Unformatted text preview: Problem 5.6
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 5.6 What type of crystal direction corresponds to the movement of interstitial carbon in αFe between equivalent (½ 0 ½ type) interstitial positions? Illustrate your answer with a sketch. SOLUTION Begin by noting that the crystal structure of αFe is bodycentered cubic, as explained in Chapter 3 (see Figure 3.4 on page 63) of the text, and the interstitial positions of the type ½, 0, ½, are found at the centers of the faces of the cube, also known as octahedral interstices. Next, choose any two of these equivalent positions, say ½, 0, ½, and 0, ½, ½; then calculate the vector connecting these two positions (from tail to head), 0 – ½ , ½ – 0 , ½ – ½ = –½, ½, 0 Clearing fractions, the lattice direction is [¯ . Generalizing, the type of crystal direction con110] necting face centers is the symmetryrelated family of directions <110>. To complete this solution, show an illustrative sketch connecting the two locations identified above, namely the center of the side face and the center of the back face, as follows. Any equivalent <110> type direction is appropriate here.
" !""#$ # ! Problem 5.6 Solution Professor R. Gronsky page 1 of 1 Problem 5.10
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 5.10 Carburization was described in Example 5.3. The decarburization of a steel can also be described by using the error function. Starting with Equation 5.11 and taking cS = 0, derive an expression to describe the concentration profile of carbon as it diffuses out of a steel with initial concentration, c0. (This situation can be produced by placing the steel in a vacuum at elevated temperature.) SOLUTION Carburization is a process by which the carbon concentration of a steel is increased by enabling the diffusion of carbon into the steel from a carbonrich environment. This problem notes that steel can also be decarburized by diffusion of carbon out of the steel, as might happen by heating the steel in a vacuum environment. The operating equations for this process is found in Fick's Laws, found on page 134 of the text. The solution to Fick's second law for the specific case of carburization is Equation 5.11 on the same page, reproduced here. cx − c0 x = 1 − erf √ cs − c0 Dt where c0 is the initial concentration of carbon, cs is the surface concentration of carbon and cx is the concentration of carbon at location x within the steel. Also, "erf" is the Gaussian error function, a tabulated solution to the integration of a Gaussian distribution curve, which is part of this specific solution, and the remaining variables are the diffusivity (D) and time (t). Rewriting this equation for the decarburization conditions in the this problem, namely that the surface concentration of carbon is 0, cx − c0 x = 1 − erf √ cs − 0 Dt rearranging,
x √ Dt cx − c0 = −c0 + c0 erf and solving, the desired expression is found. The answer is cx x = erf √ c0 Dt Problem 5.10 Solution Professor R. Gronsky page 1 of 1 Problem 5.12
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 5.12 A diffusion couple is formed when two different materials are allowed to interdiffuse at an elevated temperature (see Figure 5.8). For a block of pure metal A adjacent to a block of pure metal B, the concentration profile of A (in at%) after interdiffusion is given by x √ cx = 50 1 − erf 2 Dt where x is measured from the original interface. For a diffusion couple with D = 10–14 m2/s 10–14 m2/s, plot the concentration profile of metal A over a range of 20 um on either side of the original interface (x = 0) after a time of 1 hour. [ Note that erf (–z) = – erf (z).] SOLUTION The solution to this problem is an application of the given equation describing interdiffusion in a "diffusion couple" with given diffusivity (D = 10–14 m2/s ) and time (t = 3600 s). Programming this equation into any appropriate plotting application (such as Microsoft Excel™) yields the following table of values x (µm) +20 +15 +10 +5 0 –5 –10 –15 –20 cx (at. % A) 1 4 12 29 50 72 87 96 99 Plotting these values yields the solution, as follows. Problem 5.12 Solution Professor R. Gronsky page 1 of 2 Problem 5.12
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 100 Concentration (at. % A) 75 50 25 0 20 15 10 5 0 x (microns) 5 10 15 20 Problem 5.12 Solution Professor R. Gronsky page 2 of 2 Problem 5.20
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 5.20 The diffusivity of copper in a commercial brass alloy is 10–20 m2/s at 400°C. The activation energy for diffusion of copper in this system is 195 kJ/mol. Calculate the diffusivity at 600°C. SOLUTION Diffusivity is the term applied to the proportionality constant in Fick's Laws. It is also known as the diffusion coefficient, D, and it exhibits Arrhenius behavior, as expressed in Equation 5.12 of the text (p. 135). The equation is
D = D0 e−q/kT where D0 is the preexponential constant and q is the activation energy for defect motion, the mechanism for diffusion. The text also explains on page 137 that it is "more common" to tabulate diffusivity data in terms of molar quantities, with an associated change in the expression for diffusivity, D = D0 e−Q/RT where the activation energy Q is expressed in units of energy per mole of diffusing species, and R is the universal gas constant, equal to NAVk. Because the data given in the problem statement is expressed in molar quantities, this second equation is more appropriate. Begin by noting the ratio of the diffusivity at 600°C (873K) to that at 400°C (673K),
D873K D0 e−Q/R[873 K] = D673K D0 e−Q/R [673K ] Rearranging,
D873K = D673K e− 8.314 J/mol K ( 873 − 673 )
1 1 195 kJ/mol substituting
D873K = 10−20 m
2 /s e− 8.314 J/mol K ( 873 − 673 )
1 1 195 kJ/mol and solving, the diffusivity at 600°C is obtained, D600°C = 2.93 x 10–17 m2/s. Problem 5.20 Solution Professor R. Gronsky page 1 of 1 Problem 5.30
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 5.30 The endpoints of the Arrhenius plot of Dsurface in Figure 5.18 are Dsurface = 7.9×10–10 m2/s at a temperature of 245°C and Dsurface = 6.3×10–9 m2/s at a temperature of 398°C. Using these data, calculate the activation energy for surface diffusion in silver. SOLUTION Figure 5.18 depicts the diffusivities of silver along three different diffusion paths, the lowest value associated with "volume" diffusion through the bulk, the midrange values associated with grain boundary diffusion, and the highest associated with surface diffusion, demonstrating how diffusion is fastest along paths with more "open" structures. This problem addresses surface diffusion of silver, for which two values are given, corresponding to two different endpoint temperatures. First note the governing equation for diffusion, equation 5.13 of the text on page 137,
D = D0 e−Q/RT Now write the ratio of given diffusivities at the two temperature of interest, 398°C (671 K) and 245°C (518 K) using this governing equation,
D671K D0 e−Q/R[671 K] 6.3 × 10−9 m2 /s = = D518K 7.9 × 10−10 m2 /s D0 e−Q/R [518 K] Simplifying,
e−Q/R ( 671 K − 518 K ) = e−Q/R (−4.402×10
1 1 −4 K−1 ) = 7.975 and solving,
Q= (8.314 J/mol K) ln 7.975 R ln 7.975 = −4 K−1 4.402 × 10 4.402 × 10−4 K−1 so the activation energy for surface diffusion of silver is Q = 39.2 kJ/mol. Problem 5.30 Solution Professor R. Gronsky page 1 of 1 ...
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This note was uploaded on 01/21/2011 for the course E 45 taught by Professor Gronsky during the Fall '08 term at Berkeley.
 Fall '08
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