For an higher KVA Diesel Genset. Can someone let me understand the basic working principle.
126665
Hi,\n\nYou know that if electricity is apllied to a coil over a magnet it causes mechanical movement due to the formation of magnetic fields similarly the converse is also true\nthat if mechanical energy is applied to a similar but reverse arrangement electricity can be produced\n\n\nwhen current flows magnetic fields are induced but if a fluctuating magnetic field forms in a ciol electricity is induced
158056
First of all, you cannot turn a 240 V, 200 A service into a 480 V, 200 A service with a simple 3 phase transformer. In a transformer, power out (roughly) equals power in. If you put 83 kW in, you can't get 166 kW out. You will have your electric company run a new 480 V service for you. From "your side" of the service, you will have to have a licensed, qualified electrician make sure that all breakers, wiring, fuses, and other components are rated for 600 volts, so you may not be able to use your existing breaker box, or even the wire feeding it. All of your existing wire to motors and other 240 / 480 VAC devices should be of ample ampacity, as a motor draws half of the current at 480 than at 240, but again, it may be wired with 300 V wire that does not have insulation rated for 480 V circuits. Another investment would likely be a step-down transformer for your electrical outlets. You won't be able to go phase to neutral anymore, as that would get you 277 VAC (in a 4-wire Y system). Make sure that your qualified electrician complies with all local codes, as well as the NEC. Good luck!
102791
1.Current in the 385.451 V circuit is bigger.\n\nGod is spelled B-I-G B-A-N-G\n\nPBrax@netscape.com
535838
http://www.eas.asu.edu/~holbert/ece201/kvl.html
375889
Easy!\nIf the input is the same(1.1kW) it depends on their performance.
491090
kvar stands for "kilo-volt-amps-reactive", so...... what is the voltage?
91656
How about this one?\n\n"This calculation module allows you to quickly calculate fault currents for a complete radial system. It can simultaneously handle a main distribution transformer, service lateral, second transformer, up to 40 main-feeders, with 8 main-feeders having 40 sub-feeders for a total of 360 nodal points. \n\nAccommodating both single-phase and three-phase circuits, the module will calculate from either infinite or known fault current values at either the transformer or the main-bus. \n\nThe module accommodates multiple conduit types, bus types, multiple conductors and motor contribution. The program is based on the popular LICE (point-to-point) calculation method. \n\nOur unique Help file (of which we are quite proud) includes complete practical guidance on Fault Calculations, and even includes a theoretical derivation of the LICE method. "\n\nhttp://www.edreference.com/fault.asp
165835
Hi! A transformer works on the principle of mutual induction which is valid only when the voltage (or EMF, to be more appropriate) on one of the coils of the transformer is changing with time, as in the case of A.C. voltage. As the magnitude of the voltage changes on one of the coils, the corresponding flux on the other coil also changes, thus giving rise to a voltage on the other side.\n\nAn AA battery provides a constant voltage, as a result of which there is no change in flux and no voltage on the other coil (120 V side, in this case).\n\nHowever, there will be a flux change for a very short duration of time as the current in the 24 V side rises from 0 to its full value after the battery has been connected. During this time, a voltage will be induced in the 120 V side and you'll get an A.C voltage on this side. But this would be for a very short time duration.\n\nSo, the answer: After some time, there is no voltage (A.C. or D.C.) on the 120 V side.
196332
You would have 7500 turns in the primary winding.\n\nThere would be 0.6 amp or 600 milliamp current flow in the primary.\n\nDisregarding losses in the transformer - input/output wattage (power) is 4320 watts.\n\nAll in a perfect world.........
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.
338677
because it's not referenced to ground to begin with.\nbut if you have a wye to delta transformer then the wye transformer will be referenced to ground (center tap).\n\nDelta transformers are for more of the industrial (large 3 phase motors) application anyway.  hence there really is no need to reference it to ground.
229619
Basically.... it is as simple as (Voltage X (times) amperage = Wattage) example.... \n\na 120volt motor pulling a normal 3 amperes  ....will consume 360 watts of electricity... \n\nThe more amperage the motor requires.... the higher... the horsepower... and also the cost to run or operate it...
16074
you shouldn't have to rewire anything.  a(n AC) motor is a generator and a generator is a motor.  physically they are the same thing!  attach the motor contacts to a multimeter or oscilloscope, crank on that bad boy and watch what happens!
171951
a ditisbution transformer is much larger . but they both do the same job .they either step up or step down current.
42642
Limit the load current.\nEDIT: This is often done by adjusting the stator field current if it's a generator with an electromagnet (not permanent magnet) field.
616061
Im sorry but you have not supplied enough information for me to answer this question properly. You need to be more specific as to where your measuring the 70 MW from....\n\n Heres a good reliable motor set up anyways....\nThe FT4C-3F Twin Pac generating unit consists of two Pratt & Whitney C-3F engines. The Mitsubishi generator is rated at 70 MVA at ISO. The base load power output is 58 MW at ISO conditions or 48.5 MW at 90 degrees F. \n\n..
523492
use voltage dividers with a transistorized switch.  this method lights  a number of lights matching the input voltage in this case, all the way up the tree to it, rather than just one, so note that.  (7V in gives 7 lamps on).\n\nthe first transistor comes on if the base reaches .7V (2N2222 or so will work).  the next transistor is attached to the base of the first transistor through a voltage divider, the third ....  From the collector stick in a 1V Zener for regulation (make sure you get one that withstands 12v in), and the lights then go from there straight to ground (include a resistor here to limit current, try 15 ohms to start, experiment!).\n\ndecide how many lights you want, divide your final voltage by that number, and that is the voltage per step (here, 1V).  the voltage divider then comes from the voltage per step versus the transistor base ON voltage, across a good sized resistor (say 470 ohms or 1k, whatever you want (watch the current and the power drain here).  in your case, if you use a 470 load resistor on the voltage divider, then the second leg of the divider is (470 / (1V/1V-.7V)) = 141 is about a 130 (less is more on the base!).\n\nrecap:\n\nso the voltage divider goes from the input into the 130, into both the base and the 470, the 470 goes to ground.  the emitter can tie to the input voltage, the collector goes to a 1V Zener, to a 1V lamp, a small resistor, and ground.\n\nthe other stages stack right onto this.  the input still goes to the 130, then to the base and the 470, the 470 ties to the base of the previous stage.  the emitter again goes to the input voltage, and the collector goes to a 1V Zener, a 1V lamp, and a small resistor to ground.  sorry, i can't attach a picture here.