A recent google answer piqued my curiosity. How far would a volume of
water have to drop, beit over a long spiral incline or surface area of
a turbine, for it to yield enough power to electrolyze it at the
bottom to hydrogen and oxygen?

http://answers.google.com/answers/threadview?id=189708 discused the
thought-process and math of turbines.

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Clarification of Question by scotttygett-ga on 02 May 2004 01:36 PDT

Endash, you're not clickable, does that mean I cannot pay you or just
that you don't have a profile?

I also noticed an odd thing about your first calculation, so to be
fair, I think I should either ask if you know anything about turbines
or wait to make sure someone else doesn't post that you're off by 400
miles and that turbines at Hoover get the equivalent out of 400
feet... I respect that you've put together a nice answer otherwise
though, so unless it turns out to be heinously wrong, I look forward
to rewarding your work.

Scottygett-ga,

Falling objects do not increase their speed indefinitely, no matter
how great the distance over which they fall.  The final speed they
achieve is known as the "terminal velocity".  On Earth, the
acceleration of gravity -- which would tend to constantly increase the
speed at which the water fell -- would be counteracted by friction of
the water rubbing against air, and against the surface of the incline.
 The ater soon reaches its terminal velocity, and will not go any
faster.

Bottom line -- I don't believe the water would never accelerate to a
speed to produce enough energy to self-electrolyze a large volume of
its own mass.  I'll leave it to a more adept researcher, however, to
do the math and/or to work out the specific details.

Make that:

I don't believe the water would EVER accelerate...

pafalafa is correct that under most circumstances, falling object
achieve a "terminal velocity."  However, this is only relevent if you
want to convert the water to hydrogen and oxygen based on "energy of
impact" at the bottom of the fall.  If instead you are concerned with
the amount of energy that could be extracted from the water as it
falls (perhaps by a series of waterwheels or turbines, for example)
then this can be calculated from the height of the fall and the mass
of the water.  Since the energy to split water also depends on mass,
it should be possible to calculate how far a unit of water would need
to fall for its change in potential energy to be equal to its chemical
energy.  I don't have the right reference books handy, but I think it
would be a pretty straightforward calculation.

If you want to know how high you have to lift a mass of water so that
the energy required to lift it is equal to the energy required to
split it into Hydrogen and Oxygen.

According to An Introduction to Combustion, by Turns, the enthalpy of
formation of water (i.e., the energy required turn hydrogen and oxygen
to water) at 298 Kelvin is -241,845kJ/kmol. (Negative because energy
is released when water is formed). A kmol of water has a mass of
18.016kg, so the enthalpy of formation is equivalently  -31424kJ/kg.

The enthalpy of formation in theory is equal (except for the sign) to
the energy required to split the watter back into it's elements.
According to Wikipedia, <a
href="http://en.wikipedia.org/wiki/Electrolysis">electrolysis can be
nearly 100% efficient.</a>

So now the question is, how can 31424kJ lift one kg?

Well, 1J = 1N*m. That is, pushing with a force of one Newton for one
meter requires one Joule of energy.

One Newton equals 1kg * m/s^2. Gravitational acceleration is of course
9.8m/s^2, so 1kg weighs 9.8N. (Actually, 9.80665N, according to my
TI-85 Calculator.)

So now we have:

31,424,000J = 9.8N * x

We just solve for x.

x = 1369796 J/N

By definition, 1J/N = 1m, so your answer:

1,369,796m = 1,370km = 851 miles.

Of course, at that distance above earth, acceleration due to gravity
will not be 9.8m/s^2. It makes sense since the space shuttle uses
hydrogen powered rockets

From Fundamentals of Pysics by Halliday, Resnick, and Walker, we have
gravitational potential energy: U=-GMm/r, where M and m are the masses
of the two bodies (earth and our killogram of water), r is the
distance between their centers, and G is the gravitational constant,
6.67x10^-11 N*m^2/kg^2.
The radius of earth (r on the ground) is 6.37x10^10m and the mass of
earth is 5.98x10^24kg.

Then we have 31,424,000J = G*M*1kg/r - G*M*1kg/(r+x) and solve for x.
I won't do all the algebra out, but By that works out to x =
6,412,313m, substantially more than the 1,369,796m from the initial
calculation.

Interestingly, this result seems to suggest that a hydrogen rocket
cannot escape earth's gravity.

Okay, take the money. No one else seems to be stepping up to the plate.

Does anyone else here notice that I'm REALLY used to being able to
edit after the fact? Sorry. Thanks for the nice numbers.