Question

A swimming pool whose volume is 10,000 gal contains water that is 0.01% chlorine. Starting at t = 0, city water containing 0.001% chlorine is pumped into the pool at a rate of 5 gal/min. The pool water flows out at the same rate. What is the percentage of chlorine in the pool after 1 h? When will the pool water be 0.002% chlorine?

Short answer

The percentage of chlorine in the pool after 1 h is $0.00973\%$and the percentage of chlorine in the pool water will reach 0.002% after$73.24h$.

Step-by-step solution

Analyzing the given statement 

It can view the tank as a compartment containing salt. If we let $x\left(t\right)$denote the volume of chlorine in the tank at a timet , we can determine the concentration of chlorine in the tank by dividing $x\left(t\right)$ by the volume of solution in the tank at a time t . We use the mathematical model described by the following equation to solve for role="math" localid="1664256579818" $\mathit{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$,

$\frac{dx}{dt}=$Input rate – Output rate ..................................(1)

To determine the input rate of solution into the pool

First, one must determine the rate at which the solution enters the tank. We are given that chlorine is pumped into the tank at a rate of5 gal/min.Since the solution entering the tank is 0.001%chlorine, we conclude that the input rate of solution into the tank is,

$\left(0.001\%\right)·\left(5gal/min\right)=\frac{1}{20000}$

To determine the output rate of solution from the pool

One must now determine the output rate of solution from the tank. The solution in the tank is kept well stirred, so let’s assume that the concentration of chlorine in the tank is uniform. That is, the concentration of chlorine in any part of the tank at time t is just x(t) divided by the volume of solution in the tank.Because the tank initially contains 10,000 gal and the rate of flow into the tank is the same as the rate of flow out, the volume is a constant 10,000 gal. Hence, the output rate of solution from the tank is,

$\left(5\mathrm{gal}/\min \right)\times \left(\frac{\mathrm{x}\left(\mathrm{t}\right)}{10000}\mathrm{kg}/\mathrm{L}\right)=\frac{\mathrm{x}\left(\mathrm{t}\right)}{2000}$

Determining the initial value problem

The pool initially contained10,000 gal of 0.01% chlorine. So,

$$\begin{gathered} \frac{x\left(0\right)}{10000}=0.01\% \\ \frac{x\left(0\right)}{10000}=\frac{1}{10000} \\ x\left(0\right)=1 \end{gathered}$$

So, one sets the initial value as $x\left(0\right)=1$.

Substituting the input and output rates from step 2 and step 3 into the equation (1), we will get the following initial value problem as a mathematical model for the mixing problem,

$\frac{dx}{dt}=\frac{1}{20000}-\frac{x\left(t\right)}{2000},$$x\left(0\right)=1$

To find the solution of the initial value problem obtained in step 4 to find the percentage of chlorine in the pool after 1 hour

Firstly, it will rewrite the differential equation obtained in step 4 as,

$\frac{dx}{dt}+\frac{x\left(t\right)}{2000}=\frac{1}{20000}$

…… (2)

Integrating factor, I.F.= $e^{\int \frac{1}{2000}dt}=e^{\frac{1}{2000}t}$

Multiplying both sides of equation (2) by $e^{\frac{1}{2000}t}$,

$$\begin{gathered} e^{\frac{1}{2000}t}·\frac{dx}{dt}+e^{\frac{1}{2000}t}·\frac{x\left(t\right)}{2000}=e^{\frac{1}{2000}t}·\frac{1}{20000} \\ \frac{d}{dt}\left(x·e^{\frac{1}{2000}t}\right)=e^{\frac{1}{2000}t}·\frac{1}{20000} \end{gathered}$$

Now, integrating both sides,

$x·e^{\frac{1}{2000}t}=e^{\frac{1}{2000}t}·\frac{2000}{20000}+C$

where, C is an arbitrary constant.

$x=\frac{1}{10}+C{e}^{-\frac{1}{2000}t}$ …… (3)

At t=0, x=1

Therefore, from equation (3),

$$\begin{gathered} 1=\frac{1}{10}+C \\ 1-\frac{1}{10}=C \\ C=\frac{9}{10} \end{gathered}$$

Substituting this value of C in equation (3),

$$\begin{gathered} x=\frac{1}{10}+\frac{9}{10}{e}^{-\frac{1}{2000}t} \\ x=\frac{1}{10}\left(1+9e^{-\frac{1}{2000}t}\right) \end{gathered}$$

So, the volume of chlorine in the pool after the time, t minisrole="math" localid="1664257608352" $\mathit{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{10}}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{9}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2000}}\mathbf{t}}\mathbf{)}$.

Now, the volume of chlorine in the pool after the time, t=60 min,

$$\begin{gathered} x\left(60\right)=\frac{1}{10}\left(1+9e^{-\frac{60}{2000}}\right) \\ x\left(60\right)=\frac{1}{10}\left(1+9e^{-\frac{3}{100}}\right) \\ x\left(60\right)=\frac{1}{10}\left(1+9e^{-0.03}\right) \\ x\left(60\right)=\frac{1}{10}\left(1+9e^{-0.03}\right) \\ x\left(60\right)=0.973gal \end{gathered}$$

So, the volume of chlorine in the pool after the time, , t=1 hour is 0.973 gal.

$$\begin{gathered} x\left(60\right)=\frac{0.973}{10000}\times 100 \\ x\left(60\right)=0.00973\% \end{gathered}$$

Thus, the percentage of chlorine in the pool after, t=1 hour is 0.00973%.

To find the time at which the pool water will be 0.002% chlorine

It can determine the concentration of chlorine in the pool by dividing x(t) by the volume of solution in the pool, which is given as 10000 gal,

$\frac{x\left(t\right)}{10000}=\frac{1}{100000}\left(1+9e^{-\frac{1}{2000}t}\right)$

To determine the time, when the percentage of chlorine will reach 0.002%, we set

the right-hand side of above equation, equal to 0.002%and solve for t. This gives

$$\begin{gathered} \frac{1}{100000}\left(1+9e^{-\frac{1}{2000}t}\right)=0.002\% \\ \frac{1}{100000}\left(1+9e^{-\frac{1}{2000}t}\right)=\frac{2}{100000} \\ 1+9e^{-\frac{1}{2000}t}=2 \\ 9e^{-\frac{1}{2000}t}=1 \\ {e}^{\frac{1}{2000}t}=9 \\ \frac{t}{2000}=ln9 \\ t\approx 4394.45min \\ t\approx 73.24h \end{gathered}$$

Consequently, the percentage of chlorine in the pool will reach 0.002% after 73.24 h.