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Why won't a flywheel with a 12V DC motor producing 11V for 30 seconds run through a DC to AC inverter that requires 10V? Answered

I am doing a science fair project that involves a solar panel, connected to a charge controller, connected to a 12V battery, connected to a switch system that can either go to a flywheel with a 12V DC motor or straight to a DC to AC inverter which leads to light standard light bulbs. The system worked (light bulbs lit) when I connected the panel to the controller to the battery to the inverter. However, when I had the battery power the flywheel and then have the flywheel output its energy to the inverter, the inverter shut down and would not work (red alarm went off). The flywheel is producing more than 10V (starts at about 11V) for at least 30 seconds, and the inverter's low voltage shut down in 10.5 plus or minus 5 volts. So why doesn't it work? Is the inverter messed up or is the flywheel outputing AC power? Also, when I connected the solar panel to the charge controller to the flywheel, the charge controller completely shut off. Are charge controllers not able to be directly linked to motors? And if I tried to directly connect the solar panel (which has 20V) to the flywheel (with a 12V motor), without a charge controller in between, what would happen?


Jack A Lopez

8 years ago

I think this idea will work provided you have a beefy enough DC motor, and a flywheel with rotational kinetic energy that is at least as large as the amount of work you want it to do.

I think the calculations for energy stored in the flywheel is where you want to start, because that's something that puts a hard limit on your energy storage. Equations and examples can be found here:

Anyway, calculate U= (1/2)*J*ω2, then compare that number to the power*time product for the load you want to run.

Regarding the motor's beefy-ness, what this really refers to is the  copper resistance, R, in ohms, of the winding of your motor.  The lower this resistance, the more beefy the motor.  There are a couple of ways to measure this.  You can put an ohmmeter across the terminals while the motor is stopped.  There's a little bit of complication in that you go through a commutator to measure this R.  But I am assuming that each "switched in" winding has the same R.  There might even be little gaps in the commutator, angles at which no winding is switched in, and the ohmmeter would read "open", or infinite resistance. But of course you can just ignore the gaps.

Also you can spin the motor up to some known DC voltage, Vmax, then short the spinning motor through a DC ammeter and try to measure the current that flows.  The initial, peak current, should be Imax = Vmax/R, and you can find R that way.

Another way of saying this is that I think you can model your motor as a voltage source in series with a resistor.  The voltage of this voltage source is proportional to the speed ω at which the motor is turning.  Say V= K*ω.  And the speed ω at which the motor turns depend on how much energy U= (1/2)*J*ω2 the flywheel has left.  Part of your stored energy is going to get burned up by that internal resistance R. That's unavoidable, and a big part of the reason why you want R to be small.

In terms of showmanship, if you can't get the inverter (and its load) to run from just the flywheel, then try an easier load, like just a 12 V lamp, or a resistor in series with an LED, or something.    That way you have something to show the nice people.


8 years ago

. No need to ask the same question (at least) three times. Please delete all but one.