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Chapter 14: Problem 40

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?

Short Answer

Expert verified
It takes approximately 353.25 seconds for the hydrogen density at the center of the sheet to reach a value of \(1.2 \,\mathrm{kg}\, \mathrm{m}^{-3}\).

Step by step solution

01

Calculate the initial density of hydrogen

It is given that the uniform concentration of hydrogen is \(3 \, \mathrm{kmol}\, \mathrm{m}^{-3}\), we're going to convert this concentration into density (in kg/m鲁). We know the molar mass of hydrogen is approximately \(2 \, \mathrm{g/mol}\), so first we need to convert it to kg/kmol: \(M_{H_2} = 2 \,\mathrm{g/mol} \times \frac{1 \,\mathrm{kg}}{1000 \,\mathrm{g}} \times \frac{1\,\mathrm{kmol}}{1 \,\mathrm{mol}} \approx 2 \,\mathrm{kg/kmol}\) Now we can find the initial density of hydrogen using the given concentration: \(\rho_{H_2} = \mathrm{concentration} \times M_{H_2}\) \(\rho_{H_2} = 3 \cdot 10^3\, \mathrm{mol} \, \mathrm{m}^{-3} \times 2 \, \mathrm{kg} \, \mathrm{kmol}^{-1}\) \(\rho_{H_2} = 6 \mathrm{kg} \, \mathrm{m}^{-3} \)
02

Determine the final density of hydrogen

The final density of hydrogen at the center of the sheet is given to be \(1.2 \,\mathrm{kg}\, \mathrm{m}^{-3}\).
03

Use Fick's second law of diffusion

As the sheet is symmetric, we can consider a semi-infinite slab, apply the boundary conditions and express the solution with the error function erf(x), which we will solve in terms of time: \(\rho (t) = \rho_0 \cdot \mathrm{erf}\left(\frac{L}{2 \sqrt{D_{H_2}\cdot t}}\right) \) Here, \(\rho (t)\) - The final density of hydrogen at the center of the sheet (\(1.2 \, \mathrm{kg} \, \mathrm{m}^{-3}\)) \(\rho_0\) - The initial density of hydrogen (\(6 \, \mathrm{kg} \, \mathrm{m}^{-3}\)) \(D_{H_2}\) - The mass diffusivity of hydrogen (\(9 \times 10^{-7} \, \mathrm{m}^2\, \mathrm{s}^{-1}\)) \(L\) - The thickness of the sheet (\(40 \mathrm{mm}\))
04

Calculate the required time

Now, we plug in the values and solve for \(t\): \(1.2 = 6 \mathrm{erf}\left(\frac{0.04}{2\sqrt{9\times10^{-7}\cdot t}}\right)\) \(0.2 = \mathrm{erf}\left(\frac{0.04}{2\sqrt{9\times10^{-7}\cdot t}}\right)\) Using the inverse error function erf鈦宦�(x), we can solve for t: \(t = \frac{\left(0.04\,\mathrm{erf}^{-1}(0.2)\right)^{2}}{4\cdot9\times10^{-7}}\) By calculating the value of erf鈦宦�(0.2) 鈮� 0.1791: \(t \approx \frac{\left(0.04\cdot0.1791\right)^{2}}{4\cdot9\times10^{-7}}\) \(t \approx 353.25 \, \mathrm{s}\) It takes approximately 353.25 seconds for the hydrogen density at the center of the sheet to reach a value of \(1.2 \,\mathrm{kg}\, \mathrm{m}^{-3}\).

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Key Concepts

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

Mass Diffusivity
Mass diffusivity, also known as diffusivity, is a property that quantifies the rate at which particles of a substance, such as hydrogen in our exercise, move through a medium due to molecular diffusion. In the context of Fick's laws, mass diffusivity (\(D\textsubscript{H2}\) in our exercise) is a measure of how easily molecules spread out over time.

Understanding mass diffusivity is essential when considering any process involving the transfer of particles from one location to another, for instance, when a concentration gradient is present. The diffusivity value is affected by factors like temperature, pressure, and the medium through which diffusion occurs. In our exercise example, the mass diffusivity of hydrogen is provided (\(9 \times 10^{-7} \text{ m}^{2}/\text{s}\)), which allows students to calculate the time required for the density of hydrogen to reach a certain value at the center of the material.
Error Function
The error function, commonly denoted as erf(x), is a special mathematical function that arises frequently in probability, statistics, and partial differential equations solving related to diffusion processes. The function is non-elementary, which means it cannot be expressed in terms of simple algebraic operations and therefore is often found in tabulated forms or can be approximated through numerical methods.

In our exercise, we use the error function to relate concentration (expressed as density) changes within the material to time, through Fick's second law of diffusion. The inverse error function, erf鈦宦�(x), is used when we need to solve for time, given a change in concentration, as illustrated in the steps to resolve the exercise problem. Understanding the use of the error function is integral when solving problems involving unsteady-state diffusion, where conditions change with time.
Unsteady-State Diffusion
Unsteady-state or transient diffusion is the process where the concentration of particles within a medium changes with time. It contrasts with steady-state diffusion, where concentrations remain constant over time. Fick's second law of diffusion guides us through these types of problems, providing a mathematical framework to predict how substances will distribute over time and space within a system.

In the context of our textbook problem, we're dealing with unsteady-state diffusion because the concentration of hydrogen within the material decreases from an initial uniform value to the prescribed concentration at the surface, all while the system seeks a new equilibrium. Solving unsteady-state diffusion problems requires understanding of boundary conditions, initial conditions, and how these influence the concentration profile within the material under analysis. This dynamic concept is crucial for scientists and engineers in fields such as materials science, environmental engineering, and pharmacology, where controlled diffusion is often necessary.

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Most popular questions from this chapter

A platinum catalytic reactor in an automobile is used to convert carbon monoxide to carbon dioxide in an oxidation reaction of the form \(2 \mathrm{CO}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2}\). Species transfer between the catalytic surface and the exhaust gases may be assumed to occur by diffusion in a film of thickness \(L=10 \mathrm{~mm}\). Consider an exhaust gas that has a pressure of \(1.2\) bars, a temperature of \(500^{\circ} \mathrm{C}\), and a \(\mathrm{CO}\) mole fraction of \(0.0012\). If the reaction rate constant of the catalyst is \(k_{1}^{\prime \prime}=0.005 \mathrm{~m} / \mathrm{s}\) and the diffusion coefficient of \(\mathrm{CO}\) in the mixture is \(10^{-4} \mathrm{~m}^{2} / \mathrm{s}\), what is the molar concentration of \(\mathrm{CO}\) at the catalytic surface? What is the rate of removal of \(\mathrm{CO}\) per unit area of the catalyst? What is the removal rate if \(k_{1}^{\prime \prime}\) is adjusted to render the process diffusion limited?

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.

Consider a spherical organism of radius \(r_{o}\) within which respiration occurs at a uniform volumetric rate of \(\dot{N}_{\mathrm{A}}=-k_{0}\). That is, oxygen (species A) consumption is governed by a zero-order, homogeneous chemical reaction. (a) If a molar concentration of \(C_{\mathrm{A}}\left(r_{o}\right)=C_{\mathrm{A}, o}\) is maintained at the surface of the organism, obtain an expression for the radial distribution of oxygen, \(C_{\mathrm{A}}(r)\), within the organism. From your solution, can you discern any limits on applicability of the result? (b) Obtain an expression for the rate of oxygen consumption within the organism. (c) Consider an organism of radius \(r_{o}=0.10 \mathrm{~mm}\) and a diffusion coefficient for oxygen transfer of \(D_{\mathrm{AB}}=10^{-8} \mathrm{~m}^{2} / \mathrm{s}\). If \(C_{\mathrm{A}, o}=5 \times 10^{-5} \mathrm{kmol} / \mathrm{m}^{3}\)and \(k_{0}=1.2 \times 10^{-4} \mathrm{kmol} / \mathrm{s}^{*} \mathrm{~m}^{3}\), what is the molar concentration of \(\mathrm{O}_{2}\) at the center of the organism?

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 \(1.71 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\).

Referring to Problem 14.34, a more representative model of respiration in a spherical organism is one for which oxygen consumption is governed by a firstorder reaction of the form \(\dot{N}_{\mathrm{A}}=-k_{1} C_{\mathrm{A}}\). (a) If a molar concentration of \(C_{\mathrm{A}}\left(r_{o}\right)=C_{\mathrm{A}, o}\) is maintained at the surface of the organism, obtain an expression for the radial distribution of oxygen, \(C_{\mathrm{A}}(r)\), within the organism. Hint: To simplify solution of the species diffusion equation, invoke the transformation \(y \equiv r C_{\mathrm{A}}\). (b) Obtain an expression for the rate of oxygen consumption within the organism. (c) Consider an organism of radius \(r_{o}=0.10 \mathrm{~mm}\) and a diffusion coefficient of \(D_{\mathrm{AB}}=10^{-8} \mathrm{~m}^{2} / \mathrm{s}\). If \(C_{\mathrm{A}, o}=5 \times 10^{-5} \mathrm{kmol} / \mathrm{m}^{3}\) and \(k_{1}=20 \mathrm{~s}^{-1}\), estimate the corresponding value of the molar concentration at the center of the organism. What is the rate of oxygen consumption by the organism?
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