Materials Science and Engineering Department. MSE 360, Test #3

Transcription

1 Materials Science and Engineering Departent MSE 360, est #3 ID nuber Nae: No notes, books, or inforation stored in calculator eories ay be used. Cheating will be punished seerely. All of your work ust be written on these pages and turned in. Constants, equations, and other data are gien on the last page of the exa. Proble 1-18: points each (Key:1CB3B4B5A6D 7B8D9B10B11A1A 13A14A15A16B17A18A) 1. A regular solution will likely for an ordered atoic structure if: d.) Ω 0 b.) Ω > 0 c.) Ω < 0. With increasing teperature and constant pressure, the Gibbs free energy of a single phase A) increase, B) decrease 3. During annealing, the grain # represented in the right figure will A) be stable B) grow C) shrink 4. he echanis of diffusion for a dilute solute ato of sall atoic radius copared to the host aterial would be best described by: a.) Substitutional diffusion b.) Interstitial diffusion c.) Vacancy diffusion d.) all of the aboe 5. When L f / > 4R, where the L f is the latent heat of fusion, is the elting teperature, the liquid/solid interface is A) Sooth, B) Rough 6. he inhoogeneous nucleation is easier than the hoogeneous nucleation because the inhoogeneous nucleation A. Has a critical nucleus with saller diaeter B. Has a critical nucleus with saller olue C. Has a critical nucleus with saller critical Gibbs free energy D. Both B and C 7. For heterogeneous nucleation in the grain interior, grain boundaries, grain edges and grain corners, their critical energy barriers can be described as A. Grain boundaries > Grain edges > Grain corners > Grain interior B. Grain interior > Grain boundaries > Grain edges > Grain corners C. Grain corners > Grain edges > Grain boundaries > Grain interior D. Grain boundaries > Grain edges > Grain corners > Grain interior 8. Dislocation can help with the nucleation of a new phase by A. Reducing nucleation energy barrier B. Increasing diffusion rate C. Gliding to produce plastic strain D. Both A and B 1

2 9. In an age-hardening Al-Cu alloy, the θʹ ʹ phase is A. fored by Cu ato clustering B. Fully coherent with the atrix C. Sei-coherent with the atrix D. Incoherent with the atrix 10. See the figure on the right. During the cooling process, an alloy with a coposite of X3 will decopose by A. A spinodal transforation B. A nucleation and growth transforation C. A ixture of A and B A 11. See the CC diagra. Cooling cure D will result in A. artensite B. coarse pearlite C. fine pearlite D. artensite + pearlite 1. he artensitic transforation is A. diffusionless B. diffusion controlled C. none of the aboe 13. he crystal structure of austenite is A. fcc; B. bcc; C. bct; D. hcp 14. he figure on the right represents A. high c% artensite B. low carbon artensite C. ediu carbon artensite 15. With increasing carbon content, the c axis of artensite will A. increase, B. decrease, C. does not change. 16. With increasing carbon content, the a axis of artensite A. increase, B. decrease, C. does not change. 17. he low carbon artensite has A. dislocation structure, B. twin structure.

3 18. he habit plane of a low carbon artensite is usually A. [111] γ, B. [59] γ, C. [5] γ. 19 (6 points): See the Ag-Cu phase diagra. Draw scheatically the Gibbs free energy cures of all phases at 900 C, 779 C, and 700 C. 0 (5 points). For the carbonization of steel at 800 C, it takes hours to reach a C concentration of 0.4% at a depth of 100 µ. Assuing that the diffusion coefficient does not change with coposition. If the carbonization tie is increased to 6 hours, Calculate the depth at which the C concentration reaches 0.4%. 1. (6 points) If a grain intersection in a -phase (A, and B, phases) aterial is as shown below, and it is at equilibriu, calculate the ratio of the interface energy between A-A grains (γ AA ) and the A-B interfaces (γ AB ). (Calculate (γ AA / γ AB )). A 110 A 110 B 140 3

4 . (5 points, bonus) he figure on the right represents the cooling of an alloy with a coposition X o. Assue the liquid alloy is in a longitudinal container, and the cooling starts fro the left side. Draw the coposition profile after coplete solidification under equilibriu cooling condition. 3. (8 points) For a solid to solid phase transforation during cooling, draw scheatically the nucleation rate as a function of teperature. Explain the physical reason for the teperature dependence of the nucleation rate. Will a siilar teperature dependence occur in a solid to solid transforation during heating? Explain. 4. (3 points) See the figure on the right. In an age-hardening Al-Cu alloy, the precipitation sequence at < < 3 is 5. (3 points) Explain the role of excessie acancies in the foration of precipitates during the aging process. 6. (6 points) Define coherent spinodal. What caused the difference between the cheical spinodal and coherent spinodal? 7. (5 points) Define incubation tie in a diagra. 4

5 8. (6 points) What is a CC diagra. 9. (8 points) Define habit plane of artensite. What is the habit plane of low carbon, ediu carbon and high carbon artensite? 30. (8 points) What interstice sites does carbon atos occupy in fcc austenite? Explain why? 31. (5 points) Explain why a sall austenite grain size is desired before the artensitic transforation. 5

6 Constants and Equations N A 6.0 x 10 3 ole -1 k (or R) 8.6 x 10-5 ev/ato-k (or 8.31 J/ol K) K C n 1 10 c 1 P + F C + P + F C + 1 G H S HU+PV du ds - PdV G X aga + X bgb ΔGix H C p C p d S d 0 dp d ΔH V 0 eq eqδ Δ ΔH S L Δ G LΔ ω conf ( N a N a + N b )!! N! b 1 ΔS ix R(X a ln X a + X b ln X b ) Δ ε ε AB ( ε AA ε BB ) Ω N a zδε Δ H ix ΩX a X b ΔG ix ΔH ix ΔS ix γv γv γv Δ Gγ ΔPV X r X exp X (1 + ) r Rr Rr X e B Ae Q R n ΔG exp k Q N exp k nuber J area tie x C(x,t) C s ΔC erf ( Dt ) dc d C D dt dx dc J D dx $ C C + β 0 sin πx ' % & ( ) exp t τ l [ ] where: τ π D Qd ΔS D D0 exp( ) D 1 d d 0 zν exp k 6 K N ho f 1 C 0 exp{ ΔG ho k N het A } N ho f0c0 exp{ } Δ ΔG het f1c1 exp{ } k 1 Δi k 16πγ A 3L 3 s k & k C exp $ % Δ i #! " 6

7 k ( Δ ) 3 i δ d d β d α α D d β b δ δ γ coh γ che γ sei coh γ che + γ strain γ strain δ γ 3 γ 13 γ 1 D b/sinθ b/θ, F γ/r, ΔG γv, sinθ 1 sinθ sinθ 3 r MFMΔG/V X b X 0 exp ΔG b R D D 0 + Kt D K " t n F ax πrγ F ax 3 fγ r D ax 4r 3 f L V Δ r γ % γ ( ' 1 & ) Δ L V ΔG 16πγ 3 3 % 16πγ 3 ( 1 ' & 3L V ) Δ, ( ) ( + cosθ ) 1 cosθ S θ ( ) 4 ΔG het ΔG ho S θ ( ) 16πγ 3 SL 3 ( ) 16πγ SL S θ & ( ' 3L V 3 ) 1 + Δ S( θ) x K Dt X S kx 0 ( 1 f S ) k 1 k 1 X L X 0 f L N ho fc 0 e ΔG k # X L X 0 % 1+ 1 k $ % k x e D & ( '( γ r + ΔG S 16πγ 3 ΔG 3 + ΔG S ( ) r γ, ΔG het f ωe ΔG k N ho C 0 ωe ΔG k e ΔG k, " f 1 exp π % $ 3 N3 t 4 ' # & f 1 exp( kt n ) ΔG ho S( θ), S( θ) + cosθ ( )( 1 cosθ) DΔX 0 k X β X r x Kt ( ) 1# r 1 r & % ( $ r ' f π 3 N3 t 4 ΔG c 1 d G dx r 3 r 3 ( ΔX) 0 kt k Dγ X e f Σ 0 t CC g cc Δt CC τ ( ) dt CC 0 τ 0 $ g CC τ ( )d ( ) 7