Line efficiency increases with voltage

Transmission efficiency rises with voltage because, for the same power, a higher voltage means a smaller current and hence much smaller $I^2R$ loss.

Consider a 3-phase line delivering power $P$ at line voltage $V$ and power factor $\cos\phi$. Let $R$ be the resistance per phase, $R = \rho l / a$.

Line current

$$I = \frac{P}{\sqrt3\, V\cos\phi}$$

Line loss

$$\begin{aligned} P_{loss} &= 3I^2R = 3\left(\frac{P}{\sqrt3 V\cos\phi}\right)^2 \frac{\rho l}{a} \\ &= \frac{P^2 \rho l}{V^2 \cos^2\phi\, a} \end{aligned}$$

Efficiency

$$\begin{aligned} \eta &= \frac{P}{P + P_{loss}} = \frac{1}{1 + \dfrac{P\rho l}{a V^2\cos^2\phi}} \\ &\approx 1 - \frac{P\rho l}{a\, V^2\cos^2\phi} \end{aligned}$$

For given $P$, $l$, $a$, $\rho$ and $\cos\phi$, the loss term is proportional to $1/V^2$. So as $V$ increases, the loss falls and $\eta$ increases.

Example

A line with $R = 10\ \Omega$/phase carries 10 MW at 0.8 pf.

Doubling the voltage cuts the loss to one quarter.

Related results

• For a fixed percentage loss, the conductor area needed is $a \propto 1/V^2$, so the conductor volume (and cost) falls as $1/V^2$.
• The percentage voltage drop $\approx I(R\cos\phi + X\sin\phi)/V$ also falls as $1/V^2$, so regulation improves.
• The limit is the cost of insulation, towers and switchgear, which rises with voltage; this sets the economic transmission voltage.