Flywheel/Dynamo behavior under variable load

Question

I have a special effects project in mind where I simulate a bellows and coal fire without, you know, actual fire.

My vision thusfar involves rigging a flywheel to a short-stroke lever (the upper arm of a modified bellows), and a bicycle dynamo (or low power alternator, if I can find one for free). This question is how the generator will act on the rotor connecting it to the flywheel?

Obviously a spinning flywheel without any power being put into it will suffer from two sources of friction loss, the axle's losses to the bearing, and the drag from the surface of the flywheel interacting with the air around it.

My expectation is that whatever load the generator is experiencing is also being felt to the rotor as a sort of drag.

I see three possible interactions here, but don't know which one I'm likely to end up with (or if it's another I haven't even thought of yet):

1. Any power in the flywheel is being fully tapped by the dynamo, meaning that regardless of load or rpm the flywheel is being 'rapidly' decelerated by the dynamo; as a consequence, someone going ham on the lever means you've got a metric shit-ton of excess power you need to sink somewhere or you're gonna get a real fire anyway.

2. The dynamo only takes power off the flywheel to match the load it's under, meaning that the flywheel will only lose power to friction otherwise, and you can control the deceleration of the flywheel by adding or removing load.

3. The dynamo only lets power out to the load its connected to as needed, but takes power from the flywheel as fast as it can, leading to some kind of thermal failure in the dynamo itself.

All of these can be designed around, obviously, but I'm not clear on what problem I'm going to need to be confronting.

The load in this case is negligible: an LED array with more strings being switched on as more power is available to run them.

The desired behavior of the LED array is that a brief period of manual input to the flywheel results in a rapid ramp-up of light from the 'coal bed' which then tapers off much more slowly. Too slowly is better than too fast: I can always add drag to the flywheel.

So, to reiterate the question: what is the behavior of a flywheel/dynamo system in relation to the electrical load being served? What force does the rotor experience as a result of the dynamo's presence? Does that change as the load changes?

Answer 1

what is the behaviour of a flywheel/dynamo system in relation to the electrical load being served?

Assuming 100% power efficiency in the dynamo, whatever electrical power is being removed from the dynamo to drive an electrical load is seen as a mechanical power burden on the flywheel and that of course will gradually slow down the flywheel.

That mechanical power burden equals the electrical power taken from the dynamo and, as the speed slows, if the electrical power take-off remains constant then more load torque is felt by the flywheel and this might rapidly bring it to a halt.

What force does the rotor experience as a result of the dynamo's presence?

$$\text{Power} = 2\pi n T$$

Where T is torque and n is the revolutions per second of the rotor. So, the force it experiences is a torque proportional to the electrical take off power (plus mechanical losses).

Does that change as the load changes?

It has to change else you would have free energy for nothing.