The purpose of this admittedly rough calculation is to estimate the amount of pumped storage that would be required if wind power provided all the electricity the US uses.
First, consider the amount of electricity the US uses, a total of just over 4 billion MWH/year.[1] On an average day, not a high-demand day, that's 11,000,000 MWH/day.
We have to make up a fictitious example because there aren't any real examples. For our fictitious example, we will assume that Hoover Dam connects two lakes, each the size of Lake Mead, 30,000,000 acre-feet. Hoover Dam has an average head of 520 feet. We shall calculate how much energy the dam would generate if all the water drained through the turbines. This assumes 100% efficiency, obviously highly optimistic.
Energy = volume x pressure = volume x head x weight-density
= 30,000,000 acre-feet x 43560 sq-ft/acre x 520 feet x 62.4 lb/cu-ft
= 4.24 x 10^16 ft-lb
= 16 million MWH
Since the US uses 11 million MWH average per day, our fictitious pumped-storage system would power the US for less than 1.5 days.
How much storage would be required? Looking at data for all of the US, we see that, with surprising consistency, low-wind months have average speeds about 70% of the average speeds for the high-wind months.[4] Based on the cubic relationship between wind speed and power, we would expect seasonal variations in energy generation of around 0.35 to 1, so the average would be something like the rated capacity times (1 + 0.35)/2 or 0.675. Conversely, for the generation to equal the load, the rated capacity of the wind farms would have to be the yearly average load divided by 0.675, or multiplied by 1.48. Since we can expect the generation in a slow-wind month to be 0.35 times the rated capacity, the storage capacity would have to be about 1 - 0.35 X 1.48 (= 0.48) times the yearly average load times the length of the low-wind period.
1.5 days' capacity would be good for 1.5/0.48 = 3 days of low winds. But low-wind seasons last longer than 100 days. So the US would need more than 30 pairs of lakes, each lake the size of Lake Mead and each pair connected by a dam the size of Hoover. Clearly that many locations could never be found, and if they were the projects would not be permitted because of the high ecological cost.
Another scheme that sometimes is mentioned is storing compressed air in caves. There is a facility in Huntorf, Germany that we can use for an example.[5] It compresses air to 1000 pounds per square inch pressure.
The data show that it stores 3 x 290 = 870 MWH of energy and the cave volume is 310,000 cubic meters.
For one day of electricity storage for the US, the volume needed would be
11,000,000/870 X 310,000 = 3.92 billion cubic meters = 138.4 billion cubic feet.
Suppose a cave had an average cross-section of 50 ft X 50 ft = 2500 sq ft.
For one day's electricity storage, the cave's length would have to be 138.4 billion / 2500 = 55.34 million feet = 10,490 miles. Granted that most big caves have never been surveyed, it's still safe to say that there aren't ten-thousand miles of caves in the US. So there is no possible way enough energy could be stored to see the country through 100 days of low winds.
As our reference battery, we shall use the vanadium-redox type of flow battery, which has an energy density of up to 40 watt-hour/Kg. This is the best battery currently available for bulk storage.
One day's electricity storage for the US would be 11 million MWH or 11 billion KWH or 11,000 billion watt-hours. So the combined size of battery required would be 275 billion Kg or 275 million metric tonnes. Sulphuric acid, the main constituent, has almost twice the density of water, so the volume required would be 135 million cubic meters. Let's say that batteries are limited to 10 meters diameter and 5 meters height. That's roughly the volume of a house. The volume of each is 393 cubic meters. 344,000 batteries would be required. To get the country through 100 days of low winds would require 50 days' supply, so 17,200,000 of these house-size batteries would be required. That works out to more than one such battery for every twenty persons. But two tanks are required for each battery (the electrolyte flows from one tank to the other), so two tanks each the size of a house are required for every twenty persons.
A company called Ausra is proposing to build concentrated-solar power plants in the deserts of the American Southwest. They intend to get around the problem of night-time electrical demand by storing hot water.
It would be helpful if Ausra would provide some technical information. In its absence, we'll have to make some assumptions. To avoid accusations of unfairness, we'll make all the assumptions favorable to the solar plant.
We don't know, for example, what the storage temperature will be. Higher temperatures allow more energy storage but require thicker container walls because of the higher pressure. We'll assume 600ºF because density starts dropping off quickly at higher temperatures, and pressure rises quickly. Even at that temperature the collector efficiency will be very low and the pressure would be about 1550 pounds per square inch. At that temperature, a pound of water holds 618 BTU, compared to 70 BTU at 100º.[6]
A modest thermal plant would be rated at 1000 MW of electricity. Imagine we wished to average that power over 16 hours, for a total of 16000 MWHe. Assuming a generous 45% efficiency, the energy required would be 16000 x 3,413,000 / 0.45 = 121 billion BTU of heat. Each pound of water can give up 618 - 70 = 548 BTU, so that means the storage has to be 221 million pounds. Density at 600º is about 42 lbs / cubic foot, so the volume required would be 5,262,000 cubic feet. Suppose the tanks were 20 feet in diameter, so their area would be 3.14 x 20 ft x 20 ft / 4 = 314 sq feet. To hold the required amount of water would require a total tank length of 16,750 feet, or 3.2 miles. By the way, if we say steel is good for 50,000 psi stress, the tank thickness would have to be 1550 / 50000 x 20 ft / 2 = 0.31 feet or 3.7 inches.[6]
So we need 3.2 miles of tank for 16 hours of storage. But what if the sky is cloudy for one day? Then we need 40 hours of storage, or 8 miles of tank. And that's for just one smallish thermal plant. In fact, there could be weeks of cloudy days, or days with only a few hours of sunshine, even in Arizona.
[1] http://www.eia.doe.gov/cneaf/electricity/epa/epat1p1.html
[2] http://www.nypa.gov/facilities/niagara.htm
[3] http://www.great-lakes.net/lakes/ref/eriefact.html
[4] http://www5.ncdc.noaa.gov/documentlibrary/pdf/wind1996.pdf
[5] http://www1.ceit.es/asignaturas/tecener1/Lesson6.pdf
[6] Baumeister and Marks, Standard Handbook for Mechanical Engineers