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Calculating the Power Output of an AC Generator.

  1. Jun 11, 2012 #1
    Hello everyone!
    This is my first time posting here and I hope I can make a good first impression, and I apologize for the lack of links because of this being my first post.

    Currently I am working on a project for my grandfather's engineering business, and part of this project is designing a generator that will be built inside of an air tool that will power an Arduino as well as other electronic components. I have been working on it for maybe 2-3 months now and have learned a lot. I am looking for some guidance in doing the calculations for this generator that I am building. I must let you know that I consider myself under-qualified for this job as I lack the mathematics required so far, so please be patient.

    So far I have designed a general idea in SolidWorks that I believe will be sufficient for what we need.

    Using 4 x .5" Neodymium Magnets ~14,500 Gauss [Also have some SmCo magnets (unsure on rating for these)]
    The pieces between the magnets are made of Cast Iron. The rest of the assembly is Alum.

    The stator is going to be made of an 0.007" thick Electrical Steel whose relative permeability [μ/μ0] we will assume is 4,000-8,000 [Not 100% sure what the actual Perm. is].

    First here are my main variables:
    • The air-tool will be spinning at 5,000-7,000 RPM
    • Our wire is 22GA Enameled Copper Wire.
    • Core Material Relative Permeability 4,000-8,000.
    • The cross section of the core is .150" x .750"
    • I'd expect to be able to get 200+ winds per coil.
    • It's going to be in three phase.
    • The rotor has 4 poles
    • We will assume that the is 1 tesla at the stators surface.
    • The stator has 12 coils
    • There is a gap between the stator and rotor of about .030"

    So what I'd like to know is with the above set-up what kind of output am I looking at? I have looked closely at Maxwell's equations [I'm not even 100% sure if I am on the right track]. But after much effort I have still failed to grasp how I am to apply these laws and equations to my situation.

    We need around 15v @ 2A, 35W would be ideal.

    Please let me know if I am forgetting anything. I hope I organized things well enough that all of this is easily understood.
  2. jcsd
  3. Jun 11, 2012 #2


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    Staff: Mentor

    Welcome to the PF.

    What other electronics do you need to power? 35V at 2A is way more than you need for a microcontroller (uC). The power from the generator will obviously go away when the unit is de-activated -- what is the power plan for then? The uC will likely need to be battery backed up for some amount of time. How long?
  4. Jun 11, 2012 #3
    I will also be powering a 24v solenoid and a linear position transducer, as well as various other components. We want some headroom. We were also thinking of using some super caps in series to store some temporary power. The tool spins constantly so power is always being generated. Some super caps will power the arduino for a long enough period of time to save data etc.
  5. Jun 11, 2012 #4


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    Staff: Mentor

    Can you just use air power for the solenoid? As you design things like this, you want to try to optimize how you use power. If you can use an air-powered solenoid, there is no reason to put a big heavy expensive generator in the tool anymore. The uC and position sensor can almost be battery powered at low voltage, with perhaps a very small generator to keep the batteries charged...
  6. Jun 11, 2012 #5
    An air powered solenoid wouldn't work since we need to cycle the solenoid very quickly and precisely.

    One thing I must also mention is that reliability is extremely important. I can't stress that enough.

    I am working alongside my grandfathers engineer and he is against using a battery (reliability related a far as I know).
  7. Jun 12, 2012 #6
    Have you looked to see what is available on the market? It would almost certainly be quicker and cheaper to buy in rather then design and build. As this is going to need an umbilical connection to provide an air supply could you provide "mains" hook up as well
  8. Jun 12, 2012 #7


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    Gold Member

    I agree with berkeman that you need to optimize your power usage. Advantage of the caps is that they charge fast and output high power - perfect for your solenoid use. disadvantage is that they have low energy density per weight. Using them as a backup for the electronics seems overkill in my opinion and battery backup is generably preferable, no.
    You could then cut down on your 2A output from the generator and its size.

    By the way, once air supply is cut off the reliability of your system goes out the window anyway as you have no way of predicting how long that will be.
  9. Jun 12, 2012 #8
    I honestly prefer to use a battery, but my mentor/supervisor is against it. I only need enough reserve power to have the UC save some variables. So when the air supply is cut [physically disconnecting the air line] it will only need fractions of a seconds worth of power, the generator will spin as long as the air line is connected. There is no need for it to have a sustained source of power after the tool's air supply is cut, as the UC is the only thing that will need power. The power needed isnt going to be 35W and doesnt necessarily have to be at 15W 2A. An amp should even work.
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