ipish:
Can you guys check if my understanding is correct?
OK, let's take it point by point.
ipish:
- Determine the energy consumed in a 24 hour period.
RPi usage (load) = 5.25W * 24 Hour = 126 watt-hours
- Add in the Fudge Factor for energy consumed (50% losses)
Lets multiply the total 24 hour load energy by 1.5 to account for several factors such as the system efficiencies, including wiring and interconnection losses as well as the efficiency of the battery charging and discharging cycle, and allowing extra capacity for the system to recharge the batteries after they have been drained.
New load = 126*1.5 = 189 watt-hours
So that's really 189 W-hr/day. I might argue w/the fudge factor. You need something to account for added RPi "accessories" and to transform the battery voltage into the RPi's input voltage (5V). Don't double count the battery related stuff, just try to get a good average but conservative number for the energy used in a day.
ipish:
3. Determine Solar Insolation in Hours (disregard 9 hours of sunlight stated earlier)
Most solar map data are given in terms of energy per surface area per day (kWh/m2/day) which can be read directly as "Sun Hour Day” (hours of a day with sunlight). Kuala Lumpur receives in average around 6.0 kWh/m2/day.
OK so at the end of an average day, a 1m^2 panel would have 6 kW-hr fall on it. Actual panel output would be less as no panel is 100% efficient.
ipish:
4. Determine the size of the Solar Panel
The size of the solar panel array is determined by having the adjusted daily load divided by the Sun Hour Day. So for Kuala Lumpur;
189 /6.0 = 31.5 watts
No and yes. At first it appears that you've mixed units. 6 kW-hr is 6000 W-hr and that's for a 1 m square panel at a mythical 100% efficiency. So it's really;
189 W-hr/day / 6000 W-hr/m^2/day
… which gives an answer in m^2, not W, assuming 100% efficiency. That’s 0.0315 m^2.
So what’s a reasonable efficiency ? I’ll SWAG 12% for a panel you can afford.
So you need a panel that’s 0.0315/0.12 = 0.26 m^2 or ~ 50cm by 50 cm.
Alternately you can look at the peak power rating of a panel. This is supposed to be the power output when under an “illumination” (irradiance) of 1000 W/m^2, what you might get at high noon at the equator. If you get 6000 W-hr/day then multiply the power rating in W by 6 (=6000 W-hr/day/1000 W/m^2) to get the panel output in W-hr/day. Conversely divide the energy needed by 6 to get the panel power rating. (perhaps this is how you got it ?)
A 100 W panel, whatever it’s size, will output 100 W if it’s illumination is 1000 W/m^2. If it got that for 6 hrs (= 6000 W-hr) then it would output 100 W during those 6 hrs for an energy total of 600 W-hr.
Since you only need 189 W-hr/day, you need 189/6 = 31.5 W panel. (this is the yes part of no and yes)
I’m unsure if the Sun Hour Day includes a factor for the Sun not being perpendicular to the surface over the whole day. I think it does.
http://pveducation.org/pvcdrom/properti … -radiation
But wait there’s more. The panels output voltage won’t match the battery voltage. There will be a loss in converting one to the other and so the panel needs to output more power to account for that loss or you won’t be refilling your battery. A good efficiency number to use would be 85%, so divide the size or power rating by 0.85.
ipish:
5. Determine Battery size
Now redo using the above and don't forget to include extra capacity for days w/o any Sun. After all you can't charge it up to more than 100%. Multiple very sunny days can't make up for 2 days of pouring rain (ie : no sunlight). I also think your 50% efficiency is really conservative. A real number might be 95% for the charging cycle and 95% during the discharge cycle (= 0.95 * 0.95 = 0.9).