MSE2160 Problem Set 6 Chapters 10 and 11: Phase Diagrams and Phase Change via Canvas on Tuesday, March 26th 1. A 1.5-kg specimen of a 90 wt% Pb-10 wt% Sn alloy is heated to 250 °C (480 F), at which temperature it is entirely an α-phase solid solution (Figure 10.7). The alloy is to be melted to the extent that 50% of the specimen is liquid, the remainder being the a-phase. This may be accomplished either by heating the alloy or changing its composition while holding the temperature constant. a. To what temperature must the specimen be heated? b. How much tin must be added to the 1.5-kg specimen at 250 °C to achieve this state? Temperature (°C) 300 200 100 (Pb) 327°C α + L Liquid 600 500 232°C 183°C B+L 400 β 18.3 61.9 97.8 α + β 20 40 60 80 100 Composition (wt% Sn) (Sn) 300 200 100 Temperature (°F) 2. A 45 wt% Pb-55 wt% Mg alloy is rapidly quenched to room temperature from an elevated temperature in such a way that the high-temperature microstructure is preserved. This microstructure is found to consist of the a phase and Mg2Pb, having respective weight fractions of 0.65 and 0.35. Determine Ca and the temperature from which the alloy was quenched. Temperature (°C) 700 0 5 10 600 α + L 500 400 300 200 100 0 0 20 40 (Mg) L α + Mg2Pb 60 Composition (wt% Pb) 20 30 40 70 100 1200 L Mg2Pb M 1000 800 В L + Mg2Pb 600 β 400 ẞ + Mg2Pb 200 Mg2Pb 80 100 (Pb) Temperature (°F) 3. The temperatures of solidus and liquidus of the silicon-germanium system are listed here. a. Draw the phase diagram and label each region. b. Is this alloy isomorphous? Explain why. c. An alloy with 50%wt is held at 1300 °C. Composition (wt% Si) Solidus Temperature Liquidus Temperature (°C) (°C) Determine the relative amounts of the two phases. 0 938 938 10 1005 1147 20 1065 1226 30 1123 1278 40 1178 1315 50 1232 1346 60 1282 1367 70 1326 1385 80 1359 1397 90 1390 1408 100 1414 1414 4. A specimen of hypoeutectoid (meaning composition is below the eutectoid) steel having a composition of 0.5% C and a mass of 1 kg is slowly cooled from a temperature 850 °C. a. If we stop cooling at a temperature just above the eutectoid isotherm, determine the fractions of the two phases and the total weight of the ferrite a phase in the specimen. This phase is called proeutectoid (meaning pre- or before eutectoid) ferrite. b. Next, the specimen is further cooled slowly to room temperature, which transforms the rest of the austenite into pearlite. Determine the total weight fraction of ferrite and the total mass of ferrite in the specimen at room temperature. c. The weight fraction of protoeutectoid ferrite (part a) is not affected by the additional cooling to room temperature. Calculate the weight fraction of the total specimen that is eutectoid ferrite at room temperature. 5. Make a copy of the isothermal transformation diagram for an iron-carbon alloy of eutectoid composition (Figure 11.14, relevant portion reproduced below). a. Sketch the following transformations on it and specify the nature of the final microstructure (in terms of microconstituents present and approximate percentages of each) of a small specimen that has been subjected to the following time-temperature treatments. In each case assume that the specimen begins at 760 °C and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. i. Cool rapidly to 700 °C, hold for 104 s, then quench to room temperature. ii. Reheat the specimen in part a to 700 °C for 20 h. iii. Determine and sketch the time- temperature path that produces 100% coarse pearlite microstructure. b. The transformation of austenite to pearlite in an iron-carbon alloy at 675 °C (dashed line) was found to have a rate of 0.01 s¹. Determine the time it takes to transform 80% of austenite to pearlite. Assume the exponent n=1. Temperature (°C) 1400 Austenite (stable) Eutectoid temperature -+ 700 Austenite (unstable) 600 500 400 Pearlite -50% Completion curve Completion curve (~100% pearlite) Begin curve (~0% pearlite) T 1200 1 10 10² 103 104 105 Time (s) 1000 800 Temperature (°F)