From the MANHATTAN CONTRARIAN
July 29, 2021 / Francis Menton
Yesterday’s post made it clear that states or countries aiming for 100% “renewable” electricity cannot seem to go over the 50% mark, no matter how many wind turbines and solar panels they build. The reason for this is that in practical operation, due to the so-called “skip time”, there is no power available from the sun and wind sources at many times of high demand; therefore, other sources must supply the juice during these times. This practical problem is most evident in California, where the “renewable” strategy is based almost entirely on solar panels with very little wind. Daily graphs published by the California Independent System Operator (CAISO) show a clear and obvious pattern where solar production drops to zero every night just as the peak demand time starts from about 6 p.m. to 9 p.m.
Commenter Sean believes he has the answer: “Given the predictable daily power generation cycle of solar power in sunny locations like California and the predictable daily demand that peaks in the evening, solar generators should potentially have a storage capacity equivalent to the daily generation of their PV system . “
I thought it might be instructive to play through Sean’s idea to see how much solar capacity and storage it would take to build a system from these two elements that is sufficient to meet California’s current electricity needs. Note: This is a math exercise. It’s not complicated arithmetic. There is nothing here beyond what you learned in elementary school. Few, on the other hand, seem willing to bother to make these calculations or see the consequences.
We start with the current consumption that needs to be delivered. Consumption is currently between a low of around 30 GW and a high of around 40 GW over the course of the day. For this exercise, let’s assume an average usage of 35 GW. Multiply by 24, the rough estimate is that the system has to deliver 840 GWh of electricity per day.
How much solar collector capacity do we need to provide the 840 GWH? We start with the sunniest day of the year, June 21st. California currently has approximately 14 GW of solar capacity. Go over to these CAISO charts and we find that on June 21, 2021, which was apparently a very sunny day, these 14 GW solar panels were being produced at a rate of around 12 GW maximum from around 8 a.m. to 6 p.m., roughly that Half of it from 7 a.m. to 8 a.m. and 6 p.m. to 7 p.m., and the rest of the time basically nothing. Optimistically, they produced around 140 GWH for the day (10 hours x 12 GW plus 2 hours x 6 GW plus a little more for dawn and dusk). This means that you will need to produce your 840 GWH of electricity on a sunny 21st. When it is 7:00 p.m., you will need enough energy in the storage to get it to the next morning around 8:00 a.m. when production resumes exceeds consumption. This is around 13-14 hours with an average of 35 GW or around 475 GWH storage.
This is June 21st, your most beautiful day of the year. Now let’s look at a bad day. A good example for the past year would be December 24th, 2020, which was not only one of the shortest days of the year, but must also have been quite cloudy. The production from the existing 14 GW solar capacity averaged only around 3 GW and that only from 9 a.m. to 3 p.m. That is 18 GWH in this window (3 GW x 6 hours). Then another 1 GWH was produced from 8 a.m. to 9 a.m. and another 1 GWH from 3 p.m. to 4 p.m. About 20 GWH for the whole day. You need 840 GWH. If 14 GW solar panels produced only 20 GWH for the day, you would have needed 588 GW modules to produce your 840 GWH. (14/20 x 840) That 588 GW of solar panels is roughly 42 times your existing 14 GW of solar panels. And if this 588 GW capacity doesn’t produce anything at all around 4:00 p.m., you will also need at least 16 hours of average usage in storage to get to 8:00 a.m. the next morning. That would be around 560 GWH of storage.
So you can easily see that Sean’s idea of providing storage “according to the daily generation of the PV system” doesn’t really get to the heart of the problem. Your main problem is that you need a capacity nearly 15 times peak usage (nearly 600 GW capacity for a peak usage of about 40 GW) to handle your slowest production days of the year.
Costs? If you (non-profit) assume that the “tiered costs” of the energy from the solar collectors are the same as the “tiered costs” of the energy from a natural gas system, then this system with 15 times the capacity costs 15 times as much. Plus storage costs. In this scenario, that’s relatively modest. At current prices of around US $ 200 / KWh, the 560 GWH storage will cost around US $ 112 billion, which is about half of the California government’s annual budget.
But you could say that nobody would build the system like this, with gigantic overcapacities, just to cover the few days of the year with the lowest solar output. So why not build a lot less solar capacity and save electricity in the summer to cover the winter? Since the average output of solar panels in California is about 20% of the annual averaged capacity, you should be able to get enough power for the year with a capacity about 5 times the peak load rather than 15 times a year to generate scenario above. All you have to do is save electricity from summer to winter. Oh, and you will need many times more storage space than the day-to-day scenario. If there is less production than is consumed on 180 days per year and the average loss of production on each of those days is 300 GWH, you need batteries worth 54,000 GWh (180 x 300). At $ 200 per GWH, you get over $ 10 trillion. That would be roughly three times the annual GDP of the state of California.
But don’t worry, there are no batteries to store electricity for six months or more and to sell it on the exchange without losses. Maybe someone will invent them in time for California to meet its renewable electricity targets by 2030.
Full article here.