Distribution loss and voltage drop, four LT feeders

Assumptions:

• The 40 kW peak is shared equally by the four feeders (10 kW each) and the load is uniformly distributed along each feeder.
• The reactance is 0.25 Ω/km per phase.
• The transformer iron loss (200 W) and copper loss (480 W) are the losses at this peak load (the transformer rating is not given).
• % distribution loss is taken as peak power loss as a percentage of peak demand.

Step 1: Current in each feeder

$$I = \frac{P}{\sqrt{3} V \cos\phi} = \frac{10{,}000}{\sqrt{3} \times 380 \times 0.85} = 17.875 \text{ A}$$

Step 2: Feeder impedance

$$\begin{aligned} R &= 0.9 \times 1.5 = 1.35\ \Omega, \quad X = 0.25 \times 1.5 = 0.375\ \Omega \\ \sin\phi &= \sqrt{1 - 0.85^2} = 0.5268 \end{aligned}$$

Step 3: LT line loss

For uniformly distributed load, loss = 1/3 of the end-loaded value:

$$\begin{aligned} P_{loss,feeder} &= \frac{3 I^2 R}{3} = I^2 R = 17.875^2 \times 1.35 = 431.33 \text{ W} \\ P_{loss,LT} &= 4 \times 431.33 = 1725.31 \text{ W} \end{aligned}$$

Step 4: Total distribution loss

$$\begin{aligned} P_{loss} &= 1725.31 + 200 + 480 = 2405.31 \text{ W} \\ \%\text{ loss} &= \frac{2405.31}{40{,}000} \times 100 = 6.01\% \end{aligned}$$

(LT line alone: 4.31%; transformer: 1.70%. On the basis of input power, $2405.31/42{,}405.31 = 5.67\%$.)

Step 5: Voltage drop in LT line

For uniformly distributed load, drop = 1/2 of the end-loaded value:

$$\begin{aligned} \Delta V_L &= \frac{\sqrt{3}\, I (R\cos\phi + X\sin\phi)}{2} \\ &= \frac{\sqrt{3} \times 17.875 \times (1.35 \times 0.85 + 0.375 \times 0.5268)}{2} \\ &= \frac{\sqrt{3} \times 17.875 \times 1.3450}{2} = 20.82 \text{ V} \\ \%\Delta V &= \frac{20.82}{380} \times 100 = 5.48\% \end{aligned}$$

Answer:

• (i) % distribution loss ≈ 6.01% of peak demand (2.41 kW)
• (ii) % voltage drop in each LT feeder ≈ 5.48% (20.82 V)

The voltage drop is within the usual ±10% LT limit (and close to a 5% limit), and losses are acceptable.