Phase Diagrams and Alloy Microstructures, Exams of Mechanics

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MECH221 Assignment mid #7 – Concordia University

9.11 Is it possible to have a copper–silver alloy that, at equilibrium, consists of a β phase of composition 92 wt% Ag–8 wt% Cu and also a liquid phase of composition 76 wt% Ag–24 wt% Cu? If so, what will be the approximate temperature of the alloy? If this is not possible, explain why.

Solution It is possible to have a Cu-Ag alloy, which at equilibrium consists of a phase of composition 92 wt% Ag- 8 wt% Cu and a liquid phase of composition 77 wt% Ag-23 wt% Cu. From Figure 9.7 a horizontal tie line can be constructed across the + L phase region at about 800 C which intersects the L – ( + L ) phase boundary at 76 wt% Ag, and also the ( + L )– phase boundary at 92 wt% Ag.

9.13 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? (b) What is the composition of this liquid phase? (c) At what temperature does complete melting of the alloy occur? (d) What is the composition of the last solid remaining prior to complete melting?

Solution Upon heating a lead-tin alloy of composition 30 wt% Sn-70 wt% Pb from 150 C and utilizing Figure 10. as shown below:

(a) The first liquid forms at the temperature at which a vertical line at this composition intersects the eutectic isotherm--i.e., at 183 C.

W C^0 C^^52 49 0.

C C 58 49

W = 1.00 – 0.33 = 0.

At 300 C, the and phases are present, and from the tie line constructed above

W C^ C^0

C C^59

W = 1.00 – 0.78 = 0.

9.18 A 2.0-kg specimen of an 85 wt% Pb–15 wt% Sn alloy is heated to 200°C (390°F); at this temperature it is entirely an α-phase solid solution (Figure 9.8). The alloy is to be melted to the extent that 50% of the specimen is liquid, the remainder being the α phase. This may be accomplished 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 2.0-kg specimen at 200°C to achieve this state? Solution (a) This part of the problem calls for us to cite the temperature to which a 85 wt% Pb-15 wt% Sn alloy must be heated in order to have 50% liquid. Probably the easiest way to solve this problem is by trial and error— that is, on the Pb-Sn phase diagram (Figure 9.8), moving vertically at the given composition, through the + L region until the tie-line lengths on both sides of the given composition are the same. This occurs at approximately 280 C (535 F). (b) We can also produce a 50% liquid solution at 200 C, by adding Sn to the alloy. At 200 C and within the + L phase region, compositions of and liquid phases are determined using a tie line constructed at this temperature, as follows

C = 17 wt% Sn-83 wt% Pb CL = 57 wt% Sn-43 wt% Pb Let

C 0 be the new alloy composition to give W = WL = 0.5. Then,

W = 0.5 = CL^ C^0 CL C

=^5757 C 170

And solving for C 0 gives 37 wt% Sn. Now, let m Sn be the mass of Sn added to the alloy to achieve this new

composition. The amount of Sn in the original alloy is

(0.15)(2.0 kg) = 0.30 kg

Then, using a modified form of Equation 4.3a

0.30 kg m Sn

2.0 kg m Sn

And, solving for m Sn (the mass of tin to be added), yields m Sn = 0.698 kg.

9.21 A 65 wt% Ni–35 wt% Cu alloy 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 (b) The composition of the liquid phase (c) The mass fractions of both phases

Solution (a) In order to determine the temperature of a 65 wt% Ni-35 wt% Cu alloy for which and liquid phases are present with the phase of composition 70 wt% Ni, we need to construct a tie line across the + L phase region of Figure 10.3a that intersects the solidus line at 70 wt% Ni; this is possible at about 1345 C. (b) The composition of the liquid phase at this temperature is determined from the intersection of this same tie line with liquidus line, which corresponds to about 59 wt% Ni. (c) The mass fractions of the two phases are determined using the lever rule, Equations 9.1 and 9.2 with C 0 = 65 wt% Ni, CL = 59 wt% Ni, and C = 70 wt% Ni, as

W = C^0 CL

C CL

=^65 59 = 0.

W = C^ C^0 =^70 65 = 0.

L C CL 70 59

9.24 For alloys of two hypothetical metals A and B, there exist an α, A-rich phase and a β, B-rich phase. From the mass fractions of both phases for two different alloys provided in the following table (which are at the same temperature), determine the composition of the phase boundary (or solubility limit) for both α and β phases at this

Alloy Composition Fraction α Phase Fraction β Phase 70 wt% A– 30 wt% B 0.78 0. 35 wt% A– 65 wt% B 0.36 0.

Solution The problem is to solve for compositions at the phase boundaries for both and phases (i.e., C and C ). We may set up two independent lever-rule expressions, one for each composition, in terms of C and C ; these lever-rule equations may be for either W or W ; we have chosen to use expressions in terms of W , which are as follows:

W (^) 1

W 2

= 0.78 = C^ C^01

C C

= 0.36 = C^ C^02

C C

= C C^^70 C

= C C^^35 C

In these expressions, compositions are given in wt% of A. Solving for C and C from these equations, yield

C = 88.3 (or 88.3 wt% A-11.7 wt% B)

C = 5.0 (or 5.0 wt% A-95.0 wt% B)

9.37 For 2.8 kg of a lead–tin alloy, is it possible to have the masses of primary β and total β of 2.21 and 2.53 kg, respectively, at 180°C (355°F)? Why or why not?

Solution In order to decide we first need to compute the mass fraction of each of primary and total. It is then necessary to set up the appropriate lever rule expression for the determination of each of these two microconstituents and then solve for the alloy composition C 0. If the value of C 0 is the same for both primary and total , then the alloy is possible.

Mass fractions of primary and total (designed by W ' and W , respectively) are computed as follows:

W = 2.21 kg = 0. ' (^) 2.8 kg

2

W = 2.53 kg = 0. 2.8 kg

Next it is necessary to set up the appropriate lever rule expression for each of these quantities. From Figure 9. and at 180 C, C = 18.3 wt% Sn, C = 97.8 wt% Sn, and C eutectic = 61.9 wt% Sn.

The lever-rule expression for primary is as follows:

W = C^0 C eutectic^ = C^0 61.9^ = 0. ' (^) C C eutectic 97.8 61.

And solving for C 0 gives C 0 = 90.2 wt% Sn. The analogous expression for total is W = C^0 C^ = C^0 18.3 = 0. C C 97.8 18.

which also yields a value for C 0 of 90.2 wt% Sn. Therefore, since these two C 0 values are identical, this alloy is

possible.

9.40 A magnesium-lead alloy is cooled from 600 C to 450 C and is found to consist of primary Mg 2 Pb and eutectic

microconstituents. If the mass fraction of the eutectic microconstituent is 0.28, determine the alloy composition.

Solution Because primary Mg 2 Pb and the eutectic structure are the microconstituents, we know that the alloy

composition must lie between the eutectic composition (67 wt% Pb-33 wt% Mg, Figure 9.20). The lever rule expression for the mass fraction of the eutectic microconstituent is as follows:

W^ C Mg Pb^ C^0

eutectic 2

C Mg Pb C eutectic

Here C 0 = the alloy composition. Inasmuch as

W eutectic = 0.

C Mg Pb = 81 wt% Pb

C eutectic = 67 wt% Pb

We solve for the value of C 0 by rearrangement of the above equation as follows:

C 0 C Mg 2 Pb W eutectic( C Mg 2 Pb C eutectic)

9.50 Specify the liquidus, solidus, and solvus temperatures for the following alloys: (a) 30 wt% Ni-70 wt% Cu (b) 5 wt% Ag-95 wt% Cu (c) 20 wt% Zn-80 wt% Cu (d) 30 wt% Pb-70 wt% Mg (e) 3 wt% C-97 wt% Fe

Solution Definitions of liquidus, solidus, and solvus lines are as follows: Liquidus— boundary between liquid and (liquid + solid) phase regions Solidus—boundary between solid and (liquid + solid) phase regions Solvus—boundary between a single solid phase region and a two solid phase region.

For this problem, a liquidus, solidus, or solvus temperature corresponds to the intersection between a vertical line (constructed at a specific composition) with the respective liquidus, solidus, or solvus line.

(a) The liquidus and solidus temperatures for a 30 wt% Ni-70 wt% Cu alloy (1195 C and 1240 C, respectively) are noted on the Cu-Ni phase diagram shown below. For this alloy, there is no solvus line inasmuch as

no two-solid-phase region exists for this system.

(d) The solvus, solidus, and liquidus temperatures for a 30 wt% Pb-70 wt% Mg alloy are 390 C, 520 C, and 615 C, respectively, as noted on the Mg-Pb phase diagram shown below.

(e) The liquidus temperature for 3 wt% C-97 wt% Fe is 1290 C as noted on the Fe-Fe 3 C phase diagram shown below. For this alloy, there is neither a solidus line or a solvus line.

9.53 Figure 9.37 is a portion of the copper–aluminum phase diagram for which only single-phase regions are labeled. Specify temperature–composition points at which all eutectics, eutectoids, peritectics, and congruent phase transformations occur. Also, for each, write the reaction upon cooling.

(^33)

There is a single congruent melting point that exists at 12.5 wt% Al-87.5 wt% Cu and 1049 C. The reaction upon cooling is

L

9.61 Consider 3.5 kg of austenite containing 0.95 wt% C and cooled to below 727°C (1341°F). (a) What is the proeutectoid phase? (b) How many kilograms each of total ferrite and cementite form? (c) How many kilograms each of pearlite and the proeutectoid phase form? (d) Schematically sketch and label the resulting microstructure.

Solution

wt% C).

(a) The proeutectoid phase will be Fe 3 C since 0.95 wt% C is greater than the eutectoid composition (0.

(b) For this portion of the problem, we are asked to determine how much total ferrite and cementite form. Application of the appropriate lever rule expression for the mass fraction of total yields

C Fe C C 0 W = 3 = 3

which, when multiplied by the total mass of the alloy, gives (0.86)(3.5 kg) = 3.01 kg of total ferrite. Similarly, for the mass fraction of total cementite,

W Fe C =^ CC Fe C^0 CC = 0.95 6.70^ 0.0220.022 = 0.

And the mass of total cementite that forms is (0.14)(3.5 kg) = 0.49 kg. (c) Now we are asked to calculate how much pearlite and the proeutectoid phase (cementite) form.

To determine the mass fraction of pearlite that forms ( Wp ), we apply Equation 9.22, in which C 1 0.95 wt% C:

W p = 6.70^ C^1 = 6.70^ 0.95 = 0. 6.70 0.76 6.70 0.

which corresponds to a mass of (0.97)(3.5 kg) = 3.4 kg. Likewise, from Equation 9.23, we calculate the mass fraction of proeutectoid cementite as follows:

W Fe (^) 3 C = C^1 5.940.76 = 0.955.94^ 0.76 = 0.

which is equivalent to (0.03)(3.5 kg) = 0.11 kg of the total 3.5 kg mass. (d) Schematically, the microstructure would appear as follows:

9.65 Compute the mass fractions of proeutectoid ferrite and pearlite that form in an iron–carbon alloy containing 0.35 wt% C.

Solution The mass fractions of proeutectoid ferrite and pearlite that form in a 0.35 wt% C iron-carbon alloy are considered in this problem. To solve this problem we use Equation 9.20, which is the lever-rule expression for computation of the mass fraction of pearlite in a hypoeutectoid alloy. This computation is as follows:

Wp = C^0 0.740.022 = 0.350.74^ 0.022 = 0.

Likewise, the mass fraction of proeutectoid ferrite is determined using Equation 9.21:

W = 0.76^ C^0 =0.76^ 0. 0.74 0.74 = 0.

9.70 Consider 1.5 kg of a 99.7 wt% Fe–0.3 wt% C alloy that is cooled to a temperature just below the eutectoid. (a) How many kilograms of proeutectoid ferrite form? (b) How many kilograms of eutectoid ferrite form? (c) How many kilograms of cementite form?