Your questions are normal ones for someone who is just starting to learn
about solar energy. However, as you will see by the below analysis, you'd
be well served by doing some more research on the topic, in order to better
decide what tradeoffs you are willing to make.
But I'll try to go with your assumptions and the information you've
provided, to demonstrate what a system with those assumptions might look
like.
At another message, you wrote that you are at 50N 115W. I'm not sure
exactly what the insolation is there, and I get conflicting results from
different data sources for Edmonton, which is 53N 113W. Using the most
optimistic estimate, you have perhaps 2.5 effective sun hours (ESH) during
the worst month. Although you may have 5 hours of sunlight per day, your
panels cannot produce full power for that time. Hence the difference
between the time from sunrise to sunset, and the ESH or solar insolation.
I also assume that when you write 300 k watts per month you really mean 300
kWh/month.
That is 10 kWh/day.
To go further, is your 10 kWh/day what you want to PRODUCE, or is it what
you are CONSUMING. I suspect the latter, which means that you will need to
PRODUCE more than 10 kWh/day since no system is 100% efficient.
Also, are your loads AC or DC. Since you mention an inverter in another
post, I will assume your loads are all AC. The significance is that you
have to account for inverter inefficiencies.
A system which might supply your needs given the above assumptions:
1. Daily AC load = 10 kWh
2. Inverter losses = 15% (It might be even higher for your Brutus
inverter at low power levels, though).
3. Required production = 10/(1-15%) = 11.765 kWh/day
We're going to look at a 12V system since that's what your inverter is and
that is what you have mentioned. But a higher voltage system would be
preferable as your losses would be less.
4. 11.765 kWh/day @ 12V converted to daily ampere-hrs = 980Ah
5. System losses and safety factor multiplier = 1.2
6. Adjusted load = 1.2 * 980 = 1176 Ah/day
7. ESH = 2.5 (see above)
8. Required array current = 1176 / 2.5 = 471 amperes
You didn't supply the specifics on your solar panel, but assuming they have
an output current at maximum power of 0.9 amperes, you would require five
hundred twenty three (523) of your fifteen watt modules to reliably supply
your power needs.
At your latitude, some would recommend 10-12 days of battery storage.
Since even a high-quality deep discharge battery should not be discharged
to less than a 20% state of charge, you need, at 12V, a battery capacity of
about 14,700 ampere-hours. "How Many" batteries that would require depends
on the capacity of each one. Obviously, you need a large cell to supply
that much. Surrette makes a 2 Volt cell rated at about 2000Ah, so you
could probably get away with about 42 of them.
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Cutting your load by 50% would also reduce your PV panel and battery
requirements a similar amount; although you would have to check on the
losses in the inverter at your lower power consumption to be sure of that.
================================
If I were setting up a PV system at your location, I would use a higher
nominal system voltage (probably 48V); larger PV panels (150-200 watts); a
more efficient inverter; and probably still use the heavy duty 2V cells for
the battery bank. I would also look at economics of a shorter battery
storage time vs longer running time of the backup generator.
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-- ron (off the grid in Downeast Maine)
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