A new energy source-maybe--maybe-not

[justcurioustwo]

I am bringing this drawing back into the discussion just to ask one last question.
In the drawing there are twelve (12) buckets on the right side. Just for discussion each bucket has a lifting force of 100-foot pounds. 12 buckets times 100 = 1200-foot pounds of lifting force.
1200-foot pounds of lifting force can produce more energy at any one moment in time than 100-foot pounds;
Once all the buckets are full and this machine is running, the process continues to produce 1200-foot pounds of force if you continue to fill one (1) bucket at the bottom in sequence with the rest.
YES or NO?

 

[justcurioustwo]

I completely explained the operation of the machine in post #23.


You missed my answer as well, so I'll say it again:

The 'SeaPower' machine is a pneumatic motor, a.k.a air-motor, in which air expands from one volume to another while doing mechanical work on a linkage. The SeaPower differs in constructional details from conventional air-motors by having a balloon (or inverted bucket) and liquid medium to confine the air, rather than a piston sliding within a cylinder or a rotor with sliding vanes mounted eccentrically in a cylindrical housing. The liquid serves as a mechanical linkage to transfer to the output shaft the force exerted by the boundary surface of the air, just as the piston and connecting rod, or vanes and rotor do, in the normal type of air-motor.

Intuitively, we can see that driving the shaft (using an external prime mover) so that the balloons descend, will cause the machine to function as a compressor. Absent any losses, it will compress the same volume of air to the same pressure as would be required to make an identical machine operate as a motor and deliver the same shaft torque and speed as output. Intuitively again, we can see that coupling the two machine shafts and air pipes together would result in equilibrium and all would remain stationary.

If, however, you want to ignore the obvious symmetry and prove the result numerically, you will first need to learn the basics of integral calculus (and I do mean just the basics.)
1. Learn calculus.
2. Find an expression in terms of depth for the force exerted on the belt by a balloon containing a unit volume of air.
3. Integrate this expression with respect to depth to obtain the work done on the belt by each rising balloon.
4. Compute the work done by a unit volume of air injected at arbitrary depth, and divide by the work done compressing that air against the head of water at that depth.
5. If the result is over unity, you probably made a mistake!

Anyone who is still reading at this point might have sufficient interest in water-pistons as to be familiar with the Humphrey pump. And if you aren't, you should be. I don't know where in Texas @justcurioustwo is located but one of the few Humphrey pump installations in the world was in Del Rio TX. The Chingford installation is just ten miles from here.
Humphrey Pumps Wikipedia article

You have provided a detailed description of the seasengine. There is one aspect of the design you might have missed. In the example there are ten (10) rising buckets of air tied to each other. To sustain this rising force; you need to refile the lowest bucket with air to sustain the total lifting force. This compressed air source could come from compressed air tanks. At the top each discharged tank is replaced with a recharged one sustaining the cycle. At the bottom the air tank is opened to release the compressed air.

One basic question remains, is the total energy output greater than the energy needed to fill one compressed tank of air-?

 

[static zap]

If thermal energy is removed from the ocean ,there is a useful angle here.
..Will we get the Energy police on our tails for radiating unwanted thermal energy into space on non cloudy nights.
@justcurioustwo It is nice to have a good latteral thinking session.

 

[justcurioustwo]

If you know it all why are you on this forum.

Yes, I do not know much so that is why I am here, asking you-

 

[DPG]

Yes, I do not know much so that is why I am here, asking you-

I dont think a forum full of electricians is the best place for your questions to be honest. Have you tried other avenues?

 

[Lucien Nunes]

To sustain this rising force; you need to refile the lowest bucket with air to sustain the total lifting force.

I've no idea what information you think you are adding here. The relationships between injected air mass, torque, and speed in this machine are unusual in that it has an elastic working volume, like a hypothetical piston engine with an infinitely stretchy cylinder. Therefore it behaves neither strictly volumetrically (where shaft rotation angle is proportional to air mass input) nor inertially (where static torque is proportional to rate of mass input). Instead, the torque and speed are inversely proportional, since for a given air delivery rate, the slower the shaft is allowed to turn, the more contained air is attached to the rising side of the belt at a given instant and therefore the higher is the torque. What should then be clear is that the product of torque and angular velocity (which is the mechanical output power) is constant for a given air delivery rate at a given depth, therefore energy is conserved.

If you would take the trouble to study integral and differential calculus, the relationships between these quantities will become crystal clear and you will easily be able to visualise how it behaves.

Anyway, I've said enough on the subject so likely won't contribute further to the thread, in case I too end up going round and round on an endless belt.

 

[DPG]

I suspect we may have to wait a while for any reply, and also I suspect any questions will remain unanswered by the OP.

 

[justcurioustwo]

I dont think a forum full of electricians is the best place for your questions to be honest. Have you tried other avenues?

The output of the seaengine is electric power generated from the rising pull of the circular cable.

In my example the pulling force of the cable is 118,428 pounds rising force at 3 feet per second. Actually I believe the speed will accelerate but to what speed I can not determine

I posted the idea here because I need to convert this force to electrical output.
NOTE: use "paint" to open the attachment and lastly the drawing open at actual scale, which is huge, so shrink the image after loading
Once done, someone here please tell me that you actually opened the drawing and can see it.
Thanks in advance

 

[justcurioustwo]

 

[DPG]

I can confirm I can see (sea?) the drawing.

 

[justcurioustwo]

A bucket rises three (3) feet per second.

Buckets are thirty-three (33) feet apart.

Time between refills is eleven (11) seconds.

@ 1atm the volume of the bubble is 40,000 cubic feet.

Compressed down to 15 atm or 2,666.666 cubic feet.

Above is the energy required to refill a tank [Bucket]

Convert the above into electrical power; call this (X)

Convert 118,428 lbs. moving a 3 feet per second into electrical power; call this (Y)

I cannot do the math above, that’s why I am here-

Is (X) greater than (Y) or visa versa
fust asking for some help

 

[justcurioustwo]

I can confirm I can see (sea?) the drawing.

do you see the drawing above your post-?

 

[justcurioustwo]

I can confirm I can see (sea?) the drawing.

If you hold down your [CTRL] key while rolling up and down your mouse roller the image will enlarge or shrink.
Does this work on your computer-?

 

[justcurioustwo]

I've not looked too deep (pun-intended),
but suspect your bubbles may not be as big , if the source "gasses are above sea temperature , they will contract some on cooling ..
(even though plenty of expanding

When I read your post a light bulb popped on, thanks.
The temperature needs to be considered.
Sea water averages about 67 degrees Fahrenheit
I am not sure how much change taker place.

Do you think this will slow it down or speed it up?

 

[justcurioustwo]

bubble in seawater rises around 220 mds and almost same as that in artificial seawater and distilled water

MT03014FU.pdf (witpress.com)

Speed of rising bubble

220 mds

1 inch = 25.4 millimeters

220mm = 8.66142 inches

12/8.66= 1.3854 feet per second

33/1.385= 45.7182 seconds

Between refiles is therefore every 45 sec+/-

Is that even close?

 

[justcurioustwo]

Bubble equation @

The lifting force of an enclosed container, (bubble); is equal to the volume of water being displaced. To push one cubic foot of enclosed air under water takes a force of 64 pounds at one ATM.

This is the equation that determines the volume of air at different depths.

Formula used (ATM/V1) X V = bubble size

V1 is the volume at 1 ATM. Assuming the volume at 1 ATM is 100 cubic feet; at 2 ATM’s the volume is compressed to half its size or 50 cubic feet.

At 18 ATM’s that 100 cubic feet is compressed down to (18/100) X 100 = 18 cubic feet

Formula used (ATM/V1) x V = bubble size

ATM = atmosphere pressure at sea level. The combined lifting force is the volume of each bubble in cubic feet times 64 pounds.

A cubic foot of seawater weighs 64 pounds

If you displace a cubic foot of seawater the force of the water surrounding the cubic force to rise equals 64 pounds

If a boat displaces 1,000 cubic feet of seawater, the boat can weigh as much as

64 X 1,000 = 64,000 pounds or 32 tons without sinking

I call the 64,000 pounds of force to be in a static state or .

If that 32 tons was set into motion, the speed of that motion times the weight measured in foot pounds = torque equation

Please, someone correct me