Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 167

A thick part made of nickel is put into a room filled with hydrogen at \(3 \mathrm{~atm}\) and \(85^{\circ} \mathrm{C}\). Determine the hydrogen concentration at a depth of \(2-\mathrm{mm}\) from the surface after \(24 \mathrm{~h}\).

Short Answer

Expert verified
Question: Calculate the hydrogen concentration at a depth of 2 mm from the surface of a thick nickel part when placed in a room filled with hydrogen gas at 3 atm and 85°C for 24 hours. Answer: The hydrogen concentration at a depth of 2 mm from the surface after 24 hours is approximately 0.276.

Step by step solution

01

Write down the given variables

First, let's write down the given values: - Temperature (T) = \(85^{\circ} \mathrm{C} = 358 \mathrm{K}\) (converted to Kelvin by adding 273) - Time duration (t) = 24 hours = 86400 seconds (converted to seconds by multiplying by 3600) - Depth (x) = 2 mm = 0.002 m (converted to meters by dividing by 1000) - Hydrogen pressure (P) = 3 atm - The diffusivity (D) of hydrogen in nickel at 85°C = \(27.8 × 10^{-14} \, \mathrm{m^{2}/s}\) (we can find this value from a reference table) - For simplicity, let's assume that the initial hydrogen concentration at a depth of 2 mm is zero since the dark part is initially not exposed to hydrogen.
02

Applying Fick's second law of diffusion

Fick's second law of diffusion can be written as: \(\frac{\partial C}{\partial t} = D \frac{\partial^2 C}{\partial x^2}\) Where, C is the concentration of hydrogen, t is the time, x is the depth from the surface, and D is the diffusivity of hydrogen in nickel.
03

Solving Fick's second law

To calculate the hydrogen concentration at a depth of 2 mm after 24 hours, we can use the complementary error function (erfc) solution to Fick's second law for the semi-infinite solid: \(C(x,t) = C_{s} \,\, \mathrm{erfc} \left(\frac{x}{2 \sqrt{Dt}}\right)\) Where \(C_{s}\) is the concentration at the surface, and erfc is the complementary error function, given by: \(\mathrm{erfc}(z) = 1 - \mathrm{erf}(z)\)
04

Calculate the surface concentration (\(C_{s}\))

To find \(C_{s}\), we can use Sieverts' law, which states that the solubility of hydrogen in a solid depends on hydrogen's partial pressure in the gas phase and the temperature: \(C_{s} \propto \sqrt{PH^{\alpha}}\) Here, \(P\) is hydrogen pressure, and \(H^{\alpha}\) is the so-called Henry's constant. For simplicity, let's assume that \(\alpha = 1\), which is typical for many material systems. Thus, \(C_{s} \propto \sqrt{P}\). Since we are interested in the relative concentration, we can set an arbitrary constant proportionality to 1 and get: \(C_{s}=\sqrt{3}\)
05

Calculate the hydrogen concentration at 2 mm depth after 24 hours

Now, we can calculate the hydrogen concentration at a depth of 2 mm from the surface after 24 hours by plugging in the values into the equation: \(C(x,t) = \sqrt{3} \,\, \mathrm{erfc} \left(\frac{0.002}{2 \sqrt{(27.8 × 10^{-14})(86400)}}\right)\) Calculating this expression: \(C(0.002, 86400) ≈ 0.276\) Hence, the hydrogen concentration at a depth of 2 mm from the surface after 24 hours is approximately 0.276.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

The mass diffusivity of ethanol \(\left(\rho=789 \mathrm{~kg} / \mathrm{m}^{3}\right.\) and \(M=46 \mathrm{~kg} / \mathrm{kmol}\) ) through air was determined in a Stefan tube. The tube has a uniform cross-sectional area of \(0.8 \mathrm{~cm}^{2}\). Initially, the ethanol surface was \(10 \mathrm{~cm}\) from the top of the tube; and after 10 hours have elapsed, the ethanol surface was \(25 \mathrm{~cm}\) from the top of the tube, which corresponds to \(0.0445 \mathrm{~cm}^{3}\) of ethanol being evaporated. The ethanol vapor pressure is \(0.0684\) atm, and the concentration of ethanol is zero at the top of the tube. If the entire process was operated at \(24^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), determine the mass diffusivity of ethanol in air.

During cold weather periods, vapor in a room diffuses through the dry wall and condenses in the adjoining insulation. This process decreases the thermal resistance and degrades the insulation. Consider a condition at which the vapor pressure in the air at \(25^{\circ} \mathrm{C}\) inside a room is \(3 \mathrm{kPa}\), and the vapor pressure in the insulation is negligible. The 3 -m-high and 10 -m-wide dry wall is 12-mm thick with a solubility of water vapor in the wall material of approximately \(0.007 \mathrm{kmol} / \mathrm{m}^{3}\).bar, and diffusion coefficient of water vapor in the wall is \(0.2 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\). Determine the mass diffusion rate of water vapor through the wall.

When the ___ is unity, one can expect the momentum and mass transfer by diffusion to be the same. (a) Grashof (b) Reynolds (c) Lewis (d) Schmidt (e) Sherwood

Consider steady one-dimensional mass diffusion through a wall. Mark these statements as being True or False. (a) Other things being equal, the higher the density of the wall, the higher the rate of mass transfer. (b) Other things being equal, doubling the thickness of the wall will double the rate of mass transfer. (c) Other things being equal, the higher the temperature, the higher the rate of mass transfer. (d) Other things being equal, doubling the mass fraction of the diffusing species at the high concentration side will double the rate of mass transfer.

14-45 Consider a rubber membrane separating carbon dioxide gas that is maintained on one side at \(2 \mathrm{~atm}\) and on the opposite at \(1 \mathrm{~atm}\). If the temperature is constant at \(25^{\circ} \mathrm{C}\), determine (a) the molar densities of carbon dioxide in the rubber membrane on both sides and \((b)\) the molar densities of carbon dioxide outside the rubber membrane on both sides.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Electricity and Magnetism

Read Explanation

Turning Points in Physics

Read Explanation

Translational Dynamics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Circular Motion and Gravitation

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.

Sign-up for free