Seaengine design to use combined rising air bubbles as force

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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;

[2] connecting 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);

[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.

This mechanical process can be converted to electrical output.

If it takes less power to keep the system running than the output created; then this is a positive idea. If not; this is a dead horse with nowhere to go.

See attachment. NOTE: open with “paint”

Comments pro or con are welcome

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I need to know if someone has seen this topic and likes/dislikes it for whatever reason

 

 

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21 hours ago, tooldtocare said:

I need to know if someone has seen this topic and likes/dislikes it for whatever reason

still waiting for an answer

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maybe the readers here have a reading comprehension problem?

[G B Reid MIMechE, SIMarEST 77]

This is a system that has been used for many years to provide an upthrust in dredging. This displaced water at the top provides a negative pressure at the bottom which is automatically pressure adjusted, and very - often dangerously - powerful!

The pressure needed to create the bubbles at depth is slightly greater than the depth of the water, so there is still a net loss...but that does not mean that the system cannot be utilised do a specific end...but it should not be seen as a route to perpetual motion...e.g....if you can get compressed air bottled at a sufficiently low cost, you could use this system to displace water to power a hydro turbine.

There are methods that do produce a net gain in energy utlilsiable...heat pumps utilise a phase transition to give an energy boost and a "net gain" against input energy.  e.g. an aircon unit is an example of a heat pump...to quote one I've just looked ups...it takes an ~800W input and can give a heating or cooling output of about 3.3-3.4kW, a significant net gain!

 

{Note: Not typos...British-English spellings!}

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On 3/16/2021 at 6:02 AM, G B Reid MIMechE, SIMarEST said:

This is a system that has been used for many years to provide an upthrust in dredging. This displaced water at the top provides a negative pressure at the bottom which is automatically pressure adjusted, and very - often dangerously - powerful!

The pressure needed to create the bubbles at depth is slightly greater than the depth of the water, so there is still a net loss...but that does not mean that the system cannot be utilised do a specific end...but it should not be seen as a route to perpetual motion...e.g....if you can get compressed air bottled at a sufficiently low cost, you could use this system to displace water to power a hydro turbine.

There are methods that do produce a net gain in energy utlilsiable...heat pumps utilise a phase transition to give an energy boost and a "net gain" against input energy.  e.g. an aircon unit is an example of a heat pump...to quote one I've just looked ups...it takes an ~800W input and can give a heating or cooling output of about 3.3-3.4kW, a significant net gain!

 

{Note: Not typos...British-English spellings!}

Thank you for your input

On 3/16/2021 at 6:02 AM, G B Reid MIMechE, SIMarEST said:

The pressure needed to create the bubbles at depth is slightly greater than the depth of the water, so there is still a net loss...but that does not mean that the system cannot be utilised do a specific end...but it should not be seen as a route to perpetual motion...e.g....if you can get compressed air bottled at a sufficiently low cost, you could use this system to displace water to power a hydro turbine.

I sincerely appreciate your response. I knew there would be losses in the efficiency of the machine but because of my limited understanding of the machine’s inner workings I could not pinpoint where these losses would come from. Having said that even with these losses the machine would still operate and produce some useful work.

To calculate how much useful work the seaengine can produce a few properties need to be determined.

[1] what is the speed of the rising air bubbles?

[2] what is the lifting force of the rising bubbles?

[3] how much energy is required to fill the lowest balloon with air?

[4] other operating facts that need to be considered.

I calculated that the seaengine has a combined lifting force of 118,428 pounds of lifting force.

That is a static force.  There is no motion involved, it is equivalent to a large weight.

Questions 1-4 above need to be determined to evaluate the energy output.

Personally, I am enjoying the challenges involved in determining the above regardless of the ultimate conclusion is that it will not work, end of story.

 why will it not work?

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I am finding the answers on this forum interesting because I posed this same discussion on an electric forum and the response I got were in mechanical terms, here in a mechanical forum the response I am getting is electrical.

[G B Reid MIMechE, SIMarEST 77]

I believe this is known as a "hydraulic bouyancy lifting engine" (the Hindenburg was an Air-buoyancy lifting engine)

The easiest way to calculate the usable force is to try it out with a tube set vertically in water and catch the water that comes out over a set period of time (say 10 seconds)...the usable energy (Work) can then be calculated back from the volume of water that has been ejected. Work done is Force times distance (W = Fs). The volume of the water lifted would be linked to the mass of the water x G...this would be calculated as energy gained (PE) = mgh.

With regards to the amount of force each balloon applies, it is a buoyancy, so it is the volume of the balloon that displaces water - the volume of water displaced (less the mass of the balloon and whatever is tethered to it) is the upward force. Compared to air, the water is very viscous, so the terminal velocity of the rising balloon will be limited suite quickly by the resistance of the water it is pushing through (your diagram states 3ft per second...?)

The energy for the lowest balloon will be near as dammit equal to the pressure at the point of origin - very slightly higher - though it may have to be significantly higher depending on the elasticity of the balloon...the "danger" would be the balloons bursting as they near the surface due to the expansion of the air as the pressure drops.

For reference "Electrical answers " is not quite accurate. Energy is measured in Joules. Watts is a measure of energy applied per second, so 1 Watt would be one Joule per second.  800 Watts is 800 Joules per second. in my first paragraph, the "mgh" is measured in joules.  In the air-con unit, it benefits from a "free" energy boost as a result of an evaporative (or condensing) phase resulting in a significantly greater usable energy output than that input. Work done is also measured in Joules. The difference between the usable work done and the energy used is the efficiency.

(Note for equations "s" is "Spacial displacement" - ie distance. Time is always "t". I know many new to the field get "s" and seconds confused, so its worth mentioning.)

Unfortunately, this is nowhere near enough information for you to go on - I understand - but hopefully it is a start in the right direction. Unfortunately, there is only so much time that can be spent on "interest items" when work beckons! If I can spare some more time and expand on this, I will do, but my visits of late tend to be sporadic and short-lived due to other commitments, so there are no promises made!

 

Good luck...

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G B Reid, I used 3 feet per second because I read an article that stated that an air bubble in water will rise at 3 fs. Looking further into this I have discovered that it is a bit more complicated than I first thought.

Since the air in the balloons are expanding as they rise the enlarging balloons will accelerate the rising speed.

Based on your post, I need to convert everything into Joules but first I need to come up with a speed and from there I can work on determining if I am using more energy to keep the system running than I am getting out of it.

I have a land surveying company in Austin, Texas which has nothing to do with the seaengine. I dreamed this up one night and have not been able to let it go since. As you are aware, land surveyors are not engineers or electricians and this idea is way over my pay grade.

Having said that, I am willing to pay $1,000us to anyone who can take this idea, look at it and then write an epitaph for the seaengine so that I can insert it into the drawing before putting it in the dead drawer for good.

Thanks for your input

thomas@waterloosurveyors.com

wish you and yours the best today and for evermore.

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I came up with an idea that could possibly produce more useful energy than is required to keep the system running & no it is not a perpetual motion machine.

The output is mechanical to electric and that is why I am introducing this idea here.

SeaPower description

Attached is a diagram that details a new energy generating power source using the expanding rise of air underwater as a lifting force.

This is the same principal that keeps a boat afloat. A cubic foot of air under water has a lifting force of 67 pounds. A ship/boat that weighs 2,000 pounds must displace (2,000/67)=29.85  cubic feet of water to stay afloat.

In the diagram, there is a vertical row of balloons. The lower balloon or inverted umbrella; is injected with 40,000 cubic feet of air compressed to 15 ATM resulting in a volume of 2,666.66 cubic feet of air.

When the first balloon rises 99 feet to 12 ATM it will expand to 3,333.33 cubic feet

As each balloon rises it will expand from 2,666.66 feet until it reaches the surface with a volume of 40,000 cubic feet.

There are fifteen (15) balloons each tied together in a vertical row.

The combined lifting force of the 15 balloons is 118,428 pounds of continuous lifting force.

In order to maintain this lifting force, the lowest balloon must be injected with 40,000 cubic feet of air compressed to 2,666.66 cubic feet. When it rises a new balloon replaces it in a circular motion, as the diagram attached shows.

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;

[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)

[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.

That is why I am here. I need to calculate:

[1] the electrical power of a pulling force of 118,428 foot pounds can produce.

[2] the electric power needed to compress 40,000 cf down to 2,666.666 cubic feet

[3] the speed of the rising balloons

Comments please

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Machine

an apparatus using or applying mechanical power and having several parts, each with a definite function and together performing a particular task.

tinyurl.com/7e4v5vuk

aka: SeaEngine

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As a side note:

Machine

an apparatus using or applying mechanical power and having several parts, each with a definite function and together performing a particular task.

tinyurl.com/7e4v5vuk

 air pressure at sea level about 14.7 pounds per square inch

One atmosphere (101.325 kPa or 14.7 psi) is also the pressure caused by the weight of a column of fresh water of approximately 10.3 m (33.8 ft). Thus, a diver 10.3 m underwater experiences a pressure of about 2 atmospheres (1 atm of air plus 1 atm of water). Conversely, 10.3 m is the maximum height to which water can be raised using suction under standard atmospheric conditions.

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On 3/22/2021 at 7:49 AM, G B Reid MIMechE, SIMarEST said:

With regards to the amount of force each balloon applies, it is a buoyancy, so it is the volume of the balloon that displaces water - the volume of water displaced (less the mass of the balloon and whatever is tethered to it) is the upward force

Can we then agree that a cubic foot of sea water weighs 67 lbs. It takes a force of 67 pounds on an empty cube to submerge the cube from surface to one (1) foot deep into the water.

Do we agree-?

On 3/22/2021 at 7:49 AM, G B Reid MIMechE, SIMarEST said:

For reference "Electrical answers " is not quite accurate. Energy is measured in Joules. Watts is a measure of energy applied per second, so 1 Watt would be one Joule per second.  800 Watts is 800 Joules per second. in my first paragraph, the "mgh" is measured in joules.  In the air-con unit, it benefits from a "free" energy boost as a result of an evaporative (or condensing) phase resulting in a significantly greater usable energy output than that input. Work done is also measured in Joules.

Can you convert a constant force of 118,428 pounds of force traveling at 3 feet per second-?

On 3/22/2021 at 7:49 AM, G B Reid MIMechE, SIMarEST said:

Unfortunately, this is nowhere near enough information for you to go on - I understand - but hopefully it is a start in the right direction.

Unfortunately, this is nowhere near enough information for you to go on - I understand - but hopefully it is a start in the right direction.

On 3/22/2021 at 7:49 AM, G B Reid MIMechE, SIMarEST said:

Unfortunately, this is nowhere near enough information for you to go on - I understand - but hopefully it is a start in the right direction. Unfortunately, there is only so much time that can be spent on "interest items" when work beckons! If I can spare some more time and expand on this, I will do, but my visits of late tend to be sporadic and short-lived due to other commitments, so there are no promises made!

Thanks, I enjoyed your thoughts, again thanks,

May you and yours have a great day and beyond.

On 3/22/2021 at 7:36 PM, tooldtocare said:

G B Reid,

I gave you my company email address. if you sent anything my spam blocker may have bounced your address. 

If so, sorry I will try to fix it

If not, ignore, continue on