How to figure out how much Solar and inverter is needed to charge a battery

svetz

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These steps will help you to figure out how much solar you need to recharge batteries. Say, for example, you're going hiking for a few days and want to know how much solar you'll need to charge your critical devices.
  1. Measure how much power your devices use under actual conditions with a watt meter. So for a laptop or cell phone battery, let the device discharge and then recharge it with the meter measuring it. Record that number and how long it took to recharge. For this exercise, let's assume it took 150 watts and two hours to charge; you guessed it, that's 150W x 2 hrs = 300 Wh.
  2. Next, you need to know how much usable sunlight you have. The sun might shine for 12 hours a day, but you'll need an insolation map to get the usable hours. 5 isn't too bad a number in the summer at low latitudes on clear days. If you slept in and are hiking around a couple of hours everyday during 9-3 to get to the next camp site you'll have to subtract that out of your usable hours.
  3. Now take the watt-hours you need and divide by the number of hours of sunlight, so say 300wh/5h = 60w.
  4. Inverters have losses, so divide that by 80%. So, for 300wh you'd need at least 75W of solar panel and that assumes a perfectly clear day all day. But you probably won't have that, perhaps you'll have some clouds or rain. You'll either have to do without under those conditions or budget a little extra -how much is up to you.
  5. Next you'll need an inverter to convert the DC power from the panel into AC so your laptop's charger can convert it back into DC. You know your device draws 150W but you'll want to add a little more to be safe. So you'll need an inverter that can continuously supply at least 150 x 120% = 180 watts.
How much power do my panels generate?
Suppose you already have panels? How much power do they generate? Panels are rated in watts at a very specific set of lab conditions you'll probably never see. Without getting into really complicate equations, folks use an insolation map. This will give you the amount of "usable" sunlight in hours per day for the average day throughout the year, here's the forumla: Number of Panels x Watts/panel * hours from insolation map * losses.

So, if you have 2 100W panels and insolation map says 5 hours for your area, that's 2 x 100 x 5 x .8 = 800 watt hours per day

See the Battery FAQ to get information on properly sizing batteries to store your solar power.

Not in the U.S.? Get your insolation map here (Link courtesy of brassmonkey001).
 
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svetz

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There's a lot of sunshine here, that Insolation Map can't be right, can it?

The wattage rating on a solar panel is measure in a lab using standard test conditions (STC) of 1000 watts of solar power per meter squared at 77 degrees F with an air mass of 1.5.

If the panel is tilted perfectly perpendicular to the sun and the temperature and air density all align, then you should see about the rated wattage at solar noon if it's 77°F.

There are two negative effects going on in the morning and evening: atmospheric absorption from the distance the energy passes through the atmosphere and incidence angle as the solar energy is spread out over a wider area.

Incidence Angle
In the image below you can see how the incidence angle is important. The energy at 90° is more focused on a smaller area than when at 45°. The graph to the right shows the % intensity for a fixed system based on the incident angle of the sunlight.

1617893374495.png

Atmospheric Absorption
In earth orbit, the solar energy is about 1380 W/m², but the Earth's atmosphere absorbs a lot of that energy, at sea level it's about 1000 W/m². Similarly, at solar noon, there's the minimum amount of atmosphere between the panel and therefore the maximum amount of energy from the sun. As the sun continues in its track across the sky away from solar noon, more and more atmosphere gets in the way reducing the total amount of power available to the panel.

The chart below shows how the power available from the sun falls off based on the atmospheric absorption:

1617896425465.png


Insolation Maps
The insolation map takes many factors into consideration include seasonal variations to give you an average amount of power for the year. Considering how complex it is, the insolation map does an amazing job.

The way the insolation map works is to give you the amount of time per day in your area you receive the equivalent of the maximum possible solar power (e.g., 1000W/m2). That is around noon you get 100% for 15 minutes, then for 1 hour ~95%, then 2 hours at ~80%, then 2 hours at ~60%, 4 hrs at 40%. If we add that up 1x.25 + .95x1 + .8x2 + .6x2 +.4x4 = 5.6 hours at 1000W/m2 over a 10 hour day.

Rather than sum up discrete numbers, the insolation map uses calculus to calculate the total. So, if it says 5.5, that means during the average 12 hour day you'll see the equivalent of 5.5 hours at the full rate of 1000W/m2. That's useful because solar panels are rated at that level of solar power, so a 200W panel generates 200W of power when the sun is providing 1000W/m2.

By knowing the number of equivalent hours at 1000 W/m²; you can calculate the total number of watts your panel can generate in a day. So, if the insolation number is 5.5 hours and you have a 200W panel, the panel should on average generate 5.5x200 = 1100W/d

What month gets the maximum output from a solar array?
The earth has an elliptical orbit, it is closest
to the sun and receives the maximum energy at Perihelion (early January for the next few centuries).

This is also demonstrated by solar radiation measurements from orbit.

But if you live in the northern hemisphere, that's probably not what you see.
M1U6-Fig.6.13-ElipticalEarthOrbit.png

Extraterrestrial Solar Radiation (Earth orbit)
1596912887546.png

If you live in the southern hemisphere you get more solar energy in hot months (January) as the sun is higher in the sky and the Earth is closer to the sun then your northern hemisphere counterparts do in their hot months (July). But, you pay for it in winter as the Earth's axial tilt works against you and the Earth is farther away.

The tilt angle of your panels is typically optimized for the season when you need the most power, if you have net-metering it would be set to get the highest yearly total.

The last factor is the weather. If there's a bad weather season, regardless of the energy output from the other factors, it will negatively impact the performance of your solar panels. Tools like NREL's SAM can help you factor all of these to get the best tilt angle for your situation.

My Panels aren't new, are they any good?
Panels usually lose a couple percent efficiency the first year then 1/2 to 1% per year; you can get a better estimate from the datasheet for the solar panels. At 11 years, they might be ~90%; so if they were 250W panels new, they might be 225W now.

If you have a few months of data you can compare the actual output to the output calculated by SAM (or the subset PVWatts) to get the actual efficiency of your panels.

If you don't have a lot of data you can find a nearby weather underground weather site that has solar irradiance, you can get actual measured watts/m² and then compare that to your panels. Check around a few stations to make sure you're not looking at the only nearby one with bad data and use a day clear enough that both locations will likely see the same data.

For the panel's datasheet use the efficiency and the wattage to calculate the actual area of the panels. You'll also have to do some math to add in the temperature correction factor. Here's some example math, but substitute for real numbers:

Datasheet
  • 250 W
  • 20% efficient
  • Pmax [%/°C] = -0.36
Calculated Data
  • Area = 250/1000/.20 = 1.25 m²
Measured Data
  • Solar Irradiance: 800 W/m²
  • Panel: 180 W
  • T=32°C

The equation is: Watts measured x panel area x Panel Efficiency x (temperature correction) = panel output; so

800 W/M^2 * 1.25 m^2 x PERCENT x (100 + (32 - 25) x (-0.36))/100 = 180 W​

Solving for PERCENT will give you the panel's current efficiency. In this case, 18.5%.
 
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D

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So, other than a tracker, what is the best way to maximize this? I see solar panels on roofs and all I think is, "well that is optimum for a few days a year". Then I see large arrays that are all on one grid, and adjustable, but that's a pain, and anything that's a pain won't get done as often as it should. My solution is to have 3 strings of panels. One will point slightly to the east, one dead south, one slightly to the west. I think that's the best solution other than a tracker.
 

svetz

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what is the best way to maximize this?
The only way I know of to minimize atmospheric loss is to relocate to a higher altitude and/or location with drier air; essentially reduce the amount of atmosphere (in Earth orbit it's something like 1366 watts/m2). Most, like you, just add a few more panels to make up the loss and call it a day.

For spacecraft, distance from the sun has a big impact too, just in case you're planning solar for your extra terrestrial summer home:
PlanetDistance (x 10^9 m)Mean Solar Irradiance (W/m2)
Mercury
57​
9116.4​
Venus
108​
2611.0​
Earth
150​
1366.1​
Mars
227​
588.6​
Jupiter
778​
50.5​
Saturn
1426​
15.04​
Uranus
2868​
3.72​
Neptune
4497​
1.51​
Pluto
5806​
0.878​
Trackers do get more power than fixed tilt systems, but that gain is based on the panels always being at the optimum tilt. Don't have experience with trackers but have heard the economics rarely pay out despite getting +20 to 30%.
 
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watson118

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These steps will help you to figure out how much solar you need to recharge batteries. Say, for example, you're going hiking for a few days and want to know how much solar you'll need to charge your critical devices.

Hi Does this work for rechanging a battery too?
I need some help on working out what solar watt input I need to keep a system running 24 hours a day, 7 days a week.

I have been involved with a customer on installing some cameras on a small outbuilding (Barn) that has no power but she told me she had a solar system installed with a 130Ah battery.

So I have worked out the wattage of the cameras and the Wireless bridge, and converted them into amps,
So I have:
3 x Cameras (12vDC) 6.1w each
1 x Network switch (5vDC) 2.4w
1 x 12vDC to 5V Converter
1 x 12vDC to 24cDC convertor
1 x Wireless Bridge (24v) 7w

I worked all this out to be about 1.9amps at max.
so on 130 battery, I should get around 60 hours of run time without a charge (we actually get about 85 on a full charge without any solar)

I have been having lots of issues with the customer saying she’s having to change the battery out every few days (and charge it at home), Cut a long story short, I have gone back out to take a look at the solar system and the solar cables into the controller was wired backwards so it was never getting a charge. I fixed this and tested the panel it, it looks to be only a 50watt panel, at the time I tested it, I was only getting 26watts. It turns out that the kit she bought (second hand) only had a 7AH battery.

So it looks to me that the solar is not powerful enough to recharge the 130AH battery during the day. My question is, can I calculate how many solar panel and wattage is needed? to change the battery during the day, enough so the battery can power the CCTV during the night till the next sunlight?

I guessed 200watt kit but I am so lost with all this. I was going to add more panels but I think it's just a cheap controller and as it's only got a 50watt, I think I can only add more 50watts? The customer is on a budget but I have suggested she gets a new controller and solar panels too.

Thank you
 

svetz

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I need some help on working out what solar watt input I need to keep a system running 24 hours a day, 7 days a week.

So it looks to me that the solar is not powerful enough to recharge the 130AH battery during the day. My question is, can I calculate how many solar panel and wattage is needed? to change the battery during the day, enough so the battery can power the CCTV during the night till the next sunlight?
The process is:
  1. determine the number of what-hours the equipment consumes in a day (say 800 wh/d).
  2. Divide that number by .8 for losses (800 / .8 = 1000 wh/d)
  3. Divide that value by .5 to get the base battery size if lead acid or .9 if LiFePO4 batteries (100 / .5 = 2000 wh/d). Divide by system voltage to get battery amp-hours (e.g., a 12V system would be 2000 / 12 = 166 Ah battery).
  4. Multiply the watts in step 3 by the number of days of backup (e.g., 3 days is 3x2000=6000 wh, or 500 Ah).
  5. Get the value from an insolation map for your location that represents hours-of-sun-at-100%-output (say 4).
  6. Divide step 3 by the value from step 5 and again by .8 for losses (2000 / 4 / .8 = 625 watts) to get the minimum number of watts the panels must supply to meet your needs (minimum as clouds/shade will reduce output). If you expect 50% clouds, double the panel wattage. Keep in mind that in winter the cold weather will negatively impact the battery and solar output, see the battery FAQ for more information. If you get a larger inverter than you'll think you need (or use microinverters) you'll be able to add more panels later if needed.
 

nosys70

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for a 24/7 that is easy, you just need to know how long it should take and how much recharge you need.
for a 100Ah battery, you need 100Ah for one hour.
Obviously that is not the best, because usually you try to charge a battery slowly to avoid killing it too fast and getting 100A is not that easy.
so you will need to charge slower let's say at 0.3C that is 33A for about 3 hours. It give you some slack in case the sun is not so shiny and it would take longer.
You need to add that to your load calculation, because if you expect to use the solar power during the day, you know you will need some to charge the battery. (at least 33A for 3 hours).
For the charger, you know that a 40A should be ok

This is the worst case, when battery is empty, but it is possible the battery size is ok to fullfil the load overnight without goind completely down.
So you can do a 2nd calculation for best case, for example if you know that you need only to charge 40% of the battery.
For a small battery , you better will ignore this, since it is pretty sure you will heavily drain the battery most of the time.
But for a big battery that largely exceed the load, it could give more realistic numbers make you spare money.

if you see that you need longer time than half a day to charge your battery or there is no power left for the load , it means you must increase power on the solar panels and then also charger and eventually the charge rate.
if you are in a really sunny place, you can also spread the charging time on the full day.
 
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kenkoh

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In NE PA I get about 3.5-4hrs peak sun. Will a 400w system charge a Battleborn 100amp hr battery everyday or should I get 500w to be safe?
 

nosys70

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supposing your batttery is 12V, 400W under 12V is 33A, that is what your panel gives at best each hour.
so if you have a 100A batt, you need at least 3 hours to achive full charge. that fits the peak sun period.
even if in reality the charge will spread over 5 hours or more , this means 100A/5=20A.
So you will keep 20Ax12V=240W for charging battery most of the day, that leaves 400-240=160Wh for your load.
if you need more than this you better will upgrade to a 500W PV.
it means also, that on a 24Hours day, you will get 100/(24-5)=5A , 5x12=60W available for each hour you run on battery only.
and this is with full capacity of battery, that is not recommended. clearly 100A, is a bit short for an off-grid situation.
You would hardly run more than a few led light and small TV or laptop computer (a business laptop is about 50Wh)
 
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kenkoh

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I plan to keep it simple and run 3x 400w seperate systems each with a Battleborn 100amp because I'm a redundancy nut. Is 40amp controller enough for 500w panels? Plus do you know if the discharge for the Battleborn is 90%? Thanks.
 
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nosys70

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500W panel at 12V is 500/12= 41A, so a 40A controller is a bit short, you should target at least 60A.
strangely , power surge on panel do not happens when sky is blue and sun shining, but with cloudy sky.
You also have to take in account that controller are usually rated for more Amps in input, than in output for charger.
So a 60A controller could possibly have only 20A charge, and you know that for your 100A batt, 30A would be better.
 

Supervstech

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In NE PA I get about 3.5-4hrs peak sun. Will a 400w system charge a Battleborn 100amp hr battery everyday or should I get 500w to be safe?
This depends a lot on what your use of the battery draws during daylight hours...

if you have 500watts of draw during the day, you will need 750ish PV to feed the draw, AND charge the battery...
 

Supervstech

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500W panel at 12V is 500/12= 41A, so a 40A controller is a bit short, you should target at least 60A.
strangely , power surge on panel do not happens when sky is blue and sun shining, but with cloudy sky.
You also have to take in account that controller are usually rated for more Amps in input, than in output for charger.
So a 60A controller could possibly have only 20A charge, and you know that for your 100A batt, 30A would be better.
The cc’s I have seen all rate their output amps... they rate the input watts, and depending on voltage, it equals the output amp rating...
 

Supervstech

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This depends a lot on what your use of the battery draws during daylight hours...

if you have 500watts of draw during the day, you will need 750ish PV to feed the draw, AND charge the battery...
Of course... if you have 500w draw during the night... you need more than one 100Ah BB
 

nosys70

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that 's easy, you read the Amps or Watt witten on the charger.
for sure you will be way under the rating.
so if your charger is 60W, it means you need 12x60W
if your solar sytem runs at 12V, 60W/12V=5A
So you need a a 100w 12v 8A panel with a converter able to sustain 10A, and a battery that is at least 12V 20A.
this will cover cloudy days. I would even consider a 120 or 150W solar panel.
 
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Hi, I'm not sure which value to read to get the usable hours. Can you help please?

1580341214620.png
 

svetz

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Hi, I'm not sure which value to read to get the usable hours. Can you help please?
Your data says between 4.95 and 5.88. From this document is looks like the very north (above the Port of Spain) is >5; but most is < 5. NREL's map is this:
Capture.PNG 1580349669151.png

I'd say you're probably safe with 5.0, if you get more it's like a bonus ;-)
 

kenkoh

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This depends a lot on what your use of the battery draws during daylight hours...

if you have 500watts of draw during the day, you will need 750ish PV to feed the draw, AND charge the battery...
Will this do?

& as I understand a single string series to SCC needs no fuse correct?
 

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