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Chapter 5: Problem 19

For a steel alloy, it has been determined that a carburizing heat treatment of \(15 \mathrm{~h}\) duration will raise the carbon concentration to \(0.35 \mathrm{wt} \%\) at a point \(2.0 \mathrm{~mm}\) from the surface. Estimate the time necessary to achieve the same concentration at a 6.0-mm position for an identical steel and at the same carburizing temperature.

Short Answer

Expert verified
Based on Fick's second law of diffusion, it would take approximately 135 hours to achieve a carbon concentration of 0.35 wt% at a 6.0 mm position for an identical steel, given that a 0.35 wt% carbon concentration has been achieved at a 2.0 mm position after 15 hours of carburizing heat treatment.

Step by step solution

01

List the given information

From the problem statement, we know the following: - duration of carburizing heat treatment: 15 h - carbon concentration at 2.0 mm: 0.35 wt% - desired carbon concentration: 0.35 wt% - distance from the surface at which we want to achieve the desired concentration: 6.0 mm
02

Use Fick's second law of diffusion to derive the formula for the time

Fick's second law of diffusion is given by: \( \frac{\partial C}{\partial t} = D \frac{\partial^2 C}{\partial x^2} \) Given both the instances have the same steel and temperature, we can assume they have the same diffusion coefficient. Therefore, we can write diffusion equation as: \( \frac{\partial C}{\partial t} = D_1 \frac{\partial^2 C}{\partial x^2} = D_2 \frac{\partial^2 C}{\partial x^2} \) We are given data for one instance, where after 15 h, at 2.0 mm position, the concentration achieved is 0.35 wt%. We need to determine the time for the same concentration to be achieved at 6.0 mm position.
03

Determine the relative concentration

Since we are dealing with the same concentration and diffusion coefficients, we can rewrite the diffusion equation in terms of relative concentration as follows: \( \frac{\partial (\frac{C-C_\text{surface}}{C_\text{s}-C_\text{surface}})}{\partial t} = D_1 \frac{\partial^2 (\frac{C-C_\text{surface}}{C_\text{s}-C_\text{surface}})}{\partial x^2} \) We can use the boundary conditions from both instances to solve for the unknown time (t_2).
04

Use the given data to calculate t_2

For the first instance (2.0 mm and 15 h duration): \( \frac{\partial (\frac{C_1-C_\text{surface}}{C_\text{s}-C_\text{surface}})}{\partial t} = D_1 \frac{\partial^2 (\frac{C_1-C_\text{surface}}{C_\text{s}-C_\text{surface}})}{\partial x^2} = \frac{15}{x_1^2} \) For the second instance (6.0 mm and unknown time t_2): \( \frac{\partial (\frac{C_2-C_\text{surface}}{C_\text{s}-C_\text{surface}})}{\partial t} = D_2 \frac{\partial^2 (\frac{C_2-C_\text{surface}}{C_\text{s}-C_\text{surface}})}{\partial x^2} = \frac{t_2}{x_2^2} \) We can simplify these expressions to create a formula for the time t_2 as follows: \( \frac{t_2}{x_2^2} = \frac{15}{x_1^2} \) Plug in the known values for x_1 and x_2: \( x_1 = 2.0 \mathrm{~mm} \) \( x_2 = 6.0 \mathrm{~mm} \)
05

Calculate the unknown time t_2

Now we can find the value of t_2 by solving the equation: \( t_2 = \frac{15}{x_1^2} \times x_2^2 \) \(t_2 = \frac{15}{(2.0 \mathrm{~mm})^2} \times (6.0 \mathrm{~mm})^2 \) \(t_2 = 135\ \mathrm{h}\) Thus, it would take approximately 135 hours to achieve the same carbon concentration of 0.35 wt% at 6.0 mm position for the identical steel at the same carburizing temperature.

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Key Concepts

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

Fick's Second Law of Diffusion
When it comes to understanding how atoms move within materials, Fick's second law of diffusion is a central principle. It describes the rate at which particles, such as carbon atoms during carburizing, spread out over time. This law takes on the form of a partial differential equation, which is represented mathematically as:
\[ \frac{\partial C}{\partial t} = D \frac{\partial^2 C}{\partial x^2} \]
Here, \(C\) represents the concentration of the diffusing species at any given point, \(t\) is time, and \(x\) is position. The term \(D\) stands for the diffusion coefficient, a value that quantifies how quickly atoms are spreading out within the material. In the context of heat treatments like carburizing, understanding and applying Fick's law is crucial for predicting how long it will take for a certain concentration of carbon to reach a specific depth within a steel alloy.
Carbon Concentration
In the metallurgical process of carburizing, the term 'carbon concentration' refers to the amount of carbon that has diffused into the surface layer of steel. It's usually expressed in weight percent (wt%). Establishing an accurate carbon concentration is essential because it determines the final properties of the material, such as hardness and wear resistance.
In the exercise, we're given a target carbon concentration of 0.35 wt% at a certain depth from the surface. This precise measurement dictates the extent and duration of heat treatment required. By calculating the diffusion process over time, engineers can control the level of carbon within the steel to achieve the desired characteristics for specific industrial applications.
Diffusion Coefficient
The diffusion coefficient, symbolized by \(D\), is a pivotal factor in the heat treatment process. It represents how easily atoms move through a solid material and is influenced by temperature, the nature of the diffusing atoms, and the host material. The coefficient generally increases with temperature, implying that atoms can move more freely at higher temperatures.
The value of the diffusion coefficient is derived from empirical measurements and is specific to the material and diffusing species pair. In our carburizing example, we assume the diffusion coefficient remains constant since the steel type and temperature are unchanged. Understanding the coefficient's role allows us to manipulate Fick's law for practical uses, as is demonstrated in the exercise to determine the necessary time for achieving the required carbon concentration at a new depth.

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Most popular questions from this chapter

Consider the diffusion of some hypothetical metal Y into another hypothetical metal \(Z\) at \(950^{\circ} \mathrm{C}\); after \(10 \mathrm{~h}\) the concentration at the \(0.5 \mathrm{~mm}\) position (in metal \(Z\) ) is \(2.0 \mathrm{wt} \% \mathrm{Y}\). At what position will the concentration also be \(2.0 \mathrm{wt} \% \mathrm{Y}\) after a \(17.5-\mathrm{h}\) heat treatment again at \(950^{\circ} \mathrm{C}\) ? Assume preexponential and activation energy values of \(4.3 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\) and \(180,000 \mathrm{~J} / \mathrm{mol}\), respectively, for this diffusion system.

The diffusion coefficients for nickel in iron are given at two temperatures, as follows: \begin{tabular}{cc} \hline \(\boldsymbol{T}(\boldsymbol{K})\) & \(\boldsymbol{D}\left(\mathrm{m}^{2} / \mathrm{s}\right)\) \\ \hline 1473 & \(2.2 \times 10^{-15}\) \\ \hline 1673 & \(4.8 \times 10^{-14}\) \\ \hline \end{tabular} (a) Determine the values of \(D_{0}\) and the activation energy \(Q_{d^{\prime}}\) (b) What is the magnitude of \(D\) at \(1300^{\circ} \mathrm{C}(1573 \mathrm{~K}) ?\)

For the predeposition heat treatment of a semiconducting device, gallium atoms are to be diffused into silicon at a temperature of \(1150^{\circ} \mathrm{C}\) for \(2.5 \mathrm{~h}\). If the required concentration of \(\mathrm{Ga}\) at a position \(2 \mu \mathrm{m}\) below the surface is \(8 \times 10^{23}\) atoms \(/ \mathrm{m}^{3}\), compute the required surface concentration of \(\mathrm{Ga}\). Assume the following: (i) The surface concentration remains constant (ii) The background concentration is \(2 \times 10^{19} \mathrm{Ga}\) atoms \(/ \mathrm{m}^{3}\) (iii) Preexponential and activation energy values are \(3.74 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) and \(3.39 \mathrm{eV} /\) atom, respectively.

A sheet of BCC iron \(2-\mathrm{mm}\) thick was exposed to a carburizing gas atmosphere on one side and a decarburizing atmosphere on the other side at \(675^{\circ} \mathrm{C}\). 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 \(0.015\) and \(0.0068\) wt \(\%\), respectively. Compute the diffusion coefficient ifthe diffusion flux is \(7.36 \times 10^{-9} \mathrm{~kg} / \mathrm{m}^{2}+\mathrm{s}\). Hint: Use Equation \(4.9\) to convert the concentrations from weight percent to kilograms of carbon per cubic meter of iron.

Carbon is allowed to diffuse through a steel plate 10 -mm thick. The concentrations of carbon at the two faces are \(0.85\) and \(0.40 \mathrm{~kg} \mathrm{C} / \mathrm{cm}^{3} \mathrm{Fe}\), which are maintained constant. If the preexponential and activation energy are \(5.0 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) and 77,000 \(\mathrm{J} / \mathrm{mol}\), respectively, compute the temperature at which the diffusion flux is \(6.3 \times 10^{-10} \mathrm{~kg} / \mathrm{m}^{2} \mathrm{~s}\).
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