3 ms·

Well, it depends on whether we are thinking of making grid fossil-free on current scale of electricity use, or we are planning to replace all energy with electr

• ben_w 11mo ago
  > In the latter case, i can't see how it can happen because no new nuclear can be politically built, no new hydro can be physically built, and wind potential is rather limited - increase wind output 2x and you are down to really marginal wind areas where ROI falls, big time. Which leaves solar, and that is a totally different level of variability and amount of lands it will take, skyrockets (if you want to reliably supply all of Germany for all it's final energy consumption with solar alone, having "only" 24-hour battery storage, all of land surface of Germany will not be enough). Nuclear, perhaps. Hydro comes in two flavours, damming a river and damming a valley. You dam a river to get a constant power from the watershed, you dam a valley if all you need is a battery-equivalent; we're only short of rivers, there's a lot of valleys. Wind, you're way off. The UK's estimated wind potential is about 2.2 terawatts, which would be almost enough to double global electricity production by itself. This is because the North Sea is amazing; Germany also has some access to the North Sea. Solar, again, you're way off. The worst parts of Germany are around 1000 kWh/m^2/year as per https://upload.wikimedia.org/wikipedia/commons/1/19/Germany_GHI_Solar-resource-map_GlobalSolarAtlas_World-Bank-Esmap-Solargis.png https://upload.wikimedia.org/wikipedia/commons/1/19/Germany_... which I believe doesn't account for cell efficiency, which we can say is 22% at this point given what the cheap modules on my patio do; the area of Germany is 357,022 km^2 = 3.57e11 m^2, so "all of land surface of Germany" is 3.57e11 m^2 * 1000 kWh/m^2/year * 22% ~= 8.9 TW, which isn't just enough to fully-electrify all of Germany, it's enough to fully electrify all of continental Europe, by a factor of a bit more than 3.8 (given https://www.iea.org/regions/europe/energy-mix https://www.iea.org/regions/europe/energy-mix). Germany's 2023 consumption of primary energy (as a reminder, that's everything not just what's currently electricity) was just under 11 exajoules, averaging 342.2 gigawatts — if Germany was trying to convert itself alone (i.e. becoming weirdly isolationist), using just PV alone as the source (why?), with cheap modules (not the best), using the worst regions as the average for the whole country (low-balling the result), even without the boost you get from electrification because e.g. primary energy consumption is the full energy in ICE fuel tanks before the thermodynamic losses that BEVs don't have, it has 26 times as much land as it needs for that task. That's more than enough of a margin for a surprise climactic shift causing continuous all-year-round Dunkelflaute, even with just 24 hour storage (battery, or as per previous point, hydro). Germany's land area, by itself, even despite the location, is only insufficient if you want to use it (alone) to power more than half of the entire planet. Which of course Germany shouldn't be doing: if humanity created a power grid sufficient to do that (it is doable but only by China right now) there's plenty of better places to put the PV, enough so that no batteries at all would be needed.
  • anovikov 11mo ago
    Trick is not the total amount of insolation, but how much solar is available in the worst periods of year. Germany has about 100 GW of solar installed; or worst weeks it produces about 280 GWh in solar power... we can average over a week because we may assume 2 hours' storage (so for 100 GW nominal, 200 GWh) and that's about 4-5 worst days in a row. Now electricity consumption is about 10TWh a week. Which means, 35x of the current amount of solar is required to cover it entirely if the solar has to supply everything even in the period when there is the lowest amount of sunlight is available (because storage over even longer periods is unrealistic). Which is for the case of "electricity only, in current consumption volumes", means 104,000 km2 or over 1/4 German land surface, with the average solar park density of 1 MW per 7.3 acres. If you try for entire amount of all energy consumed in Germany, not just electricity, then you need to fill entire German territory and that won't be enough. In particular (for your figure of 342.2 GW average), 610,000km2 which is 1.5x+ German surface area. Because 280GWh over a week averages to 1.67 GW average or 1/60 of nominal installed power. Yes, ON AVERAGE, solar panels give about 1/9 to 1/10 of their nominal installed power over the year, but it's 6-7x less during worst weeks. This number can be slightly improved by installing panels over the angle that optimises performance in winter (in summer there will be more than enough anyway), but no more than by 1.5x - still entire Germany will not be enough to supply itself with power from solar that works year round. So your calculation is wrong in this, and also in the fact that solar park is not fully packed with panels because they need pathways to service them and they need to not shade one another. Average area per 1 MW installed power is 7.3 acres in the USA.
    • ben_w 11mo ago
      > 280 GWh > we can average over a week because we may assume 2 hours' storage (so for 100 GW nominal, 200 GWh) and that's about 4-5 worst days in a row […] because storage over even longer periods is unrealistic I don't see any particular logic to either part of this. Better question for "realistic" storage is how much do people actually want to spend, how do the costs change as quantities scale up from hours to days to weeks to seasons. (4kW * 24h * €50-€70/kWh) / 10 year battery life from daily cycles is an unsurprising part of a monthly energy bill (especially as this would basically substitute most of the grid's role in the hypothetical where Germany for whatever reason wasn't still buying nuclear power from France etc.); pumped hydro is estimated around €100/kWh installed, but different maintenance cost and lifetime. > Now electricity consumption is about 10TWh a week. Which means, 35x 9.3, I think you're using old numbers. I find no search results for 280 GWh claim; I will assume the number itself is correct, but given the rate of growth of PV it is of material relevance which year this happened in, e.g. if that 280 GWh was in was in 2020 it's more like 18x, if it was in 2015 it's 13x. > with the average solar park density of 1 MW per 7.3 acres You're double-counting here. 33.9 W/m^2 with 22% cells means the capacity factor is 15% of the name-plate capacity. This is a standard value in much of the world, varies from about 10%-20%, you can't do better than about 32% from anything fixed to the ground even on a completely clear sky on the equator. The map I linked to already accounts for all of that stuff, it's *internal to* how they reach the ~1000kWh/m^2/year for the worst bits of Germany: https://upload.wikimedia.org/wikipedia/commons/1/19/Germany_GHI_Solar-resource-map_GlobalSolarAtlas_World-Bank-Esmap-Solargis.png https://upload.wikimedia.org/wikipedia/commons/1/19/Germany_... Anyway, point is 1 MW/7.3 acres *already* accounts for "winter is bad". > This number can be slightly improved by installing panels over the angle that optimises performance in winter Opposite error! Not when you're talking about covering an entire country it doesn't. Panels would shade each other. > and also in the fact that solar park is not fully packed with panels because they need pathways to service them and they need to not shade one another And again, now you're double-counting. Precisely because 1 MW/7.3 acres number is already accounting for gaps, you've mixed up the optimal output per m^2 of *panel* with the optimal output per m^2 of *land*, and it's land that matters on this scale, the optimal number per unit of land is what you get from laying the panels flat on the ground.