So
@AZRoadrunner 's musings bring up an interesting question, for which I've not found a simple rule of thumb on the net. And that question is, what is the ratio of solar panel power to battery power that one should have for a usable solar station? (I suppose that "intended load" should be the actual metric - but since the battery bank has to supply more than the intended load, and perhaps with a safety factor of 2, I used battery bank capacity instead of load.)
Based on what I'm finding now with my own little experiment, I'm thinking the ratio is 3 or 4:1, to be really safe.
So here's what I'm currently doing. (Pun intended.) I am powering a mother-in-law sweet with my solar station. With one light on continuously, and this desktop computer, I draw about 170 Watts 24/7. It will be a bit more when I hook up the small refrigerator too.
Let's call it 200 watts/hr. That means that I will use 4.8 kW a day. My battery bank is 4.8 kW.
Just on its face, that doesn't sound good. A day without sun, and I have no more power.
So I wonder if there's a Rule of Thumb about the ratio of one's battery bank to the intended load?
It seems 2:1 at least.
Fortunately, it's been sunny here in AZ. So I'm going to focus now on just getting through a solar day, to get a feel for this problem.
As a baseline, at the equinox, I harvested 3.7 kW.
Naturally, that will increase as summer gets here. (Although maybe not as much as simple math would suggest, since the panels will be at 180 degrees F (or more?) in June, July and August and won't produce as much power.) I have to extrapolate to Dec 21, but I think I could have made 2 kW.
I should be using the shortest day worst case number of Dec 21 for my musings here. But for now, in real life, on the equinox, generating 3.7 kW doesn't fully recharge my battery bank with my 200 W load running continuously, I have to top off the bank at night via my Multiplus. (When electricity rates are lower.)
I haven't been able to wrap my head around a full day's calculations. But let's just focus on a solar day of 10 hours as a subset to see where I stand.
I harvest 3.7 kW and use about 2 kW for my load during the solar day (200 Watts x 10 hours). That leaves me with 1.7 kW to charge my batteries. Assuming that I started the morning fully charged (NOT a valid assumption, as below), that leaves me 1.4 kW short in the evening.
But in reality, I will never been fully charged in the morning if I were only charging with solar. So I will be even more discharged in the evening. I'll come back to this later.
If I had another 330 W panel, then (all things being equal) I could harvest 5.5 kW a day.
That should give me about 1.8 kW extra, which would fully charge my bank by the end of a 100% sunny day. (Again, assuming that I started with a fully charged battery in the morning.)
But this doesn't leave much of a reserve for partly cloudy days.
AND - my Victon MPPT maxes out at 1000 Watts. So if make more than 1000 Watts at peak day, then all things aren't equal.
(Since the best I've seen so far is 680 W from my two panels, I should never be much over 1000 watts. But this means that I am limited to three panels now. Unless I buy another MPPT.)
So all this means that, in the best case of a fully charged battery in the morning, I would break even if I had 1 kW of solar for my 4.8 kW battery bank.
That's a best case, with no reserve. And all this is with a crummy 200 W load.
If I keep my load at 200 W, it seems that I should have 2 kW of solar panels for rainy days. (Which means two MPPT's.)
Now, the reality is that, from the end of the solar day until the beginning the next morning, I will have used about 2 kW powering my load. (200 watts x 10 hours.) If I had started with a full charge at the end of the solar day (which will not happen, because I'll be losing power as the sun goes down) then already my battery bank will be down to 50% by morning.
In fact, here's an interesting aspect to this 'equation:' Even if I had a bazillion watts of solar, having more than enough solar power does not change this part of the analysis.
While, during the day, the batteries would reach full charge sooner with a bazillion watts of solar, they will begin to discharge at the same rate after the sun goes down as they would if I only had 2 kW of solar. So regardless of how much solar I have, I always lose the same power to my load overnight.
This tells me that there is a point of diminishing returns here. That you reach a point where adding more solar doesn't change this part of the equation.
And again this brings me back again to the thought that there should be a Rule of Thumb for the ratio of battery bank to load. Because, fundamentally, if your bank capacity is too small, you'll run out of power before the night is over.
Anyway, having lost 2 kW overnight means that I need to harvest an extra 2 kW in solar to fully charge my battery by the end of the day. (And I think "to the end of the day" is a clue here to optimizing how much solar power one needs.)
So, since I currently harvest 3.7 kW with 660 watts of panels, I would need yet another 330 W panel minimum (for a total minimal solar array of 1.3 kW) to fully charge my batteries by the end of the day.
Double 1.3 kW for reserve, is 2.6 kW. (And three MPPT's.)
So that's a solar power to battery powre ratio of 0.5.
And all this for a measly 200 watt load.
If my load doubles (think running a freezer and refrigerator, a window AC, a hot plate to cook food) I think that the math is NOT going to be linear (due to that overnight usage) and I would need maybe 2x the solar power.
I'll have to do that math for that case and see what I get.
But I'm too tired now from writing this all out.
Hopefully someone smarter than I am has the answer already.
