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Mixing Problem Question

  1. Jan 24, 2014 #1
    Hi all. I'm trying to help a friend work a differential equation problem since I've already taken this course and gotten an A in it, but this other problem has me a little confused.

    The book is ODE's by Tenenbaum and Pollard, chapter 3 lesson 15a number 2.

    #1) Tank initially holds 100 gal of brine containing 30 lb of dissolved salt. Fresh water flows into the tank at a rate of 3 gal/min and brine flows out at the same rate. (a) Find the salt content of the brine at the end of 10 minutes and (b) When will the salt content be 15 lb?

    This one I've solved easily but number 2 says,

    #2) Solve problem 1 if 2gal/min of fresh water enter the tank instead of 3 gal/min.

    The solution on the next page says it should be (a) x=30(1-.01t)^3 which, when 10 is plugged for t gives 21.87 lb.

    Can someone explain what I'm missing? Shouldn't the rate in still be zero?
     
  2. jcsd
  3. Jan 24, 2014 #2

    Office_Shredder

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    It is very difficult to explain what you are missing when you haven't explained what you did to try to solve the problem. Can you tell us what your solution is for both parts?

    And it is unclear what you mean when you say "the rate in should still be zero". The rate of water flowing into the tank? This is clearly not zero.
     
  4. Jan 24, 2014 #3
    Well for #1 I got a) x=30e^(-.03t), x(10)= 22.2 lb. b) 23.1 minutes

    For number I've tried finding the integrating factor but that still leaves me with the question of whether the rate of fresh water in changing from 3gal/min to just 2gal/min has anything to do with the amount of solution in the tank. When I use rate in= 2gal/min and try the integrating factor method I get a solution that looks like this: x=-2/.03+(30+2/.03)e^(-.03t) which is a ridiculous result. Any clues?

    And the rate of solution coming in is zero because the problem says it is fresh water flowing in. So rate of solution flowing in is zero but fresh water flowing in is 2 gal/min in problem 2 and 3 gal/min in problem 1.
     
    Last edited: Jan 24, 2014
  5. Jan 24, 2014 #4
    I guess my real question isn't "can someone help me solve this?" It's more "can someone explain how the rate of fresh water coming in affects the solution inside the tank?"
     
  6. Jan 24, 2014 #5

    Mark44

    Staff: Mentor

    The fresh water flowing into the tank dilutes the brine solution in the tank. Brine is flowing out at the same rate as fresh water is flowing into the tank, so in time, what's in the tank will be just water with no salt.
     
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