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Chapter 14: Problem 50

polycarbonate to reduce mamufacturing cosis. Assume that a firs-onder 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 oxyeen- proof pouch, for a DVD of thickness \(2 \mathrm{~L}\). 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, \(\bar{C}_{\text {moe }}\) as a function of time. Write an expression for \(\bar{C}_{\text {pued }}\) in terms of the oxygen molar concentration, assuming that every mole of oxygen that reacts with the polymer results in \(p\) moles of product.

Short Answer

Expert verified
The governing equation is a diffusion-reaction type with specific boundary and initial conditions, and the volume-averaged product concentration is based on integrating the oxygen concentration.

Step by step solution

01

Governing Equation

For a first-order homogeneous chemical reaction, the rate of reaction is proportional to the oxygen molar concentration, denoted as \( C_O(x,t) \). The governing equation can be expressed as the diffusion-reaction equation: \ \[ \frac{\partial C_O}{\partial t} = D \frac{\partial^2 C_O}{\partial x^2} - k C_O \] where \( D \) is the diffusion coefficient, and \( k \) is the reaction rate constant.
02

Boundary Conditions

The boundary conditions for the oxygen concentration at the surface of the DVD and at the center (since symmetry at center can be assumed) are: \ \( C_O(-L, t) = C_s \) and \( C_O(L, t) = C_s \) at the surfaces. \For symmetry at the center point: \ \( \frac{\partial C_O}{\partial x}(0, t) = 0 \).
03

Initial Condition

Initially, when the DVD is removed from the oxygen-proof pouch, it is assumed that there is no oxygen present inside the DVD. Thus, the initial condition is: \ \( C_O(x, 0) = 0 \) for \( -L \leq x \leq L \).
04

Expression for Volume-averaged Molar Concentration of Product

The volume-averaged molar concentration of product, \( \bar{C}_{\text{prod}} \), can be found by integrating the molar concentration of the product over the DVD volume and dividing by the volume. Since one mole of oxygen produces \( p \) moles of product: \ \[ \bar{C}_{\text{prod}}(t) = \frac{p}{2L} \int_{-L}^{L} C_O(x,t) \, dx \] This expression provides the average product concentration as a function of time.

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Key Concepts

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

Diffusion-Reaction Equation
The diffusion-reaction equation is a cornerstone of chemical reaction engineering, especially in scenarios where both diffusion and chemical reactions occur simultaneously. In this context, we're looking at a DVD exposed to oxygen, starting a chemical reaction within its structure. The primary equation governing this process is:\[ \frac{\partial C_O}{\partial t} = D \frac{\partial^2 C_O}{\partial x^2} - k C_O \]In this equation, \(C_O\) is the oxygen molar concentration at a given position \(x\) and time \(t\), \(D\) is the diffusion coefficient representing how fast oxygen moves within the DVD, and \(k\) is the reaction rate constant that describes the speed of the chemical reaction. This partial differential equation demonstrates how oxygen concentration changes over time due to diffusion and reaction. This balance between diffusion allowing oxygen to penetrate deeper and the reaction consuming the oxygen defines the internal chemistry of the DVD.
Oxygen Molar Concentration
The oxygen molar concentration, \( C_O(x,t) \), represents the amount of oxygen available at any point within the DVD. Understanding how this concentration varies across the DVD is crucial for predicting chemical reactions. Oxygen molar concentration depends on the diffusion rate and reaction rate, with the reaction rate being directly proportional to this concentration. Since the reaction is first-order, each increment in oxygen concentration increases the reaction rate proportionally. This function largely influences how quickly the DVD becomes opaque. In practical terms, it highlights where oxygen is more concentrated and therefore where the reaction will proceed faster, leading to potential opaqueness in these regions first.
Boundary and Initial Conditions
To effectively model and simulate a physical process, boundary and initial conditions must be defined clearly. For the DVD subjected to oxygen exposure, the boundary conditions stipulate that the oxygen concentration on the surfaces of the DVD is constant, expressed as \( C_O(-L, t) = C_s \) and \( C_O(L, t) = C_s \). This reflects a scenario where the DVD surfaces are exposed to an environment with a steady oxygen concentration.The symmetry condition at the center of the DVD, \( \frac{\partial C_O}{\partial x}(0, t) = 0 \), signifies that there is no net flow of oxygen at the center, a common assumption in problems with symmetrical geometry. Initially, when the DVD is just removed from the oxygen-proof pouch, it starts with zero oxygen: \( C_O(x, 0) = 0 \) for \( -L \leq x \leq L \). This baseline is crucial for simulating how oxygen invades from the surface to the center, setting the stage for reaction initiation.
Volume-averaged Molar Concentration
The concept of volume-averaged molar concentration is important when assessing how uniformly a product is formed throughout an entire volume. In this case, the product is formed by the reaction of oxygen with the polymer. The volume-averaged molar concentration, \( \bar{C}_{\text{prod}} \), gives insight into the overall degree of reaction within the DVD.It is calculated using the expression:\[ \bar{C}_{\text{prod}}(t) = \frac{p}{2L} \int_{-L}^{L} C_O(x,t) \, dx \]Here, \(p\) represents the number of moles of product formed per mole of oxygen, and the integral calculates the total concentration over the DVD's thickness from \(-L\) to \(L\). By dividing by the total DVD volume \(2L\), this expression reveals the average concentration of the reaction product. This measurement helps predict how changes occur over time and assess the overall performance of the DVD in terms of readability and opacity.

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Most popular questions from this chapter

Steel is carburized in a high-temperature process that depends on the transfer of carbon by diffusion. The value of the diffusion coefficient is strongly temperature dependent and may be approximated as \(D_{c-3}\) \(\left(\mathrm{m}^{2} / \mathrm{s}\right)=2 \times 10^{-5} \operatorname{cxp}[-17,000 / T(\mathrm{~K})]\). If the process is effected at \(1000^{\circ} \mathrm{C}\) and a carbon mole fraction of \(0.02\) is maintained at the surface of the steel. how much time is required to elevate the carbon content of the steel from an initial value of \(0.1 \%\) to a value of \(1.0 \%\) at a depth of \(1 \mathrm{~mm}\) ?

Hydrogen gas is used in a process to manufacture a sheet material of 6-mm thickness. At the end of the process, \(\mathrm{H}_{2}\) remains in solution in the material with a uniform concentration of \(320 \mathrm{kmol} / \mathrm{m}^{3}\). To remove \(\mathrm{H}_{2}\) from the material, both surfaces of the sheet are exposed to an air stream at \(500 \mathrm{~K}\) and a total pressure of 3 atm. Due to contamination, the hydrogen partial pressure is \(0.1 \mathrm{~atm}\) in the air stream, which provides a convection mass transfer coefficient of \(1.5 \mathrm{~m} / \mathrm{h}\). The mass diffusivity and solubility of hydrogen (A) in the sheet material \((B)\) are \(D_{A B}=2.6 \times 10^{-1} \mathrm{~m}^{2} / \mathrm{s}\) and \(S_{\mathrm{Am}}=160 \mathrm{kmol} / \mathrm{m}^{3} \cdot \mathrm{atm}_{\text {, respectively. }}\) (a) If the sheet material is left exposed to the air stream for a long lime, determine the final content of hydrogen in the material \(\left(\mathrm{kg} / \mathrm{m}^{3}\right)\). (b) Identify and cvaluate the parameter that can be used to determine whether the transient mass diffusion process in the sheet can be assumed to be characterized by a unifom concentration at any time during the process. Hint: This situation is analogous to that used to determine the validity of the lumped-capacitance method for a transient heat transfer analysis. (c) Determine the time required to reduce the hydrogen mass density at the center of the sheet to twice the limiting value calculated in part (a).

Consider the problem of oxygen transfer from the interior lung cavity, across the lung tissue, to the network of blood vessels on the opposite side. The lung tissue (species B) may be approximated as a plane wall of thickness \(L\). The inhalation process may be assumed to maintain a constant molar concentration \(C_{A}(0)\) of oxygen (species A) in the tissue at its inner surface \((x=0)\). and assimilation of oxygen by the blood may be assumed to maintain a constant molar concentration \(C_{\mathrm{A}}(L)\) of oxygen in the tissue at its outer surface \((x=L)\). There is oxygen consumption in the tissue due to metabolic processes, and the reaction is zero order, with \(\dot{N}_{A}=-k_{0}\) Obtain expressions for the distribution of the exygen concentration in the tissue and for the rate of assimilation of oxygen by the blood per unit tissue surface area.

Gaseous bydrogen at 10 bars and \(27^{\circ} \mathrm{C}\) is stored in a 100 -mm-diameter spherical tank having a steel wall \(2 \mathrm{~mm}\) thick. The molar concentration of hydrogen in the steel is \(1.50 \mathrm{kmol} / \mathrm{m}^{3}\) at the inner surface and negligible at the outer surface, while the diffusion coeffcient of hydrogen in steel is approximately \(0.3 \times\) \(10^{-12} \mathrm{~m}^{2} / \mathrm{s}\). What is the initial rate of mass loss of hydrogen by diffusion through the tank wall? What is the initial rate of pressure drop within the tank?

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 liters, whut is the mass of the coolant?
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