HVDC power loss is 2/3 of HVAC

The claim compares a three-phase HVAC line (3 conductors) with a bipolar HVDC line (2 conductors) of the same power capacity, assuming each conductor has the same loss.

Step 1: Show that both lines carry the same power

Let both lines use the same conductors (same rms current $I$, resistance $R$ per conductor) and the same insulation (same peak voltage to earth $V_m$). For DC: $V_d = V_m$, $I_d = I$.

$$\begin{aligned} P_{ac} &= 3\cdot\frac{V_m}{\sqrt{2}}\cdot I\cos\phi = 2.121\,V_mI\cos\phi \\ P_{dc} &= 2V_dI_d = 2V_mI \end{aligned}$$

Equating, $P_{dc} = P_{ac}$ when $\cos\phi = \dfrac{2\sqrt{2}}{3} = 0.943$, which is a typical operating power factor. So a 2-conductor DC line has practically the same capacity as a 3-conductor AC line.

Step 2: Loss per conductor

Since both carry the same current $I$ (rms for AC, DC value for DC) through the same resistance $R$:

$$p_{ac} = I^2R = p_{dc} = p$$

(This is the given assumption.)

Step 3: Total losses

$$\begin{aligned} \text{Loss}_{ac} &= 3I^2R = 3p \\ \text{Loss}_{dc} &= 2I^2R = 2p \end{aligned}$$ $$\frac{\text{Loss}_{dc}}{\text{Loss}_{ac}} = \frac{2p}{3p} = \frac{2}{3}$$

Hence the total power loss of the HVDC line is only 2/3 of that of the HVAC line of the same capacity.

 HVAC:  ===== a  (loss p)      HVDC: ===== +Vd (loss p)
        ===== b  (loss p)            ===== -Vd (loss p)
        ===== c  (loss p)
 Total: 3p                    Total: 2p

Further points

• Efficiency of DC is also higher because there is no skin effect (AC resistance > DC resistance) and corona loss is lower.
• No reactive (charging) current flows in a DC line, so the full current rating carries real power.
• Converter station losses (about 0.5–1% per station) must be added, which is why HVDC pays off only for long distances.