Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Magnetic Generator Power Rating

  1. Nov 20, 2012 #1

    I am very new to this, so if my questions sound a bit stupid please point me in the right direction.

    I want to build my own 2.5 kW magnetic generator, but would like to understand a bit more how this works. Please see attached picture. I will be using the 24 magnet / 18 coil configuration, and the coils will be connected in a star formation. So, I have 18 coils, or 9 pairs. When I move the magnets in a clockwise direction, in one turn, 22 magnets will cause a changing magnetic field on each coil. Lets say I will rotate it 3 times per second.

    I hope this does not sound confusing, but which one is my power rating:

    1. 18 coils x 1 magnet moving over it (1 changing magnetic field)

    2. 18 coils x 22 magnets x 3 revolutions

    Attached Files:

  2. jcsd
  3. Nov 20, 2012 #2
    None of these are power ratings. The number of magnets and coils and RPM will give you the frequency of the generator output. The power depends on the EMF generated by coils, and that depends on their construction and the magnetic field in the generator. It also depends on the resistance of the wiring. And, finally, it must be less than the mechanical power fed to the generator's shaft.
  4. Nov 20, 2012 #3
    Thanks for the reply. I understand everything you say, however I am still a bit unclear here on how to determine / calculate my power rating. Let's say the EMF induced in each coil every time there is a change in magnetic field is (purely example) 1 volt, my wire gives a resistance of 3 ohm, and my magnets has a magnetic flux of 1.4 Tesla (this will change depending in the size of my magnets, Neodymium magnets). What would be my power rating at 180 rpm?
    Last edited: Nov 20, 2012
  5. Nov 20, 2012 #4
    I am not sure what you mean by this: "EMF induced in each coil every time there is a change in magnetic field is (purely example) 1 volt".

    Much depends on the geometry of the machine, too. I'd suggest you find a decent book on the theory of electrical machines, otherwise it is unlikely you will succeed.
  6. Nov 20, 2012 #5
    I think you are trying to go to deep with this. What I am after is a simple direction - if I have 18 coils, and I move the magnets once in a clockwise direction, so each coil "experiences" 1 magnetic field change and an EMF is induced, is this my power rating (and yes, I understand there are many variables)? Or is my power rating 18 coils x 22 magnetic field changes with an EMF induced (1 rotation)? Or is my power rating 18 coils x 22 magnetic field changes x 3 rotations with an EMF induced?

    I am after what I should aim for with induced EMF per coil per magnetic field change to achieve a power rating of 2.5 kW.
  7. Nov 20, 2012 #6
    The EMF of one coil is about ## 2fn NB_0S ##, where ## f ## is the number of revolutions per second, ## n ## is the number of magnetic poles, ## N ## is the number of turns in the coil, ## B_0 ## is the max. magnetic flux density, and ## S ## is the crosssection of the coil.

    You have got 3 phases, so each phase must generate 2.5 kW / 3 = 830 W. Total resistance is 3 Ohm, thus 1 Ohm per phase, so total EMF per phase must be about 30 V, or 5 V per coil, if they are connected in series.

    Since N and S are not specified, I cannot determine if your setup will really generate 5 V EMF per coil. But you could use the formula above to figure out N and S required for that.

    Note this is a very rough estimate, because much depends on the true geometry of the magnetic field.
  8. Nov 20, 2012 #7
    Hi voko,

    Perfect, thank you, this is exactly what I was after! With regards to poles, I will use the layout in the picture in my first post. Blue will be North and Red will be South.

    I do apologise but have 1 last question, again this might be a bit up in the air. When there is movement, the magnet will induce an EMF in the copper wire coil, which in turn will generate a magnetic field equal in size to the EMF but opposite to the magnet's field. At what kg force will I need to push the magnets forward to generate continuous EMF? Or is it a bit more complicated than that, and what factors must I keep in mind?
  9. Nov 20, 2012 #8
    Well, this is actually quite simple. The mechanical power fed to the generator must be somewhat greater than power produced by it. ## P = \tau \cdot \omega ##, where ## P ## is power, ## \tau ## is torque, and ## \omega ## is angular velocity. Given the required power and frequency, you can obtain the torque.
  10. Nov 20, 2012 #9
  11. Nov 20, 2012 #10
    The EMF frequency is the mechanical frequency multiplied by the number of magnets. With 24 magnets and 3 revolutions per second, you get 72 Hz.

    Angular velocity (of the rotor) is the mechanical frequency multiplied by 2 pi. For 3 Hz, that is about 20 rad/second.
  12. Nov 21, 2012 #11
    Thanks voko, you are a great help!

    I have found this site : http://www.6pie.com/faradayslaw.php, and have used this to calculate how many windings I propose I should have.

    N = -1 * (-V/ change in (( tesla * area meters squared)/ seconds))

    Volt per coil = 5 V

    Magnetic Flux (I want to use Neodymium magnets), so 1.4 Tesla,. and I will divide this by 4 to play safe, giving me 0.35 Tesla.

    I propose a magnet size of Width (0.15 m) and Height (0.1 m), giving me an area of 0.015 m2.

    Turns per second will be 3, so time will be 0.333333333 s.

    Using the above formula, I will need N = -1 * (-5/((0.35*0.015)/0.333333333))) = 317.4603175 windings.

    This seems a bit different from the formula you supplied, so is this one correct for me to try and find my windings?

    Or if I try yours:


    2 * 3 revolutions * 22 magnetic poles * 317 windings * 0.35 Tesla * cross sectional area (here I am stuck again, I tried to look at http://en.wikipedia.org/wiki/Cross_section_(geometry [Broken]) but it makes my head spin).

    On another point, am I right in saying that if I reduce my resistance, I will need to increase my EMF per coil to achieve my 2.5 kW power rating?
    Last edited by a moderator: May 6, 2017
  13. Nov 21, 2012 #12
    The formulas are about the same, but using the web site's formula, you need to factor in the number of magnetic poles into "seconds". This is because the more poles you have, the more frequently the field alternates. Note it is number of "poles", which is twice the number of magnets.

    "Cross-section" is the same as "area". I think your coils will be roughly rectangular or perhaps trapezoidal in shape, and it is easy to compute the area of these shapes.

    Reducing resistance, you increase the power. The formula is ## P = \frac {E^2} {R} ##. Note this is max power, which is achieved when the generator's output is shorted.
  14. Nov 21, 2012 #13
    To the untrained man this can get complicated very quickly!

    OK, so using your formula, I need to adjust it to:

    2 * 3 revolutions * 22 magnetic poles (is this right?) * 317 windings * 0.35 Tesla * 0.015 m2 (approximate) = 219.681 (this is volt right?).

    As per the above, am I right in saying total EMF is 219.681, divided by 3 revolutions, divided by 22 magnetic poles, gives me 3.318 V per coil?
  15. Nov 21, 2012 #14
    You have been warned, you need to get a book on some theory!

    First of all, is that 22 or 24 magnets? You started with 24. I think this is the correct number for a three-phase generator with 18 coils.

    Second, 24 magnets have 48 poles - this is the number you should plug into the formula.

    Lastly, 0.015 sq. meters as a coil cross-section is suspicious. This is the area of a square with about 0.12 m a side; you have 18 coils, and you must have gaps between them; so 0.12 m x 35 = 4.3 m; this is the circumference of your stator, meaning the diameter of the generator is greater than 1 meter.

    I would assume you want your stator about half a meter wide, then its circumference will be 1.6 meters, dividing by 35 we get 0.05 m the coil diameter, and 0.0025 sq. m its area.

    So: 2 * 3 Hz * 48 poles * 317 turns * 0.35 Tesla * 0.0025 sq. m = 80 V per coil. That's 16 times greater than you need, so you need to reduce the number of turns accordingly.
  16. Nov 21, 2012 #15
    By the way, assuming your coils are square with 0.05 m a side, and 20 turns per coil, you get 4 m of wire per coil, and 72 m total wiring. That has to be less than 3 Ohm, so that means the wire must have no more than 0.04 Ohms per meter. Assuming you are in the States, that means gauge 20 copper wire.
  17. Nov 21, 2012 #16
    Will see what I can do with a book, however working with you is great as you give real time solutions, and if I get stuck you help.

    I see where I went wrong. From the picture in my first post, at the top I have two North poles next to each other, and at the bottom I have 2 South poles, so I assumed that these wont induce an EMF as moving 2 north poles over the coil, only 1 would induce an EMF. For my cross section, I just took the area of my magnet, again a bit wrong there!

    Awesome, another step closer. I have adjusted the formulation to 2 * 3 Hz * 48 poles * 20 turns * 0.35 Tesla * 0.0025 sq. m = 5.04 V.
  18. Nov 21, 2012 #17
    Are your magnets really that huge? 0.1 m x 0.15 m? That's a tad bigger than two iPhone's side by side!
  19. Nov 21, 2012 #18
    Great minds think alike! You replied just as I typed mine. I am based in the UK, but will check USA gauge and find one similar this side. Now I have my coils, windings and size, pretty chuffed, thank you!

    Last one, now I need the torque. As per your comments "Well, this is actually quite simple. The mechanical power fed to the generator must be somewhat greater than power produced by it. P=τ⋅ω, where P is power, τ is torque, and ω is angular velocity. Given the required power and frequency, you can obtain the torque.", P is 2.5 kW, right? Struggling a bit to get my head around angular velocity. "Mechanical frequency multiplied by 2 pi. For 3 Hz, that is about 20 rad/second." What will my angular velocity be?
  20. Nov 21, 2012 #19
    Magnets are still in the development stage. Wow, your right, two iPhones together might be a bit much, will adjust the settings to see if I can get a smaller area with more turns?
  21. Nov 21, 2012 #20
    voko, I think I just confused myself again. Lets say I get Neodymium magnets, magnetized to 1.4 Tesla. If I change my magnet size to 8 cm x 10 cm as an example, what Tesla can I use in my formulation? Should I use 0.08 m x 0.1 m = 0.008 sq. m x 1.4 Tesla, or have I got the wrong end of the stick again?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook