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The value of the constant c

  1. Nov 12, 2007 #1
    1. The problem statement, all variables and given/known data

    A tank contains 80 gallons of pure water. A brine solution with 2lb/gal of salt enters at 2 gal/min, and the well stirred mixture leaves at the same rate. Find a.) the amount of salt in the tank at any time and b.) the time at which the brine leaving will contain 1 lb/gal of salt.

    2. Relevant equations

    i've enclosed a copy of my solution
    [​IMG]


    3. The attempt at a solution
     
  2. jcsd
  3. Nov 12, 2007 #2
    i dunno if my solution is correct since the tank initially contains only pure water...
     
  4. Nov 13, 2007 #3

    HallsofIvy

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    It's hard to tell what you have done! You have a lot of parameters, called v1, c1, q1, q0 which you have not identified- very bad practice.

    If you let X be the amount of salt in the vat at time t, with X measured in pounds and t measured in minutes, then dX/dt is how much salt enters of leaves the vat in pounds per minute. According to the information, 2 gallons of water enter the vat each minute, carrying 2 pounds of salt per gallon: that brings 4 pounds of salt in per minute. Since the same amount of water goes out, the total amount of water stays the same and the concentration is X(t)/80. That is, water leaves at 2 gallons per minute, carrying X/80 pounds per gallon: salt goes out of the tank at X/40 pounds per minute. Putting those together, dX/dt= 4- X/40= (160- X)/40. You have that equation but seem to have separated dX and dt in a rather peculiar way. It is true that dX= 4dt- (X/40)dt but I don't see any reason to do that. dX/dt=(160- X)/40 is a separable equation:
    [tex]\frac{dX}{160-X}= \frac{dt}{40}[/itex]
    Integrating both sides, -ln(160-X)= t/40+ C or 160-X= [itex]ce^{-t/40}[/itex] where c= eC.

    Since the tank originally contains only pure water, X(0)= 0 so 160- 0= c or c= 160.
     
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