bus C. Only the angle is updated further D. The Jacobian is recomputed each time until all mismatches are below 1e-5 Answer: D. N-R recalculates the Jacobian with updated voltages; the mismatches fall to zero by iteration 4. (Key point: Iterate to convergence) 4. How is QG2 obtained and why does QG2 equal Q2? A. From converged voltages through the Q equation; bus 2 has no load B. From the first iteration only; bus 2 is the slack C. From the specified value; it is given in the problem D. From the line charging only; bus 2 has no losses Answer: A. Q2 is evaluated from the converged voltages (1.2571 pu); with no load at bus 2, QG2 = Q2. (Key point: Finding QG2) ## Related in Power System Analysis II - [G-S load flow, j0.12 lines, losses](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-three-bus-j012-losses) — Power System Analysis II, chapter 2, asked 2× - [G-S first iteration with line charging](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-first-iteration-line-charging) — Power System Analysis II, chapter 2 - [Ybus for 3-bus lines with shunt susceptance](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-three-bus-shunt-2071) — Power System Analysis II, chapter 1 - [G-S two iterations, lines 0.026+j0.11](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ags-two-iterations-2075) — Power System Analysis II, chapter 2 - [Ybus and bus types (parallel lines)](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-bus-types-2074) — Power System Analysis II, chapter 1 - [Ybus with nominal π model, line lengths](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-km-lines-pi-model) — Power System Analysis II, chapter 1, asked 2× ## Related in other subjects - [300 km line: short line vs nominal-π](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A300km-short-vs-nominal-pi) — Power System Analysis I, chapter 5, asked 2× - [Nominal-T 100 km line, 20 MW at 110 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-20mw) — Power System Analysis I, chapter 5 - [Nominal-T 100 km line, 50 MVA at 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-50mva) — Power System Analysis I, chapter 6 - [280 km, 60 Hz line: nominal-π vs long line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A280km-60hz-nominal-pi-long) — Power System Analysis I, chapter 5, asked 2×