"Gieniu" <warendag@telus.net> wrote in message
news:5bN3i.40884$Xh3.15207@edtnps90...

"Gieniu" <warendag@telus.net> wrote in message
news:rpL3i.30385$V75.5672@edtnps89...

"Gieniu" <warendag@telus.net> wrote in message
news:HTp3i.22970$V75.18159@edtnps89...

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.
Nextly, when the tube is without the water,we remove
preasured air
From outside the tube ,from the bottom end of the
tube, under hydrostatic preasure,water flow into
the tube.
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 see also that 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,

*******************************************

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

We must reply on the question - why velocity of water in the
tube become lesson when the hight h of column of water
in tube is biger then H ?
We must remowe this mass of water above the level H.
using the moto-pump which is instaled on the end of tube
.

Sincerely E.W.

************************

If we remove water with velocity v =sqrt ( gH]

then this quick water does not acts on the water column in
the tube.
Inside of tube, velosity of water is v =sqrt {gH) = constant.
So we get energy
E =(1/2)mv^2 +mgh1=mgH/2 +mgh1
where m mass of icrease water
on the hight h1.with velocity
v = sqrt(gH)
Istead of we give energy E = mgh1 only


Sincerely yours E.W.



























E = m/2 *v^2 +mgh1

In sci.physics, Gieniu
<warendag@telus.net>
wrote
on Fri, 25 May 2007 05:50:00 GMT
<cOu5i.56400$Xh3.5122@edtnps90>:

"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
message
news:245di4-20j.ln1@sirius.tg00suus7038.net...

In sci.physics, Gieniu
<warendag@telus.net>
wrote
on Fri, 18 May 2007 22:36:23 GMT
<HTp3i.22970$V75.18159@edtnps89>:

Hi
Straight tube, opened on both its ends ,is verticaly put intofilled
with 1 atm air,

and connected via a hose to our rather drenched variety.

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,

********************
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,

In sci.physics, Gieniu
<waren...@telus.net>
wrote
on Fri, 25 May 2007 05:50:00 GMT
<cOu5i.56400$Xh3.5122@edtnps90>:

"The Ghost In The Machine" <e...@sirius.tg00suus7038.net> wrote in message
news:245di4-20j.ln1@sirius.tg00suus7038.net...

In sci.physics, Gieniu
<waren...@telus.net>
wrote
on Fri, 18 May 2007 22:36:23 GMT
<HTp3i.22970$V75.18159@edtnps89>:

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.

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;
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.

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
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,

--
Posted via a free Usenet account fromhttp://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 itse