Capacitor size, feeder with 500 kVA end load

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 --(3+j2)-- A --(2+j4)-- B --(1+j2)-- C
              |            |            |
          200 kVA      100 kVA      500 kVA
          0.8 lag      0.7 lag      0.6 lag  [Qc]

Reactive loads

Section flows

Optimum capacitor

$$Q_c = \frac{1774.24+942.83+400}{3+2+1} = \frac{3117.07}{6} = 519.5\ \text{kVAR}$$

Because the largest resistance (S–A) carries the largest reactive flow, the optimum capacitor exceeds the 400 kVAR load at C; it slightly over-compensates section B–C but gives the largest total loss reduction.

Answer: Capacitor at node C ≈ 519.5 kVAR (≈ 520 kVAR).