| Topic: |
Science > Physics |
| User: |
"EugeniuszW" |
| Date: |
23 Oct 2005 11:33:02 PM |
| Object: |
5 00 MWatt free energy |
Welcome
We put perpendiculary in the ocean a long tube
The heigh of this tube is H =1000m,
the crossection of this tube ic S=1m^2.
The both ends of this tube are open completly.
We instal the moto-pump on the tube .
The moto-pump is instaled on height equal the level
of water in the ocean.
The moto-pump is pumping the water from ocean ,
on.heigh h =10 m
over the level of water in ocean.
The speed of water in the tube become bigger and bigger,
The theoretical limit of this speed is V=sqrt (gH).
And theoretical limit of this speed is V=~ sqrt( 10*10^3)
and V=100m/s !!!!?
Power P in this case is;
P= Ekinet/10s= (MV^2)/20s
P = ((1000000kg)* (10000[m^2]/s^2))/20s =
5 00 MWatt
Because :
1((kgm^2)/s^3 = 1 Watt
1 J = 1 W * s
..
Sincerelly E.W.
.
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| User: "Jan Panteltje" |
|
| Title: Re: 5 00 MWatt free energy |
24 Oct 2005 05:42:30 AM |
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On a sunny day (Mon, 24 Oct 2005 04:33:02 GMT) it happened "EugeniuszW"
<EugeniuszW@shaw.ca> wrote in <2uZ6f.277418$1i.262343@pd7tw2no>:
Welcome
We put perpendiculary in the ocean a long tube
The heigh of this tube is H =1000m,
the crossection of this tube ic S=1m^2.
The both ends of this tube are open completly.
We instal the moto-pump on the tube .
The moto-pump is instaled on height equal the level
of water in the ocean.
The moto-pump is pumping the water from ocean ,
on.heigh h =10 m
over the level of water in ocean.
The speed of water in the tube become bigger and bigger,
The theoretical limit of this speed is V=sqrt (gH).
And theoretical limit of this speed is V=~ sqrt( 10*10^3)
and V=100m/s !!!!?
Power P in this case is;
P= Ekinet/10s= (MV^2)/20s
P = ((1000000kg)* (10000[m^2]/s^2))/20s =
5 00 MWatt
Because :
1((kgm^2)/s^3 = 1 Watt
1 J = 1 W * s
.
Sincerelly E.W.
You forgot the part where all the water molecules are converted to energy
using E=m.c^2
And even then your answer of 500MW is wrong.
_________________________________________
Usenet Zone Free Binaries Usenet Server
More than 140,000 groups
Unlimited download
http://www.usenetzone.com to open account
.
|
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| User: "EugeniuszW" |
|
| Title: Re: 5 00 MWatt free energy |
24 Oct 2005 09:54:40 PM |
|
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"Jan Panteltje" <pNaonStpealmtje@yahoo.com> wrote in message
news:435cbd0e$1@news.usenetzone.com...
On a sunny day (Mon, 24 Oct 2005 04:33:02 GMT) it happened "EugeniuszW"
<EugeniuszW@shaw.ca> wrote in <2uZ6f.277418$1i.262343@pd7tw2no>:
Welcome
We put perpendiculary in the ocean a long tube
The heigh of this tube is H =1000m,
the crossection of this tube ic S=1m^2.
The both ends of this tube are open completly.
We instal the moto-pump on the tube .
The moto-pump is instaled on height equal the level
of water in the ocean.
The moto-pump is pumping the water from ocean ,
on.heigh h =10 m
over the level of water in ocean.
The speed of water in the tube become bigger and bigger,
The theoretical limit of this speed is V=sqrt (gH).
And theoretical limit of this speed is V=~ sqrt( 10*10^3)
and V=100m/s !!!!?
Power P in this case is;
P= Ekinet/10s= (MV^2)/20s
P = ((1000000kg)* (10000[m^2]/s^2))/20s =
5 00 MWatt
Because :
1((kgm^2)/s^3 = 1 Watt
1 J = 1 W * s
.
Sincerelly E.W.
You forgot the part where all the water molecules are converted to energy
using E=m.c^2
And even then your answer of 500MW is wrong.
_________________________________________
Usenet Zone Free Binaries Usenet Server
More than 140,000 groups
Unlimited download
***********
*************
Welcome
The continous speed of water in the tube,when the moto-pompe is woking is
V=100 m/s.
What the power Pp of motopomp must to be ,to get
500MWatt of the water from this tube ??
We have to lift this mass of water (10^6kg) on 10 m height over the level of
the water in ocean.
The speed of this water should bo constant in this tupe.
In this cause speed of water is 100m/s.
It is moving in the gravitational field.
It is not important for us ,what velocity of water is in
tube; the power Pp of motopumpe is same.
Pp=(Mgh)/t t=H/V = 10s
Pp =(1000kg) *(9.81m/.s^2 )*(10m}/10s
Pp =10MW
We see that power of motopumpe is less than
power created by the generator of electricity.
P =500MW > Pp = 10MW
Sncerely yours E.W.
.
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| User: "EugeniuszW" |
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| Title: Re: 5 00 MWatt free energy |
25 Oct 2005 03:28:31 PM |
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"EugeniuszW" <EugeniuszW@shaw.ca> wrote in message
news:Q7h7f.290433$1i.233556@pd7tw2no...
"Jan Panteltje" <pNaonStpealmtje@yahoo.com> wrote in message
news:435cbd0e$1@news.usenetzone.com...
On a sunny day (Mon, 24 Oct 2005 04:33:02 GMT) it happened "EugeniuszW"
<EugeniuszW@shaw.ca> wrote in <2uZ6f.277418$1i.262343@pd7tw2no>:
Welcome
We put perpendiculary in the ocean a long tube
The heigh of this tube is H =1000m,
the crossection of this tube ic S=1m^2.
The both ends of this tube are open completly.
We instal the moto-pump on the tube .
The moto-pump is instaled on height equal the level
of water in the ocean.
The moto-pump is pumping the water from ocean ,
on.heigh h =10 m
over the level of water in ocean.
The speed of water in the tube become bigger and bigger,
The theoretical limit of this speed is V=sqrt (gH).
And theoretical limit of this speed is V=~ sqrt( 10*10^3)
and V=100m/s !!!!?
Power P in this case is;
P= Ekinet/10s= (MV^2)/20s
P = ((1000000kg)* (10000[m^2]/s^2))/20s =
5 00 MWatt
Because :
1((kgm^2)/s^3 = 1 Watt
1 J = 1 W * s
.
Sincerelly E.W.
You forgot the part where all the water molecules are converted to
energy
using E=m.c^2
And even then your answer of 500MW is wrong.
_________________________________________
Usenet Zone Free Binaries Usenet Server
More than 140,000 groups
Unlimited download
***********
*************
Welcome
The continous speed of water in the tube,when the moto-pompe is woking is
V=100 m/s.
What the power Pp of motopomp must to be ,to get
500MWatt of the water from this tube ??
We have to lift this mass of water (10^6kg) on 10 m height over the level
of
the water in ocean.
The speed of this water should bo constant in this tupe.
In this cause speed of water is 100m/s.
It is moving in the gravitational field.
The work W =Mgh of motopumpe is same.
Power Pp of motopump
Pp=(Mgh)/t t=H/V = 10s
Pp =(1000kg) *(9.81m/.s^2 )*(10m}/10s
Pp =10MW
We see that power of motopumpe is less than
power created by the generator of electricity.
P =500MW > Pp = 10MW
Sncerely yours E.W.
.
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| User: "Puppet_Sock" |
|
| Title: Re: 5 00 MWatt free energy |
25 Oct 2005 02:58:54 PM |
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So, the farmer down the road has a tunnel under the road.
His cows walk between two fields by walking through this
tunnel. And he's quite sure that walking through the tunnel
does not change the number of cows. So, when some city
slicker drives up and starts talking about ordered hooves
and bovine algebras and horn operators, the farmer simply
listens and laughs. As long as all that is happening is
cows walking through the tunnel, the number in equals the
number out. Tunnel-walking is a process that does not change
the number of cows. Cow number is invariant under moving
cows through the tunnel. So, you can't fool the farmer with
funny tunnel arrangements. If he counts cows and gets a
number different from yesterday, he knows something besides
walking through the tunnel has happened.
EugeniuszW wrote:
We put perpendiculary in the ocean a long tube
The heigh of this tube is H =1000m,
the crossection of this tube ic S=1m^2.
The both ends of this tube are open completly.
We instal the moto-pump on the tube .
The moto-pump is instaled on height equal the level
of water in the ocean.
The moto-pump is pumping the water from ocean ,
on.heigh h =10 m
over the level of water in ocean.
Ok, you are lifting water 10 m. Later you've
approximated 1g = 10m/s^2. Fine.
The speed of water in the tube become bigger and bigger,
The theoretical limit of this speed is V=sqrt (gH).
And theoretical limit of this speed is V=~ sqrt( 10*10^3)
and V=100m/s !!!!?
Where does this limit come from?
Also, why is H here 1E3 meters? The pressure difference
only arises due to the last 10 meters where you've taken
water out.
After all, the pressure difference due to the other 990
meters of pipe is zero. Or, to put it another way, if you
do this in a part of the ocean that is 9 km deep, you don't
get the water spraying out 3 times as fast.
Power P in this case is;
P= Ekinet/10s= (MV^2)/20s
P = ((1000000kg)* (10000[m^2]/s^2))/20s =
5 00 MWatt
Because :
1((kgm^2)/s^3 = 1 Watt
1 J = 1 W * s
Well, you are lifting some mass M of water per second.
And you are lifting it a height h against a force g.
That needs an energy per second of:
M g h
Now, how fast are you speeding up water? Well, you've got
a column of water moving at v. But how much water do
you actually get moving that fast each second? Why, quite
obviously, just the amount of water you take out of the
top of the pipe, M per second. So you get a kinetic energy
each second of 1/2 M v^2. But what is v?
Well, suppose it's 100 % efficient. Then the power you
put in lifting a mass M of water up a height of h
against g would be the same as the power added to the
moving water. Or:
1/2 M v^2 = M g h
Or solving for v gives v=sqrt(2gh). Hey, that looks familiar.
You had v = sqrt(gh). You've dropped a factor. And you've
mis-interpreted the height of the pipe as the height over
which a pressure difference occurs.
But you should have known better. Gravity is a conservative
force. If you move stuff around in a gravity field, you
can show quite mathematically that the total energy is
unchanged. With a quite general theorm. So, if you don't
bring any new physics, only gravity, you can be quite sure
from the start that you won't get more out than you put
in at the start.
No matter how the cows walk through the tunnel, you get
the same number.
Socks
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| User: "EugeniuszW" |
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| Title: Re: 5 00 MWatt free energy |
25 Oct 2005 08:32:17 PM |
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"Puppet_Sock" <puppet_sock@hotmail.com> wrote in message
news:1130270334.828282.5900@z14g2000cwz.gglegro.
EugeniuszW wrote:
We put perpendiculary in the ocean a long tube
The heigh of this tube is H =1000m,
the crossection of this tube ic S=1m^2.
The both ends of this tube are open completly.
We instal the moto-pump on the tube .
The moto-pump is instaled on height equal the level
of water in the ocean.
The moto-pump is pumping the water from ocean ,
on.heigh h =10 m
over the level of water in ocean.
Ok, you are lifting water 10 m. Later you've
approximated 1g = 10m/s^2. Fine.
The speed of water in the tube become bigger and bigger,
The theoretical limit of this speed is V=sqrt (gH).
And theoretical limit of this speed is V=~ sqrt( 10*10^3)
and V=100m/s !!!!?
Where does this limit from ?
Also, why is H here 1E3 meters? The pressure difference
only arises due to the last 10 meters where you've taken
water out.
""""""""""""
To push out the all water from the tube, we can
use the compressed air.
Therefore is performed work W1 = MgH/2.
And MgH /2= M/2(V^2) and V=sqrt(gH)
&&&&
That needs an energy per second of:
M g h
No ,it must be Mgh/10s
Welcome
E.W.
.
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| User: "Puppet_Sock" |
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| Title: Re: 5 00 MWatt free energy |
26 Oct 2005 09:13:00 AM |
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EugeniuszW wrote:
"Puppet_Sock" <puppet_sock@hotmail.com> wrote in message
news:1130270334.828282.5900@z14g2000cwz.gglegro.
EugeniuszW wrote:
We put perpendiculary in the ocean a long tube
The heigh of this tube is H =1000m,
the crossection of this tube ic S=1m^2.
The both ends of this tube are open completly.
We instal the moto-pump on the tube .
The moto-pump is instaled on height equal the level
of water in the ocean.
The moto-pump is pumping the water from ocean ,
on.heigh h =10 m
over the level of water in ocean.
Ok, you are lifting water 10 m. Later you've
approximated 1g = 10m/s^2. Fine.
The speed of water in the tube become bigger and bigger,
The theoretical limit of this speed is V=sqrt (gH).
And theoretical limit of this speed is V=~ sqrt( 10*10^3)
and V=100m/s !!!!?
Where does this limit from ?
Also, why is H here 1E3 meters? The pressure difference
only arises due to the last 10 meters where you've taken
water out.
""""""""""""
To push out the all water from the tube, we can
use the compressed air.
Therefore is performed work W1 = MgH/2.
And MgH /2= M/2(V^2) and V=sqrt(gH)
Um. No. As I worked out in the parts you snipped,
it's Mgh on the left, so V = sqrt(2gh). If you
raise a mass M through a distance h against a force g
then the work is M g h. And kinetic energy is 1/2 M v^2.
So you've still dropped that factor of 2. And the limit
comes from conserving energy.
In other words, you've used that gravity is a conservative
force in order to derive this limit. (Well, to fail to
derive this limit, as you've stubornly refused to correct
your error of that missing 2.)
That needs an energy per second of:
M g h
No ,it must be Mgh/10s
Where did the 10 come from? Plus, you snipped the part
where I said M was the mass per second removed from the
pipe. So, in my notation M g h is a power.
EW the thing you need to work on is also in the part of
my post you snipped. Gravity is a conservative force.
If you don't introduce any new physics other than gravity
then you don't get free energy. This is a mathematical
theorm. It does not matter how clever you get, you can't
find a way to make the cows walk through the tunnel and
get more on the other side.
Stop and take that on board. You might stop looking silly.
Socks
.
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| User: "EugeniuszW" |
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| Title: Re: 5 00 MWatt free energy |
26 Oct 2005 05:30:47 PM |
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"Puppet_Sock" <puppet_sock@hotmail.com> wrote in message
news:1130335980.375752.293530@f14g2000cwb.googlegroups.com...
EugeniuszW wrote:
"Puppet_Sock" <puppet_sock@hotmail.com> wrote in message
news:1130270334.828282.5900@z14g2000cwz.gglegro.
EugeniuszW wrote:
We put perpendiculary in the ocean a long tube
The heigh of this tube is H =1000m,
the crossection of this tube ic S=1m^2.
The both ends of this tube are open completly.
We instal the moto-pump on the tube .
The moto-pump is instaled on height equal the level
of water in the ocean.
The moto-pump is pumping the water from ocean ,
on.heigh h =10 m
over the level of water in ocean.
Ok, you are lifting water 10 m. Later you've
approximated 1g = 10m/s^2. Fine.
The speed of water in the tube become bigger and bigger,
The theoretical limit of this speed is V=sqrt (gH).
And theoretical limit of this speed is V=~ sqrt( 10*10^3)
and V=100m/s !!!!?
Where does this limit from ?
Also, why is H here 1E3 meters? The pressure difference
only arises due to the last 10 meters where you've taken
water out.
""""""""""""
To push out the all water from the tube, we can
use the compressed air.
Therefore is performed work W1 = MgH/2.
And MgH /2= M/2(V^2) and V=sqrt(gH)
Um. No. As I worked out in the parts you snipped,
it's Mgh on the left, so V = sqrt(2gh). If you
raise a mass M through a distance h against a force g
then the work is M g h. And kinetic energy is 1/2 M v^2.
So you've still dropped that factor of 2. And the limit
comes from conserving energy.
In other words, you've used that gravity is a conservative
force in order to derive this limit. (Well, to fail to
derive this limit, as you've stubornly refused to correct
your error of that missing 2.)
That needs an energy per second of:
M g h
No ,it must be Mgh/10s
Where did the 10 come from? Plus, you snipped the part
where I said M was the mass per second removed from the
pipe. So, in my notation M g h is a power.
EW the thing you need to work on is also in the part of
my post you snipped. Gravity is a conservative force.
If you don't introduce any new physics other than gravity
then you don't get free energy. This is a mathematical
theorm. It does not matter how clever you get, you can't
find a way to make the cows walk through the tunnel and
get more on the other side.
Stop and take that on board. You might stop looking silly.
Socks
&&&&&&&&&
H =1000m , V=100m/s , t = H/V =10s !!
============
M = (1000m *1m^2)*1000kg/m^3=(10^6)kg
1/2MgH > Mgh
500000*10*1000=5.000.000.000W=5 GWatt
Mgh = 1000000*10*10=100MW
1/2MV^2 = 500000*10000=
= 5.000.000.000W=5GWatt
EW.
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| User: "EugeniuszW" |
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| Title: Re: 5 00 MWatt free energy |
27 Oct 2005 01:23:54 AM |
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"EugeniuszW" <EugeniuszW@shaw.ca> wrote in message
news:rsT7f.304423$oW2.18756@pd7tw1no...
"Puppet_Sock" <puppet_sock@hotmail.com> wrote in message
news:1130335980.375752.293530@f14g2000cwb.googlegroups.com...
EugeniuszW wrote:
"Puppet_Sock" <puppet_sock@hotmail.com> wrote in message
news:1130270334.828282.5900@z14g2000cwz.gglegro.
EugeniuszW wrote:
We put perpendiculary in the ocean a long tube
The heigh of this tube is H =1000m,
the crossection of this tube ic S=1m^2.
The both ends of this tube are open completly.
We instal the moto-pump on the tube .
The moto-pump is instaled on height equal the level
of water in the ocean.
The moto-pump is pumping the water from ocean ,
on.heigh h =10 m
over the level of water in ocean.
Ok, you are lifting water 10 m. Later you've
approximated 1g = 10m/s^2. Fine.
The speed of water in the tube become bigger and bigger,
The theoretical limit of this speed is V=sqrt (gH).
And theoretical limit of this speed is V=~ sqrt( 10*10^3)
and V=100m/s !!!!?
Where does this limit from ?
Also, why is H here 1E3 meters? The pressure difference
only arises due to the last 10 meters where you've taken
water out.
""""""""""""
To push out the all water from the tube, we can
use the compressed air.
Therefore is performed work W1 = MgH/2.
And MgH /2= M/2(V^2) and V=sqrt(gH)
Um. No. As I worked out in the parts you snipped,
it's Mgh on the left, so V = sqrt(2gh). If you
raise a mass M through a distance h against a force g
then the work is M g h. And kinetic energy is 1/2 M v^2.
So you've still dropped that factor of 2. And the limit
comes from conserving energy.
In other words, you've used that gravity is a conservative
force in order to derive this limit. (Well, to fail to
derive this limit, as you've stubornly refused to correct
your error of that missing 2.)
That needs an energy per second of:
M g h
No ,it must be Mgh/10s
Where did the 10 come from? Plus, you snipped the part
where I said M was the mass per second removed from the
pipe. So, in my notation M g h is a power.
EW the thing you need to work on is also in the part of
my post you snipped. Gravity is a conservative force.
If you don't introduce any new physics other than gravity
then you don't get free energy. This is a mathematical
theorm. It does not matter how clever you get, you can't
find a way to make the cows walk through the tunnel and
get more on the other side.
Stop and take that on board. You might stop looking silly.
Socks
&&&&&&&&&
H =1000m , V=100m/s , t = H/V =10s !!
============
M = (1000m *1m^2)*1000kg/m^3=(10^6}kg
1/2MgH= 1/2(1000.000kg )(10 m/s^2)(1000m)=
= 5000MJ
Mgh =1000000 * 10*10=100MJ
Power of pump Pp =Mgh/10s= 10Mwatt
Power of the water P =1/2MgH/10s=500MWatt
P/Pp = 500MW/10MW = 50
E.W.
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| User: "Boris Mohar" |
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| Title: Re: 5 00 MWatt free energy |
26 Oct 2005 07:41:17 AM |
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|
On Wed, 26 Oct 2005 01:32:17 GMT, "EugeniuszW" <EugeniuszW@shaw.ca> wrote:
"Puppet_Sock" <puppet_sock@hotmail.com> wrote in message
news:1130270334.828282.5900@z14g2000cwz.gglegro.
EugeniuszW wrote:
We put perpendiculary in the ocean a long tube
The heigh of this tube is H =1000m,
the crossection of this tube ic S=1m^2.
The both ends of this tube are open completly.
We instal the moto-pump on the tube .
The moto-pump is instaled on height equal the level
of water in the ocean.
The moto-pump is pumping the water from ocean ,
on.heigh h =10 m
over the level of water in ocean.
Ok, you are lifting water 10 m. Later you've
approximated 1g = 10m/s^2. Fine.
The speed of water in the tube become bigger and bigger,
The theoretical limit of this speed is V=sqrt (gH).
And theoretical limit of this speed is V=~ sqrt( 10*10^3)
and V=100m/s !!!!?
Where does this limit from ?
Also, why is H here 1E3 meters? The pressure difference
only arises due to the last 10 meters where you've taken
water out.
""""""""""""
To push out the all water from the tube, we can
use the compressed air.
Therefore is performed work W1 = MgH/2.
And MgH /2= M/2(V^2) and V=sqrt(gH)
&&&&
That needs an energy per second of:
M g h
No ,it must be Mgh/10s
Welcome
E.W.
And the energy to compress air will come from?
--
Boris Mohar
.
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| User: "EugeniuszW" |
|
| Title: Re: 5 00 MWatt free energy |
26 Oct 2005 06:47:00 PM |
|
|
"Boris Mohar" <borism_-void-_@sympatico.ca> wrote in message
news:5auul1dlao5p7fjo65t5f704ge9djhejrl@4ax.com...
On Wed, 26 Oct 2005 01:32:17 GMT, "EugeniuszW" <EugeniuszW@shaw.ca> wrote:
"Puppet_Sock" <puppet_sock@hotmail.com> wrote in message
news:1130270334.828282.5900@z14g2000cwz.gglegro.
EugeniuszW wrote:
We put perpendiculary in the ocean a long tube
The heigh of this tube is H =1000m,
the crossection of this tube ic S=1m^2.
The both ends of this tube are open completly.
We instal the moto-pump on the tube .
The moto-pump is instaled on height equal the level
of water in the ocean.
The moto-pump is pumping the water from ocean ,
on.heigh h =10 m
over the level of water in ocean.
Ok, you are lifting water 10 m. Later you've
approximated 1g = 10m/s^2. Fine.
The speed of water in the tube become bigger and bigger,
The theoretical limit of this speed is V=sqrt (gH).
And theoretical limit of this speed is V=~ sqrt( 10*10^3)
and V=100m/s !!!!?
Where does this limit from ?
Also, why is H here 1E3 meters? The pressure difference
only arises due to the last 10 meters where you've taken
water out.
""""""""""""
To push out the all water from the tube, we can
use the compressed air.
Therefore is performed work W1 = MgH/2.
And MgH /2= M/2(V^2) and V=sqrt(gH)
&&&&
That needs an energy per second of:
M g h
No ,it must be Mgh/10s
Welcome
E.W.
And the energy to compress air will come from?
--
Boris Mohar
^^^^^^^^^
^^^^^^^^^^^
Welcome
To start such power station, we can use the compressed air or pumping the
water from ocean .
Power station consist of electrical motopump ,turbine,
generator of elecricity.
All these devices work together.
The motopump is installed on the top of tube,on the level
of water in the ocean.
To start this power station use energy from out side, but power station can
work many years without such necesserity
Sincerely yours E.W.
.
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