HW11 Solutions

HW11 Solutions - ENMA 300 Homework 11 Due Dec 7 2010 1 This...

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1 ENMA 300 Homework 11 Due: Dec 7, 2010 1. This is a combination of problems 10.5 and 10.8 because as a combination they provide important practice for interpreting phase diagrams. Remember that while book gives formulas for converting weight percent to mole percent and vice versa, it is good to understand how to convert compositions using first principles (including the atomic or molecular weight) so that you can work with compositions that contain 3 or more components (this is highly likely in ‘real life’). For each of the following alloys: i) Cite the phases that are present and the phase compositions for each phases and ii) determine the relative amounts (in terms of mass fractions) of the phases for the alloys at the given temperature: (a) 25 wt% Pb–75 wt% Mg at 425°C (800°F) i) For an alloy composed of 25 wt% Pb-75 wt% Mg and at 425 ° C, from Figure 10.20, only the α phase is present; its composition is 25 wt% Pb-75 wt% Mg. ii) For an alloy composed of 25 wt% Pb-75 wt% Mg and at 425 ° C, only the α phase is present; therefore W α = 1.0. (b) 55 wt% Zn–45 wt% Cu at 600°C (1110°F) i) ) For an alloy composed of 55 wt% Zn-45 wt% Cu and at 600 ° C, from Figure 10.19, β and γ phases are present, and C β = 51 wt% Zn-49 wt% Cu C γ = 58 wt% Zn-42 wt% Cu ii) Because the composition of the alloy, C 0 = 55 wt% Zn-45 wt% Cu, then, using the appropriate lever rule expressions and taking compositions in weight percent zinc W β = C γ - C 0 C γ - C β = 58 - 55 58 - 51 = 0.43 W γ = C 0 - C β C γ - C β = 55 - 51 58 - 51 = 0.57

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2 (c) 7.6 lb m Cu and 144.4 lb m Zn at 600°C (1110°F) i) For an alloy composed of 7.6 lb m Cu and 144.4 lb m Zn and at 600 ° C, we must first determine the Cu and Zn concentrations (using Equation 5.6 OR first principles!), as C Cu = 7.6 lb m 7.6 lb m + 144.4 lb m × 100 = 5.0 wt% C Zn = 144.4 lb m 7.6 lb m + 144.4 lb m × 100 = 95.0 wt% From Figure 10.19, only the L phase is present; its composition is 95.0 wt% Zn-5.0 wt% Cu ii) For an alloy composed of 7.6 lb m Cu and 144.4 lb m Zn (95.0 wt% Zn-5.0 wt% Cu) and at 600 ° C, only the liquid phase is present; therefore, W L = 1.0 (d) 4.2 mol Cu and 1.1 mol Ag at 900°C (1650°F) i) For an alloy composed of 4.2 mol Cu and 1.1 mol Ag and at 900 ° C, it is necessary to determine the Cu and Ag concentrations in weight percent. However, we must first compute the masses of Cu and Ag (in grams) using a rearranged form of Equation 5.7: m Cu ' = n m Cu A Cu = (4.2 mol)(63.55 g/mol) = 266.9 g m Ag ' = n m Ag A Ag = (1.1 mol)(107.87 g/mol) = 118.7 g Now, using Equation 5.6, concentrations of Cu and Ag are determined as follows: C Cu = 266.9 g 266.9 g + 118.7 g × 100 = 69.2 wt% C Ag = 118.7 g 266.9 g + 118.7 g × 100 = 30.8 wt% From Figure 10.7, α and liquid phases are present; and C α = 8 wt% Ag-92 w% Cu C L = 45 wt% Ag-55 wt% Cu
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