Alex's question at Yahoo Answers regarding a mixing problem

[MarkFL]

Here is the question:

How do I solve this linear equation problem?


A tank with a capacity of 1600 L is full of a mixture of water and chlorine with a concentration of 0.0125 g of chlorine per liter. In order to reduce the concentration of chlorine, fresh water is pumped into the tank at a rate of 16 L/s. The mixture is kept stirred and is pumped out at a rate of 40 L/s. Find the amount of chlorine in the tank as a function of time. (Let y be the amount of chlorine in grams and t be the time in seconds.)

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Alex,

We want to determine the amount $y(t)$ in grams of chlorine present in the tank at time $t$ in seconds. So, let's begin with:

\(\displaystyle \text{time rate of change of chlorine = time rate of chlorine coming in minus time rate of chlorine going out}\)

\(\displaystyle \text{time rate of chlorine coming in=concentration of solution coming in times the time rate of volume coming in}\)

\(\displaystyle \text{time rate of chlorine going out=concentration of solution going out times the time rate of volume going out}\)

Let:

\(\displaystyle C_I\) = the constant concentration of solution coming in.

\(\displaystyle R_I\) - the constant rate of incoming solution.

\(\displaystyle C_O\) = the variable concentration of solution going out.

\(\displaystyle R_O\) - the constant rate of outgoing solution.

Hence, stated mathematically, we may write:

\(\displaystyle \frac{dy}{dt}=C_IR_I-C_OR_O\)

Concentration equals amount per volume, and the volume of solution is a function of $t$, let's call it $V(t)$. So, we have:

\(\displaystyle \frac{dy}{dt}=C_IR_I-\frac{y(t)}{V(t)}R_O\)

To find the volume of solution present in the tank at time $t$, consider the IVP:

\(\displaystyle \frac{dV}{dt}=R_I-R_O\) where \(\displaystyle V(0)=V_0\)

Separating variables, and integrating, we have:

\(\displaystyle \int\,dV=\left(R_I-R_O \right)\int\,dt\)

\(\displaystyle V(t)=\left(R_I-R_O \right)t+C\)

Using the initial values, we may determine the parameter $C$:

\(\displaystyle V(0)=C=V_0\)

Hence:

\(\displaystyle V(t)=\left(R_I-R_O \right)t+V_0\)

Thus, we may now write:

\(\displaystyle \frac{dy}{dt}=C_IR_I-\frac{y(t)}{\left(R_I-R_O \right)t+V_0}R_O\)

Arranging the ODE in standard linear form, we may model the amount of chlorine present in the tank at time $t$ with the following IVP:

\(\displaystyle \frac{dy}{dt}+\frac{R_O}{\left(R_I-R_O \right)t+V_0}y(t)=C_IR_I\) where $y(0)=y_0$

Computing the integrating factor, we find:

\(\displaystyle \mu(t)=e^{\int \frac{R_O}{\left(R_I-R_O \right)t+V_0}\,dt}=\left(\left(R_I-R_O \right)t+V_0 \right)^{\frac{R_O}{R_I-R_O}}\)

Multiplying the ODE by this integrating factor, we obtain:

\(\displaystyle \left(\left(R_I-R_O \right)t+V_0 \right)^{\frac{R_O}{R_I-R_O}}\frac{dy}{dt}+R_O\left(\left(R_I-R_O \right)t+V_0 \right)^{\frac{R_O}{R_I-R_O}-1}y(t)=C_IR_I\left(\left(R_I-R_O \right)t+V_0 \right)^{\frac{R_O}{R_I-R_O}}\)

Observing that the left side of the equation is now the differentiation of the product $\mu(t)\cdot y(t)$, we may write:

\(\displaystyle \frac{d}{dt}\left(\left(\left(R_I-R_O \right)t+V_0 \right)^{\frac{R_O}{R_I-R_O}}\cdot y(t) \right)=C_IR_I\left(\left(R_I-R_O \right)t+V_0 \right)^{\frac{R_O}{R_I-R_O}}\)

Integrating with respect to $t$, we have:

\(\displaystyle \int\,d\left(\left(\left(R_I-R_O \right)t+V_0 \right)^{\frac{R_O}{R_I-R_O}}\cdot y(t) \right)=C_IR_I\int\left(\left(R_I-R_O \right)t+V_0 \right)^{\frac{R_O}{R_I-R_O}}\,dt\)

\(\displaystyle \left(\left(R_I-R_O \right)t+V_0 \right)^{\frac{R_O}{R_I-R_O}}\cdot y(t)=\frac{C_IR_I}{R_I-R_O}\left(\left(R_I-R_O \right)t+V_0 \right)^{\frac{R_O}{R_I-R_O}+1}+C\)

Solving for $y(t)$ we find:

\(\displaystyle y(t)=\frac{C_IR_I}{R_I-R_O}\left(\left(R_I-R_O \right)t+V_0 \right)+C\left(\left(R_I-R_O \right)t+V_0 \right)^{-\frac{R_O}{R_I-R_O}}\)

Using the initial conditions, we may determine the parameter $C$:

\(\displaystyle y(0)=\frac{C_IR_I}{R_I-R_O}\left(V_0 \right)+C\left(V_0 \right)^{-\frac{R_O}{R_I-R_O}}=y_0\)

\(\displaystyle C=V_0^{\frac{R_O}{R_I-R_O}}\left(y_0-\frac{C_IR_IV_0}{R_I-R_O} \right)\)

Thus, the solution satisfying the IVP is:

\(\displaystyle y(t)=\frac{C_IR_I}{R_I-R_O}\left(\left(R_I-R_O \right)t+V_0 \right)+V_0^{\frac{R_O}{R_I-R_O}}\left(y_0-\frac{C_IR_IV_0}{R_I-R_O} \right)\left(\left(R_I-R_O \right)t+V_0 \right)^{-\frac{R_O}{R_I-R_O}}\)

\(\displaystyle y(t)=\frac{C_IR_I}{R_I-R_O}\left(\left(R_I-R_O \right)t+V_0 \right)+\left(y_0-\frac{C_IR_IV_0}{R_I-R_O} \right)\left(\frac{V_0}{\left(R_I-R_O \right)t+V_0} \right)^{\frac{R_O}{R_I-R_O}}\)

Using the given data for this problem:

\(\displaystyle C_I=0\,\frac{\text{g}}{\text{L}},\,R_I=16\, \frac{\text{L}}{\text{s}},\,R_O=40\, \frac{\text{L}}{\text{s}},\,V_0=1600\text{ L},\,y_0=0.0125\, \frac{\text{g}}{\text{L}}\cdot1600\text{ L}=20\text{ g}\)

we have:

\(\displaystyle y(t)=20\left(1-\frac{3}{200}t \right)^{\frac{5}{3}}\)

Here is a plot of the solution on the relevant domain \(\displaystyle 0\le t\le\frac{200}{3}\):

View attachment 1601