MSE 230 Solutions for Assignment 8 Fall 2014 1. a) For Li+ substituting for Ca2+ in CaO, oxygen vacancies would be created. For each Li+ substituting for Ca2+, one positive charge is removed; in order to maintain charge neutrality, a single negative charge may be removed. Negative charges are eliminated by creating oxygen vacancies, and for every two Li+ ions added, a single oxygen vacancy is formed.

b) For Cl- substituting for O2- in CaO, calcium vacancies would be created. For each Cl- substituting for an O2-, one negative charge is removed; in order to maintain charge neutrality, two Cl- ions will lead to the formation of one calcium vacancy. c) In general it takes less energy to form a vacancy than an interstitial. This can be seen by looking at close packed structures (see Fig. 12.8 for example). For an interstitial, a significant amount of lattice strain would be produced by stuffing an extra atom or ion into the already crowded structure. In contrast, there is relatively little lattice strain involved in vacancy formation (although we still need to break few bonds). 2.* a) When the two Al2O3 cubes are brought into contact, there is a sharp Mn4+ cation

concentration gradient due to the doping. Since we are at room temperature diffusion in Al2O3 will be so slow that essentially no diffusion will occur and the concentration gradient will remain.

b) Heating the bicrystal to 1300°C increases the diffusivity such that the Mn4+ cations diffuse from the doped Al2O3 cube into the high purity Al2O3 cube. This is reflected by the concentration profile in the diagram. If we wait long enough the Mn4+ cation concentration should be uniform throughout the two cubes.

c) Because the cations have a similar ionic radius, Mn4+ ions will substitute for Al3+ cations in the Al2O3 structure, resulting in a net +1 charge for each Mn4+ substituting for Al3+. To compensate for charge imbalance, an Al3+ vacancy must form for every three Mn4+ substituting for Al3+. O2- anion interstitials are also possible but highly unlikely.

d) At 2071°C, just below the melting temperature of Al2O3. This is where the rate of diffusion is the highest but we have not yet melted the crystal.

3.* In this problem we are asked to cite which of the elements listed form with Cu the three possible solid

solution types. For complete substitutional solubility the following criteria must be met: 1) the

difference in atomic radii between Ni and the other element (ΔR%) must be less than ±15%, 2) the

crystal structures must be the same, 3) the electronegativities must be similar, and 4) the valences

should be the same, or nearly the same. Below are tabulated, for the various elements, these

criteria.

Crystal ΔElectro-

Element ΔR% Structure negativity Valence

Cu FCC 2+

C -44

H -64

O -53

Ag +13 FCC 0 1+

Al +12 FCC -0.4 3+

Al2O3 dopedwith Mn4+ Al2O3

Mn4+

concentation(atm. %)

Position

10-5After bonding at

room temperature

Mn4+

concentation(atm. %)

Position

10-5After heating at

1300°C for 24 hours.

Co -2 HCP -0.1 2+

Cr -2 BCC -0.3 3+

Fe -3 BCC -0.1 2+

Ni -3 FCC -0.1 2+

Pd +8 FCC +0.3 2+

Pt +9 FCC +0.3 2+

Zn +4 HCP -0.3 2+

(a) Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having complete

solubility.

(b) Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these

metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii

and that for Ni are greater than ±15%, and/or have a valence different than 2+.

(c) C, H, and O form interstitial solid solutions. These elements have atomic radii that are

significantly smaller than the atomic radius of Cu. 4. Please refer to the Ni-Cu phase diagram in figure 9.3 of Callister. Making an enlarged

photocopy of the phase diagram will also help. Draw a vertical line at the 50 wt.% Ni composition on the x-axis. Start at 1400°C and

follow the line down. a) The first solid forms at approximately 1310°C. This is where the vertical line

intersects the liquidus line. We are at the boundary between the liquid phase and the α + L two phase region.

b) The composition is determined by drawing a tie line across the α + L two phase

region. Where the tie line intersects the solidus line gives the composition of the solid phase. The solid phase contains approximately 61 wt.% Ni and 39 wt.% Cu.

c) Dropping further down the vertical line, the last liquid solidifies where the vertical

line intersects the solidus line. Now we are at the boundary between the α + L two phase region and the solid α phase. The last liquid solidifies at approximately 1270°C.

d) As in part b), draw a tie line spanning the α + L two phase region. Where the tie line

intersects the liquidus line gives the composition of the last remaining liquid which is approximately 37 wt.% Ni and 63 wt.% Cu.

5.* From Figure 9.6a, a tensile strength greater than 350 MPa is possible for compositions between about 22.5 and 98 wt% Ni. On the other hand, according to Figure 9.6b, a ductility greater than 48%EL exists for compositions less than about 8 wt% and greater than about 98 wt% Ni. Therefore, the stipulated criteria are met only at a composition of 98 wt% Ni. 6. Please refer to the Cu-Ag phase diagram below to see how tie lines are drawn in two phase

regions. Our alloy contains 40 wt.% Ag and 60 wt.% Cu → Co = 40 wt.% Ag. a) 1000°C i) Liquid ii) same as Co iii) WL = 1 850°C i) α + L ii) a → 9 wt.% Ag, 91 wt.% Cu L → 53 wt.% Ag, 47 wt.% Cu iii) Use the lever rule (use just wt.% silver for composition)

Wα =CL − C0CL − Cα

=53 − 4053 − 9

= 0.295

WL =C0 − CαCL − Cα

=40 − 953 − 9

= 0.705

600°C i) α + ß ii) α → 3 wt.% Ag, 97 wt.% Cu ß → 97 wt.% Ag, 3 wt.% Cu iii)

Wα =Cβ − C0Cβ − Cα

=97 − 4097 −3

= 0.606

Wβ =C0 − CαCβ − Cα

=40 − 397 − 3

= 0.394

200°C i) α + ß ii) α → 0 wt.% Ag

ß →100 wt.% Ag iii)

Wα =Cβ − C0Cβ − Cα

=100 − 40100 − 0

= 0.60

Wβ =C0 − CαCβ − Cα

=40 − 0100 − 0

= 0.40

b) The eutectic temperature for the Ag-Cu system is 779°C. At 780°C we are just in the

α+L two phase region, and at 778°C we are just in the α + ß two phase region. 780°C i) α + L ii) α → 8.0 wt.%Ag, 92 wt.% Cu L → 71.9 wt.% Ag, 28.1 wt.% Cu iii)

Wα =CL − C0CL − Cα

=71.9 − 4071.9 − 8

= 0.499

WL =C0 − CαCL − Cα

=40 − 871.9 −8

= 0.501

778°C i) α + ß

ii) α → 8 wt.% Ag, 92 wt.% Cu ß → 91.2 wt.% Ag, 8.8 wt.% Cu iii)

Wα =Cβ − C0Cβ − Cα

=91.2 − 4091.2 − 8

= 0.615

Wβ =C0 − CαCβ − Cα

=40 − 891.2 − 8

= 0.385