Thanks for updating the OP with more information.
Facts: Run a .37 kW pump without batteries, 2x 165W panels
Question: How long can the panels power the pump on an average day.
Theory: Without a battery/capacitor to act as a buffer, the pump can only run while the panels output more power than are are used. Panels only provide their maximum rating at solar noon. Using the chart to the right you can use the SRU as a percentage of the panels maximum.
Answer: 0 hours; 2x165*.8 = 264 Watts which is less than 370 watts.
Question 2: What gear should I use?
Practical considerations like that need real experience, hopefully one of the other embers can help you out. |
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So that's no fun. Lets make it 4 panels so we can run through the math. At 4 panels it's 2x264=528 watts. As long as the panels are over 370 watts the pump should run. So, 375/528 = 71%. At 71% the sun is 45 degrees in the sky. So that's 45 degrees before solar noon plus 45 after solar noon, or 90 degrees.
The earth rotates at 15 degrees per hour, so the pump should run 90/15 = 6 hours. That assumes the panels are always tilted perpendicular to the sun (i.e., you're out there adjusting them every day) In practice probably 4 to 5 hours per day, plus without buffering for steady operation the pump will get more wear and tear.
How much longer would the pump run if I added a battery to 4 panels?
(Aside:
@MrNatural22's
post on batteries might interest you)
Let's assume a solar insolation of 5, so 528x5=2525 W. Which if you captured all of it via a battery could run the pump 2525 / 370 = 6.8 hours. But the round trip on a lithium might only be 90% efficient, so 6.8 x .9 = 6 hours. Since that uses the insolation map it's an average throughout the year without adjusting the panels.
With a battery, could 2 panels run it?
264x5 = 1320 watt hours. Pump takes 370 watt hours and battery 90% efficient round trip, so 1320 *.9 / 370 = 3.2 hours