--- title: "Capacitor size, feeder with three equal loads" subject: "Transmission and Distribution System Design" chapter: 7 key: "semester_8/transmission_and_distribution_system_design:capacitor-size-equal-loads-ohms" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/capacitor-size-equal-loads-ohms site: BCT Engineering --- # Capacitor size, feeder with three equal loads Transmission and Distribution System Design · Chapter 7: Distribution System Design - Asked in: 2067 Mangsir (old course), 6 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-equal-loads-ohms - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7 ## Question For an 11 kV radial primary distribution feeder shown in figure below, determine the size of the capacitor (kVAR) at the stated location in the feeder to achieve the maximum loss reduction. For simplifying the analysis assumes voltage at each node is 1 p.u. Justify your answer with a valid logic. [Figure: radial feeder from a 0.5 MVA, 33/11 kV source substation S through three sections each of impedance 1+j2 Ω to three successive nodes; each node supplies 100 kVA at 0.8 p.f. lag through a distribution transformer. The capacitor is placed at the last (third) node.] ## Short answer The optimal capacitor size is the resistance-weighted average of reactive power flows in the relieved sections. With each section R = 1 Ω and flows 180, 120, 60 kVAR, Q_c = 120 kVAR. ## 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)-- N1 --(1+j2)-- N2 --(1+j2)-- N3 | | | 100 kVA 100 kVA 100 kVA 0.8 lag 0.8 lag 0.8 lag [Qc] ``` #### Reactive flows Each load: $Q = 100 \times 0.6 = 60$ kVAR. | Section | R (Ω) | Q flow (kVAR) | R × Q | |---|---|---|---| | S–N1 | 1 | 180 | 180 | | N1–N2 | 1 | 120 | 120 | | N2–N3 | 1 | 60 | 60 | | Total | 3 | | 360 | #### Optimum capacitor $$ Q_c = \frac{180+120+60}{3} = 120\ \text{kVAR} $$ #### Justification - With $V=1$ p.u., the reactive current in each section is proportional to its reactive flow, so the loss due to reactive current in a section is $R(Q-Q_c)^2$. - If $Q_c$ were only 60 kVAR (just the local load), sections S–N1 and N1–N2 would still carry 120 and 60 kVAR. If $Q_c$ were 180 kVAR, section N2–N3 would carry 120 kVAR leading (and N1–N2 60 kVAR leading), raising their losses. - 120 kVAR is the value at which the extra loss in the over-compensated last section exactly balances the saving in the upstream sections ($d\Delta P/dQ_c=0$). With equal resistances it is simply the average of the section flows. - The result equals 2/3 of the total reactive load (2/3 × 180 = 120 kVAR). This matches the size given by the "2/3 rule", but that rule is for a uniformly distributed reactive load and places the capacitor at 2/3 of the feeder length; here the size follows from the derivation above for a capacitor at the last node. **Answer: 120 kVAR at the third node.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Each 100 kVA, 0.8 p.f. lag load has reactive demand of: A. 80 kVAR B. 100 kVAR C. 75 kVAR D. 60 kVAR Answer: D. Q = 100 x sinφ = 100 x 0.6 = 60 kVAR. (Key point: Reactive load per node) 2. With equal section resistances and flows 180, 120, 60 kVAR, the optimum capacitor is: A. 180 kVAR B. 120 kVAR C. 60 kVAR D. 90 kVAR Answer: B. Average of the flows: (180+120+60)/3 = 120 kVAR. (Key point: Optimum value) 3. Why not choose 180 kVAR (the full reactive load)? A. It exceeds the source rating B. Last section would carry leading reactive flow, raising its loss more than the upstream saving C. Capacitors come only in 120 kVAR D. The voltage would drop Answer: B. At 180 kVAR the last section carries 120 kVAR leading, so extra loss outweighs the gain. (Key point: Over-compensation) 4. For equal sections and loads, the optimum capacitor is about what fraction of total reactive load? A. 1/3 B. 1/2 C. 2/3 D. All of it Answer: C. 2/3 x 180 = 120 kVAR. 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