/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 The presence of \(\mathrm{CO}_{2... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 48

The presence of \(\mathrm{CO}_{2}\) in solution is essential to the growth of aquatic plant life, with \(\mathrm{CO}_{2}\) used as a reactant in the photosynthesis. Consider a stagnant body of water in which the concentration of \(\mathrm{CO}_{2}\left(\rho_{\mathrm{A}}\right)\) is everywhere zero. At time \(t=0\), the water is exposed to a source of \(\mathrm{CO}_{2}\), which maintains the surface \((x=0)\) concentration at a fixed value \(\rho_{\mathrm{A}, 0}\). For time \(t>0, \mathrm{CO}_{2}\) will begin to accumulate in the water, but the accumulation is inhibited by \(\mathrm{CO}_{2}\) consumption due to photosynthesis. The time rate at which this consumption occurs per unit volume is equal to the product of a reaction rate constant \(k_{1}\) and the local \(\mathrm{CO}_{2}\) concentration \(\rho_{\mathrm{A}}(x, t)\). (a) Write (do not derive) a differential equation that could be used to determine \(\rho_{\mathrm{A}}(x, t)\) in the water. What does each term in the equation represent physically? (b) Write appropriate boundary conditions that could be used to obtain a particular solution, assuming a "deep" body of water. What would be the form of this solution for the special case of negligible \(\mathrm{CO}_{2}\) consumption \(\left(k_{1} \approx 0\right)\) ?

Short Answer

Expert verified
The differential equation representing the CO鈧� concentration in the water is: \[\frac{\partial \rho_{\mathrm{A}}(x, t)}{\partial t} = D \frac{\partial^2 \rho_{\mathrm{A}}(x, t)}{\partial x^2} - k_{1}\rho_{\mathrm{A}}(x, t)\] The boundary conditions are: 1. \(\rho_{\mathrm{A}}(0,t) = \rho_{\mathrm{A}, 0}\) 2. \(\rho_{\mathrm{A}}(x,t) \to 0\) as \(x \to \infty\) For the special case of negligible CO鈧� consumption \(\left(k_{1} \approx 0\right)\), the solution is: \[\rho_{\mathrm{A}}(x, t) = \rho_{\mathrm{A}, 0} \cdot \operatorname{erfc} \left(\frac{x}{2 \sqrt{D t}}\right)\]

Step by step solution

01

a) Differential equation for \(\rho_{\mathrm{A}}(x,t)\)

We will have a diffusion term and a rate term as contributors to the time rate of change of the \(\mathrm{CO}_{2}\) concentration. The general form of a diffusion equation with reaction terms can be written as: \[\frac{\partial \rho_{\mathrm{A}}(x, t)}{\partial t} = D \frac{\partial^2 \rho_{\mathrm{A}}(x, t)}{\partial x^2} - k_{1}\rho_{\mathrm{A}}(x, t)\] Here, \(\rho_{\mathrm{A}}(x,t)\) represents the CO鈧� concentration in the water as a function of depth (x) and time (t), D represents the diffusion coefficient, \(k_{1}\) is the reaction rate constant, and \(\frac{\partial^2 \rho_{\mathrm{A}}(x, t)}{\partial x^2}\) is the second derivative of the concentration with respect to the depth. The first term in the equation \(D \frac{\partial^2 \rho_{\mathrm{A}}(x, t)}{\partial x^2}\) represents the diffusion of the \(\mathrm{CO}_{2}\) in the water, and the second term, \(-k_{1}\rho_{\mathrm{A}}(x, t)\), represents the consumption of \(\mathrm{CO}_{2}\) due to photosynthesis.
02

b) Boundary conditions and solution for negligible consumption

To obtain a particular solution for the given problem, we need to provide appropriate boundary conditions. The problem states that the surface (x=0) concentration of CO鈧� is fixed at a value of \(\rho_{\mathrm{A}, 0}\). For a "deep" body of water, we can assume that the concentration of CO鈧� becomes negligibly small as x becomes large. Thus, we have the following boundary conditions: Boundary condition 1: \(\rho_{\mathrm{A}}(0,t) = \rho_{\mathrm{A}, 0}\) Boundary condition 2: \(\rho_{\mathrm{A}}(x,t) \to 0\) as \(x \to \infty\) For the special case of negligible \(\mathrm{CO}_{2}\) consumption \(\left(k_{1} \approx 0\right)\), the reaction term in the differential equation vanishes. Therefore, the equation becomes: \[\frac{\partial \rho_{\mathrm{A}}(x, t)}{\partial t} = D \frac{\partial^2 \rho_{\mathrm{A}}(x, t)}{\partial x^2}\] In this case, the solution results in a simple diffusion equation (Fick's second law) which, accounting for the boundary conditions, assumes the error function's form: \[\rho_{\mathrm{A}}(x, t) = \rho_{\mathrm{A}, 0} \cdot \operatorname{erfc} \left(\frac{x}{2 \sqrt{D t}}\right)\] where \(\operatorname{erfc}\) is the complementary error function. This expression is the solution for the concentration of \(\mathrm{CO}_{2}\) in the water as a function of depth and time when the consumption term is negligible.

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Key Concepts

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

Diffusion Coefficient
The diffusion coefficient, denoted as D, is a crucial parameter in the diffusion-reaction equation. It quantifies the rate at which a substance, such as \text{CO}_2, spreads out or diffuses from an area of high concentration to an area of low concentration. The diffusion process is driven by the random motion of particles and is influenced by factors such as temperature and the medium through which diffusion occurs.

In the exercise, the diffusion coefficient D is a part of the term \(D \frac{\partial^2 \rho_{\mathrm{A}}(x, t)}{\partial x^2}\), which represents the change in \text{CO}_2 concentration with respect to depth in the water. This term accounts for the movement of \text{CO}_2 molecules as they diffuse downward due to the concentration gradient. The higher the value of D, the faster the \text{CO}_2 will disperse into the water.
Reaction Rate Constant
The reaction rate constant, represented as \(k_1\) in the homework exercise, is another pivotal component of the diffusion-reaction equation. It defines the speed at which a chemical reaction, such as photosynthesis consuming \text{CO}_2, proceeds. The reaction rate constant not only depends on the nature of the reaction but also on environmental conditions like temperature and pH.

In the context of the given problem, \(-k_1\rho_{\mathrm{A}}(x, t)\) is the term that models the rate of \text{CO}_2 consumption by photosynthesis. A higher \(k_1\) indicates a faster consumption rate, leading to a quicker decrease in \text{CO}_2 concentration over time at any given point in the water. When the reaction is negligible (\(k_1 \approx 0\)), this term vanishes, simplifying the equation to model pure diffusion without chemical reaction.
Boundary Conditions
Boundary conditions are vital for solving differential equations in physics and engineering. They describe the values of the function being investigated at the edges or boundaries of the domain where the problem is defined. In the exercise about aquatic plant life and \text{CO}_2, boundary conditions help us to find a particular solution for the \text{CO}_2 concentration over time and space.

The first boundary condition given is \(\rho_{\mathrm{A}}(0,t) = \rho_{\mathrm{A}, 0}\), indicating that the \text{CO}_2 concentration at the surface, where \(x=0\), remains constant. The second boundary condition is \(\rho_{\mathrm{A}}(x,t) \to 0\) as \(x \to \infty\), assuming the \text{CO}_2 concentration decreases to negligible levels as the depth increases. These conditions reflect the physical setup: a sustained source of \text{CO}_2 at the surface of a deep body of water. For negligible \text{CO}_2 consumption, the boundary conditions, coupled with the simplified diffusion equation, yield a solution describing the gradual diffusion of \text{CO}_2 into deeper water layers.

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Most popular questions from this chapter

A 1-mm-thick square \((100 \mathrm{~mm} \times 100 \mathrm{~mm})\) sheet of polymer is suspended from a precision scale in a chamber characterized by a temperature and relative humidity of \(T=300 \mathrm{~K}\) and \(\phi=0\), respectively. Suddenly, at time \(t=0\), the chamber's relative humidity is raised to \(\phi=0.95\). The measured mass of the sheet increases by \(0.012 \mathrm{mg}\) over \(24 \mathrm{~h}\) and by \(0.016 \mathrm{mg}\) over \(48 \mathrm{~h}\). Determine the solubility and mass diffusivity of water vapor in the polymer. Preliminary experiments have indicated that the mass diffusivity is greater than \(7 \times 10^{-13} \mathrm{~m}^{2} / \mathrm{s}\).

A person applies an insect repellent onto an exposed area of \(A=0.5 \mathrm{~m}^{2}\) of their body. The mass of spray used is \(M=10\) grams, and the spray contains \(25 \%\) (by mass) active ingredient. The inactive ingredient quickly evaporates from the skin surface. (a) If the spray is applied uniformly and the density of the dried active ingredient is \(\rho=2000 \mathrm{~kg} / \mathrm{m}^{3}\), determine the initial thickness of the film of active ingredient on the skin surface. The temperature, molecular weight, and saturation pressure of the active ingredient are \(32^{\circ} \mathrm{C}, 152 \mathrm{~kg} / \mathrm{kmol}\), and \(1.2 \times 10^{-5}\) bars, respectively. (b) If the convection mass transfer coefficient associated with sublimation of the active ingredient to the air is \(\bar{h}_{m}=5 \times 10^{-3} \mathrm{~m} / \mathrm{s}\), the partition coefficient associated with the ingredient-skin interface is \(K=0.05\), and the mass diffusivity of the active ingredient in the skin is \(D_{\mathrm{AB}}=1 \times 10^{-13} \mathrm{~m}^{2} / \mathrm{s}\), determine how long the insect repellent remains effective. The partition coefficient is the ratio of the ingredient density in the skin to the ingredient density outside the skin. (c) If the spray is reformulated so that the partition coefficient becomes very small, how long does the insect repellent remain effective?

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.

Consider an ideal gas mixture of \(n\) species. (a) Derive an equation for determining the mass fraction of species \(i\) from knowledge of the mole fraction and the molecular weight of each of the \(n\) species. Derive an equation for determining the mole fraction of species \(i\) from knowledge of the mass fraction and the molecular weight of each of the \(n\) species. (b) In a mixture containing equal mole fractions of \(\mathrm{O}_{2}\), \(\mathrm{N}_{2}\), and \(\mathrm{CO}_{2}\), what is the mass fraction of each species? In a mixture containing equal mass fractions of \(\mathrm{O}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{CO}_{2}\), what is the mole fraction of each species?

The presence of a small amount of air may cause a significant reduction in the heat rate to a water-cooled steam condenser surface. For a clean surface with pure steam and the prescribed conditions, the condensate rate is \(0.020 \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\). With the presence of stagnant air in the steam, the condensate surface temperature drops from 28 to \(24^{\circ} \mathrm{C}\) and the condensate rate is reduced by a factor of 2 . For the air-steam mixture, determine the partial pressure of air as a function of distance from the condensate film.
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