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8 February, 04:09

A fish tank initially contains 10 liters of pure water. Brine of constant, but unknown, concentration of salt is flowing in at 4 liters per minute. The solution is mixed well and drained at 4 liters per minute.

a. Let x be the amount of salt, in grams, in the fish tank after t minutes have elapsed. Find a formula for the rate of change in the amount of salt, dx/dt, in terms of the amount of salt in the solution x and the unknown concentration of incoming brine c.

dx/dt =

b. Find a formula for the amount of salt, in grams, after t minutes have elapsed. Your answer should be in terms of c and t.

x (t) =

c. In 25 minutes there are 25 grams of salt in the fish tank. What is the concentration of salt in the incoming brine?

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  1. 8 February, 05:36
    0
    a) dx/dt = 4*c - (4*x / 10) = 4*c - 0.4*x

    b) x = 10*c - 10*c*e^ (-0.4*t)

    c) c = 2.500 g / Liters

    Step-by-step explanation:

    Given:

    - Initial volume of pure water V_o = 10 Liters

    - Inflow of brine Q_in = 4 Liters / min

    - Outflow of mixture Q_out = 4 liters / min

    Find:

    a) Find a formula for the rate of change in the amount of salt, dx/dt, in terms of the amount of salt in the solution x and the unknown concentration of incoming brine c.

    b) Find a formula for the amount of salt, in grams, after t minutes have elapsed. Your answer should be in terms of c and t.

    c) In 25 minutes there are 25 grams of salt in the fish tank. What is the concentration of salt in the incoming brine?

    Solution:

    - We will define a constant (c) as concentration of salt as amount of salt in grams per liter.

    - dx / dt is the amount of salt in grams per unit time t.

    - We will construct a net flow balance:

    dx / dt = Q_in*c - (amount of salt / V_o) * Q_out

    - Denoting the amount of salt in grams to be x. then we have:

    dx/dt = 4*c - (4*x / 10) = 4*c - 0.4*x

    - Once we have formulated the ODE, we will solve it to get the amount of salt x in grams at any time t, as follows:

    dx/dt = 4*c - 0.4*x

    - Separate variables:

    dx / (4*c - 0.4*x) = dt

    - Integrate both sides:

    - (10 / 4) * Ln | 4*c - 0.4*x | = t + K

    - Evaluate constant K, when t = 0, x = 0.

    - (10 / 4) * Ln|4c| = K

    - Input constant K back in solution:

    Ln |4*c - 0.4*x| = - 0.4*t + Ln|4c|

    4*c - 0.4*x = 4c*e^ (-0.4*t)

    x = 10*c - 10*c*e^ (-0.4*t)

    - Evaluate using the derived expression t = 25 mins and x = 25, compute c?

    x = 10*c - 10*c*e^ (-0.4*t)

    25/10 = c (1 - e^ (-0.4*25))

    c = 2.5 / (1 - e^ (-0.4*25))

    c = 2.500 g / Liters
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