1 Purdue University School of Materials Engineering MSE 230 Fall 2007 Exam II November 8 2007 Show All Work and Put Units on Answers Name KEY Unique name or email Recitation Day and Time Recitation Instructor s Name 1 15 2 20 3 20 4 15 5 15 6 15 Total 100 2 Short Answer 15 pts 1a Put the best answer in the blank F 1080 Steel A Al Cu E Sensitized Stainless Steel B Silicon A 2024 C VdW Bonds J White Cast Iron D Cherios B Fe3C 3 and Graphite E Slow Cool F 100 Pearlite G Magnesium H Chromium I Frosted Flakes J 70 Fe3C 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 727oC Show your work W Pearlite 0 5 0 76 x 0 76 0 022 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 Fe Fe3C Phase Diagram 20 pts 2 Use the provided Fe Fe3C phase diagram to answer the following questions for a 10100 steel Assume the steel has been austenitized for 1 hour at 850oC then slow cooled to 700oC a What is the proeutectoid phase and what is its composition in terms of at carbon Fe 3C 25at C 10100 steel has 1 wt C b What is the wt fraction of Pearlite W Wp 6 7 1 0 0 96 6 7 76 c What is the wt fraction of ferrite and the total weight fraction of cementite in the microstructure W 6 7 1 0 0 85 6 7 0 022 WFeTotal 3C 1 0 0 022 0 15 6 7 0 022 d What is the wt fraction of cementite in the Pearlite WFeTotal W pro Fe3C W pearlitic Fe3C 3C 0 15 0 04 W pearlitic Fe3C W pearlitic Fe3C 0 11 P 2 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 0 76 0 8 wt C b Using a straight edge indicate the path to form coarse pearlite after austenitization for 1 hour at 800oC See Schematic transform above about 650oC c Coarse pearlite is limited by Growth or Nucleation or Both circle one d What microstructure s would be formed for path A on the image Tempered Martensite e What is unique about Martensite compared to Pearlite Speroidite Bainite and Tempered Martensite Short Answer Martensite is the BCT Phase all the others are Fe3 C P 3 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 x C x C0 1 erf 2 Dt C s C0 x 0 2 1 erf z where z 2 Dt erf z 0 8 So 0 9 From Table z 0 9 3 3 5 x10 m 2 D 172 800 s 1 2 Therefore D 2 2 x10 11 m 2 s For C diffusing in Austenite from table Q 148 KJ mol D0 2 3 x10 5 Q D D0 exp RT 2 2 x10 11 148 10 3 2 3x10 exp 8 314 T 5 148 10 T 1283 K 8 314 T 3 13 86 T 1010 D C 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 Why 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 What 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 Why did you draw them this way Why Modifier breaks corner sharing of SiO4 tetrahedron Why Little or no affect due to replacing Si with Pb P 5 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 4000 8000 7000 2000 Total 21 000 Mi g mol 2500 7500 12 500 17 500 x i 4000 21000 0 19 0 38 0 33 0 095 x i 1 0 w i 0 05 0 313 0 453 0 182 w i 1 0 a Calculate the number average molecular weight M N x i M i 19 2500 g mol 0 38 7500g mol 0 33 12 500 g mol 0 095 17 500 g mol 9112 g mol b Calculate the weight average molecular weight You can assume you have one mole of PE 19 moles 2500g mol 38 moles 7500 g mol 33 moles 12500 g mol 095 moles 17 500 g mol W TOTAL 9112 g W TOTAL x i M i Mw w i M i 0 05 2500g mol 0 313 7500 0 453 12 500 0 182 17 500 M W 11320 g mol P 6 8 R 8 314 J mol K z 0 0 025 0 05 0 1 0 15 0 2 0 25 0 3 0 35 0 4 0 45 0 5 0 55 0 6 0 65 0 7 0 75 0 8 erf z 0 0 0282 0 0564 0 1125 0 168 0 2227 0 2763 0 3286 0 3794 0 4284 0 4755 0 5205 0 5633 0 6039 0 642 0 6778 0 7112 0 7421 z 0 85 0 9 0 95 1 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 2 2 2 4 erf z 0 7707 0 797 0 8209 0 8427 0 8802 0 9103 0 934 0 9523 0 9661 0 9763 0 9838 0 9891 0 9928 0 9953 0 9981 0 9993
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