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Exam3 Practice Materials Science and Engineering Department MSE 360, Test #3 ID number _______________________________ Name: _____________________________________ No notes, books, or information stored in calculator memories may be used. Cheating will be punished severely. All of your work must be written on these pages and turned in. Constants, equations, and other data are given on the last page of the exam. _______________________________________________________________________________ Problem 1-18: 2 points each (Key:1C2B3B4B5A6D 7B8D9B10B11A12A 13A14A15A16B17A18A) 1. A regular solution will likely form an ordered atomic structure if: d.) Ω = 0 b.) Ω > 0 c.) Ω < 0 2. With increasing temperature and constant pressure, the Gibbs free energy of a single phase A) increase, B) decrease 3. During annealing, the grain #2 represented in the right figure will A) be stable B) grow C) shrink 4. The mechanism of diffusion for a dilute solute atom of small atomic radius compared to the host material would be best described by: a.) Substitutional diffusion b.) Interstitial diffusion c.) Vacancy diffusion d.) all of the above 5. When Lf/Tm > 4R, where the Lf is the latent heat of fusion, Tm is the melting temperature, the liquid/solid interface is A) Smooth, B) Rough 6. The inhomogeneous nucleation is easier than the homogeneous nucleation because the inhomogeneous nucleation A. Has a critical nucleus with smaller diameter B. Has a critical nucleus with smaller volume C. Has a critical nucleus with smaller 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 9. In an age-hardening Al-Cu alloy, the θʹʹ phase is A. formed by Cu atom clustering B. Fully coherent with the matrix C. Semi-coherent with the matrix D. Incoherent with the matrix 10. See the figure on the right. During the cooling process, an alloy with a composite of X3 will decompose by A. A spinodal transformation B. A nucleation and growth transformation C. A mixture of A and B __A__ 11. See the CCT diagram. Cooling curve D will result in A. martensite B. coarse pearlite C. fine pearlite D. martensite + pearlite 12. The martensitic transformation is A. diffusionless B. diffusion controlled C. none of the above 13. The crystal structure of austenite is A. fcc; B. bcc; C. bct; D. hcp 14. The figure on the right represents A. high c% martensite B. low carbon martensite C. medium carbon martensite 15. With increasing carbon content, the c axis of martensite will A. increase, B. decrease, C. does not change. 16. With increasing carbon content, the a axis of martensite A. increase, B. decrease, C. does not change. 17. The low carbon martensite has A. dislocation structure, B. twin structure. 2 18. The habit plane of a low carbon martensite is usually A. [111]γ, B. [259]γ, C. [225]γ. 19 (6 points): See the Ag-Cu phase diagram. Draw schematically the Gibbs free energy curves of all phases at 900°C, 779 °C, and 700 °C. 20 (5 points). For the carbonization of steel at 800°C, it takes 2 hours to reach a C concentration of 0.4% at a depth of 100 µm. Assuming that the diffusion coefficient does not change with composition. If the carbonization time is increased to 6 hours, Calculate the depth at which the C concentration reaches 0.4% . 21. (6 points) If a grain intersection in a 2-phase (A, and B, phases) material is as shown below, and it is at equilibrium, calculate the ratio of the interface energy between A-A grains (γAA) and the A-B interfaces (γAB). (Calculate (γAA / γAB)). A 110° B 140° A 110° 3 22. (5 points, bonus) The figure on the right represents the cooling of an alloy with a composition Xo. Assume the liquid alloy is in a longitudinal container, and the cooling starts from the left side. Draw the composition profile after complete solidification under equilibrium cooling condition. 23. (8 points) For a solid to solid phase transformation during cooling, draw schematically the nucleation rate as a function of temperature. Explain the physical reason for the temperature dependence of the nucleation rate. Will a similar temperature dependence occur in a solid to solid transformation during heating? Explain. 24. (3 points) See the figure on the right. In an age-hardening Al-Cu alloy, the precipitation sequence at T2 <T < T3 is 25. (3 points) Explain the role of excessive vacancies in the formation of precipitates during the aging process. 26. (6 points) Define coherent spinodal. What caused the difference between the chemical spinodal and coherent spinodal? 27. (5 points) Define incubation time in a TTT diagram. 4 28. (6 points) What is a CCT diagram. 29. (8 points) Define habit plane of martensite. What is the habit plane of low carbon, medium carbon and high carbon martensite? 30. (8 points) What interstice sites does carbon atoms occupy in fcc austenite? Explain why? 31. (5 points) Explain why a small austenite grain size is desired before the martensitic transformation. 5 Constants and Equations 23 -1 -5 NA = 6.02 x 10 mole k (or R) = 8.62 x 10 eV/atom-K (or 8.31 J/mol•K) K = C +273 109 nm = 1 m 102 cm = 1 m P + F = C + 2 P + F = C + 1 G = H – TS H=U+PV dU =TdS - PdV G = X aGa + X bGb − ΔGmix T T C p dP ΔH H = C dT S dT = ∫ p = ∫ 0 0 T dTeq Teq ΔV ΔH L LΔT (N + N )! ΔS = m = ΔG ≅ ω = a b T T T conf N !⋅N ! € m m m a b 1 ΔS = −R(X ln X + X ln X ) Δε = ε − (ε − ε ) mix a a b b AB 2 AA BB N z H X X G H T S Ω = a Δε Δ mix = Ω a b Δ mix = Δ mix − Δ mix € 2γVm 2γVm 2γVm ⎛ − ΔGv ⎞ ⎛ − Qv ⎞ ΔGγ = ΔPVm = X r = X ∞ exp ≅ X ∞ (1 + ) nv = exp⎜ ⎟ = N exp⎜ ⎟ r RTr € RTr ⎝ kT ⎠ ⎝ kT ⎠ Q − e RT X B = Ae number dC d 2C dC J = = D 2 J = −D area ⋅ time dt dx dx € x C(x,t) = Cs − ΔC • erf (− ) 2 Dt $ πx ' l 2 C = C + β sin ⋅ exp −t where: τ = 0 %& () [ τ ] 2 D π € − Qd 1 2 ⎛ ΔSd ⎞ D = D0 exp( ) D0 = d zν exp⎜ ⎟ kT 6 ⎝ K ⎠ € 3 2 ΔG *hom A 16πγ sv Tm Nhom = f1C0 exp{− } N hom = f0C0 exp{− 2 } A = 2 kT ΔT 3Lv kT ΔG * & k # N f C exp{ het } v k T v C exp 2 het = 1 1 − = 1Δ i = $− ! kT % ΔTi " 6 2 d β − dα d β b v = k3 (ΔTi ) δ = D = ≈ dα δ δ γ coh = γ chem γ semi−coh = γ chem + γ strain γ strain ∝ δ γ γ γ 2γV 23 = 13 = 12 D = b/sinθ ≈ b/θ, F = 2γ/r, ΔG = m , sinθ1 sinθ 2 sinθ 3 r 2 2 ΔGb D = D0 + Kt v=MF=MΔG/Vm X = X exp Fmax = πrγ € b 0 RT n € € D = K "t 3 fγ 4r L ΔT F = D = ΔG = − V max max 3 f V T 2r € m € % ( 3 % 3 2 ( 2 2γ 2γTm 1 16πγ 16πγ Tm 1 (2 + cosθ)(1− cosθ) r* = − = ' * ΔG* = = ' * , S(θ) = G L T 3 G2 3L T 2 4 € Δ V €& V ) Δ € Δ V & V ) Δ 16πγ 3 & 16πγ 3 T 2 ) 1 ΔG* = ΔG* S(θ) = SL S(θ) = ( SL m + S(θ) x = K Dt het hom 3ΔG2 3L2 ΔT 2 € V€ ' V * € x # − & k−1 k−1 1− k D v X S = kX0 (1− f S ) X L = X 0 fL X L = X 0 %1 + e ( € $% k '( ΔG* − kT Nhom = fC0e € € 2γ r* = − ΔG € − m " π 3 4 % ΔG + ΔG kT D X 1# r *& V S f = ωe f =1− exp$− Nv t ' Δ 0 # 3 & v = %1− ( k(Xβ − Xr ) r $ r ' 16πγ 3 ΔGm ΔG* ΔG* = 2 − − n kT kT f 1 exp kt 3 ΔGV + ΔGS N = C ωe e = − − ( ) hom 0 , ( ) x = Kt 2 2γ * * (2 + cosθ)(1− cosθ) r* = − , ΔGhet = ΔGhomS(θ) , S(θ) = ΔGV 2 3 3 2 r − r0 = kt π 3 4 1 d G 2 f = Nv t ΔG = ΔX c 2 ( ) k ∝ Dγ X 3 2 dX e T T T Δt dt gC$ C T dT f = Σ CC = CC = ( ) T0 ∫ ∫ T τ TTT T0 τ TTT T0 τ TTT ( ) t = g T CC cc ( ) 7 .
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