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Acceleration of Submarine when Ejecting Water from its Tanks

  1. Jan 29, 2015 #1
    Brief Summary of Question:

    One submarine starts with a neutral buoyancy 100m below the surface when it displaces 8.9m^3 of water from its ballast tanks. If the mass of the submarine is 69 metric tons what is the initial, vertical acceleration of the submarine immediately after displacing the water?

    [ρ(water)=1,000 kgm-3, 1 atm=105 N/m2, g=9.8ms-2]

    Related Equations: F(buoyancy) = pgV, where p=density of fluid, g=acc., V = displacement of the water due to sub.
    F = ma
    Neutral buoyancy --> F(buoyancy) = Weight​

    Work done:

    Used the buoyancy formula to calculate the displacement of the water by the sub. = 69m^3. (Not sure how the 100m below the surface played into this equation, as well it would seem we need the difference in contained water to calculate the acceleration upwards after ejection)

    Guessing we need to use this in tandem with the 8.9m^3 that gets ejected from its tanks.
     
  2. jcsd
  3. Jan 29, 2015 #2

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    What's vertical acceleration before ejecting water? From the problem statement, consider the ejection to be instantaneous, and that there is no change in volume of the submarine. What is the initial vertical acceleration?
     
  4. Jan 29, 2015 #3
    Well there is no vertical acceleration before the ejection of water of the sub. Its neutral buoyancy, so the buoyancy force is equal and opposite to the weight of the submarine.
     
  5. Jan 29, 2015 #4

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    Okay. Now, what happens to those forces when the water is ejected from the submarine?
     
  6. Jan 29, 2015 #5
    Well Air gets replaced where the water was inside of the sub, makes the buoyancy force greater, allowing the submarine to rise?
     
  7. Jan 29, 2015 #6

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    Okay, we'll back up a little further. You have a 69 ton submarine. It has a constant volume. If you eject 8.9 m3 of water that were in the submarine's ballast tanks, what has happened to the mass of the submarine? And to the volume of the submarine?
     
  8. Jan 30, 2015 #7
    Well mass would go down, because it is now filled with 8.9m^3 of air instead of water, but the volume would be the same, because the size of the sub has not changed.
     
  9. Jan 30, 2015 #8

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    Okay. Has anything happened to the buoyancy force experienced when submerged?
     
  10. Jan 30, 2015 #9
    Well at second glanced, it only seems now that the reason why it is acc. upwards is because of less weight, rather than more buoyancy, because buoyancy only has to do with the amount of water being displaced around the vessel, which is directly related to the volume of the sub which does not change.
     
  11. Jan 30, 2015 #10

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    Correct. You have all the information you need to calculate the difference in the two forces. What is that difference?
     
  12. Jan 30, 2015 #11
    Well I would use the gravitational acc. with the buoyancy formula to get the Volume displaced by the ship, then I would use the amount of water ejected with its density to find its weight. Subtract that amount from the ships weight, and then work backwards with the same volume, to find the acceleration, then subtract the gravitational acc. to find its upwards acc.

    I think this is correct?
     
  13. Jan 30, 2015 #12
    The initial condition of neutral buoyancy will give you the buoyant force acting on the submarine. Once the water is displaced, the weight of the submarine will reduce but the buoyant force will remain the same. Using Newton's second law, you can get the acceleration.
     
  14. Jan 30, 2015 #13

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    Check.
    Check.
    Check.
    I don't know what you said here, but do the calculation, and we'll see what you get.
     
  15. Jan 30, 2015 #14
    Simplified: V = m/density = 69000kg/(1000kg/m^3) = 69m^3
    Weight of water displaced: 8.9m^3 * 1000kg/m^3 = 8900 kg
    Weight of ship (post-ejection) = 69000-8900 = 60100 kg (* 9.81ms^-2) = 589581N (Fw)
    Find acc. --> F(w) = pgV --> F(w)/pV = g --> 589581N/((1000kg/m^3)(69m^3)) = 8.5446 ms^-2

    9.81 ms^-2 - 8.5446 ms^-2 = 1.265 ms^-2 acceleration upwards?
     
  16. Jan 30, 2015 #15
    Or is it that its a difference in force, which I imagine to be true.

    Weight of Ship = 589581 N

    Buoyancy Force = (1000kg/m^3)(9.81ms^-2)(69m^3) = 676890 N

    Difference in the two: 87309 N --> divide this by the new ships mass

    87309N/60100kg = 1.4527 ms^-2 upwards acceleration.

    I believe this one to be correct.
     
  17. Jan 30, 2015 #16

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    As do I. Haven't checked the arithmetic integer by integer, but the method is correct.
     
  18. Jan 30, 2015 #17
    The assignment has told me it is incorrect.
     
  19. Jan 30, 2015 #18
    That is the same answer I got.
     
  20. Jan 30, 2015 #19
    As the assignment told me that this value was incorrect. Unfortunately.
     
  21. Jan 30, 2015 #20
    Maybe the mistake lies in the initial mass. Maybe 69 tonnes is the mass after the water has been displaced.
     
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