Problem 53 - Physics

Question

A room has a volume of \(120 \mathrm{~m}^{3}\). An air-conditioning system is to replace the air in this room every twenty minutes, using ducts that have a square cross section. Assuming that air can be treated as an incompressible fluid, find the length of a side of the square if the air speed within the ducts is (a) \(3.0 \mathrm{~m} / \mathrm{s}\) and \((\mathrm{b}) 5.0 \mathrm{~m} / \mathrm{s}\).

Step-by-Step Solution

Answer

For air speeds of (a) 3.0 m/s, the side length is 0.1826 m, and (b) 5.0 m/s, the side length is 0.1414 m.

Step 1: Calculate Air Flow Rate

The volume flow rate, or the volume of air to be replaced every minute, is determined by dividing the room's volume by the replacement time. Given, the room volume is \(120 \, \mathrm{m}^3\) and it needs to be replaced every 20 minutes. Thus, the flow rate \(Q\) is: \[ Q = \frac{120 \, \mathrm{m}^3}{20 \, \mathrm{minutes}} = 6 \, \mathrm{m}^3/\mathrm{minute} \]Convert the flow rate to \(\mathrm{m}^3/\mathrm{second}\): \[ Q = \frac{6 \, \mathrm{m}^3}{1 \, \mathrm{minute}} \times \frac{1 \mathrm{minute}}{60 \mathrm{seconds}} = 0.1 \, \mathrm{m}^3/\mathrm{s} \]

Step 2: Relate Cross-Sectional Area and Air Speed

The relationship between volume flow rate \(Q\), cross-sectional area \(A\), and air speed \(v\) is given by: \[ Q = A \times v \]The cross-sectional area \(A\) for a square duct of side length \(s\) is \(s^2\). Thus, the formula becomes: \[ Q = s^2 \times v \]

Step 3(a): Calculate Square Side Length for \(v = 3.0 \, \mathrm{m}/\mathrm{s}\)

Using the relation \(Q = s^2 \times v\) and substituting the given values: \[ 0.1 = s^2 \times 3.0 \]Solve for \(s^2\): \[ s^2 = \frac{0.1}{3.0} = 0.0333 \ bar s = \sqrt{0.0333} \approx 0.1826 \, \mathrm{m} \]

Step 4(b): Calculate Square Side Length for \(v = 5.0 \, \mathrm{m}/\mathrm{s}\)

Similarly, substitute \(v = 5.0\) into the equation: \[ 0.1 = s^2 \times 5.0 \]Solve for \(s^2\): \[ s^2 = \frac{0.1}{5.0} = 0.02 \] \[ s = \sqrt{0.02} \approx 0.1414 \, \mathrm{m} \]

Key Concepts

Incompressible Fluid

In fluid mechanics, the term "incompressible fluid" describes a fluid whose density does not change significantly with varying pressure. This is a common assumption for many liquid flows and is also applied to gases under certain conditions, such as low-speed flows. For air, being a compressible gas, treating it as incompressible generally applies when flow speeds are much less than the speed of sound. The assumption simplifies many problems by maintaining a constant density, and as such, makes it applicable for scenarios like air-conditioning systems in enclosed environments.
This means the mass of air entering a space equals the mass of air exiting, allowing calculations to focus on volume rather than mass. These models help us answer questions about duct sizes and flow rates without delving into complex changes in air pressure.

Volume Flow Rate

The volume flow rate, often symbolized as \(Q\), is a measure of the volume of fluid that passes through a specific section of a duct or pipe per unit of time. This is key in fluid dynamics problems involving air movement in air-conditioning systems.
To determine \(Q\), you simply divide the volume of the space the air needs to fill by the time it takes to replace the air. In this exercise, the room's volume is 120 \(\text{m}^3\) and the time for replacement is 20 minutes, resulting in a flow rate of \(6 \text{ m}^3/\text{minute}\) or \(0.1 \text{ m}^3/\text{s}\).
This value is crucial when trying to ensure that air is efficiently distributed throughout a space, as it dictates how large or small your ducts should be to handle the expected airflow.

Cross-Sectional Area

Cross-sectional area is a fundamental concept when considering how ducts transport air in systems like HVAC units. In mathematics, this is the area of a two-dimensional shape as seen from a perpendicular side view.
For a square duct, this is given by \(A = s^2\), where \(s\) is the side length of the square cross-section. In relation to volume flow rate \(Q\) and air speed \(v\), the formula \(Q = A \times v\) applies. This means to maintain a constant flow rate, if the duct's cross-sectional area changes, the air speed must adjust accordingly.
By specifying \(Q\) and the desired air speed, you can calculate \(A\), helping you choose appropriate duct dimensions to ensure proper air distribution.

Air Speed

Air speed within ducts is a critical factor for ensuring an efficient air conditioning system. It refers to how quickly air particles are moving through the ductwork. This speed is often adjusted to meet desired temperature and air distribution standards.
From the formula \(Q = A \times v\), altering air speed affects both flow rate and duct dimensions. For the given exercise, two different air speeds of \(3.0 \text{ m/s}\) and \(5.0 \text{ m/s}\) are explored. By calculating \(s\), the side length of the square, we see

• Lower air speed requires a larger cross-sectional area, which translates to larger ducts.
• Higher air speed can manage with a smaller duct size.

Thus, selecting the optimal air speed involves balancing energy efficiency with physical space constraints in building infrastructure design.