Building a Small System (some problems)

[Sunking]

If half that energy is wasted and I use the maximum value to size the battery, why would I need anything larger than 43Ah? It seems if the battery was any larger, it would never be able to get a full charge.

Your battery needs sized to 5 day capacity, which in practice gives you about 3 days with clouds before you must shut down and fully recharged. 5 day sis used to get the most bang from your battery dollars, and CYA for a couple of cloudy days. Ar 4t AH is only 516 wat hours, not enough to CYA for 1 cloudy day. You never ever want to discharge your battery more than 50%.

Solar panel power must meet 2 conditions.

1. To replace what was used in a day under worse case conditions of winter short days. In your case 360 watt hours. That means the panels must generate either 720 watt hours with a PWM controller, or 550 watt hours for a MPPT system.

2. Must meet minimum Charge Current requirement of a battery. For FLA roughly C/12, with C/10 being ideal.

So you use the larger panel wattage of the two.

If you use a PWM controller on a 150 AH battery requires 270 watts to generate 15 amps of charge current or, 200 watts on a MPPT system. That takes care of the minimum requirement. Now if you live in an area with say 2 Sun Hours would require a 275 watt panel with a MPPT 20 AMP controller.

Batteries also have a Max Charger Current they can accept, and on FLA is roughly C/8. On a 43 AH battery is only 5 amps to 6 amps. Any faster and you would risk burning your batteries up. See a problem yet? With a 43 AH battery and you pulling 360 WH/day would require you to charge them with 15 to 20 amps of C/3 to C/2 current to replace what you use in a day. . You would cook your battery.

If you design based on standard practices, everything falls into place.

[Sunking]

But it may give some reasonable estimate of long term average array output that may be useful.

JPM PV Watts can give you good info to design a battery system if you know how to tweak the numbers. I do it all the time.

I enter 1, 10, 100, or 1000 watt panel or how many ever zeros you want as a panel wattage. Place the model in the area of use with tilt and orientation angles and assume 100% efficiency.Then look at the months of Dec/Jan to see what the Watt Hours were produced, So if you used 1 watt panel an din December low days is 3.2 watt hours, is directly converted to 3.2 Sun Hours. So if you need 1 Kwh usable on a MPPT system requires the panel to generate 1500 watt hours. Thus panel wattage = 1500 wh / 3.2 Sun Hours = 468 watt panel. Go shop for a 500 watt panel.

[J.P.M.]

JPM PV Watts can give you good info to design a battery system if you know how to tweak the numbers. I do it all the time.

I enter 1, 10, 100, or 1000 watt panel or how many ever zeros you want as a panel wattage. Place the model in the area of use with tilt and orientation angles and assume 100% efficiency.Then look at the months of Dec/Jan to see what the Watt Hours were produced, So if you used 1 watt panel an din December low days is 3.2 watt hours, is directly converted to 3.2 Sun Hours. So if you need 1 Kwh usable on a MPPT system requires the panel to generate 1500 watt hours. Thus panel wattage = 1500 wh / 3.2 Sun Hours = 468 watt panel. Go shop for a 500 watt panel.

Not being experienced in battery system design, but knowing just enough to be dangerous, I think I understand. But maybe not.

PVWatts (Dubuque IA - maybe near Bruce ?): .468 kW system in Dec./Jan., 45 deg. tilt, 180 az., zero system losses, ave. daily P.O.A. irrad. ~ 2.7-3.1 kWh/m^2, monthly system output ~~ 45 kWh/month ~ 45/30 = 1,500 Watt-hr./day if every day is average. I'm maybe missing something.

I'd guess one would need, among other things, some information such as a probability distribution of irradiance availability to take a SWAG at matching battery system size and capabilities with resource availability and expected duty requirements.

But, and not talking about our difference of opinion on the anachronistic nature of the term sun hrs., I'm not sure on your units. Assuming your referring to 1.5kWh./day: (kWh/day)/(sun hr.) = kWh/day/[kWh/(m^2*day)] = m^2 ?

Not tryin' to bust your ass, just think I might not be getting what you're writing, but willing and curious to learn.

[Sunking]

JPM what I am saying is enter a 1000 watt panel into PV watts, area, orientation,,tilt angle and 100 % efficient. PV Watts will spit out how many Kwh hrs for each day of the year in months. Look for the lowest say January might say 3. Kwh = 3.0 Sun Hours, and July might say 6.0 Kwh = 6.0 Sun Hours.

Use January. If you need 1 Kwh per day would require a Panel Wattage = 1 Kwh x 1,5 / 3.0 Sun Hours = 500 watts. If you used July, you are SCREWED with a 250 watt panel.

I know you are not busting my chops unless i deserve it. FWIW you are smart enough and read enough to know the basics of a battery system. Each component has to be matched to work with all the pieces.

[J.P.M.]

JPM what I am saying is enter a 1000 watt panel into PV watts, area, orientation,,tilt angle and 100 % efficient. PV Watts will spit out how many Kwh hrs for each day of the year in months. Look for the lowest say January might say 3. Kwh = 3.0 Sun Hours, and July might say 6.0 Kwh = 6.0 Sun Hours.

Use January. If you need 1 Kwh per day would require a Panel Wattage = 1 Kwh x 1,5 / 3.0 Sun Hours = 500 watts. If you used July, you are SCREWED with a 250 watt panel.

I know you are not busting my chops unless i deserve it. FWIW you are smart enough and read enough to know the basics of a battery system. Each component has to be matched to work with all the pieces.

Understood. I wouldn't have thought of doing it that way, but battery systems are not an area where I consider myself proficient. If I'm lucky, I learn something every day. Looks like today was a good day.

[Sunking]

Understood. I wouldn't have thought of doing it that way

Wel if you do it to make a few coins enough times you figure it out. If you think about it, know how PV Watts works, you can work it to tell you what you want. In my case I am looking for Sun Hours.

Like I said you can enter, 1, 10, 100, 1000, or 1 million watts. Whatever floats your boat as long as you subtract the least significant digits when done. If you use say 1 watt, 100% efficient you get Sun Hours. You are just forcing PV Watts not to derate anything.

I stole the idea from John Wiles years ago. The way John determined Sun Hours years ago the MacGyver way was with a 10 watt panel and MPPT tracker, and data logger. Put the panel outside, and let it log. So if you got 30 Watt Hours in a day, just use simple high school algebra. Sun Hours = Watt Hours / Watts.

I just put my spin on it using PV Watts. John did not have that available to him decades ago. . When it comes to train driving, I like to KISS. In this case it is cheap, fast, and accurate using JPL data. What is not to like? FWIW PV Watts and JPL data is the work of John Wiles. It was John and his grad students that created it. Just do not let John do the actual design or code work.

[bcroe]

Throwing up a single panel works. Rather than attach an MPPT system, I just shorted the output and divide the
observed current by 1.25A to get the fraction of rated power. PVwatts should cover the range of weather over
many days immediately. It takes a run for each orientation here. Bruce Roe