LT loss and voltage drop, four 2 km feeders

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: $P = 40$ kW, p.f. 0.8 ($\sin\phi = 0.6$), 4 feeders, $V_L = 380$ V, $l = 2$ km, $r = 1.1$, $x = 0.25\ \Omega$/km.

Current per feeder

$$P_{feeder} = \frac{40}{4} = 10\ \text{kW},\quad I = \frac{10\,000}{\sqrt3\times380\times0.8} = 18.99\ \text{A}$$

i) Percentage peak power loss

$$\begin{aligned} P_{L,feeder} &= I^2 r l = 18.99^2\times1.1\times2 = 793.5\ \text{W}\\ P_{L,total} &= 4\times793.5 = 3174.1\ \text{W}\\ \%P_L &= \frac{3174.1}{40\,000}\times100 = 7.94\ \% \end{aligned}$$

ii) Maximum percentage voltage drop

The maximum drop occurs at the far end of each feeder.

$$\begin{aligned} r\cos\phi + x\sin\phi &= 1.1(0.8)+0.25(0.6) = 1.03\ \Omega/\text{km}\\ V_{D,ph} &= \frac12\times18.99\times1.03\times2 = 19.56\ \text{V}\\ V_{ph} &= 380/\sqrt3 = 219.39\ \text{V}\\ \%V_D &= \frac{19.56}{219.39}\times100 = 8.92\ \% \end{aligned}$$

(Line value of drop $= \sqrt3\times19.56 = 33.88$ V.)

The drop is above the common LT limit of about 5–8 %, so a larger conductor or a shorter LT reach would be needed in practice.

Answer: i) 7.94 % ii) 8.92 %.