For a terrestrial PV system, we must have at least 18 hours of energy storage, since we get 6 sun-hours a day, at best in the USA. It’s only a little better further south, where 8 to 10 useful sun-hours are obtained. I’ll look at two types of storage. The most likely type to be used for large scale storage is a reservoir, but I’ll include batteries also. The reservoir is about $0.10/Wh to install and the batteries are about $0.50/Wh to install Large PV power plants, slated for installation in 1997 will have an installed cost of about $2/W. Since this is still a hard-to-believe number in the solar energy industry, I assumed $3/W, which is not too difficult to imagine now. With inefficiency of storage, we need 2x terrestrial PV, phis we need 6x additional PV to charge storage for the other 18 hours per day when there is no sun. Thus we need about 12x more terrestrial PV than SSP PV. In areas where there is not 6 hours of sun per day, even more PV is needed. Since there will be some times of year when we have 5 cloudy days in a row, we must have 5 days worth of either extra storage or back-up power from another source. While the area of the PV is not too bad, the area for the storage system is prohibitive. One of the problems utilities have now is finding land to put power plants, storage systems and transmission line upon. Land is quite expensive in some places. The average cost of land in the US is about $2000/acre. Even for much cheaper land, the costs are still in the billions for large scale storage. Notice, though that the cost per kilowatt-hour is within reason at less than $0.01/ kWh. Note that, if the grid itself is considered part of the storage, then the PV and the reservoir would not have to be in the city at all Nor would the rectenna. If there is no existing grid, then proximity is more important (this might be the case, for example, for Mexico City). Comparison #2 Solar-hydrogen-Fuel Cell Let’s suppose that we turn to hydrogen generation and fuel cells. Does this alleviate the storage problems? A typical fuel cell uses about 700 Liters per kilowatt-hour. A demonstrated solar electrolyzer (PV-electrolyzer) can produce 25 liters/min of hydrogen. Supposing we want to supply 5 GW to New York City. The PV-electrolyzer which produces 25 1/min. requires a 10 kW PV array, which takes up 100 square meters. Assuming 6 hours of sun per day (best case) I’ll scale this system up to produce enough hydrogen for 20 GWh per day (or 4x the amount generated on a 6 sun-hour day with a 5 GW PV array). The inefficiency of the fuel cell and the hydrogen generation are already accounted for by the empirical performance data. To generate 20 GWh, we would need 1.4E10 liters of hydrogen. This can be produced in one day by a 1.56E7 kW PV array. This array would take up a space of 156 sq. km.(60 sq. miles). It would cost $124 billion ($8/W) to install at today’s costs (electrolyzer included, PV is about 70% of the cost) and $46 billion (S3/W) in the future (when PV is $1.00/W). Such an installed cost leads to a price per liter (over system’s 20 year life) of $0.0018 (@$8/W) and $0.0007 (@$3/W). Since fuel cells require about 700 liters per kilowatt-hour, this implies that, even in bulk generation, the cost per kilowatt-hour in hydrogen alone is $1.24 to $0.47, which is competitive in some developing regions, but not in developed regions. I did not figure the cost of the fuel cell into this, but they are unlikely to be less per Watt than PV anytime in the next 10 years. Space Solar Power If a space solar power station could provide nearly continuous power and either feed the grid at 10 cents/kWh or generate hydrogen for 1/lOOth of a cent per liter, it would be a better investment. The land
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