Capacitor size for maximum loss reduction (variant)

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+j1)-- A --(1+j1)-- B --(1+j2)-- C
              |            |            |
          100 kVA       50 kVA      100 kVA
          0.8 lag      0.85 lag     0.8 lag   [Qc]

Reactive load at each node

Reactive flow in each section

Optimum capacitor

$$Q_c = \frac{146.34+86.34+60}{1+1+1} = \frac{292.68}{3} = 97.56\ \text{kVAR}$$

In p.u. on 1000 kVA: $Q_c = 0.0976$ p.u.

Check of the loss reduction (p.u. on 1000 kVA):

$$\begin{aligned} \Delta P_L &= \sum R_i(2Q_iQ_c - Q_c^2)\\ &= 2(0.09756)(0.29268) - 3(0.09756)^2 = 0.02855\ \text{p.u.} \approx 28.6\ \text{kW} \end{aligned}$$

The capacitor is larger than the 60 kVAR load at C because it also relieves the heavier flows in S–A and A–B. In practice the nearest standard bank (about 100 kVAR) would be used.

Answer: Optimum capacitor at node C ≈ 97.6 kVAR (≈ 100 kVAR standard).