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Playing with the numbers for the capacitor size I think there is a flaw in the cap voltage. If you use 3 100 watt bulbs in parallel you have a resistance of 48 ohms and making approx 2.5 amps, assuming the generator is producing the 240 volts there will be a 120 volt drop across the resistance leaving 120 volts for the cap. In order to get the full 6Joule charge + you would need a cap in the range of 833uf. Is that correct?
The cap discharge into an inductor( motor windings ) would be in the range of 200+ amps according to the formula V*(C/L)^.5 . I measured 151uh on the motor I'm using so.. 120 volts * ( 833uf / 151uh )^.5 = 281 amps. The normal run with the motor provides aproximately 9.5 ft lbs, if it is enhanced to 3 times that then 28.5 ft lbs might be injected into the flywheel per pulse resulting in a 2.85 ft lb continuous assuming a 10% rotation or 1.42 ft lb at 5%. The required continuous torque to produce 300 watts including losses in the generator would be approximately .85 ft lbs which the 5% mark would clearly provide enough of a boost to maintain itself.
I'm machining a flywheel that is 6.75 inches in diameter and I've approximated the finished weight to be at or around 22 lbs. At 3000 rpm it should store 942 ft lbs per sec of energy ( about 1.6 hp ). Getting the energy into it may need a "pre starter" to bring it up to speed then flip some switches to start the motor/generator action.
One thing I seem to have confused myself on is the actual input requirement to the loads ( bulbs and cap ). Is it only 300 watts or is it much higher including both loads ?
I'm still a week away from getting all the parts made to assemble the test unit to begin the tweaking or balancing process.
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Zoloft Class Action
The cap discharge into an inductor( motor windings ) would be in the range of 200+ amps according to the formula V*(C/L)^.5 . I measured 151uh on the motor I'm using so.. 120 volts * ( 833uf / 151uh )^.5 = 281 amps. The normal run with the motor provides aproximately 9.5 ft lbs, if it is enhanced to 3 times that then 28.5 ft lbs might be injected into the flywheel per pulse resulting in a 2.85 ft lb continuous assuming a 10% rotation or 1.42 ft lb at 5%. The required continuous torque to produce 300 watts including losses in the generator would be approximately .85 ft lbs which the 5% mark would clearly provide enough of a boost to maintain itself.
I'm machining a flywheel that is 6.75 inches in diameter and I've approximated the finished weight to be at or around 22 lbs. At 3000 rpm it should store 942 ft lbs per sec of energy ( about 1.6 hp ). Getting the energy into it may need a "pre starter" to bring it up to speed then flip some switches to start the motor/generator action.
One thing I seem to have confused myself on is the actual input requirement to the loads ( bulbs and cap ). Is it only 300 watts or is it much higher including both loads ?
I'm still a week away from getting all the parts made to assemble the test unit to begin the tweaking or balancing process.
________
Zoloft Class Action
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