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Solar electricity: Stand-alone photovoltaic systems

Wherever a power grid is not or not at reasonable costs available, a stand-alone photovoltaic system can be used to generate the needed electric energy. Examples for such an application are alpine huts or cabins in remote areas, solar-powered water pumps, emergency telephones, but also systems for boats or recreational vehicles (camper vans).

Since the solar modules only produce electric energy during daytime, it is necessary to store energy for the night or for cloudy days. Such storage systems mostly use rechargeable lead batteries, due to their ability to accept with high efficiency both low and high input voltages. A battery regulator prevents overcharging, a load shedding circuit prevents deep discharges. Because of the high differences in energy harvest between winter and summer it is recommended to use hybrid systems for year-round applications. Such can use diesel or biogas generators as well as wind turbines; in most cases they will include a storage battery also.

Picture source: SMART Powersystems.

Picture: Stand-alone photovoltaic system for autonomous electricity supply of a summerhouse or weekend home including a 220Wp PV generator and maintenance free accumulators (100Ah/24V) as well as a 50A charge regulator. The built in inverter supplies 1.200 W continuous power and 3.300W temporary peak power.

The photovoltaic system will provide a battery output voltage of (in most cases) 12 or 24 volts DC. To supply devices which are only available for AC voltages a power inverter can be used.
One of the most important tasks in planning such a stand-alone PV system is to match the prospective energy consumption with the local average solar irradiation, the resulting energy production and the required storage capacity. As a example: A weekend home near Montelimar (France), occupied only during summer, is to supply with electric energy by a PV system. The energy consuming devices are a few (energy-saving or low-voltage halogen) lamps, a small TV, a water pump and a energy-saving refrigerator. Multiplied with the respective power-on time the daily energy demand sums up to about 680 kWh. At this location the daily energy production per 1 kWp module capacity reaches in summer leastwise 4 kWh; so a 0.18-kWp-plant would be fully sufficient, supplemented by a rechargeable battery storage with a capacity of about 280 Ah (at 12 V DC), enough to feed on for about 2.5 days.

Picture source: SMA Technologie AG

Sunlight instead of Oil: stand alone photovoltaic system in the United Arab Emirates. Rural electrification of emerging nations and developing countries is an economically and technical proven application for stand alone PV systems.

Three steps to size a stand-alone photovoltaic system

Some preliminaries: Generally one should utilize only power-saving devices to be fed by a stand-alone PV system. Additionally, by utilizing devices operating at 12 or 24 V DC (since PV systems provide originally DC voltage) some conversion losses can be avoided.

Step 1: Estimating the daily energy consumption:

Picture source: Deutsche Bundesstiftung Umwelt (DB)

For every device multiply the power input (measured in Watt) with the hours of power-on time. Sum up the results and add some buffer (depending on the uncertainty of your forecast).Since the consumption will differ with the season, you should calculate this independently for summer and winter season (at least).

The PV system of the "Rabenkopf" cabin in the Bavarian Alps nearly replaces a diesel generator for electricity production Picture Source: Deutsche Bundesstiftung Umwelt (DB)


Step 2: Determine the size (energy output) of the PV system:

The averaged daily energy yield by the PV system should be sufficient to cover the daily consumption (calculated per season, since between summer and winter the »harvest« differs widely).

To forecast the daily energy yield we need data about the daily irradiance at the location of the PV modules. Such data is available from different sources at the web (e.g. http://www.nrel.gov/ for the USA or http://re.jrc.ec.europa.eu/pvgis/ for Europe and Africa). To get the energy yield provided by the PV system the radiation (measured in kWh/m2/day) has to be multiplied with the module capacity (nominal output, given in Kilowatt Peak, kWp) and the result corrected by factors including the deviation of the optimal orientation and inclination of the modules. (Some of the sources named above offer such pre-calculated and corrected data, too.)

Now we have to discount the transmission losses caused by the electrical resistance in the cables and during the charging/discharging process of the rechargeable battery storage. Such losses can typically sum up to about 24%.

For instance: A PV system near Cambridge (UK), providing 1 kWp nominal output, would generate 3.6 kWh of electrical power per day on average in July. So it could meet a consumption of about 3.6 * 0.76 = 2.7 kWh/day. However, to plan our PV system, we will use the month with the least irradiation of the season as base – in a summer at Cambridge this would be the September (2.7*0.76 = 2.05 kWh/day), in the winter the December (0.7*0.76 = 0.5 kWh/day).

So, if we need only about 500 Wh per day, this system would suffice even for the winter season – but it could produce four times our needs in summer and would be totally oversized for more than half of the year. If we use the facility only in summer, a system with 0.5/2.05 = 0.24 kWp (consumption divided by production per kWp installed nominal output) would be fully sufficient. It would therefore be economically advisable to install an hybrid system including a PV system of about 0.25 to 0.3 kWp and an additional generator to bridge the winter season.

Step 3: Dimensioning the storage capacity

Since the PV system generates electricity when the sun is shining, which is in many cases not the time we need the energy, we use rechargeable batteries to store electrical energy. The capacity of such batteries is measured in ampere-hours (Ah). If we divide the assumed consumption per day (in Wh) by the output voltage (in V DC) of the storage system (mostly 12 V DC or 24 V DC, depending on the interconnection of the batteries), we get the capacity we need to bridge one day, e.g. with a daily consumption rate of 0.5 kWh: 500Wh/12V = 41.7 Ah.To avoid damages by deep discharge, we should double this value to 84 Ah per day. If the facility is used only in summer, we calculate 2.5 days at max to bridge, resulting in a total capacity of about 210 Ah; in winter we have to calculate with up to 5 days to bridge, so the total capacity would be 420 Ah.