--- title: "N-R first iteration, line reactance 0.1" subject: "Power System Analysis II" chapter: 2 key: "semester_5/power_system_analysis_ii:nr-first-iteration-x0-1" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/nr-first-iteration-x0-1 site: BCT Engineering --- # N-R first iteration, line reactance 0.1 Power System Analysis II · Chapter 2: Load Flow Analysis - Asked in: 2074 Chaitra, 12 marks - Video: [Newton Raphson Method Solved Problem -1| NR method | Power flow Analysis | Newton Raphson numerical](https://www.youtube.com/watch?v=mm982Ecu7Do) — 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/nr-first-iteration-x0-1 - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-2 ## Question A three bus power system network is shown in figure below. The series reactance of each line is 0.1 per unit. Line resistances and shunt admittances are negligible. The bus specification and power input, etc, at the buses is as under: | Bus No. | PGen | QGen | PLoad | QLoad | Bus Voltage | Bus Type | |---|---|---|---|---|---|---| | 1 | ? | ? | 1 | 0.5 | 1.03+j0 | Slack | | 2 | 1.5 | ? | 0 | 0 | 1.04 | PV | | 3 | 0 | 0 | 1.2 | 0.5 | ? | PQ | [Figure: lines 1-2, 1-3 and 2-3; generators at buses 1 and 2; load 1 + j0.5 at bus 1 and 1.2 + j0.5 at bus 3] (i) Form [Ybus] (ii) Find the mismatch matrix [M°] and Jacobian Matrix [J°] (iii) Perform 1st iteration of load flow Analysis by Newton-Raphson method and calculate the above unknown quantities. ## Short answer Ybus diagonal -j20, off-diagonal j10. Mismatch: ΔP2=1.5, ΔP3=-1.2, ΔQ3=0.2 pu. Jacobian: ∂P2/∂δ2=21.112, ∂P2/∂δ3=-10.4, ∂P3/∂δ2=-10.4, ∂P3/∂δ3=20.7, ∂Q3/∂|V3|=10.3; other elements zero. ## Answer Bus 1 is the slack bus ($1.03\angle 0^\circ$), bus 2 a PV bus ($P_2 = 1.5$ pu, $|V_2| = 1.04$ pu) and bus 3 a PQ bus ($P_3 = -1.2$, $Q_3 = -0.5$ pu). Lines are pure reactances $j0.1$ pu. Flat start: $V_2^{0} = 1.04\angle 0^\circ$, $V_3^{0} = 1.0\angle 0^\circ$. No Q-limits are given for bus 2. #### (i) Y-bus Each line: $y = 1/(j0.1) = -j10$ pu. Each bus has two lines, so $Y_{ii} = -j20$ and $Y_{ij} = j10$: $$ Y_{bus} = \begin{bmatrix} -j20.0000 & j10.0000 & j10.0000 \\ j10.0000 & -j20.0000 & j10.0000 \\ j10.0000 & j10.0000 & -j20.0000 \end{bmatrix} = \begin{bmatrix} 20\angle -90^\circ & 10\angle 90^\circ & 10\angle 90^\circ \\ 10\angle 90^\circ & 20\angle -90^\circ & 10\angle 90^\circ \\ 10\angle 90^\circ & 10\angle 90^\circ & 20\angle -90^\circ \end{bmatrix}\ \text{pu} $$ #### (ii) Mismatch matrix and Jacobian 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, $j \ne i$): $$ \begin{aligned} \frac{\partial P_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial |V_i|} &= 2|V_i||Y_{ii}|\cos\theta_{ii} + \sum_{j\ne i} |V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial P_i}{\partial |V_j|} &= |V_i||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_i|} &= -2|V_i||Y_{ii}|\sin\theta_{ii} - \sum_{j\ne i} |V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_j|} &= -|V_i||Y_{ij}|\sin\phi_{ij} \end{aligned} $$ Since all $\theta_{ij} = \pm 90^\circ$ and all angles are zero at the start, the $\cos$ terms vanish, so $P_i^0 = 0$: $$ \begin{aligned} P_2^{0} &= 0.0000, \quad Q_2^{0} = 0.5200 \\ P_3^{0} &= 0.0000, \quad Q_3^{0} = -0.7000 \end{aligned} $$ $$ \begin{aligned} \Delta P_2 &= 1.5000 - (0.0000) = 1.5000 \\ \Delta P_3 &= -1.2000 - (0.0000) = -1.2000 \\ \Delta Q_3 &= -0.5000 - (-0.7000) = 0.2000 \end{aligned} $$ $$ [M^{0}] = \begin{bmatrix} \Delta P_2 \\ \Delta P_3 \\ \Delta Q_3 \end{bmatrix} = \begin{bmatrix} 1.5000 \\ -1.2000 \\ 0.2000 \end{bmatrix} $$ Jacobian elements: $$ \begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (1.0400)(1.0300)(10.0000)\sin(90.00^\circ) + (1.0400)(1.0000)(10.0000)\sin(90.00^\circ) = 21.1120 \\ \frac{\partial P_2}{\partial \delta_3} &= -(1.0400)(1.0000)(10.0000)\sin(90.00^\circ) = -10.4000 \\ \frac{\partial P_2}{\partial |V_3|} &= (1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial P_3}{\partial \delta_2} &= -(1.0000)(1.0400)(10.0000)\sin(90.00^\circ) = -10.4000 \\ \frac{\partial P_3}{\partial \delta_3} &= (1.0000)(1.0300)(10.0000)\sin(90.00^\circ) + (1.0000)(1.0400)(10.0000)\sin(90.00^\circ) = 20.7000 \\ \frac{\partial P_3}{\partial |V_3|} &= 2(1.0000)(20.0000)\cos(-90.00^\circ) + (1.0300)(10.0000)\cos(90.00^\circ) + (1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_2} &= -(1.0000)(1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_3} &= (1.0000)(1.0300)(10.0000)\cos(90.00^\circ) + (1.0000)(1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial |V_3|} &= -2(1.0000)(20.0000)\sin(-90.00^\circ) - (1.0300)(10.0000)\sin(90.00^\circ) - (1.0400)(10.0000)\sin(90.00^\circ) = 19.3000 \end{aligned} $$ $$ [J^{0}] = \begin{bmatrix} 21.1120 & -10.4000 & 0.0000 \\ -10.4000 & 20.7000 & 0.0000 \\ 0.0000 & 0.0000 & 19.3000 \end{bmatrix} $$ (The off-diagonal blocks are zero at the flat start because the lines have no resistance.) #### (iii) First iteration $$ \begin{bmatrix} 1.5000 \\ -1.2000 \\ 0.2000 \end{bmatrix} = \begin{bmatrix} 21.1120 & -10.4000 & 0.0000 \\ -10.4000 & 20.7000 & 0.0000 \\ 0.0000 & 0.0000 & 19.3000 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix} $$ $$ \begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix} = [J^{0}]^{-1}\begin{bmatrix} 1.5000 \\ -1.2000 \\ 0.2000 \end{bmatrix} = \begin{bmatrix} 0.0565 \\ -0.0296 \\ 0.0104 \end{bmatrix} $$ Updated values: $$ \begin{aligned} \delta_2^{1} &= 0 + (0.0565) = 0.0565\ \text{rad} = 3.235^\circ \\ \delta_3^{1} &= 0 + (-0.0296) = -0.0296\ \text{rad} = -1.696^\circ \\ |V_3|^{1} &= 1.0000 + (0.0104) = 1.0104\ \text{pu} \end{aligned} $$ So $V_2^{1} = 1.0400\angle 3.235^\circ$ pu and $V_3^{1} = 1.0104\angle -1.696^\circ$ pu. **Reactive power at bus 2** (with the new voltages): $$ \begin{aligned} Q_2^{1} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.4682\ \text{pu} \\ Q_{G2} &= Q_2 + Q_{L2} = 0.4682\ \text{pu} \end{aligned} $$ **Slack bus power:** $$ \begin{aligned} P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0300)(-0.2879 - j0.1174) \\ &= -0.2966 - j0.1209 \\ \Rightarrow P_1 &= -0.2966\ \text{pu},\quad Q_1 = 0.1209\ \text{pu} \\ P_{G1} &= P_1 + P_{L1} = -0.2966 + 1.00 = 0.7034\ \text{pu} \\ Q_{G1} &= Q_1 + Q_{L1} = 0.1209 + 0.50 = 0.6209\ \text{pu} \end{aligned} $$ 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 | -0.2966 | 0.1209 | | 2 | 1.5078 | 0.4682 | | 3 | -1.2113 | -0.4544 | | Sum | 0.0000 | 0.1347 | $$ \begin{aligned} P_{loss} &= \textstyle\sum P_i = 0.0000\ \text{pu} \\ Q_{loss} &= \textstyle\sum Q_i = 0.1347\ \text{pu} \end{aligned} $$ **Answer (after the 1st N-R iteration):** $\delta_2 = 3.235^\circ$, $V_3 = 1.0104\angle -1.696^\circ$ pu, $Q_{G2} = 0.4682$ pu, $P_{G1} = 0.7034$ pu, $Q_{G1} = 0.6209$ pu. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Each line is j0.1 pu. What is Y22 and Y21? A. Y22 = -j20, Y21 = -j10 B. Y22 = j20, Y21 = -j10 C. Y22 = -j10, Y21 = +j10 D. Y22 = -j20, Y21 = +j10 Answer: D. y = 1/(j0.1) = -j10; Y22 is the sum of two lines (-j20) and Y21 is the negative of the line admittance (+j10). (Key point: Y-bus for reactive lines) 2. Why is P2 and P3 calculated at the flat start equal to zero here? A. The loads are zero B. All angles are zero and theta_ij = +/-90 degrees, so the cosine terms vanish C. The slack bus supplies all power D. Voltages are all 1.0 pu Answer: B. P = sum |Vi||Vj||Yij|cos(phi) with phi = +/-90 degrees gives zero, since lines have no resistance. (Key point: Flat-start real power) 3. Why are the off-diagonal blocks (dP/d|V| and dQ/d delta) zero at the flat start? A. Lines have no resistance, so they contain cos(90 degrees) terms B. Because Q is not specified C. Because the system has only 3 buses D. Because bus 2 is PV Answer: A. Those elements are proportional to cos(phi) (or cos for Q-delta), which vanishes for purely reactive lines at zero angle. (Key point: Jacobian decoupling) 4. What is the mismatch vector [DeltaP2, DeltaP3, DeltaQ3] at the start? A. [1.0, -1.2, 0.2] B. [1.5, -1.2, 0.2] C. [1.5, -1.2, -0.2] D. [0, 0, 0] Answer: B. DeltaP2 = 1.5 - 0, DeltaP3 = -1.2 - 0, DeltaQ3 = -0.5 - (-0.7) = 0.2. 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