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What California is learning from solar panels built over irrigation canals

372 points · 744 comments · Jtsummers

  1. nbf_1995 · · focus · HN ↗
    It seems like it would be more effective to simply put all the solar panels in a field, and construct a cheap shade over the entire canal no? The supports in the picture are massive and don't look cheap. Also I would imagine the solar array uses more copper than a similar capacity array just built in a field. You can't daisy chain multiple miles of solar panels together, so you need an extra power line to run alongside the whole thing.

    Put the solar panels in a field: The solar array uses less copper. The shade supports don't have to hold up solar panels: Shade supports cost less.

    The best reasoning they give is that California has insane permitting requirements, and it takes 1/6 the time to build on developed land compared to undeveloped land.

    1. sophacles · · focus · HN ↗
      Power lines are made of aluminum and steel.

      I seriously doubt that you can chain even 1/4 mile of panels together without destroying things. Thats already puting thousands of volts and hundreds or thousands of amps through the silicon. If you are making it all parallel you still need wiring between panels that can handle that. The cabling doesn't go away, it just moves.

      Wouldn't there still be supports in a feild of solar panels? Are you sure that those supports + the supports for the shade material are going to be less material than the supports for this?

      Most shade material wears out pretty quickly. Will their replacement result in more expense, more waste, etc than just putting the solar panels?

      Shade material is generally pretty heavy, is it really going to need significantly less robust support? Weight aside, how much of the load those supports are rated for is due to the actual weight of the panels, and how much is for forces from things like wind?

      1. lightedman · · focus · HN ↗
        "Thats already puting thousands of volts and hundreds or thousands of amps through the silicon."

        In a typical string of solar panels design, you'll get tons of volts but not a lot in amps - current cell maximums top out at ~11A and the connective MC4 wiring can't handle too much more current than that, so what you end up with is like a 1,000V 10A string on one MPPT connection into the inverter.

        1. sophacles · · focus · HN ↗
          A solar panel is 4 feet or so on the long side. Theres 250+ of them in a string 1/4 mile long.... at 48 V/panel, you get to 12KV. even if it's topped out at 10A thats still 120KW... you need a hefty cable to carry that panel to panel. Which is the core of the point I was making.
          1. cogman10 · · focus · HN ↗
            Amps determine how hefty a cable needs to be, not volts. Volts mostly determine how thick the insulation needs to be.

            14 gauge wire is basically all you need to carry 10A safely for an extended period of time regardless the voltage. It doesn't matter that you are carrying 120KW.

            The proof of this is in EV charge cables. Those bad boys can carry up to 350kW. Yet the cables are often thinner than you might expect. How do they do this? It's by using high voltages (around 900V) which cuts back the amps to around 300->400.

            Tesla's chargers peak (or used to) around 600V which has required them to have much beefier cables to handle the high current.

            1. dotancohen · · focus · HN ↗
              Just imagine that GP stated wattage, or better yet energy. Nitpicking the details of how electricity is distributed does not refute his point.
              1. cogman10 · · focus · HN ↗
                It's not a question of wattage or energy.

                His point, which I directly responded to, is that these panels need giant cables due to the transported power (wattage).

                Watts don't matter when talking about cable sizing. Power loss from a cable is equal to the current^2 on the cable multiplied by the resistance of the cable (P = I^2 * R).

                Power delivered is equal to the voltage * current (P = E * I). What these two facts mean is that if you raise the voltage, you can have a cable with a higher resistance (smaller cable) without worrying about the heat from power loss causing the cable to melt. How "hefty" a cable needs to be isn't related to power transported by the cable, but rather the current on the cable.

                This isn't "nitpicking". It's correcting a (common) misunderstanding about how electricity works. It's a direct refute because he said

                > even if it's topped out at 10A thats still 120KW... you need a hefty cable to carry that panel to panel. Which is the core of the point I was making.

                The cable sizing was his "core" point. Which is why these "nitpicks" on cable sizing and energy distribution are directly responsive and refute his point.

                Also, a nitpick for you, wattage measures power, not energy.

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