NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/sci/electron/21709

Internet Message Format | 1992-12-29 | 3.3 KB

Xref: sparky sci.electronics:21709 sci.energy:6494 rec.autos:30700
Path: sparky!uunet!portal!cup.portal.com!lordSnooty
From: [e-mail] (Andrew - Palfreyman)
Newsgroups: sci.electronics,sci.energy,rec.autos
Subject: Re: Flywheel batteries as EV power source
Message-ID: <[e-mail]>
Date: Tue, 29 Dec 92 18:20:47 PST
Organization: The Portal System (TM)
Distribution: world
References: <[e-mail]> <[e-mail]>
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Richard Stead hypothesises that the specific energy (J/Kg) of
a practical flywheel will not be able to exceed the specific
energy of gasoline, when both the shielding mass and realistic
materials' stress characteristics are taken into account.
This is a key parameter in this whole discussion, independent
of the safety debate which has been raging here; i.e. if this
assertion is true, then flywheels look really unattractive in
any case, modulo political and environmental arguments of course.
Let's look at that.

Let the flywheel be of mass mf, density df, radius r, thickness f
and rotational speed (rad/sec) of w.
Its energy E = 0.5*mf*(r*w)^2 and mf = pi*df*f*r^2.

Model the shield as a sphere of radius r (more or less), density ds,
mass ms and thickness s, where ms = (4*pi/3)*ds*[ (r+s)^3 - r^3 ].

We need a relationship between s and E.
Generally we could write s = k*E^n, but let's intuit (probably gravely
wrong!) that, like viscous drag (m*d2x/dt2 = -a*dx/dt), s will need to
vary as k*sqrt(E), where k is determined by the fragments' maximum
expected mass (mx), the viscosity (a) of the shield material, and the
proportion (x) of the total energy (E) carried by a worst-case fragment.
Classically k= sqrt(2*mx*x)/a. So assume that s = k*sqrt(E).

The specific energy of the system comprising flywheel plus shielding is
A = E / (mf + ms). Since s is a function of E, there will be an optimal
A, which we find by partially differentiating A wrt E.
I find this to be at the critical flywheel mass

mfc = (2*pi*ds*k/3)*(k^2*E*sqrt(E) - 3*r^2/sqrt(E))

Hopefully, for the large E-values we're playing with, this is a positive
number. Assuming further that it's realisable, the substitution into A
gives
Amax = 1/( 4*pi*ds*k *(k^2*sqrt(E) + 2*k*r + 2*r^2/sqrt(E) - r^2/(E*sqrt(E)) )

Note that Amax depends inversely on k, which makes sense when you realise
that a small k-value translates into a thinner shield for a given E.
However, note also that Amax depends inversely on ds, which is naively
obvious, but disregards the relationship between k and ds for real
materials; one would expect ds to be inversely related to k. This implies
that there is a further optimisation on the shield material to be made.

If you've tagged along so far, I expect that you're expecting a conclusion.
Well, there ain't one :-). Although the specific energy for gasoline
burned in an IC engine is known, and one could estimate ds = 10^4 Kg/m^3,
I still don't know k, the shielding parameter with units metres/sqrt(Joule).
So let me know what you come up with yourselves!
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