If a solar panel (using 36 3x6 solar cells) puts out 60 watts, how many panels would I need (for a house) if I use a max of 2300 kWh in a month? I checked my electric usage for the past 2 years and the kWh is usually lower but I wanted to use the most kWh used (2300) for my question. Thanks!
Good question. Something to keep in mind is that modern solar electric works _alongside_ the grid, so you don't need to displace your entire 2300 kWh. The best financial return, if any, might come at a point where you generate only 500, 000, or 500 kWh per month. How many panels you need strongly depends on your location. Speaking for the location of my house in San Jose, Ca, that amount of usage would be about a 2 kW array, and if you divide 60 into that, you get 200 panels. In a perfect location, you would need slightly less, in a northern state, you might need double that. If this was a real project, you would probably not use 60-watt panels. The choice of panels is not simply a matter of getting enough watts. 2 kW is a large system today for residential. The usual advice is to go for energy efficiency first, then consider solar after the usage is down.
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The latter one has some strategies to consider for cutting your power consumption even more by timing your freezers and refrigerators..
One key piece of information missing is the average number of peak sun hours per day. This will vary by location. It is latitude, regional weather, and site shading must be considered. Another consideration is the loses. The 60 watt solar panel is a DC voltage. The 2300 kWh is for the AC power consumed. There are loses in the inverter to change the electricity from DC to AC. There are also loses for wire resistance, soiling of the panels, wire resistance, etc. To work out an example of the math, I'll use 5 peak sun hours and a total efficiency of 80%. The math can be worked out as a single formula but I'll break it into steps for you. (2300 kwh/month)(month/30.4 days) = 75.65 kWh/day (60 W)(5 psh)(80%) = 240 Wh/day = 0.24 kWh/day/panel # of panels needed = Energy use / Energy generated (75.65 kWh/day) / (0.24 kWh/day/panel) = 36 panels.
