Kallog
the below system and its energy input and energy output (recovery) is opperating
durring a time frame of 1 minute.
if Im burning 1800 lpm at the top of the system.
and generating 1800 lpm at the bottom of the system
then I am putting 1 lpm of water in the pipe at the top each minute.
Michael Faraday has shown that 1 litre of HHO gas is produced
in one hour using 2.34 watts for a period of 1 hour.
2.34 watts X 1 hour = 2.34 watt hours each litre of HHO.
1800 x 2.34 watt hours = 4212 watt hours
if I use the 4212 watt hours in one minute this is my energy input
---------------------- ENERGY IN PER MINUTE --------------------------
252720 watts / 252720 joules / 4212 watt hours
---------------------- POWER IN PER MINUTE -------------------------
252720 watts / 252.7 kw
-------------------------------------------------------------------------
that part is now out of the way !!!!! we know our input energy.
---------- NOW FOR THE GRAVITY PART ------------------------------------
1 litre of water weighs 2.205 lbs.
the volume of 1 litre of water = 61.02374 cu/inches
if Im using a water pipe that has a inside cross sectional area of 61 sq/inches
then the water level in the water pipe will only drop 1 inch every minute because
I am only allowing 1 litre of water to pass through the hydraulic piston at the
bottom each minute , however I am putting back 1 litre of water in the top each minute.
to keep this simple as I like to do , we know we only have 1 litre a minute to work with
and I have to recover 4212 watt hours each minute to balance the recovery system out
(energy in = energy recovered) LOL.
so I will use a piston at the bottom and stroke it out once each minute.
if the piston stroke is 12 inches then the piston surface area will need to be
61.02374 / 12 = 5.002
5.002 sq/inches
this way every minute when it strokes it will remove 1 litre of water
from the water pipe at the bottom.
and when the piston strokes it will provide a force that can be applied over a 12 inch distance.
if the water pipe is 1000 ft tall , the water pressure at the bottom of the water pipe
will be 1000 X .433 psi = 433 psi
433 psi X 5.002 piston area = 2165 lbf applied to the piston
and applied for a distance of 12 inches or 1 ft in 1 minute = 2165 ft-lbf/minute
1 watt = 44.253 ft-lbf/minute.
2165 / 44.253 = 48.99
so if I apply the 2165 lbf over the 12 inches distance in one minute
I would recover a constant 48.99 watts during that one minute.
since my energy recovery system recovers 48.99 watts per 1000 ft in height
the ratio of energy recovery per foot in height would be
48.99 watts / 1000 ft = 0.0489 watts per ft in height.
I need to recover 252700 watts each minute as that is what it cost to produce the HHO.
so
252700 / 0.0489 = 5167689.16 ft high.
or 978.72 miles
since the energy out (recovery) is 252 kW/minute the energy that can be produced at the generator would be apx the same.
but the above is a way to produce more energy than you need , it would be much shorter if you were
focussing on just getting back the 22% you were loosing in normal water - HHO conversion.
and if its only 10% like you say , then thats even better.
and of course not everybody needs a 252 kW/minute power system.
the ISS is apx 199 miles high.
so that puts it right in the 22% range.
if I build a system this tall ((( ISS ))) ((( MARS ))) hint hint nanocarbons maybe !!!
I have recovered all of the energy I used to generate the 1800 lpm HHO
plus I still have all the energy that I produced at the top using the 1800 lpm HHO as a fuel in the power plant.
and if there are any inefficientcies found , just add a foot or so in height to compensate.
Im sure all of the stuff you write is important to you , and I do read it all , and you have found many
interesting obstructions that need to be considered , I just only reply to the stuff that you write that
applies to the discussion , I try not to focus on the energy conservation beliefs you have that you have
been taught and the what ifs that wont really accomplish anything.
the important thing about this is that it shows that the current usage of thermodynamics is a load of CR@P.
.
