test3 - Home Introduction Policies Requirements & Grading...

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Home Introduction Policies Requirements & Grading Resources Syllabus Assignments Handouts Discussion Web Links Distance Learners Contact TEST#3 MSE 200 Test #3 S. S. Number: Read the question carefully and answer. Use the iron carbon diagram given at the end. Q1. Show that for a steel of highest carbon percentage, fraction of cementite at room temperature is no greater than 22%. Points :4 Highest carbon steel :1.4%weight carbon Fraction of cementite=(1.4-0.005)/(6.7-0.005)=1.4/6.7=0.208 Q2. Determine the fraction of proeuetctoid and eutectoid phases in 0.4%carbon steel and dra
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Above 727 o C, fa=(0.76 -0.4)/(0.76-0.022)=0.36/0.76=0.473 - proeutectoid ferrite Below 727 o C, fa=(6.67-0.4)/(6.67-0.022)=6.27/6.67=0.94 - total ferrite (eutectoid plus proeutectoid) Eutectoid ferrite= total-proeutectoid=0.94-0.473=0.466 Eutectoid cementite=0.06 Q3: a. Determine the fraction of cementite present above eutectoid temperature in 4.3%carbo cast iron. b. Determine the fraction formed below the eutectic temperature 1143 o C. c. Explai the difference in the values calculated fron (a) to (b). Points:6 Fraction of cementite above 727 o C=(4.3-0.76)/(6.67-0.76) =3.54/5.91=0.599 Fraction of cementite below 1143 o C=(4.3-2.14)/(6.67-2.14) =2.16/4.53=0.477 Difference=0.599-0.477=0.122 arises because of the sloping solvus line and solubility of carbon in austenite decreases from 2.14% to 0.76% Q4. a. Label the TTT diagram given below by showing different microstructures obtained in the eutectoid steel by isothermal transformation. b. The steel is subjected to isothermal transformation given by the line in the diagram. What is the resultant microstructure. Points: 50% fine pearlite and 50% bainite. Q5. What fraction by moles of Na
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This note was uploaded on 10/11/2009 for the course MSE 200 taught by Professor Wholedepartment during the Spring '08 term at N.C. State.

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test3 - Home Introduction Policies Requirements & Grading...

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