Applications of Differential Equations

Concentration of a Brine Solution

Salt is dissolved in a tank filled with 100 gallons of water. Originally 25 lb of salt was dissolved in the tank. Salt water containing 2 lb of salt per gallon is poured in at the rate of 4 gal/min. The solution is well stirred. The mixture is poured out at a rate of 4 gal/min. Find the amount of salt as a function of time Q(t). How much salt remains after 3 hours?

Start by summarizing what is known.

   t = 0      Qo = 25 lb


We need an equation. Here’s a hint: the rate of change of Q in the tank is the difference between what enters the tank and what exits the tank. See if you can write this a differential equation.

        dQ/dt = (4gal/min)2 lb/gal – (4gal/min)Q/100gal

        dQ/dt = 8 – 0.04Q

Now separate the variables and integrate:

    dQ/(8-0.04Q) = dt


    -25ln(8-0.04Q) = t + C

    ln(8-0.04Q) = -0.04t + C

    8 – 0.04Q = Ce-0.04

    Then solve for the constant, C and finish the problem..

    Use    t = 0    Qo = 25 lb

    8 – 0.04(25) = Ce-0.04(0)

    C = 7

    The equation is then      Q = 200 – 175e-0.04t

    And after 3 hours (180min):    

    Q =  200 – 175e-0.04(180) =  199.9 lb

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