I have accurate statistics about the total length of high-voltage (110-400 kV), medium-voltage (1-70 kV) and low-voltage (0-1 kV) transmission lines in my country, and they are:

  • High-voltage: 22 500 km
  • Medium-voltage: 140 000 km
  • Low-voltage: 240 000 km

I would like to obtain an accurate estimate of how much aluminum is used for the transmission lines, but so far have failed to find any. I assume all lines have approximately 3 conductors due to the 3-phase system (in some cases there may be a fourth one, but 3 is a good approximation). Thus, to calculate the amount of aluminum, I would need an accurate estimate of conductor radius for high, medium and low voltage networks. Obviously, I cannot climb to reach the transmission line and measure its size by a vernier caliper!

One source I found is http://large.stanford.edu/courses/2010/ph240/harting1/ where a figure says that optimal radius (one that minimizes total costs) for a 2.25 GW 3-phase 765 kV line is 4 centimeters.

Thus, for a total length of 402 500 km of transmission lines, there would be 402500000*pi*0.04^2 = 2023186 square meters of aluminum, or by weight 2023186 m3 * 2700 kg/m3 = 5462602200 kg of aluminum, per phase. Three phases would require 16.4 million metric tons of aluminum. Per capita, that would be approximately 3 metric tons, or 3000 kg.

I feel that estimate may be a bit on the high side, because I found a statistic that about 25 kg of aluminum is used per year per capita. Thus, it would require 120 years of production to manufacture 3 metric tons of aluminum per capita. I don't believe that all lines, including the low-voltage (and presumably low-current) lines would use 4 cm radius conductors.

So, how much aluminum does a kilometer of high voltage (HV), medium voltage (MV) or low voltage (LV) transmission line use?

  • $\begingroup$ The amperage, not the voltage is key to the cable size. For 3 phase , I believe a fourth wire is used on top as a lightening arrestor.. $\endgroup$ Commented Jul 28, 2019 at 20:00

1 Answer 1


Ok, since nobody answered, I investigated this more on my own. Note: the answer is just an estimate and might have incorrect assumptions. In particular, I'm assuming 400 kV means 400 kV line-to-line voltage and not 400 kV line-to-neutral voltage, and I'm also assuming the voltages and amperages are RMS and not peak. These assumptions, if incorrect, change the estimate only in a very minor way.

According to http://large.stanford.edu/courses/2010/ph240/harting1/ there is a corona discharge threshold, which is given by the equation:

U / 1.73 - 21.1 kV/cm * 0.95 * x * 1 * ln(d/x) = 0

where U is the line-to-line voltage in kV, 1/1.73 converts line-to-line voltage to line-to-neutral voltage, x is the conductor radius in centimeters, and d is the conductor spacing in centimeters.

Example: if U = 400 kV, d = 550 cm, this equation gives x = 2.0655 cm, so approximately 2 cm conductor radius is needed for 400 kV lines with 5.5 meters of spacing or else corona discharge occurs and steals a lot of power.

For high voltages, the conductor size is determined not by amperage, but rather by voltage, because of this corona discharge.

According to a thesis, about 1.5 meter spacing is used for 110 kV and 5.5 meter of spacing is used for 400 kV. We can assume 220 kV uses about 3.5 meters of spacing therefore. Plugging in these values, we get:

  • 400 kV: 2 cm radius needed
  • 220 kV: 1 cm radius needed
  • 110 kV: 0.5 cm radius needed

The thesis also gives in Appendix 1 typical aluminum cross sections:

  • Duck: 305 mm2: 0.99 cm radius
  • Finch: 565mm2: 1.34 cm radius

We can probably be safe in saying that for 110 kV, it's not the corona discharge that limits the transmission line radius, but rather the resistive losses. The thesis gave no example of 0.5 cm radius conductor. So, we probably can bump up the 110 kV radius to 1.0 cm.

I also have the information that the national grid operator has these transmission lines:

  • 400 kV: 4600 km
  • 220 kV: 2200 km
  • 110 kV: 7600 km

The rest is 22500 km - 4600 km - 2200 km - 7600 km = 8100 km, which are the high voltage lines not operated by the national grid operator. We can probably be safe in saying that the lines not operated by the national grid operator are either 220 kV or 110 kV, so we get:

  • 400 kV: 4600 km @ 2 cm radius => 46.8 kilotons of aluminum used
  • 220 kV: 2200 km @ 1 cm radius => 5.6 kilotons of aluminum used
  • 110 kV: 7600 km @ 1 cm radius => 19.3 kilotons of aluminum used
  • 110 or 220 kV: 8100 km @ 1 cm radius => 20.6 kilotons of aluminum used

So, we can safely say the HV lines use about 92 kilotons of aluminum.

How about MV and LV, then? Well, the amperage isn't probably reduced much from 110 kV, so I assume they have conductor radius between 0.5 cm (242 kilotons of aluminum) and 1.0 cm (967 kilotons of aluminum). The thesis (page 25) says 1 mm2 cross section shouldn't carry more than 1 ampere, so a 30 kV medium voltage line would carry per phase 30 kV * 79 A = 2370 kW, or 30 kV * 314 A = 9420 kW, depending on the conductor radius (0.5 cm or 1.0 cm). This would give a power value between 7 MW - 28 MW, reasonable to power quite many homes.

So, in summary, the transmission lines in this country are:

  • HV: 92 kilotons of aluminum
  • MV and LV: between 242 and 967 kilotons of aluminum
  • Total: between 334 and 1059 kilotons of aluminum (corresponding to between 61 and 193 kg per capita, which is between 2.4 and 7.7 years worth of aluminum production)

Now we seem to be in the correct order of magnitude in the estimate.

  • $\begingroup$ I wouldn't ignore the 4th wire used to provide lightning protection. It will add a substantial amount of metal. $\endgroup$
    – user16
    Commented Jul 30, 2019 at 21:22

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