LT loss and voltage drop, 60 kW four 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.

Data

Current per feeder

$$P_{feeder} = \frac{60}{4} = 15\ \text{kW},\quad I = \frac{15\,000}{\sqrt3\times380\times0.9} = 25.32\ \text{A}$$

i) LT percentage peak power loss

Resistance per phase per feeder $R = 0.9\times1.5 = 1.35\ \Omega$.

$$\begin{aligned} P_{L,feeder} &= I^2R = 25.32^2\times1.35 = 865.65\ \text{W}\\ P_{L,total} &= 4\times865.65 = 3462.6\ \text{W}\\ \%P_L &= \frac{3462.6}{60\,000}\times100 = 5.77\ \% \end{aligned}$$

ii) Maximum percentage voltage drop

$$\begin{aligned} r\cos\phi + x\sin\phi &= 0.9(0.9) + 0.25(0.4359) = 0.919\ \Omega/\text{km}\\ V_{D,ph} &= \frac12\times25.32\times0.919\times1.5 = 17.45\ \text{V}\\ V_{ph} &= 380/\sqrt3 = 219.39\ \text{V}\\ \%V_D &= \frac{17.45}{219.39}\times100 = 7.96\ \% \end{aligned}$$

Line-to-line drop $= \sqrt3\times17.45 = 30.23$ V.

Why the UDL factors

• The current falls linearly from $I$ at the transformer to zero at the end, so the drop equals that of the full current flowing over half the length ($\frac12 Il$).
• Loss integrates $I^2$, giving one-third of the lumped value; for three phases, $3\times\frac13 I^2 r l = I^2 r l$.

The drop (≈ 8 %) is at the upper end of the usual LT limit, and the loss (≈ 5.8 %) is acceptable but could be reduced with a larger conductor.

Answer: i) LT peak power loss = 5.77 % ii) Maximum voltage drop = 7.96 %.