Solar design for the serious boondocker

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solarman

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Joined
Feb 8, 2018
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Location
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Solar design and how we get to spend lot's of cash..


Ok, this is one of several posts I intend to make on aspects of solar design.
The calculations presented are typical of what we use in the professional world for
off grid systems. These differ somewhat from a typical RV as they are fixed
installations and have very well defined operating conditions.
I'll address RV differences as we proceed.

So, the first thing we have to know is the power use over a 24 hour period.
In stationary applications such as radio towers and remote monitoring stations
the daily wattage is very well defined and therefore makes it much easier to design.
In RV use, there are a considerable number of variables so more effort and time is required
to establish an accurate figure.

Daily use is measured in Watt Hours, not amps or Amp hours. we do this so everyone
is on the same page. watt hours are simply the number of watts consumed in an hour.
and daily use is simply that total over a 24 hr period.

eg: you have 10 LED lights and they consume 6 watts each then the total power is 10*6 = 60 W
if you run these for 8 hours then the consumption is 60 * 8 = 480 Whr
we have to do this for every appliance/load and it will take time..

For devices powered from an inverter there is an additional parameter, efficiency,
so we fudge the results with a figure to best approximate actual consumption.
Inverters can be around 90% the rest goes out as heat.. yeh.. sorry..!
90% can be written as 100/90 or 1.11 save this in your spreadsheet.

eg: Coffee pot with a 700 Watt rating will cause the inverter to draw 700 * 1.11 = 777 W

I suggest you setup a spreadsheet with all your appliances and loads in two sections,
the first is DC loads and the second is AC loads ( don't forget to multiply with the fudgefactor )
then total the daily subtotals and you will get a final daily Watt/hr figure.

Let's assume boondocker Bob likes his coffee too much and has a daily total
of 1700 Watt hours or 1.7 Kwh - what are his panel requirements ?

Panel size:

Panel Watts = (Daily Watt Hours x 1.5) / Sun Hours

To account for overall system losses in the wiring, charge controller and battery charge
efficiency, we multiply by a fudge factor of 1.5 ( empirical )

next we have to know how many hours of sun we might get.. this is called Insolation and tables
are available for the entire USA, ( http://solarinsolation.org/ ).
we need to lookup the sun hours for our location. For this example i'm going
to be in Denver Colorado. We also need the time of year, so if you boondock in winter, your
sun hours will be less than summer..doh!
I'm going to use February at 3.2 hours for a worst case estimate. you may choose
to use a different value instead.. In a real situation we calculate winter and
summer and use the worst one for our installation.


so Bob gets to buy ( 1700 * 1.5 ) / 3.2 = 796 watts of panels.
we now round this up to best fit an even number of panels. say 800 Watts
we need to keep the number of panels even and not use a prime number
as the total panel voltage has to fit the charge controller, so we will use four 200 Watt panels.
This assumption is based on residential panels with high VOC values.
It is preferable to use these as the cost per watt is the lowest.


Next we determine battery size and voltage:

Battery Amp Hours = [5 x Daily Watt Hours] / Battery Voltage

To keep a reasonable balance between charge controller cost and cabling, we must
have limits on voltage. so...

from 0 to 600 watts = we use 12 Volts or higher
from 601 to 2000 watts = 24 volts or higher
anything over 2000 = 48 volts

The standard practice for lead acid batteries is a minimum 5 Day reserve capacity.
This sounds like a lot but is necessary so we can keep the depth of discharge within limits to maximize
service life. You do not want to take your batteries down to more than 50 to 60 % Depth of Discharge.
Limiting daily discharge to say 20% will maximize your battery life.
reserve here is called "autonomy"

so Bob needs ( 1700 * 5 ) / 24 = 354 Ah or ( 1700 * 5 ) / 48 = 177 Ah

bending the rules for RV ! we could use 3 days

( 1700 * 3 ) / 48 = 106 Ah
( 1700 * 3 ) / 24 = 212 Ah
( 1700 * 3 ) / 12 = 425 Ah


Charge controller ( MPPT ):

Controller amps = Panel Wattage / Battery Voltage

800 / 12 = 66 Amps
800 / 24 = 33  Amps
800 / 48 = 16  Amps

Next select the best fit for battery and controller:

Ideally, the lowest amperage is preferred, controllers get very expensive as
the amperage increases.

for 12 Volts, we could use a 60 Amp controller and 425 Ah of battery
for 24 Volts, we could use a 35 Amp controller and 212 Ah of battery
for 48 Volts, we could use a 20 Amp controller and 106 Ah of battery


Battery selection:

This depends on the charge controller amps and also your preference for FLA or AGM
I'll talk about Lithium in a dedicated post.

Now, batteries have minimum and maximum charge and discharge rates that should be
observed in order to maximize service life. we denote capacity with the letter "C"

Charging for FLA is typically C/8 to C/12
Discharging is typically no more than C/4

Charging for AGM is typically C/4 to C/8
Discharging is typically no more than C/2

To observe those C constraints for FLA:

for 12 V we have a range of 66*8 = 528 Ah to 66*12 = 792 Ah
we have to use a min 528 Ah in this case to meet the C value

for 24 V we have a range of 33*8 = 264 Ah to 33*12 = 396 Ah
we have to use a min 264 Ah in this case to meet the C value

for 48 V we have a range of 16*8 = 128 Ah to 16*12 = 192 Ah
we have to use a min 128 Ah in this case to meet the C value

ideally we have to ensure we don't exceed battery manufacturers maximum
and minimum charge rates. NOW, in practice we can abuse batteries a little
so it may be possible to raise the Ah a bit depending on what's available
else we have to use AGM for higher charging rates.


To observe those C constraints for AGM:

for 12 V we have a range of 66*4 = 264 Ah to 66*8 = 528 Ah
we have to use a min 425 Ah in this case to meet autonomy

for 24 V we have a range of 33*4 = 132 Ah to 33*8 = 264 Ah
we have to use a min 212 Ah in this case to meet autonomy

for 48 V we have a range of 16*4 = 64 Ah to 16*8 = 128 Ah
we have to use a min 106 Ah in this case to meet autonomy


based on those figures we can now make our battery choice.
In my world we never parallel unless there is absolutley NO choice.

For RV you can and will parallel because you're stuck at 12 v unless
you design your solar setup right from the beginning.

for RV there is also a compromise on available space, so that
further complicates selection.

so that's all there is to basic solar calculation.


next post will have very specific details on Lithium batteries for
those with deep pockets.. :)


 
Paul 1950 said:
Why do you believe it is necessary to have an even number of panels?
Thanks


This comes from voltage issues when mixing high voc residential panels with
non residential controllers.

non res controllers come in 100,150 and 250 volt max input limits
so quite often it's not possible to use an odd number in series because the combined voc exceeds the controllers input limit.
the recommendation is to stick with even pairs. this also makes it easier to expand a system.
adding one odd panel is often not going to work, but adding a pair in series to an existing array is..

just something to be aware of, however, if the voltages do fit then there is nothing to stop you putting 3 or 5 in series.
when summing panel Voc voltages, multiply the total by 1.2 as a safety factor for cold conditions. Panels produce
more output when cold..




 
Solarman, am I correct in multiplying the battery banks ah by the voltage to get the kw.
say a battery bank is 240ahx 48 volts = about 11.5kw but you say that only about 20-25%
should be drawn down on any given day so that leaves on 2-3kw to use. Is this along the
right track?
 
butchiiii said:
Solarman, am I correct in multiplying the battery banks ah by the voltage to get the kw.
say a battery bank is 240ahx 48 volts = about 11.5kw but you say that only about 20-25%
should be drawn down on any given day so that leaves on 2-3kw to use. Is this along the
right track?

Yes Sir...

V * I = W

48 * 240 = 11520 W or 11.5 kW

if you use 20% per day then you could use 2300 W per day for three days and that's acceptable..

 
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