A tank contains 100 kg of salt and 1000 L of water. A solution of a concentration 0.05 kg of salt per liter enters a tank at the rate 10 L/m

Question

A tank contains 100 kg of salt and 1000 L of water. A solution of a concentration 0.05 kg of salt per liter enters a tank at the rate 10 L/min. The solution is mixed and drains from the tank at the same rate.

(a) What is the concentration of our solution in the tank initially?
concentration = (kg/L)

(b) Find the amount of salt in the tank after 1 hours.
amount = (kg)

(c) Find the concentration of salt in the solution in the tank as time approaches infinity.
concentration = (kg/L)

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Anna 2 weeks 2022-01-07T15:22:42+00:00 1 Answer 0 views 0

Answers ( )

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    2022-01-07T15:24:17+00:00

    Answer:

    Step-by-step explanation:

    Initial Quantity of salt Q(0) = 100 kg

    Capacity of tank = 1000 L

    Inflow of liquid = 10 l /minute and outflow = same rate thus leaving the content of volume 1000 L at any time

    Assume after entering constantly mixes and drains out.

    Concentration flow = inflow – outflow

    = 0.05(10) – \frac{Q(t)}{1000} *10

    i.e. Q'(t) = 0.5-0.01Q(t), Q(0) = 100

    Q'(t) = -0.01(Q(t)-50), Q(0) = 100\\\frac{dQ}{Q-50} =-0.01 dt\\ln |Q-5| = -0.01t +C\\Q-50 = Ae^{-0.01t}

    Use Q(0) = 5

    -45 = A  

    So the equation would be

    Q-50 = -45 e^{-0.01t} \\Q(t) = 50-45 e^{-0.01t} \\

    a) Initial concentratin = \frac{100}{1000} =0.10

    b) Use the solution fo rDE substitute t =1

    Q(1) = 50-45e^(-0.01) = 5.4477 kg

    c) As t becomes large Q = 50

    So concentration limit = 50/1000 = 0.05

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