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Setting up differential equation

  1. Apr 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt
    in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal/min, and
    the mixture is allowed to flow out of the tank at a rate of 2 gal/min. Find the amount
    of salt in the tank at any time prior to the instant when the solution begins to overflow.
    Find the concentration (in pounds per gallon) of salt in the tank when it is on the point of
    overflowing. Compare this concentration with the theoretical limiting concentration if the
    tank had infinite capacity.

    2. The attempt at a solution
    I get stuck in the very beginning where you have to set up the differential equation.

    dQ/dt = rate of flow in - rate of flow out

    rate of flow in = 3gal/min × 1 lb/gal = 3 lb/min

    (C = concentration, Q = quantity of salt, V = Volume)
    rate of flow out = C r = (Q / V) r
    The volume changes; starts at 200 and 3-2=1 gallon per minute extra:
    V = 200 + t
    And r=2

    So: dQ/dt = 3 - (Q / 200+t)2, which I would then have to solve.

    But according to my answer book the equation that has to be solved is:
    dQ/dt = 3 - Q(t)

    Does anyone understand how they get there?
  2. jcsd
  3. Apr 21, 2009 #2
    The first sentence gives the initial conditions of the tank. What does Q(t) represent? It's the amount of salt leaving the tank as a function of time and this is not explicitly known because of the mixing that is occurring with differing amounts of salt concentration, whereas the input concentration is explicitly known and is a constant.
  4. Apr 21, 2009 #3
    Yes, I had figured that out... but why would that give you the equation:

    dQ/dt = 3 - Q(t) ?
  5. Apr 21, 2009 #4
    I got what you got. Are you sure you're not reading the answer to a different problem? The equation dQ/dt=3-Q is the sort of equation that would result if the inflow and outflow rates are equal.
  6. Apr 21, 2009 #5
    That's what I thought as well... I checked and I really am looking at the correct corresponding answer to the problem. It says:

    Salt flows into the tank at the rate of (1)(3) lb/min. and
    it flows out of the tank at the rate of (2) lb/min. since
    the volume of water in the tank at any time t is 200 +
    (1)(t) gallons (due to the fact that water flows into the
    tank faster than it flows out). Thus the I.V.P. is dQ/dt =
    3 - Q(t), Q(0) = 100.

    Strange isn't it?

    I thought it was me, but if I'm right I'll just skip this one.
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