Capacitor size, four-node 11 kV feeder

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.

 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

Reactive flow in each section

(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).