# ENGR 321: Homework Set 6 Fall 2002

### Prof. W. H. Warnes Office: Rogers Hall 308 Phone and Voice Mail: (541) 737-7016 FAX: (541) 737-2600

email: warnesw@engr.orst.edu

### PHASE DIAGRAMS

1) Callister 9.5, b, d, f, and g.

2) Callister 9.15

3) Identify all the invarient and congruent points on the Mg-Pb binary phase diagram (p. 274). List them by composition, temperature of reaction, type of reaction, and the phase changes that occur on cooling through the reaction temperature.

4) Use the phase diagram for the Mg-Pb binary system on page 274 of Callister to answer the following questions.

a) Describe the equilibrium cooling process and likely microstructure for an alloy of Mg-20wt%Pb.

b) Describe the equilibrium cooling process and likely microstructure for an alloy of Mg-45wt%Pb.

c) At 400 deg C, what changes would happen to the system as we slowly add Pb to pure Mg?

5) Refering to figure 10.9 for the transformation of austenite, briefly explain why it takes a long time for the transformation to begin at high temperatures (near the eutectoid temperature) and also at low temperatures (around 250 deg C), but less time at intermediate temperatures.

6) Briefly explain the difference between a EUTECTIC, EUTECTOID, and PERITECTIC phase transformation.

7) Refering to Figure 10.9, describe the microstructure of the final steel using the following cooling processes:

1. from 750 deg C rapidly cool to 600 deg C, hold for 1 second, and rapidly cool to room temperature;
2. from 750 deg C rapidly cool to 600 deg C, hold for 10 second, and rapidly cool to room temperature;
3. from 750 deg C rapidly cool to 600 deg C, hold for 100 seconds, and rapidly cool to room temperature;
4. from 750 deg C rapidly cool to 300 deg C, hold for 10 seconds, and rapidly cool to room temperature;
5. from 750 deg C rapidly cool to 300 deg C, hold for 100 seconds, and rapidly cool to room temperature.

NOTE: For this problem, do not consider the martensitic phase transformation in giving your answers.

### SOLUTIONS

1) Callister 9.5, b, d, f, and g.

b) Use the Mg-Pb binary phase diagram on page 274. At a composition of Pb-75w%Mg, at a temperature of T = 425 deg C, we are in the single phase ALPHA solid region. The answer is:

100% ALPHA solid of composition Pb-75w%Mg.

d) Use the Cu-Zn binary phase diagram on page 273. At a composition of Cu-55w%Zn, at a temperature of T = 600 deg C, we are in the two phase BETA + GAMMA solid region. From the isotherm at 600 deg C, we find that the composition of the two phases in equilibrium are about Cu-50.5w%Zn for BETA, and Cu-58w%Zn for GAMMA. We can use the lever rule to find the mass fraction of each phase: M% of BETA = (58-55)/(58-50.5) = 40%. The answer is:

40% BETA solid of composition Cu-50.5w%Zn, plus
60% GAMMA solid of composition Cu-58w%Zn.

f) Use the Cu-Zn binary phase diagram on page 273. Convert the mass composition to weight % composition. At a composition of Cu-95w%Zn, at a temperature of T = 600 deg C, we are in the single phase Liquid region. The answer is:

100% Liquid of composition Cu-95w%Zn.

g) Use the Cu-Ag binary phase diagram on page 261. The composition is given as Cu-20.8a%Ag. Convert the atomic percent composition to weight percent cdomposition to get the overall composition to be: Cu-30.8w%Ag. At a composition of Cu-30.8w%Ag, at a temperature of T = 900 deg C, we are in the two phase ALPHA + liquid region. From the isotherm at 900 deg C, we find that the composition of the two phases in equilibrium are about Cu-8w%Ag for ALPHA, and Cu-43.5w%Ag for Liquid. We can use the lever rule to find the mass fraction of each phase: M% of ALPHA = (43.5-30.8)/(43.5-8) = 35.8%. The answer is:

35.8% ALPHA solid of composition Cu-8w%Ag, plus
64.2% Liquid of composition Cu-43.5w%Ag.

2) Callister 9.15

 a) The composition of the alloy at the solubility limit at 300 deg C (see point a on the phase diagram) is about Mg-16.7w%Pb. Since the total mass of the alloy is 7.5kg, this means there are 1.25kg of Pb in the alloy. b) At 400 deg C (point b), the solubility limit is about Mg-31w%Pb. The amount of Mg doesn't change from that in part a, so there are 6.25kg of Mg. Since this mass of Mg is 69% of the total alloy, the alloy now weighs (6.25kg)/(0.69) = 9.06kg. The 2.81kg that is not Mg must be Pb, which means we've added 2.81kg. - 1.25kg = 1.56kg of Pb to the alloy.

3) Identify all the invarient and congruent points on the Mg-Pb binary phase diagram.

There are FIVE invarient points in this phase diagram. listing them from left to right (as marked with orange dots on the phase diagram) we have:

 No. Composition Temp. [deg C] Type Phase change 1 Mg-0w%Pb 640 melting temp L to ALPHA 2 Mg-65w%Pb 455 eutectic L to ALPHA + Mg2Pb 3 Mg-81w%Pb 550 congruent melting L to Mg2Pb 4 Mg-97w%Pb 240 eutectic L to Mg2Pb + BETA 5 Mg-100w%Pb 330 melting temp L to BETA

4) Use the phase diagram for the Mg-Pb binary system on page 274 of Callister to answer the following questions.

 a) Describe the equilibrium cooling process and likely microstructure for an alloy of Mg-20wt%Pb. T>T1: 100% L of Mg-20w%Pb. See microstructure (a). T1>T>T2: Two-phase mixture of L and ALPHA-solid. See microstructure (b). T2>T>T3: Single phase ALPHA of Mg-20w%Pb. See microstructure (c). T3>T: Two-phase mixture of ALPHA and Mg2Pb. See microstructure (d). MICROSTRUCTURES: a) b) c) d) b) Describe the equilibrium cooling process and likely microstructure for an alloy of Mg-45wt%Pb. T>T1: 100% L of Mg-45w%Pb. See microstructure (a). T1>T>T2: Two-phase mixture of L and ALPHA-solid. See microstructure (b). T2>T: Two-phase mixture of ALPHA and Mg2Pb. There is primary ALPHA combined with the eutectic structure of ALPHA + Mg2Pb. See microstructure (c). a) b) c) c) At 400 deg C, what changes would happen to the system as we slowly add Pb to pure Mg? For compositions < C1, he alloy will be in the single phase ALPHA solid region, until the amount of Pb added increases the composition to C1. For compositions C1< C < C2 the alloy will be in the two-phase region of ALPHA and Mg2Pb, with increasing mass fractions of Mg2Pb as Pb is added. At C2 the alloy will be 100% Mg2Pb solid phase. For compositions C2< C C3, the alloy will be 100% liquid.

5) Refering to figure 10.9 for the transformation of austenite, briefly explain why it takes a long time for the transformation to begin at high temperatures (near the eutectoid temperature) and also at low temperatures (around 250 deg C), but less time at intermediate temperatures.

At higher temperatures, just below the eutectoid temperature, the amount of undercooling is very small, and there is not enough undercooling (or driving force) to have rapid nucleation of the ferrite and cementite phases. At high temperatures the transformation is NUCLEATION LIMITED. At lower temperatures, the nucleation rate increases, so the reaction moves to shorter times. At still lower temperatures, however, the diffusion of the atoms slows down because of the decrease in the diffusivity with temperature. This again slows down the reaction rate to occur at longer times. At low temperatures the reaction is DIFFUSION LIMITED.

(NOTE: Nucleation Theory will NOT be a part of the final exam for Fall 2002.)

6) Briefly explain the difference between a EUTECTIC, EUTECTOID, and PERITECTIC phase transformation.

An EUTECTIC is defined as a transformation in which a LIQUID phase changes to TWO DIFFERENT SOLIDS on cooling.

An EUTECTOID is defined as a transformation in which a SOLID phase changes to TWO DIFFERENT SOLIDS on cooling.

A PERITECTIC is defined as a transformation in which a LIQUID and a SOLID phase change to A SINGLE DIFFERENT SOLID phase on cooling.

7) Refering to Figure 10.9, describe the microstructure of the final steel using the following cooling processes:

 a) from 750 deg C rapidly cool to 600 deg C, hold for 1 second, and rapidly cool to room temperature; 100% metastable austenite (A). b) from 750 deg C rapidly cool to 600 deg C, hold for 10 second, and rapidly cool to room temperature; 100% pearlitic microstructure (lamellar composite of ferrite and iron carbide.) c) from 750 deg C rapidly cool to 600 deg C, hold for 100 seconds, and rapidly cool to room temperature; 100% pearlitic microstructure (lamellar composite of ferrite and iron carbide.) d) from 750 deg C rapidly cool to 300 deg C, hold for 10 seconds, and rapidly cool to room temperature; 100% metastable austenite (A). e) from 750 deg C rapidly cool to 300 deg C, hold for 100 seconds, and rapidly cool to room temperature. 10% Bainite and 90% metastable austenite.

(NOTE: Transformation Kinetics will NOT be a part of the final exam for Fall 2002.)

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