Question

During cold weather periods, vapor in a room diffuses through the dry wall and condenses in the adjoining insulation. This process decreases the thermal resistance and degrades the insulation. Consider a condition at which the vapor pressure in the air at \(25^{\circ} \mathrm{C}\) inside a room is \(3 \mathrm{kPa}\), and the vapor pressure in the insulation is negligible. The 3 -m-high and 10 -m-wide dry wall is 12-mm thick with a solubility of water vapor in the wall material of approximately \(0.007 \mathrm{kmol} / \mathrm{m}^{3}\).bar, and diffusion coefficient of water vapor in the wall is \(0.2 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\). Determine the mass diffusion rate of water vapor through the wall.

Short answer

The mass diffusion rate of water vapor through the drywall is approximately \(2.975 \times 10^{-4} \mathrm{kg/s}\).

Step-by-step solution

Understand the given quantities and Fick's law of diffusion

Given quantities: - Ambient vapor pressure in the room, \(P_A = 3 \mathrm{kPa}\) - Vapor pressure in insulation, \(P_I = 0 \mathrm{kPa}\) (negligible) - Drywall dimensions: height \(H = 3 \mathrm{m}\), width \(W = 10\mathrm{m}\), and thickness \(L = 12 \mathrm{mm} = 0.012 \mathrm{m}\) - Solubility of water vapor in the wall material, \(C = 0.007 \mathrm{kmol / m^3}.\mathrm{bar}\) - Diffusion coefficient of water vapor in the wall, \(D = 0.2 \times 10^{-9} \mathrm{m^2/s}\) Fick's law of diffusion states that the mass diffusion rate \(N\) is proportional to the concentration difference and the diffusion area divided by the diffusion distance: $$N = -D \times \frac{A \times (C_A - C_B)}{L}$$ where, \(N\) is the mass diffusion rate (kg/s) \(-D\) means that diffusion occurs from higher to lower concentration \(A\) is the diffusion area (m²) \(C_A\) and \(C_B\) are the concentrations at points A and B respectively \(L\) is the diffusion distance (m) We will use this formula to find the mass diffusion rate of water vapor through the drywall.

Calculate the concentrations at points A and B

First, we need to find the concentrations at points A and B via the given solubility. $$C_A = C \times P_A$$ $$C_A = (0.007\, \mathrm{kmol/m^3.\, bar}) \times (3\, \mathrm{kPa})$$ Since 1 bar equals to 100 kPa, we can then convert kPa to bar: $$C_A = (0.007\, \mathrm{kmol/m^3.\, bar}) \times (0.03\, \mathrm{bar})$$ $$C_A = 0.00021\, \mathrm{kmol/m^3}$$ As the vapor pressure in the insulation is negligible, \(C_B\) will also be \(0\, \mathrm{kmol/m^3}\).

Calculate the diffusion area

The diffusion area, \(A\), is equal to the product of the height \(H\) and width \(W\) of the wall. $$A = H \times W$$ $$A = (3\, \mathrm{m}) \times (10\, \mathrm{m})$$ $$A = 30\, \mathrm{m^2}$$

Calculate the mass diffusion rate using Fick's law

Now we have all the required values to find the mass diffusion rate using Fick's law. $$N = -D \times \frac{A\times (C_A - C_B)}{L}$$ $$N = - (0.2 \times 10^{-9}\, \mathrm{m^2/s}) \times \frac{(30\, \mathrm{m^2})\times (0.00021\, \mathrm{kmol/m^3})}{0.012\, \mathrm{m}}$$ Now, we need to convert kmol to kg. Since 1 kmol of water is equal to \(18.015 \mathrm{kg}\): $$N = - (0.2 \times 10^{-9}\, \mathrm{m^2/s}) \times \frac{(30\, \mathrm{m^2})\times (0.00021 \times 18.015\, \mathrm{kg/m^3})}{0.012\, \mathrm{m}}$$ $$N = - (0.2 \times 10^{-9}\, \mathrm{m^2/s}) \times \frac{(30\, \mathrm{m^2})\times (0.00378\, \mathrm{kg/m^3})}{0.012\, \mathrm{m}}$$ $$N \approx -2.975 \times 10^{-4}\, \mathrm{kg/s}$$ Since the negative sign indicates that the mass diffusion is in the direction from higher to lower concentration, and we only want the magnitude: $$N \approx 2.975 \times 10^{-4}\, \mathrm{kg/s}$$ Therefore, the mass diffusion rate of water vapor through the drywall is approximately \(2.975 \times 10^{-4} \mathrm{kg/s}\).

Key concepts

Fick's Law of Diffusion

Fick's Law of Diffusion is a principle used to describe how particles such as gases or liquids spread from an area of high concentration to an area of low concentration. The law states that the rate of diffusive transfer of a substance is proportional to the gradient of concentration. This means that substances move more quickly when the concentration difference between two points is greater.

In simple terms, if you imagine a cup of tea where sugar is being dissolved, diffusion helps describe how that sugar moves. If there’s a lot of sugar in one area, it will start spreading out evenly throughout the tea until it is all distributed equally. Similarly, water vapor in the problem moves through the drywall from where there is more vapor (inside the room) to where there is less (the insulation).

To find the mass diffusion rate ( N), we use the equation:

• diffusion coefficient ( D): It tells you how easily a substance diffuses through another.
  
• diffusion area ( A): The surface through which the substance is spreading.
  
• length ( L): The distance over which diffusion takes place.
  
• concentration difference ( C_A - C_B): The driving force behind diffusion.

Mass diffusion rates show how fast or slow a substance like water vapor spreads, which is crucial in situations like insulation materials reducing thermal resistance.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its liquid or solid phase at a given temperature. It is a critical factor when discussing diffusion because it influences how molecules move and spread.

In our scenario, when air inside a room at 25°C holds vapor at 3 kPa, this represents a certain amount of water vapor present. This vapor tries to diffuse through the drywall into areas where vapor pressure is much less, like the adjoining insulation zone with negligible vapor pressure.

Understanding vapor pressure helps us grasp why molecules move from one place to another. It's like understanding why steam rises from a hot cup of coffee; it’s the vapor pressure trying to equalize with its surroundings. In technical terms, the water vapor molecules are "escaping" into the insulation due to the pressure difference, overcoming any barriers in their path.

Thermal Resistance

Thermal resistance is the property of a material that restricts heat flow through it. It's crucial in building and construction because proper insulation keeps homes warm in winter and cool in summer.

When discussing the insulation next to the drywall, thermal resistance becomes interesting. As vapor diffuses through the drywall into insulation, it condenses. This reduces the thermal resistance, degrading the insulation's efficiency and leading to higher energy costs to maintain temperature.

In homes, if vapor is continuously condensing within the insulation, it can lead to problems like mold or water damage, apart from reduced insulation effectiveness. Hence, understanding both diffusion and thermal resistance aids in constructing buildings that are both energy-efficient and durable.

Diffusion Coefficient

The diffusion coefficient, often symbolized as ( D), is a crucial parameter in the diffusion process which measures how easily molecules such as water vapor diffuse through a material.

In our example, the diffusion coefficient of water vapor in the wall is given as 0.2 × 10⁻⁹ m²/s. It quantifies the speed with which vapor can penetrate and move through the drywall material.

Think of it as a measure of a material’s permeability to vapor. A higher value means vapor can move more quickly through the material, while a lower value suggests a barrier to vapor movement. This coefficient is vital in designing materials meant to trap or control weather elements, ensuring they perform optimally in controlling vapor diffusion and maintaining thermal comfort indoors.