Autonomous Photovoltaic Calculation


1. Consumption: {{watthours = watt * hours * 1.5}}

2. Batteries(Ah) :{{sizebatteries = watthours * 2}}

  2a. Sunshine : {{sunshine = sizebatteries * 3}}

  2b. Inverter loss : {{inverterloss = sunshine * 0.8}}

3. Solar panels : {{watthours / 3}}

1.

consumption

Each device has a small label on it that indicates the electrical consumption.
For example, a 21-inch TV can write 220 volts and 0.5 amps (A).
This means that it can consume 220 x 0.5 = 110 Watt.

Multiply the Watts of each device by the number of hours it will operate.
The sum of all these products will be our total daily consumption in Wh.

Because there are losses in our system but also hidden consumptions from devices that we did not calculate (eg devices that consume power even when closed or on hold), we multiply the previous sum by 1.5. So, if after the above 3 steps we have come to the conclusion that we need a total of 600 Wh for all our devices per 24 hours, then we must install a system of photovoltaics (collectors – panels – solar energy) and batteries (batteries) that can provide us at least 600 Wh every day.

2.

batteries

Accumulators (batteries) indicate their capacity in Ah (amperes per hour).
Thus, a 12 volt and 100 Ah battery provides 12 X 100 = 1,200 watts of direct current (DC) for 1 hour or 120 watts for 10 hours or 12 watts for 100 hours.
Another important indicator is the one that provides us with information about the discharge rate based on which the battery can give the listed Ah.

Thus, a battery that says it has a capacity of 100 Ah in C20 means that 100 Ah is achieved when the gradual discharge takes 20 hours. For less hours (eg C10, 10 hours) we get less Ah, while in gradual discharge of more hours (eg C100, 100 hours) we get significantly more Ah.

The C10, C20 and C100 index is not arbitrary. It is described by specific specifications and is standardized. ALL photovoltaic batteries (ie all lead batteries) have different capacities to give a discharge rate of 10 hours (C10), higher at C20 and even higher at C100. It is preferable during operation to provide a few watts for more hours than many watts for a few, because in the latter case their life time is drastically reduced. We never completely discharge the batteries as this can damage them.

There are different types of accumulators with different degrees of allowable discharge. The general rule is during normal use not to allow more than about 50% discharge and only in exceptional cases to reach 80%.

So, when we buy batteries for the photovoltaic system, we choose a capacity at least twice what we estimated would meet our needs (including the days of system autonomy). The longer the better for battery life.

So if we calculated that we need 600 Wh per 24 hours, we choose batteries with double capacity (1,200 Wh), ie 12 volts and at least 100Ah to have a range of one day.

But we usually predict for 3 to 5 days without any sunshine, so we multiply the previous value by 3 or by 5: For example, 100Ah X 5 = 500Ah at 12 volts (or equivalent, 24 volts and 250Ah).

When a device requires 220 volts – 1 A and we use a 12 volt to 220 volt inverter (inverter) to operate it from the battery, then it will draw 18.33 A from the battery and not 1 A, because the 220 watts in operation with alternating current (220v X 1A = 220 watts) translates to 12 volts X 18.33 A (= 220 watts) when operating with an inverter and 12 volt battery power. The same applies to the case when we use a 24 volt battery, where it will “pull” 9.16 A (24v X 9.16 = 220 watts). Because the use of a voltage inverter (inverter) entails losses of 10% to 20% the final consumption will be higher than indicated in full operation.

3.

Solar panels

So if we have reached the size of the batteries, then we can only calculate the size of the solar panels that will be able to charge the batteries. A solar collector of 50 watts / p nominally (per hour of sunshine) will give a day with 5 hours of sunshine (eg in April) 250 watts / h theoretically (due to losses it will be 10% to 20% less) while in a day with 7 hours of sunshine (eg in July) 350 watts / h.

It will take 4 days in April and 3 days in July to charge completely empty batteries (theoretically, because they will never be completely empty as we said above) of 12 volts and 100 Ah (1,200 watts / h). If we install three such solar panels of 50 watts / p each (or one of 150 watts / p), then it will take one day in July and almost two days in April.

When designing a large photovoltaic system for the home, it is good to have as a basis the worst case scenario, which is the winter hours of sunshine (on average), which for Greece is 3 hours a day (in December). If we are planning for a holiday home that we visit ONLY in the summer (May to September), the sunshine hours we estimate are 6 (AV).

So, for the previous example that we calculated that we will consume 600Wh per 24 hours, we need photovoltaic panels of power 600/3 = 200Wp to cover us winter-summer.

If we wanted to be covered ONLY for the summer, we would need photovoltaic panels with a total power of 600/6 = 100Wp. In this case, in fact, we would need smaller batteries, since in summer no autonomy is required for 5 days without sunshine that we calculated in the 2d step.

A simple way to calculate the current consumption of batteries and the size of a photovoltaic system.