I am trying to make a generator and before making it, I would need to know how many kWh it generates, in order to make sure it is actually feasible or if I should redesign it. I am just a highschooler however, and I don't really understand how to estimate how much energy production this would cause. My design is an axial flux turbine (several groups of copper coils in a circle, and then a powered wheels on both side, both moving at the same speed. The wheels have neodymium magnets, which alternate in poles, inducing an alternating current).

This is the information I know:

  • Type of magnet (neodymium)
  • Amount of time before it switches from north pole to south
  • distance between magnets and copper
  • length of wire in the copper coil
  • number of copper coils
  • number of magnets

From here, I need to calculate voltage and wattage in terms of the above factors. Is there a formula to calculate for each? Do I need more information?


EDIT: I cannot use input energy to calculate this.

  • 1
    \$\begingroup\$ kWh is a unit of energy, not power. Your generator could generate any number of kWh given enough time. \$\endgroup\$ – DKNguyen May 3 '20 at 4:40
  • \$\begingroup\$ @DKNguyen Yes, i meant in for that one how long it would take, but I guess I can figure that out with wattage. Thank you for feed back, I edited the question to clarify it. \$\endgroup\$ – Ankit May 3 '20 at 4:45
  • \$\begingroup\$ What is powering the "powered wheels"? \$\endgroup\$ – The Photon May 3 '20 at 5:12
  • \$\begingroup\$ @ThePhoton A mixture of several things, including wind, human power, momentum and a few others. I'm a bit hesatant to post the exact details online as this is something new I have created, and I have not yet patented the IP. This is why I am asking to get it from those factors rather than using horsepower input. \$\endgroup\$ – Ankit May 3 '20 at 5:53
  • \$\begingroup\$ What you are describing is an Axial Flux generator. For plans to build one, in the 0.5 to 1 kW (not kWh) range, see this answer electronics.stackexchange.com/questions/327442/… . If you don't copy that design, the information in those plans- and on that website may help calculate the power available from yours. \$\endgroup\$ – user_1818839 May 3 '20 at 14:58

The drawing below shows the basic construction of an axial-flux, permanent-magnet, synchronous generator.

In order to calculate how much power can be generated by permanent-magnet generator, it is necessary to have the following:

  1. Detailed magnetic strength specifications and dimensions of the magnets.

  2. The generator construction and geometry details.

  3. The rotor iron dimensions and magnetic characteristics.

From the construction details, determine the length of the path of the magnetic field. From the path of the magnetic field and the magnetic characteristics of the materials, determine the strength of the magnetic field in the air gap.

For any instant in time, the voltage can be calculated in each segment of wire that the magnetic field passes through from E = v x B x L, where: E is the electromotive force (voltage); v is the velocity of the field; B is the strength of the magnetic field; and L is the length of the segment of wire. For any segment of wire that is not at a right angle to the direction of field motion, multiply by the sine of the angle of the wire with respect to the direction of motion.

To get the total voltage a coil at any instant in time, add all of the voltages of the wire segments. Then determine the waveform from the calculations for an appropriate number of instants in time.

To determine the output voltage of the generator, add the coil voltages considering the scheme selected for connecting the coils together.

There will be some voltage drop and power loss due to the wire resistance. Since the generator will have an AC output, there will be some voltage drop in the inductance of the coils.

The amount current that can safely be drawn will be determined by the losses, the rate at which heat is rejected to the surrounding air and the temperature rating of the winding insulation.

The maximum amount of power that can be generated will be determine by the voltage and current and the available input mechanical power from the system driving the generator.

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  • \$\begingroup\$ Thanks! This is helpful \$\endgroup\$ – Ankit May 5 '20 at 21:02

If your energy input is 550 foot * pounds per seconds, which for example would be 55 pounds on a 10 foot lever, or 275 pounds on a 2' lever (crank on the generator shaft), then you are providing 1 Horsepower of energy rate.

That is also 746 watts.

I can hike uphill at high elevation (cool climate), with about 3" elevation increase every second (the path is a moderate slope in fir forests). With my weight, that is about 1/10 horsepower. I can do this for hours.

The guy who pedaled that ultra-lightweight plane across the English Channel could provide 1/2 to 1Horsepower for those required minutes. He trained a lot.

  • \$\begingroup\$ Hi, thanks for writing that but i'm not sure how it answers my question. \$\endgroup\$ – Ankit May 3 '20 at 4:55

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