| Topic: |
Science > Physics |
| User: |
"Eugeniusz Wareda" |
| Date: |
22 Jun 2004 01:58:30 PM |
| Object: |
The ecological power station of 5(five) GigaWatt |
Welcome All!
You should vetically put in to the sea water on the deep
equal 1000meters,the tube of the crossection
equal 1(m)(m).
..The tube is connected to the input of water pump which is located on the
level of the sea .
This pump is pumping the water on hight equal 10m over the level of sea
water, for examply.
It is the water turbine on which the water is fallen.
from hight 10 meter over the sea level.The work task
of pump is to keep velocity V of water in pipe as constat
only,
The velocity of this water cannot be lessen than velocity in the tube
suppling the pump of the water.
The second pipe is connected to the outside of the pump.
The task of second tube is to supply a water to the
hight equal h ,and from this to over the water turbin.
The velocity V of the water falling on turbin is equal
V = sqrt(VoVo +2gh) =~120m/s ??!!
Ekin =~5GigaWatt = it is Power of this power station
Power P of the moto-pomp should be equal P=Mgh/t.
P =~0.8 Gwatt ...it is needed power of the motopomp.
It is power needed to keep velocity of water in the tube as constant.
We see that in the time t get Pkin >>Ppomp
I wrote it many tmes. I would like to ask for the answer
from you.I send plenty of best greetings for you
E.Warenda
More , energy losted for pump is get back on the turbine
E.W.
.
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| User: "Uncle Al" |
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| Title: Re: The ecological power station of 5(five) GigaWatt |
22 Jun 2004 05:41:42 PM |
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Eugeniusz Wareda wrote:
Welcome All!
You should vetically put in to the sea water on the deep
equal 1000meters,the tube of the crossection
equal 1(m)(m).
Idiot. Calculate the energy necessary to raise a tonne of water 1000
meters. Add friction losses. Thres no net to harvest.
[snip]
More , energy losted for pump is get back on the turbine
Hook a generator to a motor and you never need pay for electricity
again.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
.
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| User: "Eugeniusz Wareda" |
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| Title: Re: The ecological power station of 5(five) GigaWatt |
22 Jun 2004 07:57:11 PM |
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I am idiot and what are you?
If you a long tube put into the sea water perpendiculary,
and a hight of this tube is 1000 and the water is removed from this tube
using the compressed air (100atm), then you do a work W;
dW = FdH=( ro)gHSdH F = F (h) =(ro)ghs
W = (ro)gHHS/2 =MgH/2
So,we have the potential energy Epot=MgH/2>.0
And this all energy is chaged for the kinetical energy.
If the level of sea water and a level of water meniscus in pipe will be
same, the kinetical energy is equal max and
the potential energy is equal zero.... ... ...
E,W.
Uzytkownik "Uncle Al" <UncleAl0@hate.spam.net> napisal w wiadomosci
news:40D8B5A6.D88FC7BA@hate.spam.net...
Eugeniusz Wareda wrote:
Welcome All!
You should vetically put in to the sea water on the deep
equal 1000meters,the tube of the crossection
equal 1(m)(m).
Calculate the energy necessary to raise a tonne of water 1000
meters. Add friction losses. Thres no net to harvest.
[snip]
More , energy losted for pump is get back on the turbine
Hook a generator to a motor and you never need pay for electricity
again.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
.
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| User: "The Ghost In The Machine" |
|
| Title: Re: The ecological power station of 5(five) GigaWatt |
23 Jun 2004 07:00:24 AM |
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In sci.physics, Eugeniusz Wareda
<lala@magma-net.pl>
wrote
on Wed, 23 Jun 2004 02:57:11 +0200
<cbalpv$et4$1@korweta.task.gda.pl>:
I am idiot and what are you?
If you a long tube put into the sea water perpendiculary,
and a hight of this tube is 1000 and the water is removed from this tube
using the compressed air (100atm), then you do a work W;
It takes work to compress air. Surprise!
If one sets up a movable sealing barrier within the pipe
(which makes it a bit of a piston) at the bottom, then
introduces compressed air, how much air, energy, etc. would
be required to lift the *entire* column of water 10 m
(treating the column of water as a solid slug)?
First, we need to work out the pressure. That's easy
enough: the pressure of that column is g * mass / area =
9.8048 * 1000000 kg = 9.8 million Pascal. Compared to
that, air pressure at 101,500 Pascal (give or take) is
hardly noticeable. And yes, it's a little shy of 100 atm.
For comparison, scuba tanks are commonly pressurized to
about 3000 psi, which is about 20 atm.
Now, the amount of air. We need 10 m^3 of 9.8 million Pascal air.
PV = nRT by Guy-Lussac. We're not all that interested in nR
and T is ignorable for now.
So PV = P'V', or V' = PV/P'. V' = 9.8 million / 101,500 * 10 m^3
= 965.5 m^3 of sea-pressure air.
At this point we might reverse the problem and allow the
column of water to fall 955.5 m onto our sea-level air,
which would be the same amount of work as to compress
the air using, say, a piston compressor. We then get our
requisite compressed air, but we also have expended some
energy doing it.
Quite a bit, in fact.
An interesting subproblem: how much will T increase?
I'm not sure how to compute that, but you're already
over budget, energywise. A common bicycle pump gets
rather warm while pumping a tire, that much I know.
(The computations are admittedly approximate. One might
see a small benefit -- but that benefit will easily be
eaten up by frictional losses and computational errors.
For example, that aforementioned T will creep in while
compressing the air, which may help for awhile until
one realizes that the main tube holding the water being
lifted, and the tube feeding the bottom of the main tube,
aren't exactly the world's best heat holders, and the
surrounding water carries all of that not-so-free heat
away, decreasing the pressure. Whoops!)
dW = FdH=( ro)gHSdH F = F (h) =(ro)ghs
W = (ro)gHHS/2 =MgH/2
So,we have the potential energy Epot=MgH/2>.0
And this all energy is chaged for the kinetical energy.
If the level of sea water and a level of water meniscus in pipe will be
same, the kinetical energy is equal max and
the potential energy is equal zero.... ... ...
E,W.
Uzytkownik "Uncle Al" <UncleAl0@hate.spam.net> napisal w wiadomosci
news:40D8B5A6.D88FC7BA@hate.spam.net...
Eugeniusz Wareda wrote:
Welcome All!
You should vetically put in to the sea water on the deep
equal 1000meters,the tube of the crossection
equal 1(m)(m).
Calculate the energy necessary to raise a tonne of water 1000
meters. Add friction losses. Thres no net to harvest.
[snip]
More , energy losted for pump is get back on the turbine
Hook a generator to a motor and you never need pay for electricity
again.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
--
#191,
It's still legal to go .sigless.
.
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| User: "tj Frazir" |
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| Title: Re: The ecological power station of 5(five) GigaWatt |
23 Jun 2004 02:05:51 AM |
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The sea turbine is just an underwater windmill.
It just sits in the curent off greenland and runs.
the curent is a steady 14 knots .
Run that pipe 1000 m up a river to get 200 foot fall put a
hydroeectric at the low end and you got somthing.
.
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| User: "The Ghost In The Machine" |
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| Title: Re: The ecological power station of 5(five) GigaWatt |
23 Jun 2004 07:00:20 AM |
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In sci.physics, Eugeniusz Wareda
<lala@magma-net.pl>
wrote
on Tue, 22 Jun 2004 20:58:30 +0200
<cba0bh$7hl$1@korweta.task.gda.pl>:
Welcome All!
You should vetically put in to the sea water on the deep
equal 1000meters,the tube of the crossection
equal 1(m)(m).
Fine. Put a 1 km tube into the sea with a 1m^2 cross-section,
and one gets 1,000 metric tonnes or so of seawater.
(See also "caissons".)
I'd have to see how deep the Troll's "feet" are (the Troll
is a gas platform dropped in the sea off Norway some time
back); one of the more interesting construction artifacts
was that the tube was constructed "dry" within the water,
and then flooded later when the unit was being towed out
to sea.
http://www.structurae.de/en/structures/data/s0003286/index.cfm
suggests a 472 m height but that probably includes the platform itself.
.The tube is connected to the input of water pump which is located on the
level of the sea .
OK.
This pump is pumping the water on hight equal 10m over the level of sea
water, for examply.
Bear in mind that the tube will be losing water as you're pumping
it out, and that therefore the pump will have to work harder.
If one lets water into the bottom of the tube then one need
really not have the tube at all (unless one really wants deepwater
pumped to the surface, which may interest certain hydrologically-
inclined scientists but is a different problem).
It is the water turbine on which the water is fallen.
Erm....are you postulating a new piece of equipment or
simply trying to have the pump double as a turbine for
the purposes of your computation?
Either way, you're not going to get very far; to pump that
very last liter of water out of your tank (for that's
what the tube has become) will take far more energy than
pumping the first, even if one allows for the small decrease
in g (from 9.8048 to about 9.8033) at the bottom of the tank.
E = mgh, approximately, where h is taken from the center
of the water cube (this isn't quite right but is close enough):
m = 1 kg (1 liter) of water
g = 9.8032 (and that's being optimistic)
h = 1 km + 10 m - 0.5 m = 1009.5 m
E = 9896 J
That first liter, by contrast, only has to be pumped over
a 10.5 m distance:
g = 9.8048
h = 10.5 m
E = 102.9 J
from hight 10 meter over the sea level.The work task
of pump is to keep velocity V of water in pipe as constat
only,
An additional energy requirement is to move the water.
I'm not up on turbulent flows and such but to accelerate
a 1 m^3 (1 metric tonne) "slug" of water at a velocity of
1 m/s is going to require 500 J in a frictionless pipe.
To have a continuous motion of water through a pipe
of 1 m/s of 1 m^3 from a holding-tank requires 500 W.
(Since the feed is generally at the bottom of the
holding tank one gets a substantial chunk of that power
for free -- until the tank is nearly empty, of course.
Of course in your case you've got the opposite problem:
you're taking the output from the *top*. There will be
interesting issues if we ever get a significant amount of
fluids into space; presumably NASA et al have worked this
out already, though, at least for small ammonia compressor
cooling units.)
The velocity of this water cannot be lessen than velocity in the tube
suppling the pump of the water.
Incompressibility, yes.
The second pipe is connected to the outside of the pump.
The task of second tube is to supply a water to the
hight equal h ,and from this to over the water turbin.
The velocity V of the water falling on turbin is equal
V = sqrt(VoVo +2gh) =~120m/s ??!!
Ekin =~5GigaWatt = it is Power of this power station
Now you're just being silly.
[rest snipped]
--
#191,
It's still legal to go .sigless.
.
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| User: "Eugeniusz Wareda" |
|
| Title: Re: The ecological power station of 5(five) GigaWatt |
23 Jun 2004 12:27:43 PM |
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Użytkownik "The Ghost In The Machine" <ewill@aurigae.athghost7038suus.net>
napisał w wiadomości news:oe4oq1-6g4.ln1@lexi2.athghost7038suus.net...
In sci.physics, Eugeniusz Wareda
<lala@magma-net.pl>
wrote
on Tue, 22 Jun 2004 20:58:30 +0200
<cba0bh$7hl$1@korweta.task.gda.pl>:
Welcome All!
You should vetically put in to the sea water on the deep
equal 1000meters,the tube of the crossection
equal 1(m)(m).
Fine. Put a 1 km tube into the sea with a 1m^2 cross-section,
and one gets 1,000 metric tonnes or so of seawater.
(See also "caissons".)
I'd have to see how deep the Troll's "feet" are (the Troll
is a gas platform dropped in the sea off Norway some time
back); one of the more interesting construction artifacts
was that the tube was constructed "dry" within the water,
and then flooded later when the unit was being towed out
to sea.
http://www.structurae.de/en/structures/data/s0003286/index.cfm
suggests a 472 m height but that probably includes the platform itself.
.The tube is connected to the input of water pump which is located on
the
level of the sea .
OK.
This pump is pumping the water on hight equal 10m over the level of sea
water, for examply.
Bear in mind that the tube will be losing water as you're pumping
it out, and that therefore the pump will have to work harder.
If one lets water into the bottom of the tube then one need
really not have the tube at all (unless one really wants deepwater
pumped to the surface, which may interest certain hydrologically-
inclined scientists but is a different problem).
It is the water turbine on which the water is fallen.
Erm....are you postulating a new piece of equipment or
simply trying to have the pump double as a turbine for
the purposes of your computation?
Either way, you're not going to get very far; to pump that
very last liter of water out of your tank (for that's
what the tube has become) will take far more energy than
pumping the first, even if one allows for the small decrease
in g (from 9.8048 to about 9.8033) at the bottom of the tank.
E = mgh, approximately, where h is taken from the center
of the water cube (this isn't quite right but is close enough):
m = 1 kg (1 liter) of water
g = 9.8032 (and that's being optimistic)
h = 1 km + 10 m - 0.5 m = 1009.5 m
E = 9896 J
That first liter, by contrast, only has to be pumped over
a 10.5 m distance:
g = 9.8048
h = 10.5 m
E = 102.9 J
from hight 10 meter over the sea level.The work task
of pump is to keep velocity V of water in pipe as constat
only,
An additional energy requirement is to move the water.
I'm not up on turbulent flows and such but to accelerate
a 1 m^3 (1 metric tonne) "slug" of water at a velocity of
1 m/s is going to require 500 J in a frictionless pipe.
To have a continuous motion of water through a pipe
of 1 m/s of 1 m^3 from a holding-tank requires 500 W.
(Since the feed is generally at the bottom of the
holding tank one gets a substantial chunk of that power
for free -- until the tank is nearly empty, of course.
Of course in your case you've got the opposite problem:
you're taking the output from the *top*. There will be
interesting issues if we ever get a significant amount of
fluids into space; presumably NASA et al have worked this
out already, though, at least for small ammonia compressor
cooling units.)
The velocity of this water cannot be lessen than velocity in the tube
suppling the pump of the water.
Incompressibility, yes.
The second pipe is connected to the outside of the pump.
The task of second tube is to supply a water to the
hight equal h ,and from this to over the water turbin.
The velocity V of the water falling on turbin is equal
V = sqrt(VoVo +2gh) =~120m/s ??!!
Ekin =~5GigaWatt = it is Power of this power station
Now you're just being silly.
[rest snipped]
--
#191,
It's still legal to go .sigless.
***************
^^^^^^^^^^^^^^
&&&&&&&&&
Welcome,
The both ends of a pipe are open. The crossection of this pipe is equal 1
(m)(m),and the lenght of this pipe is
equal 1000meters.The one of two ends of opipe is put
in the sea water to the deep equal 1000meters under level
of sea.
The second of two open ends of this pipe is connected
to input of pump.
If the level of meniscus inside is equal as level outside of
the tube then ,in this time the our energetical system is in
the balance of power.
But our system have to be in the oscilating state, first time only .
So we have to do a work W;
dW =Fdh
F = (ro)ghS
W = {(ro)SgHH/2}= MgH/2
This energy must be changed on kinetikal energy and
on the level of sea it is equal Ekin=MVV/2
and the velocity Vmax of water i pipe is equal;
Vmax=sqrt(gH) =sqrt( 10m/ss1000m)=100m/s
E kin= Epot=MgH/2
Ekin=1(m^2)(10m/ss)(1000/2)(m) =~5 Gigawatt
From this time, from now the water i s going
to the input of pomp automaticali.
The task for pump is to keep velocity of water as constant.
It is the work in the potential field so it is constant for all velocity of
equal constant= sqrt(2gh).only.
..This work used by the pump is giwing back on the pump.
Amen. cordial greetings for all.
E.W.
.
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| User: "MorituriMax" |
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| Title: Re: The ecological power station of 5(five) GigaWatt |
23 Jun 2004 11:43:19 PM |
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Eugeniusz Wareda wrote:
.This work used by the pump is giwing back on the pump.
Even assuming one can take any of the rest of this without a big fat grain of
salt, are you really saying that your pump is 100% efficient with no loss of
energy in the process you "described?"
.
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| User: "Eugeniusz Wareda" |
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| Title: Re: The ecological power station of 5(five) GigaWatt |
24 Jun 2004 09:01:16 AM |
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Uzytkownik "MorituriMax" <newage@sendarico.net> napisal w wiadomosci
news:HZsCc.4112$OX2.1231@fe2.texas.rr.com...
Eugeniusz Wareda wrote:
.This work used by the pump is giwing back on the pump.
Even assuming one can take any of the rest of this without a big fat grain
of
salt, are you really saying that your pump is 100% efficient with no loss
of
energy in the process you "described?"
*********
*********
Yes ,I described it 2 yers ago the application, and I have sent it toPatent
Offiice in the Poland and Germany.
Sincerelly Yours E.W.
.
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| User: "MorituriMax" |
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| Title: Re: The ecological power station of 5(five) GigaWatt |
24 Jun 2004 12:15:19 PM |
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Eugeniusz Wareda wrote:
Yes ,I described it 2 yers ago the application, and I have sent it toPatent
Offiice in the Poland and Germany.
Sincerelly Yours E.W.
Hmmm, have you gotten the Nobel award yet for making a 100% efficient pump?
Have you patented the new material you are using for your pump?
.
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| User: "The Ghost In The Machine" |
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| Title: Re: The ecological power station of 5(five) GigaWatt |
25 Jun 2004 08:34:11 AM |
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In sci.physics, Eugeniusz Wareda
<lala@magma-net.pl>
wrote
on Wed, 23 Jun 2004 19:27:43 +0200
<cbcfbd$nk6$1@korweta.task.gda.pl>:
Użytkownik "The Ghost In The Machine" <ewill@aurigae.athghost7038suus.net>
napisał w wiadomości news:oe4oq1-6g4.ln1@lexi2.athghost7038suus.net...
In sci.physics, Eugeniusz Wareda
<lala@magma-net.pl>
wrote
on Tue, 22 Jun 2004 20:58:30 +0200
<cba0bh$7hl$1@korweta.task.gda.pl>:
Welcome All!
You should vetically put in to the sea water on the deep
equal 1000meters,the tube of the crossection
equal 1(m)(m).
Fine. Put a 1 km tube into the sea with a 1m^2 cross-section,
and one gets 1,000 metric tonnes or so of seawater.
(See also "caissons".)
I'd have to see how deep the Troll's "feet" are (the Troll
is a gas platform dropped in the sea off Norway some time
back); one of the more interesting construction artifacts
was that the tube was constructed "dry" within the water,
and then flooded later when the unit was being towed out
to sea.
http://www.structurae.de/en/structures/data/s0003286/index.cfm
suggests a 472 m height but that probably includes the platform itself.
.The tube is connected to the input of water pump which is located on
the
level of the sea .
OK.
This pump is pumping the water on hight equal 10m over the level of sea
water, for examply.
Bear in mind that the tube will be losing water as you're pumping
it out, and that therefore the pump will have to work harder.
If one lets water into the bottom of the tube then one need
really not have the tube at all (unless one really wants deepwater
pumped to the surface, which may interest certain hydrologically-
inclined scientists but is a different problem).
It is the water turbine on which the water is fallen.
Erm....are you postulating a new piece of equipment or
simply trying to have the pump double as a turbine for
the purposes of your computation?
Either way, you're not going to get very far; to pump that
very last liter of water out of your tank (for that's
what the tube has become) will take far more energy than
pumping the first, even if one allows for the small decrease
in g (from 9.8048 to about 9.8033) at the bottom of the tank.
E = mgh, approximately, where h is taken from the center
of the water cube (this isn't quite right but is close enough):
m = 1 kg (1 liter) of water
g = 9.8032 (and that's being optimistic)
h = 1 km + 10 m - 0.5 m = 1009.5 m
E = 9896 J
That first liter, by contrast, only has to be pumped over
a 10.5 m distance:
g = 9.8048
h = 10.5 m
E = 102.9 J
from hight 10 meter over the sea level.The work task
of pump is to keep velocity V of water in pipe as constat
only,
An additional energy requirement is to move the water.
I'm not up on turbulent flows and such but to accelerate
a 1 m^3 (1 metric tonne) "slug" of water at a velocity of
1 m/s is going to require 500 J in a frictionless pipe.
To have a continuous motion of water through a pipe
of 1 m/s of 1 m^3 from a holding-tank requires 500 W.
(Since the feed is generally at the bottom of the
holding tank one gets a substantial chunk of that power
for free -- until the tank is nearly empty, of course.
Of course in your case you've got the opposite problem:
you're taking the output from the *top*. There will be
interesting issues if we ever get a significant amount of
fluids into space; presumably NASA et al have worked this
out already, though, at least for small ammonia compressor
cooling units.)
The velocity of this water cannot be lessen than velocity in the tube
suppling the pump of the water.
Incompressibility, yes.
The second pipe is connected to the outside of the pump.
The task of second tube is to supply a water to the
hight equal h ,and from this to over the water turbin.
The velocity V of the water falling on turbin is equal
V = sqrt(VoVo +2gh) =~120m/s ??!!
Ekin =~5GigaWatt = it is Power of this power station
Now you're just being silly.
[rest snipped]
--
#191,
It's still legal to go .sigless.
***************
^^^^^^^^^^^^^^
&&&&&&&&&
Welcome,
The both ends of a pipe are open. The crossection of this pipe is equal 1
(m)(m),and the lenght of this pipe is
equal 1000meters.The one of two ends of opipe is put
in the sea water to the deep equal 1000meters under level
of sea.
Hmm...well, now you're simply pumping water from out of the deeps.
While that might be interesting, it's hardly a guarantee of "free
energy" (though one might be able to do something with respect
to the temperature difference).
The second of two open ends of this pipe is connected
to input of pump.
If the level of meniscus inside is equal as level outside of
the tube then ,in this time the our energetical system is in
the balance of power.
But our system have to be in the oscilating state, first time only .
Oscillating state?
The whales are going to hate you...if this works at all.
Oscillations = sound, sound can mean trouble.
So we have to do a work W;
dW =Fdh
F = (ro)ghS
W = {(ro)SgHH/2}= MgH/2
This energy must be changed on kinetikal energy and
on the level of sea it is equal Ekin=MVV/2
and the velocity Vmax of water i pipe is equal;
Vmax=sqrt(gH) =sqrt( 10m/ss1000m)=100m/s
E kin= Epot=MgH/2
Ekin=1(m^2)(10m/ss)(1000/2)(m) =~5 Gigawatt
Unfortunately for you, the equation you're using isn't
quite the right one; what you're postulating is the
energy from a column of water 1 km high and 1m^2 in
cross-section falling through a turbine in 1 second.
This is quite impossible, as g = 9.805 m/s/s, which
means that, if a small ball were dropped from the
top of that water column (ignoring frictional losses),
which has been very conveniently teleported to, say,
Miami, the ball would take sqrt(1000 m / g) = 10.1 s
to fall to the bottom. Therefore the absolute maximum
power one can push from that column would be around
500 megawatts, as the column empties -- and even that's
not quite right, as the pressure, and therefore the power,
lessens as the column drops -- not to mention frictional
issues with respect to turbulent fluid flow.
But here's a thought for you. Next time you're drinking
from something using a soda straw, contemplate on why you
have to suck in order to get soda into the straw.
Using your methods, the amount of power one should be able
to get from the soda straw, assuming a straw length of 10 cm
and a cross-sectional area of about 25 mm^2, would be
Ekin=(25*10^-6)(10m/ss)(0.10)(m) ~= 12.5 watts
which should be more than enough to have it jump into your mouth.
All your sucking would do is maintain a constant velocity.
(Unfortunately, your equation didn't multiply out correctly, so
I simply proportionalized the problem.)
From this time, from now the water i s going
to the input of pomp automaticali.
The task for pump is to keep velocity of water as constant.
You don't need a pump for that. All you need is a hole of
the correct size.
It is the work in the potential field so it is constant for all velocity of
equal constant= sqrt(2gh).only.
.This work used by the pump is giwing back on the pump.
Amen. cordial greetings for all.
E.W.
--
#191,
It's still legal to go .sigless.
.
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| User: "Eugeniusz Wareda" |
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| Title: Re: The ecological power station of 5(five) GigaWatt |
23 Jun 2004 04:44:28 PM |
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Użytkownik "Eugeniusz Wareda" <lala@magma-net.pl> napisał w wiadomości
news:cbcfbd$nk6$1@korweta.task.gda.pl...
Użytkownik "The Ghost In The Machine" <ewill@aurigae.athghost7038suus.net>
napisał w wiadomości news:oe4oq1-6g4.ln1@lexi2.athghost7038suus.net...
In sci.physics, Eugeniusz Wareda
<lala@magma-net.pl>
wrote
on Tue, 22 Jun 2004 20:58:30 +0200
<cba0bh$7hl$1@korweta.task.gda.pl>:
Welcome All!
You should vetically put in to the sea water on the deep
equal 1000meters,the tube of the crossection
equal 1(m)(m).
Fine. Put a 1 km tube into the sea with a 1m^2 cross-section,
and one gets 1,000 metric tonnes or so of seawater.
(See also "caissons".)
I'd have to see how deep the Troll's "feet" are (the Troll
is a gas platform dropped in the sea off Norway some time
back); one of the more interesting construction artifacts
was that the tube was constructed "dry" within the water,
and then flooded later when the unit was being towed out
to sea.
http://www.structurae.de/en/structures/data/s0003286/index.cfm
suggests a 472 m height but that probably includes the platform itself.
.The tube is connected to the input of water pump which is located on
the
level of the sea .
OK.
This pump is pumping the water on hight equal 10m over the level of
sea
water, for examply.
Bear in mind that the tube will be losing water as you're pumping
it out, and that therefore the pump will have to work harder.
If one lets water into the bottom of the tube then one need
really not have the tube at all (unless one really wants deepwater
pumped to the surface, which may interest certain hydrologically-
inclined scientists but is a different problem).
It is the water turbine on which the water is fallen.
Erm....are you postulating a new piece of equipment or
simply trying to have the pump double as a turbine for
the purposes of your computation?
Either way, you're not going to get very far; to pump that
very last liter of water out of your tank (for that's
what the tube has become) will take far more energy than
pumping the first, even if one allows for the small decrease
in g (from 9.8048 to about 9.8033) at the bottom of the tank.
E = mgh, approximately, where h is taken from the center
of the water cube (this isn't quite right but is close enough):
m = 1 kg (1 liter) of water
g = 9.8032 (and that's being optimistic)
h = 1 km + 10 m - 0.5 m = 1009.5 m
E = 9896 J
That first liter, by contrast, only has to be pumped over
a 10.5 m distance:
g = 9.8048
h = 10.5 m
E = 102.9 J
from hight 10 meter over the sea level.The work task
of pump is to keep velocity V of water in pipe as constat
only,
An additional energy requirement is to move the water.
I'm not up on turbulent flows and such but to accelerate
a 1 m^3 (1 metric tonne) "slug" of water at a velocity of
1 m/s is going to require 500 J in a frictionless pipe.
To have a continuous motion of water through a pipe
of 1 m/s of 1 m^3 from a holding-tank requires 500 W.
(Since the feed is generally at the bottom of the
holding tank one gets a substantial chunk of that power
for free -- until the tank is nearly empty, of course.
Of course in your case you've got the opposite problem:
you're taking the output from the *top*. There will be
interesting issues if we ever get a significant amount of
fluids into space; presumably NASA et al have worked this
out already, though, at least for small ammonia compressor
cooling units.)
The velocity of this water cannot be lessen than velocity in the tube
suppling the pump of the water.
Incompressibility, yes.
The second pipe is connected to the outside of the pump.
The task of second tube is to supply a water to the
hight equal h ,and from this to over the water turbin.
The velocity V of the water falling on turbin is equal
V = sqrt(VoVo +2gh) =~120m/s ??!!
Ekin =~5GigaWatt = it is Power of this power station
Now you're just being silly.
[rest snipped]
--
#191,
It's still legal to go .sigless.
***************
^^^^^^^^^^^^^^
&&&&&&&&&
Welcome,
The both ends of a pipe are open. The crossection S of this pipe is equal
1m^2
,and the lenght H of this pipe is
equal 1000meters.The one of two ends of opipe is put
in the sea water to the deep equal H=1000meters under level
of the sea.
The second of two open ends of this pipe is connected
to input of pump.
If the level of meniscus inside is equal as level outside of
the tube then ,in this time the our energetical system is in
the balance of power.
But our system have to be in the oscilating state, first time only .
So we have to do a work W;
dW =Fdh
F = (ro)ghS
W = {(ro)SgHH/2}= MgH/2
This energy must be changed on kinetikal energy and
on the level of sea it is equal Ekin=MVV/2
and the velocity Vmax of water i pipe is equal;
Vmax=sqrt(gH) ~ sqrt( 10m/ss1000m)=100m/s
E kin= Epot=MgH/2
Ekin=1(m^2)(10m/ss)(1000/2)(m) =~5 Gigawatt
From this time, from now, the water is going
to the input of pump automaticaly.
The task for pump is to keep velocity of water as constant ,only.
It is the work in the potential field so it is constant for
all velocity
equal constant= sqrt(2gh).only.,if h= constant
.This work used by the pump is giving back falling on the turbine, because
Vtur =sqrt(V^2+2gh)=sqrt(gH+2gh)
Amen. cordial greetings for all.
E.W.
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