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Rates of change in volume.

  1. May 23, 2007 #1
    1. The problem statement, all variables and given/known data
    A tank is in a shape of a cylinder with a circular cross-section of area A. Initially the depth of water in the tank (the head) is h0. At time t = 0 water is allowed to leave the tank ffrom a valve at the bottom. The rate at which the water leaves the tank is proportional to the head of the water at that instant; the constant of proportionality, k, is related to the discharge coefficient or the tank.

    Derive an expression for the head of water in the tank at any time t after it is allowed to empty.

    How long does it take for the tank to lose 1/2 its water? How long does it take for the tank to empty.


    2. Relevant equations
    Code (Text):
    dy/dx = -ky

    y = Ce^(-kt)

    3. The attempt at a solution

    dh/dt = -k(h - h0)

    h(time) = (h - h0)e^(-kt)

    ln ((h - h0) / h(time)) = kt

    and this is where i get stuck :confused: thats even if i've got the first bit right???

    any help would be great!

    :)
     
  2. jcsd
  3. May 23, 2007 #2
    Read the question carefully, it says that the rate at which water leaves the tank is proportional to the height of the water at the the time.
     
  4. May 23, 2007 #3
    ok so:

    the rate at which water leaves the tank = -kh

    i'm a little confused about teh "discharge coefficient"
     
  5. May 23, 2007 #4
    So what does "the rate at which water leaves the tank" actually mean?


    Just a bunch of fancy words to make the problem look interesting. :wink: If the complete problem reads as you have stated it here, then there's nothing to be confused about.
     
  6. May 23, 2007 #5
    the speed at which the height of the water decreases

    well the diffrence of the height from the initial value is h - h0

    so the rate of change is

    dh / dt = (h - h0) / (t - t0)

    ?
     
  7. May 23, 2007 #6
    Ok i think i might have cracked the 1st bit now,

    dh/dt = -kh

    h = Ce^-kt

    ln h = -kt + A



    but i dont understand how to work out how long it will take to empty 1/2 of the water, there are no figures to work with?


    thanks for your help so far neutrino.
     
  8. May 24, 2007 #7
    Not exactly. The answer is in the title of this thread!

    At any instant the rate at which the volume decreases is proportional to the height of the water at that instant. That would give you,

    [tex]\frac{dV}{dt} = -kh[/tex]
     
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