Power loss, 10 km 11 kV UDL 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.

Why $1/3$

Each small element $dx$ at distance $x$ carries current $I(x)$. Integrating $r I(x)^2$ gives

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

so the loss is only one-third of what it would be with the whole load at the far end.

Substitution

$I_s = 100$ A, $r = 0.4\ \Omega$/km, $l = 10$ km, so $R = 4\ \Omega$ per phase.

$$\begin{aligned} P_L &= 3\times\frac{1}{3}I_s^2 R = I_s^2 R\\ &= 100^2\times0.4\times10 = 40{,}000\ \text{W} \end{aligned}$$

(For comparison, with the same current lumped at the end, loss $=3\times100^2\times4 = 120$ kW.)

The reactance $X = 0.3\ \Omega$/km does not affect real power loss; it only enters the voltage drop, $V_D = \frac12 I_s(R\cos\phi + X\sin\phi)$.

Answer: Power loss = 40 kW.