Loss and drop, 15 km 11 kV feeder

Derivation for uniformly distributed load

Let a feeder of length $l$ carry sending-end current $I_s$, with load tapped uniformly so the current falls linearly to zero at the far end:

$$I(x) = I_s\left(1-\frac{x}{l}\right)$$

Per-phase voltage drop, with $z = r + jx$ per km:

$$V_D = \int_0^l z I_s\left(1-\frac{x}{l}\right)dx = \frac{1}{2}I_s z l$$

Three-phase power loss:

$$P_L = 3\int_0^l r I_s^2\left(1-\frac{x}{l}\right)^2dx = 3\cdot\frac{1}{3}I_s^2 r l = I_s^2 r l$$

So a UDL feeder behaves like the whole load lumped at the middle for voltage drop, and at one-third of the length for loss.

Substitution

$I_s = 100$ A, $l = 15$ km, $z = 0.6 + j3\ \Omega$/km, $|z| = \sqrt{0.6^2+3^2} = 3.059\ \Omega$/km. No load p.f. is given, so the drop is taken as $\frac12 I_s|z|l$ (the maximum value).

$$\begin{aligned} P_L &= 100^2\times0.6\times15 = 90{,}000\ \text{W} = 90\ \text{kW}\\ V_{D,ph} &= \frac{1}{2}\times100\times3.059\times15 = 2294.6\ \text{V}\\ V_{D,L} &= \sqrt3\times2294.6 = 3974.3\ \text{V}\ (36.1\ \%\ \text{of } 11\ \text{kV}) \end{aligned}$$

Note: a reactance of 3 Ω/km is unusually high for an 11 kV line (typical is about 0.3–0.4 Ω/km). If $z = 0.6 + j0.3$ was intended, $|z| = 0.6708$, $V_{D,ph} = 503.1$ V, $V_{D,L} = 871.4$ V (7.92 %); the loss is unchanged at 90 kW since it depends only on $r$.

Answer: Power loss = 90 kW; voltage drop ≈ 2294.6 V per phase (3974 V line) with the given z.