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Homework Help: Fluid Mechanics of a weather balloon

  1. Jan 24, 2006 #1
    Here are a couple problems I would like to check and a couple I'm not sure how to do:

    1.3)A 650kg weather balloon is designed to lift a 4600kg package. What volume should the balloon have after being inflated with helium at 0C and 1 atm pressure to lift the total load?

    Density of helium = .179 pa.
    .179 = (650 + 4600)/Volume.
    Volume = 29000m^3?

    1.4) A submerged submarine alters its buoyancy so that it initially accelerates upward at 0.325m/s^2. What is the submarines average density at this time?

    The density of seawater is 1.025x10^3 kg/m^3. From here I have no idea where to go. Can someone give me a hint?

    2.2) Water is to be pumped to the top of the Empire State Building, which is 366m tall. What gauge pressure is needed in the water line at the base of the building to raise the water to this height?

    Based on a formula from my book, Absolute pressure = atmospherice pressure + (density x free-fall acceleration x depth). So,
    1.01x10^5 pa + (1.00x10^3 x 9.81 x 366) = 3.69x10^6 pa?

    3.3) The water supply of a building is fed through a main entrance pipe that is 6.0cm in diameter. A 2.0cm diameter faucet tap position 2.00m above the main pipe fills a 2.5x10^-2 m^3 container in 30.0s. What is the speed of the faucet and What is the gauge pressure in the main pipe?

    The cross-sectional area of the main pipe is .0113 meters^2. The cross-sectional area of the faucet tap is .0013m^2. The flow rate from the faucet is .00083m^3/s. From here I do not know where to go, can someone guide me in the right direction?

    Thank you.
  2. jcsd
  3. Jan 24, 2006 #2
    fluid mechanics reply

    that's funny. i have the exact same problems in my text book!
  4. Jan 24, 2006 #3
    Holt Physics 2002. You wouldnt happen to have the answers too? :smile:
  5. Jan 24, 2006 #4
    bump please
  6. Jan 25, 2006 #5
    bump again
  7. Jan 25, 2006 #6
    First, density can't be measured in pascals, which measures pressure.

    You have to find ratios of the density of the fluid displaced vs the density of the fluid used, and when you multiply a density by volume you get either a force or mass, depending on whether you use mass or force density.
  8. Jan 25, 2006 #7
    The .179 pa on the first problem was a typo. So for the first problem i dont simply use density = mass/volume? I dont believe I know what the displacement will be do I? I only know the mass of the objects, the density of helium and the density of air. So how should I set up my equation?
  9. Jan 26, 2006 #8
    Draw a force diagram. You have the weight of the balloon and package directed down. You have the necessary Volume * (force density of air - force density of helium) as the bouyant force directed up.
  10. Jan 26, 2006 #9
    So, 5250 kg = (1.29kg/m3-.179kg/m3)V
    Volume = 4700m^3
  11. Jan 26, 2006 #10
    Looks close, 4725 m^3, assuming your densities are correct. That would be about an 18 meter diameter balloon.
  12. Jan 26, 2006 #11
    Yes thats what I had before rounding to 2 significant digits, on 1.1. Thank you.
    Any insight on 1.4 or 3.3?
    Last edited: Jan 26, 2006
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