Flywheel Diameter: Calculating Capacity for Energy Storage

[moo]

Hey all,

I realize that when weight is moved from the center of a flywheel toward the outer edge (or a larger diameter wheel is used), the capacity for energy storage increases. But I have no idea how to calculate how much that increase actually is, or whether the increase remains constant (proportionately) between the two as RPMs increase.

Anyone have a formula handy for this? Hopefully with enough footnotes for a "non-engineer" who doesn't know many standardized formula abbreviations...

Thanks, moo
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moo (moo') adj. Of no practical importance; irrelevant, such as a moo point (i.e. a cow's opinion).

 

[berkeman]

Welcome to PF. wikipedia.org is often a good site for these kinds of questions:

http://en.wikipedia.org/wiki/Rotational_energy

 

[brewnog]

Sure, a common problem.

For a solid disc:

I = mk^2 = (mr^2) / 2 = (p*pi**b*r^4) / 2

Where I is the moment of inertia of your wheel, m its mass, p the material's density, b the disc's thickness, r its radius, and k the radius of gyration.

k = r / (sqrt 2)

Now, energy stored:

E = (Iw^2)/2

Where w is the angular velocity of your wheel, (2*pi*N) where N is the number of revolutions per second.

Sorry about the lack of fancy text, write it out on paper!

For non-disc-like flywheels, you will easily be able to find I geometrically; if you struggle give us a shout or have a quick look in a mechanics book.

You following?

 

[moo]

Thanks guys, that should do the trick.

moo
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moo (moo') adj. Of no practical importance; irrelevant, such as a moo point (i.e. a cow's opinion).