5 ms·
It is inefficient to distribute electricity over very long distances currently, and that inefficiency increases exponentially. Perhaps in the future a global el
by gnode 7y ago
It is inefficient to distribute electricity over very long distances currently, and that inefficiency increases exponentially. Perhaps in the future a global electrical grid backed primarily with solar may be practical with superconducting transmission lines, but for now, generation is relatively local.
While a fossil fuel or nuclear plant is not reliable, their unreliability is uncorrelated. Solar and wind have geographically correlated unreliability, which is a major problem. To make them practical, large scale energy storage is necessary.
As energy storage is expensive, fossil fuel and nuclear redundant generation is going to be necessary for some time. Scaling demand could also help ease the problem. Electric storage heaters, aluminium smelters, and air conditioning for example can respond dynamically to supply through a real time market, or demand-side response agreement.
- astrodust 7y ago> It is inefficient to distribute electricity over very long distances currently... Québec built a 735kV AC very-long-haul transmission system in 1965 to address the efficiency concerns, carrying power over 1000km efficiently and reliably. That's almost the width of Texas. It's also been using renewable hydro-electric power.
- pintxo 7y agoTrue. Today, and probably in the next couple of years. But in 10, 20, 40 years? Hopefully not so much any longer. I firmly believe in the power of engineering research and given humanities history in solving engineering challenges, we will also solve this one. If we will end up with reliable long distance cables, or some form of energy storage I have no opinion on, only the future will tell us.
- mrfusion 7y agoI’ve never heard of it increasing exponentially. Also with very high voltage AC we can actually transmit quite far.
- bluGill 7y agoWith high voltage DC we can transmit farther then high voltage AC (for reasons I don't understand - ask a high voltage engineer), and it avoids all the phase problems with long distance AC.
- mrfusion 7y agoI don’t understand how you change between voltages with dc? Is it as efficient as AC using transformers?
- bluGill 7y agoAsk an electrical engineer not me. The basic idea is you go to high voltage DC at the generation end, run a long ways (across states), turn the DC to AC and then run that through a transformer. It isn't as efficient as a transformer, but it isn't far behind so you more than get it back by the smaller losses in transmission.
- pjc50 7y agoEffectively a giant version of your laptop power supply. The conversion is not quite as efficient as a transformer, and certainly more expensive, but for long distances the whole system can be more efficient than AC. (The magic component is the "insulated gate bipolar transistor", sometimes using silicon carbide as the semiconducting element. These are capable of multi-kilovolt switching.)
- gnode 7y agoCapacitance is a major reason; the capacitance of the very large line doesn't need to be charged and discharged at very high voltage 100 or 120 times a second (per half wave of 50 or 60Hz). Also, an X volts RMS AC line is 2.8X volts peak-to-peak. The peak voltage is limiting with regards to breakdown voltages, and so DC can reach a higher equivalent voltage on a line with the same maximum voltage.
- Nasrudith 7y agoIt makes sense if you confuse electrical fields with current. Since the exponential increase for transmission is geometrical essentially - which is why we prefer wires for transmission so strongly.
- dgacmu 7y agoIt does. There's also the somewhat complicated issue of reactive power, but there are ways to solve that, such as DC transmission, or capacitor banks.
- gnode 7y agoI should have elaborated on my exponential claim. Resistive losses are indeed linear, however in the context of a maximum transmission voltage, efficiency is bounded exponentially. If you consider a voltage source of 1V, a line of 1ohm, with a load of 1ohm (0.25W), the transmission is 50% efficient. With a load of 0.5ohm (0.33W) the line is 33% efficient. with a load of 0ohm it is 0% efficient. > Also with very high voltage AC we can actually transmit quite far. We can't transmit efficiently at a quarter the circumference of the globe, which would be required for a continually directly solar powered grid.
- dgacmu 7y agoThis is not correct. Resistive losses are linear with distance. There are some other issues such as getting out of phase, which can require a transformer (and its associated losses), but there's absolutely nothing exponential. For the basics: http://www.learnabout-electronics.org/Resistors/resistors_03.php http://www.learnabout-electronics.org/Resistors/resistors_03...
- gnode 7y agoNaively that's true, but if you consider a distribution grid with a series of transmission line segments at a given voltage, after each segment the voltage is increased to normalise it to the target transmission voltage. Each stage results in a proportional loss of power transmitted, resulting in an exponential loss as distance increases. Achieving a linear transmission loss ignores bounds on the source voltage. In this case, you may as well choose a transmission voltage of infinity and say resistive losses are constant because current tends to zero.
- dgacmu 7y agoNo? If I lose a fixed percent at each stage, my losses as a function of distance are slightly sublinear. Ex: 10% per stage, I lose 10% at first loss, then 9%, and so on. Further, if I have intermediate transformers, my current on later segments is lower, which means that my resistive heating is lower, which means my resistance (and therefore loss) is lower.
- gnode 7y agoConsider supplying a fixed amount of power P to a load. If each stage loses 10% of the power transmitted, after one stage you need to supply 1.1P, after the second stage, you must supply that power plus ten percent, and so on, so you need to supply 1.1^N P after N stages.