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Bubble rising underwater. What is the time taken for it to reach the surface?

  1. Aug 9, 2009 #1
    1. The problem statement, all variables and given/known data
    The initial volume and depth of the gas bubble is V0 and H respectively. The bubble has a mass M and the gas can be assumed to be ideal. The gravitational field strength is a constant g. And the density of water is Dw.

    What then, is the time taken for a bubble initially at rest at a depth H underwater rise to the surface? Assuming that there is no resistive force on the bubble and the temperature is constant during the rise. Try to avoid the presence of an atmosphere above the water surface, however if needed then take the atmospheric pressure at the water surface be Patm.

    2. Relevant equations
    F = ma
    Hydrostatic Pressure = height x density of liquid x g
    Archimedes' Principle
    Ideal gas law

    3. The attempt at a solution

    I tried finding the net force from the buoyant force and the weight of the bubble with the aid of Archimedes. But since when the bubble rises the pressure of its surroundings decreases and its volume should rise. So the buoyant force is not constant. I use the ideal gas law to find the volume in terms of H, V0 and h(t). Where h(t) is the depth at some time t. After some work I arrived at:
    d2h/dt2 + [V0HDwg/M](1/h) -g = 0

    The plan was to find h(t) then equate it to 0 and solve for t which will give the time for it to rise to the surface. But it is so complicated that I believe I must have made a mistake somewhere. So I am here to seek help. :wink:
     
  2. jcsd
  3. Aug 9, 2009 #2

    tiny-tim

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    Homework Helper

    Hi Nanyang! :smile:

    I can't check your equation without seeing your full calculations, but if it's right, then the standard way of solving it would be to write dh/dt = v, and then substitute d2h/dt2 = dv/dt = dv/dh dh/dt = v dv/dh :wink:
     
  4. Aug 9, 2009 #3
    Hi! Thanks for that tip though I still can't solve it as I'm a rookie. :redface:

    These are the steps I took:

    Using Archimedes and Newton,

    F = m a = DwVg - mg

    a = DwVg/m -g

    Then using P1V1=P2V2 and P=hDwg,

    V= V0H/h

    Substituting this into a, and since a= d2(H-h)/dt2...

    d2h/dt2 + [V0HDwg/M](1/h) -g = 0

    Which I believe to have a big mistake somewhere. But never mind my attempt, what would be another (even easier) way to find the time taken for the bubble to rise up?
     
  5. Aug 10, 2009 #4

    ideasrule

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    P isn't just equal to rho*gh; it's equal to rho*gh+p_atm, so V=V0H/h doesn't hold. (If it did, V would be infinite at the surface.)
     
  6. Aug 10, 2009 #5
    Is it necessary to include P_atm to solve the problem? Can someone give a clue to find the time?:redface:
     
    Last edited: Aug 10, 2009
  7. Aug 10, 2009 #6
    I think there is more than one problem in that derivation.

    Consider this; the downward forces on the bubble is the bubbles weight, the upward force is the buoyancy of the bubble.

    So you will then have;
    ma = buoyancy -mg (bubble)

    ma = rho(water)*V(gas)*g - m(gas)*g
    This would be my starting point. You can express the mass in terms of P, V, R, T and molecular mass of air ~28g.

    Some terms should cancel and a relationship between a and h should be possible and with a bit of calculus a relationship with time should be possible. It will still include h but I hope this works. Is this a viable method? I think it has if it legitimate to cancel the volume terms.
     
    Last edited: Aug 10, 2009
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