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Chapter 5: Problem 9

When \(\alpha\)-iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, \(C_{\mathrm{H}}\) (in weight percent), is a function of hydrogen pressure, \(p_{\mathrm{H}_{2}}\) (in MPa), and absolute temperature \((T)\) according to $$ C_{\mathrm{H}}=1.34 \times 10^{-2} \sqrt{p_{\mathrm{H}}_{2}} \exp \left(-\frac{27.2 \mathrm{~kJ} / \mathrm{mol}}{R T}\right) $$ Furthermore, the values of \(D_{0}\) and \(Q_{d}\) for this diffusion system are \(1.4 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) and \(13,400 \mathrm{~J} / \mathrm{mol}\), respectively. Consider a thin iron membrane \(1 \mathrm{~mm}\) thick that is at \(250^{\circ} \mathrm{C}\). Compute the diffusion flux through this membrane if the hydrogen pressure on one side of the membrane is \(0.15 \mathrm{MPa}\) (1.48 atm), and on the other side \(7.5 \mathrm{MPa}\) (74 atm).

Short Answer

Expert verified
Question: Calculate the diffusion flux of hydrogen through a thin iron membrane with a thickness of 1 mm, given that the hydrogen concentration \(C_H\) depends on H2 pressure and temperature according to the formula \(C_H = 1.34 \times 10^{-2} \sqrt{p_{H2}}\exp(-\frac{27.2 \mathrm{~kJ/mol}}{R T})\), the diffusion constant \(D_0 = 1.4 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\), and the activation energy \(Q_d = 13,400 \mathrm{~J} / \mathrm{mol}\). The pressures on either side of the membrane are \(0.15 \mathrm{MPa}\) and \(7.5 \mathrm{MPa}\), and the temperature is \(250^{\circ} \mathrm{C}\). Answer: To find the diffusion flux of hydrogen through the membrane, we need to follow these steps: 1. Calculate the hydrogen concentration on both sides of the membrane using the given formula and pressures. 2. Calculate the diffusion coefficient at the given temperature using the Arrhenius equation and the given values for \(D_0\) and \(Q_d\). 3. Use Fick's first law to find the diffusion flux by inserting the calculated values for hydrogen concentrations, diffusion coefficient, and membrane thickness. After calculating these values, you can substitute them into Fick's first law formula, \(J = -D \frac{C_H2 - C_H1}{\Delta x}\), to determine the diffusion flux of hydrogen through the iron membrane.

Step by step solution

01

Calculate hydrogen concentration on both sides of the membrane

To determine the hydrogen concentration on both sides of the membrane, we will use the given formula for \(C_H\) as a function of pressure and temperature: $$ C_{\mathrm{H}}=1.34 \times 10^{-2} \sqrt{p_{\mathrm{H}}_{2}} \exp \left(-\frac{27.2 \mathrm{~kJ} / \mathrm{mol}}{R T}\right) $$ The pressure on one side of the membrane is \(0.15 \mathrm{MPa}\), and on the other side is \(7.5 \mathrm{MPa}\). The temperature is given as \(250^{\circ} \mathrm{C}\), which we need to convert to Kelvin by adding \(273\): $$ T = 250 + 273 = 523 \mathrm{K} $$ Now we can plug in the values to find \(C_H\) on both sides: $$ C_{\mathrm{H1}}=1.34 \times 10^{-2} \sqrt{0.15} \exp \left(-\frac{27.2 \times 10^3}{8.314 \times 523}\right) $$ $$ C_{\mathrm{H2}}=1.34 \times 10^{-2} \sqrt{7.5} \exp \left(-\frac{27.2 \times 10^3}{8.314 \times 523}\right) $$
02

Calculate the diffusion coefficient

Now, we will find the diffusion coefficient \(D\) using the Arrhenius equation: $$ D = D_0 \exp \left( -\frac{Q_d}{RT} \right) $$ The given values for \(D_0\) and \(Q_d\) are \(1.4 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) and \(13,400 \mathrm{~J} / \mathrm{mol}\), respectively. The temperature is already in Kelvin (\(523 \mathrm{K}\)). We can now find the diffusion coefficient \(D\): $$ D = (1.4 \times 10^{-7}) \exp \left( -\frac{13,400}{8.314 \times 523} \right) $$
03

Determine the diffusion flux using Fick's first law

Finally, we will use Fick's first law to compute the diffusion flux through the membrane: $$ J = -D \frac{C_H2 - C_H1}{\Delta x} $$ The thickness of the membrane is given as \(1 \mathrm{~mm}\) which we need to convert to meters before we can use it in the equation: $$ \Delta x = 1 \mathrm{~mm} \times \frac{1}{1,000} = 1 \times 10^{-3} \mathrm{m} $$ Now, we can use the calculated values for \(C_H1\), \(C_H2\), \(D\), and \(\Delta x\) to find the diffusion flux \(J\): $$ J = -D \frac{C_H2 - C_H1}{1 \times 10^{-3}} $$

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Key Concepts

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

Fick's First Law
Fick's First Law is fundamental in the field of materials science, helping us understand how atoms and molecules move through various substances. This law states that the rate of diffusion, or the movement of particles, is proportional to the concentration gradient.

In simpler terms, it relates to how stuff moves from an area where there's a lot of it to an area with less of it. The formula to describe this phenomenon is: $$ J = -D \frac{\Delta C}{\Delta x} $$ where \( J \) represents the diffusion flux, \( D \) is the diffusion coefficient, \( \Delta C \) is the change in concentration, and \( \Delta x \) is the distance over which the concentration changes.

The negative sign indicates that diffusion occurs in the opposite direction of increasing concentration. Understanding this law is key to many processes, such as the treatment of iron and steel to change their properties, where controlling the movement of atoms like carbon or hydrogen is essential.
Arrhenius Equation
The Arrhenius equation is another cornerstone concept in materials science, especially in the context of diffusion. It provides a quantitative basis for the effect of temperature on the rate of a chemical reaction—or in the case of diffusion, the mobility of atoms and molecules in a solid. The Arrhenius equation is given by: $$ k = A \exp \left( -\frac{E_a}{RT} \right) $$ where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the absolute temperature in kelvin.

This equation shows that as the temperature increases, the rate constant also increases, making diffusion faster. It's crucial for understanding how to control processes like hardening steel or manufacturing semiconductors, where precise temperatures are required to achieve the desired rate of diffusion.
Hydrogen Diffusion in Iron
Finally, let's focus on a specific example using the concepts we've learned: hydrogen diffusion in iron. This process is significant in metallurgy and materials science because hydrogen can affect the properties of iron, including its mechanical strength and ductility.

The rate at which hydrogen diffuses into iron depends on factors like temperature and hydrogen concentration. Higher temperatures or higher hydrogen concentrations will generally increase the rate of diffusion. For instance, when manufacturing high-strength steel, controlling hydrogen diffusion is essential to prevent 'hydrogen embrittlement,' a phenomenon that leads to the premature failure of metal parts.

In exercises like the one from the textbook, we use these concepts to calculate how much hydrogen moves through an iron membrane under specific conditions. Such calculations are critical in designing industrial processes like gas purification systems or hydrogen storage materials, where the control of hydrogen diffusion is vital for efficiency and safety.

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Most popular questions from this chapter

Briefly explain the difference between selfdiffusion and interdiffusion.

Phosphorus atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat treatments; the background concentration of \(\mathrm{P}\) in this silicon material is known to be \(5 \times 10^{19}\) atoms \(/ \mathrm{m}^{3}\). The predeposition treatment is to be conducted at \(950^{\circ} \mathrm{C}\). for 45 minutes; the surface concentration of \(P\) is to be maintained at a constant level of \(1.5 \times 10^{26}\) atoms \(/ \mathrm{m}^{3}\). Drive-in diffusion will be carried out at \(1200^{\circ} \mathrm{C}\) for a period of \(2.5 \mathrm{~h}\). For the diffusion of \(\mathrm{P}\) in \(\mathrm{Si}\), values of \(Q_{d}\) and \(D_{0}\) are \(3.40 \mathrm{eV} /\) atom and \(1.1 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\), respectively. (a) Calculate the value of \(Q_{0}\). (b) Determine the value of \(x_{j}\) for the drive-in diffusion treatment. (c) Also for the drive-in treatment, compute the position \(x\) at which the concentration of \(P\) atoms is \(10^{24} \mathrm{~m}^{-3}\).

An FCC iron-carbon alloy initially containing \(0.35 \mathrm{wt} \% \mathrm{C}\) is exposed to an oxygen-rich and virtually carbon-free atmosphere at \(1400 \mathrm{~K}\) (1127 \(\left.^{\circ} \mathrm{C}\right)\). Under these circumstances the carbon diffuses from the alloy and reacts at the surface, with the oxygen in the atmosphere; that is, the carbon concentration at the surface position is maintained essentially at \(0 \mathrm{wt} \%\) C. (This process of carbon depletion is termed decarburization.) At what position will the carbon concentration be \(0.15 \mathrm{wt} \%\) after a 10 -h treatment? The value of \(D\) at \(1400 \mathrm{~K}\) is \(6.9 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\).

A sheet of steel \(1.5 \mathrm{~mm}\) thick has nitrogen atmospheres on both sides at \(1200^{\circ} \mathrm{C}\) and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is \(6 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\), and the diffusion flux is found to be \(1.2 \times\) \(10^{-7} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\). Also, it is known that the concentration of nitrogen in the steel at the highpressure surface is \(4 \mathrm{~kg} / \mathrm{m}^{3} .\) How far into the sheet from this high-pressure side will the concentration be \(2.0 \mathrm{~kg} / \mathrm{m}^{3}\) ? Assume a linear concentration profile.

A sheet of BCC iron \(1 \mathrm{~mm}\) thick was exposed to a carburizing gas atmosphere on one side and a decarburizing atmosphere on the other side at \(725^{\circ} \mathrm{C}\). After reaching steady state, the iron was quickly cooled to room temperature. The carbon concentrations at the two surfaces of the sheet were determined to be \(0.012\) and \(0.0075 \mathrm{wt} \%\). Compute the diffusion coefficient if the diffusion flux is \(1.4 \times 10^{-8} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\). Hint: Use Equation \(4.9\) to convert the concentrations from weight percent to kilograms of carbon per cubic meter of iron.
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