--- title: "Capacitor size, four-node 11 kV feeder" subject: "Transmission and Distribution System Design" chapter: 7 key: "semester_8/transmission_and_distribution_system_design:capacitor-size-four-node-feeder" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/capacitor-size-four-node-feeder site: BCT Engineering --- # Capacitor size, four-node 11 kV feeder Transmission and Distribution System Design · Chapter 7: Distribution System Design - Asked in: 2070 Magh, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/capacitor-size-four-node-feeder - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7 ## Question For the 11 kV primary distribution network shown below, determine the size (kVAR) of the capacitor to be placed at the location shown in diagram to achieve the maximum loss reduction. For simplifying the analysis assume voltage at each node is 1 p.u. All impedances are in p.u. at 1000 kVA. [Figure: radial feeder from a 1000 kVA, 33/11 kV source substation S to node A through 1+j2, A to B through 2+j3, B to C through 2+j4, C to D through 3+j4. Loads through distribution transformers: 100 kVA at 0.6 p.f. lag at A, 100 kVA at 0.8 p.f. lag at B, 150 kVA at 0.9 p.f. lag at C, 200 kVA at 0.6 p.f. lag at D. The capacitor is placed at node D.] ## Short answer Optimum capacitor size equals the resistance-weighted average of reactive flows in the sections it relieves. Sum of R_i Q_i = 1866.92, sum of R_i = 8, so Q_c = 233.4 kVAR for maximum loss reduction. ## Answer #### Method (optimum single capacitor) With all node voltages taken as 1 p.u., the reactive current in a section equals its reactive power flow $Q_i$ (p.u.). A shunt capacitor $Q_c$ at the end node reduces the reactive flow in **every section between the source and that node** from $Q_i$ to $(Q_i - Q_c)$. The loss reduction is $$ \begin{aligned} \Delta P_L &= \sum_i R_i\left[Q_i^2-(Q_i-Q_c)^2\right] = \sum_i R_i\left(2Q_iQ_c-Q_c^2\right) \end{aligned} $$ For maximum loss reduction, $\dfrac{d(\Delta P_L)}{dQ_c} = \sum_i R_i(2Q_i-2Q_c)=0$, so $$ Q_{c,opt} = \frac{\sum_i R_iQ_i}{\sum_i R_i} $$ i.e. the optimum capacitor equals the **resistance-weighted average** of the reactive flows in the sections it relieves. The ratio does not depend on whether $R$ is in ohms or p.u., so kVAR can be used directly. ```text S-(1+j2)-A-(2+j3)-B-(2+j4)-C-(3+j4)-D | | | | 100kVA 100kVA 150kVA 200kVA 0.6 lag 0.8 lag 0.9 lag 0.6 lag [Qc] ``` #### Reactive load at each node | Node | S (kVA) | p.f. | $\sin\phi$ | Q (kVAR) | |---|---|---|---|---| | A | 100 | 0.6 | 0.8 | 80.00 | | B | 100 | 0.8 | 0.6 | 60.00 | | C | 150 | 0.9 | 0.4359 | 65.38 | | D | 200 | 0.6 | 0.8 | 160.00 | #### Reactive flow in each section | Section | R (p.u.) | Q flow (kVAR) | R × Q | |---|---|---|---| | S–A | 1 | 365.38 | 365.38 | | A–B | 2 | 285.38 | 570.77 | | B–C | 2 | 225.38 | 450.77 | | C–D | 3 | 160.00 | 480.00 | | Total | 8 | | 1866.92 | (Section flows: S–A carries all four loads, A–B carries B+C+D, and so on.) #### Optimum capacitor $$ Q_c = \frac{\sum R_iQ_i}{\sum R_i} = \frac{365.38+570.77+450.77+480}{1+2+2+3} = \frac{1866.92}{8} = 233.4\ \text{kVAR} $$ In p.u. on 1000 kVA, $Q_c = 0.233$ p.u. Only the resistances matter for loss; the reactances affect voltage drop, not $I^2R$ loss. **Answer: Capacitor at node D ≈ 233 kVAR for maximum loss reduction (nearest standard bank ≈ 240–250 kVAR).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A capacitor at node D changes the reactive flow in which sections? A. Only section C-D B. Only the section adjacent to the source C. Every section between the source and D D. Every section beyond D Answer: C. Reactive flow Q_i in each section from source to D reduces to Q_i - Q_c. (Key point: Effect of capacitor on flows) 2. The optimum capacitor size for maximum loss reduction is: A. Sum of R_i Q_i divided by sum of R_i B. Sum of X_i Q_i divided by sum of X_i C. Total reactive load D. Sum of Q_i divided by number of sections Answer: A. Setting d(ΔP)/dQc = 0 gives the resistance-weighted average of section reactive flows. (Key point: Optimum formula) 3. Why are the reactances (j2, j3...) not used? A. They are in p.u. so cancel B. They only affect voltage drop, not I^2R loss C. They are too small D. They equal the resistances Answer: B. Loss depends on resistance only. (Key point: Reactance role) 4. The reactive flow in section A-B equals: A. Q of node B only B. Q of A plus B C. Q of B plus C plus D D. Q of all four loads Answer: C. A-B carries everything downstream of A: 60 + 65.38 + 160 = 285.38 kVAR. (Key point: Section flow) 5. The optimum capacitor at D is about: A. 365 kVAR B. 160 kVAR C. 305 kVAR D. 233 kVAR Answer: D. 1866.92/8 = 233.4 kVAR. (Key point: Result) ## Close variants set in other exams - [Capacitor size for maximum loss reduction (variant)](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Acapacitor-size-11kv-network-b) — Transmission and Distribution System Design, chapter 7, asked 5× - [Capacitor size for maximum loss reduction](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Acapacitor-size-11kv-network) — Transmission and Distribution System Design, chapter 7, asked 5× - [Capacitor size, feeder with 500 kVA end load](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Acapacitor-size-500kva-load-c) — Transmission and Distribution System Design, chapter 7, asked 5× ## Related in Transmission and Distribution System Design - [Capacitor size, feeder with three equal loads](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Acapacitor-size-equal-loads-ohms) — Transmission and Distribution System Design, chapter 7 - [Loss and drop, 15 km 11 kV feeder](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Audl-feeder-15km-11kv) — Transmission and Distribution System Design, chapter 7, asked 2× - [Power loss, 10 km 11 kV UDL feeder](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Audl-feeder-10km-11kv) — Transmission and Distribution System Design, chapter 7, asked 2×