If you plan to go "off the grid," then the question becomes a bit more complex. Although economics is another section of this document, it's not excessive to say that solar costs A LOT. The first best thing you can do is conserve whenever and wherever possible.
Ideally, you would discard all of your present electrical equipment, replace electric equipment and appliances with alternatives (like propane powered refrigerators), go with DC lighting and appliances where possible, and so on. If you are starting a home from scratch that you plan to power with PV, you should do all of these things.
Most of us are unlikely to go to such lengths (although if you plan to go off grid completely, you really ought to consider all of these choices!)
How much power you need is determined by a pair of maximums. The first is the total amount of energy you use annually. You can get a figure for this from your utility company or from your utility stubs. The sum of each month's kilowatt hour usage will give you your annual power use.
Of course, some of your electricity usage varies with weather. If, for example, you use electric powered air conditioning (which, I must tell you, spells DOOM for your possibilities of living off-grid for any reasonable amount of money) and the past year was a mild one, then you must allow for a higher amount (or live with the heat.)
Let's take an example. In my small rented house in Minneapolis in January through December 1995 I used 9,126 kWh of electric power. This figure could be cut in half simply by not using air conditioning. At any rate, this means that I would have to have enough solar panels to deliver that many kilowatt hours of usable power over the course of a year.
There's more to it than that (a lot more as it turns out, but we'll go slowly). Getting enough PV to provide that much power over the course of a year is not the only maximum. Sometimes we use very little electric power (on a cool night with no lights on while we sleep, for example). At other times we use a lot. Say when vacuuming while the stereo is on with both air conditioners going, something heating in the microwave, the refrigerator compressor running, and every damned light in the house on while someone fires up the air compressor.
Even if these average out to a load that can be handled by an array of a given size, we must have sufficient panels and batteries to handle short periods of extremely heavy loads. This is the peak demand and would be measured in watts. We must have a system that can deliver our peak demand for at least a short time. When heavy inductive loads come on (electric motors like air conditioners, vacuums, power tools, etc.) there is a brief but very strong power demand. This must also be allowed for (the air compressor starts while the system is already under heavy load).
Let's do a rather simple analysis (we'll get more detailed later) based on the following assumptions:
This is just to give you an idea. (All calculations performed on a solar powered calculator! ;-)
4.6 kilowatt hours per square meter per day on a panel that is 10% efficient yields 0.46 kWh/sq. m./day. 80% battery efficiency leaves us with 0.368 kWh/sq. m./day. 92% inverter efficiency leaves us with 0.339 kWh/sq. m./day. This means that 1 square meter provides about 124 kWh annually. Thus, to supply the 9,126 kWh I used in 1995, I would have needed just over 73 square meters of solar panels.
This is a square about 8.6 meters on a side, or a square roughly 28 feet on a side (~ 795 square feet). Note: The math above is correct since 12/13/96 when I finally got around to fixing a mistake pointed out to me by Mike Volberding. Thanks for catching my error, Mike!
At the price of solar panels, this is clearly not a viable option for persons of other than, how shall I put it, advanced means.
Even so, this is not an impossible proposition. A south facing roof pitched at the latitude angle and covered with panels could quite conceivably have such an area. But this 9,126 kWh figure was in my old rental property. This was not and is not an energy efficient dwelling. It has outdated appliances. I had two window mount air conditioners (and 1995 was an unusually hot summer as US readers will recall!). These energy hogs were further overdrawing their feeds (the wires became noticably hot -- fire marshalls need not call!) resulting in still greater energy waste.
Replacement of all lights with compact fluorescents would cut about 600 kWh. A new refrigerator would save another 400-600 (we'll say 400). Elimination of the A/C would save a lot more, but let's assume I'm a comfort junkie. Careful attention to conservation would probably shave another 100 kWh off without any substantial reduction in lifestyle. The annual use would drop to 8026, which would drop the area to 64.7 sq. meters, or a square 8.05 meters on a side or about 26.4 feet (696 square feet).
Drop the air conditioning (1100 kWh guesstimated) and it gets better yet: 6,926 kWh/yr. This works out to about 55.85 square meters, or a square 7.47 meters on a side or about 24.5 feet (600 square feet).
What if you did a grid intertie instead of a stand alone? You would be able to get rid of the battery inefficiency. Then the calc runs 4.6 kWh times 10% panel efficiency, or 0.46 kWh/sq. m./day. 0.46 kWh times 92% inverter efficiency, or 0.42 kWh/sq. m./day. 0.42 kWh times 365.25 days per year is an annual energy per square meter of 154.5. If we use this figure with the 6,926 kWh/yr figure, we get 44.8 sq. meters. This is a square 6.7 meters or 22 feet on a side (482.5 square feet). That's quite a variation from 795 square feet!
This is just a thought example using some actual numbers thrown in with reasonable guesses (the panel efficiency is an average, the battery efficiency is average, and the inverter efficiency is an average). The points it intends to illustrate are:
So far we have only considered the first maximum, which is total annual power consumption. The array sizes we arrived at were based purely on meeting this number. Obviously, you must allow some slack. You will probably add additional loads in the future. You might have a child and start using more electricity simply as a result of that. Expect to have to build an array with "slush factor."
The peak load question is quite a bit more complicated. In any system, the limiting factor is most likely the maximum output of the inverter(s) in the system. This is because in an SA system electrolyte batteries are, in terms of domestic loads, practically infinite current sources. Batteries will ramp up output to meet demand of almost any size (until they are discharged, of course). In this case, the maximum output of the panels should not be terribly important, since the inverter feeds off the batteries. In a UI system, the grid supplies current through the inverter and thus the panel output does not matter here either.
Basically, then, you size the panels for your total power use and you size the inverter (and battery array in an SA system) for peak demand.
Also note that there are really two peak demand figures. The peak operating load is the largest sustained demand. However, devices like air conditioners and power tools and other large inductive loads draw a surge of current when they start. A 3,000 watt inverter might very well experience a surge to 6,000 watts for a fraction of a second when, say, an air conditioner starts. As a ballpark figure, you can guesstimate that loads with AC motors will draw about 3 times their rated amperage at startup. You will want to know your inverter's maximum operating load and its peak or surge limit.
The data is also available in text form from NREL. See the resource listings at the end of the FAQ.