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March 26, 2017

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A tank contains 50 gallons of a solution composed of 90% water and 10% alcohol. A second solution containing 50% water and 50% alcohol is added to the tank at rate of 4 gallons per minute. As the second solution is being added, the tank is being drained at the rate of 5 gallons per minute. Assuming the solution in the tank is stirred constantly, how much alcohol is in the tank after 10 minutes?

  • math - ,

    Would it not be a similar to a previous problem?
    http://www.jiskha.com/display.cgi?id=1344476111

    You can try the same approach and post your attempt for a check.

  • math - ,

    oke thanks, i'ill try it

  • math - ,

    y(t)=amount of alcohol in the tank at time t
    dy/dt
    =y’(t)
    = alcohol inflow rate – alcohol outflow rate
    = 4 - 5(y/50)
    =200 - 5y/50
    = 5 (40-y)/50

    dy/dt = 5 (40-y)/50
    Separate variable and integrate :
    dy/(40-y) = 5/50dt
    -ln(40-y) = 0.1t + C
    ln(40-y) = -0.1t + C
    40-y= Ce^-0.1t
    Y= 40(1-Ce^-0.1t) ….. eq. 1

    Initially, the tank contains 50 gallon (50% alcohol) = 25 gallon of alcohol,
    y(0) = 25
    substituting y=25 and t=0 in to eq.1 gives

    25 = 40(1-C)
    25-40 = -40C
    C = 0.375

    When t=10 minutes
    y(10) = 40(1-0.375*e^-0.1(10))
    y(10) = 40(1-0.375*0.367)
    y(10) = 34.48 gallon

    at t=10’, amount of alcohol in the tank
    = 50% x 34.48 gallon = 17.24 gallon or 34.48%


    please correct the result of my work. thanks

  • math - ,

    Yes, you have done a fantastic job of setting up and solving the problem.

    If you reread the question carefully, you will probably discover a few traps.

    If you don't find the traps, here are a few hints to improve the solution:

    1. Alcohol coming into the tank is 50%, so intake rate is 4*0.5=2 gal/min.
    2. Initial concentration is 90%, so initial alcohol content is 50*0.9=45 gal.
    3. Initially, output volume is y/50 (as you had it).
    However, since the input is 4 gal/min, and output is 5 gal/min, you will need to divide by the total volume as a function of time, which turns out to be 50-t.
    4. finally, do not forget required parentheses. Otherwise, your work will be very hard to read, such as:
    "= 4 - 5(y/50)
    =200 - 5y/50
    ..."
    instead of
    = 4 - 5(y/50)
    =(200-5y) / 50


    Good job, keep up the good work.

  • math - ,

    thanks for your help :)

  • math - ,

    You're welcome!

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