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Experimental oxidation data for a titanium alloy, used to sheath hypersonic vehicle nose cones, indicates that the total weight gain can be described by a parabolic law of the form \(w^{2}=C t\), where \(C=480 \exp \left(-E_{a} / \mathscr{R T}\right) \mathrm{g}^{2} / \mathrm{cm}^{4} \mathrm{~s}\), and \(E_{a}=61,800\) \(\mathrm{cal} / \mathrm{mol}\). Examination of oxidized samples shows that about \(20 \%\) of the oxygen goes to form a rutile \(\left(\mathrm{Ti} \mathrm{O}_{2}\right)\) surface scale. Assuming that the diffusion coefficient for oxygen in alpha titanium can be expressed as \(\mathscr{D}_{12}=\mathscr{D}_{0} \exp \left(-E_{a} / \mathscr{R} T\right)\), with \(E_{a}\) again \(61,800 \mathrm{cal} / \mathrm{mol}\), estimate \(\mathscr{D}_{0}\). Show also that the parabolic oxidation law implies diffusion across the rutile scale characterized by a linear concentration gradient. Use an alloy density of \(4400 \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
Using the laws of diffusion and parabolic oxidation, we find \(\mathscr{D}_{0}\) by aligning exponential and diffusion coefficients; confirms linear gradient.

Step by step solution

01

Understand the Parabolic Law

The weight gain formula given is a parabolic law: \(w^2 = Ct\). Here, \(w\) is the weight gain per unit area of the titanium alloy, \(t\) is time, and \(C\) is a constant dependent on temperature and activation energy.
02

Analyze Oxygen Contribution to Rutile Scale

Approximately 20% of the oxygen turns into a rutile (\(\text{TiO}_2\)) scale. This means that if \(w\) is the total weight gain due to oxidation, \(0.2w\) accounts for the rutile scale formed on the surface.
03

Relate Parabolic Law to Diffusion

The parabolic law \(w^2 = Ct\) suggests that diffusion (oxygen diffusion in this case) is the rate-controlling step, characterized by a linear concentration gradient across the rutile scale. The linear gradient implies the differential equation for diffusion in steady state \(\frac{dC}{dx} = \text{constant}\), where \(C\) is concentration and \(x\) is the position across the scale.
04

Calculate the Diffusion Coefficient \(\mathscr{D}_{12}\)

The diffusion coefficient \(\mathscr{D}_{12}\) for oxygen in the alpha titanium is given by \(\mathscr{D}_{12} = \mathscr{D}_{0}\exp\left(-\frac{E_{a}}{\mathscr{R} T}\right)\). Given, \(E_{a} = 61800 \ \text{cal/mol}\). The task is to find \(\mathscr{D}_{0}\).
05

Determine \(\mathscr{D}_{0}\) Using Weight Gain Expression

Equate the expressions for \(C\) related to both diffusion and parabolic law since they share the same activation energy \(E_{a}\). Given that \(C=480 \exp \left(\frac{-E_{a}}{\mathscr{R} T}\right)\), realize that \(C\propto \mathscr{D}_{12}\). Since they share the same exponential, solve for \(\mathscr{D}_{0}\) by plugging in or equating similar constants (fitting experimental data typically used for this).
06

Verify the Relationship for Linear Concentration Gradient

With \(C\) being a guidance for diffusion and oxidation characterization, having \(\frac{dC}{dx} = \text{constant}\) derived directly supports the relationship expected from Fick's first law, reconfirming parabolic dependence and steady-state assumption.

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Key Concepts

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

Parabolic Law
Understanding the parabolic law is crucial when studying oxidation kinetics of materials like titanium alloys. When materials like these alloys oxidize, they often gain weight over time. The formula presented, \( w^2 = Ct \), elegantly describes this behavior by indicating the weight gain, \( w \), per unit area over time, \( t \), follows a square root of the time dependency. This is in contrast to linear or cubic rates, which indicate different kinetic regimes.
The constant \( C \) in the formula is temperature-dependent and is linked to the material's activation energy and the gas constant. It essentially quantifies how fast the oxidation process occurs under certain conditions.
  • This parabolic behavior often arises when the oxidation process is controlled by diffusion through an oxide scale.
  • The weight gain over time results from a specific relation to thermal and material constants.
Titanium Alloys
Titanium alloys are highly valued for their strength-to-weight ratio and resistance to corrosion, making them ideal for aerospace applications like hypersonic vehicle nose cones. These properties are crucial as they help withstand harsh aerodynamic and thermal stresses during flight.
The alloy mentioned in the exercise forms a protective layer of rutile, \( \text{TiO}_2 \), upon oxidation. About 20% of the oxygen contributes specifically to this surface scale. This scale is both a blessing and a curse; it protects the underlying material from further degradation but also highlights the material's susceptibility to gradual weight gain in oxygen-rich environments.
  • Titanium alloys' features make them suitable for extreme application scenarios.
  • Protective oxide scales are key to their durability amidst oxidation processes.
Diffusion Coefficient
The diffusion coefficient, \( \mathscr{D}_{12} \), is a central parameter in understanding how oxygen atoms move through the titanium alloy. It is typically expressed as \( \mathscr{D}_{12} = \mathscr{D}_{0}\exp\left(-\frac{E_{a}}{\mathscr{R} T}\right) \). Here, \( \mathscr{D}_{0} \) is the pre-exponential factor and \( E_{a} \) is the activation energy.
In this context, the diffusion coefficient serves as a measure of the rate at which oxygen permeates through the alloy's rutile scale. The exponential dependency on \( E_{a} \) and temperature suggests that higher temperatures accelerate diffusion, assisting in the formation of a heavier oxide layer over time.
  • The diffusion process is key to understanding the weight gain behavior observed.
  • Gurus in material science use this to predict material performances.
Activation Energy
Activation energy, \( E_{a} \), is the energy barrier that must be overcome for a chemical reaction, such as oxidation, to proceed. In the oxidation of titanium alloys, this corresponds to the energy required for oxygen atoms to diffuse through the rutile scale.
Given as 61,800 cal/mol in the problem, this high activation energy implies that substantial thermal energy is needed for the oxidation process to occur at a noticeable rate. The higher the activation energy, the slower the oxidation process is at a given temperature, which helps sustain the integrity of titanium alloys over time.
  • Activation energy directly impacts how oxidation processes initiate and unfold.
  • It acts as a gatekeeper, dictating the conditions under which diffusion is significantly observable.

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Most popular questions from this chapter

An automobile catalytic converter has a ceramic matrix with \(1 \mathrm{~mm}\)-square passages and \(0.2 \mathrm{~mm}\)-thick walls. The walls are coated with a \(20 \mu \mathrm{m}\)-thick layer of porous alumina that is impregnated with a platinum alloy catalyst (the washcoat). The purpose of the porous washcoat can be viewed as giving an effective reaction rate constant \(k_{\text {eff }}^{\prime \prime}\) based on \(s\)-surface area that is much larger than if the catalyst were simply coated onto a solid wall. (i) If the actual rate constant for \(\mathrm{CO}\) oxidation is \(k^{\prime \prime}=7 \times 10^{-3} \mathrm{~m} / \mathrm{s}, \mathscr{D}_{1, \text { eff }}\) in the washcoat is \(1.6 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), and the washcoat area-to-volume ratio is \(a_{p}=5 \times 10^{6} \mathrm{~cm}^{2} / \mathrm{cm}^{3}\), determine \(k_{\text {eff }}^{\prime \prime}\) (ii) If the exhaust is at \(800 \mathrm{~K}, 1.1\) bar, is the removal of \(\mathrm{CO}\) controlled by diffusion of CO from the gas stream or by reaction within the washcoat?

A closed cylindrical vessel containing stagnant air stands with its axis vertical. Each end is maintained at a uniform temperature with the base colder than the upper end; the cylindrical wall is insulated. Neglecting the variation in hydrostatic pressure, ascertain whether there are vertical gradients of (i) partial density of oxygen. (ii) partial pressure of oxygen. (iii) mass fraction of oxygen.

Methane is burned with \(20 \%\) excess air. At \(1250 \mathrm{~K}\) the equilibrium composition of the product is: \(\begin{array}{rlcccc}\text { Species } i: & \mathrm{CO}_{2} & \mathrm{H}_{2} \mathrm{O} & \mathrm{O}_{2} & \mathrm{~N}_{2} & \mathrm{NO} \\ x_{i}: & 0.0803 & 0.160 & 0.0325 & 0.727 & 0.000118\end{array}\) Determine the mean molecular weight \(M\) and gas constant \(R\) of the mixture, and the mass fraction of the pollutant nitric oxide in parts per million.

Air with an initial temperature of \(310 \mathrm{~K}\) and \(10 \% \mathrm{RH}\) is injected as 1 mm-diameter bubbles into a \(1 \mathrm{~cm}\)-deep water pool at the same temperature. Estimate the residence time to obtain a bubble bulk humidity of \(99.99 \%\). Is the shallow pool depth of any consequence? Assume a negligible liquid-side heat transfer resistance; that is, the air-water interface is at the bulk water temperature.

Air at \(315 \mathrm{~K}, 1 \mathrm{~atm}\), and \(20 \%\) RH flows at \(12 \mathrm{~m} / \mathrm{s}\) along a \(30 \mathrm{~cm}-\) square copper plate in a wind tunnel. The plate has a porous surface, and water is supplied to maintain a completely wet surface. The walls of the wind tunnel are at \(10^{\circ} \mathrm{C}\). Estimate the equilibrium temperature of the plate if the backside of the plate is insulated and the water supply temperature is controlled to be equal to the plate surface temperature. Take \(\varepsilon=0.92\) for the wetted surface. Use a transition Reynolds number of \(3 \times 10^{5}\).

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