A tank contains 150 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into th

Question

A tank contains 150 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 3 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

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Tryphena 3 years 2021-09-05T16:58:32+00:00 1 Answers 188 views 0

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  1. minhkhoi
    1
    2021-09-05T17:00:10+00:00
    Reply

    Answer:

    The number A(t) of grams of salt in the tank at time t is A(t) = 150 - 110 e^{-\frac{t}{50} }

    Explanation:

    Knowing

    \frac{dA}{dt} = Rin - Rout

    First we have to find the Rin and Rout

    Rin = (concentration of the salt inflow) * (input rate of brine)

    Rin = 1 g/L * 3 L/min = 3 g/L

    Rout = (concentration of the salt outflow) * (output rate of brine)

    Rout = (\frac{A(t)}{150} g/L) * (3 L/min) = \frac{A(t)}{50} g/min

    Substituting this results

    \frac{dA}{dt} = 3 – \frac{A(t)}{50} –> \frac{dA}{dt} + \frac{1}{50} A(t) = 4

    Thus, integration factors is

    e^{ \int\limits^._. {\frac{1}{50} } \, dt } = e^{\frac{t}{50} }

    e^{\frac{t}{50} } \frac{dA}{dt} + \frac{1}{50} e^{\frac{t}{50} A(t) = 4 e^{\frac{t}{50} }

    e^{\frac{t}{50} } A(t) = \int\limits^._. {3 e^{\frac{t}{50} } } \, dt\\ \\A(t) = 150 + c e^{-\frac{t}{50} }

    Applying the initial conditions

    A(0) = 40

    c = 150 – 40 = 110

    Now, substitute this result in the solution to get

    A(t) = 150 - 110 e^{-\frac{t}{50} }

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