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11 March, 15:40

Initially 15 grams of salt are dissolved into 25 liters of water. Brine with concentration of salt 4 grams per liter is added at a rate of 6 liters per minute. The tank is well mixed and drained at 6 liters per minute. a) Let x be the amount of salt, in grams, in the solution 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. dxdt = grams/minute b) Find a formula for the amount of salt, in grams, after t minutes have elapsed. x (t) = grams c) How long must the process continue until there are exactly 20 grams of salt in the tank?

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  1. 11 March, 19:17
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    a) dx/dt = 600 - 6x

    b) x = 100 - 4.12 ((e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

    c) The mass of salt in the tank attains the value of 20 g at time, t = 0.227 min = 13.62 s

    Explanation:

    Taking the overall balance, since the total Volume of the setup is constant, then flowrate in = flowrate out

    Let the concentration of salt in the tank at anytime be C

    Let the concentration of salt entering the tank be Cᵢ

    Let the concentration of salt leaving the tank be C₀ = C (Since it's a well stirred tank)

    Let the flowrate in be represented by Fᵢ

    Let the flowrate out = F₀ = F

    Fᵢ = F₀ = F = 6 L/min

    a) Then the component balance for the salt

    Rate of accumulation = rate of flow into the tank - rate of flow out of the tank

    dx/dt = Fᵢxᵢ - Fx

    Fᵢ = 6 L/min, C = 4 g/L, F = 6 L/min

    dC/dt = 24 - 6C

    dx/dt = 25 (dC/dt), (dC/dt) = (1/25) (dx/dt) and C = x/25

    (1/25) (dx/dt) = 24 - (6/25) x

    dx/dt = 600 - 6x

    b) dC/dt = 24 - 6C

    dC / (24 - 6C) = dt

    ∫ dC / (24 - 6C) = ∫ dt

    (-1/6) In (24 - C) = t + k (k = constant of integration)

    In (24 - 6C) = - 6t - 6k

    -6k = K

    In (24 - 6C) = K - 6t

    At t = 0, C = 15 g/25 L = 0.6 g/L

    In (24 - 6 (0.6)) = K

    In 20.4 = K

    K = 3.02

    So, the equation describing concentration of salt at anytime in the tank is

    In (24 - 6C) = K - 6t

    In (24 - 6C) = 3.02 - 6t

    24 - 6C = e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾

    6C = 24 - (⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)

    C = 4 - ((e⁻⁽⁶ᵗ ⁻ ³•⁰²⁾) / 6

    C = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾) / 6)

    x/25 = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾) / 6)

    x = 100 - 4.12 ((e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

    c) when x = 20 g

    20 = 100 - 4.12 (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

    80 = (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

    - (6t - 3.02) = In 80

    - (6t - 3.02) = 4.382

    (6t - 3.02) = - 4.382

    6t = - 4.382 + 3.02

    t = - 1.362/6 = 0.227 min = 13.62 s
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