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Number of turns in coil to get an output of 8V

  1. May 11, 2015 #1
    Hello!

    First of all excuse my English. I'm Swedish.

    I'm doing an exprement in which I utilize the energy from the rotation in a stroller wheel to charge a 5V battery .

    I plan to make a magnetic generator with a disc placed on a 10 " wheel, consisting of the 20 Neodymium magmeter on the radius of 52 mm from the wheel axis. 3 coils with iron core will be placed at a distance of 2 mm from the magnets . The stroller will roll at a speed of 5.3 km / h and I need to get about 8 V output from the generator.

    I need help with setting up an equation to get the number of turns of the coil and also help with the choice of magnets. The magnets diameter should be about 10 mm.

    Thanks in advance
    Siri
     
  2. jcsd
  3. May 11, 2015 #2

    Hesch

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    Gold Member

    Hello Sweden, Denmark here (also excuse my english).

    I don't know if this is of any help, but the number of turns in a transformer is calculated ( by a sinusoidal flux ) by:

    Veff = 4.44*f*N*A*B, (f = frquency [Hz], N = turns, A = cross section area of coil [m2], B = magnetic induction [Tesla] ).

    Probably you cannot regard the flux to be sinusoidal, but then you can use:

    Emf = dΨv/dt, ( Ψv = flux * (number of turns) ).
     
  4. May 11, 2015 #3
    Hi Denmark!

    Thanks for the quick reply !
    How will the distance between the coil and magnet fit into the equation?
     
  5. May 11, 2015 #4

    Hesch

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    Gold Member

    You say that iron core will be placed at a distance of 2 mm from the magnets. If the area of the magnets/iron core is large as compared to the distance, you can regard the B-field being the same in the magnet and the iron core.

    I'm not calling for a smaller distance as you must have some volume in the air-gap between magnet and iron-core. That's because the power conversion ( mecanical energy → electric energy ) is due to change of energy in said airgap just outside the ironcore.

    The energy density in this airgap = ½*B*H [ J/m3 ]. So the energy = ( energy density ) * ( volume of airgap ), and if the volume = 0, so is the energy. And if volume is very high (large distance), the B-field and the H-field becomes low. An optimal distance is very hard to calculate, but it exists, and it is not 0 mm.
     
    Last edited: May 11, 2015
  6. May 16, 2015 #5
    Conceptually you are building an electromagnet that exactly matches the induced flux. Your geometry is not clear to me. I'm assuming the magnets and solenoids are beside the wheels?

    Build a preprototype. The difference between the ideal and the real world is likely large enough that you will need to do this anyway. (Your magnets won't have the nominal flux, etc.) The actual equations are hard (non-linear) for an unsaturated iron core. There are core losses (hysteresis and eddy current losses).

    Also, if this is a commercial type product, do not limit yourself to 8V. Switching supplies are cheap and efficient and work on a wide range of voltages. If this is for a USB type charger you will want a non-linear voltage regulator anyway. (Linear regulators are inefficient, and hand powered devices cry out for efficiency.)

    For a sinusoidal output you will want the peak-to-peak (pp) output, rather than the RMS (root mean squared). Lots of equations measure total energy or the like and give the answer in RMS. (PP is about 1.4 times RMS) You then need to calculate the rectifier loss of about 0.6V before the switcher.

    Remember magnets are current driven, so the voltage you get depends on the impedance of the driven circuit. An open circuit will give a different voltage reading than a short.

    If this is more of a homework type experiment, try posting in the homework forum. Post a new post after reading the FAQs.
     
  7. May 16, 2015 #6
    You will need to apply Faraday's law.
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html#c2

    For B,
    For that you need the strength of your magnets, which you may have to obtain from its specifications.

    Try searching for 'axial flux generator' and you may get some hits.
    such as
    http://www.academia.edu/7610343/Axi...sign_for_Low_Cost_Manufacturing_of_Small_Wind
    or
    http://depts.washington.edu/nnmrec/docs/Generator Final Report ME 495 Autumn 2013.pdf

    Of course you will also have to determine the torque on the wheel from your generator and consider the fact that if you try to pull too much power, the wheel may skid.
     
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