HW Assignment #7 MS201 Sp24 (due Tu Mar 26) (1)

MS201 Sp 24 Homework Assignment #7 Due Tue March 26, 2024 2 p.m.. Ch.9 (Topic 8) PHASE DIAGRAMS #1. (C&R9.1) Consider the sugar–water phase diagram of Figure 9.1. (a) How much sugar will dissolve in 1000 g of water at 80°C (176°F)? (wt%) = x 100 𝐶 ??𝑔𝑎? 𝑚 ??𝑔𝑎? 𝑚 ??𝑔𝑎? +𝑚 𝑤𝑎?𝑒? → 77(wt%) = x 100 = 5022 grams of sugar 𝑚 ??𝑔𝑎? 𝑚 ??𝑔𝑎? +1500 (b) If the saturated liquid solution in part (a) is cooled to 20°C (68°F), some of the sugar will precipitate out as a solid. What will be the composition of the saturated liquid solution (in wt% sugar) at 20°C? = 64 wt% sugar. (c) How much of the solid sugar will come out of solution upon cooling to 20°C? (wt%) = x 100 𝐶 ??𝑔𝑎? 𝑚 ??𝑔𝑎? 𝑚 ??𝑔𝑎? +𝑚 𝑤𝑎?𝑒? → 64(wt%) = x 100 = 2667g → 5022 – 2667 = 2355 grams of sugar 𝑚 ??𝑔𝑎? 𝑚 ??𝑔𝑎? +1500 #2. (C&R9.13) A lead–tin alloy of composition 30 wt% Sn–70 wt% Pb is slowly heated from a temperature of 150°C (300°F). (a) At what temperature does the first liquid phase form? – At 183C (b) What is the composition of this liquid phase? – 61.9 wt% Sn (c) At what temperature does complete melting of the alloy occur? – At 260C (d) What is the composition of the last solid remaining? – 13 wt% Sn 1

#3. (C&R9.14) A 50 wt% Ni–50 wt% Cu alloy is slowly cooled from 1400°C (2550°F) to 1200°C (2190°F). (a) At what temperature does the first solid phase form? – 1320C (b) What is the composition of this solid phase? – 38% Sn (c) At what temperature does the liquid solidify? – 1270C (d) What is the composition of this last remaining liquid phase? – 63% Sn 2

1000 – L (100%, completely liquid) 800 – (100%, completely ) β β 500 – + y β → Fraction of y = = 33.33% weight 52−49 58−49 → Fraction of = = 67.67% weight β 58−52 58−49 300 – + y β → Fraction of y = = 22.22% weight 52−50 59−50 → Fraction of = = 77.78% weight β 59−52 59−50 #6. (C&R9.21) A 65 wt% Ni–35 wt% Cu alloy(Fig.9.3a) is heated to a temperature within the (α + liquid) phase region. If the composition of the α phase is 70 wt% Ni, determine: (a) The temperature of the alloy – ~1340C (b) The composition of the liquid phase – ~59 wt% Ni (c) The mass fractions of both phases – = → = 0.55% 𝑊 𝑎 𝐶 𝐿 − 𝐶 𝐿 𝐶 𝑎 − 𝐶 𝐿 65−59 70−59 = → = 0.45% 𝑊 𝐿 𝐶 𝑎 − 𝐶 0 𝐶 𝑎 − 𝐶 𝐿 70−65 70−59 4

#7. (C&R9.43) For a 76 wt% Pb–24 wt% Mg alloy (Fig. 9.20), make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 575°C (1070°F), 500°C (930°F), 450°C (840°F), and 300°C (570°F). Label all phases and indicate their approximate compositions. #8. (C&R9.44) For a 52 wt% Zn–48 wt% Cu alloy, make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 950°C (1740°F), 860°C (1580°F), 800°C (1470°F), and 600°C (1100°F). Label all phases and indicate their approximate compositions. 5

#9. (C&R9.57) Compute the mass fractions of α-ferrite and cementite in pearlite. The percentages are as follows. Cα = 0.022%. Cc = 6.70%. Co = 0.77% According to the lever rule, Wc = 11.2 #10. (C&R9.58) ( a) What is the distinction between hypoeutectoid and hypereutectoid steels? (b) In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explain the difference between them. What will be the carbon concentration in each? Hypoeutectoid Steel: ● Contains less carbon than the eutectoid composition (typically less than 0.76% carbon). ● During cooling, the first phase to solidify is ferrite. ● Austenite transforms into a mixture of ferrite and cementite during cooling. ● Examples include plain carbon steels with carbon content below 0.76% Hypereutectoid Steel: ● Contains more carbon than the eutectoid composition (typically more than 0.76% carbon). ● During cooling, the first phase to solidify is cementite (iron carbide). ● Austenite transforms into a mixture of cementite and pearlite during cooling. ● Examples include plain carbon steels with carbon content above 0.76% Proeutectoid Ferrite: ● Forms during the cooling of the austenitic phase before reaching the eutectoid temperature. ● Carbon concentration: Less than the eutectoid composition (usually around 0.022%). Eutectoid Ferrite: ● Forms during the eutectoid transformation at the eutectoid temperature. ● Carbon concentration: Equal to the eutectoid composition (0.76%) #11. (C&R9.67) The microstructure of an iron–carbon alloy consists of proeutectoid ferrite and pearlite; the mass fractions of these two microconstituents are 0.174 and 0.826, respectively. Determine the concentration of carbon in this alloy. Wp = Co’ – 0.022 / 0.74 = 0.826 → Co’ – 0.022 = 0.61124 = 0.63324 → 0.633 wt% C 7

#12. (C&R9.68) The mass fractions of total ferrite and total cementite in an iron–carbon alloy are 0.91 and 0.09, respectively. Is this a hypoeutectoid or hypereutectoid alloy? Why? The eutectoid composition for iron-carbon alloy is approximately 0.76% carbon. The given composition of 0.09% carbon in the alloy indicates it is hypoeutectoid, as it's less than the eutectoid composition (using the lever rule) #13. (C&R9.78) For an iron–carbon alloy of composition 3 wt% C–97 wt% Fe, make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 1250°C (2280°F), 1145°C (2095°F), and 700°C (1290°F). Label the phases and indicate their compositions (approximate). 8