Newbie question

Gene.243

I'm just getting into solar power and was wondering how much usable power can be collected by say a 1 KW collector array charging a bank of flooded acid batteries.

TIA
Gene

icarus

Re: Newbie question

1 kw of PV connected to a FLA battery bank and providing power through an inverter will look like this.

Take the name plate rating of the PV, divide in half to account for all cumulative system loses, then multiply that by4 to represent the AVERAGE hours of good sun you can reasonably expect, per day over the course of the year.

So, 1000/2*4=2000/WH/day.

Welcome to the forum, good luck and keep in touch,

Tony

Gene.243

Re: Newbie question

Then to use the 2000WH/day I would need a battery bank where 2kwh/per day is 75% of my battery bank capacity?

Gene.243

Re: Newbie question

No 2kwh/day is 25% of bank capacity. So I would need 8 12V 100 ah batteries.

Cariboocoot

Re: Newbie question

Gene.243 wrote: »

Then to use the 2000WH/day I would need a battery bank where 2kwh/per day is 75% of my battery bank capacity?

That is a better starting point: you need 2 kW hours per day, how do your get it?

On 12 Volts it's about 167 Amp hours. If you want to limit the depth of discharge to 25% (recommended; no more than 50%) you need a 12 Volt battery bank of (167 * 4) 668 Amp hours. That can be done, but it's a bit large. Double the system Voltage to 24 and the battery bank in Amp hours is half: 334.

To recharge that you'd want an array capable of peak charge rate that is 10% of the capacity, or 33 Amps @ 24 Volts (792 Watts). Adjust for efficiency losses of panel and controller: 1028 Watt array.

Now let's look at how that works with Tony's formula:

1028 Watts * 4 hours of equivalent good sun * 52% over-all efficiency = 2139 Watt hours AC.

This is what we call a balanced system.
Of course with longer sun hours you can get more power (which may not be usable unless you can time loads to take advantage of panel output when the batteries are charged). Shorter sunlight means less power.

Gene.243

Re: Newbie question

If I am only 34 degrees from the equator can I count on more hours per day of sunshine than you?

icarus

Re: Newbie question

I would only add, the people at once and the same time over estimate thier solar harvest potential (by adding hours primarily) and underestimate their loads.

For example, my little 400 watt system does great (on average) for my loading of 5-800wh/day. On an ideal day, I can get 2 kwh, but in most places (us SW might be an example) ideal days (and times of year) are not ideal, hence the four hours. The mistake people make is seeing sun on thier roof frm 8-4, then assuming they get 8 hours of sun. The reality is the insolation rises to a peak near noon, and falls off in the PM. Additionally, the 4 hour rule (generally) accounts for averge cloud cover, grey days etc. Partly cloudy for weather cumulous can have a dramatic effect in net harves. You can see huge numbers at edge of cloud events, but the puffy clouds can drop harvest to near zero for much of a day, just to cite an example.

Tony

Cariboocoot

Re: Newbie question

Gene.243 wrote: »

If I am only 34 degrees from the equator can I count on more hours per day of sunshine than you?

Almost everyone can count on more hours of sunshine than me!

Check out your harvest potential using PV Watts: http://www.nrel.gov/rredc/pvwatts/grid.html

Gene.243

Re: Newbie question

Thanks for the inputs guys. I'll go sleep on it for the night.
Gene

ggunn

Re: Newbie question

Gene.243 wrote: »

If I am only 34 degrees from the equator can I count on more hours per day of sunshine than you?

If you are asking 'Coot, that would depend on the time of year. Since he lives way north, he will have a lot fewer hours of sun per day than you in the winter and a lot more than you in the summer. That said, at the north pole there is sun 24 hours a day in the summer, but you'd need a tracking system to take advantage of it.

Gene.243

Re: Newbie question

Actually the sun shines on just over half of the planet all the time.

inetdog

Re: Newbie question

Gene.243 wrote: »

Actually the sun shines on just over half of the planet all the time.

I am so glad I do not live in that half. :-)

NorthGuy

Re: Newbie question

Gene.243 wrote: »

Actually the sun shines on just over half of the planet all the time.

It would've, if not for the clouds.

Gene.243

Re: Newbie question

Back to the thread. How much better would it be to use DC power?

inetdog

Re: Newbie question

Gene.243 wrote: »

Back to the thread. How much better would it be to use DC power?

If you charge batteries and then use the DC directly, a rough approximation would be:

4 solar hours times 1Kw = 4KwH.
Battery charging efficiency, about 90 %. (The claimed and calculated numbers vary widely and this is only a starting number.)
MPPT charger efficiency, about 95% if you operate it within its best range.
4 x .90 x .95 => 3.42 KwH.
But you have to derate that more to account for the fact that your batteries may not be capable of accepting the full power output of the cells the entire time. (If they could, they are undercharged). That brings us back closer to the previous 50% utilization. The efficiency of a good inverter, in its sweet spot of power output, can be over 95%, so skipping the conversion to AC does not gain you a lot.

waynefromnscanada

Re: Newbie question

Gene.243 wrote: »

So I would need 8 12V 100 ah batteries.

Surprised no one picked up on the EIGHT 12 volt batteries.
Do you intend to wire these in parallel?

NorthGuy

Re: Newbie question

inetdog wrote: »

... skipping the conversion to AC does not gain you a lot.

Also, with 12V wiring, the losses in wires will be approximately 100 times bigger than with 120V AC, which is likely to exceed losses in the inverter.

Cariboocoot

Re: Newbie question

NorthGuy wrote: »

Also, with 12V wiring, the losses in wires will be approximately 100 times bigger than with 120V AC, which is likely to exceed losses in the inverter.

Assuming long wire runs and/or high power demands.

Otherwise it is a matter of the DC use avoiding conversion losses (up to 15%) and inverter power consumption (could be 30 Watts right there if its a large, inefficient inverter).

Of course 12 VDC is simply not always interchangeable with 120 VAC as some things just aren't available in both versions and any high Wattage items will draw large amounts of current which exacerbates the V-drop problem as per NorthGuy's conclusion.

Eight 12 Volt batteries in parallel will be a problem either way.

Gene.243

Re: Newbie question

My well pump is rated for 12 to 40 volts DC. If I go with a 36 volt system how would I set up my batteries?

Cariboocoot

Re: Newbie question

Gene.243 wrote: »

My well pump is rated for 12 to 40 volts DC. If I go with a 36 volt system how would I set up my batteries?

To get 36 Volts from eight 12 Volt batteries you use two parallel strings of three in series, and have two left over.

You might regret choosing 36 Volts. Many charge controllers can not be configured for it and if you should decide to use an inverter there's not much choice either.

Gene.243

Re: Newbie question

What's a guy to do? Go to 48 volts, That removes my well pump from the DC system. Do I give it it's own panel and just run the well separately? Then I can install a separate 1KW PV array to power my other electrical needs. Depending on spacing between panels I can fit 6 at 0 space between panels, 5 gives me plenty of room to reach between panels.

Cariboocoot

Re: Newbie question

24 Volts is your friend.
It reduces the current problems found on 12 Volt systems, does not have component issues like 36 or 48 Volt, and will run your pump.

Using your eight 12 Volt batteries placed two in series per string and four strings in parallel you're sort of at the practical limit for parallel connections and should use bus bars to connect them via equal length wires. This minimizes the resistance differences that lead to imbalance.

Gene.243

Re: Newbie question

Are there better battery choices?

Cariboocoot

Re: Newbie question

Gene.243 wrote: »

Are there better battery choices?

Yes: string made up of higher Amp hour 6 Volt batteries. For example the basic "golf cart" type battery is 220 Amp hours @ 6 Volts. Four in series gives you one string capable of 220 Amp hours @ 24 Volts, up to about 2.6 kW hours of stored power.

Compare that to your original proposal of eight 100 Amp hour 12 Volt batteries in parallel which would yield about 4.8 kW hours of stored power.

I don't know if I've pointed you to this thread before, but it's a good explanation of the whys of choosing system Voltage: http://forum.solar-electric.com/showthread.php?15989-Battery-System-Voltages-and-equivalent-power

Gene.243

Re: Newbie question

So if I go with a traction battery I can routinely discharge further so a smaller amp hour battery will work? That was a good read last night, thanks.

Gene.243

Re: Newbie question

So to review my system we have the given 1kw PV array that collects 4 kwh per day. We store that energy in four 220 ah golf cart or traction batteries wired in series that has a nominal rating of 220ah*24v=5280 wh. We can use one half of that so I can use 2640 ah per day. Can I assume that I can draw 110 watts continuously?

BB.

Re: Newbie question

First, regarding the traction battery--They are designed to discharge to 50% or deeper (if you routinely only discharge by 25% to 50%, and avoid going deeper than 50% very often, and never below ~20%, your batteries will last longer). Automotive/truck batteries are designed to discharge by only 15% or so. Marine are supposed to be somewhere between starting and deep cycle--But they are not recommended for deep cycling.

There are lots of ways of looking at power from the solar array and how it gets to your loads... But from a 50,000 foot point of view, the equation would look something like this for a flooded cell, AC Inverter system (conservative power estimate):

1,000 watt panels * 0.52 overall system efficiency * 4 hours of sun per day = 2,080 Watts minimum of "useful" AC power per day (typically ~9 months of the year)

So, if you have 5,280 WH of storage and discharge a maximum of 50%, the amount of "useful" AC power would be:

5,280 watts * 0.50 discharge * 0.85 inverter efficiency = 2,244 WH per day

Of course, if you do not have enough panels/sun to fully recharge the bank for the next day's use--Then you will have less energy for the next day's use (unless you take the battery further into discharge--"deficit charging", or have a backup generator to assist).

Note that, typically, we use the C/20 (20 hour discharge rate) for most of our work here... However, if you have heavy loads (well pump, etc.), the battery when discharged at C/10, C/8, C/5 discharge rates will appear to have less AH capacity. You can read about Peukert below:

Peukert's Equation. A quick explanation of the equation.
Peukert's Equation (Peukert's Law) properly explained (and correctly).
A Peukert Calculator spreadhseet (extremely useful program).
The 50% rule for deep cycle batteries.
More on the effects of Peukert.

The general rules of thumbs work pretty well because they give conservative results, and most people/homes, on average, are a mix of loads... Not a set of loads that run 100 watts * 24 hours per day, or 4,800 watts for 30 minutes a day.

For folks with "unusual loads and profiles", then we do need to look closer to see that the system will meet expectations.

So, yes, your 110 Watt load * 24 hours per day = 2,640 Watt*Hours per day. But when you start looking at the details (inverter losses, and overall losses from panel ratings of 81%, controller at 95%, flooded cell battery at 80%, and inverter at 85% efficiency--You multiply all this together and get 0.52 or 52% efficiency--It turns out that you don't get exactly that amount. Also, for most people, they will see a mix of ~3-5+ hours of sun per day (seasonal). And there will be days/weeks of bad weather where you may see less than a 1/2 hour equivalent sun per day. So--Either you need to be flexible (i.e., postpone washing) or have to have a back up genset to cover your loads on days with poor power production.

-Bill

Cariboocoot

Re: Newbie question

Gene.243 wrote: »

So to review my system we have the given 1kw PV array that collects 4 kwh per day. We store that energy in four 220 ah golf cart or traction batteries wired in series that has a nominal rating of 220ah*24v=5280 wh. We can use one half of that so I can use 2640 ah per day. Can I assume that I can draw 110 watts continuously?

110 Watts on 24 Volts is only 4.6 Amps. That's 2% of the battery capacity and a continuous draw of that amount won't adversely affect the batteries. However, as Bill pointed out that is based on the nominal Voltage. In reality the Voltage will shift from a high of around 28 Volts where the current will be 3.9 Amps to a low that may be less than 24 Volts depending on the system settings for low Voltage. In addition the Peukart effect Bill link to says the power curve of the battery is not linear along that Voltage. Nor does battery capacity remain constant throughout its life or under all temperature and load conditions.

My preferred short cut here is to use the nominal system Voltage as the low point, as it pretty much ensures the battery will not be discharged below 50% under load.

People who try to design systems with minimal tolerances end up disappointed.

NorthGuy

Re: Newbie question

Gene.243 wrote: »

So to review my system we have the given 1kw PV array that collects 4 kwh per day. We store that energy in four 220 ah golf cart or traction batteries wired in series that has a nominal rating of 220ah*24v=5280 wh. We can use one half of that so I can use 2640 ah per day. Can I assume that I can draw 110 watts continuously?

On avarage.

There will be times when every day is sunny and you could've drawn much more.

And, there will be times when you don't see sun for a long time and you won't be able to get your 110 W.

Gene.243

Re: Newbie question

My system is coming along slowly. I found a battery supplier not too far out of town. He has golf cart batteries at 205 or 250 ah. My guess is that I should go with the smaller ones to avoid deficit charging.
TIA again
Gene

waynefromnscanada

Re: Newbie question

Gene.243 wrote: »

My system is coming along slowly. I found a battery supplier not too far out of town. He has golf cart batteries at 205 or 250 ah. My guess is that I should go with the smaller ones to avoid deficit charging.
TIA again
Gene

Easiest way to think of deficit charging:
Deficit charging happens when over an extended time, you continue to take more OUT of the batteries than you put back IN, regardless of the size of the batteries.