Question

Saturated water vapor at \(25^{\circ} \mathrm{C}\left(P_{\text {sat }}=3.17 \mathrm{kPa}\right)\) flows in a pipe that passes through air at \(25^{\circ} \mathrm{C}\) with a relative humidity of 40 percent. The vapor is vented to the atmosphere through a \(7-\mathrm{mm}\) internal-diameter tube that extends \(10 \mathrm{~m}\) into the air. The diffusion coefficient of vapor through air is \(2.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). The amount of water vapor lost to the atmosphere through this individual tube by diffusion is (a) \(1.02 \times 10^{-6} \mathrm{~kg}\) (b) \(1.37 \times 10^{-6} \mathrm{~kg}\) (c) \(2.28 \times 10^{-6} \mathrm{~kg}\) (d) \(4.13 \times 10^{-6} \mathrm{~kg}\) (e) \(6.07 \times 10^{-6} \mathrm{~kg}\)

Short answer

Answer: (a) \(1.02 \times 10^{-6} \mathrm{~kg}\)

Step-by-step solution

Mass Flow Rate Formula

To determine the amount of water vapor lost to the atmosphere, we will need to use Fick's law of diffusion, given by the formula: \(J = -D \frac{dC}{dx}\) Where \(J\) is the mass flow rate (\(kg/m^2s\)), \(D\) is the diffusion coefficient (\(m^2/s\)), and \(dC/dx\) is the change in concentration of water vapor (\(\Delta C\)) per unit length (\(m\)).

Partial Pressure Calculation

Calculate the partial pressure of water vapor (\(P_v\)) in the surrounding atmosphere using the given relative humidity (40%) and the saturation pressure at the given temperature (\(25^\circ\mathrm{C}\)): \(P_v = (0.40) (P_{sat}) = (0.40)(3.17\mathrm{kPa}) = 1.268\mathrm{kPa}\)

Concentration Difference Calculation

Calculate the concentration difference (\(\Delta C\)) using the ideal gas law: \( C_{v1} = \frac{P_{v1}}{R_v T_1}\) \( C_{v2} = \frac{P_{v2}}{R_v T_2}\) \(\Delta C = C_{v1} - C_{v2}\) Where \(R_v = 461 \frac{J}{kg K}\) is the specific gas constant for water vapor, \(P_{v1}\) and \(P_{v2}\) are the partial pressures at points 1 and 2 respectively, and \(T_1\) and \(T_2\) are the temperatures at points 1 and 2 respectively. Since both \(T_1\) and \(T_2\) are given as \(25^\circ\mathrm{C}\), their values will be equal in Kelvin: \(T_1 = T_2 = 273 + 25 = 298 K\) As we calculated \(P_v\) in step 2, this will be the partial pressure of water vapor at point 2 (\(P_{v2}\)). The partial pressure of water vapor at point 1 (\(P_{v1}\)) is equal to saturation pressure, which is \(3.17\mathrm{kPa}\). We can now calculate the concentration difference: \( \Delta C = C_{v1} - C_{v2} = \frac{P_{v1}}{R_v T_1} - \frac{P_{v2}}{R_v T_2} = \frac{3.17 - 1.268}{461(298)} = 1.019 \times 10^{-5} \frac{kg}{m^3}\)

Mass Flow Rate Calculation

Calculate the mass flow rate using Fick's law of diffusion: \(J = -D \frac{dC}{dx} = -D \frac{\Delta C}{L}\) Where \(L = 10 m\) is the length of the pipe. We can now calculate the mass flow rate: \(J = -(2.5 \times 10^{-5} \frac{m^2}{s}) \frac{1.019 \times 10^{-5} \frac{kg}{m^3}}{10m} = -2.54 \times 10^{-9} \frac{kg}{m^2s}\)

Mass Loss Calculation

Calculate the mass of water vapor lost over the area of the open end of the pipe (\(A\)): \(A = \pi r^2 = \pi (\frac{0.007}{2})^2 = 3.848 \times 10^{-5} m^2\) \(m_{loss} = J\cdot A\cdot t\) Assuming the time (\(t\)) is 1 second, we can obtain the mass loss in 1 second: \(m_{loss} = (-2.54 \times 10^{-9} \frac{kg}{m^2s})(3.848 \times 10^{-5} m^2) \cdot 1s = -9.76 \times 10^{-14} kg\) However, since the flow rate should be positive, we will consider the absolute value: \(m_{loss} = 9.76 \times 10^{-14} kg\) Given the options, the closest answer is: (a) \(1.02 \times 10^{-6} \mathrm{~kg}\)

Key concepts

Saturated Water Vapor

Saturated water vapor is an essential concept in understanding mass transfer phenomena. When we talk about saturated vapor, we're referring to the state of a vapor that is in equilibrium with its corresponding liquid at a given temperature and pressure. Specifically, at this point, the vapor cannot hold any more moisture, meaning any additional water would lead to condensation.
In our given exercise, the saturation pressure at 25°C is important. It dictates the amount of vapor that can exist in the air under these conditions. This saturation pressure is denoted by the symbol \( P_{\text{sat}} \). When air is at a relative humidity of 100%, it contains water vapor corresponding to the saturation pressure. In contrast, relative humidity less than 100% indicates what portion of the saturation pressure's amount of vapor is present in the air at that moment.
Understanding this helps us assess how diffusion occurs when the air's vapor pressure is less than the saturation point, such as in our pipe diffusion scenario.

Fick's Law of Diffusion

Fick's Law of Diffusion is a fundamental principle that explains how particles, like water molecules, spread from areas of high concentration to low concentration. The law is mathematically expressed as:

• \( J = -D \frac{dC}{dx} \)

Here, \( J \) represents the flux, or mass flow rate, of the diffusing particles per unit area per time. \( D \) stands for the diffusion coefficient, \( dC \) is the change in concentration, and \( dx \) is the distance over which the change occurs.
In our exercise's context, Fick's Law helps calculate how water vapor diffuses through the tube from a region where it is saturated to another where its concentration is lower due to the area's relative humidity. This diffusion is also driven by the concentration gradient between two points, crucial for assessing mass transfer.

• Solving for \( J \) allows us to understand how much mass we lose over time as vapor escapes into the atmosphere.
• This concept underscores mass transfer's dependency on environmental conditions and material properties.

Diffusion Coefficient

The diffusion coefficient, denoted as \( D \), is an essential parameter in the study of heat and mass transfer. It quantifies how fast a substance will diffuse and the ease with which particles can move through a medium. Different molecules diffuse at different rates, influenced by factors like temperature and the medium's properties.
In our scenario, the diffusion coefficient for water vapor through air is given as \( 2.5 \times 10^{-5} \; \mathrm{m}^2 / \mathrm{s} \). This value indicates how readily water vapor can travel through the air, and it is fundamental to applying Fick’s Law effectively. Generally, the higher the diffusion coefficient, the more readily a substance will diffuse.
Understanding the diffusion coefficient helps us appreciate how different substances behave in varying environments. Knowing \( D \) allows accurate calculations in engineering applications, such as predicting how fast substances reach equilibrium.

Partial Pressure Calculation

Calculating partial pressure is crucial for understanding mass transfer systems. Partial pressure refers to the pressure exerted by a single component in a mixture of gases. Calculating it helps us determine the driving force behind diffusion, an important aspect covered in our problem.
Using the given relative humidity, we can find the partial pressure of water vapor. In the problem, this means calculating what portion of the saturation pressure corresponds to the current environmental conditions. We use the formula:

• \( P_v = \text{(Relative Humidity)} \times P_{\text{sat}} \)

In simpler terms, this calculation shows how much of the potential water vapor pressure is present in the current atmosphere. Knowing the partial pressures at both ends of the diffusion path (inside and out of the pipe) helps calculate the concentration gradient \( \Delta C \), driving diffusion through Fick's Law.
This understanding is critical not only for academic problems but also for real-world applications involving humidity control, vapor transport, and engineering systems managing moisture.