# what should I be measuring?

Options
Solar Expert Posts: 37 ✭✭✭
Hi,

I have a basic 12V solar set up:

120W panel
2 x 6V 225 amp hour batteries
10amp regulator
digital meter

I'm running:

laptop 7 amp max, but mostly I'm keeping it fully charged so it draws between 1.5 - 3 amps
stereo ~2 amp
cell ph charger <1 amp
water pump 5 amp max, but very low daily use.

It's winter here and I don't have enough power generation. I've been trouble shooting that, but I keep coming up against the limits of my understanding.

I feel I have a reasonable understanding of energy use audits. And I know that W/V=amps. I also understand that energy in doesn't equate to energy out in a linear way. But I'm struggling to get my head around how to keep track of what I am using relative to how much is being generated.

I have a back up generator, and am using the 12V output on that. It's rated at 8.3 amps but I am guessing I am getting less than that (don't have a working multimeter at the moment).

The digital meter that came with the panel shows readings for the panel, the battery and the draw, in volts, amps and percentage.

I have two charts (charge and discharge) that show SOC % depending on the C/XX rating, but am not sure which C/XX rating I should be using. I don't think that the % on the meter is very accurate, so am wanting to understand SOC, and how volts and amps interrelate.

Can someone please explain what the C/XX is? Or tell me which I should be using? I've read a few different explanations and I still don't get it.

thanks.

• Solar Expert Posts: 3,741 ✭✭✭✭
Options
Re: what should I be measuring?
leaf wrote: »
I have a back up generator, and am using the 12V output on that. It's rated at 8.3 amps but I am guessing I am getting less than that (don't have a working multimeter at the moment).

As you realize, you don't have enough panel to charge your battery. The DC output on your generator puts out less power than your panel. Your generator would be more effective if you got a battery charger to use with the generator. It should be at least a 20 amp (output) charger.

leaf wrote: »
I have two charts (charge and discharge) that show SOC % depending on the C/XX rating, but am not sure which C/XX rating I should be using. I don't think that the % on the meter is very accurate, so am wanting to understand SOC, and how volts and amps interrelate.

Can someone please explain what the C/XX is? Or tell me which I should be using? I've read a few different explanations and I still don't get it.

C is the capacity of your battery (225 ampHours). XX is a number that you divide into C to get the amps you are drawing from the battery. For example, if you are drawing 2.25 amps from your battery, that means you are drawing at a C/100 rate.

The best way to know what is going on with your battery is to get a hydrometer or refractometer. If you get a hydrometer, make sure it reads out in numbers (not colors, the ones with 4 little balls are not accurate enough).

You need to spend some money (more panels, bigger charge controller, etc) to take care of your batteries. Over time, the batteries will be the most expensive part of your system. The less you take care of them, the more expensive they will be.

--vtMaps
4 X 235watt Samsung, Midnite ePanel, Outback VFX3524 FM60 & mate, 4 Interstate L16, trimetric, Honda eu2000i
Options
Re: what should I be measuring?

Leaf,

leaf wrote: »
• laptop 7 amp max, but mostly I'm keeping it fully charged so it draws between 1.5 - 3 amps
• stereo ~2 amp
• cell ph charger <1 amp
• water pump 5 amp max, but very low daily use.

At what voltage? 120 VAC/12 VDC/???

It looks like you may be running it all at 12 VDC?
• 120W panel
• 2 x 6V 225 amp hour batteries
• 10amp regulator
• digital meter

Very roughly, like to see 5% to 13% rate of charge for the battery bank. For a 12 volt 225 AH battery bank those solar array sizes would be:
• 14.5 volts charging * 225 AH * 1/0.77 panel+controller losses * 0.05 rate of charge = 212 Watt minimum
• 14.5 volts charging * 225 AH * 1/0.77 panel+controller losses * 0.10 rate of charge = 424 Watt nominal
• 14.5 volts charging * 225 AH * 1/0.77 panel+controller losses * 0.13 rate of charge = 551 Watt "cost effective maximum"
It's winter here and I don't have enough power generation. I've been trouble shooting that, but I keep coming up against the limits of my understanding.

Do you have any information on the amount of sun (hours of sun per day, MJ/Day, etc.) for your place? Assuming New Zealand, it looks like you have a great amount of variability based on location (and mountain ranges/marine layer effects and such).
I feel I have a reasonable understanding of energy use audits. And I know that W/V=amps. I also understand that energy in doesn't equate to energy out in a linear way. But I'm struggling to get my head around how to keep track of what I am using relative to how much is being generated.

• 24 AH = 3 amps * 8 hours per day Laptop
• 20 AH = 2 amps * 10 hours per day Stereo
• 1 AH = 1 amp * 1 hour per day (or less) Cell Charger
• 1.2 AH = 5 amps * 1/4 hour per day Pump
==============================================
26.2 AH per day @ 12 volts

26.2 AH / 225 AH battery bank = 0.12 = 12% of battery capacity--Could be a fairly light load (~25% load per day of battery capacity is a good middle ground).
I have a back up generator, and am using the 12V output on that. It's rated at 8.3 amps but I am guessing I am getting less than that (don't have a working multimeter at the moment).
This sounds like the 8 amp output of a typical Honda euX000i family generator--And probably others.

In general, the 8 amp output from these generators needs the genset to be operating at full speed (ECO throttle off for Honda's at least).

I would suggest that you get a "true AC to DC battery charger" instead. 8amps*14.5volts=116Watts -- That is not a very heavy load for most generators and will be a waste of fuel unless you have other AC loads running at the same time (most gensets are fuel efficient from ~50% to 100% rated load, the farther down from 50%, the worse the fuel economy becomes--because most petrol generators run at ~50% fuel flow for 0-50% of rated load).

The Honda euX000i (and other brands) family of inverter/generators run a bit more fuel efficient at lower rated loads (25% or so), with ECO throttle turned on.
The digital meter that came with the panel shows readings for the panel, the battery and the draw, in volts, amps and percentage.

This does not sound like a battery monitor--But more of a volt meter calibrated to estimate battery capacity based on voltage?
I have two charts (charge and discharge) that show SOC % depending on the C/XX rating, but am not sure which C/XX rating I should be using. I don't think that the % on the meter is very accurate, so am wanting to understand SOC, and how volts and amps interrelate.

The C is "capacity". And for batteries, the "capacity" of the battery appears to vary. The heavier the load current, the less the apparent battery capacity-- (also colder batteries have lower capacity too).

SmartGauge Electronics - Peukert's Equation - what it means to you

So, the apparent battery capacity is based on how much current you use. Your 225 AH capacity is probably based on the "20 Hour" discharge rate. Or:
• 225 AH / 20 Hours = 11.25 average current for 20 hours to "flatten" your battery bank

We use the "20 Hour" rate for most of our "rules of thumb" here. And in your case, it is "close enough" for our work (your average loads appear to be less than 11.25 amps--so your battery's apparent capacity will be a bit higher than 225 AH).
Can someone please explain what the C/XX is? Or tell me which I should be using? I've read a few different explanations and I still don't get it.

I am not sure what context your C/XX question is here... We use some rules of thumb for various loading and charging based on C/XX numbers (where "C" is the battery bank 20 Hour rated capacity) and the XX is the Number of Hours of discharge to "dead".

For example C/20 would be a 20 Hour discharge rate. C/10 would be 10 hour rate, etc... And C/20 = 5% rate of discharge (or charging). C/8 = 12.5% rate etc.

For example, I would suggest C/8 (or 12.5% or ~13%) is the highest sustained rate of continuous discharge or charge rate. Above that rate, the battery bank can start to overheat. Also, when you get to C/5 rate of discharge, for example, the battery apparent capacity is much less (battery is less efficient--Although AGM's are much better than flooded cell at high discharge rates).

And I will use C/2.5 as the maximum current you should draw from a flooded cell battery bank to avoid "battery voltage collapse" from too much current draw.

Any way, a little math. Say you use 8.3 amps from your genset (have you measured the current?) to recharge 26.2 AH per day... Then, roughly would would need:
• 26.2 AH / 8.3 amps = 3.16 hours of recharging

But, another good rule of thumb would be to ~double that time, or ~6.3 hours of charging to replace the energy used (charger should taper down from 8.3 amps to 2 amps or so once the battery is fully charged or ~1% or so of rated capacity).

If you can use a true AC battery charger with ~20-30 amps or so of rated DC output, your battery bank would charge much faster.

To try and do the charging with just sunlight, lets assume you have 3 hours of "good sun" in winter:
• 26.2 AH * 14.5 volts charging * 1/0.77 panels+controller losses * 1/0.80 battery eff * 1/3 hours of sun per day = 206 Watt solar array minimum

So, with all of my guesses, my suggested solar array size would be in the 212-551 Watt Array size (over 206 Watt array should recharge your daily use based on 3 hours of noon time equivalent sun per day, and 212-515 watt array would keep the battery "happy" while being charged). A 424 Watt array would give you a 10% rate of charge and should keep the batteries "electro-chemically" happy and get them quickly recharged (plus give you more power and/or allow for poorer sun conditions).

Does this help? My numbers are pure guesswork--So you will probably have to supply the "real numbers" that apply to your needs/location.

-Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
• Solar Expert Posts: 3,741 ✭✭✭✭
Options
Re: what should I be measuring?
BB. wrote: »
For example C/20 would be a 20 Hour discharge rate. C/10 would be 10 hour rate, etc... And C/20 = 5% rate of discharge (or charging). C/8 = 12.5% rate etc.

Doesn't C/20 usually mean a 225/20 = 11.25 amps rate of discharge? --vtMaps

4 X 235watt Samsung, Midnite ePanel, Outback VFX3524 FM60 & mate, 4 Interstate L16, trimetric, Honda eu2000i
Options
Re: what should I be measuring?

Yep--Assuming "C" is the 20 Hour Rate battery capacity.

-Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
• Solar Expert Posts: 37 ✭✭✭
Options
Re: what should I be measuring?

Yes, it's a 12VDC system. It was set up in a hurry in the summer just to get me up and running. The intention has always been to add more panels (I want to run lights as well eventually, and very occasional 240V appliances). The generator is on loan (an older model Honda EM2200, and the AC output is 240V). I only have a small 2 amp charger, so am using the 12V output instead.

I don't want to spend any more money until I know what I am doing. My preference is to buy panels rather than rely on the generator. I can probably get an extra panel for less than a decent 20amp charger (a charger is on the longer term wish list).

Until I understand the technicals I don't want to buy more panels either (I'm in an RV so there are other issues there as well. Also I bought the 120W panel without really knowing what I was doing and assumed I would get 10amp output at optimal sun (120W/12V=10amp) which I know now is not how panels work). I can hire a 25amp charger occasionally to boost the batteries if they get really low.

vtMaps:
C is the capacity of your battery (225 ampHours). XX is a number that you divide into C to get the amps you are drawing from the battery. For example, if you are drawing 2.25 amps from your battery, that means you are drawing at a C/10 rate.

How does that work? 225/100 = 2.25, which suggests C100 not C10.

BB:
This does not sound like a battery monitor--But more of a volt meter calibrated to estimate battery capacity based on voltage?

I don't know the difference. It's a budget meter that came with the regulator and panel.

http://www.yoosmart.com/epip20-d-20a-solar-charge-controller-with-meter.html

Very roughly, like to see 5% to 13% rate of charge for the battery bank. For a 12 volt 225 AH battery bank those solar array sizes would be:

14.5 volts charging * 225 AH * 1/0.77 panel+controller losses * 0.05 rate of charge = 212 Watt minimum
14.5 volts charging * 225 AH * 1/0.77 panel+controller losses * 0.10 rate of charge = 424 Watt nominal
14.5 volts charging * 225 AH * 1/0.77 panel+controller losses * 0.13 rate of charge = 551 Watt "cost effective maximum"

Sorry, I don't know what all that means. Is it directly related to my questions?

24 AH = 3 amps * 8 hours per day Laptop
20 AH = 2 amps * 10 hours per day Stereo
1 AH = 1 amp * 1 hour per day (or less) Cell Charger
1.2 AH = 5 amps * 1/4 hour per day Pump
==============================================
26.2 AH per day @ 12 volts

26.2 AH / 225 AH battery bank = 0.12 = 12% of battery capacity--Could be a fairly light load (~25% load per day of battery capacity is a good middle ground).

Ok, this I understand. That's probably not a bad estimate of actual use, although on grey days I use less and/or charge the laptop and cell when I am driving.

What I don't get is how do I tell when my batteries are full, or what % they are at? This is why I've started looking at SOC and trying to understand that.

So, the apparent battery capacity is based on how much current you use. Your 225 AH capacity is probably based on the "20 Hour" discharge rate. Or:

◦ 225 AH / 20 Hours = 11.25 average current for 20 hours to "flatten" your battery bank

We use the "20 Hour" rate for most of our "rules of thumb" here. And in your case, it is "close enough" for our work (your average loads appear to be less than 11.25 amps--so your battery's apparent capacity will be a bit higher than 225 AH).

Ok, so do I ignore the calculation then? eg if I am only using say 3amps on average.

Sunshine hours here are pretty variable at the moment. Some days it's full sun all day, but the panel is nearly flat mounted so actual output from the panel is low. However, on the full sun days the batteries appear to become fully charged. Other days (full cloud or partial) I have to be more careful with usage and/or use the gennie.

Right now, it's dark. After I turned the gennie off the meter showed 12.6 volts. I had the gennie going for maybe 3 hours and have been using the laptop mostly at 2 amp draw. The batteries never got fully charged today (not enough sun and I had the gennie on for 3 hours in the middle of the day).

On the SOC chart I have the C/100 rate for 12.6V shows anywhere between 70% and 100%. At C/10 12.5V is 100%. The C/20 12/6V is 100%.

Now, 45 minutes later, using the laptop at 3amps, the voltage is 12.4 according to the meter. At C/20 that's 60%. Can I really have used up 40% in such a short time?

I'm confused, in part because the meter shows SOC 37%. The man who did the solar install is suggesting I use the SOC chart rather than the cheapo meter, but it still doesn't appear to make sense.

What I want to know at this stage is how to tell when I am overusing the batteries.

A couple of generator questions:

Can I run the AC and DC outputs at the same time to the batteries?

Is it likely that I am getting close to 8amps from the DC at the moment?

When you say I would get better efficiency from a higher load, do you mean the DC output will increase if I'm also using say a heater off the AC?
• Solar Expert Posts: 37 ✭✭✭
Options
Re: what should I be measuring?

Am trying to upload the charts I am using...

• Solar Expert Posts: 3,741 ✭✭✭✭
Options
Re: what should I be measuring?
leaf wrote: »
What I don't get is how do I tell when my batteries are full, or what % they are at? This is why I've started looking at SOC and trying to understand that.
<snip>
Now, 45 minutes later, using the laptop at 3amps, the voltage is 12.4 according to the meter. At C/20 that's 60%. Can I really have used up 40% in such a short time?

A hydrometer or refractometer is the gold standard for determining the SOC. If your batteries appear to have used 40% of their capacity with a small load, it may be because they no longer have much capacity due to sulfation. A hydrometer or refractometer will tell you what's going on. If the specific gravity doesn't rise to about 1.275 after charging it means the batteries are sulfated.

--vtMaps
4 X 235watt Samsung, Midnite ePanel, Outback VFX3524 FM60 & mate, 4 Interstate L16, trimetric, Honda eu2000i
• Banned Posts: 17,615 ✭✭✭
Options
Re: what should I be measuring?

What vtMaps said in his first post is all there is to it: you do not have enough power to recharge those batteries properly.

You want a peak charge current potential of C/10: 10% of the battery's capacity at the 20 hour rate. In this case 225 Amp hours / 10 = 22.5 Amps.
At 12 Volts, that is 270 Watts and since panels do not average 100% of their rated output you need more than that. Your array should be in the neighbourhood of 350 Watts total.

What you've got: one 120 Watt panel capable of about 6 Amps; one generator capable of about 8 Amps. Together your peak charging current would be 14 Amps on a good day. That would be a 6% charge rate; just above minimum. The good news is this means there is the possibility your batteries haven't been ruined. Yet.

Proper charging will be a problem anyway, because that deep cycle set needs to be brought to ~14.8 Volts and kept there for a couple of hours. The generator's 12 Volt output is not designed to reach that Voltage.

Never mind the usage numbers, the battery monitor, the hydrometer, et cetera. The batteries are not getting properly charged. There is no way that they can be given what there is to charge them.

You need a decent battery charger designed for deep cycle batteries like this: http://www.solar-electric.com/ioen12vo15am.html
And/or you need at least one (two would be better) more of those 120 Watt panels (or their equivalent) and a proper charge controller.

Beyond that an inexpensive hydrometer will get you in the ballpark on the specific gravity readings of the cells and thus their actual state of charge.
Voltage-wise you can really only tell charge rate with the batteries at rest; no current going in or out. Even then it can be inaccurate as abused batteries may have lost capacity even though they read proper Voltage level.

Once you have the potential to recharge properly and the batteries go through the full Bulk/Absorb/Float cycle you can learn to judge your SOC by the Voltage readings. Please read the deep cycle battery FAQ's for a better understanding of what should be happening: http://www.windsun.com/Batteries/Battery_FAQ.htm
Options
Re: what should I be measuring?

Part 1
leaf wrote: »
Yes, it's a 12VDC system. It was set up in a hurry in the summer just to get me up and running. The intention has always been to add more panels (I want to run lights as well eventually, and very occasional 240V appliances). The generator is on loan (an older model Honda EM2200, and the AC output is 240V). I only have a small 2 amp charger, so am using the 12V output instead.

I don't want to spend any more money until I know what I am doing. My preference is to buy panels rather than rely on the generator. I can probably get an extra panel for less than a decent 20amp charger (a charger is on the longer term wish list).

You really need both... A battery bank not "treated correctly" (under charged, too long between full charges, over discharging, even over charging, etc. will take years of the life of a battery--And can destroy a bank in days/weeks).
Until I understand the technicals I don't want to buy more panels either (I'm in an RV so there are other issues there as well. Also I bought the 120W panel without really knowing what I was doing and assumed I would get 10amp output at optimal sun (120W/12V=10amp) which I know now is not how panels work). I can hire a 25amp charger occasionally to boost the batteries if they get really low.

Agree 100%--We need to get the basics of your loads and needs down first--Then reconfigure/buy equipment that meets those needs.

We try to error on the "conservative" side here. So the system will perform at least as well as predicted--And allow for aging and less than perfect installation/equipment.
vtMaps:

How does that work? 225/100 = 2.25, which suggests C100 not C10.

It was late last night. You are correct.
BB:

I don't know the difference. It's a budget meter that came with the regulator and panel.

http://www.yoosmart.com/epip20-d-20a-solar-charge-controller-with-meter.html

I am not sure what the meter is... A Battery Monitor (Victron is another good brand) is, sort of, like putting an electronic fuel level gauge on your car's gas tank. It monitors the current going into/out of the battery bank by a large power resistor (current shunt) in the negative lead to the battery.

It measures the current flow and the time. For example if you draw
• 2.25 amps * 10 hours = 22.5 Amp*Hours used from battery bank
• (225 AH battery capacity - 22.5 AH) / 225 AH battery capacity = 0.90 = 90% State of Charge for battery bank

So, you have a simple digital display where you can monitor the state of charge of the battery, plus how much current you are using, etc.... Also the Battery Monitor works backwards and will count backup up as you recharge the battery bank. It is not perfect, but is usually saves somebody from having to measure battery voltage (technically you need to have the battery bank rest for 3+ hours and then measure the resting voltage to estimate the state of charge) or use a hydrometer to measure specific gravity.

In the US, a good meter costs around \$155 for a nice, lower cost, unit. May not be worth it for a small system--But can really help you understand your system/loads and make sure the battery bank is properly recharged.

Deep Cycle Battery FAQ
www.batteryfaq.org
Very roughly, like to see 5% to 13% rate of charge for the battery bank. For a 12 volt 225 AH battery bank those solar array sizes would be:

14.5 volts charging * 225 AH * 1/0.77 panel+controller losses * 0.05 rate of charge = 212 Watt minimum
14.5 volts charging * 225 AH * 1/0.77 panel+controller losses * 0.10 rate of charge = 424 Watt nominal
14.5 volts charging * 225 AH * 1/0.77 panel+controller losses * 0.13 rate of charge = 551 Watt "cost effective maximum"

Sorry, I don't know what all that means. Is it directly related to my questions?

Sorry--A battery needs a certain amount of charging current for proper recharging. Too low of current and the battery will not satisfactorly recharge (electro-chemical reactions), not charge quickly enough (batteries sulfate if left less than full charged for days/weeks/months at a time), and other losses (battery self discharge, etc.) will use a larger fraction of the charging current.

More or less:
• C/100 (1%) rate of charge == "trickle charging" (small solar panel to keep battery charged in storage)
• C/20 (5%) rate of charge == Minimum rate of charge we recommend for solar array or AC battery charger (less than this, batteries do not get fully charged, takes too long to recharge--solar panels are "dark" for ~18 hours per day--need enough charging current to "quickly" recharge battery bank over the next day or three for longer battery life)
• C/10 (10%) rate of charge == a good sized battery charger/solar array. Some battery vendors (such as Trojan) recommend a 10% rate of charge as "ideal" for their batteries.
• C/8 (~13%) rate of charge == For standard flooded cell lead acid batteries, about the maximum current for recharging a deeply cycled battery bank. Higher rates can tend to over heat the battery bank (and will be less efficient). For Solar Arrays, higher than C/8 rates of charge are sort of a "waste" of money. The battery bank will be quickly recharged by or before noon--and the rest of the day's energy will be "wasted". Plus, high charging currents can reduce battery life a bit. Note--This is assuming a standard system that recharges during the day and battery supplies loads at night--Every system is different and if you have significant loads during the day (water pumping for irrigation, running an office during they day, etc.), those loads will reduce charging current to the battery bank and could use a large bank.
• C/4 (25% rate of charge) == More or less the maximum rate of charge recommended. At these levels, you should monitor battery bank temperature and, if supported by your charge controller, use a remote battery temperature sensor. Folks will sometimes get a large AC battery charger to quickly recharge the battery bank to reduce generator run time (recharge to 80-90% capacity, then let solar array recharge the rest of the way--conserve fuel/reduce generator run time).

So, that above calculation was sizing the solar array purely on battery bank capacity and rate of charge. This is different than sizing the array based on the amount of power you use. The system needs to run both calculations to ensure both needs are met (calculation below is based on hours of sun and loads--So the range of minimum to maximum recommend solar array size--Note that solar power is a series of estimates and "actual" weather--So while I carry out 3-4 digit calculations so you can repeat my math, if your numbers/hardware is within ~10% f those numbers--that is about the same).

We like to have a "balanced system" design. Battery bank is sized to loads, chargers/arrays are sized to battery bank and loads, etc... Your present loads are probably a bit "small/light" for that size battery bank--But not bad.

The 1/0.77 is a typical derating number that accounts for solar panel losses (solar panels have "marketing" ratings--Real ratings in sunny/hot weather is less, wiring losses, dusty panels, and charge controller losses, etc.). It all works out to ~77% of solar panel name plate is a good, conservative, amount of power that most people will achieve from their panels in normal operation.
Ok, this I understand. That's probably not a bad estimate of actual use, although on grey days I use less and/or charge the laptop and cell when I am driving
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
Options
Re: what should I be measuring?

Part 2
What I don't get is how do I tell when my batteries are full, or what % they are at? This is why I've started looking at SOC and trying to understand that.

With a volt meter, if the battery bank reaches ~14.5 volts and holds that voltage for 2-4 hours or so, then the battery bank is probably "full".

If you measure resting voltage (no charging/no discharging for 3+ hours), a 25C battery bank will show a minimum of 12.7-12.8 volts. Around 12.4 volts is 75% state of charge, etc...

As a rough estimate, I would suggest you never draw your bank below ~11.5 volts (under light to moderate loads) as going too low (below ~20% state of charge), you run the risk of damaging the battery bank (a weak cell in a battery could go to zero volts and begin to reverse charge--this will kill virtually any rechargeable battery).
Ok, so do I ignore the calculation then? eg if I am only using say 3amps on average.

As a first estimate:
• 225 AH / 3 amps = 75 hours to flatten your battery bank from full charge

A second estimate:
• Estimate the rate of discharge as 75-100 hour rate, look up battery capacity for 100 Hour rate (say it is 270 AH), then "actual" discharge would be:
• 270 AH / 3 amp = 90 hours to flatten battery bank

Now we normally suggest 1-3 days and 50% maximum discharge rate. So a 2 day and 50% maximum discharge (no sun) is a 1/4 of battery capacity would be:
• 225 AH * 4 = 56.25 AH typical "recommend" nominal daily load
• 225 AH * 0.50 max discharge = 112.5 AH recommended "maximum" daily/weekly discharge for long battery life.
• 225 AH * 0.80 max discharge = 180 AH draw from battery bank maximum... More will probably ruin battery bank

Basics rules of thumb to give you a long life for your battery bank. For RV/caravan installations, some people will discharge 50% or even close to 80% because they don't have room for more batteries--And just replace them every 2 years instead of every 5-8 years or so for a larger AH battery bank. Also, many folks only use a caravan for limited times per year so batteries tend to die from age vs cycle life.

And pretty much everyone has killed their first battery banks from over discharging, deficit charging, or even over charging. Why we suggest your first bank or two be "cheaper" batteries (like golf cart batteries) until you have everything working well and know what size bank you will need.
Sunshine hours here are pretty variable at the moment. Some days it's full sun all day, but the panel is nearly flat mounted so actual output from the panel is low. However, on the full sun days the batteries appear to become fully charged. Other days (full cloud or partial) I have to be more careful with usage and/or use the gennie.

In winter, tilting the panel can add significant amount of charging current. It is a pain for RV's, but may be worth it to you. Also, "flat" mounted panels need to be cleaned (tilted panels tend to be self cleaning from rain, etc.).
Right now, it's dark. After I turned the gennie off the meter showed 12.6 volts. I had the gennie going for maybe 3 hours and have been using the laptop mostly at 2 amp draw. The batteries never got fully charged today (not enough sun and I had the gennie on for 3 hours in the middle of the day).

In your next post, that is a very nice graph for your battery bank. I may "steal" that for our forum FAQ thread.
On the SOC chart I have the C/100 rate for 12.6V shows anywhere between 70% and 100%. At C/10 12.5V is 100%. The C/20 12.6V is 100%.

I am not sure you are reading the graph correctly. Here is how I would do it.

Say you are charging at 8.3 amps with your generator:
• 8.3 amp charging / 225 AH battery capacity = 0.037 ~ 1/27 rate of charge

Using the C/40 line (really somewhere between C/20 and C/40) of charging current, 12.6 volts @ C/40 rate of charge would indicate a present battery state of charge of just over 30%--Pretty close to near maximum recommended discharge state of 20% SOC.

With either generator or solar array charging at ~C/40 rate, you would want to see ~13.2 volts for 90% minimum state of charge.

If you had a C/10 Rate of charge, the battery votlage would be ~15.2 volts (note, we usually recommend around 14.5 to 14.75 volts as the bulk to absorb transision voltage--So instead the charger would hold ~14.5 volts for 2-4 hours to full recharge the battery bank). Using this chart, we would see a C/10 rate of charge transition to Absorb voltage at ~94% state of charge (above 90% SOC several times a week is usually "good enough" for a battery bank--Charging to 100% every day is actually a bit "hard" on a battery bank and not needed).
Now, 45 minutes later, using the laptop at 3amps, the voltage is 12.4 according to the meter. At C/20 that's 60%. Can I really have used up 40% in such a short time?

Yep--that is what the chart says... But the curve is very "flat" and meter error, battery bank not at 25oC, etc. gives a fair amount of wiggle room in reading the graph. But, more than likely, you are not starting with a 100% SOC battery bank.
I'm confused, in part because the meter shows SOC 37%. The man who did the solar install is suggesting I use the SOC chart rather than the cheapo meter, but it still doesn't appear to make sense.

I am not sure how the meter works--But many attempt to use those graphs and just by measuring battery bank voltage, estimate the State of Charge. I have not used those types of meters, so I cannot comment on their accuracy.

But, you are probably not fully recharging the battery bank in the first place. Plus, as batteries operate at less than full charge (below ~75% SOC especially), batteries begin to sulfate more quickly. This is the conversion of "fluffy" lead sulfate into a hard crystal form which does not participate in the charge/discharge cycle... Locking away both lead and sulfur from the acid (reducing full charge specific gravity) and the ability of the battery to perform to expectations.
What I want to know at this stage is how to tell when I am overusing the batteries.

I believe you are undecharging the batteries. Which will, eventually, lead to early life battery failure (from sulfation).
A couple of generator questions:

Can I run the AC and DC outputs at the same time to the batteries?

It depends, my eu2000i does not say anything one way or another. Somebody else here as another eu2000i (or maybe smaller eu1000i) that says not to run both at same time (overheats alternator windings?).
Is it likely that I am getting close to 8amps from the DC at the moment?

Don't know--but probaby not. Besides recommending a battery monitor, there are DC Curent Clamp Meters. Sears (in the US) sells an inexpensive one that is "good enough" for our needs for ~\$60. I would guess in your part of the world, you will pay ~2-3x as much for a similar meter.

I highly recommend that you get a DC clamp meter--They are great for debugging any DC system by simply placing the clamp around ONE WIRE and measure the current. However--I do understand that not everyone can justify the expense (or the expense of a battery monitor).

You might be able to find a decient DC current meter you can wire into the battery bank negative lead (keep leads short and heavy copper wire). At least you can then monitor the current into/out of the battery. Check an RV/Caravan or autoparts store for a meter than can read +/- current flow.

Also, there are current shunts that can be combined with a remote reading meter face (really just a sensitive volt meter measuring voltage drop across a "power resistor").

Deltec 100 amp, 100 millivolt current shunt

AEE Digital DC Amp Meter
When you say I would get better efficiency from a higher load, do you mean the DC output will increase if I'm also using say a heater off the AC?

I think I am refering to runnin the generator. Just using some guesses (probaby close enough for our discussion)...

Lets say your generator uses ~1 liter per hour of fuel at 2,000 watts and 0.5 liter per hour at 1,000 watts... It probably uses around 0.5 liters per hour at Zero Watts output.

So, if you run the genset with a ~1,000 watt to 2,000 watt load, you are getting around 2,000 WH per liter of fuel.

If you run the generator at ~500 watts, it still draws around 0.5 liters per hour... So its fuel economy would be:
• 500 watt*hours / 0.5 liters = 1,000 WH per liter

And at 116 watts from your 8.3 amp charger:
• 116 WH / 0.5 liter per hour = 232 WH per liter of fuel

Or almost 10x the amount of fuel per kWH/WH vs running the genset at > 1,000 Watt load. Very similar to calculating vehicle MPG or kM per liter or liter per 100 kM...

If you run a 1,000 watt input AC battery charger, you will be much more efficient (there are other losses, and my model above is "simple"--but represents roughly what you are probably seeing).

If you can get a Kill-a-Watt type meter in your location--You can connect to the genset and monitor your AC power usage with the genset--when the AC battery charger starts tapering off, you will shut the genset down and let the solar array recharge the rest of the day (run genset early in the AM).

-Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset