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Pressure on a submarine window

  1. May 7, 2004 #1
    I have this fluids problem I've been working on for a while, but I can't seem to get the correct answer. The problem is:

    A circular window with radius 25 cm in a submarine can withstand a maximum force of 1.23E6 N. If the interior of the submarine is maintained at a pressure of 1 atm, approximately how deep the submarine can dive without rupturing the window? (You can assume sea water density of 1000 kg/m3.)

    The pressure underwater can be found using P = P[0] + pgh, where P[0] is 1 atm (101325 Pa). So what I did was this:

    1.23E6 N = 101325 Pa + 1000*9.8*h

    and solved for h. The answer I got was 115 m. This is wrong, and I think it's because I didn't take into account the radius of the window. Since the pressure would be different at different points on the window, I can't just treat the window as a point. I think I need to integrate the pressure over the height of the window, but how would I do this? Would it look something like this:

    [tex]\int_{0}^{0.25} \left(\rho*g*h\right) dh[/tex]

    Thanks for any help you can give me.
     
  2. jcsd
  3. May 7, 2004 #2

    arildno

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    No it won't be correct to do so either.

    Force=pressure*area,

    hence, you need to integrate the pressure over the circular area in order to find the net force.
     
  4. May 7, 2004 #3

    Doc Al

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    approximate, don't integrate

    Your main mistake, as arildno explained, is confusing force with pressure ( P = F/area).

    First figure out the maximum pressure that the window can withstand. I would ignore pressure variation over the height of the window--just assume a uniform pressure. (Don't bother trying to actually integrate the pressure over the window surface including its variation with height: too hard!)

    Once you have that maximum pressure, then figure out how deep the sub can go.
     
  5. May 7, 2004 #4

    arildno

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    If you are unfamiliar with integration in polar coordinates, you should definitely use Doc Al's procedure!
     
  6. May 7, 2004 #5

    enigma

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    Don't forget that the pressures add up. You need to account for the pressure on the inside pushing outwards as well.

    As was mentioned above, anytime you have an equality sort of like:

    X Pa = Y N

    You know you've done something wrong. The units need to add up on both sides of the equality sign, or else they're not equal.
     
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