{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

chap6_5

# chap6_5 - 材材材材材材 Fundamental of Materials Prof...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 材材材材材材 Fundamental of Materials Prof: Tian Min Bo Prof: Tian Tel: 62795426 材 62772851 E-mail: [email protected] [email protected] Department of Material Science and Engineering Tsinghua University. Beijing 100084 Lesson twenty­eight §6.5 concentration dependence of D. Matano method Matano Ⅰ.D-C dependence Ⅱ.Matano method ∂C ∂ ∂C = (D ) ∂t ∂x ∂x (1) D = D(C ) at t = 0 C = C1 C = C2 for x > 0 for x < 0 ( 2) x Let λ = t ( 3) ∂C dC x dC λ = (− )= (− ) ∂t dλ 2t t dλ 2t ( 4) ∂C dC ∂λ 1 dC 1 = ⋅ = ∂x dλ ∂x t dλ 材t ∂ ∂C d 1 dC ∂λ 1 d dC (D )= (D ⋅ )⋅ = 材 材 ∂x ∂x dλ t dλ ∂x t dλ dλ put ( 4) in (1) ­ λ dC 1 d dC = (D ) 2t dλ t dt dλ C 1C dC − ∫C λdC =∫C d ( D ) 1 21 dλ For points in C-x curve, t = const 11 − 2t 1 − 2t dC ∫C1 xdC = t ∫C1 d( D dx ) C C dC dC dC ∫C1 xdC =D dx − D dx = D dx C C1 C C 材材材材材材 =0 1 dx C ∴ D = − ∫C xdC 2t dC C 1 x → xM C2 ∫C 1 xM dC = 0 1 dxM C ∴ D(C ) = − ∫C xM dC 2t dC C 1 dC dx M C C2 A1 A C1 C = CM B 0M 0 xM §6.6 Kirkendall effect and Partial diffusion coefficients coefficients Ⅰ.K-experimental results: 1. The markers move toward brass slice. 2. ∆l = k t Kirkendall–Smigelskas ∆l : k shift Ⅱ.K-effect: .K-effect: In substitutional solid solutions, the markers placed on original interface move parabolically with time toward the zinc-rich zone. toward This phenomenon is called K-effect The meaning of K-effect: The a. Denied the diffusion mechanism about transposition b. Proved the vacancy mechanism c. Partial diffusion coefficients is different d. Built relationship between micro and macro e. Universality Accessary effect: Accessary Superfluous vacancy Ⅲ.Partial diffusion coefficients. referred to the fixed interface referred ∂C A ∂C A J A = −D = − DA + C Av ∂x ∂x ∂C B ∂C B J B = −D = − DB + CB v ∂x ∂x (1) ( 2) C A + C B = ρ = const. ∂C A ∂C B ⇒ =− ∂x ∂x ∂C A ∂C A ∴ DB + CBv = D ∂x ∂x (3) d( ∆l ) k ∆l ν= = = dt 2 t 2t from 1, 3 D = D A N B + DB N A ( 4) ( 5) NA, NB : mole fraction CB NB = C A + CB CA NA = C A + CB ∆l ∆l = b t ⇒ v = = 2 t 2t b 材 l 材材材 t 材材材材 v 材材材材材材材材 D 材 材 D 材 v 材材材材材 DA 材 DB Examples and Discussions Examples Exercise Exercise Thank you ! Thank you ! 28 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online