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Basic summary Formula for Battery sizing and PV Sizing.

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  • Basic summary Formula for Battery sizing and PV Sizing.

    Hi everyone.

    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

    [B]Battery Sizing[/B] (From Sunking -

    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
    6. 1.11
    7. 2000 wh x 1.11 = 2220 wh
    8. 2220 wh / 12 volts = [B]185 Amp Hours[/B] @ 12 volts. (Rounding off may come into play here)

    [B]PV Sizing[/B] 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.)


  • #2
    Can no-one confirm if these calculations are correct?


    • #3
      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.
      MSEE, PE


      • #4
        One question

        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?


        • #5
          Originally posted by Offgrid View Post
          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.

          With a PWM system you multiply your daily watt hour usage by 2, for MPPT 1.5, and grid tied 1.2
          MSEE, PE


          • #6
            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!


            • #7
              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.



              • #8
                Originally posted by russ View Post
                Actually you should be looking at insolation levels (watts/m2/day) for your area - the hours are misleading.
                Hi Russ. I see what you mean. have a look here on page 22 .This a a goverment doc and it looks like the range is 4.5 to 7.

                I am a little confused with this (watts/m2/day) though...So does 4.5kWh/m2/day mean 4.5 hours of ACTUAL solar radiation/sunlight? I mean if it isnt,how do I equate the radiation back to hours of sunlight for calcualtions?


                • #9
                  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


                  • #10
                    Originally posted by Offgrid View Post
                    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!
                    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.
                    MSEE, PE


                    • #11
                      Most solar map data are given in terms of energy per surface area per day. No matter the original unit used, it can be converted into kWh/m2/day. Because of a few convenient factors, this can be read directly as "Sun Hour Day” The number you want to use in this example is for December since December days are the shortest. San Diego is shown to receive 4.6 kWh/m2/day in December.. For Seattle, the number is 1.2 Kwh/m2/day. So we need to note 4.6 and 1.2 for our Sun Hour Day as it will be used to determine the solar panel array wattage
                      I qouted Sunking from my "homework" he has given me lol. SO....because our lowest insolation month is 4.5kwh/m2/day (South Africa,Johannesburg) This equates to 4.5 hours thats should be used in my calculations? I noted Russ's comment about only 85% being usable so I am taking that I need to again reduce this figure by 15%. I think I get the insolation more or less now.Its basically the amount of usable solar energy that hits the earth because not all sunlight is equal in terms of the season and time of day?


                      • #12
                        Originally posted by Offgrid View Post
                        I qouted Sunking from my "homework" he has given me lol. SO....because our lowest insolation month is 4.5kwh/m2/day (South Africa,Johannesburg) This equates to 4.5 hours thats should be used in my calculations?
                        You are getting there now.

                        So let's say you need 2000 wh or 2 Kwh per day, and using a MPPT controller. Formula is [WH x 1.5] / SH, so [2000 x 1.5] / 4.5 = 667 watt solar panel.
                        MSEE, PE


                        • #13
                          Ok guys

                          I have decided to forget everything I "think" I know and rewrite Sunkings post in my own words to give myself a better understanding but not forgetting what was taught in this post.I hope this helps other people in some or any way.

                          [B]Solar Sizing Guide Line[/B]

                          Workout daily loads (Small example of just 1 room)

                          1. 15W light bulb for 6 hours daily = 90WH
                          2. 90W Laptop for 2 hours = 180WH
                          3. 5W Fish tank pump for 8 hours (off when I sleep) = 40WH
                          5. 5W Bedside Alarm for 24 hours = 120WH
                          Total WH = 430WH

                          We now add the inefficiency factor of 2 for system losses like wiring and charge controller and in my case a PWM controller,this would be less and I figure cheaper in the long run if I had a MPPT(Less Panels).

                          [B]430AH x 2 = 860WH[/B]

                          Next we need to work out our Solar Insolation which will be used for our sunlight hours.I learnt an important lesson here.DONT just use the hours of sunlight in your part of the world,not all sunlight is the same in terms of solar radiation.Remember even though the sun may rise really early or set really late,those begining and end hours suck at charging and cant properly count.Also the figure used here must be the WORST insolation month or you wont produce enough power in that month.The worst isolation WINTER month should be used.Look at a solar map for your area.The data is measured as kWh/m2/day.For me my area has 4.5 kWh/m2/day in June and the worst isolation month of the year (Winter Solstice).This 4.5 kWH wil become my Sun Hour Day or the amount of hours with the best charging.More sun hours would obviously always be better.

                          So we devide the 860WH by the worst solar insolation month's hours to give us the size PV array we need.

                          [B]860WH / 4.5 hours = 200 Watts [/B] (Rounded off) So I will need 200 Watts of panels.

                          We now need to [B]Determine the Battery Size needed[/B]

                          Batterys last longer when only shallow discharged also the deeper/quicker the disharge the worse something called Peukert's law becomes.In simple terms a battery rated at say 100AH can be reduced to a 80AH battery if it is dischaged deeply/quickly and you basically get robbed of AH.You can see this on the battery manual if you look at the discharge rate section.Ideally we do not want to drain our batterys lower than 80% total capacity (Use only 20%).So we need to make sure we have 5 times our daily load in battery capacity (20% x 5 = 100% our daily load needs)

                          Since we already know our daily load when we were working out what loads we had in our 1st step,we take this figure and we multiply it by 1.5.The reason for this is there are inefficiencies when it comes to batteries too.Things like only 80% of a battery is available,the fact that rarely do batteries charge at their ideal tempretures.Also the are inefficiencies in the discharge of a battery like Peukerts law as mentioned above.So the daily load we had in our 1st step was 430WH

                          [B]430WH x 1.5 = 645WH[/B] taken daily from the batteries.

                          We now need to multiply this by 5 as we mentioned to ensure only 20% is taken from the batteries.

                          [B]645WH x 5 = 3225WH needed.[/B]

                          Batteries are usually given in amp hours so we need to convert Watts hours to Amp hours so we devide by the volts AH = WH/V.My system 12 Volts so...

                          [B]3225WH / 12 = 270AH[/B] (Rounded off) So I will need 270AH of batteries or above for my system for ONE day

                          There is something we need to look at and that is days of autonomy.Since we dont know when there will be periods of rain,cloud cover etc we need to plan for this so basically days of autonomy is backup power for those days so we dont drain out batteries to nothing.So we multiply our battery for one day above by the amount of "backup" days we need.I amd going to use 5 days as and example but wouldnt suggest anything lower than 3.

                          [B]270AH x 5 = 1350AH[/B] of batteries for my system with [B]5 days autonomy.[/B]

                          These calculations now provide a ball park as to the sizing of your system.

                          That concludes my small tutorial.Sunking and the other senior members can validate my calculations and make corrections.I would like to just give a massive thanks to [B]Sunking[/B] and [B]Russ[/B] for all their assistance provided here.Oscars for both of you lol.If the above information is correct,could you point me in the direction of working out what compromises the system inefficiency in the 1st step above (I used 2 for PWM) ie invertor inefficiency,charge controller inefficiency etc?


                          • uisraan
                            uisraan commented
                            Editing a comment

                            lets go over this problem please :
                            If I have a:
                            80 watts LED
                            runs 8 hours per nights
                            3.2 hours sun a day
                            3 days back up
                            12 volts battery
                            discharge limit of battery is 75%
                            temperature( min -7' c) and (max +37' c)

                            80 * 8 = 640 watts per night
                            640/0.744= 860 w (0.744= losses " wiring, charge controller,system availability..."

                            860/3.2 = 268 watts panel
                            panels information :
                            135 w panel
                            Vmp 18.14v
                            Imp 7.44a
                            Voc 21.74v
                            Isc 8.04a

                            2*18.14= 36.28 v
                            36.28 V *7.44 A = 270 w ( 2 135 watts panel in series )
                            first question : as we know voltage will be changed at hot and cold weather .maximum voltage 24.19 V and minimum voltage 14.41 V. can you enplane it to me please ?

                            800WH*1.59 =1272 WH taken daily from the batteries
                            1272w/24v= 53 AH
                            53AH * 3 days / 75% discharge = 212 AH

                            second question:
                            ambient temperature multiple

                            80' F 1
                            70' F 1.04

                            20' F 1.59

                            the temperature is changing every season , what factor we have to use , factor from max temp or min temp ?

                            thank you

                        • #14
                          Nice job! It really helps when you run through it step by step like that does it not?


                          • #15
                            Yeah for sure!It's so important to understand the design before implementation.So to you and Sunking,everything looks correct?