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I A Fluid Dynamics Problem with Confusing Results

  1. Apr 16, 2017 #1
    Hi all,

    I encountered a fairly straightforward problem involving fluid dynamics, but the result I got is very confusing and I seem not to understand where I got wrong.

    This is the question:

    A rectangular flat-bottom barge with a bottom area A=50 m2A=50~\text{m}^2A=50 m2 is loaded so that the bottom is at H=1 m below the surface. The density of water is ρ=103 kg/m3\rho=10^3~\text{kg/m}^3ρ=103 kg/m3 , and the water surface is perfectly still. A round hole with radius r=1 cm is made in the bottom of the barge, and the water starts leaking in. When the water level reaches h=5 cm, a bilge alarm will alert the barge operator. How long will it take for the water to reach the level 5 cm? (Assume that the Bernoulli’s equation is applicable.)

    My attempt to the question:

    First, at the surface of the water outside the boat: P_0 + rho*g* H = const
    Second, inside the small hole at the barge where I call point 2: P_2 + 1/2*rho*(v_2)^2 = const
    Third, at the surface of water inside the boat above the barge, where I call point 1:
    P_0 + 1/2*rho*(v_1)^2 + rho*g*h = const


    Applying Bernoulli's equation, I now set first equation equal to the second,

    P_0 + rho*g* H = P_2 + 1/2*rho*(v_2)^2

    and for second and third:

    P_2 + 1/2*rho*(v_2)^2 = P_0 + 1/2*rho*(v_1)^2 + rho*g*h

    but what I find strange is that there is no way to get P_2 and v_2 and they get substituted as an intermediate value, and solving this equation set I can only get equivalence of first and third equations:

    P_0 + rho*g* H = P_0 + 1/2*rho*(v_1)^2 + rho*g*h

    rho*g*H = 1/2*rho*(v_1)^2 + rho*g*h

    solving for v_1 = dh/dt = sqrt(2*g*(H-h))

    and this gives me an extremely small time, smaller than 1, which is obviously physically not plausible.

    I asked a lot of people and they don't seem to figure out why this is not correct, so I am posting to ask for help.

    Thanks
     
  2. jcsd
  3. Apr 16, 2017 #2

    Nidum

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    There are two possible solutions to this problem depending on whether you consider the sinking of the barge and the variation of the head of water driving the flow to be a significant factors in the analysis or not .

    If you consider - or better can demonstrate - that these factors are not important then you can use a very simple Bernoulli's principle analysis .

    A more complete analysis taking those factors into account would need to use both Bernoulli's principle and Archimedes principle . The neatest way to get an actual solution would be to set the problem up as a differential equation and solve it .

    if you are not familiar with differential equations then there are other ways of getting an approximate solution .

    First thing to do in any case is to post a clear diagram showing the basic geometry of the problem . It helps in problems like this to draw diagrams with relevant dimensions shown in correct proportion to each other .
     
    Last edited: Apr 16, 2017
  4. Apr 16, 2017 #3

    boneh3ad

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    To be honest, under no condition is Bernoulli's equation valid during a problem like this because the depth (and therefore the flow) is varying in time. This is why the problem says to simply assume Bernoulli's equation is valid. That said, as @Nidum pointed out, there are a couple levels of difficulty here depending on how much of the physics you want to consider. If you don't consider the slow sinking of the barge that accompanies the filling, then this takes one form, and if you do consider that effect, the problem becomes harder. However, unless I am missing something, I don't see how you do this without a differential equation.

    Either way you have a time-varying quantity in both the height of the water and the flow rate through the hole (and possibly the depth of the barge), so you are going to have to do some calculus.
     
  5. Apr 16, 2017 #4

    Nidum

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    The barge only fills with 5 cm depth of water . That is a small change of level compared to the bottom of the barge being initially submerged by 1 metre .

    Also the hole is very small compared to the bottom area of the barge . That suggests slow filling and slow sinking of the barge .

    If I was doing this problem for some real purpose I would probably just do a very simple analysis using mean values .
     
  6. Apr 16, 2017 #5

    boneh3ad

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    I suppose you could take it to its logical conclusions, say that since 1 m is 20 times larger than 5 cm, then the flow through the hole is effectively constant (even though it isn't really) and just do volume divided by flow rate to estimate the time. It would probably be pretty interesting to look at the error incurred by those two approaches as a function of the height ratio of the water levels.
     
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