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Ordinary Differential Equation - tank with inflow and outflo

  1. Oct 6, 2016 #1
    1. The problem statement, all variables and given/known data
    A tank contains 60 kg of salt and 2000 L of water. A solution of a concentration 0.015 kg of salt per liter enters a tank at the rate 6 L/min. The solution is mixed and drains from the tank at the same rate.

    Find the amount of salt in kg at t = 3 hours
    Find the concentration of salt in the solution in the tank as time approaches infinity.
    2. Relevant equations


    3. The attempt at a solution
    My attempt:


    C(t) = C = The concentration of salt in (kg/L) at time t in hours.



    dC/dt = (flow_in - flow_out + 60kg/2000L)



    flow_in = (0.015kg/L *6L/min*60min/hour) / 2000L // the concentration that flows in



    flow_out = C*6L/min*60min/hour)/ 2000L // the concentration that flows out



    2000L dC/dt = (0.015kg/L *6L/min*60min/hour) - C*6L/min*60min/hour) + 60kg



    2000 dC/dt = 5.4 - 360C + 60



    2000 dC/dt = (65.4 - 360C)



    2000/ (65.4-360C) dC = dt


    *Integrate both sides*


    Left side :

    let u = 65-360C
    du = -360 dC

    du / -360 = dC

    2000/-360 ∫1/u du

    -----------

    (-2000/360) ln(65.4-360C) = t + c // where c is a constant.



    ln(65.4 - 360C) = 360t/-2000 + c



    e^ln(65.4 - 360C) = e^ 360t/-2000 + c



    65.4 - 360C = ce^(360t/-2000)



    -360C = ce^(360t/-2000) - 65.4



    360C = 65.4 - ce^(360t/-2000)



    C = (65.4/360) - ce^(-360t/2000)





    C(0) = 60/2000 Plug in this to find the constant c



    60/2000 = (65.4/360) - c



    c = 65.4/360 - 60/2000



    C = (65.4/360) - (65.4/360 - 60/2000)e^(-360t/2000)
     
  2. jcsd
  3. Oct 6, 2016 #2
    Got it. We can delete this post!
     
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