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Energy created by a balloon released underwater

  1. Dec 28, 2009 #1
    How much energy (Watts?) would be created by a balloon that was released under water? I am picturing a scuba diver's inflatable vest (a BC) that is suddenly inflated (assume zero air inside to begin with). Or more accurately if you have ever seen a free-diver who swims down a few hundred feet and then inflates the balloon that rockets them back up to the surface. In both cases, compressed air is released into a balloon-like vessel all at once. Assume the following:

    Depth released: 200ft below sea level
    Length of travel: 100ft (stops 100ft below surface)
    Volume of balloon: 3 cubic ft
    Balloon holds this 3 cubic ft of air at 100psi so it will inflate at 200ft- assumed .5psi per ft. (don't know if this is relevant)

    I am interested in how much energy is generated in the 100ft of travel towards the surface. Does the energy increase in direct relation to the length of travel or along some sort of a curve? A formula would be great so I could input different numbers. Thanks in advance for the help!
     
  2. jcsd
  3. Dec 29, 2009 #2
    Hi.
    Mass of that volume of water * g * ( depth2 - depth1 ), I think.
    Regards.
     
  4. Dec 29, 2009 #3
    Thanks, Springs. I take it that is the volume of water displaced by the balloon (ie 3 cubic ft)?

    If it is the volume displaced then:

    Kw=(Mass of Water per cubic Ft)*(Gravity)*(Depth2 - Depth1)

    Kw=(62.4 lbs/cu ft*volume)*(9.8m/s^2)*(Depth2 - Depth1)


    Kw=(187.2 Lbs)*(31.36 FT/s^2)*(200ft-100ft)

    Kw=587,059 Ft Lbs/Sec^2

    Kw=766.19 Ft Lbs/Sec

    Kw=1.039

    Watts=1038.96

    Correct?

    Also, that formula seems to say to me "energy produced by a volume of water free falling a given distance" as opposed to the same volume of air floating a given distance through water, but I am certainly a novice. What is the theory there?
     
    Last edited: Dec 29, 2009
  5. Dec 29, 2009 #4
    Hi.

    Unit of the last formula is
    Kw=587,059 Lbs Ft^2 /Sec^2
    By multiplying the constant ratio of 1kg/1Lbs * (1m/1Ft)^2 , this will give Joule = Watt sec , MKSA unit of energy.

    You are right. the difference of heights of water ( or minus water mass excluded by the balloon ) generates kinetic energy of balloon.
    Some approximations done are
    - weight of balloon including air was neglected to zero. You should replace water mass to the mass difference of water and balloon in the formula for correction.
    - there is no water flow, heat, sound or any other energy dissipation by motion of balloon.
    Regards.
     
    Last edited: Dec 29, 2009
  6. Dec 29, 2009 #5
    I'm sure this is well understood by the OP and Sweet, but for others keep in mind that the total energy gained is less than the total energy required to enable the event.
     
  7. Dec 30, 2009 #6
    Deleted post which didn't make much sense!
     
    Last edited: Dec 30, 2009
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