Re: Free Energy
Re: Free Energy
Source: http://sci.tech−archive.net/Archive/sci.physics/2007−05/msg02393.html
• From: "Gieniu"
• Date: Fri, 25 May 2007 05:50:00 GMT
"The Ghost In The Machine" wrote in message
news:245di4−20j.ln1@xxxxxxxxxxxxxxxxxxxxxxxxxx
In sci.physics, Gieniu
wrote
on Fri, 18 May 2007 22:36:23 GMT
:
Hi
Straight tube, opened on both its ends ,is verticaly put into
the water.
One of these ends of tube is on deep H, second on the surface of the
water.
Using of the preasured air, we remove the water from this tube.
OK, but it'll take a little more than a pair of lungs and a one−way. :−)
Nextly, when the tube is without the water,we remove
preasured air
From outside the tube
Uh...how'd the air get out there? The tube's surrounded
by water....
,from the bottom end of the
tube, under hydrostatic preasure,water flow into
the tube.
Ah...well, it seems to me that the air will naturally exhaust itself out
of the top of the tube, so there's no real need to do much at the other
end except let the water flow in.
Re: Free Energy 1
Re: Free Energy
On the mass of water in tube act difference two forces:
F1 =ro g H S
F2 =ro g h S where h is current hight of column
of water in the tube.
F1 −F2 = F = ro g S ( H−h)
Aceeleration a = F/m m =roSh
From hire
a = ( H −h }g/h
We see, thet acceleration is maximal
when h=0 and acceleration is eqal zero ,if h=H.
From hire we cee also thet velocity of the water
in column h is increasing in the scope (0;H)
For h=H velocity of ideal liquid in the column
is maximal.
Sincerely E.W,
Ah, so you're suggesting a large expenditure of energy in order to
recover a small amount of energy through the moving water, by
forcing it through a turbine at the bottom of the tube, then.
Not exactly the most efficient method by which to proceed.
There's a few other issues as well −− the most obvious one being water
friction as the tube fills.
In any event, removing the water from the tube −− however
one does it −− will take more energy than the amount one
can extract from the tube later by letting it refill.
Let's see if I can come up with some specific numbers.
I'll assume the tube is 1 meter in inside diameter, and 1
km long. (Presumably it would be fabricated in sections.)
The tube is initially capped at the bottom and heavily
weighted. We'll see how heavy that weight has to be
later on.
The tube, of course, will have a weight of its own; let's
say that it's made out of iron, 1 cm thick. I'm not sure
if that'll be strong enough or no, but assuming it is, that
tube will contain a volume of iron of
(0.6^2 * Pi − 0.5^2 * Pi) * 1000 = 346 m^3,
or 2700 metric tonnes of iron (each m^3 of iron or steel
weighs 7,874 kg). The tube, of course, would contain
785 m^3 of usable volume, and displace
1131 m^3 or 1131 metric tonnes of water. This is good;
Re: Free Energy 2
Re: Free Energy
it will sink under its own weight though a stabilizing
weight might still be needed to keep things straight.
So now we have a tube full of relatively rarefied 1 atm
air. We let in the water through a turbine. We assume,
for the sake of argument, perfect efficiency of extraction
in this turbine (real turbines might get half, but I'd
have to look). The initial pressure of that incoming water
will be enormous: about 100 atmospheres or 10 megaPascal.
There is, however, a little problem; once the water gets
into the tube the pressure will presumably drop to near
zero, at least until the tube starts to fill. But pressure
doesn't generate energy by itself.
For purposes of calculation it's probably easier just
to open the top end (if it's not open already), put a
floating pinwheel near the bottom, and just let the water
spill in. Initially, the pinwheel will be 1 km down,
and the water will hit it at a certain velocity, which
can be algebraically calculated:
h = 1/2 g t^2
t = (2h/g)^(1/2)
v = gt = (2gh)^(1/2)
where h starts out at 1 km, and g = 9.805 N/kg
is the acceleration of gravity as usual.
The total amount of energy available is then simply
E = integral(h=0,1km) (M/2 v^2 dh) = integral(h=0,1km) M/2 ( 2gh ) dh)
= Mgh^2/2, where M is the mass of a 1−meter disc of water −− about
785 kg. Total extractable energy: about 3.8 GJ.
Now we have a rather useless tube full of water.
We open the bottom valve, close off the top, and start a
compressor, which simply takes ambient air and compresses
it, pushing the water out under pressure through the
bottom.
Or is it that simple?
Compressing air takes work −− anyone who's pumped up
a bicycle tire can attest to that, and furthermore one
can observe that the pump gets rather warm, if it's a
hand−operated affair. One can model the problem as a
very long tube [*] with a frictionless sealed piston, and
since we know the pressure −− it's just under 100 atm −−
we'll need a tube, conceptually, 100 km long, also 1
m in inside diameter, initially filled with 1 atm air,
and connected via a hose to our rather drenched variety.
Re: Free Energy 3
Re: Free Energy
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,
V = x * (0.5^2 * Pi) + P * 0.5^2 * Pi / (9805 Pascal/m)
where 9805 Pascal is the approximate pressure of a 1 meter
column of water.
Since K = PV = 100000 m * 100000 Pascal * 0.5^2 * Pi is a constant,
we can replace V by K/P:
K/P = x * (0.5^2 * Pi) + P * 0.5^2 * Pi / (9805 Pascal/m)
Of course this is a quadratic equation in P, with roots
(−x * 0.5^2 * Pi +/− sqrt(x^2 * 0.5^4 * Pi^2 + 4 * K * 0.5^2 * Pi/9805))
/ (2 * 0.5^2 * Pi / 9805)
We discard the negative root and grind it out.
P(x) = (−0.785 * x + sqrt(x^2 * 0.617 + 2.516 * 10^7)) / (1.602 * 10^−4)
We can now integrate 0.5^2 * Pi * (P(x) − 100000)
dx to get the total energy. (The −100000 is because
the atmosphere on the one side is helping. To be sure,
it's not helping much.) That sqrt() is going to give us
headaches and may require specialized techniques, but we
can get a quickie estimate by simply pushing the plunger
halfway in and keeping the water immobile −− the water only
will move 10 meters anyway as we move the plunger 50 km. [+]
So lessee. With this revised estimate
K/P = x * (0.5^2 * Pi)
and now P/K = 1/(x * (0.5^2 * Pi)) or P = K / (x * (0.5^2 * Pi)).
Since F = P * A, F = K / x, and dE is simply F * dx, so we
can now integrate a far simpler expression:
E' = integral(x = 0,50000) (K/x − 100000 * 0.5^2 * Pi) dx
= K * log(50000) − 100000 * 50000 * 0.5^2 * Pi
= 81 Gigajoules −− and we're not even halfway there!
One might contemplate lifting out the tube and letting the water drain.
The mass of the tube of course is 2700 metric tonnes, and we'd have to
lift it 1000 meters against the gravitational force. Bouyancy will
help but it only goes so far, and one can get a snap estimate by
Re: Free Energy 4
Re: Free Energy
assuming one has to lift half the tube out −− but even so one
still requires 13 GJ just for that.
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:
char * f(char *p) {char *q = malloc(strlen(p)); strcpy(q,p); return q; }
−−
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********************
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..
Sincer E.W.
.
Re: Free Energy 5