Capacitor size, feeder with three equal loads

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)-- 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.

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.