A new energy source-maybe--maybe-not

[DPG]

I need help in calculating the numbers
Help !!!!!!!!!!!!!

You previously said you were able to do the calculations (in Basic was it? - can't remember), but you no longer had that software. You asked for advice on a program which could do the calculations for you. The spreadsheet we advised about will able to do mathematical calculations.

However, if you don't know what calculations you are wanting to calculate then I can't see how you're going to progress.

 

[littlespark]

I thought by #76 you had given up on this idea.

 

[static zap]

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 will occur , also some chemical reactions with the sea water ,may dissolve some chemicals present, and pose some corrosion problems with matterials ) These factors are going to eat into you gains a little. (Nothing planet earth does is stable when dealing with lava !)

 

[justcurioustwo]

No one seems to want to directly address the design but instead dance around the subject.

The principle is simple enough so address the design/ principle behind the design.

Thanks in advance

 

[DPG]

I think you summed things up nicely in post #76.

 

[justcurioustwo]

I think you summed things up nicely in post #76.

Thank you for your reply.

Having said that, I am still not getting any feedback on the principle behind the design itself.

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;

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

Output of this machine is 118,428 pounds of lifting force moving at 33 feet per second at any one moment in time

 

[Baddegg]

I’ve missed him ?.......I seldom meet people more mental than me ?

 

[static zap]

I'm a proud British eccentric ... but I have my identity on here ..
To keep that info from "potential" Employers ...

 

[Megawatt]

No one seems to want to directly address the design but instead dance around the subject.

The principle is simple enough so address the design/ principle behind the design.

Thanks in advance

If you know it all why are you on this forum. This is an electrical forum NOT rising bubbles forum. You’ve been on here before asking stupid questions. If you want advice on electrical then come back. These folks have better things to do than play your game

 

[Lucien Nunes]

No one seems to want to directly address the design but instead dance around the subject.

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

Where’s @Lucien at when you need him

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

 

[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