MCEN 2024, Spring 2008 The week of Apr 21 HW 10 Final

Transcription

1 MCEN 2024, Spring 2008 The week of Apr 21 HW 10 Final The Quiz questions based upon HW10 will open on Thursday, Apr. 24 and close on Monday, Apr 28 at 1:30 PM. References to A&J: Topic: High Temperature Phenomena in Materials Chapters 20-23, but especially Chapter 21 on "Kinetic Theory of Diffusion". Overview: The mechanical properties of materials at high temperatures differ from room temperature behavior for the following (interrelated) reasons: 1) While the "room temperature" behavior is essentially time independent, the high temperature mechanical behavior involves "time" as an essential variable. Thus, instead of just strain, which is a consequence of applying a stress, now becomes time dependent, making it necessary to describe the mechanical properties in terms of the strain rate. 2) The time dependency arises from the fact that atoms can move within the substance, whether it may be a solid or a liquid, as the temperature increases. This movement of atoms is called diffusion. Many applications of materials, ranging from microelectronics to turbine blades are constrained by diffusion. 3) The rate of diffusion scales with the "melting points" of materials. In crystals, like iron based materials, the melting points are well defined. (In glasses and polymers, melting occurs over a range of temperatures.) Therefore, processes that occur in lead near room temperature, occur at much higher temperatures in nickel base alloys: because the melting temperature for lead is 610K, while it is ~1600K for nickel base alloys. Homologous temperature is the ratio of the actual temperature divided by the melting temperature. The homologous temperature is helpful in the selection of materials for high temperature applications.

2 Explanation of the Figures and the Tables (These are now appended at the end of this HW) (i) Table and chart for the melting temperature of various substances. (ii) Table for the diffusion coefficients of various substances. (iii) Properties of ceramics for high temperature applications. Figure 1: Sagging of lead pipes (used in England for water) over a period of time.

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4 1. Assuming that the radius of curvature of the sagging lead pipe in Fig is equal to approximately 1 meter, and assuming the pipe to be approximately 1.5 inches in diameter, calculate the average strain rate on the outer surface of the the pipe assuming that it has been in service for 5 years. 2. The stress in the pipe can be calculated from the following equation, which derived from the information given in HW9: (1) where σ is the maximum tensile stress exerted in the bending configuration. F is the force, in N, exerted by the weight of the pipe, R is the pipe radius and t is the wall thickness of the pipe. Calculate the stress (the density of lead is equal to 7.85 g cm 3 ). 3. Now calculate the (uniaxial tensile) viscosity, of the lead near the room temperature, assuming the following equation: (tensile stress) = uniaxial tensile viscosity * tensile strain rate. Answer: ~1*10 15 Pa s 4. The diffusion coefficient of molecules or atoms in a liquid or a glass is related to viscosity in the following way: (2) where k is the Boltzmann's constant (1.38*10-23 J K -1 mol -1 ), T is the temperature in Kelvin, and a is the effective size of the diffusing atom or the molecule (note that, where Ω is the volume occupied by one atom/molecule in the material). (i) What are the units for D? (ii) The viscosity of water at ambient temperature is 1 mpa s. Calculate the diffusivity of the water molecules under these conditions. (Answer 7*10-10 m 2 s -1 )

5 5. The effective diffusion distance, L, of a marked atom is related to the time of diffusion, t, by the following equation: (3) where D is the diffusion coefficient. The constant α depends on the dimensionality of the problem: it is equal to 2 for a one-dimensional problem (like diffusion down a fiber), 4 for a two dimensional problem (like lateral diffusion in a thin film), and 6 for a three dimensional problem. 6. A dab of red dye molecules is placed on the surface of a bowl of water. Assuming the diffusion coefficient for the dye molecules to be the same as that of the water molecules, that is, equal to 7*10-10 m 2 s -1, calculate the effective distance to which the red dye will diffuse in a period of 1 minute, and 10 minutes. (Answer is 0.5 mm and 1.6 mm: this seems counterintuitive which may be because the size of the red dye molecule is smaller). 7. You need to design a process for diffusion of the dopant aluminum into silicon from the surface to a certain distance. The website: calculates the diffusion to a certain depth as a function of time and temperature. (i) Let us define the "diffusion distance" as the distance over which the concentration of the diffusing species falls to one half of it surface value. Assume a diffusion distance of 200nm for this problem. Calculate the time for the diffusion of Al to this distance at 1050 o C. (Answer: 15 min) (ii) Change the surface concentration to another number. Does this change the effective diffusion distance for the same time as in (i). (iii) Calculate the relative times for diffusion at 950 o C, 1000 o C, 1050 o C, 1100 o C. (iv) Using Eq. (3) calculate the values for the diffusion coefficients at the temperatures given in(iii).

6 8. The diffusion coefficient is very sensitive to temperature; this dependence is described by: m 2 s -1 (4) where, D o is a temperature dependent pre-exponential with units of m 2 s -1 Q is called the activation energy for diffusion, with units of kj mol -1 R is the gas constant with a value of 8.31 J K -1 mol -1 T is the temperature in K. Combining (3) and (4) we obtain: (5) Note that making a plot of log(t) versus (1/T) will yield a straight line having a slope of Q/R. If L is constant (as we assumed in Problem 7) then Q can be calculated by having the data for t at two temperatures: Note that base-10 is used for the logarithm. Calculate the value for Q from the following plot:

7 Answers:, 23 min. 11. The figure below (Fig in A&J) gives a mechanism for creep deformation by diffusion in polycrystalline materials:

8 The schematic below shows how the performance of turbine blades has been enhanced by making the blade from a single crystal: The picture shows the polycrystalline blade for gas turbines from the 1960's, and the right hand picture shows the single crystal blade of today. Why does the single crystal blade improve performance? 12. Diffusion is related fundamentally to the movement of atoms in a solid. In a crystal these atom jumps are discreet; the jump distance being equal to the interatomic spacing in the crystal. It can then be shown that the diffusion coefficient is given by: (6) where Γ is the jump frequency (with units of s -1 ) and a is the jump distance. The diffusion coefficient for most metals with the face centered cubic structure, just below their melting point, is given by ~10-12 m 2 s -1. Assuming the jump distance to be approximately 0.2 nm. Calculate the jump frequency of the atoms when.

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