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A new energy source; maybe/maybe not?

Skwim

Veteran Member
I wouldn't laugh too quickly people. I remember when people use to laugh at this perpetual motion waterfall

c_fit,fl_progressive,q_80,w_470.jpg



But take a look at THIS!!!



So there!

.

 
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Twilight Hue

Twilight, not bright nor dark, good nor bad.
Eh? no.

The point of applying physical science to problems such as this is to save yourself the trouble of building the bloody thing. But of course if a person can't understand physical science, then you may be right, in that the only way they can convince themselves is to actually make it. But James is obviously not going to do that. He is going to hawk his useless idea round the internet. :rolleyes:
That's because I'm a bottom line person.

The real thing will always make or break the paper fantasy* .

*Of course the saying that if something doesn't quite work as planned , the following quip usually suffices. "Oh well , back to the drawing board".
 

james dixon

Well-Known Member
Premium Member
HOW IT WORKS
I would build the thing to prove it works but to do that it would have to be built full scale, in the ocean as advertised. You can’t make a scaled model, it wouldn’t work.

A three-dimensional computer model would work if it was programed to operate with the same parameters as the real thing including depth and sea pressure; ATM’s and it would have to operate in real time..

For every ATM in depth (33') an air bubble is compressed to half its size.

This also works in the reverse. An air bubble with a radius of 10 FT at 15 ATM's will expand to twice its size (r=20) at 14 ATM's.
:)-
 

exchemist

Veteran Member
HOW IT WORKS
I would build the thing to prove it works but to do that it would have to be built full scale, in the ocean as advertised. You can’t make a scaled model, it wouldn’t work.

A three-dimensional computer model would work if it was programed to operate with the same parameters as the real thing including depth and sea pressure; ATM’s and it would have to operate in real time..

For every ATM in depth (33') an air bubble is compressed to half its size.

This also works in the reverse. An air bubble with a radius of 10 FT at 15 ATM's will expand to twice its size (r=20) at 14 ATM's.
:)-
Perhaps you are getting there, by degrees.

What you say is correct. So what that means is that the energy needed to compress the bubble is the same as the energy it gives you back by rising to the surface. Right?

So your device cannot produce more energy than it takes to run it.
 

Ouroboros

Coincidentia oppositorum
Principles to run the machine

There are a few basic principles that you cannot deny.

[1] an enclosed container (X) of air submerged in water has a lifting force (Y) equal to the volume of the water displaced minus the weight of the container;
[yes] [no]

[2] connection multiple containers one on top of the other creates a combined lifting force of (Y)+ (Y)+ (Y)+ (Y)+ (Y)+ (Y)+ (Y)+ (Y)+ (Y)+ (Y)
Which is a greater lifting force than (Y);
[yes] [no]

[3] the energy needed to fill one container is equal to the energy needed to sustain the combined lifting force of the 10 (ten) containers referenced above;
[yes] [no]


:)-
A thought I had, how about instead of getting compressed air down, why not just have a heater at the bottom boiling the water locally to release gases? Then the efficiency of the system depends more on the generator and heating elements and such.
 

james dixon

Well-Known Member
Premium Member
The lift won't generate enough power to compress the air required to run it.
Tom
I believe you are missing the core point here.

In the following energy = (E)
In the following force = (F) energy to compress

Balloon one has a (F1) that equals energy = (E1)
Balloon two has a (F2) that equals energy = (E2)
Balloon three has a (F3) that equals energy = (E3)
Balloon four has a (F4) that equals energy = (E4)

When you have them all attached to each other
they have a combined lifting force of (F1)+(F2)+(F3)+(f4)

While at the same time it only takes the force of (F1) to keep the system running.

You now have (E1)+(E2)+(E3)+(e4) combined working together

Or so I believe
 

james dixon

Well-Known Member
Premium Member
A thought I had, how about instead of getting compressed air down, why not just have a heater at the bottom boiling the water locally to release gases? Then the efficiency of the system depends more on the generator and heating elements and such.
I designed a contraption like your idea that was along the mid-Atlantic ridge where molten lava was superheating seawater that did not flash into steam because of the extreme pressure at that depth.
The idea also extracted the minerals as well.


SEAFLUM1-page-001 (1).jpg
 

exchemist

Veteran Member
I believe you are missing the core point here.

In the following energy = (E)
In the following force = (F) energy to compress

Balloon one has a (F1) that equals energy = (E1)
Balloon two has a (F2) that equals energy = (E2)
Balloon three has a (F3) that equals energy = (E3)
Balloon four has a (F4) that equals energy = (E4)

When you have them all attached to each other
they have a combined lifting force of (F1)+(F2)+(F3)+(f4)

While at the same time it only takes the force of (F1) to keep the system running.

You now have (E1)+(E2)+(E3)+(e4) combined working together

Or so I believe
Please think. You have to expend your "energy to compress" to fill each balloon, not only once.
 

james dixon

Well-Known Member
Premium Member
Air Bubble
Air is a gas
gases compress with pressure

Assuming constant temperature, Boyle's - a pressure of 19 atmospheres (1 atmosphere at sea level + 18 atmospheres for being 600ft under water)

One square food of water at 1 ATM weighs 64 lbs.

One square foot of water will compress to 1/2 square foot at 2 ATM

EXAMPLE—
Surface volume (X) = 300 CF
Upper force (Y) = 64 lbs.

Volume at 2 atm (X)/2 = 150 CF = 150 X 64 = 9,600 pounds of upward force

Volume at 3 atm (X)/3 = 100 CF = 100 X 64 = 6,400 pounds of upward force

Volume at 6 atm (X)/6 = 50 CF = 50 X 64 = + Y X CF = 3,200 lbs.

Volume at 9 atm (X)/9 = 33.33 CF; upward force = Y X CF = 2,133 lbs.

Volume at 12 atm (X)/12 = 25 CF; upward force = Y X CF = 1,600 lbs.

Volume at 15 atm (X)/15 = 20 CF; upward force = Y X CF = 1,280 lbs.

Volume at 18 atm (X)/18 = 16.66 CF; upward force = Y X CF = 1,066 lbs.

Total upward force = 25,259 lbs.

An air bubble rises at 2 feet a second

2 X 60 = 120 feet per minute
120 X 60 = 7,200 feet per hour
5,280 feet in a mile

7,200/5,280 = 1.36 miles an hour

TOTAL force 25,259 lbs. traveling at 1.36 miles an hour

globule

I'm going to plug these numbers into the contraption and see what happens next

:)-
 

exchemist

Veteran Member
Air Bubble
Air is a gas
gases compress with pressure

Assuming constant temperature, Boyle's - a pressure of 19 atmospheres (1 atmosphere at sea level + 18 atmospheres for being 600ft under water)

One square food of water at 1 ATM weighs 64 lbs.

One square foot of water will compress to 1/2 square foot at 2 ATM

EXAMPLE—
Surface volume (X) = 300 CF
Upper force (Y) = 64 lbs.

Volume at 2 atm (X)/2 = 150 CF = 150 X 64 = 9,600 pounds of upward force

Volume at 3 atm (X)/3 = 100 CF = 100 X 64 = 6,400 pounds of upward force

Volume at 6 atm (X)/6 = 50 CF = 50 X 64 = + Y X CF = 3,200 lbs.

Volume at 9 atm (X)/9 = 33.33 CF; upward force = Y X CF = 2,133 lbs.

Volume at 12 atm (X)/12 = 25 CF; upward force = Y X CF = 1,600 lbs.

Volume at 15 atm (X)/15 = 20 CF; upward force = Y X CF = 1,280 lbs.

Volume at 18 atm (X)/18 = 16.66 CF; upward force = Y X CF = 1,066 lbs.

Total upward force = 25,259 lbs.

An air bubble rises at 2 feet a second

2 X 60 = 120 feet per minute
120 X 60 = 7,200 feet per hour
5,280 feet in a mile

7,200/5,280 = 1.36 miles an hour

TOTAL force 25,259 lbs. traveling at 1.36 miles an hour

globule

I'm going to plug these numbers into the contraption and see what happens next

:)-
Square food? Surely you mean square meal?

But seriously, you might start by getting your units right. Volumes are measured in cubic feet (or cubic metres), not square feet. Also, to say an air bubble rises at 2ft/second is just rubbish. Your air bubbles have to be connected to your conveyor in order for you to extract any energy.

It is obvious you have no idea at all what you are doing and that you won't learn, since you fail to engage at all with any of the criticisms that have been made in this thread.

Time to stop reading your rubbish, I think.

[click]
 

james dixon

Well-Known Member
Premium Member
[1] The lifting force of an air bubble (balloon) is equal to the water being displaced.; call this energy [X] This you cannot deny.
[2] In water, an air bubble expands as it rises. This you cannot deny.
[3] The lifting force of multiple balloons in a vertical row, all attached to each other has a lifting force equal to the combined lifting force of all the balloons; call this energy (F)
[4] Energy [X]+[X]+[X]+[X]+[X] = [F]
[5] [F] five (5) times greater than [X] at any one moment in time. This you cannot deny
[6] To maintain this process all you have to add, at any one moment in time is [X] energy to get an output of [F]
This you cannot deny
seaengine2.jpg
 

Subduction Zone

Veteran Member
[1] The lifting force of an air bubble (balloon) is equal to the water being displaced.; call this energy [X] This you cannot deny.
[2] In water, an air bubble expands as it rises. This you cannot deny.
[3] The lifting force of multiple balloons in a vertical row, all attached to each other has a lifting force equal to the combined lifting force of all the balloons; call this energy (F)
[4] Energy [X]+[X]+[X]+[X]+[X] = [F]
[5] [F] five (5) times greater than [X] at any one moment in time. This you cannot deny
[6] To maintain this process all you have to add, at any one moment in time is [X] energy to get an output of [Y]
This you cannot deny
View attachment 33399
You forgot, it takes at least as much energy to pump the air down as you could get out. And that is assuming a perfect frictionless system. Your device whole consume energy, not
produce it.
 

james dixon

Well-Known Member
Premium Member
You forgot, it takes at least as much energy to pump the air down as you could get out.
Whatever the energy is needed to pump the air down is (X) while I am getting out the expanding, lifting force of (X}+(X)+(X)+(X) out-? **at any moment in time**

Where am I going wrong-
ok, ok, ok, I'm wrong; just don't know why ?
:)-
 

james dixon

Well-Known Member
Premium Member
A lot of people here and elsewhere have been telling me that this gadget is unworkable; they have been correct. My calculations were gust wrong. I have gone back and crunched the numbers once again.

Attached is my latest version.

This version only goes to a depth of 198 feet and it generates a pulling force of 1,016,183 pounds of upward pull, puling at an accelerating speed of over 4 feet per second.


 

Attachments

  • SEAP4.pdf
    75.7 KB · Views: 0

exchemist

Veteran Member
A lot of people here and elsewhere have been telling me that this gadget is unworkable; they have been correct. My calculations were gust wrong. I have gone back and crunched the numbers once again.

Attached is my latest version.

This version only goes to a depth of 198 feet and it generates a pulling force of 1,016,183 pounds of upward pull, puling at an accelerating speed of over 4 feet per second.


You are deliberately missing the point. Your gadget can never work as a source of energy, in any form, as to do so it would need to break the laws of thermodynamics. The whole notion is silly and pointless.

To be taken at all seriously, you need to quote, not only the force you generate but the work done per unit compressed air supplied, and set that against the energy needed to compress the air. If you do that, I guarantee the energy input to compress the air will be more than the energy output from your useless device.
 
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