MS1015-TUT3-1Q1 Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600K). Assume an energy for vacancy formation of 0.55 eV/atom MS1015-TUT3-2 A1 In order to compute the fraction of atom sites that are vacant in lead at 600 K, we must employ Equation (4.1). As stated in the problem, Qv = 0.55 eV/atom. Thus, N V N = exp − Q V kT | \ | . | = exp − 0.55 eV/atom 8.62 x 10 −5 eV/atom-K ( ) (600 K) ¸ ( ¸ ( ( ( = 2.41 x 10 -5 MS1015-TUT3-3 Q2 Calculate the energy for vacancy formation in silver, given that the equilibrium number of vacancies at 800°C (1073K) is 3.6 ×10 23 m -3 . The atomic weight and density (at 800°C) for silver are, respectively, 107.9 g/mol and 9.5 g/cm 3 MS1015-TUT3-4 A2 Use the following equation N = N A ρ Ag A Ag = 6.023 x 10 23 atoms/mol ( ) 10.49 g/cm 3 ( ) 107.87 g/mol = 5.86 x 10 22 atoms/cm 3 = 5.86 x 10 28 atoms/m 3 Taking natural logarithms of both sides of Equation (4.1), V Q = RT ln V N N | | − | \ . =exp V V N Q N kT | | − | \ . All parameters besides Qv are given except N, the total number of atomic sites. However, N is related to the density, (ρ), Avogadro's number (N A ), and the atomic weight (A) according to Equation (4.2) as = − 8.62 x 10 -5 eV/atom- K ( ) (1073 K) ln 3.60 x 10 23 m −3 5.86 x 10 28 m −3 ¸ ( ¸ ( ( = 1.11 eV/atom MS1015-TUT3-5 A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200°C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6 ×10 -11 m 2 /s, and the diffusion flux is found to be 1.2 ×10 -7 kg/m 2 -s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m 3 . How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m 3 ? Assume a linear concentration profile. (Question 5.7 in the textbook) Q3 MS1015-TUT3-6 X=0 X C A =4 KG/m -3 C B =2 KG/m -3 1.5 mm T=1,200 °C A3 MS1015-TUT3-7 A3 We are asked to determine the position at which the nitrogen concentration is 2 kg/m 3 . This problem is solved by using Equation (5.3) in the form A B A B C C J D X X − = − − EQ (5.3) If we take C A to be the point at which the concentration of nitrogen is 4 kg/m 3 , then it becomes necessary to solve for x B , as A B B A C C X X D J − | | = + | \ . x A =0 (at the surface), in which case ( ) 3 3 11 2 7 2 4 kg/m 2 kg/m 0 6 10 /sec 1.2 10 / sec B X x m x kg m − − | | − = + | \ . = 1 x 10 -3 m = 1 mm MS1015-TUT3-8 Determine the carburizing time necessary to achieve a carbon concentration of 0.45 wt%at a position 2 mminto an iron-carbon alloy that initially contains 0.20 wt%C. The surface concentration is to be maintained at 1.30 wt%C, and the treatment is to be conducted at 1000°C. Use the diffusion data for γ-Fe inTable 5.2 of the textbook. (Question 5.11 in the textbook) Q4 MS1015-TUT3-9 A4 To compute the diffusion time required for a specific nonsteady-state diffusion situation, use Equation (5.5) 1 2 X o S o C C x erf C C Dt − | | = − | − \ . EQ (5.5) where C x = 0.45, C o = 0.20, C s = 1.30, and x = 2 mm = 2 x 10 -3 m Thus, 0.45 0.20 0.2273 1 1.30 0.20 2 X o S o C C x erf C C Dt − − | | = = = − | − − \ . 1 0.2273 0.7727 2 x erf Dt | | = − = | \ . z MS1015-TUT3-10 By linear interpolation from Table 5.1 z erf(z) 0.85 0.7707 ?? 0.7727 0.90 0.7970 0.850 0.7727 0.7707 0.90 0.85 0.7970 0.7707 z − − = − − 0.854 2 x z Dt = = Now, from Table 5.2, at 1000°C (1273 K) D = 2.3 x 10 -5 m 2 /s ( ) exp − 148,000J /mol (8.31J /mol-K)(1273 K) ¸ ( ¸ ( =1.93 x 10 -11 m 2 /sec A4 MS1015-TUT3-11 ( ) ( ) ( ) 3 11 2 2 10 0.854 2 1.93 10 /sec x m x m t − − = Solving for t yields t= 7.1 x 10 4 sec = 19.7 hr A4 MS1015-TUT3-12 Q5 (Question 5.16 in the textbook) Cite the values of the diffusion coefficients for the interdiffusion of carbon in both α-iron (BCC) and γ-iron (FCC) at 900°C. Which is larger? Explain why this is the case. MS1015-TUT3-13 A5 Compute the diffusion coefficients of C in both α and γ iron at 900°C using the data in Table 5.2 D α = 6.2 x 10 -7 m 2 /s ( ) exp − 80,000 J / mol (8.31 J /mol-K)(1173 K) ¸ ( ¸ ( = 1.69 x 10 -10 m 2 /s D γ = 2.3 x 10 -5 m 2 /s ( ) exp − 148,000 J /mol (8.31 J /mol-K)(1173 K) ¸ ( ¸ ( = 5.86 x 10 -12 m 2 /s The D for diffusion of C in BCC α iron is larger, the reason being that the atomic packing factor is smaller than for FCC γ iron (0.68 versus 0.74); this means that there is slightly more interstitial void space in the BCC Fe, and, therefore, the motion of the interstitial carbon atoms occurs more easily. MS1015-TUT3-14 (Question 5.18 in the textbook) At what T will the diffusion coefficient for the diffusion of copper in nickel have a value of 6.5x10 -17 m 2 /s? Use the diffusion data in Table 5.2 …additional questions… MS1015-TUT3-15 A5.18 MS1015-TUT3-16 (Question 5.24 in the textbook) C is allowed to diffuse through a stell plate 10 mm thick. The concentrations of C at the two faces are 0.85 and 0.40 kg C/cm 3 Fe, which are maintained constant. If the preexponential and activation energy are 6.2x10 -7 m 2 /s and 80000 J /mol, respectively, compute the temperature at which the diffusion flux is 6.3x10 -10 kg/m 2 s. …additional questions… MS1015-TUT3-17 A5.24