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Can this machine create more energy than consumed?

shockmetwo

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Can this machine create more energy than consumed?
If you why & if not why not?
An energy design that may lead us to tomorrow
SeasEngine design-potential
Following is the process-
At 18 ATM, 594 feet down, the balloon is injected with air that is compressed to 36 cubic feet.
At 15 ATM that same 36 cubic feet will expand to 108 CF
At 12 ATM that same 36 cubic feet will expand to 324 CF
At 9 ATM that same 36 cubic feet will expand to 972 CF
At 6 ATM that same 36 cubic feet will expand to 2,916 CF
At 3 ATM that same 36 cubic feet will expand to 8,748 CF
At 1 ATM that same 36 cubic feet will expand to 26,244 CF
Having stated the above, the combined lifting force is---
39,312 CF X 64 lbs. = 2,515,968-foot lbs. of lifting force.
speed of rising bubble
Using his model and my assumptions, the balloon starts off with a radius of 6.2cm at depth of 100ft or 30.8m and a terminal velocity of about 8.28 cm/s. As the balloon rises to the surface it's volume will increase 4.19 times so it's final radius is 10 cm with a final velocity of 6.6cm/s.
Principles to run the machine
[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 minus the energy needed to keep it running.
; [yes] [no]
 

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wwfeldman

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i have not examined your process very much, however:

how much energy do you need to compress how much air to 36 cubic feet at 18 ATM?
how do you get the air to 594 feet below sea level? that likely takes a fair amount of energy

compressing air tends to increase its temperature
de-compressing air tend to decrease its temperature
how do you take that onto account?

how will you extract the energy?

your model at the end - if the sizes and velocities are in cm and cm/s respectively, there isn't
much energy to extract

your three items at the bottom:
1) buoyant force is related to the mass of displaced water and the mass of the stuff that displaced it,
not the volume of the stuff that displaced it, as the volume of the displaced water is the same as the
volume of the stuff that displaced the water.

2)combining several of these things on top of one another does not yield Y + Y + ...
as the Y changes as the volume of the air in the balloon changes as the balloon rises.

3)? don't follow this at all

i crunched a few of the numbers - they don't work

exotic energy schemes usually turn out to be impractical for a variety of real world issues
and engineering capabilities

i do not think this plan works
 
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    FvM

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Easy peasy

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the above is correct - you have not considered all the required energy inputs to the system
 

Kajunbee

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Maybe as a energy storage device. [video]https://www.google.com/amp/s/spectrum.ieee.org/energy/renewables/hydrostor-wants-to-stash-energy-in-underwater-bags.amp.html[/video]
 

FvM

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i crunched a few of the numbers - they don't work
Yes. I stumbled upon the first lines which obviously don't correspond to PV = nRT (ideal gas law).
At 18 ATM, 594 feet down, the balloon is injected with air that is compressed to 36 cubic feet.
At 15 ATM that same 36 cubic feet will expand to 108 CF
At 12 ATM that same 36 cubic feet will expand to 324 CF
(and so on)
When expanding 36 CF from pressure 18 to 15, the gas volume rises to 36*(18/15) = 43.2 CF (isothermal process) respectively 36*(18/15)^(5/7) = 41.0 CF (adiabatic process)
 

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