How to connect it so to push the excedent on the supply grid ?\nwhat size if wind turbine would be required for 10 kw ?
212144
You need to mechanically generate motion to turn the shaft of the motor/generator at the speed specified on the name plate. Say 1800rpm for a induction motor would then generate the given voltage at 60Hz.\n\nIt is important to maintain this motion otherwise the frequency will vary.\nThe torque to maintain this motion is related to the load on the motor. So a 1 hp motor is about 746 watts. So for a 10kw turbin you will need a 2 x 15HP motor or 30HP motor due to a 50% efficiency factor for the maximum load. Most of the mechanical energy goes into heat.
266686
I don't know, but if you find one, let me know so I can move there.  I wish to integrate all of these into my new home but the cost is unbelievable.  The wind turbines will cost 20k and the solar panels around 30k.  I haven't even begun to look into the waste system yet.  Electric companies only buy and sell electricity and service the delivery equipment.  Look at the logo on the sides of their trucks...they are all the same logo, just a different name.  What a wonderful world.  haha
206188
Ok here ya go!\n\n In 2005, worldwide capacity of wind-powered generators was 58,982 MW, their production making up less than 1% of world-wide electricity use. Although still a relatively minor source of electricity for most countries, wind power generation more than quadrupled between 1999 and 2005.\n\nMost modern wind power is generated in the form of electricity by converting the rotation of turbine blades into electrical current by means of an electrical generator. In windmills (a much older technology) wind energy is used to turn mechanical machinery to do physical work, like crushing grain or pumping water.\n\nWind power is used in large scale wind farms for national electrical grids as well as in small individual turbines for providing electricity in isolated locations.\n\nWind energy is abundant, inexhaustible, widely distributed, clean, and mitigates the greenhouse effect.\n\nAn estimated 1 to 3 % of energy from the Sun that hits the earth is converted into wind energy. This is about 50 to 100 times more energy than is converted into biomass by all the plants on earth through photosynthesis. Most of this wind energy can be found at high altitudes where continuous wind speeds of over 160 km/h (100 mph) occur. Eventually, the wind energy is converted through friction into diffuse heat all through the earth's surface and atmosphere.\n\nThe origin of wind is simple. The earth is unevenly heated by the sun resulting in the poles receiving less energy from the sun than the equator does. Also the dry land heats up (and cools down) more quickly than the seas do. The differential heating powers a global atmospheric convection system reaching from the earth's surface to the stratosphere which acts as a virtual ceiling.\n\nThe change of seasons, change of day and night, the Coriolis effect, the irregular albedo (reflectivity) of land and water, humidity, and the friction of wind over different terrain are some of the factors which complicate the flow of wind over the surface.
528679
just plug in 4 for t\n\nh=80t-5t^2\nh=80(4)-5(4)^2\nh=320-5(16)\nh=320-80\nh=240 m
522051
Just from looking at this, it seems that it can be set up as a ratio:\n\nP1/W1 = P2/W2\n\nwhere Pi and Wi are the power and wind speed, respectively. So for the first one it would be set up like:\n\n(5 kW)/(10 mph) = x/(20 mph)\n\nsolving for x gives us:\n\nx = (5 kw)*(20 mph)/(10 mph)\n= 10 kW\n\nFor the second speed we could set it up like this:\n\n(5 kW)/(10 mph) = x/(25 mph)\n\nonce again we solve for x and get:\n\nx = (5 kW)*(25 mph)/(10 mph)\n= 12.5 kW\n\nI don't know if this is correct, but from the information given this is the only way I can think of to do it.
377973
The fuel should increase to take the load. My experience is on turbine generators, one was 9 mega. WATS.
172042
i am assuming you are trying to minimize losses in a utility distribution system...when i was a system engineer at a utility..the quickest payback items were 1)correcting power factor as close to unity as possible with switched capacitor banks 2)have a policy of sizing distribution transfomers properly in relation to the load (i.e not oversized) to miminimize NL (no load) losses and not undersized too much either to minimize LL (load losses).\n3)size distribution feeder conductors to miminize both NL and LL losses.  The correct point at which you achieve max savings has a good bit to do with your %rate of capital as well as your cost of power both KW and KWH rates.
170041
The power grid is a very generic term used to describe the system which distributes electrical power in the United States.  The gri`d is composed of electrical producers (namely power plants burning things or using nuclear material, or water/wind/geothermal conversion), conversion stations to link the producer to the distribution network, and substations for more local distribution, through wires and transformers to your house.\n\nThe one thing constant on the grid is the 60 Hz (60 cycles per second) voltage--which is the same everwhere.  All producers are synchronized to the 60 Hz across the country.  If this were not true, there would be no way to make sure power flowed from the producers to the customers.  \n\n60 Hz power is generated by turning a generator--either by steam power, water turbine, wind mill, etc.  The electrical machinery generating power has a limited voltage output.  This is converted by transformers at the power station to very high voltages.  Power is the product of volts and amps.  Amperes is a measure of the actual number of electrons passing through a conductor for a time period.  The more electrons per second means more amps and more power.  One volt multiplied by one amp equals one watt.  \n\nIf you have two D batteries in a flashlight making 3 volts, and place an ammeter in series with the bulb reading 2 amps, then the bulb is burning 6 watts. \n\nIn a transmission line, a major source of loss is heat because the conductor is not perfect, and energy is lost.  This loss known as "i squared R" is the square of the current flowing in a conductor times the resistance.  Resistance goes up with length, all other things being equal.  A 2' length of copper wire at room temp is double the resistance of a 1' length of the same exact wire at the same room temp.  \n\nSo when a power plant is 100 miles outside a city, they must REDUCE the CURRENT flowing through the wires to the city.  The wires, part of the POWER GRID, carry very HIGH voltages.  The producers have transformers which convert amps and volts to lower amps and higher voltages to carry the power to the consumer.\n\nOn the consumer end, thousands of volts at lower currents are converted to 120V AC at higher currents using transformers.  There are several levels of amps/volts depending on which part of the distribution system you are.  At the telephone pole at a house, for instance, you may have 3000 V AC stepped down to 240/110 at a 200 amp service per household.  \n\nThere are long stretches of power distribution, say through the Nevada desert, that may run for hundreds of miles at 120,000V AC.\n\nSo to summarize, when people refer to the power grid, they refer to the equipment and wiring used to produce and distribute electrical power in this country.  I know the discussion of volts and amps was very brief, but I hope it helps.  The more important thing is to think of the grid as a giant spider web, where everthing is interconnected.  It is a strength as "economy of scale," but a weakness when something goes wrong.  When one part of the web is damaged, it can be very difficult to keep the damage from spreading out throughout the network.  There is a lot of swtiching and protection along the distribution network, but it can still fail.
42284
Call 1-800-HP invent for customer support.  They can probably guide you in the right direction.
618518
I would just run it at 60 Hz. Output rpm will be 20% higher. \n\nAlternately, you could get a 400V VFD and set it's output to 50 Hz.
509118
= 9u-10u + 6v = 6v - u
24144
Is there any specific reason you want to change the power supply?  Not enough connectors?  Is it failing?  Power supply upgrades usually are only done when a unit fails or if you are adding more hardware into the system than the PS can handle... If neither is the case, you may end up spending alot of cash on a good aftermarket supply and it will do absolutely nothing to make your machine run better.  \n\nPower supplies simply convert AC voltage to the required DC supply lines needed for the machine.  They do not do anything to improve the performance.
128991
well, figure that you have 24,000 Kilowatts at your disposal.  Running at 400 Volts.  Power=Current * Voltage.  This is a rough estimate! So roughly 60 amps MAX coming out.  I would refrain from anything that draws more than 40 amps.  Also I would check with the company that made the generator.  You don't want to get crossed into that kind of power.  Believe me.  I've taken hits from a fully charged 50KV capacitor.
406266
Since you're dealing with a variable energy source (wind), you're best bet is to build a DC generator...\n\nIs this a science progect?\n\nA DC Generator is the same machine as a DC motor...  So if you're building something for your home, then just use a DC motor and a simple belt an pulley to reduce the rotational resistance (or load) on the blades, so a slower wind can be used...\n\nFor a Science project...  \nBuid three loops of wire and seperate them 60° on the rotor of the wind mill.  Place a magnet on either side. north to south\nSet up "brushes" (chunks of graphite like pencil lead) on either side so that as the rotor spins with the wind, the same side is going to the same terminal + or - no matter which side of the coil it is.\n\nHave fun with it...
51733
get an uninterruptible power supply...
361811
A bicycle seems like a good base. If you connect it to a automotive alternator and 12v battery you can produce and store power using cheap available items.\n\nIf you want 220 volts, you can buy inverters that produce it from a 12v battery, again using "off the shelf" equipment.\n\nHappy cycling cause I think you are going to find the human body quite limited at power production.\n\nps, To build something using the items I suggest would mean supporting the bicycle, hopefully you could use the wheel as a pulley without its tire and drive the alternator (having the gears of a standard bicycle being better than an exercise bike to try and match the ratio you need) Since we're not talking much power, the drive belt wouldn't need to be too technical but you may want to change the pulley on the alternator to something wider to match the wheel. If your bicycle support is bolted to a piece of plywood laying on the floor, you should be able to bracket the alternator to that plywood and keep the belt tight enough.\n\nThe electrical connection is fairly simple, connect the big lug on the alternator to the car battery +ive and the casing to the -ive. You will need to connect a small indication light between the small lug and the +ive as this is part of the charge circuit. I suggest a ammeter in the circuit to prove the doubting people on here they are wrong and that you can drive a charge current into the battery but don't buy one with too big of a range :-)\n\npps Try to select a low power alternator (off a small car). \n\nppps If you want to do this without incurring a large expense, you need to make best use of parts that are readily available and do the engineering to suit. If this is not your best subject, you may need to seriously consider whether you want to waste any time any money attempting it.
451953
i would try to find two sheets of a fine metal mesh. Use the first sheet to charge the incoming air and the second sheet as the negative attracting grid. place them about an inch from eachothers and attach the positive and negative terminals each to a sheet. I would use a car battery as this gives you the most voltage of something that is portable. Whatever source you get make sure it is dc and that each sheet of wire mesh is insulated from the other or else the electrostatic effect wont work. Or if you have ever seen the ionic breeze by sharper image it uses three or four metal plates placed side by side to achieve dust collection. This may work also.