MSE 230 Solutions for Assignment 8 Fall 2013 1 12 34 a For Li substituting for Ca2 in CaO oxygen vacancies would be created For each Li substituting for Ca 2 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 2 b For Cl substituting for O in CaO calcium vacancies would be created For each Cl 2substituting for an O 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 Al2O3 doped with Mn4 Al2O3 After bonding at room temperature 10 5 Mn4 concentation atm Position After heating at 1300 C for 24 hours 10 5 Mn 4 concentation atm Position 3 4 4 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 R 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 Element R Cu Crystal Electro Structure negativity FCC Valence 2 C 44 H 64 O 53 Ag 13 FCC 0 1 Al 12 FCC 0 4 3 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 on page 288 of Callister Making an enlarged photocopy of the phase diagram will also help Also this problem is essentially the same as the solidification process described in Fig 9 4 of Callister 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 insterects the liquidus line We are at the boundary between the liquid phase and the L two phase region b The composition is determined by drawing a tie line across the L two phase region Where the tie line intersects the solidus line gives the composition of the solid phase which is 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 L two phase region and the solid phase The last liquid solidifies at approximately 1270 C d As in part b draw a tie line spanning the 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 ductilities greater than 48 EL exist 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 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 W CL C0 53 40 0 295 C L C 53 9 WL C0 C 40 9 0 705 C L C 53 9 600 C i ii 3 wt Ag 97 wt Cu 97 wt Ag 3 wt Cu iii W W C C0 C C 97 40 0 606 97 3 C 0 C 40 3 0 394 C C 97 3 200 C i ii 0 wt Ag 100 wt Ag iii C C0 100 40 W 0 60 C C 100 0 W C 0 C 40 0 0 40 C C 100 0 b The eutectic temperature for the Ag Cu system is 779 C At 780 C we are just in the L two phase region and at 778 C we are just in the two phase region 780 C i L ii 8 0 wt Ag 92 wt Cu L 71 9 wt Ag 28 1 wt Cu iii C C0 71 9 40 W L 0 499 C L C 71 9 8 C C 40 8 …
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