Peak loss and voltage drop for LT feeder options

Assumptions: load is shared equally among the feeders of each option and uniformly distributed along each feeder; LT line voltage = 400 V (transformer secondary); $\cos\phi = 0.8$, $\sin\phi = 0.6$; $z = 1.1 + j0.25$ Ω/km per phase.

Formulae (uniformly distributed load)

$$\begin{aligned} I &= \frac{P_f}{\sqrt{3} \times 400 \times 0.8} \\ P_{loss} &= n \times I^2 R \quad (R = 1.1\,l) \\ \Delta V_L &= \frac{\sqrt{3}\, I (R\cos\phi + X\sin\phi)}{2} \end{aligned}$$

where $n$ = number of feeders, $P_f = 50/n$ kW per feeder, $l$ = feeder length.

Option I: 4 feeders × 3 km

$$\begin{aligned} P_f &= 12.5 \text{ kW}, \quad I = \frac{12{,}500}{\sqrt{3}(400)(0.8)} = 22.553 \text{ A} \\ R &= 3.3\ \Omega, \quad X = 0.75\ \Omega, \quad R\cos\phi + X\sin\phi = 3.09\ \Omega \\ P_{loss} &= 4 \times 22.553^2 \times 3.3 = 6713.9 \text{ W} \Rightarrow 13.43\% \\ \Delta V &= \frac{\sqrt{3}(22.553)(3.09)}{2} = 60.35 \text{ V} \Rightarrow 15.09\% \end{aligned}$$

Option II: 2 feeders × 4 km

$$\begin{aligned} P_f &= 25 \text{ kW}, \quad I = 45.105 \text{ A} \\ R &= 4.4\ \Omega, \quad X = 1.0\ \Omega, \quad R\cos\phi + X\sin\phi = 4.12\ \Omega \\ P_{loss} &= 2 \times 45.105^2 \times 4.4 = 17{,}903.6 \text{ W} \Rightarrow 35.81\% \\ \Delta V &= \frac{\sqrt{3}(45.105)(4.12)}{2} = 160.94 \text{ V} \Rightarrow 40.23\% \end{aligned}$$

Option III: 3 feeders × 2 km

$$\begin{aligned} P_f &= 16.667 \text{ kW}, \quad I = 30.070 \text{ A} \\ R &= 2.2\ \Omega, \quad X = 0.5\ \Omega, \quad R\cos\phi + X\sin\phi = 2.06\ \Omega \\ P_{loss} &= 3 \times 30.070^2 \times 2.2 = 5967.9 \text{ W} \Rightarrow 11.94\% \\ \Delta V &= \frac{\sqrt{3}(30.070)(2.06)}{2} = 53.65 \text{ V} \Rightarrow 13.41\% \end{aligned}$$

Summary

Answer: Option III (3 feeders × 2 km) gives the lowest peak loss (11.94%) and voltage drop (13.41%); option II is the worst. All options exceed the usual LT voltage drop limit (about 5–10%), so in practice the area should be split with an additional transformer, a larger conductor, or a higher voltage (11 kV) brought closer to the loads.