Capacitor size for maximum loss reduction

A shunt capacitor supplies reactive power locally, reducing the reactive current in every section upstream of it. The optimum size minimises the total $I^2R$ loss.

Assumptions: all node voltages = 1 p.u. (given), so section current magnitude² = $P^2 + Q^2$ in p.u.; base 1000 kVA; transformer impedances neglected.

Step 1: Loads in p.u.

Step 2: Section flows

Step 3: Loss as a function of capacitor size

With capacitor $Q_c$ at C, the reactive flow in every section reduces by $Q_c$:

$$P_L = \sum R_i \left[P_i^2 + (Q_i - Q_c)^2\right]$$

For maximum loss reduction:

$$\begin{aligned} \frac{dP_L}{dQ_c} &= -2\sum R_i (Q_i - Q_c) = 0 \\ Q_c &= \frac{\sum R_i Q_i}{\sum R_i} \\ &= \frac{1(0.2854) + 2(0.2254) + 3(0.16)}{1 + 2 + 3} \\ &= \frac{1.2162}{6} = 0.2027 \text{ p.u.} \end{aligned}$$ $$Q_c = 0.2027 \times 1000 = 202.7 \text{ kVAR}$$

Step 4: Check of loss reduction (p.u.)

Loss falls from 0.545 p.u. to 0.299 p.u., a reduction of about 45%. (The given impedances are very large, so the absolute p.u. losses are only for comparison.)

Note: the optimum $Q_c$ (0.203 p.u.) is larger than the reactive load at C (0.16 p.u.) because upstream sections also carry the reactive power of A and B; the result is a weighted average of section reactive flows.

Answer: Capacitor size ≈ 203 kVAR at node C (a standard 200 kVAR bank would be used).