When HVDC is economical; equal transmission capacity

HVDC needs costly converter stations (thyristor valves, converter transformers, filters, reactive compensation) at both ends, but its line is cheaper per km and has lower losses. So HVDC becomes economical only when the line saving is large enough to pay for the terminals, i.e. for bulk power over a long distance.

Justification of the statement

• Terminal cost: HVDC terminal cost is much higher than AC substations. This is a fixed cost, independent of distance.
• Line cost: DC line needs 2 conductors instead of 3, lighter towers, less insulation and narrower right of way, so cost per km is lower. Its losses are about 2/3 of AC.
• Total cost = terminal cost + (cost per km × distance). The AC curve starts low but rises steeply; the DC curve starts high but rises slowly. They cross at the break-even distance (about 500–800 km for overhead lines, 40–80 km for cables).

 Cost
  |                      / AC
  |                    /
  |                  /    _____ DC
  |            _____X-----
  |     _____-    / :
  | DC -        /   :
  |           /     :
  |  AC     /       :
  |       /         :
  |     /           :
  +-----------------:------> km
             break-even distance

• Power level: converter cost per MW falls as rating rises, and the 2-conductor saving grows with power; for small power the fixed terminal cost cannot be recovered.
• Also, long AC lines need series and shunt compensation and are stability-limited ($P = \frac{V_1V_2}{X}\sin\delta$), adding cost; DC has no such limit.

So HVDC is cheaper than HVAC only when large power is sent beyond the break-even distance (or for long cables / asynchronous links).

Power transmission capacity of HVDC and HVAC lines are equal

Assumptions: same conductor size and current $I$, same insulation (peak voltage to earth $V_m$), DC voltage $V_d = V_m$, $I_d = I$.

$$\begin{aligned} P_{ac} &= 3\cdot\frac{V_m}{\sqrt{2}}\cdot I\cos\phi = 2.121\,V_mI\cos\phi \\ P_{dc} &= 2V_dI_d = 2V_mI \\ \frac{P_{dc}}{P_{ac}} &= \frac{0.943}{\cos\phi} \end{aligned}$$

For a normal power factor of $\cos\phi = 0.943$, $P_{dc} = P_{ac}$. Thus a bipolar DC line with 2 conductors has the same capacity as a three-phase AC line with 3 conductors, while its losses are $2I^2R$ against $3I^2R$ (2/3) and it saves one conductor and its insulators.