At the risk of being Ridiculed..

gafarmboy · May 2011

At the risk of being Ridiculed..I will ask the following questions..
I have spent a inordinate amount of time reading the archives here and elsewhere trying to sort out this question. That being said, I still have no clue HOW I should proceed on my solar project which is, at this time, to provide independent 12 volt power for lights, ceiling fans and basic functions such a recharging AA,AAA, D and C batteries.

1) my journey to energy independence will begin by purchasing 2 mono-crystalline solar cells, each in the 200+ range. Have looked at Canadian Solar and REC.
2) Have located on-line sites to purchase fans, lights and other need accessories.
3) Local Golf Cart shop will supply batteries at a fair price.

NOW the problem. I do not want to use an inverter. I want to skip that part of the equation and go with straight 12 Volt lights and fans.

I understand the need for thicker multi-strand wire for all runs, but I am not sure of what gauge to use. I understand that the Batteries must be placed as close as possible to the fuse panel (which again will not be attached in any way to the existing grid.)

So the gist of my question is "How do you figure out the gauge of wire to use for lights or fans or what-ever-I-want-to-use? And please understand that I am an electrical rock..Just don't get it..Have tried for years..but the space between the ears must be full:p

My lack of knowledge, which is great compared to most on this board, does not let me envision the HOW to wire things up. I can wire basic AC and have done so many times. Same goes with a car. But this solar DC stuff is confusing me because I cannot get my head around it.

Thanks in Advance for all help
Gafarmboy

icarus · May 2011

Re: At the risk of being Ridiculed..

No risk of ridicule here!

Wire size is a matter of current load and distance ( and how much line loss you are willing to live with). There is a wires size calculator somewhere in this forum, but I can't find it right now. I'm sure someone will chime in.

For example, #14 wire will carry ~ 15 Amos safely, regardles of the system voltage. The reality however is that it will loose much more powe at 12 volts than 120 volts for any given length or wire.

Hope this helps, and welcom,

Tony

Cariboocoot · May 2011

Re: At the risk of being Ridiculed..

You may find this Voltage Drop Calculator helpful:
www.solar-guppy.com/forum/download/voltage_drop_calculator.zip

It's important on a 12 Volt system to be able to handle the Amps and keep the Volts from falling too low. You may even find it feasible to run a 'loop' circuit so that there's fairly even distribution around the cabin (one fuse only, please, feeding both ends of the loop).

In terms of wiring, AC and DC are largely the same except for the direction of current flow and the need to use DC rated devices (like switches) appropriate to the Voltage and current.

icarus · May 2011

Re: At the risk of being Ridiculed..

Just a comment on DC rated switches. For small current loads like small bulbs, fans etc, AC switches work fine.

T.

gafarmboy · May 2011

Re: At the risk of being Ridiculed..

Thanks for the warm welcome and the information. I will use it to formulate my next inane question.:D

Does anyone have an opinion on which is better in a panel box. The Square D- Q series is suppose to be rated for DC usage and that is what I can purchase local. Or would you go with something like or similar to the Midnight Solar combiner boxes? Inquiring Minds want to know.

Thanks Again for the Welcome
gafarmboy

Cariboocoot · May 2011

Re: At the risk of being Ridiculed..

Before you start selecting equipment, get your system designed. So far you've only picked the nominal Voltage. Next you will need to know the maximum current you expect to draw and the average current over time (can also be expressed in Watts and Watt hours). This will give you the data you need for selecting a battery bank size and the necessary array/charge controller configuration to replenish it. Expensive combiner boxes and circuit breakers may not be necessary at all.

icarus · May 2011

Re: At the risk of being Ridiculed..

I have a hybrid 12 vdc/120vac system in my house. I feed all the 12 vdc circuits through QO breakers, and the 120 vac through another QO panel.

Like Marc says, don't buy anything before you have designed your system soup to nuts.

All design considerations begin with the loading. The three simple rules of off grid solar are these: People over estimate the amount of solar harvest they can actually get, while at the same time they under estimate their loads. The third rule, is that loads ALWAYS grow with time. Good design factors these rules into the calculations.

Once you get a handle on your daily/weekly loads, then you can consider how to supply them. A quick rule of thumb for off grid, is the name plate/2*4 rule. That is, take the name plate rating of the PV array, divide that number by two to account for all cumulative system loses, then multiply that number by 4 to represent the average number of hours of GOOD sun one can logically expect on a daily basis, over the course of the year. A PVwatts calc.: http://rredc.nrel.gov/solar/calculators/PVWATTS/version1/
can give you a pretty good idea of whether or not you can adjust the 4 number up or down based on your locale.

Once you have a handle on the loads, you can do your calcs to determine battery size and PV array size.

So, if you have 100 watts of PV, 100/2=50*4=200 watt hours of power per day (on average. So if your loading is ~ 200 wh, than you can see that 100 watts of PV is in the ball park.

As an FYI, we live off grid with 400 watts of PV. We routinely draw ~5-800 wh from the system per day. On an average day, we produce enough power to cover the battery draw. Our batteries ( 450 ah of T-105s) gives us 3 days of reserve before we draw the batteries down more than ~25%.

Good luck and keep in touch,

Tony

gafarmboy · May 2011

Next series of questions..

Based on the information that ya'll have given me and my own due diligence, I can expect an average of 5.25 to 5.50 of good sunlight per day. Using the posted equation ( 410/2=105*5=525), I can expect to have 525 watt hours of power per day if I round off the numbers for safeties sake. Ok, I think I got that part.

Now for the part about loads. I am working on that. Is the following the correct formula for determining the load?

(Watt Usage * Hours/Day * Days/Mo. ) / 1000 = Kilowatt Hours used that month

So if I am using 6 :MR11 10 White LED w/ G4 Base that uses 2.1 Watt it would look like this?

2.1*5/24 (0.4375) * 30/12(2.5) = [1.09375]/ 1000 = 0.0010937 kw per month.
Is this right?

As for your statements about designing the system down to the nuts and bolts. that is where I am having the problems. If it were a 120v system, no problem.

But I am learning...with your help.
Much thanks

Gafarmboy

Cariboocoot · May 2011

Re: At the risk of being Ridiculed..

Stop thinking in terms of months. Months have different numbers of days in them and aren't relevant to off grid applications. Kilowatt hours per month is something the utility uses to bill you.

For off grid you look at days, as you expect to recharge daily. So Volts * Amps * hours of use = Watt hours. It's actually fairly straight-forward on a DC only system as there are no inverter losses or power factor corrections to be concerned with (note to those who will, don't start the nit picking about PF).

Tony's rule-of-thumb of panel rating * hours of sun / 2 is for AC out from PV. On a DC only system you'll do somewhat better. There will still be a loss factor, as panels only put out an average percentage of their rating and this can vary quite a bit depending on local conditions. Heat reduces output, as do atmospheric conditions like fog or pollution.

Once you get an estimate on your total daily Watt hours you can pick some "generic" specifications for batteries. Then round up to the nearest commercial equivalent. Say it works out that you need 180 Amp hours @ 50% DOD. That means a 360 Amp hour battery bank. So you might select some L16's that are 390 Amp hours. From there you base your solar array and charge controller.

To continue the example, you're trying to supply 39 Amps of charge current @ 14.2 Volts. That's 553.8 Watts. Expect 77% efficiency and you get 719 Watt array. Which there's no such thing as. Again, you adjust to the nearest equivalent. Best to round that up too, if practical. That would be four 180 Watt panels. But you might select four 175 Watt panels if that's best on budget/availability. Then you need a charge controller, and for that much array it's worthwhile going for an MPPT type, and in that power range the Tristar 45 could be the choice.

After that you start measuring for runs and calculating Voltage drop to size wires and selecting appropriate fuses, et cetera.

BB. · May 2011

Re: Next series of questions..

I think we got lost in the math here:

gafarmboy wrote: »

I can expect an average of 5.25 to 5.50 of good sunlight per day. Using the posted equation ( 410/2=105*5=525), I can expect to have 525 watt hours of power per day if I round off the numbers for safeties sake. Ok, I think I got that part.

If you have 410 watts worth of solar panels, then, on average assuming 5.5 hours of sun per day for 120 VAC power:

410 watts * 0.5 system efficiency * 5 hours of sun = 1,025 Watt*Hours of 120 VAC power per day

Also, you need to remember that an AC inverter running with zero AC load will still consume around 5-30 to even 60 watts for larger inverters...

So picking the correct inverter (some more expensive ones have "sleep mode" which allows them to reduce standby power draw to even less than 1 watt) and/or adding a power switch to turn of the DC input power when the inverter is not needed is very important for small solar RE systems. For example a 6 watt standby inverter (actually a small/efficient inverter):

6 watts * 24 hours per day = 144 WH per day for standby losses

Leaving a small inverter running 24x7 will use almost 10% of your solar power--Not good in a small system.

Now for the part about loads. I am working on that. Is the following the correct formula for determining the load?

(Watt Usage * Hours/Day * Days/Mo. ) / 1000 = Kilowatt Hours used that month

So if I am using 6 :MR11 10 White LED w/ G4 Base that uses 2.1 Watt it would look like this?

2.1*5/24 (0.4375) * 30/12(2.5) = [1.09375]/ 1000 = 0.0010937 kw per month.
Is this right?

6 bulbs * 2.1 watts * 5 hours per day = 63 Watt*Hours per day
63 WH per day * 30 days per month = 1,890 WH per month = 1.89 kWH per month

As for your statements about designing the system down to the nuts and bolts. that is where I am having the problems. If it were a 120v system, no problem.

As long as we are working with Watts--AC or DC does not matter:

Power = Watts = Volts * Amps
Work = Energy Used = Watts * Time = Watts * Hours
Amp*Hours (battery capacity) = Watt*Hours / Volts = AH

In theory, to measure Amp*Hours, just use a current meter and a stop watch:

3.4 amps * 6 hours = 20.4 Amp*Hours

But we need to know the voltage (12 volts, 24, 48 VDC or even 120 VAC)

20.4 AH * 12 volts = 244.8 Watt*Hours
20.4 AH * 48 volts = 979.2 Watt*Hours
20.4 AH * 120 volts = 2,448 Watt*Hours

For measuring AC Watt*Hours (and kWH), a Kill-a-Watt meter or equivalent it great and cheap for measuring 120 volt plug in appliances.

For smaller DC loads, you can use something like one of these to measure Amp*Hours and Watt*Hours.

There are, of course, units for measuring larger loads--but those are much more expensive and complex to use.

-Bill

gafarmboy · May 2011

Re: Next series of questions..

Let's try this again..shall we..

Volts * Amps * hours of use = Watt hours

Use 6 led bulbs that are rated at 2.1 watts each ;
Use them for a total of 5 hours per day...

12.00 * 0.18 * 5 =10.5 watt hours per unit

6 units * 10.5 watts = 63.00 watts total

Is this correct?

Is it possible to just take the Watt Hours and multiply by the number of units?
2.1 watts * 6 = 12.6 rounded up 13.00 * 5 = 65 watts..I would be higher than the actual number but it would be an error on the plus side.

Gafarmboy

BB. · May 2011

Re: Next series of questions..

gafarmboy wrote: »

Let's try this again..shall we..

Volts * Amps * hours of use = Watt hours

Use 6 led bulbs that are rated at 2.1 watts each ;
Use them for a total of 5 hours per day...

12.00 volts * 0.18 amps * 5 hours per day =10.5 watt*hours per unit per 5 hour usage day

6 units * 10.5 watts*hours = 63.00 watt*hours total per day based on 5 hours usage per day

Is this correct? always make sure you carry the units through the various equations

And, in the following, instead mixing WH and W... This is likes mixing Miles Driven vs Miles per Hours...

The confusion is that Miles per Hour is a rate, just like Watts (actually Joules per Second).

And Miles driven and Watt*Hour is an amount (like gallons in a bucket).

Watt*Hours is not a real "SI" unit... SI units are actually Watt*Seconds--but that would make your 1,000 kWH per month power bill read as 3,600,000,000 Watt*Seconds or 3,600,000 kWS per month...

So Watt*Hours just scales to a more "human" sized set of numbers.

Is it possible to just take the Watt Hours and multiply by the number of units?
2.1 watts * 6 = 12.6 rounded up 13.00 * 5 = 65 watts..I would be higher than the actual number but it would be an error on the plus side.

Note that the "power rating" is in Watts and yes, you can have 6 units x the Watt rating of each unit.

To figure out how much energy they would use after 5 hours, muliply by 5x hours:

6 units * 2.1 watts per unit = 12.6 Watts per 6 units
12.6 Watts per 6 units * 5 hours = 63 Watt*Hours per typical day of use

Common ways of saying the same thing...

Watt*Hours (the actual SI symbol is "·" or Watt·Hours -- but I don't have that on my keyboard)
WattHour
WH
Watt Hour

Have all been used and accepted as is.

However, I have a big problem with:

Watt/Hour
W/H

Has been used--But I think it is wrong... Other countries appear to use the "/" as a separator--but here in the US, we use the "/" in equations as a divide sign...

This confuses people into thinking:

W/H or Watt/Hour is equivalent to M/H or Miles per Hour (wrong)

And that causes a whole new round of confusion.

-Bill

gafarmboy · May 2011

Re: Next series of questions..

BB. wrote: »

And, in the following, instead mixing WH and W... This is likes mixing Miles Driven vs Miles per Hours...

The confusion is that Miles per Hour is a rate, just like Watts (actually Joules per Second).

And Miles driven and Watt*Hour is an amount (like gallons in a bucket).

Watt*Hours is not a real "SI" unit... SI units are actually Watt*Seconds--but that would make your 1,000 kWH per month power bill read as 3,600,000,000 Watt*Seconds or 3,600,000 kWS per month...

So Watt*Hours just scales to a more "human" sized set of numbers.

Note that the "power rating" is in Watts and yes, you can have 6 units x the Watt rating of each unit.

To figure out how much energy they would use after 5 hours, muliply by 5x hours:
6 units * 2.1 watts per unit = 12.6 Watts per 6 units

12.6 Watts per 6 units * 5 hours = 63 Watt*Hours per typical day of use

Common ways of saying the same thing... [fixed by BB]
Watt*Hours (the actual SI symbol is "·" or Watt·Hours -- but I don't have that on my keyboard)

WattHour

WH

Watt Hour

Have all been used and accepted as is.

However, I have a big problem with: [fixed by BB]
Watt/Hour

W/H

Has been used--But I think it is wrong... Other countries appear to use the "/" as a separator--but here in the US, we use the "/" in equations as a divide sign...

This confuses people into thinking:
W/H or Watt/Hour is equivalent to M/H or Miles per Hour (wrong)

And that causes a whole new round of confusion.

-Bill

Thanks for that...I am learning on the fly here, so all comments are welcomed.
gafarmboy

gafarmboy · May 2011

Re: At the risk of being Ridiculed..

I am back..

I have done all my load calculations and have come to a semi-definite number. That number is 245 watts per hour. And that is if I am using everything at one time. With that being said, and based on the knowledge that I will have 420 watts of solar available from the panels for 5.25 hours per day at my disposal what types of wire sizes will I need from;

A) Battery storage to the QO breaker box. Distance of 45 feet,zero loss

from breaker box to 3 over head lights at 2.1 watts each through a standard 120 switch, zero loss
C) from breaker box to 2 over head 12v fans at 1.2 watts each using standard 120 switch, zero loss
D) From breaker box to various battery charging stations. zero loss.

Thank again
gafarmboy

Cariboocoot · May 2011

Re: At the risk of being Ridiculed..

You're doing it again: "245 Watts per hour".
Is that 245 Watts constant (each hour) or 245 Watt hours in a given day?
1 Watt for 1 hour is 1 Watt hour. 0.5 Watts for 2 hours is 1 Watt hour. "Per" doesn't come in to it. At least we've dropped the virgule. :roll:

Wire sizing has little to do with Watt hours consumed or generated. The first consideration is how much current is expected on the circuit. The second is how long the wire run is which will affect Voltage drop. And there's no such thing as "zero loss". You can try for it, but there are always losses. Negligible loss is the best you can hope for. 3% Voltage drop is the most you'd want to tolerate anywhere in the wiring.

The first problem I see is the 45 foot run from the batteries to the distribution panel. On a 12 Volt system, that will be murder in the Voltage drop department.

Time for you to start having fun with math!

Voltage drop calculator: www.solar-guppy.com/forum/download/voltage_drop_calculator.zip

waynefromnscanada · May 2011

Re: At the risk of being Ridiculed..

"Zero loss" in the various circuits? Low loss perhaps, but not zero loss.

Peter_V · May 2011

Re: At the risk of being Ridiculed..

gafarmboy wrote: »

Thanks for the warm welcome and the information. I will use it to formulate my next inane question.:D

Does anyone have an opinion on which is better in a panel box. The Square D- Q series is suppose to be rated for DC usage and that is what I can purchase local. Or would you go with something like or similar to the Midnight Solar combiner boxes? Inquiring Minds want to know.

Thanks Again for the Welcome
gafarmboy

The Square D QO and QOB series (10Amp to 70Amp) are UL tested and approved for systems up to 48V DC, they are what I used for both the AC and DC breakers in my workshop (separate breaker boxes).

MUCH cheaper than the Midnight Solar stuff.

At the risk of being Ridiculed..

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