Basic summary Formula for Battery sizing and PV Sizing.
I am from South Africa and enjoy an awesome 6 hours of sunlight hours (On Average). I got interested in solar about a year or 2 ago and have been hooked ever since.I have built my own 60W panel and a solar box that houses a 100AH deep cycle and the switches,fuses,meters etc.
I had bought that Earth for energy guide which helps only up to a point.The video says work on watts to size the system yet the additional pdf with the formula equates everything to Amps.So I had a bit of a confusing time.I then came across this site and browsed articles to get some info on how to size a pv system in Watts using different formulae.(Thank you Sunking,Russ,Mike etc)
After doing the browsing on this forum,I have pieced together how to work out formula for "Battery Sizing" and "PV Sizing" to charge the batterys in the correct time frame.This is JUST what I have gathered and need it to be verified by one of the forum members who know their stuff like the people mentioned above so not to mislead anyone else.
Temperature Correction Factor Table for below calculations:
80 F / 26.7 C = 1.00
70 F / 21.2 C = 1.04
60 F / 15.6 C = 1.11
50 F / 10.0 C = 1.19
40 F / 4.4 C = 1.30
30 F / -1.1 C = 1.40
20 F / -6.7 C = 1.59
Battery Sizing (From Sunking - http://www.solarpaneltalk.com/showth...ing-Calculator)
I am using 1 day autonomy
1. 1000 WH (Needed for 1 day)
2. 1 day autonomy (Just an example)
3. 1000 wh x 1 = 1000 wh (Replace "1" with days autonomy for your own calculations)
4. 50 % = .5 (Depth of discharge,remember the less discharge the longer the batteries last but the more cost)
5. 1000 wh / .5 = 2000 wh
7. 2000 wh x 1.11 = 2220 wh
8. 2220 wh / 12 volts = 185 Amp Hours @ 12 volts. (Rounding off may come into play here)
PV Sizing 2 formula can be used:
1. How long it will take to charge the above 185 AH battery with the panels you have right now with 50% DOD.
2. What PV Watts are needed to charge the battery ideally in 1 day
1. 50% battery is 93 AH @ 14V (charge volts) = 1302 WH to replace
Say you have a 120W PV Panel already.
120W panel x 80% = 96W (Actual panel size taking inefficiencies from the panel into account,including charge controller inefficiency?)
1302 WH / 96W = 13.6 hours to fully charge.
13.6 hours / 6 Sun hours = 2.26 days to fully charge. ( 6 hours of sun in South Africa)
2. 1302 WH to replace / 6 hours sunlight = 220 W panel (Rounde off)
220 W panel / .80 = 275W Panel needed to recharge the 50% DOD in 6 hours (1 Day) taking panel inefficiencies into account.
So anyways.These are the formula taken from various parts of the site.Please verify if they are correct or show me where I can tweak (Without getting ridiculously detailed,this is a guideline.)
Can no-one confirm if these calculations are correct?
OK the answer is yeas and no. Plus there is missing information and wrong assumptions.
First error is you think it will be less expensive with only 1 day autonomy. If you discharge 50% per day, you will be replacing batteries every year or less. In addition if you just have one cloudy day you will be shutting down for at least 2 days, one for the cloudy day, and 1 to recharge plus any additional cloudy days before it clear up.
Second you are using what sounds like the yearly average average insolation which will cause you to go dark in you shorter daylight hour months. For a battery system you need to use the shortest daylight month for your area.
Lastly I do not see where you determined how many watt hours you will use in a day or account for system losses for either PWM controller or MPPT controller.
So from what I can see you need to go back to the drawing board and rethink things. Use this link as a guide and come back with questions.
Thanks Sunking, I see your point and will investigate! One question though, I am aware about your last comment and know I have left off the last calculation out where I work out total Watts per day, am I correct in saying this is also the part were inverter inefficiencies need to be looked at?
Initially you ball park all inefficiency into a single factor for a bar napkin design which is as far as you are going to get in DIY. Once you get in th eball park the equipment engineer plugs in the real numbers and fine tunes with real data once equipment and materials are known.
Originally Posted by Offgrid
With a PWM system you multiply your daily watt hour usage by 2, for MPPT 1.5, and grid tied 1.2
I am so pleased,after investigations,the area of South Africa that I live in receives on average between 7 hours of sunlight in the shortest sunlight month and 10 hours in the longest!!!! So I'll use 7 and NOT 6 like I originally thought.Thanks Sunking!
Actually you should be looking at insolation levels (watts/m2/day) for your area - the hours are misleading.
PV Watts does not have any points listed for South Africa so I went to the same latitude for South America. Santiago, Chile has an annual average of 5.25 kW/m2/day and Buenos Aires, Argentina shows 5.04 kW/m2/day (annual average).
Your 7 hours per day are very far off for the winter months - maybe OK for summer.
Hi Offgrid - You are actually looking at the insolation levels (kW/m2/day) that I was running on about.
The earths surface receives x amount of solar radiation per day - that is referred to as 'insolation'.
Maximum is normally about 1000 watts per m2 for a few minutes at mid-day or early afternoon when the sunlight hits the panel at approximately a 90
You got that wrong, dead wrong. 10 hours of direct sunlight does not translate to 7 Sun Hours, more like 3 to 5. Use insolation, not how many hours the sun shines.
Originally Posted by Offgrid