Need guidance

aaronbtxnc · August 2012

Hi, total newbie here. I am looking for an emergency back-up power source that is quiet and doesn't rely on purchased fuel, so no generators. I live in Western NC and am seriously considering purchasing a SolarPod Standalone, which consists of 4-230W panels, an off-grid 1800W modified sine wave inverter with mounted charge controllers, racking w tilt adjust, cable and 230V/20A male rec w weather proof boot (connected to panel string), female receptacle w weather proof boot, and up to 100 ft cable. Kit originally comes w 2 batteries (expandable to 4) but opted to source locally to save on shipping. I want to be able to power at least my fridge and deep freeze enough to keep them cold enough, as well as a couple of lights and fans. We can heat water and cook either on wood stove in winter or outdoors in summer so no need for water heater or oven or stove, we will rely on wood stove for heat in winter and fans and windows for cooling in summer, we are purchasing a hand pump for our well so don't need to power well pump, only other possibilities I can think of is recharging small batteries for use in flashlights, radios, etc. Assuming that we would unplug deep freeze and fridge most of the time, plugging in about every 4 hours to keep cool enough to keep food. I am trying to research other kits or something I could put together myself that would either be equivalent to this kit or better suited for what we need, but would need a very detailed list of exactly everything i would need and very detailed instructions on how to hook it all up. Budget not to exceed 7K. Not at all interested in grid tied system that won't work if the grid goes down.
Any and all help and advice is greatly appreciated.
Aaron

Cariboocoot · August 2012

Re: Need guidance

Welcome to the forum, Aaron.

Okay before you spend any money figure out how much power you need to supply.
The specs you give on that kit don't mean much, because any battery-based system is, well, battery-based. "Two" batteries or "four" batteries does not tell you how much power you will have available. Amp hours @ Voltage does. The panels are there to recharge the batteries, and it's easy to have the wrong amount. The inverter's rating just means its maximum output any any given time.

You at least picked the right loads to keep running in an emergency: the refrigeration! No sense letting the food spoil, eh? But unfortunately these are bound to be massive loads (trust me; I've tested many). So the first thing you buy is a Kill-A-Watt meter (about $30). Plug in the 'frige and see what it uses over a week. Do the same for the freezer and any other loads you're likely to need. Don't scrimp: plan for keeping them running all the time. The duty cycle on refrigeration equipment is usually less than 40%.

Another problem: AC induction motors are not fond of MSW power. They run hotter because they work harder. As far as they are concerned, MSW is "low Voltage". You would be better off with a sine wave inverter.

Fortunately you have a good budget. It should be possible to put something together that will fulfill the requirements for a lot less than $7,000. My cabin system cost $8,000 CDN back in 2008 and it runs everything all the time. Panel prices have come down a lot since then. Frankly a couple of golf cart batteries, a sine wave inverter around 2kW, about 350 Watts of panel, and a charge controller capable of 30 Amps is what you're looking at (from memory; I've done many "keep the 'frige running" systems).

As a rule, "solar kits" tend to be a bad value.

But get those power numbers as they are vital for a good system design.

SolarVader · August 2012

Re: Need guidance

Hello I’m new here but let me give this a try.

The most important thing is to know what are your loads and then work from there. You mentioned a refrigerator, fan and lights. How big if the refrigerator and what will be its “duty time” (on average it’s 1/3 of the time). Also is it VDC or VAC?

Example an AC 18cf refrigerator is on averge 240W and runs 1/3 of the time. Since you need an inverter then you’ll have to factor in the DC-AC conversion. See bellow (assuming a 24VDC battery bank)

Fridge Current = 240W / 20 = 12A
Run Time = 8hours
Fridge AH needs = 12A x 8h = 96AH
Now batteries with 105AH / 20H rate the DOD is about 85% for drawing 12A
96A/.85% = 113AH , add a safety factor for kicks, say 10%, and you have 124AH

So your refrigerator alone would consume 124AH from you batteries.

Now say you don’t want to discharge your batteries more than 50% and you want 2 days of autonomy (2 days of no battery charge due to poor weather), also factor in the efficiency of your Inverter (let’s use 88% efficient) then your battery bank size should be:

124AH x 2days x 1.12 / 0.5% = 555AH. This is equivalent to twelve 12V 105AH batteries (for a 24V battery bank)

These numbers can vary depending on the size of the fridge and the year it was made. For example a 12cf refrigerator would need a 300AH battery bank or six 12V 105AH batteries. Also newer fridges are more efficient.

Cheers !

BB. · August 2012

Re: Need guidance

Generally, I highly recommend conservation and measuring your loads first. For AC loads (smaller loads that are 120 VAC 15 amp, plug in type), a Kill-a-Watt type meter is the best way to start. Don't spend one $$$ on any solar/off grid power until you have sized your loads.

Refrigerators and Freezers are among the most power hungry appliances in many homes (excluding AC, electric heat/hot water/cooking). Very good full size energy star refrigerators are around 400-500 kWH per year by themselves (500kWH per year/365 days = 1.4 kWH per day).

I use ~100kWH per month (3.3 kWH per day) as a starting point for a "cost effective" off grid system that will give a family a fairly "normal" electrical life living off grid... Picking appliances that don't use much power (energy star appliances, laptop vs desktop computer, lots of insulation in walls, ceilings, double pane windows, picking entertainment systems based on power consumption, etc., well pump, clothes washing machine). However--this really depends on the homeowners watching their loads and a commitment to conservation to keep power use "low".

Using PV Watts for Asheville NC, fixed array, tilted from horizontal to latitude, 1,000 Watt (1 kW) solar array, 0.52 end to end off grid system derating:

"Station Identification"
"City:","Asheville"
"State:","North_Carolina"
"Lat (deg N):",  35.43
"Long (deg W):",  82.53
"Elev (m): ",  661
"PV System Specifications"
"DC Rating:","   1.0 kW"
"DC to AC Derate Factor:"," 0.520"
"AC Rating:","   0.5 kW"
"Array Type: Fixed Tilt"
"Array Tilt:"," 35.4"
"Array Azimuth:","180.0"

"Energy Specifications"
"Cost of Electricity:"," 8.5 cents/kWh"

"Results"
"Month", "Solar Radiation (kWh/m^2/day)", "AC Energy (kWh)", "Energy Value ($)"
   1,  3.89,       62,       5.27
   2,  4.40,       62,       5.27
   3,  5.25,       81,       6.88
   4,  5.50,       80,       6.80
   5,  5.43,       79,       6.71
   6,  5.64,       76,       6.46
   7,  5.36,       74,       6.29
   8,  5.39,       76,       6.46
   9,  4.90,       68,       5.78
  10,  5.14,       76,       6.46
  11,  4.21,       62,       5.27
  12,  3.58,       57,       4.84
"Year",  4.89,      852,      72.42

So, you can see a 1kW array will give you around 57 to 81 kWH per month (on average). So, if you are looking for a 100+ kWH per month. Pick February as your "break even month" (use generator for darker weeks in winter):

62 kWH per Feb / 28 days per month = 2.21 kWH per day = 2,210 WH per day (Feb) per 1kW array
1,000 watt array * 3.3kWH per day desired * 1/2.21 kWH per day actual = 1,493 Watt array minimum

Battery size. Recommend 1-3 days of "no sun" storage" and 50% maximum discharge (for longer battery life). Pick 2 days and 50% maximum discharge for a "balanced" system:

3,300 Watt*hours per day * 1/0.85 inverter eff * 2 days no sun * 1/0.50 max discharge * 1/48 volt battery bank = 324 Amp*Hour at 48 volts

To double check the minimum charge rate of 5% to 13% rate of charge for a "happy" battery bank:

324 AH * 59 volts battery charging * 1/0.77 panel+charger derating * 0.05 rate of charge = 1,241 watt array minimum
324 AH * 59 volts battery charging * 1/0.77 panel+charger derating * 0.10 rate of charge = 2,483 watt array nominal
324 AH * 59 volts battery charging * 1/0.77 panel+charger derating * 0.13 rate of charge = 3,227 watt array "cost effective maximum"

So, for a 3.3 kWH per day system for western NC, I would suggest around 1,493 to 3,227 watt array (note I am carrying out numbers to 3-4 places so you can check my math--Anything within 10% is about the "same" in solar).

Anyway--You can see how loads drive your system design. Do that side of the problem first (with conservation) first, then look at the hardware to meet those needs.

Questions, corrections?

-Bill

aaronbtxnc · August 2012

Re: Need guidance

Wow, fast responses, thanks to all!

Cariboocoot wrote: »

"Two" batteries or "four" batteries does not tell you how much power you will have available. Amp hours @ Voltage does.

Sorry about that, they called for 12V 110Ah batteries. Since W=V*A, That'd give us 1320 watt hours per battery, but from all I've read we'd only want to use 660 watt hours per battery per day, assuming system fully recharged every day. Am I right on my calculations and assumptions there? Assuming 2 batteries we'd have 1320 watt hours per day?

How do i figure out if it will recharge in a day here or not? Wait, let me try on my own, 920W in panels, I'll assume maybe 6 hours per day, so 920*6=5520 watt hours recharge capacity, assuming panels are not 100% efficient, drop on lines, what, 80% actual usable power for recharging, so 4416 watt hours per day available to recharge batteries? Assuming this and calc above are correct it should recharge itself just fine in one day.

Are my calculations anywhere near right? I'm good at math and science, just don't know the formulas to use to figure all this out.

In reading other posts in reply,
"Fridge Current = 240W / 20 = 12A
Run Time = 8hours
Fridge AH needs = 12A x 8h = 96AH
Now batteries with 105AH / 20H rate the DOD is about 85% for drawing 12A
96A/.85% = 113AH , add a safety factor for kicks, say 10%, and you have 124AH"

Where did the 20 come from in the first calculation? If W=V*A, then A=W/V, what is 20 volts? If I'm using 12V batteries, then that'd be 240W/12V=20A right, runs 8 hrs, 20A*8h=160Ah, what's the 20H for, and what is DOD, ignoring that and assuming the 85% is a general rule of thumb, 160Ah/85%=188, 188*1.1(10% safety factor)=207Ah. So does that mean that 2 12V 110 Ah batteries would be sufficient for my fridge (2*110Ah=220Ah) ? I realize that's theoretical 240 W draw and theoretical 8 hour run time and doesn't include freezer, lights or fans. On the subject of fans, I'm assuming I could rig up my ceiling fans with plugs after disconnecting them from the existing home's wiring so no power goes back into those lines. How would one measure the draw on a ceiling fan with a kill a watt thing, impossible I'd think unless I went ahead and wired one with a plug and ran it that way. Is a ceiling fan or table top fan more efficient based on air movement?

Freezer and fridge are both at most 6 years old.

Thanks again for putting up with my ignorance,
Aaron

SolarVader · August 2012

Re: Need guidance

aaronbtxnc wrote: »

Wow, fast responses, thanks to all!

Where did the 20 come from in the first calculation? If W=V*A, then A=W/V, what is 20 volts? If I'm using 12V batteries, then that'd be 240W/12V=20A right, runs 8 hrs, 20A*8h=160Ah, what's the 20H for, and what is DOD,
Aaron

If your appliance runs in VAC and your battery bank is VDC then you need to use an inverter. If your appliance is 100W and the voltage is 120VAC then the current it 0.83A, but since you are going from DC to AC trough the inverter the current draw on the batteries will not by 0.83A but higher. That 20 is used (as a rule of thumb) to determine DC amps needed to run the AC appliance for a 24VDC system and 10 for a 12VDC system.

A battery rated at 20H and 105AH means that in 20H the battery will be 100% used up (in 12V batteries the voltage will go to like 10.8V or 10.5V) at a rate of 5.25A per hour (105/20). But the higher the current draw on a battery the less capacity you will have (known as Peukert's Effect). So if you look at your battery specs, if you draw around 12A your battery capacity is about 85% this is what I mean by DOD (BTW I made a mistake I meant Rate of Discharge instead of DOD, sorry about that)

DOD is Depth of Discharge, when people say a DOD of 50% means that you don't want to discharge your batteries more than 50%. So if you have a glass of water don't drink more that half the glass.

Also take into consideration that what I wrote was for a 24VDC battery bank.

Cheers !

inetdog · August 2012

Re: Need guidance

SolarVader wrote: »

but since you are going from DC to AC trough the inverter the current draw on the batteries will not by 0.83A but higher. That 20 is used (as a rule of thumb) to determine DC amps needed to run the AC appliance for a 24VDC system and 10 for a 12VDC system.

Not even close....

If the load is 120 volts AC and the inverter is working from 12 volts DC, then the current will be ten times as great with a perfect inverter, perhaps 11 times as great with are real-life inverter which is operating near its maximum load.
For a 24 volt system, the (ideal) current multiplier will be 120/24 = 5, not 20.

SolarVader · August 2012

Re: Need guidance

Actually my statement and yours are very close. That 20 I divide in the Power, I don't multiply the current times 20 (following my calculation in the original post)

What I did: 240W / 20 = 12A which is close to the 5 x current (240W x 5 / 120VAC = 10A )

This is the procedure that some inverter manuals use to determine battery bank size.

Cariboocoot · August 2012

Re: Need guidance

Maybe we could retitle the thread "How to get confused with numbers".

It's easy! It's fun! Try it at home!

Or, look at it like this:

A refrigerator that uses 1200 Watt hours per day AC.
The inverter is 90% efficient, so that becomes (1200 / 0.9) 1334 Watt hours DC.
The inverter consumes 20 Watts per hour (for example) on its own: 480 Watt hours per day.
Total DC Watt hours per day: 1814
On 12 Volts: 1814 / 12 = 151 Amp hours.
50% DOD = (151 * 2) 302 Amp hour 12 Volt battery bank minimum.

If you do the math in order you get better results. If you try to shortcut from 'A' to 'B' and twist up Amps, Watts, hours, Volts ... you get a confused mess.

aaronbtxnc · August 2012

Re: Need guidance

I like that a lot better, simple calculations. I have ordered a Kill-a-watt and will measure fridge for a week, then freezer for a week. We could probably get by with just the fridge in an emergency as it's got a freezer too, it's just not very big (the freezer part, the fridge is pretty big). This gives me a calculation for how much battery bank i need. Thank you for the clarification!

What about how big of an inverter?

What about how much in panels?

And then I have to source it all and hook it up, you see why I'm still considering the kit...

Cariboocoot · August 2012

Re: Need guidance

aaronbtxnc wrote: »

I like that a lot better, simple calculations. I have ordered a Kill-a-watt and will measure fridge for a week, then freezer for a week. We could probably get by with just the fridge in an emergency as it's got a freezer too, it's just not very big (the freezer part, the fridge is pretty big). This gives me a calculation for how much battery bank i need. Thank you for the clarification!

What about how big of an inverter?

What about how much in panels?

And then I have to source it all and hook it up, you see why I'm still considering the kit...

Inverter should be sized to meet the maximum demand at any one time. You've got another issue there with refrigeration: the immeasurable (almost) start-up surge. K-A-W will not catch it, but it's there; a momentary jump in current to get the compressor started. The inverter has to be able to handle this. It can be 5 times the running current (but not always so don't take that as an absolute). A good 2kW sine wave inverter will have no trouble. Some less expensive inverters, however, have a very short duration surge rating; being able to supply 2X the rated power for 1 second is not of any use. But think of it like this: a typical household outlet is 15 Amps and can handle a momentary surge well above that. I've not seen a standard 'frige yet that will exceed 3kW on start. Thus my statement that a good 2kW unit will handle it. By the same token there are some smaller units that may also work (Exeltech 1100 Watt for example - have had good results with those). You'll know more once you have the real world numbers from your equipment.

On that same line, for 12 Volt inverters trying to handle that load it is absolutely vital to have large, quality wiring on the DC side. What seems like a small amount of extra resistance in the wiring can be the difference between "works/doesn't work" here.

The panels should be sized to recharge the batteries, so first you have to work out how much battery is needed.

The day our host NAWS offers such a kit is the day I'll start believing it will work.

BB. · August 2012

Re: Need guidance

If if you want to go with a kit/pre-configured system--You still need to do the measurements and the math.

Once you did that, then you can start looking at kits/parts and figure out what you will need. And you will quickly see that most "kits" have a big MSW inverter, small battery bank, and about 1/2 or 1/3rd the solar panels you really need. Yes--it does not cost much, but there is a reason behind that.

Some companies (like Midnite, Xantrex, and Outback) do sell pre-wired panels. And some retails (like our host NAWS) will configure, wire, and test a system built to your needs (usually, you just add batteries and solar panels).

If this is the "first time" for a large electrical project--I would highly suggest the pre-wired e-panels at the very least. It will save you weeks to get the misc. parts and figuring out exactly what part(s) will fit your needs.

-Bill

Photowhit · August 2012

Re: Need guidance

You might look at all your loads if you want more than the fridge and freezer...

Also look at if you want to get the minimum that will do the job, or the most under your budget...

You also might want to consider how mainanance free you need this to be, flooded lead acid batteries are the best bang for the buck, but require some mainanance.

You might find that an Exeltech XP1100 and a generic E-Panel, a Midnite Classic and a couple strings of 'best deal per watt today' panels (1000 - 1500watts) and 4 golf cart or 2 L-16's purchased locally would give plenty of capacity and minimal wiring @$4K

ggunn · August 2012

Re: Need guidance

Just a general comment...

When you are deciding ball park matters of feasibility and such, rules of thumb are fine, but when it comes to plunking down your money and building the system, do the real math. It's the Goldilocks Principle; you don't want your system or components thereof to be too big or too small, but just right.

Need guidance

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