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The diffusion of water vapor through plaster boards and its condensation in the wall insulation in cold weather are of concern since they reduce the effectiveness of insulation. Consider a house that is maintained at \(20^{\circ} \mathrm{C}\) and 60 percent relative humidity at a location where the atmospheric pressure is \(97 \mathrm{kPa}\). The inside of the walls is finished with \(9.5\)-mm-thick gypsum wallboard. Taking the vapor pressure at the outer side of the wallboard to be zero, determine the maximum amount of water vapor that will diffuse through a \(3-\mathrm{m} \times 8-\mathrm{m}\) section of a wall during a 24-h period. The permeance of the \(9.5\)-mm-thick gypsum wallboard to water vapor is \(2.86 \times 10^{-9} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{Pa}\).

Short Answer

Expert verified
Answer: The maximum amount of water vapor that can diffuse through this section of the wall during a 24-hour period is approximately 2.639 kg.

Step by step solution

01

Calculate the partial pressure of vapor inside the house

To determine the partial pressure of vapor (Pi) inside the house, we need to use the given information: the temperature inside the house is \(20^{\circ} \mathrm{C}\), the relative humidity is 60% and the atmospheric pressure is \(97\ \mathrm{kPa}\). First, we need to find the saturation pressure (\(P_{sat}\)) of water vapor at the inside temperature (20°C). We can use the Antoine equation for this: \(P_{sat}=610.5\times{10^{\frac{7.5 \times T}{T + 237.3}}}\) where \(T\) is the temperature in Celsius. Substituting the temperature into the equation, we have: \(P_{sat}=610.5\times{10^{\frac{7.5\times20}{20+237.3}}}=2.338\ \mathrm{kPa}\) Now, we apply the definition of relative humidity to find the partial pressure of vapor (Pi) inside the house: \(Relative\ Humidity=\frac{P_i}{P_{sat}}\) \(P_i=Relative\ Humidity\times P_{sat}=0.6\times2.338 = 1.403\ \mathrm{kPa}\)
02

Calculate the vapor pressure difference

Now that we have calculated the partial pressure of water vapor inside the house (Pi), we can find the vapor pressure difference (ΔP) between the two sides of the wall. The vapor pressure outside the wall is given as zero, so: \(\Delta P=P_i-0=1.403\ \mathrm{kPa}\)
03

Calculate the mass flow rate of water vapor through the wall

To calculate the mass flow rate (m) of water vapor through the wall, we need to use the permeance of the wall and the vapor pressure difference. We are given that the permeance of the wall is \(2.86\times10^{-9}\ \mathrm{kg/s\cdot m^2\cdot Pa}\). To use this value, we need to convert the vapor pressure difference from kPa to Pa: \(\Delta P=1.403\ \mathrm{kPa}\times1000\ \mathrm{Pa/kPa}=1403\ \mathrm{Pa}\) Now, we can find the mass flow rate (m) per unit area of the wall: \(m=\mathrm{Permeance}\times\Delta P=2.86\times10^{-9}\ \mathrm{kg/s\cdot m^2\cdot Pa}\times1403\ \mathrm{Pa}=4.010\times10^{-6}\ \mathrm{kg/s\cdot m^2}\)
04

Calculate the maximum amount of water vapor that will diffuse through the wall during a 24-hour period

We now have the mass flow rate of water vapor through the wall per unit area. To find the maximum amount of water vapor that will diffuse through the specified area (\(3\ \mathrm{m}\times8\ \mathrm{m}\)) of the wall during a 24-hour period, we need to multiply the mass flow rate by the area of the wall and the duration (in seconds) of the 24-hour period: \(Mass\ Flow\ Rate\times Area\times Time=4.010\times10^{-6}\ \mathrm{kg/s\cdot m^2}\times(3\ \mathrm{m}\times8\ \mathrm{m})\times(24\ \mathrm{h\cdot3600\ s/h})=2.639\ \mathrm{kg}\) So, the maximum amount of water vapor that will diffuse through the \(3-\mathrm{m} \times 8-\mathrm{m}\) section of a wall during a 24-hour period under these conditions is approximately 2.639 kg.

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Key Concepts

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

Diffusion
Diffusion is the process in which molecules spread from an area of higher concentration to an area of lower concentration. It's essentially like spreading sugar in water, where the sugar molecules move from their concentrated form (the sugar granule) and mix evenly throughout the liquid. In mass transfer, especially within building materials like gypsum wallboard, diffusion represents how water vapor migrates from inside a home (where it's more humid) to the outside.
  • Concentration Gradient: This is the driving force behind diffusion. The greater the difference in concentration, the faster the diffusion process.
  • Diffusion Coefficient: This gives us an idea of how easily molecules can move through a medium. In the case of water vapor, the diffusion coefficient would be higher if there’s less resistance from the medium, such as a lightweight or less dense material.
  • Applications: Understanding diffusion is critical when assessing how moisture moves through materials, affecting building performance and structural integrity. It aids in designing better insulation systems and waterproofing solutions.
In the context of a gypsum wallboard, diffusion determines how the water vapor permeates through the material, potentially leading to condensation problems when temperatures drop outside.
Vapor Pressure
Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid phase at a given temperature. It’s crucial for understanding how easily a liquid can evaporate from or condense into the gas phase. For indoor environments, maintaining a balance in vapor pressure is key to preventing condensation and mold growth.
  • Influence of Temperature: As temperature increases, vapor pressure increases, meaning more molecules have enough energy to escape into the vapor phase.
  • Relative Humidity: This is a way of expressing how much water vapor is in the air compared to the maximum amount it can hold at that temperature. A 60% relative humidity means the air holds 60% of the water vapor it could maximally contain at the current temperature.
  • Role in Building Materials: In building physics, vapor pressure differences across a wall drive the diffusion of moisture through the wall. If one side has a high vapor pressure compared to the other, vapor will tend to migrate from the higher pressure side to the lower pressure side.
When evaluating the gypsum wallboard scenario, a significant vapor pressure exists between the inside and outside environment, ultimately driving water vapor to diffuse through the wall material.
Permeance
Permeance is a measure of a material's ability to allow liquids, gases, or vapors to pass through it. It's like a filter letting different sizes of particles through. In construction, this property is critical as it dictates how well a wall system can manage moisture and prevent issues such as mold or rot.
  • Unit of Measurement: Permeance is expressed in units of mass per time per area per pressure difference (\( ext{kg/s} imes ext{m}^2 imes ext{Pa}\)). It indicates the rate at which moisture vapor penetrates a material.
  • Permeance vs Permeability: While permeance is concerned with the rate of passage, permeability relates to the material's intrinsic property of allowing the passage of steam or liquid.
  • Practical Insights: Design standards suggest using materials with appropriate permeance levels according to environmental conditions. For instance, in colder regions, walls might require lower permeance to minimize condensation within the building envelope.
In the case of gypsum wallboard, which has a defined permeance value, understanding this allows architects and builders to predict how much moisture will migrate through the wall under varying conditions.
Gypsum Wallboard
Gypsum wallboard, commonly known as drywall, is a construction material used to line walls and ceilings. It consists of a gypsum plaster core sandwiched between two thick sheets of paper. This composition offers a balance of cost effectiveness and functionality for modern building.
  • Material Properties: Gypsum wallboard is appreciated for its simplicity in installation, fire resistance, and modest thermal insulation properties.
  • Moisture Management: Despite its advantages, gypsum wallboard can absorb moisture, which can lead to structural issues if not managed correctly. Vapor barriers and coatings can be used to enhance its resistance.
  • Application in Wet Areas: Special variants treated with water repellents are used in humid environments like bathrooms to mitigate the risk of damage.
It's essential to consider the permeance and diffusion characteristics of gypsum wallboard when it is part of an external wall, especially in climates with extreme temperature differences between inside and outside.

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Most popular questions from this chapter

You probably have noticed that balloons inflated with helium gas rise in the air the first day during a party but they fall down the next day and act like ordinary balloons filled with air. This is because the helium in the balloon slowly leaks out through the wall while air leaks in by diffusion. Consider a balloon that is made of \(0.1\)-mm-thick soft rubber and has a diameter of \(15 \mathrm{~cm}\) when inflated. The pressure and temperature inside the balloon are initially \(110 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). The permeability of rubber to helium, oxygen, and nitrogen at \(25^{\circ} \mathrm{C}\) are \(9.4 \times 10^{-13}, 7.05 \times 10^{-13}\), and \(2.6 \times 10^{-13} \mathrm{kmol} / \mathrm{m} \cdot \mathrm{s} \cdot\) bar, respectively. Determine the initial rates of diffusion of helium, oxygen, and nitrogen through the balloon wall and the mass fraction of helium that escapes the balloon during the first \(5 \mathrm{~h}\) assuming the helium pressure inside the balloon remains nearly constant. Assume air to be 21 percent oxygen and 79 percent nitrogen by mole numbers and take the room conditions to be \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\).

Write an essay on diffusion caused by effects other than the concentration gradient such as thermal diffusion, pressure diffusion, forced diffusion, knodsen diffusion, and surface diffusion.

An 8-cm-internal-diameter, 30-cm-high pitcher halffilled with water is left in a dry room at \(15^{\circ} \mathrm{C}\) and \(87 \mathrm{kPa}\) with its top open. If the water is maintained at \(15^{\circ} \mathrm{C}\) at all times also, determine how long it will take for the water to evaporate completely.

Liquid methanol is accidentally spilt on a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) laboratory bench and covered the entire bench surface. A fan is providing a \(20 \mathrm{~m} / \mathrm{s}\) air flow parallel over the bench surface. The air is maintained at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), and the concentration of methanol in the free stream is negligible. If the methanol vapor at the air-methanol interface has a pressure of \(4000 \mathrm{~Pa}\) and a temperature of \(25^{\circ} \mathrm{C}\), determine the evaporation rate of methanol in molar basis.

A steel part whose initial carbon content is \(0.12\) percent by mass is to be case-hardened in a furnace at \(1150 \mathrm{~K}\) by exposing it to a carburizing gas. The diffusion coefficient of carbon in steel is strongly temperature dependent, and at the furnace temperature it is given to be \(D_{A B}=7.2 \times 10^{-12} \mathrm{~m}^{2} / \mathrm{s}\). Also, the mass fraction of carbon at the exposed surface of the steel part is maintained at \(0.011\) by the carbon-rich environment in the furnace. If the hardening process is to continue until the mass fraction of carbon at a depth of \(0.7 \mathrm{~mm}\) is raised to \(0.32\) percent, determine how long the part should be held in the furnace.

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