Infocenter calculation of an island system

The exact calculation of an island system is a complicated matter. But since the weather is very irregular, even the calculated values ​​are only clues, ie. you never get around oversizing the system, then some information for a rough calculation.
 
 
1 Calculation of weekly electricity consumption
 
1.1 Multiply voltage times current of each consumer (e.g., 0.9A x 12V = 11W) (power of a consumer)
 
1.2 Multiply the power of each consumer by the time in hours the consumer will be turned on daily (e.g., 11W x 4h = 44Wh) (Daily energy consumption)
 
1.3 Add all watt hours of all consumers: Lamp 44Wh + Radio 60Wh + Refrigerator 360 Wh = 464 Wh (Daily energy consumption of all devices)
 
1.4 Multiply the Wh of all consumers by the number of days per week where the equipment is needed. (e.g., 464 Wh x 2 days per week = 928 Wh / week) (Weekly energy consumption of all devices)
 
2 Calculate the energy to be produced
 
2.1 Between the consumers and the generators a battery is causing the losses. A new battery has an efficiency of about 85%, an older one e.g. 70%.
Divide the work calculated in point 1 into Wh by the efficiency of the battery. (e.g., 928 Wh / week: 0.7 = 1325 Wh per week)
 
3 Calculate the size of solar panel or wind turbine
 
The size of the solar panel depends on the location, season, angle of inclination and sky orientation of the module. Decide in which months the system is in operation. (April to September, for example) From the tables at the bottom of this page, take the value from the table with the lowest number. (eg location Zurich, 45 °, orientation south, September has the lowest number in this case 1502 Wh)
 
Example calculation: A panel with 100W would produce 1502 Wh in a week, but you need only 1325 Wh, that is, 100W / 1502x1325 = 88 watts
 
Theoretically you need in the above example a panel with a power of 88 watts, but it is necessary to size the system larger, so that it works even if a September does not correspond to the metereological mean.
A factor of 1.1 to 1.4 depending on how safe your system must work is recommended.
 
That means in the above example a solar panel with approx. 100 to 125 W is needed
 
If a wind turbine or another power generator is integrated into the system, the module can be selected correspondingly smaller.
4 Calculate the size of the accumulator
 
By default, the battery is dimensioned so that the system continues to run for 5 days, even if it has no sun and wind. After these 5 days, the battery should not be discharged more than about 30%, i. 70% of the capacity should still be available.
 
Take the Wh from item 1.4 Wh / 7 days x 5 days / 3 x 10 / 12V (e.g., 928 (Wh) / 7 (days) x 5 (days) / 3 x 10/12 (volts) = 185 Ah)
 
5 Determining the cable cross sections
 
The minimum cable cross-section is determined by the heating of the cable in the event of a short circuit and by the maximum tolerable losses.
 
As a guideline, do not load the cables with more than 4A per 1 mm2
 
For small cable lengths in caravans, boats, holiday homes, etc. are usually sufficient 2.5 mm2
 
The calculation of the cable losses: Cable length (m) / cross section (mm2) x 0.0178 (ohms / m / mm2) x current (A) x current (A) = losses in watts
Example: cable 20m back and forth 4mm2 power 10A: 20 / 4x0.0178x100 = 8.9 watts
 
 
6 energy generated by a 100 watt panel per week in watt-hours (Wh)