--- title: "N-R single iteration with Q limit" subject: "Power System Analysis II" chapter: 2 key: "semester_5/power_system_analysis_ii:nr-single-iteration-j0-08" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/nr-single-iteration-j0-08 site: BCT Engineering --- # N-R single iteration with Q limit Power System Analysis II · Chapter 2: Load Flow Analysis - Asked in: 2076 Asoj, 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/nr-single-iteration-j0-08 - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2 ## Question Consider a three bus system of figure, each of the three lines has a series impedance of (0+j0.08) pu. The specified quantities at the buses are tabulated below. Perform single iteration of load flow using N-R method. (Note: Reactive power limit is 0 ≤ QG2 ≤ 1.5 pu) | Bus | Demand real power | Demand reactive power | Generator real power | Generator reactive power | Voltage | |---|---|---|---|---|---| | 1 | - | - | - | - | 1∠0° pu | | 2 | 0 | 0 | 0.5 pu | - | 1.02 pu | | 3 | 1.5 pu | 0.6 pu | - | - | - | ## Short answer Form Y-bus (series j0.08 pu). At flat start, P2=P3=0, Q2=0.51 pu (within limits), mismatches ΔP2=0.5, ΔP3=−1.5, ΔQ3=−0.35. Solving Jacobian yields V2 = 1.02∠−0.774° pu, V3 = 0.9859∠−3.79° pu after one iteration. ## Answer Bus 1 is the slack bus ($1\angle 0^\circ$), bus 2 a PV bus ($P_2 = 0.5$ pu, $|V_2| = 1.02$ pu, $0 \le Q_{G2} \le 1.5$ pu) and bus 3 a PQ bus ($P_3 = -1.5$, $Q_3 = -0.6$ pu). Flat start: $V_2^{0} = 1.02\angle 0^\circ$, $V_3^{0} = 1\angle 0^\circ$. #### Y-bus Each line: $y = 1/(j0.08) = -j12.5$ pu. Each bus has two lines: $$ Y_{bus} = \begin{bmatrix} -j25.0000 & j12.5000 & j12.5000 \\ j12.5000 & -j25.0000 & j12.5000 \\ j12.5000 & j12.5000 & -j25.0000 \end{bmatrix}\ \text{pu} $$ i.e. $|Y_{ii}| = 25$, $\theta_{ii} = -90^\circ$; $|Y_{ij}| = 12.5$, $\theta_{ij} = 90^\circ$. #### Calculated powers and Q-limit check 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: $$ \begin{aligned} P_2^{0} &= 0.0000, \quad Q_2^{0} = 0.5100 \\ P_3^{0} &= 0.0000, \quad Q_3^{0} = -0.2500 \end{aligned} $$ $Q_{G2} = Q_2^{0} = 0.5100$ pu lies within $0$ to $1.5$ pu, so bus 2 stays a PV bus. #### Mismatch vector $$ \begin{aligned} \Delta P_2 &= 0.5000 - (0.0000) = 0.5000 \\ \Delta P_3 &= -1.5000 - (0.0000) = -1.5000 \\ \Delta Q_3 &= -0.6000 - (-0.2500) = -0.3500 \end{aligned} $$ #### Jacobian $$ \begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (1.0200)(1.0000)(12.5000)\sin(90.00^\circ) + (1.0200)(1.0000)(12.5000)\sin(90.00^\circ) = 25.5000 \\ \frac{\partial P_2}{\partial \delta_3} &= -(1.0200)(1.0000)(12.5000)\sin(90.00^\circ) = -12.7500 \\ \frac{\partial P_2}{\partial |V_3|} &= (1.0200)(12.5000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial P_3}{\partial \delta_2} &= -(1.0000)(1.0200)(12.5000)\sin(90.00^\circ) = -12.7500 \\ \frac{\partial P_3}{\partial \delta_3} &= (1.0000)(1.0000)(12.5000)\sin(90.00^\circ) + (1.0000)(1.0200)(12.5000)\sin(90.00^\circ) = 25.2500 \\ \frac{\partial P_3}{\partial |V_3|} &= 2(1.0000)(25.0000)\cos(-90.00^\circ) + (1.0000)(12.5000)\cos(90.00^\circ) + (1.0200)(12.5000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_2} &= -(1.0000)(1.0200)(12.5000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_3} &= (1.0000)(1.0000)(12.5000)\cos(90.00^\circ) + (1.0000)(1.0200)(12.5000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial |V_3|} &= -2(1.0000)(25.0000)\sin(-90.00^\circ) - (1.0000)(12.5000)\sin(90.00^\circ) - (1.0200)(12.5000)\sin(90.00^\circ) = 24.7500 \end{aligned} $$ $$ [J^{0}] = \begin{bmatrix} 25.5000 & -12.7500 & 0.0000 \\ -12.7500 & 25.2500 & 0.0000 \\ 0.0000 & 0.0000 & 24.7500 \end{bmatrix} $$ #### Solution of the first iteration $$ \begin{bmatrix} 0.5000 \\ -1.5000 \\ -0.3500 \end{bmatrix} = \begin{bmatrix} 25.5000 & -12.7500 & 0.0000 \\ -12.7500 & 25.2500 & 0.0000 \\ 0.0000 & 0.0000 & 24.7500 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix} $$ $$ \begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix} = [J^{0}]^{-1}\begin{bmatrix} 0.5000 \\ -1.5000 \\ -0.3500 \end{bmatrix} = \begin{bmatrix} -0.0135 \\ -0.0662 \\ -0.0141 \end{bmatrix} $$ $$ \begin{aligned} \delta_2^{1} &= 0 + (-0.0135) = -0.0135\ \text{rad} = -0.774^\circ \\ \delta_3^{1} &= 0 + (-0.0662) = -0.0662\ \text{rad} = -3.794^\circ \\ |V_3|^{1} &= 1.0000 + (-0.0141) = 0.9859\ \text{pu} \end{aligned} $$ So $V_2^{1} = 1.0200\angle -0.774^\circ$ pu and $V_3^{1} = 0.9859\angle -3.794^\circ$ pu. **Q-limit check with the new voltages:** $$ \begin{aligned} Q_2^{1} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.7089\ \text{pu} \\ Q_{G2} &= Q_2 + Q_{L2} = 0.7089\ \text{pu} \end{aligned} $$ This is still within 0 to 1.5 pu, so bus 2 remains a PV bus. **Slack bus power:** $$ \begin{aligned} P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0000)(0.9877 + j0.0451) \\ &= 0.9877 + j0.0451 \\ \Rightarrow P_1 &= 0.9877\ \text{pu},\quad Q_1 = -0.0451\ \text{pu} \\ P_{G1} &= P_1 + P_{L1} = 0.9877 + 0.00 = 0.9877\ \text{pu} \\ Q_{G1} &= Q_1 + Q_{L1} = -0.0451 + 0.00 = -0.0451\ \text{pu} \end{aligned} $$ **Answer (after one N-R iteration):** $\delta_2 = -0.774^\circ$, $|V_3| = 0.9859$ pu, $\delta_3 = -3.794^\circ$, $Q_{G2} = 0.7089$ pu, $P_1 = 0.9877$ pu, $Q_1 = -0.0451$ pu. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which mismatch equations are solved in the N-R iteration for this system (bus 2 PV, bus 3 PQ)? A. DeltaP2, DeltaP3, DeltaQ3 B. DeltaP2, DeltaQ2, DeltaP3, DeltaQ3 C. DeltaP3 and DeltaQ3 only D. DeltaP2, DeltaP3, DeltaQ2 Answer: A. Unknowns are delta2, delta3 and |V3|; PV bus 2 has no Q equation because |V2| is fixed. (Key point: Unknowns and equations) 2. Calculated Q2 is 0.5100 pu (and 0.7089 after the iteration) with limits 0 to 1.5 pu. What follows? A. Bus 2 becomes PQ with Q = 0 B. Bus 2 becomes PQ with Q = 1.5 C. Bus 2 stays PV because Q is within limits D. The slack bus changes to bus 2 Answer: C. Only when Q violates a limit is it fixed at the limit and the bus treated as PQ; here it stays PV. (Key point: Q-limit check) 3. At the flat start, why is the Jacobian block-diagonal (dP/d|V| and dQ/d(delta) are zero)? A. Because the line resistances are large B. Because the slack bus has zero power C. Because the voltage magnitudes are all 1 pu D. Because the angles are zero and phi = +/-90 deg, so the cos terms vanish Answer: D. The lines are pure reactances (theta = +/-90 deg) and the angles are zero, so cos(phi) = 0 and those entries are zero. (Key point: Jacobian structure at flat start) 4. Why is DeltaP3 = -1.5000 pu at the first iteration? A. Specified P3 = -1.5 (load) minus calculated P3 = 0 B. Specified P3 = +1.5 minus calculated 3.0 C. Calculated P3 is -1.5 and specified is 0 D. Bus 3 generates 1.5 pu Answer: A. Mismatch = specified - calculated = -1.5 - 0, as the 1.5 pu demand is a negative injection. 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