MSE 230 Fall 2007 Exam II

Transcription

1 1 Purdue University School of Materials Engineering MSE 230 Fall 2007 Exam II November 8, 2007 Show All ork and Put Units on Answers Name: KEY Unique name or Recitation Day and Time: Recitation Instructor s Name: 1 /15 2 /20 3 /20 4 /15 5 /15 6 /15 Total /100

2 2 (15 pts) 1a. Put the best answer in the blank. Short Answer _F 1080 Steel A. Al-Cu _E_ Sensitized Stainless Steel B. Silicon _A 2024 C. Vd Bonds _J hite Cast Iron D. Cherios _B Fe 3 C 3α and Graphite E. Slow Cool F. 100% Pearlite G. Magnesium H. Chromium I. Frosted Flakes J. 70% Fe 3 C 1b. Using the phase diagram on the next page, for what steel composition would you expect to have about 50% Pearlite and 50% proeutectoid phase upon slow cooling through 727 o C? Show your work! Pearlite 0.76 x = 0.5 = Solve for x : x = 0.40 or 1040 Steel Note that the problem statement reads steel not cast iron. This guides you as to which side of the phase diagram to work. P.1

3 3 Fe-Fe 3 C Phase Diagram (20 pts) 2. Use the provided Fe-Fe 3 C phase diagram to answer the following questions for a steel. Assume the steel has been austenitized for 1 hour at 850 o C, then slow cooled to 700 o C. (a) hat is the proeutectoid phase, and what is its composition in terms of at. % carbon? Fe 3 C 25at. % C steel has 1 wt.% C. (b) hat is the wt. fraction of Pearlite? γ = p = = (c) hat is the wt. fraction of ferrite and the total weight fraction of cementite in the microstructure? = = = Total Fe C = 0.15 (d) hat is the wt. fraction of cementite in the Pearlite? + Total Fe 3 C = pro Fe C 3 pearlitic Fe3C.15 = pearlitic Fe C pearlitic Fe = C P.2

4 4 Non Equilibrium Microstructures (20 pts) 3. Use the TTT curve below for an Fe-C alloy to answer the following questions. a. Looking at the figure below, how much carbon do you think is in the Fe-based alloy: wt%c b. Using a straight edge, indicate the path to form coarse pearlite after austenitization for 1 hour at 800 o C. See Schematic transform above about 650 o C c. Coarse pearlite is limited by: Growth or Nucleation or Both? (circle one) d. hat microstructure(s) would be formed for path A on the image? Tempered Martensite e. hat is unique about Martensite compared to Pearlite, Speroidite, Bainite, and Tempered Martensite? Short Answer Martensite is the BCT Phase, all the others are α + Fe 3 C P.3

5 5 Diffusion Concepts (15 pts) 4. A 1010 steel heated to the austenitic region is carburized at an elevated temperature and in an atmosphere wherein the surface carbon concentration is maintained at 1.10 wt.%. If after 48 hr the concentration of the carbon is 0.30 wt.% at a position 3.5 mm below the surface, determine the temperature at which the treatment was carried out. Data required is at the back of the exam! Show all work! T= 48 hrs = 172,800 secs, Temperature? C x C0 = x 1 erf C C s 0 2 Dt x 0.2 = 1 erf () z where z = 2 Dt () z 0. 8 erf = From Table, z = 0.9 So, 0.9 = 2 3.5x10 3 m 1 [ D( 172,800s) ] Therefore D = 2.2x10 m /s For C diffusing in γ (Austenite), from table, D = D 0 2.2x10 Q exp RT 11 = 5 2.3x10 3 ( ) ( T) = ( ) exp 8.314T T = 1283K T = 1010 C 3 Q = 148KJ / mol, D0 = 2.3x10 5

6 6 Ceramics (15 pts) 5. (a) Beginning with sub-micron diameter aluminum oxide powders, draw the expected modulus of rupture versus sintering temperature plot. Assume sintering time is constant at 1-hr. hy did you draw it this way? Increased sintering times decreases porosity which should improve strength in the ceramic. (b) On the image below, draw a schematic of the expected diffraction patterns for cristobalite and fused silica. Label your graphs. hat processing parameter likely influenced the structure formed in both? Cooling Rate (c) On the images below, draw the expected (i) viscosity versus modifier and (ii) viscosity versus intermediate amount. hy did you draw them this way? hy? Modifier breaks corner sharing of SiO 4 tetrahedron hy? Little or no affect due to replacing Si with Pb P.5

7 7 Polymers (15 pts) 6. You are given the following information for polyethylene (C = 12 g/mol and H = 1 g/mol). Number of Chains Mi, g/mol x(i) w(i) /21000= , , Total =21,000 x i = 1.0 w i = 1. 0 (a) Calculate the number average molecular weight. M N = x im i =.19* 2500 g/mol *7500g/mol *12,500 g/mol *17,500 g/mol = 9112g/mol (b) Calculate the weight average molecular weight. You can assume you have one mole of PE. TOTAL ( 2500g/mol) ( 7500 g/mol) = x im i =.19 moles +.38 moles +.33 moles( g/mol) moles 17,500 g/mol TOTAL = 9112g ( ) Mw = 0.05( 2500g/mol) 0.313( 7500) ( 12,500) ( 17, )= = w i M i M = g/mol P.6

8 8 R = J/mol K z erf(z) z erf(z)