WPR%202%20soln

WPR%202%20soln - APPROVED SOLUTION CADET SECTION TIME OF...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: APPROVED SOLUTION CADET SECTION TIME OF DEPARTURE DEPARTMENT OF CBEMIS'IRY _ CH363 20069007 WRITTEN PARTIAL REVIEW II 55 Minutes 26 Match 2007 TEXT: Seader & Home)? SCOPE: CHAPTERS: 5, 5 References Permitted: Open note and open book; Mathematics: worksheet (no other computer resources), ruier. - {NSTRUC’I‘IONS 1. Do not mark this exam or open it until “begin work” is given. You wiii have 55 minutes for the examination. 2. Solve the problems in the 6 provided. Show all work to receive full credit. 3. "fhere are 3 probiems olgges in this writ (not including the cover page). Write your name onlthe top of eac " sheet. Answer 2111 questions. (TOTAL WEIGHT: 200 I’OINTS) {)0 NOT WE IN THIS SPACE Caéet: Problem: Wei t: APPROVED SOLUTION A 60 - A proprietary soivont ieaches caffeine fmm coffee beans. Analyses indicate that nothing eise dissoives in this sokvent except caffeine. A piiot study of equilibrium stages roveaied that 55% of the underflow from each stage was soivemt on a soiutewfree oasis. Starting with pure soivent and 20003 kgfh of coffee beans that have 20.0 g caffeineflcg beans, the goal is to reduce the caffeine content in the beans to 5.0% of the original content. 3) (20 pts) Detennine the minnan amount of solvent required to achieve this separation using a countemurrent equilibrium-staged system. Wye as I. 5;.” ~— Ia FA (mm W e 1 g is; e: 1.3: ¢- .m- ' ‘15 . g 3 53h 3? an. 9305 ")wc‘bg 1: \fl‘ chféf} Laban; .k ...‘%m k h . in f _ 0 94¢in 3. 3 g,talb 9i If: C“ \ 1' ")3 b. m gqxib k g rqo €553” )Mfl (" ? A} “ 9‘ 3k b A»: in; 0.55 2) (40 pts) Deter:nt how many equiiibrium stages this separation would require using twice the minimum soivent flow rate found in part (1) above. 6'”) xmfi L“) . “fig-4 f'~—"~ QwfiL-fi 9‘ GL5) w “:4 (an; X (“:69 Sub-fiat v4 u-Jfifix'g‘low P O‘agbhfig 1-2;- W m W ‘J («.59 de‘UJM-j‘ :A’ ufiQ—b\_“‘€{ou Shmq mix: ML.” 3947* 95W 9"“ CH363, AY06~07 1 WE’RE Cadet: Fromm; weigm: APPROVE?) SOLUTiON B 70 . A}; absorption tower uses water to remove ammonia {mm air. The tower operates at 131m anti 27°C. Its inlet water is pure and its iniet gas has 8045 mole fraciion of ._ ammonia. "Ibo tower removes 95.75% of the ammonia from the air. The feed rate is 15000 molfh and the iiqoid to gas flow rate ratio, LN, is 2.25. The equilibrium data for tho ammonia»water system at 1 atm, in terms of 121028 ratios of ammonia, X and Y, are given on page 1) (20 p13) flow many equilibrium stages does this separation require? SHOW ALL WORK TO RECEIVE FULL CREDIT 3; cares 0 W W“ ppm: EALLAIJQE. 9‘: fig} L; 1 ‘ “’70 _, \J‘ 4- K y x, r- m ~ ‘l ~ mi ‘ if. 5:. X” + y; ‘ v‘ Sfl HGMIEL ' I W L . \th—n-W‘Hamwmm— ' ft: or C‘.Lfi( 1 ML; :3 —-- 3 KW”_,:M..,fi.,.\.4...... U 1: a“ ind-«Wm V \r . Y m )ngu + 01am); ' :2: 0‘96).L.\ "*5 _%WW 2) (20 1313) What is the mole fraction of ammonia in the exiting water stream? SHOW ALL WORK TO RECEIVE FULL CREDIT " («Ag flag <0.q5}f)(0 .GW§)C‘§H‘§B9’(M&IHB I» '” W* 1:. W " w ,r Nam; _ rooflésjyfiar ©.afir)(o.m§(asyéo> :: moi??? ALSo from; new KN “no , - " ? Cadet: mm Wei t: B (cent) 70 . APPROVED SOLUTION 3) (20 1325) What is the minimum ratio of liquid~to~gas flow rates? SHOW ALL WORK TO RECEIVE FULL CREDIT - l. . on. w SW “(fl‘ww a (fidsa- MAL V M" [W . - I [M @ Y 2' 0' on?» AT If e: QbuJabfiW ‘ an XI“. w 0.0-33 ‘ O6 WIS”. \I-‘V'bn— r5 ’3. I, I-«- w 3. LG? V (“IAIJH 0-539“ 0 MAA‘S’ 4) (20 pts) Estimate the tray efficiency if the viscosity ofpure water is 0.383 centipoise at 27°C. (Y on may assmne the liquid in the 003mm has the same viscosity as pure water.) What does this result mean? SHOW ALL WQRK TO RECEIVE FULL CRED??? Ugckg‘ W __ Gng-r rum-\st Lu‘P CH363, AYOG—O‘Y 3 WE'RE Cadet: APPROVED SOLUTION 9.2.x 3.0 mmod mod . mmad mod m 36 Ed wood 5“ mw c, wad Nod $53 3.37 8.0 wad bod . Rwy “awaiafiéfiofig Cfi363, AY06~07 W731? Cadet: ' Problem: Wei ' t: _ - _ _ C 70 . - APPROVED sow‘rzou A fenncntation process produces a COrz'ich vapor cumming a small amount ofethamci. The cthazzol can be recovered by absorption with water. The {allowing process has been. proposed: 100 moZe-% £120 Lean gas _ 30°C,110k?a 30°C,1101\:Pa absorber K = 0.57 _ 2.54:1 Pall rings fiTU=2fi come}; Gas Mixture 130 kznobh 98 mole~% C02 Rich liquid 2 1120363 e EtOTfi 30°C, 110 kPa 30°C, 2 10 kPa - ' 2) (25 pts) We the minimum water flow rate, L’mm, for 99% recovery ofthe ' ethanol. ‘ mav'mul-w (“an 3 13' m) mm}: 99.5425 Vvvb‘lk WK (31-1363, AYOéuO'Y - 5 ' WPRII ' (b: 0.01; V" m V*0.98; mnmv'*K*(l-¢)(*Im m ram-5m) C)LI‘€[21¢’Z}m 99 . 5425 @281; (*Part 3*) 309: 2: I9: Im*1.5; Ia: Ip-r- (1-41)) * (0.02) 121.80; L i :33: $35] Ki? {MES}: l . @9004 mg}: yin 0.02; fix 0.0; ¢*yi*v . m m*(1—¢)*fi ' .w .. I (5" “ 1? ’ Alf ‘ (Vi " K it??? (mt “ ‘53:?) ) * €3:..ff....m fim— mg" (Ami) {A ymt GEMS}: i0 . 5232 |n[45§:2 If; m Hogi Nag M45]: 21.2463 £4». MM! CH363, AY06»07 6 WPRII ...
View Full Document

Page1 / 7

WPR%202%20soln - APPROVED SOLUTION CADET SECTION TIME OF...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online