Hello, I'm building a generator and have completed initial phase. I made coil with 300 turns, which gives me 12v when rotor is turned at 2000rpm, if I short the coil it gives me 16 amp. 1] Now I'm wondering what is the output wattage of the generator I built / described? 2] If I use 18, similar coils and I rotate the rotor at 2000 rpm as before, what could be the output/wattage I may get from the generator in ideal conditions? Any help? Thank you.
If you want the maximum wattage, you will have to multiply 12v ( if you are sure it is turned at 2000 rpm only) by the maximum current through it. But shorting will not give you the maximum wattage. Because by doing so, the voltage appears on the coil of the generator. So first measure it's internal resistance, which is equal to the load you will have to use in order to measure the max watts or current. (Just applying max power transfer theorem). Hence by that theorem if R is the int. resistance of the generator coil, the max power is v^{2}/4R which is 36/R in your case.
You mean to say 18 coils each of 300 turns? All mounted to the same shaft in series? Then the o/p voltage will increase, also the internal resistance, whose consequence is explained by V^{2}/4R. If you are connecting 18 coils parallel, the internal resistance decreases (which is favourable) and I think voltage remains same but power again increases, as is obvious from- V^{2}/4R.
Thank you Simon and Raj. I wanted to get an Idea of the output before I make another coils, as it is very hard to wind them by hand. Took almost 4-5 hours for one coil and my hands were cursing me. The resistance for 300 turns coil shows me 0.8 ohm. Also does the output wattage depend on the way I connect them [parallel or series]?
Can't we calculate something like max wattage a generator can produce based on open voltage and short circuit current output?
You can - but the results will likely be misleading. Much the same way as you get a misleading idea of output power from a battery from the short-circuit current. The short-circuit current depends on the internal impedance from all that wire in coils, ... you could work out what that power is based on ##P=I^2R_{int}## or something but that would be the power dissipated as internal losses. You need to know the power that can be delivered to a load. Try it and see - use different loads and use ##P=VI## and see what happens. You could get a ballpark figure off faraday's law with the rpm of what the generator could, in principle, supply ... but it really depends on the power that is supplied to whatever turns the handle.
Rotating this 300 turn coil at 2000 rpm generates 12v (as you have stated) so V=12v; the load you have to connect in order to draw maximum power is same as the internal resistance so R=0.8Ω; then the max wattage is P_{m}=V^{2}/4R =(12)^{2}/4*0.8 =45W. You also told about connecting 18 similar coils. Then if in series, resistance will be 18 times, that is 18*0.8=14.4Ω but o/p voltage will change, Iam not sure if that will also become 18 times so if you connect 18 coils measure the voltage. If they are connected parallel, R=0.8/18=0.0444..Ω. But then Iam not sure about the voltage which you will have to measure. Any way the formula V^{2}/4R where R is the effective internal resistance will serve you through out.
I think connecting in series will give you more voltage but the drawback is that the internal resistance increases hence power decreases, while in parallel, the voltage remains same but resistance decreases so power increases. So you will have to be sure for what purpose you are using the gen. at 12v..?
You can tell an engineer: "build me a generator to output yay much power to my load at this much turn rate." and expect to get one. The engineer may have some extra questions for you about the use of the generator. You can buy generators that are rated at a particular power output - the rating will be assuming some standard use for that kind of generator. The manual will warn you not to expect the rated power for all uses. When you build your own generator by putting bits together to see what happens, the question becomes quite open ended. You can use Faraday's Law to get you a ##V_{peak}## and thus a ##V_{rms}## for the open circuit case - then figure out the current for different loads/usages. That will give you an idea of what to expect from your generator. At some point you will need to measure things to work out the detailed performance.
Here that engineer is myself. I'm building 100% cog less generator, which one day will be used as wind generator. I was able to get around 20v out of that coil after reducing the air gap. Though wind won't be able to generate that much voltage as the revolutions will be pretty low.
<Puzzled> these are odd questions for an engineer to ask in a general community. Just to be clear: in NZ the word "engineer" is legally reserved. An engineer of the sort who would reasonably be commissioned to build a generator would be a specialist electrical engineer... I think it would help us to help you if we knew your education better, just so we don't short change you.
For simplicity let's assume it's single phase generator. All coils are in phase and in series. This gives me resistance of 0.8 * 18 = 14.4 ohm. Current = 16 Amp Now internal Power dissipated = 16*16*14.4 = 3686.4 watt when the generator coil is shorted. [Am I right?] Now let's do parallel arrangement of all coils. V = 12, R = 0.044 Ohm I = 16 Amp in each coil. Which makes total current of 16*18 = 288 amp if shorted without load. This gives 288*288*0.044 = 3649.536 watt [again is this right?] I'm taking current in calculation, as I know what is the amount of current when I short it. Both calculations are little off from each other. 3686.4 watt != 3649.536. But I think that is the power generated by either Parallel or series arrangement of the coils. Can we consider this as the max power output this generator can deliver?
I'm Software Engineer, but not Electrical Engineer, though I think I've more curiosity of Electrical Engineering [The Technology, which runs the world].
This is the power dissipated in the coil of the generator, not in the load. When you connect the load, the current reduces, hence also the power. The power drawn by the load depends on the load and the output voltage. Hence shorting is not a good way to measure wattage. The best thing is to first get an idea for what use you will put the generator to (lighting or heating etc.) and decide the output voltage. With that done you can check the wattage needs of your load and design the generator according to that. If you go the other way around, it will be trickier.
Thank you for the inputs. I guess rechargeable battery would be best load for this generator. But for calculation/design let's say the load requires 25V, 100Amp and have resistance of 0.005 ohm. How should I proceed for this case?
Do we have any book or online resource, which I can refer to build such generator as I mentioned in earlier post?
If you can produce 25v with your gen and still have the internal resistance at 0.8Ω, then P_{max}=V^{2}/4R P_{max}=195.3W That's all right, on course. The generator is upto it's mark. All depends on the load now.. Ok let's see.. So your load has a resistance 0.005Ω. But this is very much small compared to the int resistance of your generator. So, much of the voltage appears on the generator coils leaving only a small voltage on the load. By voltage divider rule, voltage across your 0.005Ω load is: V_{L}=VR_{L}/(R+R_{L}) V_{L}=0.155 volts!! The power drawn by the load is P_{L}=V_{L}^{2}/R_{L} P_{L}=4.8W! See how low the values are! Reason: load is smaller than the internal resistance. It's very important that you match the load and int resistance. See: http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem Your load resistance must be greater than or equal to the internal resistance (or source resistance).