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F2007_Exam3_Solt - (£3!ng wncwm‘l{wwsw WEIW 4L9...

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Unformatted text preview: (£3!ng : wncwm‘l-{wwsw WEIW 4L9 Wag/alum"! “‘3 NAME ‘ ’46/7/ Separations Processes CHE 363 November 6, 2007 Mid-term Exam #3 Open book, one 8.5 x 11 inch cheat sheet, caicuiator, your brain, writing utensil, 50 minutes. I Short answer (3 pts each) 1. How are concentrations of the gas going into and the liquid coming out of a total condenser related? 34W Concewtm—HWS‘) Ci) 'szxo :80/18‘ M ’UWGY" thoMS la (MIA S’EW’je ‘Hbf— cancer/ITIM’ilioi/to (kg-1L an agar/téyvam Sfm 30 Mo VLJE/ fMUo/W/ 2. What ES one reason that coold cause distl ation of two volatile species to fail? a o< ,2 rglm't‘t‘i‘ft W‘al‘f'ii ”Lia FS +00 M a calm-Mn 4,30 shawl“ +0 clotted/t 3794:427sz “' ED 7‘00 (SIM/£5 3. What is one advantage and one disadvantage of a iarge reflux ratio? we Swag/V D/Qum'i! <7! pynM+. / of we we 4. Considerr a distillation column with specified flow rates, feed and product concentrations (mp, 11;, 153). Will the overali operating fines change based on whether the column iffilled with structured packing or equilibrium stages? (Why?) 15:32“ cum/U Egret/”.173 M Irséagedomamméefm airméd ____,.._._u—-—— 46 prom/m, (er 1201,) ,- 7%.. W! W M New 7% W n” 1 0W 0’ W wm Aer?“ or as l Pm?“ ”Eff pmkr'if? Manhattan: 553% filak’l‘rhw 5. What IS one reason structured packing Is replacing stae es in disti ation columns? a 6M Wei?) /+lomalapu~ (mm—t LN /8M PW) " Lat—jaw in+cm~f¢cj p0 Surface/C W fl W5)» [(3 L 8‘3) (gt/A aoéfiu'lfi: W Wwa't'ah whiz: 5W Cfl/QM‘M‘“ ProbEems 5. What is the minimum number of stages required to separate methanol and water into a 99% pure distillate aad 1% pote bottoms? The partial pressures of methanol and water are 760 and 188 in mm Hg at methanol‘s boiling point and 2730 and 760 at water's boiling point. (10 pts) ’FDY— Nminl W FWSILC (9; OK cooLiiEL, :5 MMH 5P Z—J “<43?”- 13;; * 4 04 ,__ I"; Mal? ._ 1/ /, ~ ( it s) 2 JT 1430 “jam; 5 3 (El l-tofi‘iifl/ ES m. b Hum W 5 {a v 6 “20 6P ———l $5 il l: l}: K4 s ’1? '31” \4 ll 3% T3 ii: 6. You are distiiling enthanol and water to produce a 90% distillate and a 5% bottoms stream from a saturated liquid feed of 30% methanol. The reflux ratio you are using is 2 and the stages have a Murphree efficiency of 50%. Equiiibrium line attached. 5 "f5 Zf”. WA}? a. What is the vapor concentration leaving the reboiler? (5pts) 3" b What is the Eiquid concentration in the first stage? (5 pts) £642. jmif o. How many stages are required and on which should the feed enter? (10 pts)“; 33an .r Majfgcho’fi/Mffc ROLAi'fir‘Z-m 3,3 g ”Kb—“oft 6L) RB acts, as an (LBW/Q 811% M ; 90‘ Q. S‘l‘nfamS {gm}? 04A in w/ Mimi-.3} FWLW m1“... a 5% K Nd 6.0 md To md Nd «.0 Lllllltltlltrill'llnurl ‘ ¢ 9 NHL 333.6555: 7. You need to remove traces of propane from a feed of 95% propylene before sending the propylene to a polymerization reactor, which requires at least 99.7% purity. You plan to feed the bottom of this column with saturated vapor. You also know that the equilibrium line for the more volatile propylene is'y* = 0.20 + 0.80x at the temperature and pressure indicated. a. Using the distillation set—up shown below with no rte-boiler, you have a liquid bottoms stream of 94% propylene. What recycle ratio (RD) will you be using? (5 pts) b. Calculate the number of stages required. (5 pts) 0. if you replace the stages with structured packing such that the coturnn NTU is now 60 m, what is the new reflux ratio? (NOTE: x5 will also change in this situation.) (10 pts) we fiend-{sneer DtlrL . 3 if Q \ a; I i“ lax/L ”games? marinate ) (Rm 690'? D l” 5 0“ “l” / FLEHWW _____ .....f app? flfi M W" ) l l “mart g t ,3 ,6???“ 6M ;o-5a5 l 1 13° We it“: M (52?;— re} a ., 2A5 c (i— .225)r.D :9 , M ii fl as: W a to are, or it r W“; M “it L. to) Kmmw bait am: sway. ,__... £14 35, .751) If. m = 515.1 = _ p. wt- “3} M w z 37.8 . 4 _ lak(f§?:/) (N"4o.l ) ’ ' S a) F=D+L=1 (cM1m&) W ,CIS’: ,cflfilD—I fi+L . fie L 23> D: .lii'S‘t) v.91“ , 23’ 4.?0 ”5 , fl??~.fi§ ”(£255 5W o'fiS :' .‘lfi?(l- L)4 "K'QL’ L d a???’ (X - 2. Eng» ha Adi ll mild/1 l' / \ rvv .- CAL " NTUKT (90M: I 0.14-0.37? "0.5 {h 57!} .‘i7~%- he??? L— : 0.3(X “- (5+0 ,0“ [ofing—o 95]) C309 TMéflW J1? wi:,?3?5} L‘,3/?5./ 20: 44.46: low-W r70 “7‘7 (”07‘ M but 071667) pro/rider /wwpw Ml Possibly Useful Information Differential Contacting Operati n 0 Line I, L ‘35» RD , = — ——x + —x = + x 'J (’1)qu G (I) G Column Hsioht f : HTU - NTU y l d NTU :2 ' 41}— 3’3’ ‘J’ 1 _- 1‘ = In yr y: 1 [MG yo ~— y“ L (assuming)? = a + mx) S+fies Qperaring Line L L x R yn-l-l : (y! _— ix“) + MIN :: U + D u (I G 1+ R” I + RD Column i-leéohl . : 0.612? -N J).F.. h J 1' L L Nuln MA” J“ /ln ~— M 1 g yi —~ y(J mG‘ ““6 0r (Kremser equations, when y; Murghree efficiency ~— J yn -}n+i 7? : "Tm“ J)” — } 1‘») +1 Fenske x I—x N=ln[ D mi] /Ina’ 1—3:!) xi: Underwood L. _ a :32 xi, inxp RD 2: W = a + mxn) ...
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