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Homework Help: Height of water level in an open ended fully submerged can

  1. Jan 26, 2012 #1
    1. The problem statement, all variables and given/known data

    An open ended can, with height 0.3m is filled with air (P = Patm ~ 100kPa) and is submerged (open end downward) 3 metres below the water surface (bottom of can 3m below while top of can is 2.7m below). The pressure above the water's surface is also Patm. The air is initially at 20 C, and after it is submerged it remains at 20 C. Assume the air acts as an ideal gas. Estimate the height of the water that will fill the can. p(water) = 1000kg/m3 Where p = density; P = pressure

    2. Relevant equations

    P = pgh or P = pgΔh

    3. The attempt at a solution

    How I figure it, is that you calculate the pressure at the very bottom of the can first. P(bottom) = (1000 kg/m3)(9.81 m/s2)(3 m) = 29430 kPa. The air in the can will initially be the same as it was originally - Patm - but eventually the water pressure underneath the can will compress the air and the water will begin to rise into the can. What i'm not sure about is how the height of the can comes into play as there is no radius given to calculate volume. Would the problem involve the pressure difference between the downward pressure on the top of the can, and the upward pressure on the opening in the bottom of the can?

    This is my first post and any feedback on anything I've done wrong in my post (and problem solving logic) would be appreciated, as well, the problem statement and the diagram are attached in case my explanation wasn't clear. Thanks everyone
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Jan 26, 2012 #2
    You are on the right track with your pressure calculation.
    If the temperature of the air does not change then the trapped air obeys Boyles law......do you know the equation?
    You can take the length (0.3) of the can to represent volume because it has uniform cross section.
    Hope this helps
  4. Jan 28, 2012 #3
    Hey thanks for the reply, the formula you're talking about is P1V1 = P2V2 right? In that case would I just go (100kPa)(A)(0.3m) = (29430kPa)(A)(hf), and then hf is the height of the gas after the pressure equalizes? If thats correct, I get a really small number in the area of 0.1 cm for the gas height which seems wrong to me.
  5. Jan 29, 2012 #4
    You have done the right thing but the pressure due to the water is 29.43 kPa not 29430kPa
    This means that the total pressure on the submerged can = Patmos + 29.43kPa
    I got h to be 0.23m (not cm)
  6. Jan 29, 2012 #5
    Awesome, that makes sense. I forgot the pressure was in Pa. Thanks for all your help! :)
  7. Jan 29, 2012 #6
    Thank you...good luck
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