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Energy transmitted via power lines necessarily undergoes some degree of attenuation. The question is, how much?

This source, screencapped from this video, says:

enter image description here

But, I've heard via word-of-mouth from a source I previously trusted that losses can be as high as 50%. (This person is, however, entering the 8th decade of life.)

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  • $\begingroup$ So, that tells you... just choose the distance you want to work with and AC/DC... $\endgroup$
    – Solar Mike
    Commented Nov 15, 2021 at 20:02
  • $\begingroup$ I don't have a source but when I was involved with interruptible power at a factory I seem to recall that losses at highest demand on hottest days were ~20% $\endgroup$
    – Tiger Guy
    Commented Nov 15, 2021 at 20:49
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    $\begingroup$ power = I×V. What is the voltage drop in transmission and how much current do you draw during that drop? Also = I×I×R so you lose proportional to resistance of the transmission line (which does vary by temperature). We can pull percentages as we please, but exactly 0 guesses will exactly match reality. What error bounds are you willing to tolerate? $\endgroup$
    – Abel
    Commented Nov 16, 2021 at 1:44
  • $\begingroup$ @PhilSweet what chart? $\endgroup$
    – Solar Mike
    Commented Nov 16, 2021 at 7:17
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    $\begingroup$ I would post this on electronics.stackexchange.com. A EE who works in power will likely know this as a matter of course. $\endgroup$
    – Tiger Guy
    Commented Nov 17, 2021 at 13:25

2 Answers 2

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Losses are due to skin effect. Direct current flows through the full cross-sectional area (CSA) of the, typically aluminum, conductor, while alternating current flows on the outer surface of the conductor. This reduces the effective CSA of the conductor and increases the resistive losses to the cable.

At 4:44 of video, Real Engineering illustrates this relationship:

enter image description here

So HVDC have less losses to deal with. HVDC have additional costs due to conversion to and from DC and cheaper transmission towers, so there will be a cross-over point where it is cheaper to use HVDC transmission vs high-voltage 3-phase AC transmission.

Question image was extracted from eia U.S. Energy Information Administration - Assessing HVDC Transmission for Impacts of Non‐Dispatchable Generation

Previous paragraph explains this.

Lower reactive and “skin effect” losses: The power‐carrying capability of AC lines is affected by the reactive power component of AC power, and the “skin effect” losses, which cause a non‐uniform distribution of current over the cross‐sectional area of the conductor. HVDC lines are not affected by reactive power components nor do they experience any losses because of “skin effect.”

Same page of that document, the image from ABB shows a crossover point for the same voltage HVDC transmission vs 3-phase AC vs cost.

enter image description here

It's 300km for 1,200MW. Video states around 500 to 800km, but breakeven point would vary based upon transmitted power. More power, more current and greater the $I^2 R_{line}$ losses, so the shorter the breakeven point.

The eia report cites a Siemens report (High Voltage Direct Current Transmission – Proven Technology for Power Exchange), which provides additional details.

The advantages of a DC link over an AC link are:

■ A DC link allows power transmission between AC networks with different frequencies or networks, which can not be synchronized, for other reasons.

■ Inductive and capacitive parameters do not limit the transmission capacity or the maximum length of a DC overhead line or cable. The conductor cross section is fully utilized because there is no skin effect.

Which can be interpreted as:

  • HVDC allows going from different countries with different frequencies (60Hz to 50Hz) or slight variations in fundamental frequencies (as in generator synchronization or load variance).

  • AC current is continuously changing, so there is inductive and capacitive coupling between the cables and surrounding environment. Essentially a reactive power loss (MVAR). HVDC current would depend on load, so changes, but if you keep load constant, no inductive and capacitive coupling.

  • Current flows through full CSA of conductor, so less resistance/CSA/km.

Actual losses are as mentioned in video.

: On average, the losses on the HVDC lines are roughly 3.5% per 1000 km, contrasted with 6.7% for comparable AC lines at similar voltage levels (Siemens 2017). HVDC lines also experience losses at the converter stations, which range between 0.6 and 1% of the power delivered. In a side‐by‐side comparison, the total HVDC transmission losses are still lower than AC losses for long‐distance lines (lower by 30%–40%, typically).

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Energy loss in the cable is mainly caused by copper loss: $$I^2R$$ Where
I is the current passes through the cable.
R is the resistance along in the cable.

We could reduce copper loss by reducing the value of I, i.e., current.

Real energy transferred between two points of the cable is given by $$P×t=VI×t$$

Where
P is the power transferred between the two points of the cable.
t is the time elapsed.
V is the Voltage at the source end of the cable.
I is the current at the source end of the cable.

Electric power, P is the product of voltage, V and current, I.

We could transmit the same amount of power using a high value of V but a small value of I.

So, one of the possible ways to reduce energy loss is the transit energy at high voltage.

If we were to transmit gigawatt-hour (GWh) of energy from power station to load centres at low voltage, such as 400 V, all the energy would be lost along the tranmission line. Hence we usually transmit at ultra high voltages such as 230 kV and 400 kV. This is known as AC transmission.

AC transmission will encounter its limitations (reflection, synchronization, etc) when the trasmission line length is at a signficant length of its AC half wavelength (3000 km for 50 Hz AC). Hence High Voltage DC (HVDC) transmission system is used when transmission distance is almost 1000 km (could be lower depending on many practical issues)

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  • $\begingroup$ this answer explains the mechanism of loss but not the requested answer of how much. $\endgroup$
    – Tiger Guy
    Commented Nov 17, 2021 at 13:24
  • $\begingroup$ It depends on transmission voltage, current in the cable, resistance of the cable, surrounding temperature of the cable, number of neighbouring cables, installation method, etc. There is no standard answer. $\endgroup$
    – yoonghm
    Commented Nov 17, 2021 at 15:21

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