FinalEx-F02 - ENSC 3313. Fall 2002 PI HAL No me Answer anv...

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Unformatted text preview: ENSC 3313. Fall 2002 PI HAL No me Answer anv III} Questions «only.I 2 points each. 1. _...—_ I recollect that after that Chernobyl nuclear power plant disaster in Line Ukraine, radioactive fallout spread over Europe. One issue in Britain was radioactive strontium 91]. It can get from grass to milk (via cows}. It is absorbed by bone. The Sr substitutes for Ca in Cafil'flflg. H has a half-life of EH years. What proportion would have decoyefi in 90 years? Proportion decayed = l — {lull = fl Detennine whether ultraviolet light of wavelength soc urn would be able to promote electrons across a l e‘v' hunt] gap. The velocityr of light is 3.0 x 1135 rats. Planck’s constant = 6.63 x “III—3iI Is, and I ev=1.s x or” J. _h_c oE— l 44 3 Substituting ME: fi'fiaxm 13K“) ) = 4.14 c‘r’ toxic—Ehocsic'”) 4.14 I} 1* hence YES. Identify:r the location of the 2—fold symmetry axes in a regular tetrahedron. How many are there? You may find it helpful to begin by drawing a regular tetrahedron within a cube. Join the midpoint of opposite edges. get 3. 4. Show hu + k‘u' -r t'w = D fer the cerubiuatieu sketched. z D Frern start te euti erline in cell, we ge [l T 233] hence directien -: is 3 3 2 . [ i A 3,: With erigin at [L the rceigrecal ef the plane intereepts en the axes are {D T iii), which is equivalent te [fl 2 3). 3 1f3 The direction lies in the plane, sn he I in; + itw = If}. Checking (0-s+§~§+§-2)=e 5. Per F GE geid, the aternie radius is t]. I44 nm, se henr many geld stems are there per cubic meter? FCC, thus stems eentaet aleng the face diageual. 41' = «Ea and 4 atetust'ceii a3 =[4(U.144}>< ie'tjj J5 —.e.esrsxte'”n13 That veiumc centains 4 stems, hence numhcr per in3 4 as s -—~«~—— —.s.ssx1e m users at itrET —’rr 6. What ASTM grain size number dees this sketch [x1133] cerrespend te? N _ zit—i We need ‘n’. N is the numhcr we ceunt in a 1” x 1" square. Iceuut w l. 1=2'3'=2”'1 11=l T. What have I h lacked out? Reeolleet that the gas constant is E .3 l .I'r‘mnl- Kt. ’Ilthln 53 1|- Tlhllll-Iim of Dil'flulun Hill I. _-u—_— _ . _—___ —I' . Bar-"m Hm Artfmifulflwfifl- Cummd Phin- D 2 Due 9 RT EE New Unruly”; Ufa-Ian“ {Flam 'FQ‘L'E D Julia's a.» s Fe 2.5 x 11)" as: as: an so a tail i - .. m 1" I '” Substituting, In as 1th.- 5.I:Ix1l.'l * 234 m sea 1.1 x 1o"T ; ' in?! use rs x 19-“ 25MB :2 a—Pe I312 5r 1c"T at: us} 2w 2.4 at "3 ' . "l l3“ 733' m mug-.. TJ=d.f-_ixlt_l o 37 e rFe 2.3 a 1o"I 14s 1.5:. see 1' a e 10'“ into seatc'“ , . -21 g on or.- re x to“ in so see 4.1 a 1r" D = "in ca as we to" use 1.9:. sea as 2 tr“ at A: as a It!" 144 1.49 an: 4.2 it its—'4 ca Al as a to" 13d 1*: sea a: a- 1r“ Mr, a: 1.2 s. m" 111 1.23 sec 1.9 x tlr“ L'u _ Ni 2.7 a to“ His ass sec 1.3 w Ic'“ —"—-———_..___ E. The stress-strain plot is from a cylindrical specimen of fit] tnrn gauge length and 12.5 mm diameter. Determine a] the yield strength by the 13.1% proof stress (offset) method, is) the elastic modulus. c} the tensile strength, and d) the specimen length at fracture. a} Drawing {1 line parallel to the initial slope at an offset of ilflfll. I get an intersection at - fiflflfifl’a. b} The initial slope is ~ sentence. ~ m. c} Reading directly off plot, w W. o} Extension is {50l(fl.flfi} that, -- 3.0 mm. Hence, specimen length at fracture is 5D 1- 3 = .‘tii T't‘tt'n. acct __ t5 9. A cold drttWIt 3 mm diameter iron rod is to be drawn down to a 4 mm diameter rod. The final product must have a tensile strength 1» 325 MPa an elongation [5ft mm} :=- 4fl%, but at a hardness r» l It]I BHN. Specify a processng sequence. Trmi nun, a Tm soda; I In: lull. I ma [Ls as as jfi'mili as as us as mu as M as cal fl. g I. £- aw i . r _ Hm so“ 5:51 _ ' m IUD El 40% E? g , .- _ so " ' rig and “E E: : Fm w'fluinmsomsosometamm-iuwec o to sun: so so so mamsfi mach Cold. WEI? 'Et T3 1‘» 325 MPa Elongation 3“ Allin/l1 Hardness 1* 1 1C! Need is l 5% CW Need s: l?‘% CW Need 3* 12% CW Choose -- 16% CW. for iinal stage. 2 _42 Need a ‘d‘ before final stage of t}. J n‘ L d rs- d3 Need d = 4.3a rnm Rod is cold drawn initially, hence anneal and reduce d to 4.345 mm by cold workr'amteal stages, thence anneal and CW 16% or hot work rod from 8 mm to 4.36 min, thence CW lot/s. 3'3” ll]. ll. 12. 1n the lr-Mo system, state 2 alloy compositions that contain 1335.; at equilibrium at lflflflac. Iridium-Molybdenum Merlin-is Purcunlone Melrbmm EDS Ir~Mo We loeate E; as occurring between Elli-64 wi’e Mo. We need to go 2H3 of the way he the left or right of the 6; zone towards the next boundary. IO EU 30 4D 5-D fifit ll} Henee, or ~ se — 23:13!) — 42} —~ 43 'Wr'fU ME! we hit- — RIMS”? — 1‘34} ~ 3t} wi'o Mo hi the Ir—Mo system, state the composition of the a) first solid formed and h] the last remaining liquid when an 3t] wfo Mo alloy is cooled slowly From the melt. Iii-swing temperature horizontals at the liquidus and solirtus, we get a} r- 93 who Mn h} m so wfe Me With respeet to steel, explain the terms a) Martensite, h} fi—Ferrite, e} Fearlite, and cl} Austenite. a) The tlistoi'teti ECT product obtained by quenching Austenite to a sufficiently low temperature that it tries to transform FCC to ECC but the C is trapped in solution. b) The BCC structure that austemte reverts to on heating to as NWT. e} The euteetoitl Lleeonipeeitien product ofauatenite. ii} The FCC structure fir} stable at from 9: Wt]- | 4ttt1"t_‘.. A t3. is Clo-Cr an age-hardening system? Explain! Chi-Cr Cotumblum-Chromium . Atomic Purser-t1:th Chromium ‘3 :e so so so so so so so so Yes! We need a system where a} We can solution heat—treat at a “high” temperature, b) Quench to got a superusatnrated solid- solution, e] Reheat [age] at an intermediate temperature to get a E precipitate everywhere of a hard second phase, art intermetallie compound. This serves as a barrier to dislocation motion so hardens the alloy. Recognize that the diagram is “cut—oft” at 14DUGC1 Composition of ~ It} and 92 wto Cr satisfy the age-hardening criteria. The precipitate is [3]. 'l'he problems were a combination of" corrosion and fatigue. This was a plane that underwent many short flights between the a'aIIIIII'.'.Hunts-wJ 9». Islands — it had undergone - I'IJi'o'li-a'l-h custom-an lumen: filial-u. optic an .Mlahh .lur.'Ir-.'.-. jrl "Mn,- abflut i Mnsenirllr" :| JJ I'._._. I“... "furl-Its III: mulls and urea- ll"' ' M l-Ilowtns. d- lllil'tl Iticmam c: WIth tier death Imam the you Ian-41.: "l'h+re tam hm, hang when. II none-um:mama-sum! Fatiflc cracks developed at a line of $ Ui' int-l *r HEW JIJLtlm' all ls'ue mama] Wmnlvrflufll'lnlL-ill- ri'llrl'fll holes where crevice cnfififiin'n HIIHIII, |I| TlIL' llt-Illrtt' '.'.l-'-'. It'llll nr..'- ul ilt L'M'l- engine-t all: me and ulniu: I H 'In '-'|_r um -.-I' .-|:i r.1l-.I|| at. ' I III-MI: |"'||I|I-.1 |.|'.u: '- ale-nu. le- |_-.. Bel-rt BEflflfl ground-air— ground corresponding cabin pressurizations, hence cyclic stresses. was occurring through exposure to m. .mi salt water and chlorine ions that are : la' |iJl"'.'.:l' :I'r tern. lit-u- lot _. :u.lu5. one. ahlmlnu'ln_ IJI'H'I'WUIFJI'JH Iaitflll'N - - Fannie-lira Illa] innit-a. 1 hrs: accllon or its rmr lilown on HF 4'" in“! "'1" -'!'-'"l"- 3'- -‘|'|'--'-‘v!='|'- . :.:-..-.i-.- ai-rranc: “It'll: nri' tr. our Aloha Axum ism-phony; cg. Audit-,1 ..| Its-1.1.... ’uqdnr ._-i QuIz I-'| “lllci- out». a “.1; hum-mg Thursday MM"... .' p "I ll: :isd-u II ,1 re. ED“ ’ dlllull: Di.-IL'I:II; “In. 15. You should be familiar with epoxy. Which of the follow terms apply? Explain! a} Glass Transition Temperature No! it is a thermoset. It does not flow, become plastic, as the temperature is raised. b} 'v'iscoelastieity No! It is a thennosct. e} Condensation Polymerization Yes! Two components are mixed to make it. :21} Number Average Molecular Weight No! It is one macromolecule. lti. How many kg of sulfur would be needed to eross link 2% of the available sites in ltllltl kg of polyisoprene, C5 H3? The auto are H = l, C = 12, E: = 32. Let the S be 3t]% efficient. eis - For eomplete eross linking, one double bond per mer would open to git-e 2 linking sites per met tojoin 2 rners. J lenee for eornplete cross linkng we average 1 8 per mer. - - s - C _ C - .— For complete eross linking need ' | S - 32gS per 158 g of eis polyisoprene a-_s- O I or 4T1 g 3 per lfltltlg ofeis polyisoprene m- H.4lg S per lfifltlg ofeis polyisoprene if 2% linked At Htl'lii efficiency, need 9.41i’tlfi — 11.?fi ltg ot'S. ii". A wel|~lr You have seen the theme rnieroeells on the quiz and the tests. (MCI is initiallyr steel and graphite. The graphite is the cathode and remains; the “steel” matrix is the anode and eorrodes away. Left is gray graphite; “graphitiaation.” a“ --\_.-.'.-._- :,-. Jhnlnl my; film an arm tron about, Etmfltuu: type JL mprdee fillet»: :u. a matrix I:II' ultra- iatteruattni human” or Ila'ht-E [nil-e and flutter ardent-Ital. W “5pm H‘J _1i'II-'n|- .|.I.l.a'l.'lll‘.' war-[tilt .lll. slim .u- gl- |f.|JI.I...,,' .- .-r.-II.-;-I_l'. 's'I.:Ir-;|.-| u. 'i- r- r|:.im||.-r ='- 3""? Tail IIIIII Emu L'W 3m- mIrar I'I'I "-II !| ll'llll'l'm .n-:‘ Ilru'ufl'm. .-_rl-r.-._-.- l 3. What is barbed wire made of? Why doesn‘t it rust easily? In western Oklahoma, its lifetime is probably 4D years, unless meehanieal failure occurs {starnpeding bulli). Steel. Probably, the earbon eontent is quite low as high strength is not needed and fonnaliility is. Grades can be provided as annealed wire or with degrees of cold worl-t. Corrosion proteetion is provided by galvanizing. The wire is passed through a bath of molten zine. The zine sen-es as a barrier eoating first, thenee as a saerifieial anode, with a favorable anodei’eathode size ratio. 19. 2G. 21. in a metal, what is the probability of an energy level full e‘v’ above the Fermi energy level being occupied at WC? Reeelleet that Boltzmatm‘e constant in 8.152 x Ill—5 eW'atom-li. l FlE}=W Substituting 1 F(E}= ' i .., = 3.395 1 +£0.13]! segue 'jznj — The band gap ol‘ diamond is ~ 5 eV. At what temperanire would the intrinsic conductivity he lflx that at ltitltJDC‘? Bolthann's constant is 3.62 x 113—5 eVIatorn-K. . —E .-'2k'1 fi-v [e " —E “an. 4:17 r—Cc “-’ ' e -"2kl".-'3 5mm“C‘= m h I,” r- a 433'! definite") 1—— 111 i! =3 “I 't LT2 ["3 l 10-— GT Uitiee 2.3m _ — 5 i —L_ 2 H.62ilfl'5l T2 121% T2 - 13912] K. ~ 1131; 99.999999? The theme is electrical conductivity. Explain the relevanee of such a number. * The necessary purity level for an extrinsic semiconductor! 4‘ If the dopant level is going to he 1 or Ill parts per million {ppm} then the haae metal has. to be about twn orders of magnitude purer to minimize unwanted impurity affects the impurity level must he in the parts per billion [pph] range. Recollect that purification is achieved by zone refining. sv- ‘er .l‘ 2. 23. 24. This is an etched ci'ossvseetion of the gig] stator core oi'a large electrical that the steel has {I} negligible carbon boundaries, and {III} is these features. generator. The caption iterates to eliminate carbides, {11] large grains to minimize grain fully annealed to minimize the dislocation content. Explain the significance of Desired is as soft a magnetic steel“ as possible. The stator is repeatedly magnetized and demagnetized. Wanted is a minimal hysteresis loop, which signifies pas};r domain boundary movement. This means minimal imperfections. The area within the hysteresis loop represents an energy inefficiency — which is lost as he_at. TWhat have I marked out in this text paragraph? Molecular olarization is permanent, since it is inherent in the molecular structure. These dipoles can be oriented with the field. Fin-thennore, the molecule "flips" each half cycle of an ac field. Smce the masses involved depend on the size of the molecule the maximUm frequency of response vari cs si tilicantl from material to material. It is, however. always less than that tor electronic an Furthermore, it is hi ghlv sensitive to for example, polarization of polyt-‘inyl chloride {CgHaCDn does not reverse below the glass _“ because molecular movements are restricted See restorations in paragraph. HW question was on electronic polarization, so I expected you to malte the association to orientationimolecular polarization. I have several times in the course used the [Tu-Ni systenir’allovs to illustrate themes. t_‘.ite one example! For example: a} To illustrate a continuous solid solution — diagram T—E. h} To show the effect of alloying on electrical resistivity — diagram 23—3. c} To show the decrease in magnetism as copper is alloyed to nickel, causing the gradual titling fifthe {1 band — diagram 26—3. d] The USA 5d; coin is 275% Cu — 25% Ni. ...
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This note was uploaded on 01/10/2012 for the course ENSC 3313 taught by Professor Staff during the Fall '11 term at Oklahoma State.

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FinalEx-F02 - ENSC 3313. Fall 2002 PI HAL No me Answer anv...

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