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Author Topic: Submersible Engine Design  (Read 25780 times)

TommeyLReed

• Hero Member
• Posts: 554
Submersible Engine Design
« on: January 12, 2011, 11:01:18 PM »
This is another project I'm working on, is using displacement of water to convert it into direct rotational energy.
I should have a prototype done soon.

What do you think?

http://www.youtube.com/watch?v=AnraqeoGaNU

http://overunitynow.com/page26

Tommey Reed

exnihiloest

• Hero Member
• Posts: 715
Re: Submersible Engine Design
« Reply #1 on: January 13, 2011, 11:52:38 AM »
This is another project I'm working on, is using displacement of water to convert it into direct rotational energy.
I should have a prototype done soon.

What do you think?

http://www.youtube.com/watch?v=AnraqeoGaNU

http://overunitynow.com/page26

Tommey Reed

When you provide air to a cone at the bottom, you replace a certain water volume with air. As water is incompressible, this air volume can be made only by raising the water above, what is strictly equivalent to raise to the surface the same volume of water as the volume of air in the cone. Work is v*d*g*h where v is the air volume in a cone at the bottom, d the water volumetric mass, and h the height.
It follows that the work needed to compress air in order to fill a cone is exactly what the system will provide back when the cone will rise up, restoring towards the bottom the volume of water that had been displaced upwards.
No extra energy. Buoyancy and Archimedes principle is a well known phenomenon.

quarktoo

• Guest
Re: Submersible Engine Design
« Reply #2 on: January 13, 2011, 12:11:50 PM »
As the air rises, it expands and displaces more water assuming you did not fill the cup all the way.

Does a cubic foot of displacement have the same amount of buoyancy 100 feet below the surface as it does just below the surface? Yes because the water is incompressible.

So the energy you gain is due to expansion of the water as the cup rises. The taller the bucket line the more efficient it gets.

Sound about right?

tbird

• Sr. Member
• Posts: 317
Re: Submersible Engine Design
« Reply #3 on: January 13, 2011, 10:02:20 PM »
Tommey Reed,

i think you have overlooked something.  436.96 lb of upward force sounds like a lot, but you should look closer.

let's say the compressor is a simple box, 1 foot x 1 foot x 1 foot (1 cubic foot).  the top is moveable and when weight is added to the top, it would go inside compressing the air.  to get your 4.4 psi (pounds per square inch), you would need 144 (1 sq ft) x 4.4 = 633.6 lbs.  of course the volume inside will be reduced also.  this sounds like a bit more than your 436.96 you predict.

maybe you should have another look at this.

tom

TommeyLReed

• Hero Member
• Posts: 554
Re: Submersible Engine Design
« Reply #4 on: January 13, 2011, 10:53:17 PM »
My data say that 4.4(salt water), psi is needed to force air under water at a depth of 10ft. fill the cube (12^3) 1728 cu/in having a displacement of 1cu/ft of are. This will produce at a 10ft lift:(62.3*10)=623lb
pressure ((4.4 psi x 1728)/12)=633.6lb is what I get.
Now as you may know to compress air at a rate of 3cfm@90 psi=about 1hp, this could also much more volume at (90psi/18)=5 psi@(3*18)=54cfm.
Would not a 1hp make a total of 54cfm@5psi?

If 6.28 cu/ft in needed to turn a 2' wheel to turn 1 revolution, this will have a constant load of 62.3lb*10ft=623ft/lb.
At 60 rpm's or 376.8 cu/ft of air is need to make ((623*60)/5252)=7.12hp
If 7.11hp is used to pump the air pressure is it takes 1hp for each 54cfm@5 psi. Would this not also say that 7.11*54cfm=384.3?
384.3/376.8=1.02,102% efficient?
I do know that the less psi to compress, the more efficient the pump can be...

exnihiloest

• Hero Member
• Posts: 715
Re: Submersible Engine Design
« Reply #5 on: January 14, 2011, 06:10:29 PM »
As the air rises, it expands and displaces more water assuming you did not fill the cup all the way.
...

I agree that we would have to assume the change of the air volume in the calculation. But it doesn't really a matter. The volume expands because the water pressure is decreasing, but an expanding gas loses its energy. This wasted energy is the energy provided for raising up to the surface a increasing volume of water, that one corresponding to the increasing volume of air.

tbird

• Sr. Member
• Posts: 317
Re: Submersible Engine Design
« Reply #6 on: January 14, 2011, 06:12:14 PM »
tommey,

it looks like you are changing the numbers from your website.

Quote
This will produce at a 10ft lift:(62.3*10)=623lb
pressure ((4.4 psi x 1728)/12)=633.6lb is what I get.

other numbers just seem to appear out of nowhere...

Quote
If 6.28 cu/ft in needed to turn a 2' wheel to turn 1 revolution, this will have a constant load of 62.3lb*10ft=623ft/lb.

i don't think you meant 10ft.  maybe 10 containers?

maybe a good idea to make a list of the givens you want us to use.

tom

tbird

• Sr. Member
• Posts: 317
Re: Submersible Engine Design
« Reply #7 on: January 14, 2011, 10:07:15 PM »
exnihiloest

Quote
I agree that we would have to assume the change of the air volume in the calculation. But it doesn't really a matter. The volume expands because the water pressure is decreasing, but an expanding gas loses its energy. This wasted energy is the energy provided for raising up to the surface a increasing volume of water, that one corresponding to the increasing volume of air.

i think you are a bit confussed.  the air expanding inside the cone does make a difference, if the cone only has 1 cubic foot of air to start.  at 10 feet it will look like only 83% of the full size.  thus the lift will be less.  you would have to use the average (91.5%) over the distance to figure how much work it can do.

if the cone is in a tube and sealed so water can't pass by, then your statement....

Quote
As water is incompressible, this air volume can be made only by raising the water above, what is strictly equivalent to raise to the surface the same volume of water as the volume of air in the cone.

would have some truth to it.  you forget about the void left behind by the cone as it goes up.  the water simply fills in behind.  i will grant you that if the air is allowed to expand, that small amount of water will have to go some place.  i don't think it will make a measurable difference in the water level.

btw, the last quote, is a very poor way of explaining your point.  i'm sure not many people get it.  maybe you want to try again.

tom

TommeyLReed

• Hero Member
• Posts: 554
Re: Submersible Engine Design
« Reply #8 on: January 15, 2011, 03:08:03 AM »
This is a prototype of the belt drive system.

http://www.youtube.com/watch?v=oX6QLaYbN2E

This system will be running soon.

The theory to this system is, how much energy is need to produec 2.5 psi to fill the cans vs the output of work?
Can this re-compress enough air to fill each can while it moves upwards?
Could this turn into overunity?

Tom

spinn_MP

• Full Member
• Posts: 224
Re: Submersible Engine Design
« Reply #9 on: January 15, 2011, 09:52:35 AM »
This is a prototype of the belt drive system.

http://www.youtube.com/watch?v=oX6QLaYbN2E

This system will be running soon.

The theory to this system is, how much energy is need to produec 2.5 psi to fill the cans vs the output of work?
Can this re-compress enough air to fill each can while it moves upwards?
Could this turn into overunity?

Tom

Hey, nice engineering! I'd like to be there with you, sacrificing myself for emptying all those cans.... Beer, I presume? No? Damn..

I believe it will work, but, sadly, it won't be 'OU'.
Far from UNITY, actually.

I'd say, some 15% efficiency? Not bad, it is quite better than, e.g., Steorn's electrical Orbo from the year ago....

markdansie

• Hero Member
• Posts: 1471
Re: Submersible Engine Design
« Reply #10 on: January 15, 2011, 01:03:05 PM »
Hi Tommy
I am always in full admiration of your projects so well done.
However they are correct that what you have here will not work and has been tried many times.
If you email me markdansiebigpond,com or Skype me mark dansie (nsw austraia) I will show you two ways that should work. One is a maybe but it uses gravity to get the air in under presure the other will work and when I explain it you will see why (uses existing technology that will create a vacuum and suck the air in.
There is a third way.
I never got around to building them but your welcome to te information as I think your a decent person
Mark

TommeyLReed

• Hero Member
• Posts: 554
Re: Submersible Engine Design
« Reply #11 on: January 15, 2011, 01:37:04 PM »
This is what I have come up with, and how much energy is needed to make over unity.

If you only pump down at a depth of 10ft, .5(864cu/in)cu/ft
At 1 cu/ft of displacement of water is about 62.3lb at 1ft at depth, does this force not increase at a depth of 10 ft?
Pumping .5 cu/ft at a depth of 10 ft will increase the displacement each foot as it rise to the surfice, this will this also increase on lift too?
Its all theory, this is why it really needs to be look at.
Remember this 4.4 psi compare to a higher pressure in more efficient too make then compressing 90psi
Thanks you all the comment, now its almost ready to do real testing. This is where theory's become facts.
If i'm wrong so be it, but a prototype is better then any theory's!

Tommey Reed

exnihiloest

• Hero Member
• Posts: 715
Re: Submersible Engine Design
« Reply #12 on: January 15, 2011, 04:01:11 PM »
exnihiloest

i think you are a bit confussed.  the air expanding inside the cone does make a difference, if the cone only has 1 cubic foot of air to start.  at 10 feet it will look like only 83% of the full size.  thus the lift will be less.  you would have to use the average (91.5%) over the distance to figure how much work it can do.

if the cone is in a tube and sealed so water can't pass by, then your statement....

Do the maths.

Let V(h) the air volume of a cone, depending on the pressure. The variation of pressure is inversely proportional to the variation of height, thus we can write:
V(h)=V0+(V1-V0)*h/H
where h is any height where the cone is, V(h=0) = V0 is the inital volume at the bottom, V(h=H) = V1 is the volume near the surface at height H.

At height h, the mass of the volume of displaced water is:
M(h) = V(h)*d where d is the water volumetric mass (d=1 kg/dm3).
Its weight is P = M(h)*g = (V0+(V1-V0)*h/H)*d*g

The work of the Achimedes force is thus:
W = int[P*dh]  to be integrated from h=0 to h=H
W = int[(V0+(V1-V0)*h/H)*d*g*dh] = g*d* int[(V0+(V1-V0)*h/H)*dh]

int[h*dh] being 1/2*h2, the calculus is straightforward:
W = g*d*(V0*H + 1/2*H*(V1-V0)) = g*d*(1/2*V0 + 1/2*V1)*H

This result is obvious because the volume is proportional to the pressure thus to the height, it follows that we can replace the variable pressure and the variable volume by a constant pressure and a constant volume equal to the mean values of the variable case. We can operate at constant pressure with a constant volume V= 1/2*V0 + 1/2*V1 and get strictly the same result.

This is explained by the Archimedes force and the weight of displaced water which are always balanced. Therefore the problem reduces to a pure problem of potential energy. The energy to compress air in the cone at the bottom, being the only energy to be considered as potential energy that can be recovered when the cone rises up, without extra gain.

TommeyLReed

• Hero Member
• Posts: 554
Re: Submersible Engine Design
« Reply #13 on: January 15, 2011, 08:13:31 PM »
This is two basic system.
One using gravity and the other using displacement.

http://www.youtube.com/watch?v=rn0yt0mylW8

http://www.youtube.com/watch?v=jC2k-mKsfEQ

Tom.

powercat

• Hero Member
• Posts: 1091
Re: Submersible Engine Design
« Reply #14 on: January 16, 2011, 12:16:22 AM »
Excellent videos