# Assignment 2 Sol(1)

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Assignment #2 ENGR 2220 Structure & Properties of Materials To get part marks, show all your work 1. Calculate the concentration of vacancies in aluminum at room temperature (250C). Assume that 20,000 cal are required to produce a mole of vacancies in aluminum. Solution: The lattice parameter of FCC aluminum is 0.4044651 nm. Therefore, the number of aluminum atoms or lattice points per cm3 is n= 4 atoms / cell  6.045 x10 22 Al atoms / cm 3 8 3 (4.044651x10 cm) At room temperature, T=25 + 2
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ENGR 2220 Assignment #2Structure & Properties of Materials To get part marks, show all your work  1.Calculate the concentration of vacancies in aluminum at room temperature (25 0 C. Assume that 20!000 cal are re uire to prouce a mole of vacancies in aluminum. Solution\$%he lattice parameter of CC aluminum is 0.'0''51 nm. %herefore! the num)er of aluminum atoms or lattice points per cm *  is *22*+ ,100'5. 100''51.'( ,' -  cmatoms Al  x cm xcell atomsn = − At room temperature! %-25  2/* -2+   ( )        −=      −  K  xk mol cal  mol cal cmatoms x RT Qn  vv 2+.,+/.1 ,20000ep,100'5.(epn- *22 - 1.2  10 +  vacancies,cm * 2.Calculate the composition! in 3eight percent an atom percent! of an allo4 that contains 15+.0 g titanium! 20.* g of aluminum! an +. g of vanaium.Solution%he concentration! in 3eight percent! of an element in an allo4 ma4 )e compute using amoifie form of 6 uation '.*. or this allo4! the concentration of titanium ( C  %i is 7ust  C  Ti  = m Ti m Ti + m Al + m V   ×  100  3t8+'.5'-100  .+*.205+1  5+1 -  ×++  kg kg kg kg  Similarl4! for aluminum  3t810.+-100  .+*.205+1  0.*2 - Al  ×++  kg kg kg kg  C  An for vanaium1 of   ENGR 2220 Assignment #2Structure & Properties of Materials  3t8'.-100  .+*.205+1  .+ - 9  ×++  kg kg kg kg  C  Atom percent: Masses must net )e converte into moles (6 uation '.'! as mol**00.(-,/.+/' 5+0001 -- %i mol  g  g  Amn TiTim ’ mol/52.'-,.+2 0*002 - Al mol  g  g n m mol1+.+-,0.'5 00+ - 9 mol  g  g n m  :o3! emplo4ment of a moifie form of 6 uation '.5! gives 100 -  %i  ×++ V  Al TiTi mmmm nnnnC  ’ at8/+.1+-100  +.+1'./52*00.*  *00.* - ×++  mol mol mol mol  100 -  Al  ×++ V  Al Ti Al  mmmm nnnnC  ’ at81/.+2-100  +.+1'./52*00.*  52.'/ - ×++  mol mol mol mol  100 -  9  ×++ V  Al TiV  mmmm nnnnC  ’ at8'.0-100  +.+1'./52*00.*  +.+1 - ×++  mol mol mol mol  *. The outer surface of a steel gear is to be hardened by increasing its carbon content. The carbon is to be supplied from an external carbonrich atmosphere! hich is maintained at an elevated temperature. A diffusion heat treatment at #\$% °  C &''#( K) for '* min increases the carbon concentration to *.#* t+ at a position '.* mm belo the surface. ,stimate the diffusion time re-uired at %% °  C &(/( K) to achieve this same concentration also at a '.*mm position. Assume that the 2 of   ENGR 2220 Assignment #2Structure & Properties of Materials  surface carbon content is the same for both heat treatments! hich is maintained constant. 0se the diffusion data in Table %.\$ for C diffusion in α  1e. Solution;n orer to compute the iffusion time at 5/5 ° C to prouce a car)on concentration of 0.0 3t8at a position 1.0 mm )elo3 the surface 3e must emplo4 6 uation 5.) 3ith position (  x  constant< thatis  2t   - constant=r  5/55/52525  - t  2t  2 ;n aition! it is necessar4 to compute values for )oth  2 25 an  2 5/5 using 6 uation 5.+. rom %a)le5.2! for the iffusion of C in α >e! Qd   - +0!000 ?,mol an  2 0 - .2 ×  10>/ m2,s. %herefore! +−×  2/*25( >?,mol*1.+( ?,mol000!+0 ep,sm10.2- ( 2/>25  2 - 2.00 ×  10>10 m2,s +−×  2/*5/5( >?,mol*1.+( ?,mol000!+0 ep,sm10.2- ( 2/>5/5  2 - /.2/+ ×  10>12m2,s :o3! solving the srcinal e uation for t  5/5 gives 5/525255/5 -  2t  2t   sm ,102/+./ min10(,sm10  - 2122 10 00.2( −− ×× - 2/5. min* of   ENGR 2220 Assignment #2Structure & Properties of Materials'. %he iffusion coefficient for Cr  *  in Cr  2 = *  is   10 >15  cm 2 ,s at /2/ 0 C an is 1  10 >  cm 2 ,s at 1'00 0 C. Calculate (a the activation energ4 an () the constant @ 0 .Solution(a        −          − −− (1/*>?,mol(+.*1 ep(1000>?,mol(+.*1 ep -,101,10 002215 Q 2Q 2 scm x scm x ( ) [ ] [ ] Q x 0000'+0/.0ep0.0000/1>0.00012>ep-10  −= − [ ] Q 0000'+0/.0-02'.12  −− -2.5 10 5  ?,mol() [ ] 0+o5o 1055.1+.1/ep@ (1/*>?,mol(+.*1 ,2.510>ep@-101  2 xmol  3  x  −− =−= @o - 0.0'5 cm 2 ,s %. An 1CC ironcarbon alloy initially containing *.\$( t+ C is carburi4ed at an elevated temperature and in an atmosphere herein the surface carbon concentration is maintained at '.* t+. 5f after /#.%h the concentration of carbon is *./6 t+ at a position / mm belo the surface! determine the temperature at hich the treatment as carried out. Solution%his pro)lem ass us to compute the temperature at 3hich a nonstea4>state '.5 h iffusionanneal 3as carrie out in orer to give a car)on concentration of 0.*5 3t8 C in CC e at a position'.0 mm )elo3 the surface. rom 6 uation 5.5       −−−−−  2t  xC C C C   s x 2 erf 1-0.20+*- 2+.00.1 2+.0'*.0  - 00 =r  0./1-2erf          2t  x ' of 

Jul 23, 2017

#### Assignment_3_sol(3).doc

Jul 23, 2017
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