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Homework Help: Cylindrical tank w/ water flowing out of small tube near bottom

  1. Nov 21, 2009 #1
    1. A cylindrical open tank needs cleaning. The tank is filled with water to a height meter, so you decide to empty it by letting the water flow steadily from an opening at the side of the tank, located near the bottom. The cross-sectional area of the tank is square meters, while that of the opening is square meters

    2. I've got a bunch of calculations below. I need help with setting up and solving the final integral.

    3. A_1 = area of surface, v_1 = velocity of flow at suraface
    A_2 = area of small tube, v_2 = velocity of flow out of tube

    A_1 >> A_2

    V' = A_2 * v_2 = rate of discharge

    V' = A_1 * dh/dt, relates rate of discharge to drop in height of liquid

    v_2 = sqrt[2gh + v_1^2]

    v_1 = v_2 *(A_2/A_1)

    v_1 = sqrt[(2gh) / (1-(A_2/A_1)^2)]

    dh/dt = rate of change of height of water = -A_2/A_1 * sqrt(2gh)

    Ok, so now the strategy is to solve a separable first-order ODE like so:

    dy/dx = f(x)g(y)
    dy/g(y) = f(x)dx, and integrate both sides.

    So in my case, these are the integrals I obtain:

    dh/dt = -A_2/A_1 * sqrt[2gh]
    Plus in g = 9.81 and obtain:

    dh/dt = -4.23*(A_2/A_1)*sqrt(h)

    So I get the following integrals:

    h^(-0.5)dh = -4.23*(A_2/A_1)dt with h_0 and 0.5h_0 as the limits of integration. I have solved this a couple different ways and obtained incorrect answers.

    Can someone point out what the correct integral I need to solve is?

  2. jcsd
  3. Nov 21, 2009 #2
    Any ideas?
  4. Nov 21, 2009 #3


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    Homework Helper

    What is the question?
    I'm having trouble following your work - why do you say
    v_2 = sqrt[2gh + v_1^2]
    It seems to say that if v_1 was zero, v_2 would be nonzero which doesn't make sense. Then it looks like you dropped the v_1^2 to get
    dh/dt = = -A_2/A_1 * sqrt(2gh)
    Actually that does make sense. What did you find from this that wasn't correct? I see you can get h as a function of time.
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