Differential Equations - Mixture in an overflowing tank

  1. HOW IS THE FOLLOWING 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.


    Let x(t) be the amount of cocoa solids in the mixture


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


    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)
     
  2. jcsd
  3. haruspex

    haruspex 14,349
    Science Advisor
    Homework Helper
    Gold Member
    2014 Award

    Not sure about the = 3 in your first eqn. Shouldn't it be 6?
    The soln doesn't look quite right to me. Did you check it satisfied the differential eqn?
    Please show your attempt at the last part.
     
  4. It is 3 because proctortom is asking for cocoa solids. dx/dt = rate of cocoa solids in - rate of cocoa solids out. Which is dx/dt = 3 [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] - (x/100+3t)*6 = dx/dt + 6x/100+3t = 3

    I also get stuck in finding the differential equation satisfied by y(t). The eqn should be dy/dt (3/100)*(100-y). But I don't know how to get to this !
     
  5. dy/dt = rate in - rate out

    rate in = 3L/m (cocoa solids going into mixture)

    rate out = concentration x flow rate out
    = amount/volume x flow rate out

    where the flow rate out must equal the flow rate in as the tank is remaining full! (6L/m via tube and 3L/m via overflow tubing)

    = y/300 * 9
    = 3y/100

    so dy/dt = 3 - 3y/100
    = 3/100 (100 - y)
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?