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Chapter 14: Problem 24

Hydrogen at a pressure of 2 atm flows within a tube of diameter \(40 \mathrm{~mm}\) and wall thickness \(0.5 \mathrm{~mm}\). The outer surface is exposed to a gas stream for which the hydrogen partial pressure is \(0.1\) atm. The mass diffusivity and solubility of hydrogen in the tube material are \(1.8 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\) and \(160 \mathrm{kmol} / \mathrm{m}^{3}\). atm. respectively. When the system is at \(500 \mathrm{~K}\), what is the rate of hydrogen transfer through the tube per unit length \((\mathrm{kg} / \mathrm{s}\) 路 \(\mathrm{m})\) ?

Short Answer

Expert verified
The rate of hydrogen transfer per unit length is approximately \(2.207 \times 10^{-6} \; \text{kg/m} \cdot \text{s}\).

Step by step solution

01

Calculate the Hydrogen Concentration Difference

Using the solubility of hydrogen, convert the partial pressures inside and outside the tube into concentrations. The concentration on the inner side of the tube, where the pressure is 2 atm, is calculated as \( C_1 = 160 \times 2 = 320 \; \text{kmol/m}^3 \). The concentration on the outer side of the tube, where the pressure is 0.1 atm, is \( C_2 = 160 \times 0.1 = 16 \; \text{kmol/m}^3 \). The concentration difference \( \Delta C \) is given by \( \Delta C = C_1 - C_2 = 320 - 16 = 304 \; \text{kmol/m}^3 \).
02

Determine the Diffusive Flux

To find the rate of hydrogen transfer through the tube, use Fick's law of diffusion. The diffusive flux \( J \) can be calculated using the equation \( J = -D \frac{\Delta C}{\Delta x} \), where \( D = 1.8 \times 10^{-11} \; \text{m}^2/\text{s} \) is the diffusivity and \( \Delta x = 0.5 \; \text{mm} = 0.0005 \; \text{m} \) is the thickness of the tube. Substituting these values, we have:\[ J = -1.8 \times 10^{-11} \cdot \frac{304}{0.0005} = -1.0944 \times 10^{-6} \; \text{kmol/m}^2 \cdot \text{s}\].Since we want the rate per unit length, this becomes \( 1.0944 \times 10^{-6} \; \text{kmol/m} \cdot \text{s} \).
03

Convert the Flux to Mass Transfer Rate

Convert the flux from \( \text{kmol/m} \cdot \text{s} \) to \( \text{kg/m} \cdot \text{s} \). The molar mass of hydrogen is approximately \( 2.016 \; \text{kg/kmol} \). Thus, the mass transfer rate \( \dot{m} \) can be calculated as:\[ \dot{m} = 1.0944 \times 10^{-6} \times 2.016 = 2.207 \times 10^{-6} \; \text{kg/m} \cdot \text{s}\].

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Key Concepts

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

Fick's Law of Diffusion
Fick's Law of Diffusion is fundamental in understanding how substances move through materials. It describes the rate at which a substance, like hydrogen, spreads from regions of high concentration to low concentration. This law is expressed by the equation:

\[ J = -D \frac{\Delta C}{\Delta x} \]
Where:
  • \( J \) is the diffusive flux, representing the amount of substance that flows through a unit area per unit time.
  • \( D \) is the mass diffusivity, a measure of how easily a substance diffuses through a medium.
  • \( \Delta C \) is the concentration difference across the medium.
  • \( \Delta x \) is the thickness of the medium through which diffusion occurs.
Hydrogen diffusion in materials can be calculated efficiently using this formula, helping predict the rate at which hydrogen escapes or penetrates a structure. By measuring the concentration difference and applying the known diffusivity, you can find how much hydrogen passes through a certain area over time.
Hydrogen Transfer Rate
The hydrogen transfer rate explains how rapidly hydrogen moves through a medium like a tube or membrane. In our example, we compute the rate per unit length, showing the amount of hydrogen that transfers across a defined section of the tube each second.

To determine this rate, you first need the concentration difference, obtained using Fick's Law. This difference tells us how much more hydrogen there is on one side compared to the other, motivating the diffusion process.
The transfer rate is directly related to the material's diffusivity and structure thickness, offering a practical measure for designing safe and efficient systems.
  • High transfer rates may be desired in some applications for effective exchange.
  • In others, a lower rate might be necessary to maintain balance or reduce loss.
Mass Diffusivity
Mass diffusivity, denoted as \( D \), denotes how quickly a substance like hydrogen can move within a medium. This property depends on the material itself and the nature of the diffusing substance.

Units for mass diffusivity are square meters per second \((\text{m}^2/\text{s})\), illustrating the area a molecule can spread through over time.
The given value in our example is \( 1.8 \times 10^{-11} \text{ m}^2/\text{s}\).
Mass diffusivity factors into Fick's Law of Diffusion, affecting the rate of movement:
  • The higher the diffusivity, the faster the substance will move.
  • A lower value might indicate barriers or interactions within the medium that slow down the process.
Understanding this concept allows us to predict how different materials will react to hydrogen exposure, useful in applications from fuel cells to industrial tube manufacturing.
Solubility
Solubility explains how well a substance, such as hydrogen, dissolves in another material. It's a key factor in determining concentration differences in diffusion problems.

Often expressed in terms of concentration per unit pressure, like kmol/m鲁 atm, solubility helps convert pressures into concentrations using the formula:
\[ C = S \, P \]
Where:
  • \( C \) is the concentration of the dissolved gas.
  • \( S \) is the solubility coefficient.
  • \( P \) is the total pressure of the gas.
In our example, using a solubility of 160 kmol/m鲁 atm allows us to convert pressures inside and outside the tube into concentrations, crucial for applying Fick's Law. Knowing how a gas will dissolve in a material can predict behavior in systems where gas transfer is critical, such as pipelines or container designs.

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Most popular questions from this chapter

Consider a spherical organism of radius \(r_{k}\) 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-arder, homogeneous chemical reaction. (a) If a molar concentration of \(C_{\mathrm{A}}\left(r_{\mathrm{e}}\right)=C_{\mathrm{A}}\), is maintained at the surface of the organism, obtain an expression for the radial distribution of oxygen. \(C_{A}(r)\). within the organism. From your solution. can you discem any limits on applicability of the result? (b) Obtain an expression for the rate of oxygen consumption within the organism. (c) Consider an crganism of radius \(r_{c}=0.10 \mathrm{~mm}\) and a diffusicn coefficient for oxygen transfer of \(D_{\mathrm{AB}}=10^{-1} \mathrm{~m}^{2} / \mathrm{s}\). If \(C_{\mathrm{An}}=5 \times 10^{-5} \mathrm{kmol} / \mathrm{m}^{3}\) and \(k_{0}=1.2 \times 10^{-4} \mathrm{kmol} / \mathrm{s} \cdot \mathrm{m}^{3}\), what is the molar concentration of \(\mathrm{O}_{2}\) at the center of the organism?

Consider air in a closed, cylindrical container with its axis vertical and with opposite ends maintained at different temperatures. Assume that the total pressure of the air is uniform throughout the container. (a) If the bottom surface is colder than the top surface, what is the nature of conditions within the container? For example, will there be vertical gradients of the species \(\left(\mathrm{O}_{2}\right.\) and \(\left.\mathrm{N}_{2}\right)\) concentrations? Is there any motion of the air? Does mass transfer occur? (b) What is the nature of conditions within the container if it is inverted (i.e., the warm surface is now at the bottom)?

A spherical droplet of liquuid \(A\) and radius \(r_{e}\) evaporates into a stagnant layer of gas B. Derive an expession for the evaporation rate of species \(\mathrm{A}\) in terms of the saturation pressure of species A, \(p_{\mathrm{A}}\left(r_{\mathrm{n}}\right)=p_{\mathrm{A} \text { at }}\) the partial pressure of species \(\mathrm{A}\) at an arbitrary radius \(r_{\text {. }}\) \(p_{A}(r)\), the total pressure \(p\), and other pertinent quantities. Assume the droplet and the mixture are at a uniform pressure \(p\) and temperature \(T\).

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?

Pulverized coal pellets, which may be approximated as carton spheres of radius \(r_{v}=1 \mathrm{~mm}\), are bumed in a pure oxygen atmosphere at \(1450 \mathrm{~K}\) and \(\mathrm{I}\) 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 \(\hat{N}_{0_{2}}=\) \(-k_{1}^{*} C_{\mathrm{O}_{2}}\left(r_{e}\right)\), where \(k_{1}^{*}=0.1 \mathrm{~m} / \mathrm{s}\). Neglecting changes in \(r_{e}\) 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 \(1.71 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\).
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