Structure.homework-#10 answer

Structure.homework-#10 answer - -6-1/10x10-6 = 5x10 5 10 5...

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MTLE-2100 Structure of Engineering Materials, Homework #10 - Answer 1. Calculate the excess fractional vacancy concentration, n/n 0 = (n - n 0 )/n 0 , for the position marked A in the following diagram at 1800 ° C tungsten. Assume a 0 =2A, and γ = 1J/m 2 . R 1 = 10 μ m, R 2 = - 1 μ m, P = γ (1/R 1 +1/R 2 ) = 1(1/10x10 -6 -1/10 -6 ) = 10 5 -10 6 (J/m 2 ) (1/m) = - 10 6 N/m 2 n = n - n 0 = - n 0 Pa 0 3 /kT n /n 0 = (n - n 0 )/n 0 = - Pa 0 3 /kT = 10 6 (2x10 -10 ) 3 /[(13.81x10 -24 )(1800+273)] =8/ (13.81x2073)=0.00028 ρ = 1 μ m X= 10 μ m R= 100 μ m
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2. Which location has a higher vapor pressure, B or C? ln(p 1 /p 0 ) = γ (1/R 1 +1/R 2 )(M/RT ρ ) B: 1/R 1 +1/R 2 = 1/20x10 -6 +1/20x10 -6 = 2/20x10 -6 =10 5 m -1 C: 1/R 1 +1/R 2 = 1/2x10
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Unformatted text preview: -6-1/10x10-6 = 5x10 5- 10 5 =4x10 5 m-1 Point C has a higher vapor pressure. 3. Explain why the coarsening takes place, in which large particles get larger and small particles get smaller. ln(S 1 /S ) = γ (1/R 1 +1/R 2 )(M/RT ρ ) Smaller particles has a greater solubility than larger particles. Concentration gradient is produced and solute diffuses from the neighborhood of small particles to that of larger particles. Concentration near larger particles becomes supersaturated and precipitates on the larger particles....
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This note was uploaded on 04/29/2008 for the course MTLE 2100 taught by Professor Tomozawa during the Spring '08 term at Rensselaer Polytechnic Institute.

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Structure.homework-#10 answer - -6-1/10x10-6 = 5x10 5 10 5...

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