NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/sci/electron/21767

Internet Message Format | 1992-12-30 | 3.2 KB

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From: [e-mail] (John Whitmore)
Newsgroups: sci.electronics,sci.energy,rec.autos.tech
Subject: Re: Flywheel batteries as EV power source
Date: 30 Dec 1992 23:41:30 GMT
Organization: University of Washington, Seattle
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In article <[e-mail]> [e-mail] (Chuck Parsons) writes:

    {first point: hypothetical flywheel is more efficient than
a heat engine, approximately three times less energy required}

    Good point.  Not quite enough orders of magnitude
to please me (0.5 versus 2), but good point.

>  Second mechanical strength of materials is _not_ greater than the
>strength of the chemical bonds forming the material.
>Consider a thin hoop flywheel. Any flywheel can be thought of as being made
>up of many such hoops. Let the hoop have a fixed cross-sectional area A
>and the average mass of each atom in the material be M.
>
>  The energy stored in each atom of the hoop is E=.5 M*V**2.
>
>  When the hoop spins it tries to pull itself apart. It requires
>  on the order of .25*M*V**2/R  _per_ atom  of force to keep the loop intact.
>
>  The number of atoms in a loop is 2*pi*R/D where D is the spacing between
>atoms. Thus the force required is on the order of
>
>      .25*M*V**2/R  * 2*pi*R/D = .5*pi*M*V**2/D
>
>
>   Which is about 3*E/D! but if E is greater than about 1/3 the
>chemical binding energy then the molecules/material will come apart,
>because enrgy = force * distance
>
>   Since for really bitchin materials the heat of vaporization
>is similar the the binding energy, and because real materials have
>defects and fall apart well before the strain (or is it stress)
>aproaches the atomic limit. The flywheel will at _most_ have
>enough energy to vaporize itself

    Not a reasonable expectation, though, that in a FAULT
the failure will always deliver exactly equal strain energy
to all molecules in the whole danged flywheel.  A sensible
scenario is NOT going to be so obliging: a crack will
develop, a fragment will whizz off (and through the
containment), then the remainder of the rotor will shake
itself apart.  Not in molecule-sized pieces, but in large
ballistic chunks.  Maybe large enough to have a THOUSAND
molecules in a single fragment.

    That means that the energy absorption in bond
breaking will be roughly in the ratio of volume/area of
the one-atom and thousand-atom fragments.  This makes
90% of the energy of the rotor available to do damage.

    If you want to get back the energy in the rotor,
you'll have to argue why no chunks come off it, only
individual molecules.  And, if the chunks that come off
are LARGE, like bits of visible fluff, it's gonna
be multibillion atom fragments that one must figure on.
With a billion atoms in the chunk, the bond energy can
only take 0.1% of the energy you need to absorb.

    John Whitmore