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Chapter 14: Problem 17

Consider the radial diffusion of a gaseous species (A) through the wall of a plastic tube (B), and allow for chemical reactions that provide for the depletion of A at a rate \(\dot{N}_{\mathrm{A}}\left(\mathrm{kmol} / \mathrm{s} \cdot \mathrm{m}^{3}\right)\). Derive a differential equation that governs the molar concentration of species A in the plastic.

Short Answer

Expert verified
The differential equation that governs the molar concentration of species A in the plastic tube, considering both radial diffusion and chemical reaction, is: \[\frac{d^2C_A}{dr^2} - \frac{\dot{N}_A}{D} = 0\]

Step by step solution

01

Fick's Law for Radial Diffusion

Fick's Law states that the diffusion flux (\(J_A\)) is proportional to the concentration gradient, considering the diffusion coefficient, D: \[\ J_A = -D \frac{dC_A}{dr}\] where: - \(J_A\) is the diffusion flux of species A (in \(kmol/s \cdot m^2\)) - \(D\) is the diffusion coefficient (in \(m^2/s\)) - \(C_A\) is the concentration of species A (in \(kmol/m^3\)) - \(r\) is the radial coordinate (in meters)
02

Introduce Reaction Rate

Now, we are given that there is a chemical reaction taking place in the plastic tube wall, which can deplete species A at a rate of \(\dot{N}_A\). We can express the reaction rate, R, per unit volume (in \(kmol/s \cdot m^3\)): \[R = \dot{N}_A\]
03

Derive the Differential Equation for Concentration

At steady-state, the net flux due to diffusion and reaction is zero. Therefore: \[\frac{dJ_A}{dr} + R = 0\] Now, we substitute the expression for the radial diffusion flux from Step 1 and the reaction rate from Step 2 into the equation: \[\frac{d}{dr} \left(-D \frac{dC_A}{dr}\right) + \dot{N}_A = 0\] Now, we take the derivative with respect to 'r': \[-D\frac{d^2C_A}{dr^2} + \dot{N}_A = 0\] Finally, we have derived a differential equation that governs the molar concentration of species A in the plastic tube: \[\frac{d^2C_A}{dr^2} - \frac{\dot{N}_A}{D} = 0\] This is a second-order linear ordinary differential equation that describes the concentration of species A as a function of the radial coordinate 'r' considering both radial diffusion and chemical reaction.

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Key Concepts

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

Fick's Law
Understanding Fick's Law is crucial when it comes to analyzing the diffusion processes in materials. It states that the flow of a substance from a region of high concentration to a region of low concentration is directly proportional to the concentration gradient. Mathematically, we express Fick's Law as:
\[\begin{equation}J_A = -D \frac{dC_A}{dr}\end{equation}\]
where \(J_A\) represents the diffusion flux, indicating the amount of substance \(A\) passing through a unit area per unit time. The diffusion coefficient \(D\) reflects the ease with which \(A\) diffuses through the material and \(C_A\) is the molar concentration, which shows how many moles of substance \(A\) are present in a unit volume. The negative sign indicates the flux moves from high to low concentration. This law is vital for predicting how substances spread out over time and when we allow for chemical reactions, it becomes even more complex.
Molar Concentration
Molar concentration, typically denoted as \(C\), is a measure of the amount of a solute that is present per unit volume of solution. In the realm of diffusion, we're mostly focused on the concentration of a particular species, \(A\), in a system, described by \(C_A\). It is essential for understanding how concentrated a solution is, and its units are generally \(kmol/m^3\). The molar concentration gradient is the driving force behind diffusion; molecules move from areas of higher molar concentration to lower molar concentration, resulting in the process of equilibrium. Analysing molar concentration gradients, especially when coupled with Fick's Law, is fundamental for solving problems involving diffusion in materials.
Reaction Rate
Reaction rate plays a significant role in chemical kinetics and is indicative of how quickly a reaction consumes reactants or generates products. It's usually expressed in terms of the concentration change of a reactant or product per unit time. For our example, the depletion of species \(A\) due to a reaction within the plastic tube wall is represented by \(R\) or \(\dot{N}_A\), with units \(kmol/s \times m^3\). When the reaction rate is integrated into the study of diffusion, it can significantly alter the profile of substance concentration over time. In the case of our diffusion problem, it effectively adjusts the rate at which species \(A\) is distributed throughout the material, hence it's crucial to account for reaction rates when analyzing diffusion-reaction systems.
Differential Equation
Differential equations are mathematical tools used to describe various phenomena involving rates of change. They're particularly important in physics, engineering, and economics. The equation derived in our example is a second-order linear ordinary differential equation representing the radial diffusion with a reaction source. The equation takes the form:
\[\begin{equation}\frac{d^2C_A}{dr^2} - \frac{\dot{N}_A}{D} = 0\end{equation}\]
which signifies that the rate of change of the gradient of molar concentration with respect to the radial distance (\(r\)) is balanced by the reaction depleting the species \(A\). Solving such differential equations is fundamental as they explain how the concentration of a diffusing substance changes with both position and time, considering various reactions that might occur in the process.

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Most popular questions from this chapter

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 \(10^{-9} \mathrm{~m}^{2} / \mathrm{s}\).

A He-Xe mixture containing \(0.75\) mole fraction of helium is used for cooling of electronics in an avionics application. At a temperature of \(300 \mathrm{~K}\) and atmospheric pressure, calculate the mass fraction of helium and the mass density, molar concentration, and molecular weight of the mixture. If the cooling system capacity is \(10 \mathrm{~L}\), what is the mass of the coolant?

The surface of glass quickly develops very small microcracks when exposed to high humidity. Although microcracks can be safely ignored in most applications, they can significantly decrease the mechanical strength of very small glass structures such as optical fibers. Consider a glass optical fiber of diameter \(D_{i}=125 \mu \mathrm{m}\) that is coated with an acrylate polymer to form a coated fiber of outer diameter \(D_{o}=250 \mu \mathrm{m}\). A telecommunications engineer insists that the optical fiber be stored in a low-humidity environment prior to installation so that it is sufficiently strong to withstand rough treatment by technicians in the field. If installation of a roll of fiber requires several hot and humid days to complete, will careful storage beforehand prevent microcracking? The mass diffusivity of water vapor in the acrylate is \(D_{\mathrm{AB}}=5.5 \times 10^{-13} \mathrm{~m}^{2} / \mathrm{s}\) while the glass can be considered impermeable.

Consider the DVD of Problem 14.49, except now the reacting polymer is blended uniformly with the polycarbonate to reduce manufacturing costs. Assume that a first-order homogeneous chemical reaction takes place between the polymer and oxygen; the reaction rate is proportional to the oxygen molar concentration. (a) Write the governing equation, boundary conditions, and initial condition for the oxygen molar concentration after the DVD is removed from the oxygen- proof pouch, for a DVD of thickness \(2 L\). Do not solve. (b) The DVD will gradually become more opaque over time as the reaction proceeds. The ability to read the DVD will depend on how well the laser light can penetrate through the thickness of the DVD. Therefore, it is important to know the volume-averaged molar concentration of product, \(\bar{C}_{\text {prod }}\), as a function of time. Write an expression for \(\bar{C}_{\text {prod }}\) in terms of the oxygen molar concentration, assuming that every mole of oxygen that reacts with the polymer results in \(p\) moles of product.

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 \(D_{\mathrm{AB}}=7 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\).
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