Welcome
The pipe H = 1000m,the crosssection S equal 1m^2, their botth ends ore
opened, the one of these two ends
is verticaly put in the water1000m under the level of water.In this pipe
the water is only The level of water in
the pipe and out the pipe it same.
It is needed to remove this water out of the pipe.
We want to remove water from this pipe through
the bottom upened their end.
What the work W it is needed to do, to perform ,in order to push out this
water, to out side of the pipe, to the water.
dW = FdH (ro)= 1g/cm^3.....gestosc
woda
dW =(ro)gHSdh H =1000m wys.rury
. W= (ro)gHHS/2 S =1 m^2 powierzchnia
*************
1)
W =M g H/2 M=1000 000kg
M ...masa wody w rurze
g=9,81m/s^2
So work W = 1 0^ 6[kg] 9.81[m/s^2] 1000[m]
W= ~10^11[kgm][m/s^2]
W =10^11[kgm^2/s^2]=10^11[J}
1J/s =1watt

W/s = 10^11watt= 10^5 megawatt
W/s =Gigawatt 50 Gigawatt??
Because work
W = MgH/2
So finally W/s = Pawer = 50/2 =25 GIGAWATT !!??

It is amplituda of the power oscilation ; .
On the center of this oscilation ,velocity V is maximal.
This energy 25 Gigawat * sec it was using in order to remove water out of
the pipe, and this same energy should be changed on another energy
cntinously.
With best greetings E.W.

In sci.physics, Eugeniusz Wareda

wrote
on Sat, 26 Jun 2004 19:00:44 +0200
:
Welcome
The pipe H = 1000m,the crosssection S equal 1m^2, their botth ends ore
opened, the one of these two ends
is verticaly put in the water1000m under the level of water.In this pipe
the water is only The level of water in
the pipe and out the pipe it same.
It is needed to remove this water out of the pipe.
We want to remove water from this pipe through
the bottom upened their end.
What the work W it is needed to do, to perform ,in order to push out this
water, to out side of the pipe, to the water.
dW = FdH (ro)= 1g/cm^3.....gestosc
woda
dW =(ro)gHSdh H =1000m wys.rury
. W= (ro)gHHS/2 S =1 m^2 powierzchnia
*************
1)
W =M g H/2 M=1000 000kg
M ...masa wody w rurze
g=9,81m/s^2
So work W = 1 0^ 6[kg] 9.81[m/s^2] 1000[m]
W= ~10^11[kgm][m/s^2]
W =10^11[kgm^2/s^2]=10^11[J}
1J/s =1watt

W/s = 10^11watt= 10^5 megawatt
W/s =Gigawatt 50 Gigawatt??
Because work
W = MgH/2
So finally W/s = Pawer = 50/2 =25 GIGAWATT !!??

It is amplituda of the power oscilation ; .
On the center of this oscilation ,velocity V is maximal.
This energy 25 Gigawat * sec it was using in order to remove water out of
the pipe, and this same energy should be changed on another energy
cntinously.
With best greetings E.W.

Take a clear water-filled glass and insert a soda straw
thereinto. Notice anything happening?

Your pipe will do exactly the same thing. (Namely, nothing.)

While it might be interesting to insert a thermally insulated
pipe into the sea and extract work off the temperature difference,
that's a different problem from the one you're apparently
trying to solve, and rather badly.

Another possibility is using the natural attraction of water
with materials to do various things (that soda straw, after
all, has a meniscus, and capillary action is well known).
However, that's probably going to be far less than 5 GW.

Here's something that may be a pretty problem, though
(and unlike your contrivance would actually function -- for a time).

Take a 1 km long / 1 m^2 crosssection pipe and cap one end
thereof, then insert it into the ocean, open end first,
taking care to keep it pointed perpendicular thereto
(the bouyancy will give one problems unless the pipe is
very heavy). In its installed state the cap is just at
sea level. One can then get compressed air from the cap
(using a valve) until the air runs out (or until the weight
of the combination becomes too great for the substrata
upon which it sits and the structure collapses).

[1] How heavy a pipe is needed in order to exactly counter
the bouyancy of itself, plus the compressed air?

[2] What is the pressure of the captured air, initially?

[3] How much work is needed (work = mechanical energy)
in order to drop the pipe into the ocean, assuming
an uncapped pipe is bouyantly neutral?

[4] How much work can be extracted from the air in the pipe?

[5] What is the tension on the pipe, assuming the entire
substructure is strong enough to stand the stress
and the pipe, if uncapped, would be bouyantly neutral?

There is also one compressed-air generation facility
that utilizes falling water, with the air embedded
thereinto. (Nice dry air, too, as it turns out; the
water cools the air as well as compressing it.)

--
#191,
It's still legal to go .sigless.