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Hi,
Would you be able to tell me if my solar panel yield is suitable based on the consumption readings? Am I correct to assume today I consumed the exact amount of power my solar panel yielded? It is summer here in Australia I am just worried about winter.
My battery is 9Ah LifePo4 and my solar panel is 50W.
Kind regards!
You have a 50W panel and produced 50Wh, which would be 1 hour of your panel at max production or 2 hours at 25W.
In a typical day you should be generating much more than this, assuming you are using it.
Your peak production is 26-43W, so it looks like it is working ok.
If you are only using 40-60Wh then you should have plenty of capacity left.
Can you please elaborate on 'You have a 50W panel and produced 50Wh, which would be 1 hour of your panel at max production or 2 hours at 25W.'
Does this mean my solar panel will generate 50Watts per hour, for approximately 'x' amount per day at 100% efficiency? From what I read max production would be typically four hours of the day. Meaning I would generate 200Watts of power in a period of 4 hours?
As Albert mentioned below, my battery has a total of 115W (12.8 x 9). Does this mean I would need 2 hours to charge the battery in full from depletion?
Sorry I am new to this stuff :)
Effectively, yes. Wh (Watt hours) are the number of watts multiplied by the number of hours.
Obviously, what panels produce varies - by weather, season and temperature (the hotter the more they derate above 25C).
Conversely in winter, when there is less sunlight, they are more efficient due to cooler temps.
It is rare to get full power, especially at higher temperatures.
PV is a bell curve, like so (southern hemisphere).
If you got an hour at quarter power, an hour half power, and lets assume 75% for 4 hours, then half, then a quarter again (8 hours of light), you would get a yield of around 225Wh, or around 16AH (at 14V).
Your panel seems to be fine for what you are using. I am just wondering if your battery is large enough though. You have a 9Ah LFP battery, so 9 x 4 x 3.2 = 115Wh.
If your consumption stays at this level, you discharge about 40-50% of your battery capacity, so no problem. But what happens on days when you use more than you produce and your solar cannot fully recharge the battery?
Looking back through my (grid connected) Fronius data I saw that I had hardly any yield (daily yield of about 25% of Wp installed) for a few days earlier this year (Brisbane), when we had continuous rain.
I understand you have a 50W panel, so your Wp is 50W. If you also have 25% yield on similar days as we had, that would be 12.5Wh/day, which is not enough. Your battery would be discharged in 2 days.
You can either add panels but you probably need a larger charge controller (higher voltage/current rating) for that. Also, if you keep the 9Ah battery you would have to throttle the larger charge controller on days of full sun, otherwise you will cook your battery (too high charging amps).
Probably easier to get a larger battery (or add one or more identical ones in parallel). At 50Wh usage it will remain nearly full most of the time, but has the capacity to let you drain it on days when solar yield is low.
In Sydney, Australia we have an average sunshine hours of 6-8 hours.
Would 25% efficiency still render me below the required charge?
Thank you very much for your help!
I did not mean 25% efficiency. I said "daily yield of about 25% of Wp installed". Maybe the W, Wp and Wh were confusing you as I know a lot of people get confused by W, Wp and Wh. These are not interchangable and it is essential you understand the difference. I often think the confusion comes because it is often poorly explained. I will try to do so below (I hope it becomes clear):
W is the unit of POWER, the ENERGY produced (or consumed, etc.) at a given moment in TIME. The unit Watt actually means JOULES per SECOND. It is not the same as energy, but the rate at which the energy is changing/transferred/consumed/produced. So POWER = ENERGY divided by TIME.
You can think of it as your speed while driving (SPEED = DISTANCE divided by TIME).
We don't have a special unit for speed (we say km/hr), which makes the relationship between distance and speed more obvious that the relationship between power and energy.
When you are driving at a constant speed, say 30km/hr, for 2 hours, you have driven a DISTANCE of (SPEED 30km/hr * TIME 2 hr =) 60km DISTANCE.
If a particular solar panel in a particular installation gives a constant output of 30W for 2 hours, the energy produced is (POWER 30W * TIME 2hr =) ENERGY 60Wh.
Important to note that W is the unit of Power (like speed) and Wh is the unit of energy (like distance).
In a typical day, your solar panel power output fluctuates from zero (before first light), through varying amount from 0-50W (while there is light, depending on brightness, temperature, etc.) and to zero again (after last light). These power outputs at any given moment are measured in W.
At the end of the day, if you add up all the energy that was produced during that day, you know the total energy yield. We use Wh as the unit rather than Joules (makes the maths easier).
Wp means the peak output (in Watts) of a solar panel. It means that if a particular solar panel has a standardised amount of light shining on it, at a standardised temperature, etc., it will produce that number of W of power.
It is similar to standardised fuel consumption numbers for cars. It only tells you what will happen under specific conditions, but gives you a way of comparing different cars (or solar panels, in the case of Wp).
In your case that was 50W.
I mentioned above that earlier this year that for a number of days I only had a yield of 25% of my installed Wp.
As an aside: I have 20kWp solar panels and produced only 5kWh (25% of 20) in a whole day!
If you encounter a similar run of rain days, like me, you may produce only 25% of your Wp power output. In your case 25% of 50Wp, or 12.5Wh. And that is very little and it is why I mentioned you would drain your battery in 2 days or so.
I hope the above makes it clear.
