Horrors -- You want numbers! I don't do numbers! ;-)
But I'll try. Be warned that I'll probably get this wrong.
First of all, saying it's a "1000 pound flywheel" doesn't tell us much.
The moment of inertia of the flywheel is what matters, and that
depends on the shape, not just the mass. So we need to start with a
guess, which is how big the flywheel is and what it's made of. I'll
assume it's made of iron, and is a uniform disk 2 inches thick.
Iron is about 491 pounds/ft^3 according to some web page, so we have
about 2 cubic feet.
If it's a uniform disk 2 inches in thickness, and it has radius r, then
we have
(2/12) * pi * r^2 = 2 cubic feet
or
r = sqrt(12/pi) = 2 feet
So if the flywheel is uniformly 2 inches thick and made of iron, it's
about 4 feet across. We'll take r=2 for our working value.
Now, I don't carry the moments of inertia of various shapes around in my
head, so I'll have to derive them (sigh...). First of all, angular
momentum of a point mass "m", a distance "r" from the point where we're
measuring it (call that the origin), is
L = m * r * v(perpendicular)
and the perpendicular velocity of a point mass with rotational velocity
'w' about the origin is
v(perpendicular) = w * r
so the angular momentum of a disk of uniform density rho (mass per
area), rotating at rate "w", is
L(disk) = integral((w * x) * x * (rho * 2pi*x * dx)) from 0 -> r
= 1/2 w pi rho r^4
But the mass of the disk is pi r^2 rho, so that's
L(disk) = M r^2 w / 2
We can also tell from that that the disk's moment of inertia is
I(disk) = L/w = M r^2 / 2
and from that we have the energy of the disk,
E(disk) = 1/2 I w^2 = M r^2 w^2 / 4
which we can also obtain by direct integration of
E(disk) = integral(1/2 rho 2pi x dx x^2 w^2) from 0 -> r
but I've already belabored the point enough I think.
Now, the power is energy divided by the time to spin it up, or
power = E(disk)/T = M r^2 w^2 / 4T
and the torque is angular momentum over time, or
torque = L(disk)/T = M r^2 w / 2T
To plug in numbers we need to know a little more; from Wikipedia:
1 hp = 746 watts = 33,000 ft-lb/minute
Rotational velocity at 100 RPM is 100 * 2pi = 628 radians/minute (and
the dimensions of that are 1/minutes, btw; "radians" are dimensionless). So
power = (1000 * 4 * 394384 / 60) / 33,000 = 797 horsepower
and I don't believe that result, not for a minute. It takes a 700 horse
engine to spin up a 1000 pound flywheel to 100 RPM in 15 minutes? No
way! Oh, well, I said I'd get it wrong -- checking my numbers didn't
turn anything up; maybe somebody better connected with the
pounds-shillings-and-ounces end of mechanics can see where I blew it,
assuming, of course, that the result really is wrong.
Maybe I need some extra conversion factors in there somewhere in order
to use pounds as force? (Maybe I should have used slugs?) hmmm....
Anyhow, moving right along,
torque = 1000 * 4 * 628 / 30 = 84,000 foot-pounds
and I don't believe that result either. That's a stinkin' enormous
amount of torque.
What's more, the torque and power can't actually both be constant during
the "spin-up". If the torque is a constant value 'tau' then the power
actually varies from 0 (at 0 rpm) to a maximum of w * tau, which in this
case would be
84000 * 628 / 33000 = 1600 hp
with the average power in being 797 hp.
Regardless of the implausibility of the results, note that the result is
very heavily dependent on the size of the flywheel. If the flywheel is
4" thick rather than 2" thick, then its radius is 1.4 feet, and average
power drops almost in half, to 390 hp (which is still pretty absurd).
OrionWorks wrote:
I'd like to approach this from a different angle.
It obviously takes a long time to get Newman's flywheel to rev up to
around 100 RPM. It's also true that even with a tiny torque force
applied, given enough time (and minimal friction to contend with) any
heavy flywheel can eventually be made to spin at impressive speeds. In
this case we are told we are dealing with a 1000+ pound flywheel. IOW,
a substantial amount of inertia for which applied torque must
s-l-o-w-l-y work against to accelerate the flywheel.
Anyone care to give a reasonable guestimate as to how much torque (as
expressed in Newton Meters) is probably being demonstrated in the
turning Newman's 1000+ pound flywheel? Newman, in his google video,
states at one point there's a LOT of torque being generated, but he
gives no specific amount. We are left to guess.
Unlike me, Newman is in a position to determine exactly how much torque
is being generated. He knows all the little things we're guessing
about, like the shape and density of the flywheel, the time it actually
took to spin up the wheel, the actual rotational velocity of the
wheel... but he's apparently not saying.
Instead he says stuff like "Speed is power".
Hmmmm....
Stephen, what's the mathematical formula that would best express how
much torque is likely being generated in order to speed up Newman's
1000 pound flywheel from a dead stop to say, 100 RPM in 15 minutes?
Seems to me that if we knew how much torque is actually being
generated it would give us a much better sense as to how much actual
useful work is being applied. Seems to me that if we knew the actual
amount of Newton Meters that are being generated it would make it a
lot easier to figure out if the same amount of power would be
effectively applied to small water pump, or a large fan.
Regards,
Steven Vincent Johnson
www.OrionWorks.com
