ENG+45+SP+2011+Midterm+_2+Solutions

ENG+45+SP+2011+Midterm+_2+Solutions - I SL;(b ~~ / Name Lab...

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Unformatted text preview: I SL;(b ~~ / Name Lab Section: r' ~ L 0 T/o (\/~ L/ DEPARTMENT OF CHEMICAL ENGINEERING & MATERIALS SCIENCE University of California at Davis Professor Subhash Risbud ENG 45 Properties of Materials Spring 2011 Mid term II 12.10to J.30pm (max credit = 100) Problem 1 (20 points) A company makes solder alloys based on compositions shown in the Pb-Sn phase diagram. A 1 kg alloy of composition 50 wt% Sn-50 wt% Pb is held at 184 C for a long time to ensure stable phase equilibrium resulting in a mixture of a + Liquid at 184 C. If the only goal is to obtain an ALL solid alloy without changing the temperature of 184 C we have two options: add Pb or add Sn to the 1 kg alloy. By examining the solidus composition (s) at 184 C it is clear that two compositions can meet the ALL solid criterion at 184 C: either the 97.5 wt% Sn-2.5 wt% Pb W solid solution) OR the 19 wt% Sn-81 wt% Pb (a-solid solution). By calculating the grams ofPb or Sn needed in each of these two cases comment on which would be a better option (as an extra consideration realize that Pb price ranges from $ 1.0 to 1.25 per pound and Sn ranges from $12 to 15 per pound). I~ £ e<-oC.cj·- tr1C 0,8/ /ooo+x.. S-oo + X-:= . '. C! - 0 ·8 & 10 .-f- r) 'K::::: - '. x. = ($ / 0 0 ·8/',L. -S-o 0 :::: ~ (!b37.-:JiM- ~:o= Q.ICY (CJ ~ ?6 0 7 r /9 X I S-~'~ o r 97'S: ( O(;--.Je v . c;:::,:sl-- /e....:sT , )'/-uJI- Problem 2 ( 20 points) A plain carbon steel of hypereutectoid composition (1.0 % carbon) weighing 1000 g is held at 1100 to allow equilibrium phase austenite (y) to form and then slowly cooled under stable equilibrium conditions to room temperature. From the Fe-Fe3c phase diagram, calculate the mass (in grams) of each of the phases below: c D i) ii) iii) iv) Pearlite pro-eutectoid cementite (Fe3c) at the grain boundaries Ferrite (a) inside the pearlite grains, Cementite (Fe3c) inside the pearlite grains ,- y -; -f o~g kJ 1.0 -f x. t.7--0 To ~-L '8 J l( ( Sv.6 ty-CL-_D j/ fYO -eL~" j , -- -- - ~-- C~ - - ~~ f ( ~F7 -6- '7- 1 (. -.>< /~- 0{ - -ytik( ~ .- 9' (;C::3 (=V'o:::: ___ .u.i::- - teJ C -e r- o = &<7 ?f ~ @~ aI , C 8:;-( j (;,-7 C 1'1 <1 5'~ 0- I~ C - - 5V j~(2P.-V {-~ ~ fye . £~ cf -kJ C ~LL)~cL 9'6 I =- j 31.; .5 te-o...,,'(Jc.. I, ~ ~S-( :: //S- j I J Problem 3: (20 points) An 40 wt% Al203 - 60 wt% Si02 ceramic tile weighing 1000 grams experiences two different conditions during service: (a) held at 1500°c for months and cooled very slowly to room temperature while maintaining stable phase equilibrium throughout. (b) held at 1700 C for months to achieve stable phase equilibrium and then quenched very rapidly (assume instantaneously) to room temperature, From the stable silica-alumina phase diagram answer the following questions: Service Condition (a) (b) Phases present at Room Temperature CY!j-r1aLU<.L S-\.' 0 C. c y':':s-f-, b a..Li f--e fV1. u..it -I-.e.. Gt.<L5S' + Ml.dLZ &.1L r.l J Composition of Each Phase - 0% Ai'//' .J ..,.,.71 ~ -~ Y27 ::1 A(-zP3 / D;:) ""2-i6M. Amount of Each Phase (in grams) -z.PJ- "'" S-b3.3 b() 'l -.5 '" / ( 0-it:-% ki? f---:3' ] r.B s.:J "'::1 j j Problem 4: (20 points) Three pieces of steel of eutectoid composition (0.8% carbon) are heated at 800 C to form a homogeneous austenite (y) solid solution. List the phase(s) formed at the end ofthe cooling cycle described below for each piece:. Use the TTT diagram below. M..--L UI..£ .' PIece 1: Cooled from 800 C mstantly to 650 and held for 1 sec Piece 2: Cooled from 800 C instantly to 650 and held for 1 hour --- P sf Vc.:A.' -€ ...o-.¥ (. )' (.~ ~:;,..~y Ce e. TTT diagram for 0,8% carbon steel: 8DO ... I ·u ..... ----~ I . FIne pearlite 5un '!, ( 40U ... 100 n.77 (I TIme. sFconds c<'V '--" o-/- Pe...o-> (:'&e Lex ..+-kJc.) Piece 3: Cooled from 800 C instantly to 700 C and held for 1 hour - "(' I ;: wi 'X, C L cu.L8kv u:.k; Problem 5: (20 points) The rate of recrystallization of alloys is characterized by the rate equation: Rate of recrystallization =(1/tR) = C exp( -Q/RT) Preliminary data reveals that complete recrystallization occurs in 250 mins. at 360 K and in 75 mins, at 375 K. How long will it take for complete recrystallization to occur if we never heated the alloy and left it sitting at room temperature (300 K)? -(9 ['YLC- ---- fS (~ C ::=:- 0 (I Cj a 3( c ,~&[1;] ---- 4 91 _::::= \..---~ ~ .s- ><-8'.] /'-( 2 IS- ...
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This note was uploaded on 11/11/2011 for the course ENG 45 taught by Professor Risbud during the Spring '08 term at UC Davis.

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