|Fig. 1: Solar panels on a roof. (Source: Wikimedia Commons)|
Going off-grid often connotes wood cook-stoves, candlelight and Willie Nelson music. But how practical is it to go off-grid without reducing electrical consumption at all?
To calculate the feasibility of generating enough electricity through solar panels, we'll need to know a few things:
Energy use: In the United States, the average energy use per capita is around 17.2 megawatt-hours per year, or about 6.2×1010 joules per year, according to a 2009 paper. 
Energy from the sun: The amount of solar radiation falling onto the surface of the earth is about 1366 watts per square meter. Taking into account the fact that the amount of incident sunlight changes throughout the day, and assuming a latitude of 40°, this works out to be about 333 watts per square meter (averaged throughout the day). 
Solar Cells: Photovoltaic cells can have an efficiency of about 10% at around $4 per watt. 
Size of house: In 2003, the size of a house was, on average, 83 m2 per person. 
From the above numbers, let's calculate how much power could be generated by covering a roof in photovoltaic cells:
or in terms of generated energy, about 8.7 ×1010 joules per year. Note that this exceeds the average electrical usage of 6.2×1010 joules per year, suggesting that it is at least conceivable to generate enough electricity in this fashion (see Fig. 1). The cost for such a setup would be about $40,000 (assuming the cost of a solar panel is $4 per watt in direct sunlight).
|Fig. 2: An array of batteries. (Source: Wikimedia Commons)|
Unfortunately, generating enough electricity is not sufficient; storage is required, such that electricity may be produced and stored during sunny days, for later use at night and during cloudy days (Fig. 2). A simple method for storing electricity is to use tried-and-true lead-acid batteries (car batteries are familiar examples of lead-acid batteries). Such batteries have an energy density of around 100 kilowatt-hours (kWh) per cubic meter (about 37 watt-hours per kilogram), and cost around $100/kWh. [5-6] Exactly how much energy storage is required depends on a number of factors, such as typical number of consecutive days with little sun, energy consumption during darker days, etc. For the purposes of calculation, storing a month of electricity seems reasonable (one can easily imagine several winter months with little sunlight). Simplistically dividing the average energy use per year by twelve, we get an energy consumption of about 5.2×109 joules per month. This would require about 14.4 m3 of storage in lead-acid batteries, or roughly 3800 gallons, and would weigh the staggering amount of about 40 tons—and it would cost about $150,000!
In terms of generating abilities, solar cells—although expensive—could function effectively as an electrical power source for a home (not counting gas for heating if applicable, or fuel for transportation). In my simple calculation, many factors (such as weather, efficiency of converting DC to AC, electrical efficiency of batteries, etc.) were ignored; however, from an energy generation standpoint, this can be combated by using energy-efficient appliances, or by the brute-force addition of solar panels (this might not be possible for a homeowner in a very urban location, but for sparsely populated areas this would seem a viable possibility). When it comes to energy storage, things become problematic, both in terms of cost and physical space required to store the energy. However, if one only needs enough storage for a day or so—relying on the grid for power during extended periods of darkness—the only thing standing in the way is money.
© Brannon Klopfer. The author grants permission to copy, distribute and display this work in unaltered form, with attribution to the author, for noncommercial purposes only. All other rights, including commercial rights, are reserved to the author.
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