COMPUTATIONAL MATERIALS ENGINEERING Episode 2 pot

... have ∂ ∂N i =  j ∂ ∂X j ∂X j ∂N i (2. 24) and X i =1−  j=i X j , ∂X j ∂N j = N − N j N 2 , ∂X j ∂N k = − N j N 2 (2. 25) we finally obtain µ i = G m + ∂G m ∂X i −  X j ∂G m ∂X j (2. 26) Equation (2. 26) is of considerable value in practical CT and Computational Microstructure Evolution ... N B − N A N B (N A + N B ) 2  · ω = 1 2 ωX B 2 (2. 64) and the chemical potential of a...

Ngày tải lên: 13/08/2014, 08:21

... this: E = N  i=1 z  j=1 γ(s i ,s j )+h i (3 .22 ) 5 3 2 1 4 Node Node 180˚ (a) (b) FIGURE 3 -22 An illustration of the strong pinning by par ticles in 2D systems, (a) the case of a single particle ... performed using a square (1 ,2) lattice, Glauber dynamics, Metropolis transition probability function, and kT s =0.75. 74 COMPUTATIONAL MATERIALS ENGINEERING FIGURE 3 -21 Potts model s...

Ngày tải lên: 13/08/2014, 08:21

... grain growth simulation using the Potts model. 90 COMPUTATIONAL MATERIALS ENGINEERING Grain B D /2 DD /2 Grain A Grain C Grain B q 1 q 1 q 1 q 2 q 2 g 1 g 1 g 2 f 12 FIGURE 3-34 The boundary geometry ... A r ρ y o θ γ BP γ AP (a) 55 50 45 40 35 30 25 10 15 20 25 30 4035 X Z Hellman and Hillert [2] Static Monte Carlo θ θ = 45° (b) Boundary Detaches 1.5 0.5 -0.5 -1.5 1 -1...

Ngày tải lên: 13/08/2014, 08:21

... − µb 2 (1 −ν) x 2 (3x 2 1 + x 2 2 ) (x 2 1 + x 2 2 ) 2 σ 22 = µb 2 (1 − ν) x 2 (x 2 1 − x 2 2 ) (x 2 1 + x 2 2 ) 2 σ 12 = µb 2 (1 − ν) x 1 (x 2 1 − x 2 2 ) (x 2 1 + x 2 2 ) 2 σ 33 = ν(σ 11 + σ 22 )=− µbν π(1 ... C 23 23 )∆ (8.1 32) ∆=e 2 (e + f)+e(f 2 − 1)  (a 1 a 2 ) 2 +(a 1 a 3 ) 2 +(a 2 a 3 ) 2  +(f −1) 2 (f + 2) (a 1 a 2 a...

Ngày tải lên: 13/08/2014, 08:21

... Equilibrium 16 2. 2 Solution Thermodynamics 18 2. 2.1 Entropy of Mixing 19 2. 2 .2 The Ideal Solution 21 2. 2.3 Regular Solutions 22 2. 2.4 General Solutions in Multiphase Equilibrium 25 2. 2.5 The Dilute ... Formulation 22 4 7 .2. 2 Material and Model Parameters 22 6 7 .2. 3 Application to Dendritic Growth 22 6 7.3 Case Study 22 8 7.3.1 Phase-Field Equation 22 9 7.3 .2 Finite...

Ngày tải lên: 13/08/2014, 08:21

... the 40 COMPUTATIONAL MATERIALS ENGINEERING FUNCTION GCRFCC 2. 98150E+ 02 + 728 4+.163*T+GHSERCR#; 6.00000E+03 N ! FUNCTION GCRM23C6 2. 98150E+ 02 - 521 983+3 622 .24 *T- 620 .965*T*LN(T) 126 431*T* *2; 6.00000E+03 ... TC(BCC_A2,FE:C;0) 2. 98150E+ 02 1043; 6.00000E+03 N ! PARAMETER BMAGN(BCC_A2,FE:C;0) 2. 98150E+ 02 2 .22 ; 6.00000E+03 N ! PARAMETER TC(BCC_A2,CR:C;0) 2. 98150E+...

Ngày tải lên: 13/08/2014, 08:21

... simulations 30 ´ 30 grid 20 0 CA simulations 25 0 ´ 25 0 grid 20 0 CA simulations 25 0 ´ 25 0 grid 100 CA simulations 25 0 ´ 25 0 grid 10 CA simulations 25 0 ´ 25 0 grid 1 CA simulation FIGURE ... transformation function. 120 COMPUTATIONAL MATERIALS ENGINEERING 0 25 0 500 750 1000 1 1000 CA Increment Number of Grains 0 5 10 15 20 Normalized Average Grain Ar...

Ngày tải lên: 13/08/2014, 08:21

... (20 01) [OCSBY01], Rollett (20 01) [RR01], Xu (20 01) [XL01], Janssens (20 02) [JVR 02] , Raabe (20 02) [Raa 02] , Vandyoussefi (20 02) [VG 02] , Zhang (20 02) [ZZXLW 02] , Janssens (20 03) [Jan03], Zhang (20 03) ... Sequence (λ)(λ 2 )(  1/k  λ 2 ) 1 +1 +1 1 [(1 /2) · 2] 1 /2 −1 −1 1 = (1) 1 /2 2 +1+1 +2 4 [(1/4) · 8] 1 /2 −1+1 0 0 = (2) 1 /2 +1 − 100 −1 − 1 2 4 3 +1+1+1 +3 9 [...

Ngày tải lên: 13/08/2014, 08:21

... (au) Fraction Transformed (x) 0 0 .2 0.4 0.6 0.8 1 k = 0.1 k = 1 1 10 100 n = 3 n = 1 0 0.5 1.5 2. 0 2. 51.0 Time (au) Fraction Transformed (x) 0 0 .2 0.4 0.6 0.8 1 0 0.5 1.5 2. 0 2. 51.0 2 3 4 FIGURE 6-3 Transformation ... (Length) c(x)/c 0 0 24 6810 sqrt(4 Dt) = 100 2 1 0.5 0 0 .2 0.4 0.6 1.0 0.8 5 20 FIGURE 5-8 Solutions to Fick’s second law for semi-inifinite sample. 164 C...

Ngày tải lên: 13/08/2014, 08:21

... Density 3.0 D - 4 Log(Time[s]) - 2 0 24 6 8 10 5 15 20 25 10 Log(Nucleation Rate J [m -3 s -1 ]) Log(Number Density N [m -3 ]) N J 1 1 Log(Radius [nm]) 2 1 3 1 A B C D - 4 Log(Time [s]) - 2 0 24 6 8 0 - 1 1 2 3 10 FIGURE ... constant K with ρ =  ρ 2 0 +4K 2 Dt (6.57) where K can unfortunately only be given in implicit form as 2K 2 ·  1 − √ πKe K 2 erfc(K)  = S (6.58)...

Ngày tải lên: 13/08/2014, 08:21