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Starting a Mixture problem

  1. Oct 29, 2006 #1
    Hello,

    I think this is a pretty easy question here that I think I may be just overlooking. I am asked this huge drawn out DE problem about a tank being filled with water and brine and I am asked to solve all these different parts. My problem is that the first part of the quesiton asks to find out when the tank overfills, and I can't seem to get that. I know it is just simple formula, but for some reason I am just not seeing how it is formed. Word problems definately are not my strong point. If anyone can explain the thinking behind this, it would be a great help. Thanks.

    Here is the problem:

    A 400 gallon tank is filled with 300 gallons of fluid in which 50lbs of salt is dissolved. Brine containing 2 pounds of salt per gallon is pumped in at a rate of 3 gallons per minute. The well-mixed solution is pumped out at a rate of 2 gallons per min. When will the tank overflow?

    I was given the answer in class by my professor to be 100 minutes and the formula was 300+2t = 500, but I have no idea where this came from.

    There is a lot more to the problem, but that is the first part and I need to get past that to get to the rest.
     
  2. jcsd
  3. Oct 29, 2006 #2

    Hootenanny

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    We know that initially, that the tank contain 300 gallons of fluid and its capacity it 400 gallons. If we consider the net flux ([itex]\Phi[/itex]) into the tank then we obtain;

    [tex]300+\Phi=400[/tex]

    Now, the net flux will be the flux of liquid into the tank minus the flux of water out of the tank. Can you go from here?
     
    Last edited: Oct 29, 2006
  4. Oct 29, 2006 #3
    Sorry ... I am pretty lost here. I have a basic understanding of what flux is, but I am not 100% clear. Let me see if I can understand this correctly...

    It seems to me that [tex](\Phi)[/tex] would be the flux of liquid into the tank (3 gal/min) minus the flux of liquid out of the tank (2 gal/min) , or 3-2=1. 300+t=400, then t=100. So that means that the tank will overflow in 100 minutes. Wow, that just sort of came out while I was typing here. haha. Makes sense. So then, how come the formula given is 300+2t=500, and not 300+t=400?
     
  5. Oct 29, 2006 #4
    As I look at this equation a little closer, I see that my last post is not entirely correct. I said that
    , but in the original problem, it says that the salt solution is being pumped in at a rate of 3 gal/min and being pumped out at 2 gal/min. That being said, it also means that the the rate of the overflow of the tank will be 1 gal/min. So to me, that means that the total flux of solution going out of the tank would be 2 gal/min + 1 gal/min = 3 gal/min. Then that makes the previous equation of 300+t=400 not true since t = 0, and for sure 300 does not equal 400 :tongue2: ... So back to square one for me.

    So, here is my thinking. Is rate out (when overflowing) not the total rate out (the amount being pumped out + the amount overflowing), but just the amount overflowing? That would make the formula given to me by my professor make sense with the 300+2t = 500, except, I don't know where that 500 came from and that thinking is a little counter intuitive for me. Am I totally going off in the wrong direction here?
     
  6. Oct 29, 2006 #5

    Hootenanny

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    The reason that your equation does not 'work' at t equal to zero is that we want to find the time when it overflows, that is our condition. Look at it this way, the volume V of solution in the tank at time t will be given by;

    [tex]V=300+t[/tex]

    Now, we want to know when the tank will overflow, i.e. when [itex]V>400[/itex], therefore we have;

    [tex]V>400[/tex]

    [tex]V=300+t[/tex]

    Substituting the second equation into the inequality we obtain;

    [tex]300+t>400[/tex]

    Is that more satisfactory?
     
  7. Oct 29, 2006 #6
    ok, I get that, but how does that help me here? I mean, yes, the tank will overflow at some time t and I know that at that time t it will have to be more than 400 gallons, or V > 400, but I still do not see what that does for me here except help me visualize what is going on.
     
  8. Oct 30, 2006 #7
    Any help on this one anyone?
     
  9. Oct 30, 2006 #8

    Hootenanny

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    I don't follow what your problem is? Surely you know all you need to do now is solve for t?
     
  10. Oct 30, 2006 #9
    Yes, I solve for t and I get 100 minutes, which is the correct answer, however, I do not see how the formula given in my class (300+2t = 500)makes sense. I am sorry to make you spell this out for me and go through all this, but I really appreciate you helping me. Thank you!
     
  11. Oct 30, 2006 #10

    Hootenanny

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    I'm not sure how your Prof derived the equality but you can reach the one I derived from your Prof's by taking your equation, dividing throughout by two and adding 150 units to each side, but I have no idea why your Prof would want the equality in this form unless it is relevant for further questions.
     
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