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che210exam102 - ChE 210 Fall 2002 Name Section B Exam 1...

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Unformatted text preview: ChE 210 Fall 2002 Name: Section B Exam 1 Remember to SHOW YOUR WORK! 1. (fi points) The reaction A -—> B is carried out in a laboratory reactor. According to a published article the concentration of A should vary with time as follows: CA = CA0 exp(— kt) where C A0 is the initial concentration of A in the reactor and k is a constant. (a) If C A and C A0 are in lb—moles/ft3 and t is in minutes, what are the units of k? (b) The numerical values of C A0 and k in (a) are 1.286 and 0.414, respectively. ' Convert the formula with the calculated constants included to an expression for the molarity of A in the reaction mixture in terms of t (seconds). (0) Calculate the molarity at t = 200 s. ‘l ” . ”.34 lac g (Cl) The avguméni 0? am exponem‘ba 7cm w 4 ‘ ‘1 ,.. '+ o nnm dimensimle‘iS. thus ‘7» ““3 “m S { 5:an k \5 m m“ (b) Ortfimm “b“ “ _ 3‘ ”quotes CA [lbmgfzsj - L280 it ”‘33—“ exp(— 04M [mm w‘)tEMin]) we wan-l CA‘: gig] z (a?) a“- (stow) Canver‘il 1 5; 3 “no ”“15 , m. 1, lbmoles " = 20.6 0. 4‘4 : . 9% m Kw“ 60 5 o o 3 CAEWJ‘M]: 10 6 ex? (' “Mm-7’ ‘83) _ y M t: 200:) CA = §.15’n_~3£2 3g (0 > ,_ L— ChE 210 Fall 2002 Name: Section B Exam 1 2. (20 points) The great Boston molasses flood occurred on January 15, 1919. In it, 2.3 million gallons of crude molasses flowed from a 30—foot high storage tank that ruptured, killing 21 people and injuring 150. The estimated specific gravity of crude molasses is 1.4. (Molasses had corroded the tank wall and the tank strength became inadequate for that much force). (a). What was the mass of molasses in the tank in lbm? (b). What was the reading on the pressure gauge at the bottom of the tank in psig? = 3.3. Ragnar 3.93.. m 2...._2xro'° 3M 433 7.481 8&1 ate!— +1F+ pa. .6 W W ~ ”M (a: 431: 3.2 '7: Ft 3° Manx}... - 32 n4 393+]? ”IN-‘1 = \82 95‘s N.” MM " 1“” ChE 210 Fall 2002 Name: Section B Exam 1 3. (35 points) A gas stream consisting of 100 lb—mole/h of a SOz—air mixture containing 45 mole% SO; is contacted with liquid water in an continuous absorber. The liquid leaving the absorber is analyzed and found to contain 2.00 g 802 per 100 g of H20 (there is no air in this liquid stream). The gas stream leaving the absorber is found to contain 4.16 mole% H20, 23.2 mole % $02, and the balance air. The problem is completely specified!! You need not perform a degree of freedom analysis. (a). The flow chart is incompletely labeled. Fill it in. 5 F16 $0M).‘Calculate the required water feed rate and the fraction of entering $02 that is absorbed in the exit liquid stream. Liquid H20 0 Inlet Gas Stream ”1.. 100 lb-mole/h _2.__, Exit Gas Stream so2 45 mole% H20 4w mole ”/0 an 5:: motefi/o ' 5°; 23.? mole ”la n air 72.“? mate 0/9 3 3 Exit Liquid Stream H20 . so 501 $005507, .; 2.09 “a,“ 2, ””3“” too Um.- H20 We need 'tc; Conv9r'i wild} maSS fa‘h‘viv" 5W3 r414. {law ‘m/iaufi aid-lat)“ mam-I to A “Mi/+5 «mm ‘Siku- ‘hlq‘ ”(tart si‘W-am L62 Van'anzS, rebuked +0 moleS‘, GAS spamm ‘(oo mama/k, o‘ome lbw“ the {bu-om. 0.45 wwusfu 4. 0.7.37. lawn so; 0‘5'5 Handcuff. { m b , 0111-, lbw ’ ‘Hz anuu: $TY€W W I“, “1 Zeb 'bm 30L H10 loo [uh H10 . FLOoO VI-WL 8 EA» 3 :G'uewcow .3 aémc‘vw 1 ,Fwwmo ‘1 ,y' .4“ i I 1151‘. . 5:000) = 0.77:! n 502 : ... 0.45 (1003 = 0.232 :4} 45': [1.562 + 043055;“ . i 20,3 — “mm 319‘? 487‘? " [Rm 7 “$07.3:37'44‘E2d M 1(1— kV T410 : r11 -‘ 0‘0414. fl; + HF“; ll (helm, (7517) + 487‘? x” lf', Hi = 4 88!.gki'4i'31i ”W“ k r' _ ‘S‘r och Cm WE; ; L43 : 0‘ 60?? i 0' 6 l M15 .— "W" ..———-—_ "' ‘ / 4s; ,._——- E 55 E 5k - 2.00 lb». 502’ '-"- O.OIO{{0 1 “30+ 34:0 “0.... (4,5 To “AW/"f XE *0 Y’s , Take 1 Us.“ may: ‘ M \b—Ma‘ds mefiflfi 501 0‘010‘9 94 0.0.9033, £2.00me ’1 Hto W4 LB gwo§4§ i 0.99.41” Maw: Laws Ana: floss (mo) = 0,727 Via , .32 z 7517 (EMIe/k So ; V O‘4§(I003 = 0.232 :01 + .omorfiung 45’ 7' 17.552 + amassfims :. F13 : 46107.5 lbwo‘e M.» M 0601‘s : 31.44 Hanwh )kr M K». TBTAL: too + .31 = "'2 +115 : 7s“? + 4907.5 .‘ n itydet :L a 3 AH; 55' r 56' -- 5°; 45 " msez a7.44 Hzo — _ 4883.3 3.15 4950.1 Tm L l o 0 '4'8833 '1 S . "7 549L071 .57...- PM 00 5°11 r15013 ___ 9144 o. 640 7g b 45’ “Son ChE 210 Fall 2002 Name: Section B Exam 1 4. (25 points) Liquid acetone is fed at a rate of 5450 mols/min into a heated chamber, where it evaporates into a nitrogen stream. The gas leaving the heater/evaporator is diluted by another nitrogen stream flowing at 18700 mols/min. The combined gases are then compressed to a total pressure of 6.3 atrn gauge. This stream has a composition of 9.03 mol% acetone, 90.97% nitrogen. (21). Perform a degree of freedom analysis for each unit, the process, and for the “overall” unit. Is the process specified? YES (b). For which unit would you solve first? OV€VLApL k0“— (c). ‘ Write the material balances and key relationships around the unit you would choose to solve first (or if you have difficulty with (a), choose any unit). State the order in which you would calculate your unknown variables in this unit. Do not solve for numerical values! Acetone 2 5450 mols/min 0 III 3 5 p __6_> Gas r N2 ___.____, o 9.03 mol% acetone 1 : 90.97 mol% N2 0 4 r N2 18700 mols/min - 0' 53%: mo‘ 1: V: , Em? Mm ComPCESSotL rm ovauwg now v 4 5 4 q 5 EAL a. :2. a e a «éwm (0sz " - ‘1- :1- 1 Comes cums 2L____ L p L. a 1 a :L 95 ¢ oncw Sauna—T ‘7 _\ v); m Sol-V6 C01. ‘oveuu, uwwraAuAAQESLiB: if “'1” oveaALL. Tam, n‘1 + 5450 + \8100 = fig Acofiouz. 5450 = 0.0403 he N1 r31 + 18100 = (1-0.0‘1033 lab .Solve Fry he Feom meme EAL. The»? 2:19} TomL BM, ow; '42. .54; cure. hi U52 we RrEDOQbAr’T gAuADCfi w CHCCK VIM-MES. #er RNA P at Com P-e essok . ...
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