"Eugeniusz" <gienek31@shaw.ca> wrote in message
news:gFD0f.97375$tl2.90847@pd7tw3no...

WELCOME
It is known, that simmetrical system the connected vessels consists of
two different vessels connected together
suitable of pipe.
In these vessels is the water and the heiight of these both levels of
this water is H.

We do the work W on the height H and we remove the water from the narrow
vessel to the large vessel.
If we finished this work, we see thet all mass of the water from narrow
vessel now is in large vessel
What work W we did ? W = m(ro )gH/2.
But removed water is now on high H and its potential energg is equal
E=mgH
We see that E >W.
So we have our FREE ENERGY.

E.Wreda

In sci.physics, Eugeniusz
<gienek31@shaw.ca>
wrote
on Tue, 04 Oct 2005 22:48:44 GMT
<gFD0f.97375$tl2.90847@pd7tw3no>:

WELCOME
It is known, that asimetrical system the connected
vessels consist of two different vessels connected together
suitable of pipe.
In these wessels is the water and the highest of these
both levels of this water is H.

We do the work W on the hight H and we remove the water
from the narrow vessel to large vessel.
If we finished this work, we see that all mass of the water from narrow
vessel now is in large vessel
What work W we did ? W = m{ro} gH/2.
But removed water is now on high H and its potential energy is equal

We see that E >W.
So we have our FREE ENERGY. equal E- W.

E.Warenda
**************************

Weren't you the one suggesting that we can stick a 1-km pipe into
the ocean and leverage the resulting pressure difference from
bottom to power a generator at the top?

It ain't a-gonna work.

Problem Setup:

Assume that we have two boxy tanks, with a boxy tube.
(I do this for simplicity.) The left tank is a
1m x 1m x 1m affair. The right tank is a thin tube,
sticking up, of dimensions 100m x 0.1m x 0.1m. Both tanks
are connected with a strange flat "box tube" of dimensions
100m x 0.01m x 1m, and appropriate valving. Issues with
viscosity are ignored in this argument. The volume of water
is assumed pressure-independent and 1 metric tonne per
cubic meter. The system is initially filled with 2.01
metric tonnes of water.

Why this odd amount? Read on.

If we fill the right tank to 1 m height, that's 0.01 m^3
or 10 kg of water.
If we fill the left tank to 1 m height, it's now filled,
and has a metric tonne.
If we fill the bottom tube, that's another metric tonne.
Sum 'em up, and one gets 2010 kg or 2.01 metric tonnes.

That's a reasonably good static solution for these two tanks.

The Short Way:

Ever seen liquid spontaneously spit through a soda straw?
Me neither. Therefore, this logic sucks. :-)
So must the drinker, if he wants to get at the liquid in the cup,
or he can squeeze the cup, which tends to get messy.

(Sticking a soda straw into the neck of a water-filled
balloon gets one into issues regarding the stretching of
the rubber imparting pressure and energy to the water.
This isn't a perpetual motion machine either, of course;
one has to expend energy in refilling the balloon.)

The Long Way:

Star Trek to the rescue! Using Mr. Scott's marvelous
(and impossible) device, we instantaneously teleport
each and every molecule of the water in the left tank
so that the right tank is full, in two steps. There's a
smidge left over in the left tank; after this operation
we've moved 0.99 m^3 or 990 kg of water.

(It would be interesting to boil the water in the first tank
and use the resulting pressure to push the water up the
second tank -- but that's a more involved calculation.)

The new height in the left tank is now 0.01 m. The
right tank is now full.

How much energy did Mr. Scott have to use? At a minimum,
he had to move every water molecule in that 0.99 m^3
in some fashion. Since PE = mgh, where m is mass, g
is 9.805 N/kg, and h is the height, this move should
take energy for all of the water.

As a calculation aid, assume that the water is first
initially moved to a height of 100 m, then dropped.
We replace m by dM = (density) * (cross section) * dh.

water moving up: integral(h = 0.01 to 1) g(100 - h) dM
= 1000 * 100g(1 - 0.01) - (1/2)g(1^2 -0.01^2) = 98500.05g

We now have a flat sheen of water way up there; time to
move it into the right tank. In this case
dM = 1000 * 0.01 dh.

water moving down: integral(h = 1 to 100) g(100 - h) dM
= 1000 * 0.01 * (100g(100 - 1) - (1/2)g(100^2-1^2))=49005g

Since Mr. Scott's transporter has no thermodynamic defects
we simply subtract, and get 49495.05*g = 485298.96525 J.

This actually isn't all that much energy; the collision
of two 2 metric tonne cars moving 33.6 mph = 15 m/s is
about 45 kiloJoules.

Now let the water flow back down, extracting energy as it
does so. The detailed calculations I leave to the reader,
but it's clear that each molecule will start at a certain
height, and end up at a certain height. There are issues
regarding how long the flow takes but at the end of the
day, the intermediate heights don't matter, and we get
485298.96525 J. Note that the water in the bottom tube
is moved out, but it has to go up to allow the water in
the right tank to enter it.

Since in a real application viscosity saps energy, it's not
even a perpetual motion machine of the first order.


--
#191,