Re: Free Energy
- From: "Gieniu" <warendag@xxxxxxxxx>
- Date: Sat, 02 Jun 2007 00:57:10 GMT
"Gieniu" <warendag@xxxxxxxxx> wrote in message
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"The Ghost In The Machine" <ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote in
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In sci.physics, Gieniu************************
<warendag@xxxxxxxxx>
wrote
on Fri, 25 May 2007 05:50:00 GMT
<cOu5i.56400$Xh3.5122@edtnps90>:
"The Ghost In The Machine" <ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote in
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In sci.physics, Gieniu
<warendag@xxxxxxxxx>
wrote
on Fri, 18 May 2007 22:36:23 GMT
<HTp3i.22970$V75.18159@edtnps89>:
Hiand connected via a hose to our rather drenched variety.
Straight tube, opened on both its ends ,is verticaly put intofilled
with 1 atm air,
As we push, the volume will be at any point the position of
the plunger, multiplied by 0.5^2 * Pi (the cross-sectional
area of the compressor tube), plus whatever displaced
water is enough to equalize the pressure at the other end.
In short,
All in all, not a practical solution.
[*] many compressors actually use a pair of valves and a
reciprocating piston, but the calculation's simpler with
the long tube. Besides, as the air pressure increases the
compressor has to work harder, and may even stall if it's
not powerful enough.
[+] one can reduce the distance by using a fatter
compressor-tube, but the force is proportional to the
cross-sectional area. If the compressor-tube's area is
100 times as wide, one can get away with only traveling
1/100th the distance but the force is 100 times as great,
and the factors cancel out.
--
#191, ewill3@xxxxxxxxxxxxx
Useless C++ Programming Idea #10239993:
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Posted via a free Usenet account from http://www.teranews.com
********************
Hi
In order to receive energy (mgH/2 +mgh1) ,we must to
delivery to system energy (mgh1} only.
Energy mgh1 is used by moto-pump,
enegy of falling on turbine water is equal
(mgH/2 + mgh1),
where H - deep of plungednurzenia of tube in the water,
h1 - height of the lifting water m above level
of water in reservoir..
And the water in the bottom of the tube is removed
precisely how? Can't just pump it to the ambient ocean,
unless one has a very high-pressure pump. Pumping it up
to the surface does no good, either.
As an illustration -- let me set up another problem.
This time it's a 1 km long square tube, with cross section
1 m^2, and of sufficient strength. The bottom of this tube
has a 1 meter piston, which can push out water through a
pair of very strong flapper valves. This piston has a stroke
of 1 meter -- just enough to move across the tube bottom.
Dump 1 m^3 -- a metric tonne -- of water down the tube,
and feed it through the turbine, with this bottom piston
unit empty and waiting for water. The energy of that water
(neglecting friction) at the bottom will be such that
g = 9.805 m/s/s = 9.805 N/kg
d = 1/2 g t^2
t = sqrt(2d/g) = sqrt(2000 / 9.805) = 14.282 seconds
The kinetic energy will be 1/2 m v^2, but an easier way is
simply E = mgh, where m is 1 metric tonne and h = 1000 m.
That gives us 9.805 megaJoules. (The proof that these
are equal is left to the interested reader.)
If one wants more conventional power units, 9.805
megaJoules is about 2.72 kWh. Not that it matters.
The water is now sitting in the piston chamber. The weight
of a column of water 1 m high and 1 m^2 in area is about
9805 N (1000 kg * 9.805 N/kg). The pressure of course is
9805 N / 1 m^2 = 9805 Pascal. Multiply that by 1000 and
one gets a pressure of 9.805 megaPascal, plus atmospheric
pressure of about 0.1 megaPascal, which I can ignore since
I assume the tube top is exposed to the air.
The piston has to exert a force of 9.805 megaNewton
(since its cross-sectional area is 1 m^2 again) to push
the water through the flapper valve. That distance is
1 meter. 9.805 megaNewtons times 1 meter is energy --
work = force x distance, after all -- and therefore we're
expending 9.805 megaJoules to get rid of the water.
Wow. Who'd've thought? We need to use all of the
energy extracted from the water in order to get rid of
the waste water! And that's assuming perfect extraction
of the energy in the first place.
Oops.
To be fair, one might have a chance if one can dump things
heavier than water down the tube; the main problem is
not the mass of the water in the bottom, but the ambient
pressure. For example, one might dump liquid mercury
down the tube -- not recommended from an environmental
standpoint, since oxides of mercury are quite toxic
(elemental mercury itself is not quite as toxic; it was
used in thermometers for many years before modern alcohol
and/or solid state units, but it's still not something
one would want to introduce to fish willingly) -- but one
would at least get 133 megaJoules from the 13.6 metric
tonnes of poisonous but motile mercury poured down the
tube's throat, while spending the same amount of energy --
9.805 megaJoules -- to get rid of it afterwards.
Since turbines can operate at 90% this might work --
until the environmentalists burn one's hide. :-)
(Or until one runs out of available mercury at the surface.)
Sincer E.W.
--
#191, ewill3@xxxxxxxxxxxxx
Linux. Because vaporware only goes so far.
--
Posted via a free Usenet account from http://www.teranews.com
Hi
Let ferget this preasured air,which i used to explane
free energy.
The bottom end of tube is open , and motopump is
installed on the second .top end of tube.
Let this tube and this motopump are full of water .
Let motopump starts to work
The velocity of water is increasing in the tube at some time.and reasive
maximal velocity
v =sqrt (gH)
Pumping water this motopump used work
E1 =mgh1,
This work E1 =mgh1 motopomp must
used in the potential field in order to velocity
of water in the tube must be constant.
Energy E receive from system is
E =mgH/2 + mgh1
Energy deliver to syctem is E1 =mgh1
Free energy Ef is
Ef =mgH/2
Sincer E .W.
.
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