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1. What are the specs of the DC motor/other required to/capable of creating a Wind (or other source) Powered Generator that will run a standard home on its own (or 12,000-16,000 kWh per year)

2. What is the torque (in Newton-Metres or Foot-Pounds) required (to be generated by the wind/other) to move that shaft fast enough to generate enough power for that home.

3. What is the speed that that shaft will be spinning (in Hertz/RPM/other) because of that Torque in #2.

Bonus 4. What is the resistance of that motor/other, that is, how much inherent resitance does it have to turning the shaft. (Probably answered when 2 & 3 are answered?)

Anyone who answers all three questions will not only get the 10 points, but I will personally look through your previously answered questions and give you up to 10 thumbs up for any answers I don't hate. Runners up will get a thumbs up, identical correct answers will get the 10 bonus thumbs up as well, all BS answers will get reported. GOOD LUCK!

2007-02-22 08:08:43 · 2 answers · asked by Ray of Freaking Sunshine! 2 in Science & Mathematics Engineering

Addendum to #2. Instead of Torque, Kinetic Energy would actually be better, but either one is fine.

2007-02-22 13:54:34 · update #1

2 answers

It looks like you need help in converting electrical power requirements to mechanical power requirements. I'll try to provide you with the knowledge to do that.

1. Because the wind is variable, the windmill will not be on all the time. There are a couple of options. You could oversize the windmill to generate excess power and store it in batteries. Or you could just use the power from the windmill when it is available and rely on the grid for the rest of the time. If the windmill generates more electricity than you are using, you can send the power to the grid. But the selling price is much lower than your buying price.

2. If you use 1000 kWh in a month like me, then the average power is 1000kWh / (31*24hours) = 1.344 kW. If your generator is 80% efficient, then you'll need to supply 1.344/0.80 = 1.68kW worth of mechanical power on average.

Since power is torque times angular frequency, you can see that as the RPM increases, torque decreases. This presents a unique problem. If you put too much of a load on the generator, it will cause the windmill to stall. So, the faster the windmill turns, the less torque required for the same output power. There's some math involved here. Mechanical power can be in horsepower or watts, and you can convert between the two.

I had to get out my old physics book. This will help you in determining the input mechanical power. One horsepower is equal to 746 watts. Rotational power (in Watts) is torque (in Newton meters) times the angular frequency in radians per second (one full rotation is 2pi radians).

Setting up a home windmill is not cheap. The payback time may be longer than you're willing to invest in. You'll need several things besides the windmill and generator. Battery storage is optional, but helpful. You'll need an inverter to convert the power from the windmill to 60Hz 120V, and synchronize the frequency with the grid's power.

2007-02-24 17:04:10 · answer #1 · answered by vrrJT3 6 · 0 0

Most municipalities do not allow wind turbines to be built. This is because they're ugly, and most people do not want to look at them. So you need to check your local laws. A 3 foot blade isn't going to generate much electricity... maybe 100watts at full power... thats barely enough to keep your TV on. (TVs consume about 80w of electricity when on, 20w when off) You'll also need an electrician, because you need to turn the power off at the street level, and the electrician will need to connect the system to your powermeter. This is not a do-it-yourself project.

2016-05-23 23:46:06 · answer #2 · answered by Anonymous · 0 0

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