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Chapter 14: Problem 73

Consider a 20 -cm-thick brick wall of a house. The indoor conditions are \(25^{\circ} \mathrm{C}\) and 50 percent relative humidity while the outside conditions are \(50^{\circ} \mathrm{C}\) and 50 percent relative humidity. Assuming that there is no condensation or freezing within the wall, determine the amount of moisture flowing through a unit surface area of the wall during a \(24-\mathrm{h}\) period.

Short Answer

Expert verified
Answer: The total amount of moisture flowing through a unit surface area of the wall during the 24-hour period is approximately \(1.186 \times 10^{-3}\,\mathrm{kg/m^2}\).

Step by step solution

01

Find the saturation vapor pressure values

Using a psychrometric chart or a reference table, find the saturation vapor pressure values at the given temperatures. For this problem, we can use the following approximate values for saturation vapor pressure at the indoor (\(25^{\circ} \mathrm{C}\)) and outdoor (\(50^{\circ} \mathrm{C}\)) temperatures: For \(25^{\circ}\mathrm{C}\), saturation vapor pressure: \(e_{sat,25}=3.168\,\mathrm{kPa}\) For \(50^{\circ}\mathrm{C}\), saturation vapor pressure: \(e_{sat,50}=12.344\,\mathrm{kPa}\)
02

Calculate actual vapor pressure

Calculate the actual vapor pressure at both indoor and outdoor conditions by multiplying the saturation vapor pressure by the given relative humidity values. Indoor vapor pressure: \(e_{in} = e_{sat,25} \times 0.50 = 3.168\,\mathrm{kPa} \times 0.50 = 1.584\,\mathrm{kPa}\) Outdoor vapor pressure: \(e_{out} = e_{sat,50} \times 0.50 = 12.344\,\mathrm{kPa} \times 0.50 = 6.172\,\mathrm{kPa}\)
03

Calculate the pressure difference across the wall

Find the difference of actual vapor pressures across the wall: \(Δe = e_{out} - e_{in} = 6.172\,\mathrm{kPa} - 1.584\,\mathrm{kPa} = 4.588\,\mathrm{kPa}\)
04

Find the vapor diffusion coefficient of the brick wall

The vapor diffusion coefficient of the brick wall (\(K_{diff}\)) is a measure of how easily moisture can flow through it. This is a property of the brick material. For this problem, we can use the following approximate value for the vapor diffusion coefficient: \(K_{diff} = 6 \times 10^{-10}\,\mathrm{kg/m\cdot Pa\cdot s}\)
05

Calculate moisture flow rate through the unit surface area

Using Fick's Law of diffusion, we can calculate the moisture flow rate through the unit surface area of the wall: \(Φ = -K_{diff} \times \frac{Δe}{\Delta x}\) where \(\Delta x\) is the thickness of the wall, and \(Δe\) is the pressure difference calculated earlier. In our problem, the thickness of the wall is \(20\,\mathrm{cm} = 0.2\,\mathrm{m}\). \(Φ = -6 \times 10^{-10}\,\mathrm{kg/m\cdot Pa\cdot s} \times \frac{4.588\,\mathrm{kPa}}{0.2\,\mathrm{m}}\) \(Φ = -1.3735 \times 10^{-8}\,\mathrm{kg/m^2\cdot s}\)
06

Calculate the total moisture flow during the 24-hour period

Now, we need to find the total amount of moisture that flows through the unit surface area of the wall during a 24-hour period: \(M = Φ \times t\) where \(t\) is the time duration in seconds. \(M = -1.3735 \times 10^{-8}\,\mathrm{kg/m^2\cdot s} \times (24 \times 3600)\,\mathrm{s}\) \(M = -1.1864 \times 10^{-3}\,\mathrm{kg/m^2}\) Since we are dealing with a moisture flow rate, it is customary to use the absolute value of the result. Therefore, the total amount of moisture flowing through a unit surface area of the wall during the 24-hour period is: \(M \approx 1.186 \times 10^{-3}\,\mathrm{kg/m^2}\)

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Most popular questions from this chapter

Under what conditions will the normalized velocity, thermal, and concentration boundary layers coincide during flow over a flat plate?

Explain how vapor pressure of the ambient air is determined when the temperature, total pressure, and relative humidity of the air are given.

Consider a brick house that is maintained at \(20^{\circ} \mathrm{C}\) and 60 percent relative humidity at a location where the atmospheric pressure is \(85 \mathrm{kPa}\). The walls of the house are made of 20 -cm-thick brick whose permeance is \(23 \times 10^{-12} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{Pa}\). 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 \(3-\mathrm{m} \times 5-\mathrm{m}\) section of a wall during a 24-h period.

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.

Consider one-dimensional mass transfer in a moving medium that consists of species \(A\) and \(B\) with \(\rho=\rho_{A}+\rho_{B}=\) constant. Mark these statements as being True or False. (a) The rates of mass diffusion of species \(A\) and \(B\) are equal in magnitude and opposite in direction. (b) \(D_{A B}=D_{B A}\). (c) During equimolar counterdiffusion through a tube, equal numbers of moles of \(A\) and \(B\) move in opposite directions, and thus a velocity measurement device placed in the tube will read zero. (d) The lid of a tank containing propane gas (which is heavier than air) is left open. If the surrounding air and the propane in the tank are at the same temperature and pressure, no propane will escape the tank and no air will enter.
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