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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掳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掳C = \(27.8 脳 10^{-14} \, \mathrm{m^{2}/s}\) (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: \(C(x,t) = \sqrt{3} \,\, \mathrm{erfc} \left(\frac{0.002}{2 \sqrt{(27.8 脳 10^{-14})(86400)}}\right)\) Calculating this expression: \(C(0.002, 86400) 鈮� 0.276\) Hence, the hydrogen concentration at a depth of 2 mm from the surface after 24 hours is approximately 0.276.

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Key Concepts

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

Fick's Second Law
Fick's Second Law of Diffusion provides a mathematical description for the rate of change of concentration within a medium. It states that the rate of change of concentration within a volume element is proportional to the second spatial derivative of concentration.
This can be mathematically expressed as: \[\frac{\partial C}{\partial t} = D \frac{\partial^2 C}{\partial x^2}\]Where:
  • \( C \) is the concentration of the diffusing species (in this case, hydrogen).
  • \( t \) is time.
  • \( x \) is space or depth.
  • \( D \) is the diffusion coefficient, specific to the pair of materials under consideration (nickel and hydrogen).

Fick鈥檚 Second Law is particularly useful for predicting how the concentration of hydrogen changes over time at various depths within nickel.
Diffusion
Diffusion is the process of movement of particles from an area of high concentration to an area of low concentration. It is a fundamental concept in many scientific disciplines and describes how substances like gases, liquids, or even dissolved solids spread through a mixture.
The driving force behind diffusion is the concentration gradient, which is inherently tied to randomness and the tendency of systems to spread out in order to reach equilibrium.
In the context of this exercise, diffusion accounts for hydrogen molecules moving into the nickel metal until the concentrations balance. Understanding diffusion helps explain how materials absorb gases and how gases move within solids over time.
Concentration Gradient
A concentration gradient occurs when there is a difference in the concentration of a substance in different areas. This gradient drives the diffusion process, causing the substance to move from regions of high concentration to regions of lower concentration.
Mathematically, a concentration gradient is the rate of change of concentration with respect to distance. It is central in calculating how quickly and efficiently substances can diffuse through another medium.
In our exercise, the concentration gradient is between the surface of the nickel, which is exposed to high concentrations of hydrogen, and the internal structure at 2 mm depth, which initially had no hydrogen. Hence, maintaining the gradient encourages diffusion over the 24 hour period.
Hydrogen Solubility
Hydrogen solubility refers to how much hydrogen can be dissolved in a particular substance, like nickel, at a certain temperature and pressure.
It is commonly modeled using Sieverts' Law, which states that the solubility of hydrogen is proportional to the square root of the pressure of hydrogen gas in contact with the solid: \[ C_{s} \propto \sqrt{P} \] Where:
  • \( C_{s} \) is the surface concentration of hydrogen in the solid.
  • \( P \) is the pressure of hydrogen.
Understanding hydrogen solubility is crucial for applications where hydrogen is involved, such as in hydrogen storage materials, fuel cells, and metallurgy. In this problem, hydrogen solubility helps estimate the initial concentration at the surface, which is essential to solving for the concentration at 2 mm depth over time.

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Most popular questions from this chapter

The average heat transfer coefficient for air flow over an odd-shaped body is to be determined by mass transfer measurements and using the Chilton-Colburn analogy between heat and mass transfer. The experiment is conducted by blowing dry air at \(1 \mathrm{~atm}\) at a free stream velocity of \(2 \mathrm{~m} / \mathrm{s}\) over a body covered with a layer of naphthalene. The surface area of the body is \(0.75 \mathrm{~m}^{2}\), and it is observed that \(100 \mathrm{~g}\) of naphthalene has sublimated in \(45 \mathrm{~min}\). During the experiment, both the body and the air were kept at \(25^{\circ} \mathrm{C}\), at which the vapor pressure and mass diffusivity of naphthalene are \(11 \mathrm{~Pa}\) and \(D_{A B}=0.61 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively. Determine the heat transfer coefficient under the same flow conditions over the same geometry.

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