/*! 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 51 A vitreous silica optical fiber ... [FREE SOLUTION] | 91Ó°ÊÓ

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A vitreous silica optical fiber of diameter \(100 \mu \mathrm{m}\) is used to send optical signals from a sensor placed deep inside a hydrogen chamber. The hydrogen is at a pressure of 20 bars. The mass diffusivity and solubility of the hydrogen in the glass fiber are \(D_{A B}=2.88 \times 10^{-15} \mathrm{~m}^{2} / \mathrm{s}\) and \(S=\) \(4.15 \times 10^{-3} \mathrm{kmol} / \mathrm{m}^{3}\) - bar, respectively. Hydrogen diffusion into the fiber is undesirable, since it changes the spectral transmissivity and refractive index of the glass and can lead to failure of the detection system. (a) Determine the average hydrogen concentration in an uncoated optical fiber, \(\bar{C}\), after 100 hours of operation in the hydrogen environment. Determine the corresponding change in the refractive index, \(\Delta n\), of the fiber. For vitreous silica, \(\Delta n=\left(1.6 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{kmol}\right) \times \bar{C} .\) (b) Determine the average hydrogen concentration and change in refractive index after 1 hour and 10 hours of operation in the hydrogen environment.

Short Answer

Expert verified
After 100 hours, \( \bar{C} \) and \( \Delta n \) are calculated with exponential decay; repeat steps for 1 and 10 hours.

Step by step solution

01

Calculate Pressure Influence

First, convert the pressure from bars to volts using the solubility constant. Since solubility, \( S \), is given as \( 4.15 \times 10^{-3} \) kmol/m³-bar, at 20 bars the concentration without the time factor is \( C = P \times S = 20 \times 4.15 \times 10^{-3} = 83 \times 10^{-3} \) kmol/m³.
02

Calculate Concentration Over Time for 100 Hours

Apply Fick's second law for unsteady-state diffusion to find the concentration after 100 hours. The average concentration, \( \bar{C} \), in the fiber is calculated using the approximation for short times: \[\bar{C} = C \left(\frac{8}{\pi^2}\right) \exp\left( - \frac{D_{AB} \times t \times \pi^2}{L^2} \right)\]where \( C \) is the concentration from step 1, \( D_{AB} \) is the mass diffusivity, \( t \) is the time (100 hours = 360,000s), and \( L \) is the fiber diameter (0.0001m). Substituting leads to the calculated value.
03

Calculate Change in Refractive Index for 100 Hours

Using the previously found average concentration \( \bar{C} \), the change in the refractive index is given by:\[\Delta n = \left(1.6 \times 10^{-3} \right) \times \bar{C}\]Substitute \( \bar{C} \) from step 2 to find \( \Delta n \).
04

Calculate Concentrations for 1 Hour and 10 Hours

Repeat steps similar to Step 2 for 1 hour (3600 seconds) and 10 hours (36,000 seconds). Use the same approximation of Fick's second law with respective time values to find \( \bar{C} \) at these specific times.
05

Calculate Change in Refractive Index for 1 Hour and 10 Hours

Use the concentration values obtained from Step 4 to calculate \( \Delta n \) at 1 hour and 10 hours using the formula:\[\Delta n = \left(1.6 \times 10^{-3} \right) \times \bar{C}\]Substitute the respective \( \bar{C} \) values to find \( \Delta n \) for 1 hour and 10 hours.

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Key Concepts

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

optical fibers
Optical fibers are thin strands of specialized glass or plastic used to transmit light over long distances. These fibers are crucial for high-speed internet and communications because they can carry large amounts of data with minimal loss. Their core is made of pure silica glass, surrounded by a cladding that reflects light back into the core, ensuring efficient transmission. Optical fibers are favored for their ability to maintain signal integrity over long distances and through challenging environments.
However, these fibers are sensitive to external influences, such as temperature changes or chemical exposure. In particular, hydrogen can diffuse into the fiber, altering its physical properties. This diffusion can affect the spectral transmissivity and refractive index, leading to signal degradation or system failure. Understanding how to manage these influences through controlled environments or protective coatings is vital in maintaining the performance of optical fiber systems.
Fick's second law
Fick's second law of diffusion is a core concept in understanding how particles move through different mediums over time. It is particularly applicable to situations involving time-dependent changes, such as hydrogen diffusion in optical fibers. Mathematically, this law is expressed as: \[ \frac{\partial C}{\partial t} = D \frac{\partial^2 C}{\partial x^2} \] where:
  • \( C \) is the concentration of the diffusing substance (hydrogen in this case),
  • \( D \) is the diffusion coefficient or mass diffusivity,
  • \( t \) is time, and
  • \( x \) is position within the medium.
Fick's second law helps predict how the concentration of hydrogen changes inside the optical fiber over time. This prediction allows engineers to evaluate how the fiber's properties, like refractive index, will be affected. Being able to calculate these changes is essential for ensuring the fibers' performance and reliability in various applications.
refractive index change
The refractive index is a measure of how much light bends when it enters a material. For optical fibers, maintaining a consistent refractive index is crucial for efficiently transmitting light signals. Changes in the refractive index can disrupt the path of light, causing loss and reducing the quality of data transmission.
When hydrogen diffuses into an optical fiber, it affects the refractive index. This happens because the molecular structure of the glass changes, altering how light interacts with it. The change in refractive index, denoted \(\Delta n\), can be calculated by multiplying the hydrogen concentration by a constant that relates concentration to refractive index change. For vitreous silica, this constant is \(1.6 \times 10^{-3} \ \text{m}^3/ ext{kmol}\). Through this relation, engineers can predict and mitigate the potential effects of environmental exposure on optical fiber performance.
mass diffusivity
Mass diffusivity, often denoted as \( D_{AB} \), is a measure of how quickly particles such as hydrogen diffuses through another material, like the glass of an optical fiber. It is expressed in units of area per time, \( \text{m}^2/ ext{s} \). The higher the mass diffusivity, the faster the diffusion process occurs.
In the context of optical fibers, mass diffusivity determines how vulnerable the fiber is to changes caused by external substances such as gases. A low mass diffusivity, as in the case of hydrogen in a glass fiber \((2.88 \times 10^{-15} \ \text{m}^2/ ext{s})\), suggests that diffusion occurs slowly, which may be beneficial for maintaining fiber integrity over time. Understanding mass diffusivity helps in designing fibers for specific environments, ensuring they can withstand particular molecules' penetration, thus preserving performance and reliability.

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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?

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.

A thin plastic membrane is used to separate helium from a gas stream. Under steady-stare conditions the concentration of helium in the membrane is known to be \(0.02\) and \(0.005 \mathrm{kmol} / \mathrm{m}^{3}\) at the inner and outer surfaces, respectively. If the membrane is \(1 \mathrm{~mm}\) thick and the binary diffusion coefficient of helium with respect to the plastic is \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), what is the diffusive flux?

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 \(1 \mathrm{~mm}\) ?

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)?

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