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Concentration of paint in each bowl after certain time

  1. Jun 17, 2014 #1
    1. The problem statement, all variables and given/known data
    We are taking a closer look at three identical bowls with volume ##10l##. At the beginning only the first bowl is full with water mixed with ##0.5kg## of paint. The other two bowls are empty. Than we provide a constant flow of ##0.05 l/s## of clear water straight into the first bowl. Because the first bowl is already full, the mixture of paint and water starts flowing into the second bowl and when the second bowl is full, the mixture continues into the third bowl.

    Calculate the concentration of paint in each bowl when the bowl no. 3 is full. Water and paint are always very well mixed.


    2. Relevant equations



    3. The attempt at a solution

    The first one should be simple. I will use notation ##V## for volume, ##c## for concentration of paint and ##\phi _v## for that constant flow.

    First bowl:

    ##Vdc=-c\phi _vdt##

    ##c(t)=c_0e^{-\beta t}## where ##\beta = \frac{\phi _v}{V}## and ##c_0=\frac{m_p}{m_w}## where ##m_p## is mass of paint and ##m_w## mass of water.

    If I am not mistaken ##t=\frac{2V}{\phi _v}=400s## so finally

    ##c(t=400s)=0,00676##

    Second bowl:

    ##Vdc_2=(c_1(t)\phi _v-c_2\phi _v)dt##

    ##dc_2=(c_0e^{-\beta t}-c_2)\beta dt##

    ##c_2(t)=(\beta c_0t+D)e^{-\beta t}## , we know that ##c_2(t=0)=0## therefore:

    ##c_2(t)=\beta c_0te^{-\beta t}##

    and

    ##c_2(t=400)=0,0135## (Greater than ##c_1## ??)

    Third bowl:

    ##c_3(t)=c_0(\beta t)^2e^{-\beta t}##

    ##c_3(t=200s)=0,00184 ##
     
  2. jcsd
  3. Jun 17, 2014 #2

    verty

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    Homework Helper

    For me the only important details here are amount of paint and time. You want to think in terms of paint flow and time, ignore any other distracting details.

    So for example, the amount of paint in A is changing, let A'(t) be that rate of change at time t. Similarly, the amount of paint in B is changing, let B'(t) be that rate of change. Find A(t) and B(t).

    PS. You want to keep things simple whenever possible.
     
  4. Jun 17, 2014 #3
    The second bowl is done incorrectly. During the first 200 sec, V is a function of time for bowl 2: V=øvt, and the amount of paint in bowl 2 is øvtc2. This needs to be taken into account. During the first 200 sec, I get:
    [tex]c_2=c_0\frac{(1-e^{-βt})}{βt}[/tex]

    Chet
     
  5. Jun 18, 2014 #4
    Uhh, you are right. During the first 200 seconds the amount of paint in the second bowl is only increasing, nothing is going out of it.

    I agree, I did the calculation again and I got the same result.

    Thanks Chet!
     
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