Loss and voltage drop, 16 km uniformly varying load

Assumption: "uniformly varying load" is taken as a uniformly increasing load, i.e. the load density rises linearly from zero at the sending end to a maximum at the far end (Gonen's case). No power factor is given, so the maximum possible drop is found using $|z|$ (the drop $I(R\cos\phi+X\sin\phi)$ is largest when it equals $I|Z|$).

Current distribution

Let load density $= kx$ (A/km). The current at distance $x$ from the source is the load beyond $x$:

$$\begin{aligned} I(x) &= \int_x^{l} k\,u\,du = \frac{k}{2}(l^2-x^2),\quad I_s = I(0) = \frac{kl^2}{2}\\ I(x) &= I_s\left(1-\frac{x^2}{l^2}\right) \end{aligned}$$

Voltage drop (per phase)

$$V_D = \int_0^l z\,I_s\left(1-\frac{x^2}{l^2}\right)dx = z I_s\left(l-\frac{l}{3}\right) = \frac{2}{3}I_s z l$$

Power loss (three-phase)

$$P_L = 3\int_0^l r I_s^2\left(1-\frac{x^2}{l^2}\right)^2dx = 3 r I_s^2 l\left(1-\frac{2}{3}+\frac{1}{5}\right) = 3\cdot\frac{8}{15}I_s^2 r l$$

Substitution

$I_s = 100$ A, $l = 16$ km, $R = rl = 0.6\times16 = 9.6\ \Omega$, $|z| = \sqrt{0.6^2+0.3^2} = 0.6708\ \Omega$/km.

$$\begin{aligned} P_L &= 3\times\frac{8}{15}\times100^2\times9.6 = 153{,}600\ \text{W} = 153.6\ \text{kW}\\ V_{D,ph} &= \frac{2}{3}\times100\times0.6708\times16 = 715.5\ \text{V}\\ V_{D,L} &= \sqrt3\times715.5 = 1239.4\ \text{V}\\ \%V_D &= \frac{1239.4}{11000}\times100 = 11.27\ \% \end{aligned}$$

Comparison with other load patterns

Answer: Power loss = 153.6 kW; maximum voltage drop ≈ 715.5 V per phase = 1239 V line (≈ 11.3 % of 11 kV).