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Height of Water in Straw vs. Time

  1. Dec 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Problem Statement:
    Imagine you had a very long straw, one end is in the ocean (sea
    level) and the other end is in space. Air pressure at sea level
    is a standard atmospheric pressure and space is a perfect vacuum.

    Plot "height of water in straw" vs. time.

    h(t): height of water in straw (above sea level) at time t (in meters)
    A: cross-sectional area of straw (in meters^2)

    Starting conditions:
    At time 0, height of water is 0: h(0) = 0
    At time 0, water has no velocity or acceleration

    2. Relevant equations

    F = ma
    position(t) = position(0) + integral_0_t (velocity(0) + integral_0_t acceleration(t) dt) dt
    Standard atmospheric pressure is ~101,000 kg / (m * s^2) per unit area
    Density of water = 1,000.00 kg/m³
    g = 9.8 m/s^2

    3. The attempt at a solution

    Here is what I've tried and maybe someone can point out where I'm
    going wrong.

    net force = atmospheric pressure - force of gravity on mass of water in straw

    Atmospheric Pressure
    Standard atmospheric pressure is ~101,000 kg / (m * s^2) per unit area
    Let's call A the cross-sectional area of the straw (in meters^2).
    force of atmospheric pressure = 101,000 * A kg m/s^2

    Force of gravity on mass of water in straw
    Force of gravity on water depends on how much water is in the
    straw (above sea level). Let's call h(t) the height of the
    water in the straw at time t.

    mass-of-water-in-straw(h(t)) = volume * density = h(t) * A * 1000 kg
    force_of_gravity(h(t)) = mass-of-water-in-straw(h(t)) * g = 1000 * h(t) * A * g kg m /s^2

    F(t) = 101,000 * A - 1000 * h(t) * A * g
    m(t) = 1000 * A * h(t) (from mass of water equation above)
    So acceleration (F / m) = (101 - g * h(t)) / h(t)
    Written another way: a(t) = (101 - g * h(t)) / h(t)

    Ok, great, so I think h as a function of time should be the solution to this equation:
    h(t) = integral_0_t (integral_0_t a(t) dt) dt
    Substitution my equation for a(t), I get:
    h(t) = integral_0_t (integral_0_t (101 - g * h(t)) / h(t) dt) dt

    Unfortunately, I don't know how to solve that... so I tried to
    numerically approximate it. Let's just assume constant
    acceleration over short periods of time dt, then we can use much
    simpler equations and step through time in small dt steps.

    Simpler equations:
    a(t) = (101 - g * h(t-dt)) / h(t-dt)
    v(t) = v(t-dt) + a(t) * dt
    h(t) = h(t-dt) + v(t) * dt
    with starting conditions h(0) = 0 and v(0) = 0 and a(0) = 0

    When I try to calculate my initial acceleration
    a(1), it's infinity (or rather, undefined) since the force of air
    pressure is pushing on no mass (no water is above sea level
    at the start of the problem)! There is obviously some flaw
    in that logic, but I'm not sure what it is.

    Any help is apprecated, thank you.
    Last edited: Dec 29, 2013
  2. jcsd
  3. Dec 30, 2013 #2

    Simon Bridge

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    You appear to have neglected:
    1. the mass of air initially filling the straw;
    2. the rotation of the Earth
    3. the falloff in the acceleration of gravity with altitude
    4. distance the bottom of the straw is below the water surface - i.e. initially no water in the straw.
    5. viscosity in the water ...

    You may want to start with a related question - what height of water will be supported by a pressure difference of 1atmos?

    ... it is not really useful to mix numbers and variables in an equation.

    Setting up the DE - the acceleration of the com from Newtons laws - cancelling the area - you get something like:

    ##P-\rho gh = \rho gh \ddot{h}: h(0)=0, \dot h(0)=0##

    ##P## is 1atmos, ##\rho## is density of water, ##g## accel of gravity at sea level, ##h(t)## is the distance of the top of the column above sea level ... center of mass is at h/2.

    put it another way: $$\frac{d^2h}{dt^2} = \frac{P}{\rho gh}-1$$
  4. Dec 30, 2013 #3
    Re your points:
    1) I did neglect #1-5 as you mentioned. Do you think assuming those effects are negligible is an OK approximation?
    2) I have already worked out the related question, height of water supported by 1atmos. I agree it's an interesting problem.
    3) Fair point about mixing constants with variables.

    I think you have a small typo in your DE. I think the correct statement is as follows:
    [itex]P-ρgh = ρh\ddot{h}: h(0) = 0, \dot{h} = 0[/itex]

    Are you able to solve this (numerically or otherwise)? The problem that I am having is that if [itex]h(0) = 0[/itex], then [itex]\ddot{h}=∞[/itex].
  5. Dec 30, 2013 #4

    Simon Bridge

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    I'm pretty sure it's a standard DE ... y(ay''+by+c)=d, here b=0.

    What happens to you calculation if you start with a small initial volume of water in the straw?
    Well spotted on the typo.

    Anyway - the DE is 2nd order and non-linear for form y''=f(y,y')
    ... substitute ##v=\dot h## gives you a 1st order separable equation you can solve for v(h)
    ... which, in turn, gives you 1st order equation(s) for h(t).
    Last edited: Dec 30, 2013
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