Mathematical Modeling Problem for 500-gal Aquarium Impurities

[MarkFL]

Here is the question:

Mathematical modeling problem?

A 500-gal aquarium is cleansed by the recirculating filter system. Water containing impurities is pumped out at a rate of 15 gal/min, filtered, and returned to the aquarium at the same rate. Assume that passing through the filter reduces the concentration of impurities by a fractional amount a. In other words, if the impurity concentration upon entering the filter is c(t), the exit concentration is ac(t), where 0 < a < 1.
a. Apply the basic conservation principle (rate of change = rate in - rate out) to obtain a differential equation for the amount of impurities present in the aquarium at time t. Assume that filtering occurs instantaneously. If the outflow concentration at any time is c(t), assume that the inflow concentration at that same instant is ac(t).
b. What value of filtering constant a will reduce impurity levels to 1% of their original values in a period of 3 hr?

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Kristine,

Let's let $A(t)$ represent the total amount of impurities in the aquarium at time $t$, measured in minutes. Let's also work this problem with parameters rather than numbers, so that we will derive a general formula into which we can plug the given data. Let the parameters be as follows:

$V$ is the volume in gallons of the aquarium.

$R$ is the rate, in gallons per minute, that water is filtered.

Now, if $c(t)$ is the concentration of impurities in the tank at time $t$, which is a measure of mass per gallons, then the rate at which impurities are leaving the tank and going into the filtration system is the product of the concentration and the rate of flow, which will have units of mass per minute:

\(\displaystyle \text{rate out}=Rc(t)\)

And likewise, the rate in of impurities coming in is given by:

\(\displaystyle \text{rate in}=Rac(t)\)

Now, we want to find an expression relating $A(t)$ and $c(t)$, and using the same principle that concentration is amount per volume, we may state:

\(\displaystyle c(t)=\frac{A(t)}{V}\)

And so, using the given basic conservation principle, we may model the amount of impurities in the tank at time $t$ with the following initial value problem (IVP):

\(\displaystyle \frac{dA}{dt}=\frac{R}{V}(a-1)A\) where \(\displaystyle A(0)=A_0\)

The ordinary differential equation (ODE) associated with the IVP is separable, and we may write it as:

\(\displaystyle \frac{1}{A}\,dA=\frac{R}{V}(a-1)\,dt\)

Let's exchange the dummy variables of integration so that we may use the given boundaries as our limits of integration:

\(\displaystyle \int_{A_0}^{A(t)}\frac{1}{u}\,du=\frac{R}{V}(a-1)\int_{0}^{t}\,dv\)

Applying the rules of integration and the anti-derivative form of the fundamental theorem of calculus (FTOC), we obtain:

\(\displaystyle \left[\ln(u) \right]_{A_0}^{A(t)}=\frac{R}{V}(a-1)\left[v \right]_{0}^{t}\)

Doing the evaluations indicated, there results:

\(\displaystyle \ln\left(\frac{A(t)}{A_0} \right)=\frac{R}{V}(a-1)t\)

Converting from logarithmic to exponential form, we obtain:

\(\displaystyle \frac{A(t)}{A_0}=e^{\frac{R}{V}(a-1)t}\)

Multiplying through by $A_0$, we obtain the solution satisfying the IVP:

\(\displaystyle A(t)=A_0e^{\frac{R}{V}(a-1)t}\)

In order to determine the value of $a$ that will result in the amount of impurities being 1% of the initial amount, we will let \(\displaystyle A(t)=\frac{A_0}{100}\), and $t=180$ (since 180 minutes is 3 hours), and the solve the resulting equation for $t$. This will be easier if we use the logarithmic form of the solution:

\(\displaystyle \ln\left(\frac{A(t)}{A_0} \right)=\frac{R}{V}(a-1)t\)

Solving for $a$, we find:

\(\displaystyle a=\frac{V}{Rt}\ln\left(\frac{A(t)}{A_0} \right)+1\)

We will also plug in the given data for the parameters:

\(\displaystyle R=15,\,V=500\)

And we obtain:

\(\displaystyle a=\frac{500}{15\cdot180}\ln\left(\frac{\dfrac{A_0}{100}}{A_0} \right)+1\)

Simplification yields:

\(\displaystyle a=1-\frac{10}{27}\ln\left(10 \right)\approx0.147190706298502\)