subakumaran
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is there any procedure ( formulae ) to select the power rating of solar panels and ampere hour rating of batteries ?
The discussion revolves around the selection of solar panels and batteries for a specific application involving a 12 V 100 W DC motor. Participants explore the necessary power ratings and capacities required for effective operation, considering factors such as operating time and environmental conditions.
Participants generally agree on the need to consider specific application requirements and environmental factors when selecting solar panels and batteries. However, there are varying opinions on the exact specifications and capacities needed, indicating that the discussion remains unresolved.
Participants express uncertainty regarding the exact capacity needed for the battery, as it may depend on factors such as charging efficiency and environmental conditions. There are also unresolved considerations about the optimal voltage matching between solar panels and battery systems.
mfb said:It depends on the application.
mfb said:Well, that helps a bit, but it is still a bit vague:
100W for 5 minutes are 100W*300s= 30kJ = ~8.3Wh. This is the minimum requirement for your battery, unless you want to rely on sunshine directly at the time of the motor operation.
8.3Wh at 12V correspond to 8.3Wh/(12V) = .69 Ah.
In general, your battery won't get fully charged between cycles, so I guess you'll need more capacity. If your solar cells are sufficient to provide the power for a day even if it is cloudy, a capacity for 4 cycles could be sufficient. Otherwise, you could need more capacity.
The required peak power rating of the solar cells depends on the amount of sunshine you get, and the battery capacity. The average energy per day has to be at least the required energy for the motor operation, but there should be some additional safety factor for cloudy days/weeks.
Matching the voltage is another thing you have to take care of - if the output voltage of the solar cells is not 12 Volts, you get some conversion losses.
mfb said:20W*5h = 20J/s*5*3600s = 360kJ. Should be more than sufficient, especially with the large battery that can store enough energy for a few days.
Just keep in mind that you need some controlled way to charge the battery with the right voltage and current range.