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hw sol ch10 - 10.1 Three variables that determine the...

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Unformatted text preview: 10.1 Three variables that determine the microstructure of an alloy are 1) the alloying elements present, 2) the concentrations of these alloying elements, 3) the heat treatment of the alloy 10.10 (a) Spreading salt on ice will lower the melting temperature, since the liquidus line decreases from 0°C to the eutectic temperature at about -21°C. Thus, ice at a temperature below 0°C (and above -21°C) can be made to form a liquid phase by the addition of salt. (b) We are asked to compute the concentration of salt necessary to have a 50% ice—50% brine solution at 40°C (14°F). At 40°C, C. = 0 wt% NaCl—1DCI wt% H O Ice 2 C . = 13 wt% NaCI—B? wt% H O bnne 2 Thus, W- —0.5— Cbrine—Co _13_Co Ice Cbrine _ Cice 13 _ 0 Solving for Co (the concentration of salt) yields a value of 6.5 wt% Nam—93.5 wt% H20. 10.15 (a) This portion of the problem asks that we calculate, for a Pb—Mg alloy, the mass of lead in 5.5 kg of the solid on phase at 200°C just below the solubility limit. From Figure 9.18, the composition of an alloy at this temperature is about 5 wt% Pb. Therefore, the mass of Pb in the alloy is just (0.05)(5.5 kg) = 0.28 kg. (b) At 350°C, the solubility limit of the 0'. phase increases to approximately 25 wt% Pb. In order to determine the additional amount of Pb that may be added (mph), we utilize a modified form of Equation (4.3) as 0.28 kg + me 6P0 = 25 wt% 2m): 100 Solving for me yields me = 1.46 kg. 10.26 For this problem Co = 55 (or 55 wt% B45 wt% A) CB = 90 (or 90 wt% 3—10 wt% A) If we set up the lever rule for Wu w 2052(30'00290—55 0r. 90% 90‘00 And solving for Cu C”. = 20 (or 20 wt% 3—80 1.771% A) 10.41 The principal difference between congruent and incongruent phase transformations is that for congruent no compositional changes occur with any of the phases that are involved in the transformation. For incongruent there will be compositional alterations of the phases. 10.50 This problem asks that we compute the mass fractions of ferrite and cementite in pearlite. The lever—rule expression for ferrite is and, since C = 6.?0 1.141% C, C = 0.76 wt% C, and C = 0.022 wt% C Fe3c o 05 _ 0.70 — 0.70 _ we: ' 0.70 — 0.022 ' 0'89 Similarly, for cementite c —c _ w o a _0.70 0.022 = _ = 0.11 Fe3C CFeRC — Cu 6.70 — 0.022 10.63 This problem asks that we compute the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid iron—carbon alloy. This requires that we utilize Equation (9.23) with Ci = 2.14 wt% C, the maximum solubility of carbon in austenite. Thus, w _'34 "0-75 _ 2.14 — 0.7’6 _ 0232 FeqC'_ 5.94 ‘ 5.94 " ...
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