/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Insulation degrades (experiences... [FREE SOLUTION] | 91Ó°ÊÓ

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Insulation degrades (experiences an increase in thermal conductivity) if it is subjected to water vapor condensation. The problem may occur in home insulation during cold periods, when vapor in a humidified room diffuses through the drywall (plaster board) and condenses in the adjoining insulation. Estimate the mass diffusion rate for a \(3 \mathrm{~m} \times 5 \mathrm{~m}\) wall, under conditions for which the vapor pressure is \(0.03\) bar in the room air and \(0.0\) bar in the insulation. The drywall is \(10 \mathrm{~mm}\) thick, and the solubility of water vapor in the wall material is approximately \(5 \times 10^{-3} \mathrm{kmol} / \mathrm{m}^{3}\) - bar. The binary diffusion coefficient for water vapor in the drywall is approximately \(10^{-9} \mathrm{~m}^{2} / \mathrm{s}\).

Short Answer

Expert verified
The mass diffusion rate of water vapor through the drywall is approximately \(- 2.25 \times 10^{-10} \mathrm{~kmol/s}\).

Step by step solution

01

Find the concentration difference

To find the mass diffusion rate, first, we need to find the difference in concentration between the two sides of the wall. Since the vapor pressure in the insulation is 0.0 bar, there is no water vapor in the insulation. Therefore, the concentration difference can be calculated as follows: \[ \text{Concentration difference} = C_{1} - C_{2} = C_{1} \] where \(C_{1}\) is the concentration of water vapor in the room and \(C_{2}\) is the concentration in the insulation. We are given that the solubility of water vapor in the wall material is approximately \(5 \times 10^{-3}\mathrm{ \ kmol/m^{3}- bar}\), and the vapor pressure in the room is \(0.03~\text{bar}\), so the concentration difference is: \[ C_{1} = (\text{solubility of water vapor}) \times (\text{vapor pressure in room}) = (5 \times 10^{-3} \mathrm{~kmol/m^{3}-bar}) \times (0.03 \mathrm{~bar}) \] Calculating the concentration difference, we get: \[ C_{1} = 1.5 \times 10^{-4} \mathrm{~kmol/m^3} \]
02

Find the mass flux

Now that we have the concentration difference, we will use Fick's first law of diffusion to calculate the mass flux. Fick's first law is given by the following equation: \[ J = -D_{AB} \frac{dC}{dx} \] where \(J\) is the mass flux (\(\text{kmol/m}^{2}\text{s}\)), \(D_{AB}\) is the binary diffusion coefficient of water vapor in the drywall (\(10^{-9} \mathrm{m^2/s}\)), \(dC\) is the concentration difference, and \(dx\) is the thickness of the drywall (\(0.01~\text{m}\)). Plugging in the given values, we have: \[ J = - (10^{-9} \mathrm{m^2/s}) \times \frac{1.5 \times 10^{-4} \mathrm{~kmol/m^3}}{0.01 \mathrm{~m}} \] Calculating the mass flux, we get: \[ J = - 1.5 \times 10^{-11} \mathrm{~kmol/m^2s} \]
03

Calculate the mass diffusion rate

Finally, to find the mass diffusion rate, we need to multiply the mass flux by the wall's surface area. The dimensions of the wall are \(3 \mathrm{~m}\) by \(5\mathrm{m}\), so the surface area is: \[ \text{Area} = 3 \mathrm{~m} \times 5 \mathrm{~m} = 15 \mathrm{~m^{2}} \] The mass diffusion rate is given by: \[ \text{mass diffusion rate} = J \times \text{Area} = (- 1.5 \times 10^{-11} \mathrm{~kmol/m^2s}) \times (15 \mathrm{~m^{2}}) \] Calculating the mass diffusion rate, we get: \[ \text{mass diffusion rate} = - 2.25 \times 10^{-10} \mathrm{~kmol/s} \] Note that the mass diffusion rate is negative, which means that the vapor is flowing from the room side to the insulation side, as expected. The mass diffusion rate of water vapor through the drywall is approximately \(- 2.25 \times 10^{-10} \mathrm{~kmol/s}\).

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Key Concepts

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

Understanding Insulation Thermal Conductivity
Insulation thermal conductivity is a crucial factor when considering the efficiency of insulation materials. This term refers to the material's ability to conduct heat. Lower thermal conductivity means that the material is a better insulator, as it slows down the transfer of heat. In our exercise context, insulation degradation occurs because of water vapor condensation, leading to an increase in thermal conductivity. As moisture accumulates, the insulation becomes less effective at trapping heat, leading to higher energy costs and potential structural damage over time.

In the real scenario, once the vapor condenses, it changes the insulation’s properties by adding moisture, which is a better conductor of heat than dry insulation materials. Consequently, knowing the level of thermal conductivity in various states of an insulation material aids in designing structures that remain energy-efficient and structurally sound even in varying environmental conditions.
Vapor Pressure and Its Effect on Insulation
Vapor pressure is the pressure exerted by a vapor that is in equilibrium with its non-vapor phases at a given temperature. In simpler terms, it’s a measure of the tendency of molecules to escape from a liquid or a solid. High vapor pressure indicates a high evaporation rate of a substance.

In home insulation, as mentioned in our exercise, the vapor pressure inside a room can lead to the diffusion of water vapor through walls, reaching the insulation and potentially condensing. Condensation occurs when the vapor pressure drops, causing the water vapor to change into the liquid phase. This phenomenon can deteriorate the insulation properties because water has a higher thermal conductivity compared to air. The lower the vapor pressure in the space surrounding the insulation, the more likely water vapor will condense, as observed with the 0.0 bar vapor pressure inside the insulation in our problem.
Fick's First Law of Diffusion
Fick's First Law of Diffusion is a principle that describes the movement of particles from regions of higher concentration to regions of lower concentration. The law can be mathematically expressed as:\begin{align*}J = -D_{AB} \frac{dC}{dx}\end{\align*}where \(J\) represents the mass flux (the amount of substance that travels through a unit area per unit time), \(D_{AB}\) is the diffusion coefficient for a binary mixture, \(dC\) is the concentration difference, and \(dx\) is the distance over which the concentration changes. In our exercise, we calculate the mass flux of water vapor through drywall using Fick's First Law, which tells us how much vapor transfers per time through the wall.
Binary Diffusion Coefficient
The binary diffusion coefficient, symbolized by \(D_{AB}\), is a parameter that describes the rate at which two gases mix. It is determined by the size of the gas molecules, the temperature, and the medium through which the gases are diffusing. Larger molecules or higher viscosities result in a smaller diffusion coefficient, meaning gases will mix more slowly. The coefficient generally increases with temperature.

In the case of the insulation problem, the binary diffusion coefficient for water vapor in drywall determines how quickly the moisture will spread throughout the material. A higher diffusion coefficient would lead to a faster diffusion rate, worsening problems such as insulation degradation. By knowing the diffusion coefficient, as well as the vapor pressure and wall's properties, we can accurately estimate the rate at which water vapor will permeate the insulation, as demonstrated in our step-by-step solution.

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Most popular questions from this chapter

Ultra-pure hydrogen is required in applications ranging from the manufacturing of semiconductors to powering fuel cells. The crystalline structure of palladium allows only the transfer of atomic hydrogen (H) through its thickness, and therefore palladium membranes are used to filter hydrogen from contaminated streams containing mixtures of hydrogen and other gases. Hydrogen molecules \(\left(\mathrm{H}_{2}\right)\) are first adsorbed onto the palladium's surface and are then dissociated into atoms \((\mathrm{H})\), which subsequently diffuse through the metal. The H atoms recombine on the opposite side of the membrane, forming pure \(\mathrm{H}_{2}\). The surface concentration of H takes the form \(C_{\mathrm{H}}=K_{s} p_{\mathrm{H}_{2}}^{05}\), where \(K_{s} \approx\) \(1.4 \mathrm{kmol} / \mathrm{m}^{3} \cdot \mathrm{bar}^{0.5}\) is known as Sieverts constant. Consider an industrial hydrogen purifier consisting of an array of palladium tubes with one tube end connected to a collector plenum and the other end closed. The tube bank is inserted into a shell. Impure \(\mathrm{H}_{2}\) at \(T=\) \(600 \mathrm{~K}, p=15\) bars, \(x_{\mathrm{H}_{2}}=0.85\) is introduced into the shell while pure \(\mathrm{H}_{2}\) at \(p=6\) bars, \(T=600 \mathrm{~K}\) is extracted through the tubes. Determine the production rate of pure hydrogen \((\mathrm{kg} / \mathrm{h})\) for \(N=100\) tubes which are of inside diameter \(D_{i}=1.6 \mathrm{~mm}\), wall thickness \(t=75 \mu \mathrm{m}\), and length \(L=80 \mathrm{~mm}\). The mass diffusivity of hydrogen \((\mathrm{H})\) in palladium at \(600 \mathrm{~K}\) is approximately \(D_{\mathrm{AB}}=7 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\).

Hydrogen gas is used in a process to manufacture a sheet material of \(6-\mathrm{mm}\) thickness. At the end of the process, \(\mathrm{H}_{2}\) remains in solution in the material with a uniform concentration of \(320 \mathrm{kmol} / \mathrm{m}^{3}\). To remove \(\mathrm{H}_{2}\) from the material, both surfaces of the sheet are exposed to an airstream at \(500 \mathrm{~K}\) and a total pressure of \(3 \mathrm{~atm}\). Due to contamination, the hydrogen partial pressure is \(0.1 \mathrm{~atm}\) in the airstream, which provides a convection mass transfer coefficient of \(1.5 \mathrm{~m} / \mathrm{h}\). The mass diffusivity and solubility of hydrogen (A) in the sheet material (B) are \(D_{\mathrm{AB}}=2.6 \times 10^{-8} \mathrm{~m}^{2} / \mathrm{s}\) and \(S_{\mathrm{AB}}=160 \mathrm{kmol} / \mathrm{m}^{3} \cdot\) atm, respectively. (a) If the sheet material is left exposed to the airstream for a long time, determine the final content of hydrogen in the material \(\left(\mathrm{kg} / \mathrm{m}^{3}\right)\). (b) Identify and evaluate the parameter that can be used to determine whether the transient mass diffusion process in the sheet can be assumed to be characterized by a uniform concentration at any time during the process. Hint: This situation is analogous to that used to determine the validity of the lumped-capacitance method for a transient heat transfer analysis. (c) Determine the time required to reduce the hydrogen mass density at the center of the sheet to twice the limiting value calculated in part (a).

Consider air in a closed, cylindrical container with its axis vertical and with opposite ends maintained at different temperatures. Assume that the total pressure of the air is uniform throughout the container. (a) If the bottom surface is colder than the top surface, what is the nature of conditions within the container? For example, will there be vertical gradients of the species \(\left(\mathrm{O}_{2}\right.\) and \(\left.\mathrm{N}_{2}\right)\) concentrations? Is there any motion of the air? Does mass transfer occur? (b) What is the nature of conditions within the container if it is inverted (i.e., the warm surface is now at the bottom)?

Pulverized coal pellets, which may be approximated as carbon spheres of radius \(r_{o}=1 \mathrm{~mm}\), are burned in a pure oxygen atmosphere at \(1450 \mathrm{~K}\) and 1 atm. Oxygen is transferred to the particle surface by diffusion, where it is consumed in the reaction \(\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}\). The reaction rate is first order and of the form \(\dot{N}_{\mathrm{O}_{2}}^{\prime \prime}=\) \(-k_{1}^{\prime \prime} C_{\mathrm{O}_{2}}\left(r_{o}\right)\), where \(k_{1}^{\prime \prime}=0.1 \mathrm{~m} / \mathrm{s}\). Neglecting changes in \(r_{o}\), determine the steady-state \(\mathrm{O}_{2}\) molar consumption rate in \(\mathrm{kmol} / \mathrm{s}\). At \(1450 \mathrm{~K}\), the binary diffusion coefficient for \(\mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) is \(1.71 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\).

Assuming air to be composed exclusively of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\), with their partial pressures in the ratio \(0.21: 0.79\), what are their mass fractions?

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