What I would like to know is how to determine how much energy can be
extracted from a falling stream of steel balls. For example: If I had
a spout, with its opening at 3 meters above a "water" wheel (2 meters
in diameter), that was dropping 100 4Kg steel balls a second onto the
water wheel how much energy would the water wheel be pulling from the
stream. Would this be different from the impact force if the
steelballs hit a flat surface?

Assuming complete efficiency of energy transfer, the kinetic energy of
each ball will be 0.5mv^2.

The mass is 4kg.

The velocity, well, we have to creep up on that.

The distance an object falls is 0.5 at^2, or in this case: 3 = 0.5 x
9.81 x t^2, so t = 0.782 seconds.

It's speed will be at, that is 0.782 x 9.81 = 7.672 m/s (looking at
this, there was probably a single step formula - never mind)

Now, the kinetic energy is 0.5 x 4 x 7,672^2 = 117.7 Joules.  You are
dropping 100 per second, so that is 11772 J/s = 11.772 kW.

This is the maximum power that you could recover, but it depends on
perfect energy transfer and mechanical efficiency.

I think.  The method is OK, but I may have goofed in the calculations.

Hello,

   The maximum energy that could be extracted would be the kinetic
energies of all those falling balls. Each would, at contact with the
wheel, have kinetic energy amounting to

KE = 1/2Mv^2

KE = kinetic energy
M = mass
v = velocity

   The velocity would be equal to the local acceleration due to
gravity multiplied by the time of fall

v = at

a = acceleration
t = time


   Using the mks(meters,kilograms,seconds) system, and assuming that
the balls are dropping on Earth at sea level, then the acceleration =
9.8m/s^2, giving a velocity at capture of 29.4m/s.

   The conventional unit of energy is the Joule(J), given by a 1kg
mass at 1m/s. A 4kg ball at that velocity is 4J, and at 29.4m/s that
comes to 117.6J. The conventional unit of power is the Watt(W), which
equals one Joule per second. One hundred of your steel balls per
second equals 11,760W; a little more than one kilowatt. A typical
large nuclear power plant nowadays pumps out around one gigawatt, or
approximately one million times the wattage of your ball powered
wheel.

   Now of course, this is the energy being pumped into the wheel by
the falling spheres. The *useful* energy that comes out is necessarily
less due to heat loss in the system. There will be a small loss from
air resistance as the spheres fall. If the wheel's bearings exert a
lot of friction, then there will be considerable loss. The efficiency
can further be affected by the radius of the wheel, with larger radii
giving more torque for a given bearing friction. Without specifying
the total inefficiency in the system, you can't say how much useful
energy is extracted. But you can say that, in principle, such & such
quantity of energy is being transacted.

-Mark Martin

...excuse me, make that a little over eleven kilowatts.

-Mark Martin

Incidentally, the kinetic energy of a falling body is equal to its
potential energy across the distance of the fall, which is given by
the much simpler equation

PE = Mgh

PE = potential energy
M = mass
g = acceleration due to gravity
h = height of fall

Thus, a single 4kg sphere falling through a height of three meters,
accelerating at 9.8m/s^2 comes to 117.6J, same as in the calculation
for kinetic energy, but a lot more easily arrived at!

-Mark Martin

Yeah - that's it.  As soon as I saw 9.81 appearing twice in my
solution I knew there had to be a faster way.

I'll tell you what though - I wouldn't want to get in the way of a
machine that dropped 100 4kg steel balls every second!

"I wouldn't want to get in the way of a machine that dropped 100 4kg
steel balls every second!"

-And I'd hate to be the poor guy loading all those balls at the top!
My aching back! ;)

-Mark Martin

Hi Mark:

I'm a bit puzzled.  We both got the same answer, but I can't follow
your method.  Can you point what I'm missing please?

You calculate the terminal speed using v=at, same as me, but you don't
have the speed at that point.  You seem to have used the height
instead and ended up with 29.4 m/s versus my 7.67 m/s which I got
after I worked out the duration of the acceleration.

I'm also puzzled by your description of a J as being 1 kgm/s. If it's
energy, and calculated from mv^2, surely its dimensions have to be
kgm^2/s^2, don't they?

It looks like you know a few short cuts that never made it to
Xarqiland.  Can you share?

regards
xarqi

As exargi typed there are some problems getting an exact answer. I'll
use old units and assume the balls fall an average of 10 feet
including some dropping distance inside the spout and some initial
velocity. Mass = 2.2 times 400 = 880 pounds per second:  KE = 8800
foot pounds per second, minus the losses. That much  additional energy
is extracted from the balls as they ride the wheel downward to the
dump point of the water wheel 10 feet below the impact point, again
minus the losses. Some sound energy is produced as the ball clatters
against the wheel and each other. Some heat is produced. Air
resistance loss is small for the balls but may be considerable for the
wheel if it turns more than a few revolutions per minute. If 5500 foot
pounds per second of useful energy is extracted from the balls, that
is 10 horse power. Multiply by 746 to get watts = 7460 watts output.
If the wheel is more than about 4 meters in diameter the balls may
ride the wheel more than 10 feet average. I think the other comments
are correct. 8800 foot pounds = 16 Hoursepower = 11,936 watts which is
a bit more probably because my conversions are not exact.  Neil

Oops - after reading Neilzero's comment I realised I DID make a
blunder.  The total fall of the balls will be 5m not 3m as they will
ride the wheel down contributing energy all the way.

You'll have to multiply my earlier answer by 5/3 to correct it - makes
the power 19.62 kW.

xarqi,

You're right about that terminal velocity. That's a questionable habit
of mine. The numbers come out the same, so I just multiply the
acceleration by the distance. It's fast, but it's also kind of dirty,
and it doesn't really yield the velocity. As for Joules, the
dimensions are kgm^2/s^2. But that's one second squared divided by one
second squared, which just equals one kilogram at one meter per
second.

After I (finally!) went to bed late last night I realised like you &
neilzero that I ought to have mentioned the extra energy due to the
weight of the spheres riding the wheel part way around. But it also
occurred to me that there will be a phase relationship between the
"frequency of balls" and the frequency of capture. The balls are
undoubtedly being caught by some protrusions sticking out of the
wheel's spokes. If the wheel starts out at zero frequency, then at
first lots of balls' energies may be wasted because they scatter off
the already seated ones. And once it gets going, then the wheel
necessarily aquires some angular speed with subsequent captures. As it
speeds up the spokes may get out of phase with the stream of spheres,
and so some will arrive between spokes and will be missed,
contributing no energy to the rotation.

-Mark Martin

So much controversy!!!  The whole thing is impossible anyway!

If the rate of release is 100/s, then each ball has only 0.01s to get
out of the way.
In this time it will have fallen 0.5 x 9.91 x 0.01^2 m = 9.81 x 10^-4m
- a bit less than 1mm - sound right.
This means the maximum diameter of these spheres is 9.81 x 10^-4m
So their maximum volume is 4/3 pi (9.81 x 10^-4 / 2)^3 m^3 = 6.18 x
10^-11 m^3
If their mass is 4kg, their density must be at least 4/6.18^10^-11
kg/m^3, or 6.47 x 10^7 g/mL in the conventional units.

What you have here is *not* steel.  Anyone seen a neutron star around
here?

xarqi

'Scuse the typos. :-)

Thank You for all of the answers. To answer xarqi-ga point about it
being impossible: I never said the spout was a small round circle it
could have been a long slot :) Anyway, thanks everyone. I got the
formula's and the explanation on how to use them.

"What you have here is *not* steel.  Anyone seen a neutron star around
here?"

   Heh, heh- that's extra funny!

Hey myxlplix, I'm glad that we've been helpful to you. Do each of us
now get $5/3? JUST KIDDING!!!

I've got half a mind to work out just how wide that slot would have to
be given the density of steel.

Still - some people say I just have half a mind.

Glad it worked out.

xarqi

There is another efficiency to be considered.

Consider a second stream of balls that was falling *past* the turbine
without encountering it.  Upon reaching a point level with the bottom
of the turbine, each ball in that stream would have an energy of mgh =
(4)(10)(5) = 200 Joules.
The actual energy will be somewhat less than this because of air
friction, as has been pointed out.

(Note that I am using g = 10 m/s^2, and that the total height is the
three meters before the wheel plus the two meter diameter of the
wheel.)

Each steel ball represents a possible 200 Joules of energy that *can*
be extracted.  Some of that energy will be turned to heat, either by
friction in the turbine or Joule heating in the wires.  Also, impact
heating as the balls strike the turbine (I would recommend padding the
turbine blades.)

But the dominant factor in finding the efficiency of the turbine lies
depends on the exit speed of the steel balls.  If, upon exiting the
turbine, the steel ball is moving with a speed of, say, 5 meters per
second, then it has a kinetic energy of 0.5mv^2 = (0.5)(4)(5)^2 = 50
Joules.

Conservation of energy applies here: if the steel ball started out at
rest with 200 Joules of gravitational potential energy, and ended up
with 50 Joules of kinetic energy, then the maximum amount of energy
that it imparted to the turbine (heat losses included) is 150 Joules. 
Realize that my 50 Joule amount is *completely arbitary*--the steel
ball can have a final kinetic energy of anywhere between 0 and 200
Joules.

Consider a loadless turbine, one that is not having its rotational
kinetic energy converted into some other form and thus drained off. 
If you channel your steel balls down a sloping ramp so that the impact
occurs on the underside of the turbine, then the balls have a velocity
found by solving 0.5mv^2 = 200 Joules, thus v = 10 metres per second. 
The loadless turbine will quickly gain rotational kinetic energy until
the speed of its blades (located at a radius of one metre) matches the
speed of the balls.  (Incidentally, this leads to an angular frequency
of 10 radians per second, e.g. a frequency of about 100 revolutions
per minute--but that doesn't matter just yet.)

But the balls exiting the loadless turbine are *moving just as fast*
as the 'control' balls in the hypothetical second stream: *they are
imparting no energy at all to the turbine!*  This makes sense if you
consider that the energy of the turbine is not increasing over time;
it is simply spinning at the same speed.

It is easy to see that if you want to gain the maximum amount of
energy from the balls, you need them to be moving slowly when they
exit the turbine.  In order to do this, the turbine blades must have
as small a speed as possible; the angular frequency of the turbine
must be a minimum (nearly zero.)  This means making sure that the load
on the turbine is sufficiently great so that it never accumulates too
much excess speed.

I called this the 'dominant factor' in the efficiency of the turbine,
because it can easily cut your efficiency to zero percent: several
kiloWatts in and no power out.  But for a sufficiently slow turbine,
you can call the loss due to unclaimed kinetic energy nearly zero,
just like friction and air resistance losses.  Then your power would
be the K.E. of each ball times the number of balls per second, or
(200)(100) = 20,000 kiloWatts.