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Math Help - Salt water pumped into a tank problem.

  1. #1
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    Salt water pumped into a tank problem.

    Suppose that a large mixing tank initially holds 400 gallons of water in which 65 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 6 gal/min, and when the solution is well stirred, it is pumped out at a faster rate of 8 gal/min. If the concentration of the solution entering is 3 lb/gal, determine a differential equation for the amount of salt in the tank at time
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  2. #2
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    Re: Salt water pumped into a tank problem.

    Hey derail.

    What is the rate of change of salt? Is the density of salt the same for all forms of brine?
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  3. #3
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    Re: Salt water pumped into a tank problem.

    Since the tank initially "holds 400 gallons of water" and "brine solution is pumped into the tank at a rate of 6 gal/min" and "it is pumped out at a faster rate of 8 gal/min" so after t minutes it will hold 400+ 6t- 8t= 400- 2t gallons of water. The density of the brine is the amount of salt, A(t), divided by the volume of water, 400- 2t, is \frac{A(t)}{400- 2t}. The point of that is that 8 gallons of water being pumped out each minute will contain \frac{8A(t)}{400- 2t} pounds of salt. At the same time, a brine solution with 3 lbs/gal is coming in at 6 gal/min and so salt is coming in at (3 lbs/gal)(6 gal/min)= 18 gal/min.

    dA/dt is the rate at which salt is coming into the tank (positive for inflow, negative for outflow) so dA/dt= 18- \frac{8A}{400- 2t}.
    Last edited by HallsofIvy; January 26th 2013 at 06:51 PM.
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