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How do I find percentage of a substance in a mixture?

  1. Sep 26, 2012 #1
    How is this question done??


    Milk chocolate is being produced in a 300 litre tank, which initially contains 100 litres of milk. The following things then occur simultaneously:
    
    1. Liquid cocoa (made up of equal parts cocoa solids and cocoa butter, both in liquid form) is added at a rate of 6 litres per minute.

    2. Milk is added at a rate of 3 litres per minute.  The well-stirred mixture leaves the tank via a tube, at a rate of 6 litres per minute.

    The differential equation for dx/dt + 6x / (100 + 3t) = 3
    ......Therefore......
    x(t) = (9t^3 + 900t^2 + 30,000t) / (3t + 100)^2

    Find the time in minutes (t) for the mixture to be 25% Cocoa solids.
     
  2. jcsd
  3. Sep 26, 2012 #2

    chiro

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    Science Advisor

    Hey proctortom.

    What does the x in the x(t) refer to?
     
  4. Sep 26, 2012 #3
    I get a differential equation that does not agree with yours. I am assuming that x is the volume fraction of liquid cocoa in the tank.

    Total volume in tank at time t = (100 + 3t) liters

    Rate of liquid cocoa entering = 6 lpm

    Rate of liquid cocoa leaving = 6x lpm

    Rate of liquid cocoa accumulation = d ((100 + 3t) x )/dt

    Differential equation:

    d ((100 + 3t) x )/dt = 6 - 6x

    Or equivalently:

    (100 + 3t) dx/dt + 9x = 6

    Solution:

    x = (2/3)(1 - 106/(100 + 3t)3)
     
  5. Sep 29, 2012 #4
    He is asking for cocoa solid (from liquid cocoa) so dx/dt + 6x(100+3) = 3

    Does anyone know how to do this

    "After the tank is full, the process continues as above. However, in addition to the well-stirred mixture leaving via the tube, it also flows over the edges of the tank and is collected by overflow tubing which takes it to the cooling process. Let y(t) be the number of litres of cocoa solids present in the tank t minutes after it is full.

    Find the differential equation satisfied by y(t)"
     
  6. Sep 29, 2012 #5
    His differential equation is still wrong. In this problem, the concentration of the cocoa liquid is always twice the concentration of the cocoa solids. If I put x = 2y in my equation, I get

    dy/dt + 9y/(100+3t) = 3/(100+3t)

    where y is the concentration of cocoa solid. The coefficient of 6 in his equation is incorrect. If the volume V of liquid in the tank is changing with time, then

    d(Vy)/dt is not equal to V dy/dt. It is equal to V dy/dt + y dV/dt. He left out the second term which, in this problem, is equal to 3y. I also assert that there should be 100+3t in the denominator on the RHS.

    To get the cocoa solids concentration from the solution I gave, just divide by 2.

    In the limit of very long times, neglecting the finite capacity of the tank, the concentration of cocoa solids must approach 1/3 (consistent with my solution) since that is the concentration of cocoa solids in the feed, where, at long times, the effects of the initial liquid in the tank become negligible.

    Chet
     
    Last edited: Sep 29, 2012
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