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Which inverter for a small windmill/solar panel setup? What wire gage?
I have a choice between a variety of inverters (80 - 5000 Watt continuous). I have a 45Watt solar panel and I will find a house fan to assist as a windmill in charging a car battery. The fan will sit in a 3rd story window facing the prevailing wind. The set up is for charging batteries for drill and other small equipment and running some lamps at a house where I can't afford to have electricity installed. I would like the setup to be the best I can get for future ramping up of solar/wind electrical. What are the problems I will face with too large or too small an inverter and too large or too small of a wire gage going both to and from the battery and the inverter? Can I just clip the wires from the fan to the battery? If so, how do I tell which is the positive and which the negative connection? Will hooking up a fan this way charge the car battery? Will it charge the battery if the switch on the fan is turned off? Do I have to change something on the fan's motor in order to charge the car battery? The inverters come in different Wattages, but each individual inverter doesn't appear to have a mechanism for varying the wattage or amperage. Is this a problem? Of what other problems should I be cognizant?
4 Answers
- billrussell42Lv 78 years agoFavorite Answer
House fans won't act as a generator, first of all. The only motors that can act as a generator are permanent magnet DC motors.
Too large an inverter wastes money, plus it would have a lower efficiency at lower power levels, so it would waste power in a setup where you are using only a small fraction of it's capabilities.
Add up the power levels you will need. My guess would be a few hundred watts. Double that number for some margin and get an inverter that size.
Re wire size, that is a trivial issue compared to all the others. Wire from solar panel to charge controller and from there to battery and to inverter have to be sized so that the voltage drop is small at the max current. And that also depends on the length.
Example calculation: 100 feet total (50 feet for each of two conductors) and 10 amps current, and max of 0.5 volts drop. Resistance is E/I = 0.5/10 = 0.05 ohms. Now look up a wire table and find the wire size that has a resistance of less than 0.05 ohms per 100 feet.
A table shows that #6 copper has a resistance of 0.4 ohms per 1000 feet, which is 0.04 ohms per 100 feet, so that would do. Shorter lengths could use thinner wire, of course.
You will need a charge controller to avoid damaging the battery.
hope this helps. this is a complicated topic.
- ?Lv 78 years ago
[1] Buy 5000W output 120V ac converter is good enough to provide basic need for a small home.
[2] Your 45W solar panel is unless to power any converter,because its output current is too small.
[3] AC fan motor cannot produce electricity and it cannot be converted into generator.
[4] To power this 5000w converter directly , you need to buy a very large solar panel that can produce about 400A at 12V. Or use a smaller solar panel has output 40A to charge a big big storage 12V cell rated 400AH. Then this cell is the power source to run the 5000W converter but only good for one hour if the 120V ac total load is 5000W.
