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The air in a room \(50 \mathrm{ft}\) by \(20 \mathrm{ft}\) by \(8 \mathrm{ft}\) tests \(0.2 \%\) carbon dioxide. Starting at \(t=0\), outside air testing \(0.05 \%\) carbon dioxide is admitted to the room. How many cubic feet of this outside air must be admitted per minute in order that the air in the room test \(0.1 \%\) at the end of \(30 \mathrm{~min}\) ?

Short Answer

Expert verified
Approximately \(533.33 \mathrm{~ft^3/min}\) of outside air with 0.05% carbon dioxide must be admitted to the room in order to achieve a carbon dioxide concentration of 0.1% after 30 minutes.

Step by step solution

01

Calculate the volume of the room

To calculate the volume of the room, multiply the length, width, and height of the room together. The room has dimensions of 50 ft, 20 ft, and 8 ft: \[V_{room} = 50 ft \times 20 ft \times 8 ft = 8000 \mathrm{~ft^3}\]
02

Calculate the initial mass of carbon dioxide in the room

To calculate the initial mass of carbon dioxide in the room, first we need to find the initial amount of carbon dioxide in the room. Since the room has 0.2% carbon dioxide, we can calculate the volume of carbon dioxide: \[V_{CO2\_initial} = V_{room} \times 0.2\% = 8000 \mathrm{~ft^3} \times 0.002 = 16 \mathrm{~ft^3}\]
03

Calculate the mass of carbon dioxide in the admitted air

We want the room to test 0.1% carbon dioxide after 30 minutes. First, we will find the volume of carbon dioxide in the room after 30 minutes: \[V_{CO2\_final} = V_{room} \times 0.1\% = 8000 \mathrm{~ft^3} \times 0.001 = 8 \mathrm{~ft^3}\] To achieve this volume of carbon dioxide, we need to add outside air with 0.05% carbon dioxide. Let's denote the total volume of admitted air during 30 minutes as V_admitted: \[V_{CO2\_admitted} = V_{admitted} \times 0.05\%\]
04

Calculate the volume of air admitted per minute

Now we will create an equation representing the balance of carbon dioxide volumes before and after adding the admitted air: \[V_{CO2\_initial} + V_{CO2\_admitted} = V_{CO2\_final}\] Substituting the values and solving the equation for V_admitted: \[16 \mathrm{~ft^3} + V_{admitted} \times 0.05\% = 8 \mathrm{~ft^3}\] \[V_{admitted} = \frac{8 \mathrm{~ft^3}-16 \mathrm{~ft^3}}{0.0005} = 16000 \mathrm{~ft^3}\] Since this is the total volume of air admitted during 30 minutes, we will divide it by 30 to find the volume of air admitted per minute: \[V_{admitted\_per\_min} = \frac{16000 \mathrm{~ft^3}}{30 \mathrm{~min}} = 533.33 \mathrm{~ft^3/min}\] Thus, we need to admit approximately \(533.33 \mathrm{~ft^3/min}\) of outside air into the room to achieve the required 0.1% carbon dioxide concentration after 30 minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carbon Dioxide Concentration
Carbon dioxide concentration refers to the amount of carbon dioxide present in a given volume of air. In this problem, the initial concentration is given as 0.2% within an enclosed room. This percentage means that for every cubic foot of air in the room, 0.2% of that volume is carbon dioxide. To work with these percentages effectively, it's often converted into a pure decimal by dividing by 100, such as 0.002 for 0.2%. To change the concentration to a desired level, in this case 0.1%, specified conditions must be met which determines how much fresh air needs to replace existing air. Such calculations are crucial for environmental control systems in various settings, ensuring the air quality remains safe and comfortable for inhabitants.
Air Mixing Problem
An air mixing problem involves understanding how different air streams mix to achieve a particular component concentration, often carbon dioxide in an environmental context. This type of problem can include controlling the inflow of fresh air into a contained space to change the existing mixture. The concept revolves around achieving a balance between the starting concentration and the desired concentration after mixing. The strategy is to introduce a specific volume of outside air which has a different concentration of carbon dioxide. This reduces the overall concentration in the room to the target concentration. By knowing the room's volume and initial conditions, one can compute the needed air inflow rate to meet these requirements over a set period.
Volume Calculation
Calculating volume is essential in determining the amount of space available within a room or container. It is especially crucial in problems where mixing or changing concentrations are involved, as seen in the air mixing problem. In this situation, we calculate the room's volume by multiplying its length, width, and height together. For our specific room dimensions, the volume is calculated as:\[ V_{room} = 50 \mathrm{ft} \times 20 \mathrm{ft} \times 8 \mathrm{ft} = 8000 \mathrm{ ft^3} \]Once the room volume is known, it acts as a baseline for understanding how much air can be mixed and how concentrations will change as different volumes of air are introduced. This volume calculation becomes the groundwork for more complex calculations needed in environmental engineering or chemistry applications.
Ordinary Differential Equations
Ordinary differential equations (ODEs) are equations that involve functions of one independent variable and its derivatives. In the context of this air mixing problem, while an ODE wasn't explicitly stated, this mathematical tool is frequently used to describe how concentrations of substances (like carbon dioxide) change over time within a system. For similar problems, ODEs can model the rate at which air is exchanged and mixed within a space. They provide a method to continuously describe the change in concentration levels over a period. By using initial conditions, such as the original and desired concentrations, an ODE would help in understanding the declination path or concentration over time, allowing for precise controls and predictions.

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Most popular questions from this chapter

A boat weighing \(150 \mathrm{lb}\) with a single rider weighing \(170 \mathrm{lb}\) is being towed in a certain direction at the rate of \(20 \mathrm{mph}\). At time \(t=0\) the tow rope is suddenly cast off and the rider begins to row in the same direction, exerting a force equivalent to a constant force of \(12 \mathrm{lb}\) in this direction. The resistance (in pounds) is numerically equal to twice the velocity (in feet per second). (a) Find the velocity of the boat 15 sec after the tow rope was cast off. (b) How many seconds after the tow rope is cast off will the velocity be one- half that at which the boat was being towed?

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