**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 ##

http://ift.tt/1vAa81n