3473-10 Exam 1 solution

3473-10 Exam 1 solution - Name I CHE 3473 Chemical...

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Unformatted text preview: Name I CHE 3473 Chemical Engineering Thermodynamics September 16, 2010 Exam 1 100 points Print your name at the top of each page. Open book, closed notes. 1 hour and 50 minutes 9 pages total Cell phones must be turned off and stored out of sight. Use your time wisely. Make sure you allow time for each problem. State all assumptions. All work must be shown to receive full credit. Good luck! Page 1 of 9 Name Solmkgq Key Short Answer (30 points total) Answer in your own words, no credit for direct quotes from our text, handouts, or notes. 5 pts Provide a physical explanation fig why fluid turbulence produces lost work. Fluid. ’YLmrlamloace Came: SL186?!” 904.03% malaczules in 44% flowing "TM, Slew wear-ca 4M» «cl: rec-rum “HM alia‘lwu‘e, +l¢ +er whales “Halo ‘l-OJE'léwr“ ITFQJEAJD woth JrLE‘“ serve; no 915te Pmrfmse, “Fla wag; ré’émifem’ “lb pswgm “HM-F 5L9“ wool: is proleal 67 ‘Me inleI‘rml 9N9ij of ‘Hie flD/E’HJE’J' “Pulp 41594-me H 7% loft marl: 1s Comm-le 4o lee} and )EfiQZS in an I‘AngmB :m 4418 finial, fluid ‘lwlwlema gmclwx 'lhrlivl'en 5641an #9. mflfecoile; l" 4416 flowing} uti'ol. 5 pts Describe two unrelated si ations where calculations involving entropy may be of value to a process engineer. 6 bfi‘lPFMi/te ULE—MFF q coro'oofpcyl FpOCeJ-J‘ Or. 359% ch59 i5 FoSJiLle. If 95.3%“ 3 OJ We will: clamje LxJélea t loclmalgj Syd???“ INA Eur/bungli55_ ....__..-- “Fla :‘AQM‘DiEMniES im a: Frown an E9, MGM‘L in ‘lPr/hJ eh fail” Muscle. A FEUPFEllalE ( 614‘ Pam \p\e ) p fa (Lars jflmemles no lost Wool: EKWCQ 7*" O. PPfiQrmi/ta Em Balance I}. JIqu QIQfiQng Ming-L. (7 U3 _f i: H m T) Page 2 of 9 Name Se iriia a KFL/ Short Answer - continued 5 pts 13 the liquid mole fraction, x1. , for component “5” considered an intensive state variable, extensive state van'able, or neither? Provide an explanation for your answer. )5 am ia‘iémglva 34min Uefi'aielg, L J Liémal (male, "iyraC-iiov’t/ "XV Tim value op “if #085 ma'i' D/Eflfla' CM “‘l—‘i/‘Ci QWmK'i' mm‘iEnlel Pr85€%‘l. )9“ “8&erle flit ‘CDF an?! SKQE, Sewnle inimm #am a; ma)th Mtxag CSTerch ill? Velma 9D Xi: WDfllai aim); Lt 4%. game). mix i w 2 Mal ml: in: fine/(:85 PC; “lemme; L7 defilit‘m V" 5 pts Explain why the 1St Law for steady-flow systems is commonly expreSSed in the following form Without terms for kinetic and potential energy. AH=Q+I¢I§i The Hi (Jaw- near Shady-(Claw $7Jievtx£ if t. (14+tw‘+a) : Q Mix/s 7‘ la iCi AEiiL “"5?” 4% +9 I'm in Mnj-l- Prat?ka 5‘simwiiw‘s} iixc mgniiwie 0D "Mu! CLAng i4 lfilneiic Mai '{Bfl‘igh‘iioi HANS); me we. or“ mare 0m]ng mp: Miami/ole {en ‘F’lmm. fine clingy; M @VI‘LLQcOV‘ COMQZMWELJLQ 444959 ‘iPrms are firtfluemlly tam;er ole“ 46%;? llkfmlwuw “in sieml7_£lom pr,€95;es_ we ANA 5-1:; s afpmxdmakm 97 mtg AA g 62* MS Page 3 of 9 Name Salad-lea K131- f 5 pts Explain the differences, if any, between steady-state and equilibrium. fleet aisle is bad characlwieecl as a side Delete lie dtiuiacg “caret gr Clam/fie, LS 92ml “in He rffi‘ts‘ifi‘fict} Cheese“ olerivcdim ‘gr variables Mfmifilfigl Mid-L 4L8 9/de QFQ {Pro (—anr; is AA fitCCthmltsfio-m 00D MMS/Efiegyf fitting: 9,411 War“ (A gys-iem Pfi‘ Siaaay—Fls‘d‘ax A-l QZMiiiLrimm} Me, driving “QWTE 43K Clinij is 29m. A 575.1% in agar illgrlmm “1.5 .9135 cd' Sinai)» - gdfiia . Miguel;ng Q gysi-(‘m (Xi‘ SAEQQI/‘Hfii‘e l5 flO'i‘ newmri} (I/L QfluEiILf‘r-mq‘ 5 pts Identify two specific concepts or ideas that have been discussed in class that you aren’t comfortable with or don‘t fully understand. One or two word answers (e.g., entropy) are insufficient as they are too general and don’t help understand what needs to be done differently in the future. ' Page 4 of 9 Name gait/Jinn Keg TRUE / FALSE - 3 points each (15 points total) Circle T or F to indicate whether the statements below are true or false. T = True; F = False 1. @ F Thermodynamic efficiencies greater than 100 % are not possible. W5 71% 5 fig LC Mder Pafldaie agr- animal ghafli— yum-i; (Mac/Licle is}! a From; 4» mm +i-E 'EGi'ERi Marl, (Primitich Q FEVPrEaLie, 70055855. 2. T @ The change in entropy for a stream entering and leaving a system is path dependent. Enir‘of/ t5 0m iw‘iGns'Ne sieie uur‘iqigig. Ci/wijei t4 Sit—xii? URFEAUEJ are ""fllefenaipai— 0D iiua iaima [0941»ch sides. 3. T [5) CV and CF are referred to as “specific heats” when the denominator units are expressed in English engineering units rather than SI units. (jg-Md are FQQJTEA ‘LD Q fifleaiupe Mali: when Pxpfllfleol 04 a mass Militar- Mahar- 03am, “k \\ v .\ - . t 9.0. CF13 $.97 L5. a SPAM‘A; :DFEMLL MID/W “RE-5 laser. 4. (T) F An ideal gas can also be considered an ideal solution. H The Mbiemlps it an loi'eci 55:; gig mi i-rinmci MAL silty- r’hfliQHiEJ. ginte iniorae'heftf tOQ‘i'VUCEm 0:” Mimi/{(63.5 We 44x9 55tin (definition 0D firm idea! SOiKJfiDHSI Rh TDi‘Qpi (3M aim jMiiQ‘EJ a: FM ioiaal S’aiu‘dwizm 5. T (P) The entropy change of a heat reservoir can be greater than Q/T when the heat ‘ transfer is irreversible. " The GVYJFQF); Chanélg rib A k407i {‘EJENJAP (Kimmy; :2?“ ( regard/Mg wLa‘l-i/ter the head LURE 'imM-rié‘mfléi Feversals-l/ or irreverslhi)’. Page 5 of 9 Name SDHQM K9 it) Problem 1 -fi points Steam is throttied across a valve and then recompressed as shown in the figure below. The compressor, piping, and valve are all perfectly insulated. Assuming the compression process follows a reversible and adiabatic path, what is the temperature (°F) of the steam leaving the compressor? -! 5yslfm | (isflgfig ay/omr'ma\ . P2 = 80 psia Steam P1 = 120 psia T1 = 373.92. °F $515M Z (isaniropfr. to mflf‘és'iéoA A620 - >— Pa=150 psia ~ ‘ ‘ ‘ ‘ revers‘ua The. CfiMflFBSSiaa {sarpgfigg QSSOCWLEA torle Edam 2 Li Northrop: ( l d galla‘uehc\ Tim Mt?an SE: 63. Na OAK olt‘iefmfae T3 once We am . {Cums 33 524.22 a 1'5 alumni; ngwfl (bale mfl T3 ‘m SEQHTEUEJ at e w! 833. To a’EJngi/le 5; (WL‘LL is 9..le in $35} mo, Mire System; ill "I Lt an a‘rsaiLJFiL exflamsw acrors ‘Hne. waver Clamalarkéeal ? Mam; lit 1 Sables“ «MM. {3 F740! H. using P, W; "TI. (if PL 2\ FE“?! S1 Utéiwfi Pa, Anal H?“ Sat S} 33 #3 M9135] P3 magi ‘33,. l\ R :- [2.0 [2319» Surerkic‘l’fie-l __ :AWFGLLE 40 @Jlfiuulfl Hi T= 373.911 ‘F 5.49m ~_ __ 5mm? ‘ ‘QCAVUQEfl BED [age i“.4l (lad (F 3&0 ' F’ 37332? ago ‘1.“ i"! 137%“ 1201.4 X [3123 t 38?) ~ 3670 EL“ . _ fl ’ ‘ — o . m X1 [1in9: + {37331 Blelzfl- flowlyl I'm ‘ l2 0’ 4 1L Page 6 of 9 Nana Solujrfim K9: - Problem 1 — continued 23 PL:- 80 {ix-{q mm gyprwfibfl 55mm TOME €58 P 754 SVIWW A?) 110514 11:, 05' TL: 360‘}: 3%; NE }AJr9rfig\Q¥aq rafiirmf St? M3351 EEM'F 4’ WIS: 33 M ’3) "Pa: p30 (75;: 5“ “QM Aim Enltfifpoum Jm PSEHUL T3 ‘F’ t: Wm 3‘05! E4} 450'? X 5—0“: 9.3”..— 5} [gm-F [£3313 {@3315} 1666?. Tax: X“ Ema (see-43$ * -~ 4m fl \ u: (3 : 485M \ 1 6-5-me Mm,» Page 7 of 9 Problem 2 - 35 points An ideal gas from a reactor is cooled in a heat exchanger as shown in the figure below. The heat exchanger and piping are perfectly insulated. What is the lost work (hp) associated with this process with reference to a surroundings temperature of 70 °F‘? Water Reactor Effluent Gasu Q 5‘ M7 “R 80 "F 320 °F 77mg“: 5 . a 3 CD = 1 Btul(]bm °F) , .. n n a C5) = 0,3 Btumbm DF) f-lLJC '5 FPFIELTH/ ’Uf/N/ '5A5m\fijl"90{ - (152 e O r 41'—~ 57949,” Lama/67 Water Gas 55%??? 100°F 150°F éaqmfl . m V a n a at M :ML MN : {q 6,3, [T =(4-S’R6h703 (z - saw {2 SW (mow wtflllmm Ear 13m Sworn/7 {cal M (c 51 f0 “vbl $753.9“ alsxprm; A a $6- \ h _ . , "t ms + was» or; w + we m m» o I; “L SAQ‘A/‘s‘u’e 0, no MAL he‘lme Syjlenn em! .SLAJTE lane's \ : \ 5 ‘ g I SC; m Nplg‘r' A F J“ M35; A a Q ‘ M WAEF .1 i 01’ ago 6"— . y e Mat-15 a “kawn I 5‘ LIN-A} “£13 CAO’ICIPMMQV We M Wham a faint-The MW! W3 . ~ r 5 ‘ Tint mug”; ‘l‘ 5M £25 + 3 “52:0 nu slutfir we]; ((35595 O) PDFQCJV labm‘E‘ian Maude? L94..me M ham-1r "hang!" Qe‘lwerm 57599!" .5 summit? 57,1ch am! Surname!” 5 mew-lee kgmhr 1“ vh‘jfi [El-gm 2-: :3 my 2 h mwserer D‘AW F AH a; Page80f9 Name W— cm mm: “P” 6::er Hm “(’3 M Problem 2 - continued . W Eastman {LL «OWEN IC A! “D‘F mo‘F FA HM A"\wéfi’r l f0?“ = 01 ATI :- I fi‘Pboo—fifi‘r ; 20 TM 50'? {H10 86"F m f“ rso‘F H”, mm mm Maw = f: a? m = T; 320'; 5‘“ 33” 3a.! b A” 1% f A ~ _ 'wth‘Y—zo wk m M ' r: 3 Crzhé av. 36‘.) f m 5‘ a: “‘5 55%” W .. 53¢: 67 '(2 . W I aim rid—w- = Asmker" fag E? “ Oink 1%: - 6 ER «L ssaww 51%? ‘R 34.; ._ : O Him tm-fl‘ lag. . kg - all 1 T1 13%» “4"” 1563‘; ' [CRT "'- KJQ" ID 2 Crap” - $7: fife/a «L 776.37% P V 1 ‘1: 775%.? x. D, R-fimwa 3:? m; \h 2: m. .537 mm"! "a 0M??? :JmTJHJch O 338 M M (I 131% \ kw W m _ PM” ‘\ é : (lom%§ifié;f“)(o‘03é4 firmwax + (332;.3 L, 0.6738 My; rd c» a r - 134“ ~ iii";— _ “‘1 (m - M v m w a 135;: Nag» = (szfiev'@\(74efimr\ 1 33:5” 7‘? X[°‘°"4753\% 39:03 :- F ‘ ‘ ‘ "' WWW”me \ .1 L‘ M w .05; 151:5 {Wm w /"“”’"'_:W Page 9 of 9 f/flv—fi— ...
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This note was uploaded on 10/11/2011 for the course CHE 3473 taught by Professor Staff during the Spring '08 term at Oklahoma State.

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3473-10 Exam 1 solution - Name I CHE 3473 Chemical...

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