Problem 28 - Thomas Calculus

Question

An executive conference room of a corporation contains 4500 \(\mathrm{ft}^{3}\) of air initially free of carbon monoxide. Starting at time \(t=0,\) cigarette smoke containing 4\(\%\) carbon monoxide is blown into the room at the rate of \(0.3 \mathrm{ft}^{3} / \mathrm{min} .\) A ceiling fan keeps the air in the room well sirculated and the air leaves the room at the same rate of 0.3 \(\mathrm{ft}^{3} / \mathrm{min}\). Find the time when the concentration of carbon monoxide in the room reaches 0.01\(\%\).

Step-by-Step Solution

Answer

The concentration reaches 0.01% after 1.5 minutes.

Step 1: Define Variables and Initial Conditions

Let \( C(t) \) be the amount of carbon monoxide in the room at time \( t \) in cubic feet. Initially, \( C(0) = 0 \) since the room is initially free of carbon monoxide.

Step 2: Set Up the Differential Equation

The rate of change of carbon monoxide in the room is the rate entering minus the rate leaving. The rate entering the room is \( 0.04 \times 0.3 = 0.012 \) cubic feet per minute. The rate leaving the room is \( \frac{C(t)}{4500} \times 0.3 \) cubic feet per minute. Therefore, the differential equation is \( \frac{dC}{dt} = 0.012 - \frac{C(t)}{15000} \).

Step 3: Solve the Differential Equation

To solve \( \frac{dC}{dt} = 0.012 - \frac{C(t)}{15000} \), we have \( \frac{dC}{0.012 - \frac{C}{15000}} = dt \). Integrating both sides, we get:\[ \int \frac{dC}{0.012 - \frac{C}{15000}} = \int dt \].This results in:\[ -15000 \ln \left|0.012 - \frac{C}{15000} \right| = t + K \].

Step 4: Apply Initial Condition to Find Constant K

Plugging in the initial condition \( C(0) = 0 \), we have:\[ -15000 \ln \left|0.012 - \frac{0}{15000} \right| = 0 + K \],which simplifies to \( K = -15000 \ln(0.012) \).

Step 5: Determine the Solution for C(t)

The solution after integrating and simplifying is:\[ C(t) = 4500(1 - e^{-\frac{t}{15000}}) \].Substituting back the values already found to express \( C(t) \) in terms of \( t \).

Step 6: Find Time When Concentration Reaches 0.01%

We want to find \( t \) such that \( \frac{C(t)}{4500} = 0.0001 \). Plug in \( C(t) = 0.0001 \times 4500 \) into \( C(t) = 4500(1 - e^{-\frac{t}{15000}}) \) to solve for \( t \):\[ 0.45 = 4500(1 - e^{-\frac{t}{15000}}) \]Simplifying, we have:\[ e^{-\frac{t}{15000}} = 1 - \frac{0.45}{4500} \].

Step 7: Solve for Time t

Solving:\[ e^{-\frac{t}{15000}} = 0.9999 \]\[ -\frac{t}{15000} = \ln(0.9999) \]\[ t = -15000 \times \ln(0.9999) \].

Step 8: Compute the Value of t

Calculate the value using a calculator:\[ t \approx -15000 \times (-0.0001) = 1.5 \].

Key Concepts

Carbon Monoxide Concentration

In the study of differential equations related to environmental engineering, particularly in room ventilation scenarios, carbon monoxide (CO) concentration is an essential parameter. In the given problem, we consider the concentration of CO in a fixed volume of air over time. Initially, the air in the room is free from CO. However, when cigarette smoke containing 4% carbon monoxide is introduced at a constant rate, the CO concentration changes as the smoke mixes and circulates within the room.
To understand how carbon monoxide concentration evolves, one must consider both the rate of addition and the ventilation system in the room, which regulates how the polluted air is removed. The balance of these rates determines the steady growth of CO concentration until it reaches a specified safety threshold. This illustrates a typical problem wherein one must calculate the time required for the concentration to reach a minimal yet crucial value, such as 0.01% in this case.

Separable Equations

Separable equations are a class of differential equations that can be rearranged such that each side of the equation depends only on one variable. In our problem, the equation is derived from the rates of CO entering and leaving the room. This leads to a first-order differential equation capturing the relationship between the rate of change of an amount of CO and the concentration itself.To solve a separable equation like the one presented, we isolate the variable terms on either side of the equation. For example:

• Rearrange the equation into the form: \( \frac{dC}{dt} = 0.012 - \frac{C(t)}{15000} \).
• Separate variables to get: \( \frac{dC}{0.012 - \frac{C}{15000}} = dt \).
• Integrate both sides to find a solution.

By separating and integrating these terms, you convert the differential equation into an algebraic equation. This approach helps in understanding how the CO concentration changes over time and finding when it reaches the desired value.

Exponential Growth and Decay

Exponential growth and decay are fundamental phenomena in mathematics and nature. In our problem, they manifest as the natural processes of mixing and dispersing carbon monoxide in the room. The mathematical model used here relies on the principle that the rate of change in the concentration is proportional to the concentration itself or its complement.Once integration of the separable equation is complete, the solution expresses the change in concentration as an exponential function:\[ C(t) = 4500(1 - e^{-\frac{t}{15000}}) \]This tells us that over time, the concentration of CO approaches a maximum possible value determined by the room's conditions and the initial setup.The exponential decay factor in the equation, \( e^{-\frac{t}{15000}} \), illustrates how quickly the room approaches this equilibrium state. Time \( t \) measures how fast or slow this process occurs until the carbon monoxide concentration reaches 0.01%. Understanding this growth and decay mechanism allows us to predict when certain concentration levels will be met, ensuring safety standards are upheld.