LT feeder options: loss and voltage drop, 50 kW

Formulas (uniformly distributed load, 3-phase)

For each feeder of length $l$ carrying sending-end current $I$:

$$\begin{aligned} P_L &= 3\cdot\frac13 I^2 r l = I^2 r l\\ V_{D,ph} &= \frac12 I(r\cos\phi + x\sin\phi)\,l\\ \%V_D &= \frac{V_{D,ph}}{V_L/\sqrt3}\times100 \end{aligned}$$

The peak load is shared equally between the feeders.

Given: $P = 50$ kW, p.f. 0.8 ($\sin\phi = 0.6$), $V_L = 400$ V (secondary of 11/0.4 kV), $r = 1.1\ \Omega$/km, $x = 0.25\ \Omega$/km. $r\cos\phi + x\sin\phi = 1.1(0.8) + 0.25(0.6) = 1.03\ \Omega$/km.

Option I: 4 feeders × 2 km

$$\begin{aligned} P_{feeder} &= 50/4 = 12.5\ \text{kW},\quad I = \frac{12\,500}{\sqrt3\times400\times0.8} = 22.55\ \text{A}\\ P_L &= 4\times(22.55^2\times1.1\times2) = 4\times1118.98 = 4475.9\ \text{W}\\ \%P_L &= \frac{4475.9}{50\,000}\times100 = 8.95\ \%\\ V_{D,ph} &= \frac12\times22.55\times1.03\times2 = 23.23\ \text{V}\\ \%V_D &= \frac{23.23}{230.94}\times100 = 10.06\ \% \end{aligned}$$

Option II: 2 feeders × 4 km

$$\begin{aligned} P_{feeder} &= 25\ \text{kW},\quad I = \frac{25\,000}{\sqrt3\times400\times0.8} = 45.11\ \text{A}\\ P_L &= 2\times(45.11^2\times1.1\times4) = 2\times8951.8 = 17\,903.6\ \text{W}\\ \%P_L &= \frac{17\,903.6}{50\,000}\times100 = 35.81\ \%\\ V_{D,ph} &= \frac12\times45.11\times1.03\times4 = 92.92\ \text{V}\\ \%V_D &= \frac{92.92}{230.94}\times100 = 40.23\ \% \end{aligned}$$

Comparison

Going from option I to option II doubles both the current and the length of each feeder. The loss per feeder ($\propto I^2 l$) rises 8 times, and with half as many feeders the total loss rises 4 times; the drop ($\propto I l$) rises 4 times. Option I is clearly better, but even it exceeds the usual LT drop limit (about 5–8 %), suggesting a bigger conductor or more feeders/transformers.

Answer: Option I: 8.95 % loss, 10.06 % drop; Option II: 35.81 % loss, 40.23 % drop.