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Chapter 11: Problem 57

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 (b) \(5.0 \mathrm{m} / \mathrm{s}\).

Short Answer

Expert verified
(a) The side length is approximately 0.1823 m, (b) the side length is approximately 0.1414 m.

Step by step solution

01

Finding the Flow Rate

The room's air volume is 120 cubic meters and must be replaced every 20 minutes. To find the flow rate, we convert the time to seconds: 20 minutes is 1200 seconds. Thus, the flow rate \( Q \) of the air is given by \( Q = \frac{120 \ \mathrm{m^3}}{1200 \ \mathrm{s}} = 0.1 \ \mathrm{m^3/s} \).
02

Relating Flow Rate to Duct Properties

The flow rate \( Q \) is also given by \( Q = A \cdot v \), where \( A \) is the cross-sectional area of the duct and \( v \) is the velocity of the air through the duct. For a square duct, \( A = s^2 \), where \( s \) is the side length of the square cross-section.
03

Step 3a: Solving for Side Length when Air Speed is 3.0 m/s

Using the given air speed \( v = 3.0 \ \mathrm{m/s} \), we set up the equation \( 0.1 = s^2 \times 3 \). Solving for \( s \), we find \( s^2 = \frac{0.1}{3} = 0.0333\ \mathrm{m^2} \). Therefore, \( s = \sqrt{0.0333} \). Thus, \( s \approx 0.1823 \ \mathrm{m} \).
04

Step 3b: Solving for Side Length when Air Speed is 5.0 m/s

Using the air speed \( v = 5.0 \ \mathrm{m/s} \), we set up the equation \( 0.1 = s^2 \times 5 \). Solving for \( s \), we find \( s^2 = \frac{0.1}{5} = 0.02\ \mathrm{m^2} \). Therefore, \( s = \sqrt{0.02} \). Thus, \( s \approx 0.1414 \ \mathrm{m} \).

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Key Concepts

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

Incompressible Fluid
In the exercise, air is treated as an incompressible fluid. This concept is crucial in fluid dynamics where certain fluids, like liquids or air under specific conditions, are assumed to have constant density. This simplification makes analysis easier since changes in volume do not affect density.
For an incompressible fluid:
  • The density remains uniform over time and space.
  • The fluid occupies the same volume throughout flow, even when pressure changes.
In this problem, assuming the air is incompressible helps us apply the principle that the fluid's mass flow rate is consistent through all segments of the flow system. This assumption aligns with typical real-world HVAC scenarios, simplifying the calculation of flow rates and velocities.
Cross-Sectional Area
The cross-sectional area plays a pivotal role in determining how much fluid can pass through a duct. It is the size of the section of the duct perpendicular to the flow direction. In this problem:
  • The ducts have a square cross section, meaning both dimensions are equal.
  • The area is calculated as the side length squared, or \( s^2 \), where \( s \) is the side of the square.
  • A larger cross-sectional area generally allows for a larger flow of fluid.
Understanding the cross-sectional area is fundamental when analyzing fluid dynamics because it influences the flow rate and speed at which the fluid moves through a system.
Flow Rate
Flow rate is the volume of fluid that moves through a cross section in a given amount of time. It is commonly measured in \( ext{cubic meters per second} \, (\text{m}^3/\text{s}) \). For the air conditioning problem:
  • The flow rate \( Q \) is calculated by dividing the room's volume by the time taken for a complete air exchange.
  • Here, the flow rate is determined to be \( 0.1 \, \text{m}^3/\text{s} \).
Flow rate is a key consideration in fluid systems as it helps dictate required parameters such as duct sizes and fluid velocities, ensuring efficiency and effectiveness in operation.
Velocity
Velocity is the speed at which the fluid flows and is central to understanding fluid dynamics. In this exercise, velocity affects the calculation of the duct size.
  • Velocity \( v \) is typically measured in \( \text{meters per second} \) (\( \text{m/s} \)).
  • It is directly related to flow rate and cross-sectional area through the formula \( Q = A \cdot v \).
  • Adjusting the velocity necessitates recalibrating duct dimensions to maintain a consistent flow rate.
In practice, optimizing velocity assures effective air circulation, balancing comfort and system capacity, as shown by the different duct dimensions required for velocities of 3 m/s and 5 m/s in the problem.

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Most popular questions from this chapter

A pressure difference of \(1.8 \times 10^{3}\) Pa is needed to drive water \(\left(\eta=1.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}\right)\) through a pipe whose radius is \(5.1 \times 10^{-3} \mathrm{m}\) The volume flow rate of the water is \(2.8 \times 10^{-4} \mathrm{m}^{3} / \mathrm{s} .\) What is the length of the pipe?

Three fire hoses are connected to a fire hydrant. Each hose has a radius of \(0.020 \mathrm{m} .\) Water enters the hydrant through an underground pipe of radius \(0.080 \mathrm{m} .\) In this pipe the water has a speed of \(3.0 \mathrm{m} / \mathrm{s}\) (a) How many kilograms of water are poured onto a fire in one hour by all three hoses? (b) Find the water speed in each hose.

Neutron stars consist only of neutrons and have unbelievably high densities. A typical mass and radius for a neutron star might be \(2.7 \times 10^{28}\) \(\mathrm{kg}\) and \(1.2 \times 10^{3}\) \(\mathrm{m} .\) (a) Find the density of such a star. (b) If a dime \(\left(V=2.0 \times 10^{-7} \mathrm{m}^{3}\right)\) were made from this material, how much would it weigh (in pounds)?

A hydrometer is a device used to measure the density of a liquid. It is a cylindrical tube weighted at one end, so that it floats with the heavier end downward. The tube is contained inside a large "medicine dropper," into which the liquid is drawn using the squeeze bulb (see the drawing). For use with your car, marks are put on the tube so that the level at which it floats indicates whether the liquid is battery acid (more dense) or antifreeze (less dense). The hydrometer has a weight of \(W=5.88 \times 10^{-2} \mathrm{N}\) and a crosssectional area of \(A=7.85 \times 10^{-5} \mathrm{m}^{2} .\) How far from the bottom of the tube should the mark be put that denotes (a) battery acid \(\left(\rho=1280 \mathrm{kg} / \mathrm{m}^{3}\right)\) and \((\mathrm{b})\) antifreeze \(\left(\rho=1073 \mathrm{kg} / \mathrm{m}^{3}\right) ?\)

The drawing shows a hydraulic system used with disc brakes. The force \(\overrightarrow{\mathbf{F}}\) is applied perpendicularly to the brake pedal. The pedal rotates about the axis shown in the drawing and causes a force to be applied perpendicularly to the input piston (radius \(\left.=9.50 \times 10^{-3} \mathrm{m}\right)\) in the master cylinder. The resulting pressure is transmitted by the brake fluid to the output plungers (radii \(\left.=1.90 \times 10^{-2} \mathrm{m}\right)\), which are covered with the brake linings. The linings are pressed against both sides of a disc attached to the rotating wheel. Suppose that the magnitude of \(\overrightarrow{\mathbf{F}}\) is \(9.00\) \(\mathrm{N}\). Assume that the input piston and the output plungers are at the same vertical level, and find the force applied to each side of the rotating disc.
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