Mar 27, 2011 12:35 am
Re: Calculation Confirmation of the Solar Panel Power in the DC Lighting System
The "other" calculation is certainly wrong. You can't multiply 5A @ 24V by 34V to arrive at a sane wattage. The way to do that part would be 5A * 24V = x * 34V, and so x = 5A * 24V / 34V = 3.529A. 3.529A * 34V = 120W, not 170W. And 120W / 0.8 = 150W, which is what you got.
But anyway, you had the right idea. Don't convert to amps first, as that will throw you off completely. You got it right the way you did it and it was obviously simpler and easier to understand.
48W x 10h = 480 Wh
480Wh / 0.8 = 600Wh
600Wh / 4h = 150W
That's all correct. But that assumes that the combined efficiencies of both the load and the supply is 80%. In my experience, that's not enough. I would assume no more than 80% efficiency from the solar panels. And also assume at least 10% inefficiency (90% efficiency) in the rated consumption of the load due to losses in the wiring, conversions, battery charging, etc. You can of course directly measure the load and see if that's the case or not. If 48W is what it actually consumes, even after all of the losses, then a 150W panel will be sufficient. If it turns out that the load consumes 53W on average, then you should account for that. To be safe, I would.
150W / 0.9 = 167W, which probably means a 170W panel
This will protect you in case you don't get exactly 4 hours of sunlight as well. It's always better to oversize slightly.
BTW, where are you getting panels rated at 34V? In my experience, they are rated at 12, 24, or maybe 48V.