How does mechanical torque affect generator volts and amps?

[Thomas Conway]

Hello! today I have a question regarding the most efficient way to generate electricity by moving a magnet across a coil. The theoretical magnet would be moving in a circular motion fixed to the outer circumference of a driver wheel like in many simple generators, how does the physical torque and speed(rpm) of the flywheel affect the voltage and amperage of the outputting electricity? From what I can gather it seems the only important factors in generating current is the speed of fluctuations in the magnetic field (directly caused by the velocity of the wheel) and the number of windings in the coil. if torque power has no relation to magnetic fields theoretically could you not create a step-down gear ratio system making the output wheel spin much faster than the driving wheel at the cost of non-relevant torque which would increase the wattage/ efficiency overall? or am I missing something?? The only other outcomes I imagine is that torque somehow directly relates to outputting amperage so by improving the one you sacrifice the other making no difference in total power like a transformer? Or perhaps if the torque is too low and the rpm too high the resulting eddy currents generated will create enough friction to overcome the speedy but weak mechanical motion of the output wheel forcing it to halt. If anyone can give me a clean answer it would be much appreciated. I apologize if I was confusing or non-technical with my wording. Thank you for reading!

 

[Andrew Mason]

From what I can gather it seems the only important factors in generating current is the speed of fluctuations in the magnetic field (directly caused by the velocity of the wheel) and the number of windings in the coil. if torque power has no relation to magnetic fields theoretically could you not create a step-down gear ratio system making the output wheel spin much faster than the driving wheel at the cost of non-relevant torque which would increase the wattage/ efficiency overall? or am I missing something??

Ignoring mechanical resistance to turning the armature in the armature and coils, the torque power required to keep the armature turning at constant speed (τω) is entirely related to the electro-magnetic forces between the field magnets and the armature. If one increases torque to the generator, the generator armature undergoes angular acceleration, increasing the angular speed of the armature, which produces a proportionately higher current, which in turn produces more opposing electro-magnetic force between the field magnets and the armature opposing the angular acceleration of the armature. When the speed of the armature reaches the point at which the opposing torque is equal to the applied torque, the armature stops accelerating.

The only way to make the generator more efficient is to minimize the things that cause heat losses (eg. friction, eddy currents, hysteresis, electrical resistance).

AM

 

[GrahamN-UK]

How does the physical torque and speed(rpm) of the flywheel affect the voltage and amperage of the outputting electricity?

From what I can gather it seems the only important factors in generating current is the speed of fluctuations in the magnetic field (directly caused by the velocity of the wheel) and the number of windings in the coil. if torque power has no relation to magnetic fields theoretically could you not create a step-down gear ratio system making the output wheel spin much faster than the driving wheel at the cost of non-relevant torque which would increase the wattage/ efficiency overall? or

Am I missing something??

Yes, you're missing the principle of conservation of energy. We can't create or destroy energy.

Hence your scheme for free energy from nowhere by adding gears (which is effectively what you are suggesting) fails, as all such schemes fail.

You sound as though you are already familiar with the principles involved. To sort out what's happening we just need to consider them in the correct order, which turns out to be backwards, from the electrical load back through the system.

For each step in the generating process the energy needs to be supplied by the previous step, along with any losses in each step.

We have a generator providing electrical energy to some electrical load. The generator is getting that energy, along with a bit extra for losses, as mechanical energy from the flywheel. In order to keep turning, the flywheel is presumably receiving mechanical energy from some source, usually known as a prime mover in electrical generation circles.

As you note, the voltage, V, generated in the generator is determined by the speed of the generator, and hence the speed of the attached flywheel. (There are some other factors, but we'll assume they are constant.)

The current, I, flowing in the electrical load is then determined by the generator voltage and the impedance, Z, of the load.

These are related by Ohm's Law, of course: I = V/Z Amps

The electrical power supplied to the load, P = V * I * PF Watts
(where PF = Power Factor of the load. This will be 1 for a resistive load or DC generation.)

If the voltage, V, increases, the power P will also increase because V appears in the formula for P. However the current, I, will also increase because V also appears in the Ohm's Law formula for I, hence the power increases again as I also appears in the formula for P.
We can combine these two formulae to show both effects:

P = V * I * PF = V * (V/Z) * PF = V^2 * PF/Z Watts.

If we keep the load constant the power factor and impedance will stay the same and the electrical power supplied will vary in proportion to the square of any changes in the voltage.

The generator will need mechanical power to drive it. Ignoring the small losses in the generator the required mechanical power will equal the electrical output power.

Mechanical power in a rotating system = torque * rotational speed * k
(where k is a constant dependent on the system of units you are using.)

So
torque * rotational speed * k = P = V * I * PF (Watts)

In a practical AC mains generation system we keep the rotational speed constant so the generated supply frequency is constant and we keep V constant (via a mechanism I haven't discussed). Taking out the constant parts from the above equation we get that:

torque is proportional to I.

Note that in a mains-style generation system the torque is determined by the load current, not the other way round.

So, what happens with your proposed gearing arrangement? If you add gears to spin the generator faster (whilst, we will assume, keeping the flywheel/prime mover running at constant speed) you will increase the generated voltage (in proportion to the speed of the generator). The power supplied to the load and hence the power demanded by the generator from the flywheel will increase, in proportion to the square of the ratio of the voltage increase. The torque demanded by the generator from the flywheel will increase to supply the extra power. No free energy will be created.