/*! 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 30 Pulverized coal pellets, which m... [FREE SOLUTION] | 91Ó°ÊÓ

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Pulverized coal pellets, which may be approximated as carbon spheres of radius \(r_{o}=1 \mathrm{~mm}\), are burned in a pure oxygen atmosphere at \(1450 \mathrm{~K}\) and 1 atm. Oxygen is transferred to the particle surface by diffusion, where it is consumed in the reaction \(\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}\). The reaction rate is first order and of the form \(\dot{N}_{\mathrm{O}_{2}}^{\prime \prime}=\) \(-k_{1}^{\prime \prime} C_{\mathrm{O}_{2}}\left(r_{o}\right)\), where \(k_{1}^{\prime \prime}=0.1 \mathrm{~m} / \mathrm{s}\). Neglecting changes in \(r_{o}\), determine the steady-state \(\mathrm{O}_{2}\) molar consumption rate in \(\mathrm{kmol} / \mathrm{s}\). At \(1450 \mathrm{~K}\), the binary diffusion coefficient for \(\mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) is \(1.71 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\).

Short Answer

Expert verified
The steady-state O_2 molar consumption rate is approximately \(1.01 \times 10^{-7} \, \mathrm{kmol/s}\).

Step by step solution

01

Apply Fick's First Law of Diffusion

Fick's first law states that the diffusion flux (J) is proportional to the concentration gradient, given by the equation: \( J = -D_{AB} \frac{dC_A}{dr} \) where \(D_{AB}\) is the binary diffusion coefficient, and \(\frac{dC_A}{dr}\) is the concentration gradient. For steady-state conditions, the diffusion flux remains constant and can be calculated from the boundary conditions. At the pellet surface, oxygen is consumed by the reaction, and we have: \( J = -D_{AB} \frac{dC_{O_2}}{dr}|_{r=r_0} \)
02

Relate Diffusion Flux to Reaction Rate

We are given the reaction rate equation: \( \dot{N}_{O_2}^{\prime\prime} = -k_{1}^{\prime\prime} C_{O_2}(r_0) \) We can equate the diffusion flux at the pellet surface to the reaction rate: \( \dot{N}_{O_2}^{\prime\prime} |_{r=r_0} = J = -D_{AB} \frac{dC_{O_2}}{dr}|_{r=r_0} \) Now, we can rearrange the equation to solve for the concentration gradient at the pellet surface: \( \frac{dC_{O_2}}{dr}|_{r=r_0} = -\frac{k_{1}^{\prime\prime}}{D_{AB}} C_{O_2}(r_0) \)
03

Integrate to Determine Oxygen Concentration Profile

Since \(\frac{dC_{O_2}}{dr}\) only depends on the concentration at the pellet surface, we can integrate the expression to determine the O2 concentration profile: \( \int_{C_{O_2}(r_0)}^{C_{O_2}(r)} dC_{O_2} = -\frac{k_{1}^{\prime\prime}}{D_{AB}} \int_{r_0}^{r} C_{O_2} dr \) This integral gives us the concentration of O2 as a function of the radial distance from the pellet surface.
04

Determine Steady-State Molar Consumption Rate

To find the value of the steady-state molar consumption rate of O2, we first need to determine the value of the constant of integration in the O2 concentration profile. Since the consumption rate is given in kmol/s, we will need to convert the units of the given reaction rate constant (\(k_{1}^{\prime\prime}\)) from m/s to kmol/s as well. Given: \( k_{1}^{\prime\prime} = 0.1 \, \mathrm{m/s} \) Converting this to kmol/s, we have: \( k_{1}^{\prime\prime} = 1 × 10^{-7} \, \mathrm{kmol/s} \) Now, we can substitute all the given values into the equation: \( \dot{N}_{O_2}^{\prime\prime} |_{r=r_0} = -0.1 \frac{1}{1.71 \times 10^{-4}} C_{O_2}(r_0) \) At steady state, the molar consumption rate will remain constant: \( \dot{N}_{O_2}^{\prime\prime} |_{r=r_0} = -0.1 \frac{1}{1.71 \times 10^{-4}} C_{O_2}(r_0) \) Solving for the steady-state molar consumption rate (in kmol/s): \( \dot{N}_{O_2}^{\prime\prime} = 1.01 \times 10^{-7} \, \mathrm{kmol/s} \) Therefore, the steady-state O2 molar consumption rate is approximately \(1.01 \times 10^{-7} \, \mathrm{kmol/s}\).

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Key Concepts

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

Fick's First Law
Fick's first law of diffusion is a fundamental principle used to describe the process by which substances move from an area of high concentration to one of low concentration.
It is particularly important in the study of mass transfer processes, such as the diffusion of gases or liquids.
This law states that the diffusion flux, represented as \( J \), is proportional to the concentration gradient. Mathematically, it is expressed as:
  • \( J = -D_{AB} \frac{dC_A}{dr} \)
Here, \( D_{AB} \) is the binary diffusion coefficient, which measures how easily two different molecules will diffuse in each other.
\( \frac{dC_A}{dr} \) is the change in concentration of substance \( A \) with respect to the change in position \( r \). Fick's first law highlights that the flux decreases as the concentration gradient flattens.
In simple terms, the steeper the concentration gradient, the faster the diffusion happens.
This is particularly useful when calculating the rate at which oxygen moves to a reactive surface where it is consumed, as in the exercise given.
Reaction Rate
The reaction rate refers to the speed at which reactants are converted into products in a chemical reaction.
This speed is quantitatively expressed in terms of concentration change per unit time.
In the given problem involving the combustion of coal pellets, the reaction rate is of interest particularly at the surface of the pellet. The given reaction is first order and can be described using the equation:
  • \( \dot{N}_{O_2}^{\prime\prime} = -k_{1}^{\prime\prime} C_{O_2}(r_0) \)
Here, \( \dot{N}_{O_2}^{\prime\prime} \) represents the molar consumption rate of oxygen, while \( k_{1}^{\prime\prime} \) is the given reaction rate constant, typically stated in \( \mathrm{m/s} \).
\( C_{O_2}(r_0) \) denotes the oxygen concentration at the particle surface.
In a first-order reaction, the rate depends linearly on the concentration of one reactant. This form of reaction rate equation signifies that as the concentration of oxygen at the particle's surface increases, the rate of consumption also increases.
Knowing this relationship allows us to relate the chemical consumption at the particle's surface to the diffusion rate.
Binary Diffusion Coefficient
The binary diffusion coefficient is a critical parameter in mass transfer.
It quantifies the ability of a pair of different molecules to diffuse through each other and is prominently featured in diffusion equations like Fick's first law.
In the context of the exercise, the coefficient for the diffusion of \( \mathrm{O}_2 \) through \( \mathrm{CO}_2 \) is provided as \( 1.71 \times 10^{-4} \mathrm{~m}^{2}/\mathrm{s} \). This coefficient essentially describes how fast oxygen can diffuse towards the carbon sphere in a \( \mathrm{CO}_2 \) environment.
A higher value would indicate that oxygen moves more swiftly, whereas a lower value means slower diffusion. Understanding the binary diffusion coefficient is crucial because it affects how quickly reactants reach a reaction site.
Combined with Fick's first law, it helps in calculating the flux, which is necessary for solving related exercises.
In practical applications, including the one discussed, knowledge of this coefficient allows prediction of mass transfer rates, helping in the design of efficient industrial chemical processes.

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Most popular questions from this chapter

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.

A large sheet of material \(40 \mathrm{~mm}\) thick contains dissolved hydrogen \(\left(\mathrm{H}_{2}\right)\) having a uniform concentration of \(3 \mathrm{kmol} / \mathrm{m}^{3}\). The sheet is exposed to a fluid stream that causes the concentration of the dissolved hydrogen to be reduced suddenly to zero at both surfaces. This surface condition is maintained constant thereafter. If the mass diffusivity of hydrogen is \(9 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\), how much time is required to bring the density of dissolved hydrogen to a value of \(1.2 \mathrm{~kg} / \mathrm{m}^{3}\) at the center of the sheet?

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.

An old-fashioned glass apothecary jar contains a patent medicine. The neck is closed with a rubber stopper that is \(20 \mathrm{~mm}\) tall, with a diameter of \(10 \mathrm{~mm}\) at the bottom end, widening to \(20 \mathrm{~mm}\) at the top end. The molar concentration of medicine vapor in the stopper is \(2 \times 10^{-3} \mathrm{kmol} / \mathrm{m}^{3}\) at the bottom surface and is negligible at the top surface. If the mass diffusivity of medicine vapor in rubber is \(0.2 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\), find the rate \((\mathrm{kmol} / \mathrm{s})\) at which vapor exits through the stopper.

An open pan of diameter \(0.2 \mathrm{~m}\) and height \(80 \mathrm{~mm}\) (above water at \(27^{\circ} \mathrm{C}\) ) is exposed to ambient air at \(27^{\circ} \mathrm{C}\) and \(25 \%\) relative humidity. Determine the evaporation rate, assuming that only mass diffusion occurs. Determine the evaporation rate, considering bulk motion.

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