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A single-family residence has an estimated indoor area of \(350 \mathrm{~m}^{2}\) and an average floor-to-ceiling height of \(3 \mathrm{~m}\). The local building code requires that residential houses be ventilated at a minimum air change rate of \(0.3 \mathrm{~h}^{-1}\), which means that \(30 \%\) of the interior volume must be ventilated every hour. (a) If ventilation is to be done by an air conditioning \((\mathrm{AC})\) system, determine the minimum capacity of the \(\mathrm{AC}\) system in units of \(\mathrm{m}^{3} / \mathrm{sec}\). (b) If the air intake is to be a single square vent and the maximum intake velocity to the vent is \(4.5 \mathrm{~m} / \mathrm{s}\), what are the minimum dimensions of the vent?

Short Answer

Expert verified
Minimum AC capacity is 0.0875 m³/s; minimum vent side is 13.9 cm.

Step by step solution

01

Calculate Indoor Volume

First, we need to calculate the volume of the indoor area. This can be done by multiplying the area of the residence by the average floor-to-ceiling height. The formula is: \[ \text{Volume} = \text{Area} \times \text{Height} \] Substituting the given numbers: \[ \text{Volume} = 350 \, \text{m}^2 \times 3 \, \text{m} = 1050 \, \text{m}^3 \] This means the indoor volume is \(1050 \, \text{m}^3\).
02

Calculate Required Ventilation Rate

The building code specifies that 30% of the volume must be ventilated every hour, corresponding to an air change rate of 0.3 \(\text{h}^{-1}\). We convert this to the volume of air that must be moved per second by dividing by 3600 (the number of seconds in one hour): \[ \text{Ventilation Rate} = 0.3 \times 1050 \, \text{m}^3 / 3600 \text{ s} \] \[ \text{Ventilation Rate} = 0.0875 \, \text{m}^3/\text{s} \] This rate is the minimum capacity for the AC system.
03

Calculate Minimum Vent Dimensions

Now, using the velocity, we determine the minimum size of the vent. Use the formula: \[ \text{Air Flow Rate} = \text{Area of Vent} \times \text{Velocity} \] Rearrange to find the area: \[ \text{Area of Vent} = \frac{\text{Air Flow Rate}}{\text{Velocity}} \] Plug the known values into the equation: \[ \text{Area of Vent} = \frac{0.0875 \, \text{m}^3/\text{s}}{4.5 \, \text{m}/\text{s}} \approx 0.0194 \, \text{m}^2 \] Since the vent is square, each side should be \(\sqrt{0.0194} \approx 0.139 \text{ m} \), or 13.9 cm.

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Key Concepts

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

Ventilation Rates
Ventilation rates are crucial for maintaining healthy indoor environments. It refers to the amount of outdoor air that is brought into a space within a given amount of time. In the context of building codes, ventilation rates are expressed as air changes per hour (\(\text{h}^{-1}\)). This metric helps to determine how effectively the indoor air is refreshed, affecting comfort and air quality.

For the exercise, the building code specifies a minimum ventilation rate of 0.3 \(\text{h}^{-1}\). This means 30% of the indoor air volume needs to be replaced every hour. To comply with this requirement, we must ensure that the air conditioning or ventilation system is capable of exchanging air at this rate.

Proper ventilation rates also help to control humidity, reduce the risk of respiratory issues caused by poor air quality, and eliminate odors. Here are some benefits of maintaining appropriate ventilation rates:
  • Reduces health risks by ensuring a supply of fresh air.
  • Makes indoor environments more comfortable.
  • Helps in controlling moisture levels and preventing mold growth.
Air Conditioning Systems
Air conditioning systems (AC) play a critical role not only in cooling but also in managing ventilation rates in buildings. When using an AC system for ventilation, it is essential to determine its capacity to ensure it can handle the required air changes per hour.

For this exercise, we calculated the needed capacity by converting the ventilation requirements from hours to seconds. This helps us to establish the air flow rate necessary for compliance with the 0.3 \(\text{h}^{-1}\) requirement. The AC system must have a minimum capacity of \(0.0875 \, \text{m}^3/\text{s}\), meaning it should be capable of moving this volume of air each second.

AC systems enhance indoor air quality by:
  • Maintaining consistent air circulation.
  • Filtering and dehumidifying air, which reduces allergens.
  • Supporting health by preventing poor air quality and excessive temperatures.

When selecting an AC system, it's vital to consider both the cooling capacity and the system's ability to ventilate the space to meet the building code's requirements.
Building Codes
Building codes are regulations that specify minimum standards for construction in order to ensure health, safety, and welfare. These codes dictate various aspects of a building structure, including ventilation rates, which help maintain indoor air quality.

The given problem involves a building code stipulation requiring a ventilation rate of at least 0.3 \(\text{h}^{-1}\). This creates a guideline for architects and builders to design systems that ensure adequate fresh air intake and help prevent indoor air pollution.

Building codes serve several functions:
  • Ensure air quality standards are met, avoiding health hazards.
  • Promote energy efficiency in building systems.
  • Guide homebuilders and engineers in creating safe living environments.

It is crucial for engineers and designers to stay updated on these codes to ensure compliance during the construction and renovation of buildings.
Indoor Air Quality
Indoor air quality (IAQ) is an essential aspect of a healthy living environment. It refers to the cleanliness of the air within and around buildings and structures. High indoor air quality means fewer pollutants in the air, which is vital for preventing health issues.

Ventilation plays a key role in maintaining good IAQ by removing contaminants such as carbon dioxide, volatile organic compounds (VOCs), and microbes. With adequate ventilation rates as indicated in building codes, we're not only complying with legal requirements but also enhancing the living conditions within a space.

Key factors affecting IAQ include:
  • Rate of outdoor air supply.
  • Presence of airborne pollutants and allergens.
  • Humidity and temperature control.

By installing well-sized ventilation systems like AC units, maintaining air changes per building code recommendations, and efficiently managing air flow, we can significantly improve the health and comfort of indoor environments.

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Most popular questions from this chapter

Consider the case in which fluid of density \(\rho\) circulates around a central point such that the velocity, \(V\), at any distance \(r\) from the central point is given by $$ V=C r $$ where \(C\) is a constant. Note that the streamlines are circles and that \(V\) on each streamline is in the tangential direction. It is known that the pressure is equal to \(p_{0}\) at a distance \(r_{0}\) from the central point. Determine an expression for the pressure distribution in any horizontal plane in terms of \(r,\) where \(C, p_{0}, r_{0}\) are parameters of the pressure distribution.

A velocity field, \(\mathbf{v}\), is spatially uniform and varies with time according to the following relation: $$ \mathbf{v}=\left\\{\begin{array}{ll} 3 \mathbf{i}+\mathbf{j} \mathrm{m} / \mathrm{s}, & t \in[0 \mathrm{~s}, 8 \mathrm{~s}] \\ 5 \mathbf{i}-4 \mathbf{j} \mathrm{m} / \mathrm{s}, & t \in(8 \mathrm{~s}, 15 \mathrm{~s}] \end{array}\right. $$ If an injection point is located at the origin of a Cartesian coordinate system at ( \(0 \mathrm{~m}\), \(0 \mathrm{~m}\) ), sketch to scale the following at \(t=15 \mathrm{~s}\) along with their key coordinates: (a) the pathline of a particle released at the injection point at \(t=0 \mathrm{~s},(\mathrm{~b})\) the streakline of dye continuously released at the injection point starting at \(t=0 \mathrm{~s},\) and (c) the streamlines in the flow field at \(t=15 \mathrm{~s}\).

The velocity distribution for laminar flow between two infinite parallel plates (called Poiseuille flow) is given by $$ u(y)=\frac{1}{2 \mu} \frac{\mathrm{d} p}{\mathrm{~d} x}\left(y^{2}-h y\right) $$ where \(u(y)\) is the velocity at a distance \(y\) from the bottom plate, \(h\) is the vertical distance between the top and bottom plates, \(\mu\) is the dynamic viscosity, and \(\mathrm{d} p / \mathrm{d} x\) is the constant pressure gradient driving the flow, where \(\mathrm{d} p / \mathrm{d} x\) is negative. The density of the fluid is \(\rho .\) For a unit width perpendicular to the flow, determine (a) the volume flow rate, (b) the average flow velocity, and (c) the mass flow rate.

The velocity field in a two-dimensional flow is given by $$ \mathbf{v}=(2+1.5 x+2.1 y) \mathbf{i}+(1.8-3 x+4 y) \mathbf{j} $$ (a) Calculate the acceleration field and (b) identify any stagnation points in the flow field.

The temperature, \(T,\) and the vertical component, \(w,\) of the wind velocity on the side of a very steep cliff are approximated by the relations $$ T(z, t)=20\left(1-0.3 z^{2}\right) \sin \left(\frac{\pi t}{6}\right){ }^{\circ} \mathrm{C}, \quad w=2.1\left(1+0.5 z^{2}\right) \mathrm{m} / \mathrm{s} $$ where \(z\) is the elevation above sea level in \(\mathrm{km}\) and \(t\) is the time in seconds. The horizontal components of the wind velocity are negligible along the cliff. Estimate the rate of change of temperature in the wind at \(z=1.2 \mathrm{~km}\) and \(t=5400 \mathrm{~s}(1.5 \mathrm{~h})\).

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