What's the usual VOLTAGE and CURRENT output ratings of a 2500 kW DC Generator?
609875
first of all, 120/240 are standard voltages for AC power, not DC.  DC comes in all flavors.\n\nanyway: 2500 kilowatts DC?\n\nthat is 2.5 MEGAwatts.  \n\ni find this somehow hard to swallow.  when i worked at a particle accelerator there was ONE really big DC power supply that put out something like 1.5 megawatts.  it was the size of a small apartment, and required massive amounts of water cooling, was entirely custom built, and probably cost something on the order of a million dollars or more.\n\ni don't think there is any such thing as "usual" for a 2.5 megawatt power supply.  think about it - you're talking about nearly the power consumption of a small town.  anything liek that would have to be custom built.\n\n----\n\nok, having seen your additional details:\n\ni'm quite positive it's AC, not DC.  it also probably depends on what kind of generator it is.  to give you a rough estimate, here's an example of a 11.5MW turbodiesel generator on a cruise ship:\nhttp://www.ship-technology.com/projects/destiny/\nnote the output is at 6.6kV.  that would set the current at the output at around 1.8 amperes.  \n\nhere's some more information.  as you can see there is no real "usual".  \nhttp://www.ship-technology.com/contractors/electrical/abb2/\n\ncheers!
974255
About plus and minus ten percent. That makes it 108 minimum and 132 maximum volts.\n\nMost appliances are designed to expect this range.
1096182
It does depend on the efficiency of the motor.  The horsepower rating refers to the mechanical output of the motor.  If the motor is 80% efficient (a reasonable number), the wattage will be 746*1.25 per horsepower.
512146
Van De Graaf generator
1085514
The max sytem load for 3 phase 208v Y 30kva 100% 89 Amps, so buy a 100 amp this will give you  an upgrade if ever needed and be safe.
227723
Regardless of what size capacitor you choose, there will be a small ripple on the output voltage, which of course gets bigger as the load increases.  You need to determine what is an acceptable ripple voltage for your load to calculate out a value.\n\nIf you want a totally smooth DC output, you'll need to first rectify and filter, then use a voltage regulator with a voltage set point that's a few volts below the voltage output from the capacitor.  Follow that by a smaller capacitor after the regulator to help catch any remaining transients and response after regulation.\n\nSomething to note:   Your capacitor's voltage rating should be much higher than the actual working voltage you will subject it to.  Eletrolytics like to blow up (literally) if you run them past their maximum rating, even for a short time.   Also remember this, the AC voltage that you measure at the output of the transformer (or the output of the rectifier, unfiltered) is the RMS voltage, not the peak voltage of the AC waveform.  Once you add the filter cap, your DC voltage will appear to be higher because the capacitor is charging to the peak voltage, which is basically the RMS voltage multiplied by 1.414   Keep this in mind as well when selecting the working voltage rating for your capacitor.
873132
What sort of prints? What sort of current do you want to source? How few milliviolts do you want?
935193
Not sure but this site may help you:\nhttp://www.windturbinewarehouse.com/\nthis one is a bit more specific:\nhttp://www.gepower.com/prod_serv/products/wind_turbines/en/2xmw/specs.htm\nand this will tell you lots:\nhttp://en.wikipedia.org/wiki/Wind_turbine
1088769
P = d * Q * g * H * efficiency\nWhere,\np= power in KW\nd= density of fluid used\nQ= discharge\nH = head\nefficiency= given for a turbine
515778
British horsepower is also known as just horsepower.\nOne horsepower = 745.7 watts\nSee the howstuffworks link for an interesting look at horsepower.\n\nTo find the rated electrical input power required for a motor knowing the horsepower:\nkVA = (hp*.7457)/(eff*pf)\n\nwhere kVA = rated input power in kilo volt-amps\nhp = rated output horsepower\neff = motor efficiency (Pout/Pin)\npf = power factor of the motor\n\nIn general, a generator can take a load that is equivalent to the input power times the generator efficiency.  After all, a generator just converts energy from a mechanical form to electrical form with some losses in the process.  I admit that it isn't quite as simple as that, though, so I recommend further reading.\n\nTry Wikipedia, howstuffworks, etc for a start with general reading.  Those should lead to the more in-depth stuff.
663172
700V x 2.5A = 1750VA\n\n1750VA / 240V = 7.291666A\n\nWatts (Voltamps) Input = Watts (Voltamps) Output\n\nThe answer is 7.3 Amps @ 240V
206554
1 amp (1000 millamps) is the maximum rated operating conditions for the transformer.  You may be able to get more current than that from the transformer, but the rest of its specifications (such as output voltage, operating temperature, efficiency, etc) are no longer guaranteed to be accurate.\n\nIt would be a good idea not to exceed this current draw if you want the transformer to last, and not pose a fire danger.  Consider using a fuse in your circuit to prevent overcurrent conditions.
1058406
At my job, I maintain lager equipment and small equipment for Land O’ Frost. We have motor that we use variable frequency drives on. The VFD’s range from 480VAC 3 phases to 24VAC single phase on the input side to 1000VDC, 480 single phases, 120VAC on the output side. But when installing a VFD the thing that are considered before installing one are the size of the motor 5hp, 35hp, .25hp, how many amps are needed to run the motor and peak amps so the parts are not over current. The voltage is another thing, the amount of programming that you can do so you can set limits for everything. It can take a long time to find what you need. Have fun shopping
883686
A long time ago I found a voltage regulator for an automotive alternator that would allow output at "line" voltage (for an application similar to yours), but unfortunately I lost the information on the source.  Check at an RV (Recreational Vehicle) dealer in your area for something like this - those folks have information on stuff like that..\n\nA standard automotive alternator isn't built for the sustained high power output you're describing.  You might be able to get a higher capacity unit better suited for your project from a heavy duty truck (or an RV) out of a junkyard and get it rebuilt (complete with the new voltage regulator) for a fraction of the price of a new one.\n\nWarning:  alternators typically output full capacity at frequencies far above 50 or 60 Hertz.  Your power diodes may not work well at higher line frequencies.\n\nAlso, have you checked into getting a gasoline engine powered 120 volt power supply, like what are typically sold for emergencies power outages and campers?  You can find them for a low few hundred dollars for 1000 watts, and would a lot cheaper in the long run than risking trashing car alternators....
924875
50,000V\n\nOtherwise, if you use Kilo, you just use the K - eg: 50 KV
189503
how about looking at wind generators online and seeing how much power they generate.
508033
1.125 KW or 1125 Watt