Basic summary Formula for Battery sizing and PV Sizing.

[Mike90250]

If you have problems connecting solar panels, then this website for instructions. It describes not only the connection methods, but also the best solar panels.

OK< you are now a spammer. Next one gets you banned.

[Mike90250]

I have four pieces of 250w solar panel s(poly type).

What is the Vmp listed on the sticker on the back of the panels

With 60amps solar charge controller(PWM) and four 200 amps batteries.
How will I connect the solar panels to effectively charge my 4 batteries?

What is the battery bank voltage ? are they 6V200 ah batteries ?

Answer these questions and we can begin to help you

[Lordacquit]

I have four pieces of 250w solar panel s(poly type).With 60amps solar charge controller(PWM) and four 200 amps batteries.
How will I connect the solar panels to effectively charge my 4 batteries?

[Mike90250]

......Would sure like to know some of your secrets of space travel and aliens..

a frisbee and a towel

So I'm just about through reading the stickies which I thought I should do as a newbie. Mike I have lots of pizza and soda and a big interest in radios as a ham in Australia . Would sure like to know some of your secrets of space travel and aliens..

[uisraan]

Hi

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 &quot; wiring, charge controller,system availability...&quot;

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

[uisraan]

Hi

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 &quot; wiring, charge controller,system availability...&quot;

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

[Mike90250]

...... 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?

Well, the closer you match the Vmp of the panels to the battery with a PWM controller, the more efficient things become. But if you use 30V panels on a 12V system you will have massive losses unless you use a MPPT controller.

Otherwise, the quick glance through looks ok

[indalecio.feng]

Sometime, if you want the system more efficiency, the battery should not do such design.

[BrentEMarvin]

No one can say that it is correct or not. It's your calculation not proved it's just a measurement.

[Sun Tzu]

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


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
6. 1.11
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.)

Thanks!

I am with Sunking on this one. Need more info, adjust some assumptions, etc. I would like to throw Homer into the mix. It is a software tool that allows you to size a system. It can seem a bit daunting to at first, but it isn't that bad. The kicker is that they usually offer a free trial version of it.

[JoyDxL]

Let me share my thoughts about the real power of solar panel:
usually the manufacturer specifies the technical characteristics of a solar panel (or solar cell) for standard test conditions: 1000 W / m2, 25 C, 1.5 AM
key indicator in most cases it is the power of a panel, aka in the wild rarely reaches the nominal values. therefore, for each case you must separately calculate the real power of the panel based on the nominal value of power in STC and the actual level of insolation in your area.
for example for my city average annual insolation is about 208 W / m2 (2.92 kW / m2 / day).
so you should see a graph of the power of insolation and understand that I have to split the nominal power of about 5.25 (may vary for different panels) to find the real power panels for my condition.
in general, by my count, at 15-17% panel efficiency and loss on the conversion DC to AC, I can gather in an average about 30-35 watts from one square meter

As for me, the initial design of the solar system based on the square is more obvious, since private construction are significant limitations in the effective area on which to place the panel. maybe I'm wrong, so correct

[Tree Trimmer]

is the distinction here, using his example, that:

1. he properly sized the panel(s) for his needs
2. he properly sized the batteries to give him the desired available power, for his desired autonomy.

3. he improperly sized the panel(s) for his batteries,

[suryana]

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.

Hello Sunking, the link was broken. Could you give another link?
And also, can I use the calculation above for estimating in laptop/gadget charger? many thanks for your answer

[Mike90250]

I've seen Sunking's posts, in which he trashes newbies...

not quite. He trashes irrational thinking (which I've done my share of I.T. and Trashing) and presentation of incorrect data. What a person does off the "clock" is there own business. I talked solar off the clock, when I was designing spacecraft radios. There are about 10 people in the world who "we" did peer review with. No journals or society party's - but often all nighters in the lab with pizza and sodas.