Chapter 7.3, Problem 46E

To determine

To Find: The percentage of the carbon dioxide in the room as a function of time and what is the long run effects.

Answer to Problem 46E

The amount of carbon dioxide in the room is $0.05\%$ as time tends to infinity.

Explanation of Solution

Given:

The air in the room is with the volume $180{\text{m}}^3$ contains 0.15% carbon dioxide initially.

The fresh air contains 0.15% carbon dioxide initially.

The fresher air with only 0.05% carbon dioxide flows in the room at the rate of 2 cubic meter per minute and the mixed air flower out at the same rate.

Calculation:

Consider the function $y\left(t\right)$ is the amount of the carbon dioxide in the room at $t\ge 20$ in min.

The rate at which the new carbon dioxide enters the room is,

  $\begin{array}{c}ratein=2\left(0.05\%\right)\\ =0.001{\text{m}}^3/\text{min}\end{array}$

The rate at which the new carbon dioxide leaves the room is,

  $\begin{array}{c}rateout=\frac{y}{180}\left(2\right)\\ =\frac{y}{90}{\text{m}}^3/\text{min}\end{array}$

Then, the net rate is,

  $\frac{dy}{dt}=0.001-\frac{y}{90}$

Since, the room contains 0.15% carbon dioxide at time t=0. Then, the initial condition is,

  $\begin{array}{c}y\left(0\right)=\frac{0.15}{100}\left(180\right)\\ =0.27{\text{m}}^3\end{array}$

Then, the general solution of the differential equation is,

  $\begin{array}{l}\frac{dy}{dt}=0.01-\frac{y}{90}\\ \frac{dy}{0.09-y}=\frac{dt}{90}\\ {\displaystyle \int \frac{dy}{0.09-y}}={\displaystyle \int \frac{dt}{90}}\\ \ln \left|0.09-y\right|=-\frac{1}{90}t+c\end{array}$

Solve further as,

  $\begin{array}{l}y-0.09=e^{\left(-\frac{1}{90}t+c\right)}\\ y=0.09+e^{\left(-\frac{1}{90}t+c\right)}\end{array}$

Substitute the initial condition $y=\left(0.27{\text{m}}^3\right)$ in the above equation as,

  $\begin{array}{l}0.27{\text{m}}^3=0.09+e^{\left(-\frac{1}{90}\left(0\right)+c\right)}\\ e^c=0.18\end{array}$

Then, the required equation is

  $y=0.09+{0.18}^{\left(-\frac{1}{90}t\right)}$

Thus, the percentage of carbon dioxide in the room at time t is,

  $\begin{array}{c}P\left(t\right)=\frac{y\left(t\right)}{180}\times 100\\ =\frac{0.09+0.18e^{\left(\frac{-1}{90}\right)t}}{1.8}\\ =0.05+0.1e^{\left(\frac{-1}{90}t\right)}\end{array}$

Then, for the long run,

  $\underset{t\to \infty }{\lim }\left(0.05+0.1e^{\left(\frac{-1}{90}t\right)}\right)=0.05\%$

Thus, the amount of carbon dioxide in the room is $0.05\%$ as time tends to infinity.