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Homework Help: Differential equations capacity tank problem (chemical solutions) mixture

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A 300-gal capacity tank contains a solution of 200 gals of water and 50 lbs of salt. A solution containing 3 lbs of salt per gallon is allowed to flow into the tank at the rate of 4gal/min. The mixture flows from the tank at the rate of 4 gal/min. The mixture flows from the tank at the rate of 2 gal/min. How many pound of salt are in the tank at the end of 30 min? When does the tank start to overflow? How much salt is in the tank at the end of 60 min?

    2. Relevant equations

    [tex] \frac{dx}{dt} + \frac{F}{V_o}x = F_i C_i [/tex]

    3. The attempt at a solution

    I really don't know how to start the equation, i mean i don't know where to substitute the values on the right variable. So I just need help on where to assign the values given.
     
  2. jcsd
  3. Jul 11, 2010 #2

    HallsofIvy

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    Science Advisor

    No, you need much more help than that. This is supposed to be an exercise in understanding what a differential equation says, not just in substituting numbers into a formula.

    For one thing, the formula you have written makes no sense because you haven't said what "x", "t", etc. represent.

    If x represents the amount of salt in the tank, in pounds, and t the time, in minutes, then dx/dt is the total rate at which salt flows into and out of the tank, in "pounds per minute". Now look at what you are given:
    "A solution containing 3 lbs of salt per gallon is allowed to flow into the tank at the rate of 4gal/min." Okay, that is salt flowing into the tank at (3 lbs/gal)(4 gal/min)= 12 lbs/min. That rate is positive because it is coming in and so increasing the amount of salt in the tank.

    You also have "The mixture flows from the tank at the rate of 4 gal/min." and " The mixture flows from the tank at the rate of 2 gal/min." Are you sure you have copied the problem correctly? There is no reason to give two different statements like that. I suspec that the correct statement is the second and that the "4 gal/min" in the first sentence were accidently copied from the mixture flowing in. If it were either "4 gal/min" or both, then there would be at least as much solution flowing out as in and the tank would never overflow. So I am going to assume just " The mixture flows from the tank at the rate of 2 gal/min."

    The first thing that tells us is that solution is coming in at 4 gal per min and out at only 2 gal per minute so there is a net intake of 4- 2= 2 gal/min. Since the tank initially contained 200 gal, after t minutes, it will contain 200+ 2t gallons. (Until it reaches 300 gallons and overflows- when will that happen?) If there are x pounds of salt in the tank, then there are x/(200+ 2t) "pounds per gallon"- and that is going out at 2 gal/min so out at (x/(200+2t))(2)= x/(100+ t) pounds per minute. Since that is out, it is decreasing the amount of salt- it is negative.

    Putting those together we have dx/dt= 12- x/(100+ t) with an initial value of x(0)= 50 pounds.
     
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