/*! 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 17 Gaseous bydrogen at 10 bars and ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 17

Gaseous bydrogen at 10 bars and \(27^{\circ} \mathrm{C}\) is stored in a 100 -mm-diameter spherical tank having a steel wall \(2 \mathrm{~mm}\) thick. The molar concentration of hydrogen in the steel is \(1.50 \mathrm{kmol} / \mathrm{m}^{3}\) at the inner surface and negligible at the outer surface, while the diffusion coeffcient of hydrogen in steel is approximately \(0.3 \times\) \(10^{-12} \mathrm{~m}^{2} / \mathrm{s}\). What is the initial rate of mass loss of hydrogen by diffusion through the tank wall? What is the initial rate of pressure drop within the tank?

Short Answer

Expert verified
The initial rate of mass loss is \(-7.065 \times 10^{-12} \, \text{kmol/s}\), and pressure drops at \(-3.38 \times 10^{-5} \, \text{bar/s}\).

Step by step solution

01

Understand the problem

We are tasked with finding the initial rate of mass loss of hydrogen by diffusion through a steel tank wall, and the initial rate of pressure drop inside the tank. This is a classic diffusion problem where Fick's Law can be applied.
02

Determine the relevant formula

To find the mass loss rate of hydrogen, we can use Fick's First Law of Diffusion: \[ J = -D \frac{dC}{dx} \]where \( J \) is the flux (mol/m虏/s), \( D \) is the diffusion coefficient (m虏/s), and \( \frac{dC}{dx} \) is the concentration gradient (kmol/m鲁/m).
03

Calculate the concentration gradient

The concentration gradient across the tank wall is given by the difference in concentration divided by the wall thickness:\[ \frac{dC}{dx} = \frac{1.50 \, \text{kmol/m}^3 - 0}{0.002 \, \text{m}} = 750 \, \text{kmol/m}^4 \]
04

Calculate the flux of hydrogen

Substitute the values into Fick's Law:\[ J = - (0.3 \times 10^{-12} \, \text{m}^2/ ext{s}) \times 750 \, \text{kmol/m}^4 = -2.25 \times 10^{-10} \, \text{kmol/m}^2/ ext{s} \]
05

Find the surface area of the sphere

The surface area \( A \) of a sphere is calculated using:\[ A = 4 \pi r^2 \] The inner radius is half the diameter: \( r = 0.05 \, \text{m} \). Therefore, \[ A = 4 \pi (0.05)^2 = 0.0314 \, \text{m}^2 \]
06

Compute the mass loss rate

The rate of mass loss \( \dot{n} \) in kmol/s is the flux multiplied by the area: \[ \dot{n} = J \times A = -2.25 \times 10^{-10} \, \text{kmol/m}^2/ ext{s} \times 0.0314 \, \text{m}^2 = -7.065 \times 10^{-12} \, \text{kmol/s} \]
07

Convert mass loss rate to pressure drop rate

Use the ideal gas law in molar form to relate moles and pressure, \( PV = nRT \). Assume the gas behaves ideally and rearrange to find:\[ \frac{dP}{dt} = - \frac{RT}{V} \dot{n} \] where \( V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.05)^3 \approx 5.24 \times 10^{-4} \, \text{m}^3 \) and \( R = 8.314 \, \text{J/mol} \, \text{K} \), \( T = 273 + 27 = 300 \, \text{K} \).
08

Final computation of pressure drop

Compute the pressure drop rate using:\[ \frac{dP}{dt} = - \frac{8.314 \times 300}{5.24 \times 10^{-4}} (7.065 \times 10^{-12}) \approx -3.38 \times 10^{-5} \, \text{bar/s} \] at the initial moment.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Hydrogen Diffusion
Hydrogen diffusion refers to the process where hydrogen atoms move from regions of higher concentration to regions of lower concentration, through a medium like a steel wall. This movement is a result of the random motion of hydrogen molecules and can be described by Fick鈥檚 Law of Diffusion. The diffusion of hydrogen is of particular interest in various industrial applications where hydrogen is stored or passed through metal containers.
To predict how quickly hydrogen will diffuse through a material, it is important to consider the diffusion coefficient, which measures how easily hydrogen molecules can move within the material. In the original exercise, the diffusion coefficient for hydrogen in steel is given as approximately 0.3 x 10^{-12} m^2/s, indicating slow movement as hydrogen passes through the steel wall. For practical uses, engineers and scientists need to understand and control this diffusion to prevent unwanted loss of hydrogen and maintain safety.
Concentration Gradient
The concentration gradient is the driving force for diffusion. It represents the change in concentration of a substance, like hydrogen, over a specific distance within a material. In simpler terms, it's the difference in the amount of hydrogen on one side of the steel wall compared to the other.
Mathematically, the concentration gradient can be expressed as \( \frac{dC}{dx} \), which indicates how steeply the concentration changes across the thickness of the wall. In the exercise, the concentration of hydrogen is 1.50 kmol/m^3 at the inner surface of the tank and nearly zero at the outer surface. Therefore, the concentration gradient is computed as 750 kmol/m^4 by dividing this concentration difference by the wall thickness of 2 mm. This gradient is crucial for determining the rate at which hydrogen molecules will diffuse through the steel.
Initial Rate of Mass Loss
The initial rate of mass loss in the context of hydrogen diffusion is a measure of how fast the hydrogen gas is escaping through the steel wall over time. This can be calculated using Fick鈥檚 First Law of Diffusion, which combines the diffusion coefficient, concentration gradient, and the surface area where diffusion occurs.
In the exercise provided, the formula \( J = -D \frac{dC}{dx} \) gives the diffusion flux \( J \), or the rate of hydrogen flow through the tank wall in moles per square meter per second. By multiplying this flux by the surface area of the tank, we derive the initial rate of mass loss \( \dot{n} \), calculated as -7.065 x 10^{-12} kmol/s, meaning a very small but continual loss of hydrogen from the tank, which could impact storage efficiency.
Pressure Drop Rate
The pressure drop rate is a measure of how quickly the pressure inside the tank decreases as hydrogen diffuses through the tank wall. This is directly related to the rate of mass loss of hydrogen and can be derived using the ideal gas law.The ideal gas law \( PV = nRT \) helps connect the pressure, volume, and temperature of the gas in the tank with the number of moles lost due to diffusion. By rearranging this law, we find \( \frac{dP}{dt} = - \frac{RT}{V} \dot{n} \), which calculates how pressure changes over time. In the exercise, given that \( R = 8.314 \) J/mol/K and the volume is approximately 5.24 x 10^{-4} m^3 with a temperature at 300 K, the initial pressure drop rate is computed to be around -3.38 x 10^{-5} bar/s. This shows a slow but steady decrease in the tank's pressure, impacting how hydrogen storage and utilization are managed.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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

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?

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?

Insulation degrades (experiences an increase in thermal conductivity) if it is subjected to water vapor condensation. The problem may occur in home insulation during cold periods, when vapor in a humidified room diffuses through the dry wall (plaster board) and condenses in the adjoining insulation. Estimate the mass diffusion rate for a \(3 \mathrm{~m}\) by \(5 \mathrm{~m}\) wall, under conditions for which the vapor pressure is \(0.03\) bar in the room air and \(0.0\) bar in the insulation. The dry wall is \(10 \mathrm{~mm}\) thick, and the solubility of water vapor in the wall material is approximately \(5 \times 10^{-3} \mathrm{kmol} / \mathrm{m}^{3}\). bar. The binary diffusion coefficient for water vapor in the dry wall is approximately \(10^{-9} \mathrm{~m}^{2} / \mathrm{s}\).

Ultra-pure hydrogen is required in applications ranging from the manufacturing of semiconductors to powcring fuel cells. The crystalline structure of palladium allows only the transfer of atomic hydrogen (H) through its thickness, and therefore palladium membranes are used to filter hydrogen from contaminated streams containing mixtures of hydrogen and other gases. Hydrogen molecules \(\left(\mathrm{H}_{2}\right)\) are first adserbed onto the palladium's surface and ate then dissociated into atoms (H), which subsequently diffuse through the metal. The H atoms recombine on the opposite side of the membrane, forming pure \(\mathrm{H}_{2}\). The surface concentration of H takes the form \(C_{\mathrm{H}}=K_{S} p_{H_{1},}^{09}\), where \(K_{,}=\) \(1.4 \mathrm{kmol}^{3} \mathrm{~m}^{3} \cdot\) bar \(^{\mathrm{ns}}\) is known as Sievert's constant. Consider an industrial hydrogen purifier consisting of an array of palladium tubes with one tube cnd connected to a collector plenum and the other end closed. The tube bank is inserted into a shell. Impure \(\mathrm{H}_{2}\) at \(T=600 \mathrm{~K}, p=15\) bars, \(x_{\mathrm{H}_{2}}=0.85\) is introduced into the shell while pure \(\mathrm{H}_{2}\) at \(p=6\) bars, \(T=600 \mathrm{~K}\) is extracted through the tubes. Determine the production rate of pure hydrogen \((\mathrm{kg} / \mathrm{h})\) for \(N=100\) tubes which are of inside diameler \(D_{0}=1.6 \mathrm{~mm}\), wall thickness \(t=\) \(75 \mu \mathrm{m}\), and length \(L=80 \mathrm{~mm}\). The mass diffusivity of hydrogen (H) in palladium at \(600 \mathrm{~K}\) is approximately \(D_{N A}=7 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\).
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Energy Physics

Read Explanation

Thermodynamics

Read Explanation

Astrophysics

Read Explanation

Rotational Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.