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Viscosity Problem

  1. Jun 14, 2007 #1
    1. The problem statement, all variables and given/known data

    Two identical vertical cylindrical vessels of cross section α are joined at thir bases by a narrow horizontal tube of length l and internal radius r.A liquid of density d and co-efficient of viscosity λ is flowing slowly from one vessel to the other through the tube.Initially the liquid surfaces at the two vessels are at heights 3H and H respectively over the capillary tube.After what time will the levels be equal?

    2. Relevant equations
    3. The attempt at a solution

    Plesae check if I am correct:::

    the higher cylinder starts at height 3H while the lower starts at height H. We will call the height (at any time) in the lower cylinder "h" and find h as a function of time, ok? At any time we will call the height in the higher cylinder h'

    First, because the two cylinders have equal cross section areas, the total of the heights in both cylinders at any time is 4H, so

    h = 4H - h' at any time (note that h increases while h' decreases)

    Ok, now for Poiseuilles law:

    dV / dt = ( π r4 / 8 λ L ) Δp where L is the length of the capillary tube.

    The dV/dt refers to the flow rate, rate of volume flow. And the delta p refers to the difference in pressure from one end of the tube to the other.

    We can express dV/dt in terms of dh/dt, the rate at which the height rises in the lower cylinder, and a, the cross section area: dV/dt = a dh/dt

    And we can realize that the diff in pressure is caused by a diff in height of the two columns:

    Δp = d g ( h' - h ) (remember d is the density)

    Now put this all together:

    a dh/dt = ( π r4 / 8 λ L ) d g ( h' - h )

    dh/dt = ( π r4 d g / 8 a λ L ) ( 4H - h - h )

    dh/dt = ( π r4 d g / 8 a λ L ) ( 4H - 2h )

    dh/dt = ( π r4 d g / 4 a λ L ) ( 2H - h )

    To make life easier, I'm going to call all that mess in the parentheses "K", because its all constant stuff anyway

    dh/dt = K ( 2H - h )

    And now you can easily solve for h as a function of time:

    dh / (2H - h) = K dt integrate both sides. Use a substitution on the right

    u = 2H - h du = -dh so

    - du / u = Kt

    You end up with ln ( (2H - h ) / H ) = - Kt

    Solve for h and you get h = H ( 2 - e-Kt )

    Notice that at t = 0, h is equal to H, just as it should be. So when will h equal 2H? When

    e-Kt is equal to zero, which happens at... hmmm... well, as t goes to infinity.

    Is the analysis correct???
  2. jcsd
  3. Jun 17, 2007 #2


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    Homework Helper

    Yes, you are correct. With time the pressure difference is approaching to zero and hence rate of flow is also approaching to zero and the time taken for last bit is approaching to infinite, theoretically. That is why in the question they ask that at what time the difference in the level will be halved or so. Or they ask for the difference in level at a given time t.
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