Flywheel Energy Storage: Difference between revisions
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(Created page with "On flywheel: assume a 1 meter radius for simplicity, a flywheel in the limit of all mass on rim. Say 1000 kg wheel. E=1/2MV^2 - say it's spinning 2000 RPM = 33 rps (achievabl...") |
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=Back of the Envelope Calculation= | |||
On flywheel: assume a 1 meter radius for simplicity, a flywheel in the limit of all mass on rim. Say 1000 kg wheel. E=1/2MV^2 - say it's spinning 2000 RPM = 33 rps (achievable readily) - then you have v=209 m/s so | On flywheel: assume a 1 meter radius for simplicity, a flywheel in the limit of all mass on rim. Say 1000 kg wheel. E=1/2MV^2 - say it's spinning 2000 RPM = 33 rps (achievable readily) - then you have v=209 m/s so | ||
Revision as of 22:12, 16 August 2014
Back of the Envelope Calculation
On flywheel: assume a 1 meter radius for simplicity, a flywheel in the limit of all mass on rim. Say 1000 kg wheel. E=1/2MV^2 - say it's spinning 2000 RPM = 33 rps (achievable readily) - then you have v=209 m/s so
E= 1/2 *1000 * 40,000 = 20 megajoules = 20 megawatt seconds or driving your 5 kW generator for 4000 seconds -
That appears to be 5kW for one hour. Amazing! Take out inefficiencies, and you probably have 5kW for 1/2 hour. 90% generator eff, 90% transmission efficiency, 90% extraction efficiency, 10% power loss from friction - so 40% loss right there in rough estimate.
But, can we do a 1000kg wheel with mass-on-rim at 1 meter? That is huge. And how to drive it? With continuously variable transmission?
MJ