Buoyant Lift Creates Useful Energy?

  1. Buoyant Lift Creates Useful Energy?

    I work at a manufacturing plant. One by product of this plant is the issuance of 2000 cfm (cubic feet per minute) of clean air. It is determined that this air could be released from as small as a 3 inch pipe with no adverse harm to the manufacturing process. It would be exhausted at 100 psi. So it got me thinking, how can this be put to use?

    Please look at the attached diagram-

    My general idea is to build a 100 ft tall (x18x8) tank and fill it with water. You see three shafts; top, middle, and bottom. I have received some information from Winergy, a subsidiary of Siemens regarding gearbox/generators. They manufacture a gearbox that requires 16 kilonewton meters (11,800 ft lb) of force to operate. This, in conjunction with a matched generator, will produce 100 kilowatts of power, so that a total of 300 kw can be realized in this system.

    The input of 2000 cfm at a water depth of 100 ft would mean that each cubic foot would be compressed to approx. one-fourth its surface volume. This means that the 2000 cfm will be entering the bottom of the tank at 500 cfm, then expand to 2000 cfm as it rises to the top of the tank.

    A roller chain with “air lift containers” could be attached to the shafts as shown. These containers could be made of carbon fiber, plastic or maybe aluminum. If each were a hollow sphere of 6.5 feet diameter, then the volume would be 143 cubic feet. One factor I have not considered yet (infant stage of idea) is the volume loss of the sphere once the “top” is cut off to resemble a deep bowl. For discussions sake, ignore that at this time. One cubic foot of air will lift 62 pounds to the surface. You see numbers next to each of the rising containers. I roughly calculated what the lifting capability would be for each ascending container at points along the roller chain (snapshot of system). As the air enters the base of the tank it will only fill one-fourth of the container and expand, filling the container as it rises, becoming nearly full at the top of the system, and gaining lift capability as it rises.

    Each of the 3 shafts would need to rotate at 60 rpm to generate the 100 kw each. I do not know if this can be achieved. If the cog on the shafts were 2 ft circumference then the containers would need to rise at 2 ft per second. This seems feasible to me to accomplish. I have not located answers regarding natural rising speed of air in water, or filled containers for things such as salvage diving, etc.

    With these figures, I calculate that the total, constant lifting capability would be approx. 42,000 lb. If the three shaft require 11,800 ft lbs each, totaling 35,400, then am I close in believing that this can work? I understand that system weight, drag, and water drag need to be considered. Regarding such issues as water drag, it can be minimized once the system is up and running and the water is moving in the same direction as the system. Also, right in the middle of the tank, between the ascending and descending chain, I have drawn a vertical line that represents a separator that would lessen the commingling of water.

    One other issue I do not understand how to calculate is the cog size and how it changes things. I had someone tell me the following –

    With the chain on a sprocket radius equal to 0.32, there is only 13,370 ft-lbs of torque generated by the buoyant force of 42,000. This is confusing to me. Remember that I had calculated a circumference of 24 inches, which could be adjusted of course. I just decided on that to make calculating the 2 ft per second rise simple. Am I to assume that the larger the cog the less torque?

    Anyways, I would welcome any thoughts on this idea.

    Thanks so much for reading this. Go easy on me though. I believe that I have a decent amount of common sense, but no great intellect, especially in the area of physics and mechanical engineering.


    Attached Files:

    Last edited: Aug 28, 2008
  2. jcsd
  3. russ_watters

    Staff: Mentor

    First off, does the manufacturing process require that you release the air at a depth of 100 feet or are you doing that soley to produce energy from it? At about 60 PSI and 500 CFM, you need an enormous compressor to generate this airflow. And is the top of this tank open to atmosphere? And where do you get that 100 kW? 100 kW / 16 kN-M = 6.5 m/s, buy you said you are only getting 2 ft per second (about .5 m/s). Power is force times distance over time.

    Take a step back and remember: conservation of energy applies. Whatever you can generate you have to input in compressor power. Are you recovering energy that has to be wasted by your manufacturing process, or are you increasing the energy dissipation in order to recover more? At a bare minimum, with a 50% efficient compressor, and unknown efficiency generator and turbine, generating 300 kW of electricity will require a 600kW compressor. Does your manufacturing process include a 600 kW compressor?

    If you tell us more about the process requirement, I can probably advise you of the best way to recover energy from it (if it is possible at all).
    Last edited: Aug 28, 2008
  4. russ is right, if your air doesn't come out of the manufacturing process compressed then the energy used in your compressor will be more than the energy out. However, assuming that the air comes out compressed then you do have a large amount of energy available.

    The amount of energy is equal to the pressure times the volume: 500 cfm 100 psi = 160 kW

    Of course, you will not be able to extract all of that energy, I don't know exactly how efficient these kinds of devices can be, but I would guess no more than 75 kW out in the end.
  5. uart

    uart 2,776
    Science Advisor

    kitarey. Motors that run on compressed air already exist. They are called "pneumatic motors" and I'm sure you could get one that's a lot cheaper, not to mention more compact and efficient than what you propose.
  6. Thank you for the replies. I do follow your points regarding the air flow. The outflow of air is a fixed flow of 2000 cfm at 100 psi (measured) from the plant at surface level. Regardless of how or why this is expelled, it exists. The 2000 cfm will release into the tank at 100 ft at a rate of 500 cfm considering the depth pressure. This can be done from what I understand. The reason for the outflow is unimportant right? We are just considering a way to use this outflow.

    No Russ, the exiting airflow does not have to go into this tank. The idea is to capture the exiting airflow and make it useful.
    Yes the top of the tank is open.
    YOu ask where the 100 kw comes from. The idea is to use the airflow to operate 3 shafts connected to gearbox/generators capable of generating 100 kw each. The gearbox manufacturer told me that 16 kn-m / 60 rmp / 100 kw. I figured the 60 rpm would be 2 ft per sec on the rising chain if the cog were 24" circumference


  7. Hi kitarney,

    If it already comes out as 2000 cfm at 100 psi then the water will not compress it to 500 cfm. If you compressed it to 500 cfm then it would be closer to 400 psi.

    In any case, uart is correct, just use a pneumatic motor to run a generator. Should be more efficient and you won't have to worry about cleaning out the algae that will grow in a huge tower of water :yuck:
  8. Dalespam,

    Thx for reply, but I'm confused by your reply. At 100 feet one cubic foot of air is compressed to one-fourth its volume. With this in mind then, would it not be true that the 2000 cfm would actually enter the tank at 500 cfm then expand to 2000 cfm when it rose to the surface? I aslo am not sure what you are thinking when you talk about a pneumatic motor to run a generator. The whole idea here is to take an existing exhaust of pressured air and make it useful by introducing it into this tank and thus producing new power, in this case 300 kw of useful power.

  9. stewartcs

    stewartcs 2,244
    Science Advisor

    The motor is just the prime mover for the generator. Something has to capture the air and convert it to organized motion. They are just suggesting the simplest thing.

  10. From what I understand, the buoyant force of the air will lift the containers and rotate the system. The rotation will turn the shafts of the gearbox which connects to the generator. With this in mind and the system running at optimal, why would any motor be needed. The prime mover is the gearbox shaft is it not?

  11. stewartcs

    stewartcs 2,244
    Science Advisor

    They simply are proposing a different design than what you originally posted.

  12. ok I gotcha. I guess I don't know much about how a pneumatic motor operates. Would it provide an equal amount of power to the 42,000 lbs from this idea?
  13. 100 feetH2O = 43 psi.

    Since your air going into the tank is 2000 cfm at 100 psi it is already more pressurized than the 43 psi at the bottom of 100 feet of water. Your air will actually expand, not compress. You would only get the compression you are talking about if your 2000 cfm were at a low pressure instead of 100 psi. In that case you would be completely unable to inject your air into the bottom of the tank without first compressing the air (which would defeat the whole purpose).

    A pneumatic motor, in its simplest, is like a turbine. The compressed air expands, pushes on some blades, and makes them spin. You can also have the compressed air expand and push on a piston. In either case you attach a shaft and get some torque to run a generator.
  14. russ_watters

    Staff: Mentor

    You are saying that you have both 2000 cfm at 100 psi and 500 cfm at 100 psi. It can't be both at the same time. When the air actually comes out of your plant, what is the airflow and pressure?

    You also didn't specify if that pressure is gage or absolute (we're assuming gage). In any case, 100 psi is 6.8 atm, but 100 feet of tank is only 43 PSI or 3 atm (+1 from the atmosphere itself). In other words, you don't inject air into the tank at 100 psi, you inject air into the tank at 43 psi. To do that, you run the air through a nozzle, where it loses about half the energy being converted. Better off running it through a turbine in at 100 psi and out at 43 (or why not in at 100psi and out at atmospheric and not bother with the tank?).
    No. Air has one volume at a time. When you inject it, it is either 2000 cfm or 500 cfm, not both. Again, what is the output of your plant, in airflow and pressure?
    A pneumatic motor connected to a generator produces power - and will do it more efficiency than your buoyancy motor (it doesn't have to deal with the drag of the water).

    Better yet, if you have clean air at 100 psi (whatever the actual airflow) coming out of your plant, why isn't it just being recycled? If you acutally have energetic air coming out of your plant, there are most certainly some good ways to recover energy from it. A buoyancy motor is not one of them, though. Give us more info and we can help find the best way.
    Last edited: Aug 28, 2008
  15. Maybe there is some misunderstanding regarding how the air is exiting the building. It exits in the same manner as if you had a 2000 cfm, 100 psi compressor. My biggest confusion from your responses is regarding the air volume expansion in the water tank. If you have one cubic foot of air at a water depth of 100 feet, that one cubic foot of air will expand as it rises, becoming four cubic feet at the waters surface. With this in mind then, you can’t introduce 2000 cubic feet of air into 100 feet of water and expect it to remain 2000 cubic feet of air otherwise you would end up with 8000 cubic feet of air at the surface. If you introduce 2000 cubic feet of air into 100 feet of water the water depth pressure will compress that 2000 cubic feet to 500 cubic feet, and it will expand to 2000 cubic feet at the surface. Is this not correct?

    Also, I did learn that if the hose or pipe is at least 2” in diameter, then there is no pressure loss as it introduces the air into the tank.

    As I asked earlier, do you think that a pneumatic motor would be able to use the 2000 cfm with the same results? Would there be enough power to provide the same 42,000 lb torque that could be realized from the buoyant lift of the water tank system? The 3 shafts for the gearbox/generator could be combined to one shaft but the need would be the same, 35,400 ft lbs of torque (16 kn-m)

    Regarding the recycling of the exiting air, disregard this concern. Please just consider a way to utilize it
    Last edited: Aug 28, 2008
  16. russ_watters

    Staff: Mentor

    I agree with the confusion part, but I think as much of it is on your end as ours! Where, exactly, is that pressure measured? Fluid dynamics dictates that if you have an open pipe leading to the outside, the pressure at the beginning of the pipe can be very high, but the pressure at the outlet must be atmospheric. Only by putting a nozzle at the end could you keep a back-pressure in the pipe and get a consistent 100 psi.

    It may also help if you describe the manufacturing process that is using the air. What is the input pressure and flow rate? What is the process? Air is often used as an energy transfer medium (ie, spinning pneumatic tools) and in those cases, the outlet pressure needs to be as low as possible to keep the efficiency up. Pneumatic tools are powered by the pressure drop (and flow rate, via the equation given above) across the tool and pressure can only drop if the pressure at the outlet is zero and the pipe big enough to keep the back-pressure low.
    Yes, that's true.
    Why not? If you introduce it at the same pressure as the water, it does not expand while exiting the pipe.
    Well, no - the water can't compress the air, the air has to already be compressed, otherwise it is impossible to inject the air into the water. That's why we are asking these questions: the pressure of the air at the outlet of the plant needs to already be high enough to inject directly into the water. Ie:

    -If the water pressure is 43 psi and the air coming out is 43 psi and 2000 cfm, then the air is injected into the water at 43 psi and 2000 cfm.

    -If the water pressure is 43 psi and the air coming out of the building is 100 psi and 2000 cfm, then the pipe injecting it into the water will act like a nozzle, expanding it to around 4000 cfm at 43 psi (I don't feel like doing the calculations - you actually can't just ratio it because there is a non-trivial loss due to the cooling effect of the expansion).

    -If the water pressure is 43 psi and the air coming out of the building is 10 psi and 2000 cfm, then you cannot inject the air into the water (without running it through another compressor, of course).
    My first instinct was to say that doesn't help us at all, since you still gave us multiple different parameters. We need to know which ones are the correct ones.

    But then I thought about it some more: 2000 cfm through a 2" pipe is a velocity of 1500 feet per second, which is higher than the speed of sound and therefore impossible. Even if the pipe is a little bigger, most of the energy of the compressor is going to be dissipated as frictional heat inside the pipe. That's a horrible waste of energy if it is really configured that way.

    You may be on to something with your idea of recovering some energy from your compressed air system - and I find such applications to be the most fascinating part of my job - but we really need to know what is going on here. I honestly would love to get to the bottom of this, but so far, all you've given us is a jumbled mess of conflicting and impossible information.
    No, whatever a buoyancy motor could recover, a pneumantic motor could recover much, much more. What you are trying to do is akin to driving your car through molassas. All you're doing by adding the water is adding unnecessary resistance.
    Torque and power are not the same thing and in any case, we still have no idea how much energy is available in the air exiting the building. The answer to the question is a simple yes (with gearing, you can have whatever torque you desire), but that doesn't really tell us anything useful.
    Fair enough.
  17. russ_watters

    Staff: Mentor

    Some things that could help get us started at least - could you provide us the model number of the compressor and how the air is used in the manufacturing process?

    The largest compressor that Ingersoll Rand makes is just barely large enough to provide the specified 2000 cfm at 100 psi and that's without first using the air in a manufacturing process! http://air.ingersollrand.com/IS/modelComp.aspx/item/12816

    Those specs require a 400 hp (300 kW) compressor on their own (that's in line with Dale's estimate: 50% efficiency is pretty typical).
  18. I have a question. Is the 100 psi a static number? In other words, do you get the 100 psi reading if you plug up the pipe? And is the 2000 cfm a flow measured at atmospheric pressure?
  19. russ_watters

    Staff: Mentor

    Good questions. With the velocity so high, it makes a big difference. Choked flow (sonic velocity) is something like 65 psi of velocity pressure, iirc. Compressors, afaik, are typically rated in static pressure in a pressure vessel, which is really total pressure as velocity pressure in that case is zero.
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