give me design calculation steps and formulas to design a gearbox for wind electric generator?
146543
http://www.motherearthnews.com/library/1986_July_August/The_Basement_Built_Wind_Generator\n\nthis might help
834512
Torque is not the right question to ask in this problem!\nFirst define the generator output in terms of watts.\nAssume an efficiency of 0.8 for the generator and define the Input of the generator (In = Out / 0.8).\nThis is the output of your motor (assuming some diesel/petrol).\nTorque is expressed as N/m (force x radius) and is part of the electric generator specs.  See there to get the equivalence.\nUsually, the motor power is also 0.8 efficient, so the motor power should be W / (0.8*0.8) and that will give you the right ratio.
456954
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.
720466
I suggest you put a message on "Engineering Tips" www.eng-tips you will need to register and go to the engineering area suited to your question. I guess it will be Mechanical
151509
http://www.ncsu-engineering-resources.com/store/trans/productListingRpt.asp?CatID=15\n\nTry these modules.
898940
You can find printable schematics on the page I've linked below.\n\nGood luck.
367171
work=forcexdistance\npower=work/time\n\n1 in=2.54cm\n8x2.54=20.32cm\n\n10cm=1m\n20.32cm=2.032m\n\nwork=700Nx2.032m\n    =1422.4Nm\n\npower=1422.4Nm/10s\n     =142.24Nm/s is the power
619423
Pretty general question not knowing forces needed, space conisderations, and operating conditions. Lots of gears are stock and can simply be ordeed online IF you know the parameters.
552634
Click on the McMaster-Carr link below and type in "gears" for the search string.  Then you should be off and running to buying one.\n\nOr if you want to figure out exactly what type of gear you need, try the second link.  (probably more info on gears than you would want to know)\n\nOh, and try the engineering section for a question on gears.  Engineers are gearheads.
1038623
try here---\n         http://www.bearings.machinedesign.com/
622823
anybody that produced generators!
903919
Generator sizing becomes more difficult when electric motors are involved, due to their large starting demand.  Typically, there are two motor styles: CODE G requires 3 times running watts to start, and CODE L requires 5 to 6 times running watts to start.\n\nIt is not practical to think that all loads will start at the same time.  Nor is it practical to size a generator using all starting watts for all selected motors, as some motor loads will already be running (or OFF), when others are ready to start.  As you can see, it can be difficult to exactly size a generator for a group of specific loads, as they are constantly changing in their "ON" or "OFF" cycle.  Always try to limit your load selections to the bare essential items.\n\nTO FIND:\n1 PHASE KW = V * A * PF / 1000\n3 PHASE KW = V * A * 1.73 * 0.8 / 1000\n1 PHASE KVA = V * A * 1000\n3 PHASE KVA = V * A * 1.73 / 1000\nMOTOR - REQD 1 PHASE KW = ( MOTOR HP) * (EFF)\nMOTOR - REQD 3 PHASE KW = (HP) * (0.746) * (EFF)\nMOTOR - REQD 1 PHASE KVA = (MOTOR HP) * (EFF) * (0.8)\nMOTOR - REQD 3 PHASE KVA = (HP) * (0.746) * (EFF) * (0.8)\n1 PHASE PF = 3 PHASE PF = KW / KVA\n1 PHASE AMPS = (KW * 1000) / (VOLTS * PF) =  KVA * 1000 / VOLTS\n3 PHASE AMPS = (KW * 1000) / (1.73 * VOLTS * PF) = (KVA * 1000) / (1.73 * VOLTS)\n \nREQUIRED PRIME MOVER (ENGINE) HORSE POWER\n KW / (GENERATOR EFF. X .746)\n \nFREQUENCY (HERTZ)\n NUMBER OF POLES X R.P.M / 120\n \n ~ ~\nEXAMPLES IN GENERATOR SIZING FOR LARGE ELECTRIC MOTORS \n\nAll AC electric motors require large amounts of electric power to start up, due to it's starting winding. After approximately 3/4 - 1-1/2 seconds, the starting winding drops out, and the running winding continues, at a much lower power demand.  A generator must be sized to handle the load of the starting winding.\n\nSizing a generator for three phase motor starting LRA : \nThe formula is: Motor HP x KVA per HP x (1000 / (Volts x 1.73)) = LRA. \nGiven the size of your machines, estimate KVA/HP at approximately 6.\nTherefore, for 11 kW ... LRA = 134.6 = 135 amps. Generator must produce this current at 380 V, so (1.73 * V * A * 0.8 / 1000) = KW = 71.\nSimilarly, for 15 kW .... LRA = 184 A and KW = 97.\n\nBOTH SINGLE PHASE AND THREE PHASE MOTOR SIZING EXAMPLES ARE BASED ON INFORMATION IN NATIONAL ELECTRICAL CODE HANDBOOK, ARTICLE 430.7 AND IS INTENDED FOR MAXIMUM VOLTAGE DIP OF 25% UPON INITIAL MOTOR START-UP.  IF 35% VOLTAGE DIP IS ALLOWABLE, REDUCE LRA, THEREFORE KW SIZE, BY 25% (LRA or KW X .75). IT IS NOT RECOMMENDED TO CHOOSE ANY GENERATOR KW SIZING THAT YIELDS MORE THAN 35% VOLTAGE DIP.\n\nStarting more than one motor, at one time: make a list of all motors, totaling all running watts.  Calculate the starting demand when the largest motor is started while all others are running.  Add the largest motor starting watts, to the running watts of all smaller motors already running.
786312
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...
534773
Try checking out www.engineering.com
755704
How Does the Windmill Work?\n\nWind Energy\nWind is an indirect source of solar energy. It is caused by the uneven heating of the earth by the sun and the resulting redistribution of air to equalize energy in the atmosphere moving air from areas of high pressure to areas of lower pressure. The energy of the wind is transferred to the windmill’s sails where it is harnessed to perform work.\n\nThe cap of the windmill is turned with the winch attached to the bottom of the tail pole until the sails face the direction of the wind. Depending on the wind speed, wind boards are installed on the leading edge of the sails and canvas sail-cloths are unfurled across the sail-frames. The Miller climbs the sail-frame using the sail-bars like rungs of a ladder attaching the canvas to the cleats on the whip. High winds require less open sail than low wind. The Miller determines the amount of sail surface required based on wind speed and partially furls (reefs) the sails as required. The sails catch the wind and turn the wind-shaft.\n\nThe wind-shaft which is connected to the brake-wheel can be stopped with the brake which consists of a band of friction blocks encircling the brake-wheel attached to the brake lever and controlled via a rope near the tail pole attached to the brake lever. The turning brake-wheel turns the wallower connected to the upright shaft on the fifth level. A wood wheel on the fourth level is used to adjust any irregular spinning of the upright shaft.\n\nThe upright shaft is connected to a large gear called the spur-wheel on the third level. Power to run many things in the windmill is drawn from this wheel. Two stone-nuts turned by the spur-wheel drive quant shafts in turn rotating the runner stones; the top stones of the grinding stone pairs. The lower stone of the grinder is the bedstone; it remains stationary. The spur-wheel also turns a third lantern gear (trundle) to drive power transfer belts, pulleys and shafts used to operate the grain elevators, sifters, grain separator, corn sheller and a vacuum system.\n\nHow is Grain Ground into Flour?\nGrain and flour move through the system using grain elevators and gravity. Grain is carried into the windmill and poured into first level hoppers. From there it is cleaned in the basement with the corn sheller and grain separator. Grain elevators carry cleaned grain to the third level hoppers which hold it until it is dropped into the second level grinders. Corn & oats are ground in the west grinder and dropped into a ceiling mounted sifter on the first level. Wheat & rye are ground in the east grinder. The flour drops into an elevator at the ceiling of the first level, is carried back up to a hopper on the third level and dropped into the sifter on the second level. The sifted flour is stored in hoppers suspended from the ceiling on the first level.\n\nInteresting Notes\nThe windmill contains two scales; a floor scale in the windmill and a wagon scale outside the north end of the windmill. The scale inside the windmill can weigh individual bags of grain. The scale outside is used for weighing loaded and unloaded wagons.
726624
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.
941868
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.