Newbie Designing a System

BB. · October 2013

Re: Newbie Designing a System

RobW wrote: »

After following the discussion here:
http://forum.solar-electric.com/showthread.php?21269-24v-battery-charging-with-12v-system

And revising the load I'm wanting to supply, I have this: a daily load of 870Wh. Using a K-A-W, I've got 174W * 5 hrs. Those 5 hours will be primarily at night. Now, I want to assume I go with a 24v battery bank with a couple of the Wal Mart 27DC deep cycle batteries, rated at 109Ah. Please check my figures.

In the discussion linked to above, BB shared this formula:
Battery Ah * charging voltage * 1/0.77 * 1/0.05 = minimum watt array

BB gave 14.5 for charging voltage of 12v system. I assume (?) I can double that for a 24v system. YES
109 * 29 * 1/0.77 * 1/0.05 = 205 min watt array
109 AH * 29 volts charging * 1/0.77 panel+controller derating * 0.05 rate of charge = 205 Watt array minimum
So that's the minimum watts required to charge the 24v battery bank at a standard 5% rate.

I may be confusing you as I am not using Parens--But just reading from right to left for the math--that *1/0.05 should be * 0.05 -- But you got the right answer.

Another of BB's formulas:
Minimum array watts * 0.52 efficiency * min. hours of sun = min. watts of 120 VAC per day

Plugging my daily load of 870W into this and doing some basic algebra:
Minimum array watts = 870 watts / 0.52 * 4 (min. hours of sun)
870 Watt*Hour daily load * 1/0.52 system eff * 1/4 hours of sun per day = 418 Watt array minimum
Minimum array watts = 418 watts

So my need is roughly twice the minimum required for my batteries.

Using another formula BB used in the linked-to discussion:
Battery Ah * battery voltage * 0.85 typical inverter eff. x 1/2 days storage x 0.50 max discharge = Wh of 120 VAC battery power per day

109Ah * 24v * 0.85 * 1/2 * 0.50 = 556 Wh of 120 VAC battery power per day

That is correct. That would be the recommended or optimum daily load for two 109 AH @ 12 volt batteries connected in series.

Now, if these numbers are accurate, with a 400w array feeding a 24v battery bank consisting of 2 109Ah 12v batteries, I can power my load of 174w for 5 hours a day. In amp hours that would be:
174w/120 VAC * 5 = 7.25Ah AC * 5 = 36.25Ah DC @ 24v

How close am I with these numbers?

Yes... But that is not really an accepted way of writing the math equations: 2*3=6+3=9 [2*3<>9] -- Should be (2*3)+3=9

Or, as I would write it:

174 Watts * 5 hours per day * 1/0.85 inverter eff * 1/24 volt battery bus = 42.6 Amp*Hours @ 24 VDC

I added a 1/0.85 for typical AC inverter losses.... And you sort of did a (1/120VAC)*5 conversion factor =1/24 volts

Just do the Watts/Volts=Amps... That *5 conversion from 120 VAC to 24 VDC is correct--But kind of gets buried in the equations and is easy to miss if you want to try a 12 volt or 48 volt battery bank.

And--I have a new appreciation for how hard it is to read somebody else's equations... For the most part, you just copied my equations in a slightly different format--I had to really look at them to make sure I understood them...

I have been avoiding using [()] everywhere because (I think) it reads more like an English sentence--Perhaps I am confusing folks more than helping?

-Bill "math is hard" B. :roll:

RobW · October 2013

Re: Newbie Designing a System

Thanks for checking, Bill. No, your not confusing. What helped me finally get it was your explanations of each parameter within the equations.

Now I feel like I've got a real system design and can start to bring components together. Before I started this thread I had already ordered a few items. I'll go over those later and see how they fit into this design.

Thanks again.

Cariboocoot · October 2013

Re: Newbie Designing a System

Avoid those Wal*Mart 27DC batteries. They are Marine/RV batteries and not really suitable for RE. Get four GC2's instead.

BB. · October 2013

Re: Newbie Designing a System

Thank you Rob... I really did try and figure out how to make the equations "readable"... To compare other ways of writing equations:

174 Watts * 5 hours per day * 1/0.85 inverter eff * 1/24 volt battery bus = 42.6 Amp*Hours @ 24 VDC

Why do I use 1/0.85 and 1/24? Because we recognize the numbers... Plugging in straight numbers would look like:

174 Watts * 5 hours per day * 1.17647 inverter fudge factor * 0.04167 volt^-1 battery bus = 42.6 Amp*Hours @ 24 VDC

That is a silly representation. Another would be:

(174 Watts * 5 hours per day) / (0.85 inverter eff * 24 volt battery bus) = 42.6 Amp*Hours @ 24 VDC

Which is the correct way to write the equation--With Parens to ensure the correct order of operations. But I think it sort of "groups" variables together in non-useful relationships (i.e., is Inverter Eff and Battery Bus Voltage somehow related--Not really).

And if I took out the "words" and did it the way my Math and Science Teachers taught:

(174W * 5H) / (0.85 * 24V) = 42.6 AH

It gets the units right (always useful cross check for engineering/sciences)--But sort of difficult to really get a "feel" of what is happening in "real life" system.

Then there is the "algebra" version:

P = Average power in Watts
T = Hours per day of average power
E = Typical inverter efficiency (in percent)
B = Battery Bank Nominal Voltage
C = Bank capacity in AH @ battery voltage

C=(P*T)/[(E/100)*B]

Where:

P=174 Watts
T=5 Hours
E=85%
B=24 volts
C=Results in AH

C=(174W*5h)/[(85/100)*24V]=42.6AH

Liking my "English Math" sentences any better yet?

Equations written as above are much easier to put in a spread sheet--But I think do a poor job of conveying the ideas behind why the equation does what it does.

And I now see way I was a "C" student in middle school math.

:roll: English too... Another story (or not).

-Bill

Newbie Designing a System

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