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Can anyone help me out with fluid mechanics

  1. Jun 8, 2009 #1
    i am having some troubles with this problem. please take a look and help me if you can.

    an open-ended can 1ft long is originally full of air @ 70 F. the can is now immersed in water. assuming that the air stays at 70 F and behaves like an ideal gas,how high will the water rise in the can? the is immersed at 10 ft deep into water tank open to the atmosphere.
    thanks in advance. please show the work

    i tried to find the pressure of the water at the bottom of the tank. then i assume x to be the height of the water in the can with the air on top of it. considering the can as the system now, there is a relation between the compressed air in the can and the water inside the can. the pressure at the bottom of the can which is also equal the pressure at the bottom of the tank, will be the sum of that of the air plus dgX of the water (density* gravity* X) with X being the height iam looking for. i have this one realtionship and i can't think of another one to get the height X out of it... can anyone help?
    Last edited: Jun 8, 2009
  2. jcsd
  3. Jun 8, 2009 #2


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    Welcome to PF!

    Hello rogga! Welcome to PF! :smile:
    Well, that looks ok … I don't understand why it doesn't work. :confused:

    Show us the equations you got. :smile:
  4. Jun 10, 2009 #3
    Hi there,

    I don't know what equations you are actually using but as far as I see the air inside the can is at atmospheric pressure. At the bottom of the tank, the pressure on the air column is equal to water pressure at that depth. Now the air experiences a pressure of (water pressure - atmospheric pressure). That's what compresses it. Now the problem is reduced to finding out how much compression can be achieved by this net pressure. You have a 1 foot column of air in the can. Now when the air is compressed the pressure inside the column increases at the cost of volume. Boyle's law tells us that. And the air continues being compressed till the air pressure in the can equals the water pressure. Now the water pressure is no longer that pressure of water which it was at 10 feet depth because remember, the water has risen. I hope I have given you a lead because without further data I cannot solve this problem...
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