Problem 16 - An Introduction to Materials Engineering and Science for Chemical and Materials Engineers

Question

Composite membranes are used where the membrane of choice for a particular separation must be strengthened by addition of a second material. Assume a hollow fiber membrane of inside diameter \(1.2 \mathrm{~mm}\) and outside diameter \(1.3 \mathrm{~mm}\). Mass diffusivity of the target species through this membrane is \(3.0 \times\) \(10^{-10} \mathrm{~m}^{2} / \mathrm{s}\). A second material is layered on the outside of the first material to add strength. Mass diffusivity of the same target species through this material is \(18 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}\). What thickness of the outside material can be added if the total diffusion does not exceed the resistance of the inner material by more than \(25 \%\) ?

Step-by-Step Solution

Answer

The thickness should not exceed 7.5 mm.

Step 1: Understanding the Problem

We need to find the thickness of the second material that can be layered onto the primary membrane without increasing the total diffusion resistance by more than 25% compared to the inner membrane's resistance. We are given the diffusivities and dimensions of the hollow fiber membrane.

Step 2: Calculate the Resistance of the Inner Membrane

The resistance (R) to mass diffusion through a membrane is related to the thickness (L) of the membrane and the mass diffusivity (D) as:\[ R = \frac{L}{D} \]The thickness of the inner membrane is half of the difference between outer and inner diameters: \[ L_1 = \frac{1.3 - 1.2}{2} = 0.05 \text{ mm} = 0.00005 \text{ m} \]Therefore, its resistance is:\[ R_1 = \frac{0.00005}{3.0 \times 10^{-10}} \approx 166666.67 \text{ s/m}^2 \]

Step 3: Calculate the Allowed Total Resistance

The total resistance after adding the outer layer should not exceed by more than 25% of the inner membrane's resistance. Calculate the allowable total resistance:\[ R_{total} = R_1 + 0.25 \times R_1 = 1.25 \times R_1 \approx 208333.33 \text{ s/m}^2 \]

Step 4: Calculate the Required Resistance of the Outer Material

The resistance of the outer material should be such that the sum of both resistances equals the allowed total resistance:\[ R_2 = R_{total} - R_1 = 208333.33 - 166666.67 \approx 41666.66 \text{ s/m}^2 \]

Step 5: Calculate Thickness of the Outer Material

Using the previously calculated resistance for the outer layer, and using the formula for resistance:\[ L_2 = R_2 \times D_2 = 41666.66 \times 18 \times 10^{-10} \approx 0.0075 \text{ m} = 7.5 \text{ mm} \]

Key Concepts

Mass Diffusivity

Mass diffusivity is a measure of how easily a substance such as a gas, liquid, or solute can move through a medium. In our membrane problem, it represents how quickly the target species can pass through different layers of the membrane. Mass diffusivity varies depending on several factors, including:

• The nature of the species being diffused.
• The characteristics of the medium it moves through.
• The temperature of the environment.

In this exercise, differing mass diffusivities for the two membrane layers indicate that the target species moves more easily through the second, outer material — specifically, 6 times faster compared to the inner layer, considering the values provided: \(D_1 = 3.0 \times 10^{-10} \text{ m}^2/\text{s}\) and \(D_2 = 18 \times 10^{-10} \text{ m}^2/\text{s}\). Understanding this difference in diffusivity is critical for controlling the rate at which substances are transported through composite membranes.

Diffusion Resistance

Diffusion resistance is the obstacle faced by a substance trying to move through a membrane. It plays a crucial role in determining how effective a membrane is for separation processes. The core idea here is: the higher the resistance, the slower the diffusion.The resistance \(R\) is calculated using:\[ R = \frac{L}{D} \]where:

• \(L\) is the thickness of the membrane.
• \(D\) is the mass diffusivity.

For the inner membrane of our exercise, resistance is relatively high due to its lower mass diffusivity. It must be carefully managed when adding an outer layer. The problem here illustrates that you don't want the total resistance to increase by more than 25% when you add an external layer. This calls for planning to balance the thickness and material selection to meet performance requirements.

Membrane Thickness

Membrane thickness directly influences the diffusion resistance, as seen in the relationship \(R = \frac{L}{D}\). If a membrane is thick, diffusion would be slower since resistance is proportionally increased. The calculation of thickness in this exercise is crucial for maintaining optimal diffusion rates while strengthening the membrane.To calculate the thickness of the outer layer, rearrange the resistance formula:\[ L_2 = R_2 \times D_2 \]In the example, we need the outer material's thickness to ensure that the total resistance is only 25% greater than the inner membrane alone. With \(R_2 = 41666.66\ \text{s/m}^2\) and \(D_2 = 18 \times 10^{-10}\ \text{m}^2/\text{s}\), it results in:\[ L_2 \approx 0.0075\ \text{m} = 7.5\ \text{mm} \]This thickness allows the composite membrane to function properly without excessive slowing of the diffusion process.