che210exam307 - ChE 210 A Fall 2007 Name M“ Exam#3.w Show...

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Unformatted text preview: ChE 210 A Fall 2007 Name: M“ Exam #3 _ .w_ Show your Work for credit! Good luck!- \ #1. (25 points) VLE Heaven Below is a Txy plot of ethanol and water at 1 atm using Raoult’s Law (and Antoine’s equation for vapor pressure). W?” “WPOS . huh 0 on 0.4 0.6 as ‘1' 3°”th P“ 0+ PW‘ Exes inethanol) ' From the Txy diagram:- 4195. (a). What is the boiling point of pure ethanol? '38 -5 °C l 0 #5 (b). A liquid solution at 85°C containing 0.25 mol fraction of ethanol and 0.75 mol fraction of water is heated until the first bubble point is formed. What is the bubble point temperature? What is the composition of the liquid at this point? What ' . composition of the vapor at this point? pr’i-f QESOC .: 0,25 4m Va. = 0.34 [w 0 gangs]- \ 8P3 (c). For the 25 mol% ethanol/75% water solution, instead of usingt e tagfim, w'fiie down the equations and solution strategy that you must use to SOlVe for the bubble point temperature * but do not solve” I! You do not have to look up Antoine’s eqn data or write Antoine's eqn out. 53; attached. The experimentally determined bubble point temperature for a 25 111.01% ethanol/75% water solution is 82.5”C. 3‘05 (d). Why is the bubble point you found in (b) so different from this? Rmomfi's Law ”Sums Menu 8AA: \‘clml l.‘ w'cl mde'bm}, E’WH 3' H10 art MO'l' it“ Coupon-Lads; 50 ll (deaf urban Wcli‘hnh fails. 54195 1,05 Know , Ab‘mkefis n Tit” L ”it OLASVQ/ I; 1mm W / 3‘» “Q“ - ‘F ’ “3 ’N _\ P: 750 MM “3 ms, gut-15‘.) a. 7159' Om u: be wH'hté. 80:? too; digg- Ccalodafia. pa" k4). Fab“ Calamioi’m' r’r hawcL skit 3-9 55" 49$ {'1' 23m 1:, CWMIA SMIL CW) I-f— MT , fluxes: m Tb at awn,- 'T' (In $321M {Wt-125%) “F Giro-'13 sun-+4.3, «45¢ a»- 0w; T of old“! Way- esfiM/‘tz.:{ em»— mpMI-a 50*” Mr Mk7) -. ChE 210 A Fall 2007 Name: M Exam #3 #2. (40 points) Energy Balance Around a Boiler. Two streams of water are fed to a boiler. Process data are given here: welsh?" ‘ Silt?) i=1 Nanak .. Feed stream 1: 120 kgfmin at 34"C ‘g‘figm 3 4"! Sig"; Feed stream 2: 175 kgfmin at 56°C 13- 15 V "dd Boiler Pressure: 17 bars (absolute) 1 55°C. _ ' " I a 5 55:77.3 H‘Cffit “fill-Hr:- “1- The exiting steam emerges from the boiler through a 6-cm diameter pipe. The emerging wet steam has a quality of 0.90 at the boiler pressure. The problem is specified (no degree of freedom analysis is necessary). apt-.5 (3) Draw and label a flow chart. 10 P5 (b) Write and simplify the appropriate energy balance for this boiler. State assumptions when neglecting terms, and also state whether nonzero heat and works tenns‘are . positive or negative-PWAH «- LE... +039 = i5 "' W; fi bl: = Q Q > ¢: _ 3'- bkh‘k-O W-fim-‘J'IWW‘D Elva“) 1” A‘fi': d M L1- Ml like? “MW-9 (c) Calculate the required heat input to the boiler in Isl/min. Neglect kinetic energy M changes. ' A: 11°" Reference: ? 0 31.019) ml DM'L 3gb or H 150(6)“ new: $2. il'n'h'll p}- ‘h‘l‘Fill Pf Substance A ' I Ham H20 .9 in streaml —_—- H20 2) in stream 2 _l!-—— Wet steam, quality 0.9 ———z- l 12'. kg ((1) Calculate the volumetric flowrate of the emerging wet steam. . o ‘ 1* h C” 5:635?“ MF-rlufi.” L‘s Low-it‘d. AH = G. 51” ‘ “L as: hdfifihi: +6 .I 5.11" Jere-'34 T: . N ‘ “ '1 3'5 2:153. . fiéatlicg-‘K‘fll T1 by}; Xtofl) -- [M gm H(H,b‘£3’34q_) "' 11C?%,_o®,55‘£)-t . WV“ . A I L 01 Cd) i’rass'ura we fume cal-Rd on H ol- “LO (6'). 5: an. Tel-v“- 6:5 “"6 -° w? Sui'cl weir M .3e't z...r 55°C. :2 0.“? M32. Table 3J5. ) A 3n; ' 3"” I: G CHI—‘7’ L95 1.34%? it'd) = “2'4 ”IE3 H,(“zbl¢).5‘”‘.w'd‘ ): 2544 ”1‘11 1.1 A a»: T4 (“in fia'flJ'V ., rm c...) = 2743.4 ”jig arul'ng-lw, we “in we”) = 8W8 “it: - is. ‘ - 2,}; ii 5 0.40 {3“ + w ill = Miami-1+3 +°~‘(“‘-3)= 2901.24 UM, 3 pm, 51-64 glam wFH-x X 245%.; (nolJA %) — 120 a; 142.4 353") — ”'53 . ($554-$733 '3 Lilia- : 100126 he: {13“ :2 2;} MIN A “cad V3 fi. \fi LWJ-u 4:3, Briana») = 0-0“le walks '\ 2431-5 V3 ( WSW, x30fi3 3':- th CDJIBL» *— DJ £5.03nb33 .= 0.1050510 ”Elk-j 30m h; hurv ChE 210 A Fall 2007 Name: Exam #3 #3. (35 points) Energy Balance Around a Condensor. A stream of pure cyclOpentane vapor flowing at a rate of 44.66 mols/s at 150°C and 1 atm enters a cooler in which 55% of the feed is condensed at constant pressure. The problem is Specified (no degree offreedorn analysis is necessary). CsHloU) 9P5 (a) What is the temperature at the condenser outlet? Explain how you know (a single sentence should suffice). {If you cannot determine (a), pick a Tout less than 150 and continue with(b)}. Tka “hm“ boilt' P1" {5 m a. m P1» 5,1 ”13 Fri- a} wkfi'ok 'V andl (LI-n Mum's? M in ._ '1:me “Ta lc flat) W waywat bulimic, p1” EN cydopfinfime u: 49.3°L (b) Fill in an inlet-outlet enthalpy table and calculate the required cooling rate in kW. Choose as the reference state (351-110(3) at the temp. in (a). Reference: (331-110(6) at temp. in (a) 401.30]; Substance C5H10 E Cme V 3pc (1.3.39.th = 0.55“ (44.51.) =- 24.51, filers, figswm m- oww sin-um =- aaua - 24. 5'1,» 29.“; Loan»? 1°an dam 99" ngopm‘fitM (”melts GALA 62‘) K Luv "-' 21.30 ELIwal ””9“ CP- “"33 h 3"” Cw c n. £0 = 73.391 “0'3 +- 3918 Mo'E‘T - 2554x5313 bemafls ”Incl ”CD-1.13936. '5 O T‘ n A = C ( clT‘ lli"b“*1n:WXT3 A ”V + Salaam 1”) A ‘\ ‘YO'C. __ q?“ HCSHIO (v3 ( '57an :- AsHV + all” gogifi‘g‘t: ' 38' 3" hr)”, A N. “.99, #5 3pm 14:94.14”) (491‘3‘C) 1 AN - 31.3 Ky’mD‘ t I , . n. .. stats Q 2-. AH : i no"; EM "' 2%.. H“; - Slouch-L3) q MmflBaa) :- ”Ml. 11039" _—. -"-.tt. X103 tau} 5 L—__.. ...... .4 ...
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