ENGR 220 Homework Problems: Week 4 - Chapter 5 1. (a) Compare interstitial and vacancy atomic mechanisms for diffusion. Interstitial diffusion occurs when atoms move into the spaces between the atoms of the host lattice. Vacancy diffusion occurs when atoms exchange positions with vacancies or empty lattice sites. (b) Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion. Interstitial atoms are smaller than the host atoms and can move more easily. The probability of finding an empty adjacent interstitial site is higher than finding a vacancy adjacent to a host or substitutional impurity atom. 2. A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200? and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6 × 10-11 m2/s, and the diffusion flux is found to be 1.2 x 10-7 kg/m2-s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m3? Assume a linear concentration profile. Flux = - D dc/dx 1.2 x 10-7 kg/m2s = - 6 x 10-11 m2/s dC/dx dC/dx =- 2000 -2000 = 4 - 2/0-x B x B = 2/2000 x B = 1*10^-3m x B = 0.001m 3. Nitrogen from a gaseous phase is to be diffused into pure iron at 700°?. If the surface concentration is maintained at 0.1 wt% N, what will be the concentration 1 mm from the surface after 10 h? The diffusion coefficient for nitrogen in iron at 700? is 2.5 x 10-11 m2/s. Cx/.1= 1-(10^-3m)/(2sqrt{2}(2.5*10^11m^2sec^-1*36000sec)) Cx/.1 = 1-(0.527) Cx= 0.047%N 4. Compute the values of the diffusion coefficients for the interdiffusion of carbon in both a-iron (BCC) and y-iron (FCC) at 900°C. Which is larger? Explain why this is the case. At 900 the diffusion coefficient for interdiffusion is expected to be larger in BCC compared to FCC. This is because the crystal structure allows for more open atomic packing, facilitating easier atomic movement at the same temperature. a-iron = D = 4.4*10^-13 y-iron =D= 6.8*10^-13 5. The diffusion coefficients for silver in copper are given at two temperatures: T (?) D (m2/s) 650 5.5 ×10-16 900 1.3×10-13 (a) Determine the values of D0 and Qd. RT=(Qd/ln) 923/1173 = ln(5.5*10^-16)/(lnDo-ln(1.3*10^-13)) Do = 0.0000754 Qd = 196783.8 J/atom (b) What is the magnitude of D at 875?? 875+273 = 1148K D= 8.38*10^-14
6. The steady-state diffusion flux through a metal plate is 5.4 x 10-10 kg/m2-s at a temperature of 727°? (1000 K) and when the concentration gradient is -350 kg/m4. Calculate the diffusion flux at 1027°? (1300 K) for the same concentration gradient and assuming an activation energy for diffusion of 125,000 J/mol. Jo = j/exp(-Q/RT) ((5.4*10^-10kg/m^2s)/(-350kg/m^4))exp(12500j/mol)/(8.314*1000) Jo=5.26*10^-6m^2s (5.26*10^-6m^2s)(-350kg/m^4)e^(125000)/(8.314*1300) J= 1.7*10^-8