Renogy kit

BB.

Generally, I suggest to get a bigger AH battery (many times, these are lower voltage batteries... Not 12 volts, but 6v, 4v, or even 2v batteries (very big and heavy "single cell" 2 volt battery-for example).

Going to 2 or 3 parallel strings is OK, but you need to make sure you have equal cable lengths between each string of batteries (batter string with "short length cables" will charge/discharge faster because of low resistance allows more current to flow).

http://www.smartgauge.co.uk/batt_con.html

More than 3x parallel strings will work... But they tend to require more cable/connections (more $$$) and more maintenance (checking that batteries are sharing current properly, check that you have no shorted or open cells, etc.). If you have flooded cell batteries, also more cells to keep watered.

I am really big on trying to design/build a "balanced" system design. You define your loads/energy needs, then design a battery bank that will effectively support those needs... And then design the charging system (solar, genset, utility power) to keep the battery bank "happy".

Flooded cell Lead acid batteries are (relatively) cheap and rugged. Decent life. But, do have limits about how much current they can supply in surge conditions (like starting a deep well pump) and maximum rate of charge.

AGM batteries support higher surge currents and faster charging (in general). But tend to have a bit shorter life than similar quality FLA batteries (and AGM are more expensive).

There are other battery chemistries that have advantages and disadvantages. There are always trade offs.

More or less, for your DC Battery bus (and works well for FLA batteries)--For a 12 volt battery bus system, you should limit the maximum power (watts) to around 1,200 to 1,800 Watts. If you have more than ~800 AH battery bank (or larger inverter), the current flow becomes quite high--Needing very heavy copper cabling and more (for example) solar charge controllers (most of the high end residential solar charge controllers limit output to ~80 amps--A 10% rate of charge for an 800 AH battery bank--The "nominal" charging recommendation for off grid solar power systems).

The alternative to design for you loads is to design for some other limitation. For examples, RVs have limitations for battery weight/space and roof space for solar panels. Pick one of those limitations, then design the balance of system. And we can predict its "useful" output (and you can decide if that meets your expectations or not).

Most people way overestimate how much energy a solar power system can output... They purchase a 3,000 Watt AC inverter and put it on a 200 AH @ 12 volt battery bank (for example, 2x 6 volt @ 200 AH "golf cart" batteries)... And get:

• 3,000 Watts * 1/10.5 volts cutoff voltage * 1/0.85 inverter eff = 336 Amps
• 200 AH battery bank / 336 amps (at 3,000 Watts) = 0.6 hours (36 minutes) of operation until battery is dead
• And, in reality, a 200 AH @ 6 volt battery bank (2x 6 volt @ 200 AH batteries in series for 12 volts @ 200 AH battery bank) will "reliably" support a AC inverter a 500 Watt AC inverter (there is not much use going larger with this size FLA battery bank--The battery bus voltage will quickly collapse and shutdown the inverter).

It is really difficult to give "good answers" to questions that we have little insight to what your energy needs are. A technically correct answer could give you a practically useless (for your needs) system design.

-Bill

brownbear

Thanks all for the information. I'm going to proceed with a 2nd 100watt panel and a 2nd 12V 125ah battery and put it in parallel to get 250ah.  Hopefully this will be enough AH (125AH 50% discharge) to power a 65inch tv for a good four hours a day, modem, router, roku, cell phone, small battery bank, portable speaker. Do you think I should get a total of three 100watt panels? With a single 100 watt panel my mt50 is showing an average of 0.32kWh /day.    What does that mean anyway?  Does it mean I'm getting 0.32kWh total for the day or multiple that by the hours of sunlight I get for the day or multiple that by 24 hours?  Please explain.   

mcgivor

brownbear said:

Thanks all for the information. I'm going to proceed with a 2nd 100watt panel and a 2nd 12V 125ah battery and put it in parallel to get 250ah.  Hopefully this will be enough AH (125AH 50% discharge) to power a 65inch tv for a good four hours a day, modem, router, roku, cell phone, small battery bank, portable speaker. Do you think I should get a total of three 100watt panels? With a single 100 watt panel my mt50 is showing an average of 0.32kWh /day.    What does that mean anyway?  Does it mean I'm getting 0.32kWh total for the day or multiple that by the hours of sunlight I get for the day or multiple that by 24 hours?  Please explain.   

Sometimes it's better to do calculations using a common measure, which perhaps where you are apparently becoming confused. Here is a way to use Watt hours/ Kilo Watt hours.

Take the battery capacity of 125 Ah, to convert that to watt hours multiply 125 Ah by the nominal voltage 12V, this gives you the total capacity
.
125 × 12 = 1500 Wh or 1.5Kwh (1500 ÷ 1000) 

Now divide the total by 2, this will give you the usable capacity, to 50% depth-of-discharge

1500 ÷ 2 = 750 wh or 0.75Kwh

Now it's simple to calculate what consumption is and how much battery capacity is needed to support them. Random figures used below.
80W of TV for 4 hours = 320w
20w of router for 6 hours = 20 x 6 = 120wh

35w of Roku 3 hr = 105wh

20w of modem for 6 hr = 120wh
22w of lights for 6 hours = 150 wh

Add all the totals together 320 + 120 + 105 +120 + 150 = 815wh (0.815 Kwh) So the battery usable capacity would be exceeded, so the capacity needs to be increased. So let's say you parrallel another battery of the same capacity, the usable would now be 1500 wh. a much better amount because the DOD will now be less than 50%.

Now the battery needs to be charged, so using the panels rated capacity divided by it real world output ~75%, in the case of a 100W panel 75 W, multiply this by the average hours of sunlight, let's choose 5 hours    
75 × 5 = 375 Wh or 0.37Kwh  NOTE, this is pretty close to what your controller reports as a TOTAL for the day.

When designing it's good practice to size the array to provide enough to charge the entire capacity, this gives enough of a safety margin to compensate for bad sun days. Using the above numbers the total capacity would be 3000 Wh, so 3000 ÷ 375 = 8, therefore 8 100W panels would be required.

All of this is a rough calculation, depending on where you are located there may have to be adjustments made for winter months.

Bill @BB. has provided much more detail in post #8, which includes efficiency losses etcetera, my approach is to simplify it so you can get a grasp of the fundementals, from which you can fine tune.

littleharbor2

0.32 Kwh. sounds about right for a single 100 watt panel, not optimally oriented, for a 24 hour period. 

brownbear

So 0.32kWh x 24hr= 7.68kWh.      Is that what my one 100watt panel is producing every day?    

mcgivor

No, the 0.32Kwh is what the panel was able to produce for the day whilst the sun was shining for the 4-5 hours available.

Example 
100W × 75% = 75W
75W × 4.5 hours = 337.5 Wh or 0.3375Kwh approximately what your figure is.

BB.

Solar panels only generate (useful) electricity when there is "direct sun" on the panels (i.e., when morning direct sun first hits panels). And the current is proportional to the amount of sunlight hitting the panel.

So, say you are in Los Angeles Ca, fixed array, facing south, panel tilted 56 degrees from vertical
http://www.solarelectricityhandbook.com/solar-irradiance.html

Los Angeles
Average Solar Insolation figures

Measured in kWh/m2/day onto a solar panel set at a 56° angle:
(For best year-round performance)

Jan	Feb	Mar	Apr	May	Jun
4.50	4.82	6.05	6.78	6.83	6.80
Jul	Aug	Sep	Oct	Nov	Dec
6.69	6.67	6.40	5.85	5.07	4.41

Today is November something... "hours of sun" is really hours of 1,000 Watt*hours per sq meter (full sun) for typical November day (obviously little sun at sunrise and sunset, most sun around noon):

• 100 Watt solar panel * 0.77 derating for panels+charge controller * 5.07 hours of sun (Nov Average) = 390 Watt*Hours per day supplied

That is the energy generated by the panel (per day)... If you want to estimate how much energy an AC appliance could use per day (assuming you use 100% of the available average harvest):

• 100 Watt solar panel * 0.52 end to end AC system eff * 5.07 hours of sun (Nov Average) = 264 Watt*Hours for a 120 VAC load

That includes losses for charging a flooded cell lead acid battery (~80% eff) and AC inverter (eff ~85%).

Numbers are, at best, +/- 10% accuracy. Allow for some days that are really sunny, and other days not so much...

-Bill

littleharbor2

0.32 Kwh. is 320 watt hours. That's about all you're going to get while that panel is mounted in a fixed position, this time of year on a clear day. If you could track the sun all day, from sun up to sunset you might double that. The early and late day sun has to travel through a lot more atmosphere, haze and such than the high noon sun, so even with tracking you wouldn't get anywhere near 1000 watts per square meter. That's how your panel is rated. That, and at 25 degrees Celsius. As soon as your panel warms up the voltage will drop.  Here's an example of a Kyocera 140 watt panel. The 140 watts is the STC (standard test conditions) rating The other rating is NOCT rating. Not as intense sun, at warmer cell temps. Now your 140 watt panel just became a 100 watt panel, in more , real world conditions.