HW_Soln_Ch4 - COMPUTE The temperature at wtfich the...

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Unformatted text preview: COMPUTE: The temperature at wtfich the vacancy concentration is one halfthat of 25'C. GIVEN: ca” = 2C5” Where Cv=vacmcyconcentution Qrv= activation cnergyforvtcancyinfonnafion R=gasconstant8.3l4llmote—K T=absolutetcmperature £2: RTI C: _g£ exp[ T2) -9L+g£] C3 exp[ RT} RT: In the present problemci= 005°C); CV = Cv(25°C) md T, = 35 + 273 = 308K " T2 = 25 + 273 = 298K also CV(3S'C) = 2042530) 2C;(25'C)gm;[_”Qn + (2" ] “m- , _ 005°C) _ , was) («298) , R , 91784 Solving for Q4» we get Q. = 52893.5 J/mole. Using this value of Q9, the CV(ZS°C) c’an be calculited c.(25"C)=exp( mm ] '8.31x298 cv(25°C)=5.35xm-'*' ‘ The problem requires us to calmlate the man at concentration “1,3905%”. V , - 7. an , V , l . » , 1/2 cast) = 2.575 x mi" j > ‘ 52893.5 2.675 -'°:exp( - . m 8.31”) Thus for solving T, we get: T = 288.63K or 15.63°C. comm: Cvm=3Cv(8¢C} GIVEN: ‘C.(25?C)#=§Cv(80°C) EQUATION: cw (25° C)eaxp[eégmjt * Dividing (1) by (2V)’we get: cv(2rc)=1=“p[_g(_1___1_]] Cv(80"C) 4 R 298 353 22033 56 v C = _____.__'.__ C(‘W ) 9% 8.31x353) Solving for Q, we get: Q = 22033.56 J/mole = 4.511) = 5.46x10" The problem requires computing a temperature at which Cv = BCV(SO°C). 3CV(8O°C) = 3 x 5‘46 x 104 = 1.63 x 10‘3 22033.56 1.63 -’= -—-—-—— x10 “{ 8.31xT ] solving for T, we get: T = 413.05K or 140.05°C SHOW: The extent of solid solution formation in the following systems using Hume- Rothery Rules. (a) A] in Ni fie; r(Ni) = 0.125nm; r(Al) = 0.143nm difference = 14.4% Electronegativity: A1 = 1.6]; Ni = 1.91 Most Common Valence: Al”; N” Crystal Structure: Al: FCC; NizFCC The crystal structure of Al and Ni are the same and the most common valencies are also comparable. However, the size difference is close to 15% and the difference is electronegativities is rather significant. Based on this, it appears that Ni and Al would not form a solid solution over the entire compositional range. (b) Ti in Ni Siz_e: r(Ti) = 0.147 nm, r(Ni) = 0.125nm difference = 17.6% Electronegativity: Ti: 1.54; Ni: 1.91 Valence: Ti“; Ni2+ Crystal Structure: Ti1HCP', Ni FCC Ti in Ni would not exhibit extensive solid solubility (c) Zn in Fe Siz_e r(Zn) = 0.133nm; r(Fe) - 0.124nm difference = 7.25% Electronegativity: Zn = 1.65; Fe = 1.83 Most Common Valence: Zn”; Fe2+ Cgstal Structure: An: HCP', Fe: BCC Since electronegativities and crystal structures are very different, Zn - Fe will not exhibit extensive solid solubility. (d) Si in Al _S_iz_e r(Si) = 0.117 nm; r(AI) = 0.143nm; difference = 22.2% Electronegativity: (Si) = 1.90; A] = 1.61 Valence: Si“; A? Crystal Structure: Si: Diamond Cubic; Al: FCC Since the size difference is greater than 15%, and the crystal structures are different, Si-Al would not exhibit extensive solid solubility. (e) Li in A1 SE r(li): 0.152, r(Al): 0.143, difference - 6.29% Electronegativity: Li: 0.98; A]: 1.61 Most Common Valence: Li“; Al3+ Crystal Structure: LizBCC; A1: FCC Since electronegativity and crystal structures are very different, Li-Al will not exhibit extensive solid solubility. (1‘) Cu in Au §i2_e r(Cu) = 0.125nm', r(au) = 0.144nm', difference = 12.5% Electronegativity: Cu = 1.90; Au = 1.93 Most Common Valence: Cu‘, Au+ Crystal Structure: Cu:FCC; AuzFCC Cu-Au will exhibit extensive solid solubility. (g) Mn in Fe S_ize r(Mn) = 0.112, r(Fe) = 0.124 difference = 10.71% Electronegativity: Mn 1.55; Fe 1.83 Most Common Valence: Mn”; Fe2+ Crystal Structure: anBCC; Fe BCC The difi‘erence in electronegativity is high but Mn-Fe does obey the other 3 Hume-Rothery rules. Therefore, it will form solid solutions but not over the entire compositional range. (h) Cr in Fe _S_i_z_e r(Cr) = O. 125nm, Fe = O. 144nm difference = 12.5% Electronegativity: Cr = 1.66; Fe = 1.83 Most Common Valence: Cr“; Fe2+ Crystal Structure: Cr:BCC‘, Fe:BCC Cr in Fe will exhibit extensive solid solubility but not over the entire compositional range since it obeys only 3 of 4 Hume-Rothery rules. (i) Ni in Fe _S_iz§ r(Ni) = 0.125nm, r(Fe) = O. 124nm difference = 0.8% Electronegativity: Ni: 1.91; Fe 1.83 Most Common Valence: Ni”; Fe3+ Crystal Structure: NizFCC; Fe: BCC 15. Ni and Fe obeys 3 of the 4 Hume-Rothery rules therefore, extensive solid solution will be exhibited but not over the entire compositional range. (a) When one attempts to add a small amount of Ni to Cu, Ni is the solute and Cu is the solvent. (b) Based on the relative sizes of Ni and Cu, radius of Ni = O. 128nm, radius of Cu = O. 125nm, these two are expected to form substitutional solid solutions. (c) Ni and Cu will be completely soluble in each other because they obey all four Hume- Rothery rules. COMPUTsz Relative concentration of cation vacancies, anion vacancies and cation interstitials. GIVEN: Qcy = ZOkJ/mole QAV = 40kJ/mole QC; = Bold/mole ASSUMPTION: assume room temperature T = 298K Concentration of cation vacancies, Co. is given by Co‘exp['%%] = 20.000 8. 31 x 298 co, = exp(-8. 0763) = 3.108 x m4 40 000 = expi __-___ = exp(-16.152 =9.6x -’ C” 8.311: 298] ) 10 Similarly for anion vacancies 30000 =exp|-—’————— =exp(-12.II4 =5.48x 4 C” 8.3Ix298) ) m and for cation interstitials 23. GIVEN: Ci = 0.19 at % at surface C2 = 0.18 at % act 1.2mm below the surface D = 4 x 10'“ nil/sec a, = 4.049 A° COMIUTE: Flux of copper atoms from surface to interior. We must first calculate the concentration gradient in terms of [copper atoms/cm’lcm]. It C = [(a/oCu xdensity of FCCAI) / a: wt Cu] x N4, (3, = [n 0019 x 2. 7o) / 63.54] x6.02 x m” = 4.86 x 10"atoms / my C, = [(0.0018 x 2. 7a) / 63.54] x 6.02 x 10” = 4.60 x 10'9atoms / cm! can be calculated as follows: The concentration gradient is then g: (4.60-4.86)x10” as: 0.12 0 =.M=-2,125x10”atm/cm‘ 0.12 ...
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This note was uploaded on 08/31/2010 for the course MSE 2001 taught by Professor Tannebaum during the Summer '08 term at Georgia Tech.

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HW_Soln_Ch4 - COMPUTE The temperature at wtfich the...

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