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Unformatted text preview: / /k;f/(WWW__‘\II t 5‘ KJ } .4» NAME < g y m SECTION YOU ATTEND 1 pm 2 pm Separations Processes CHE 363 October 12, 2007 Mid-term Exam #2 Open book, one 8.5 x 11 inch cheat sheet, calculator, your brain, writing utensil, 50 minutes. Short answer (2 pts each) 1. Why is the gas flux measured in moles when caiculating tower height but in mass when ; f? k 0 calculating tower width? obit mai 29: if éflddmwgi‘” ': 7‘16 mew @K/ pyjama [fife—tree 7%MV «ll/“MO Xfm)W¢ edema“ -‘. hwég r i? f u 5 , .~ ‘, 3 - e _ ' Egfltfg 1 \3 fg’ffigfififiv iii/Mai, Efikg 2., gay?!” i 2 ' {n {l 5' I “-1516; I - ‘ H I 2. What are two gases commoniy removed by absorption? £9862; 242,3) CO I NHB Lzmwmsww flmmm/i ‘ r? % @flzfit fl, TL— 3. True! False ‘ a. Separation by a porous membrane is based on different solubilities (H’s) of components. I $2; g gigs» gag-445.9313?- I .r NO Wm M Lfi/W % Masergasfewfifls/ b. Flows will be constant in concentrated separation processes (i.e., tube and shell membrane modules and absorption columns). 'NO / Mm Xfl/L WWW fits fflm. c. Using a reactive liquid during an absorption process results in a better separation. V€§’ deflate/£3 144M Ac /J%Mj 90/2013 fig/1’1 5/ a? éfld’éfizg- 4. Keeping all other parameters fixed (ie, gas flow rate, tower height and width), you increase the liguid flux rate 4—fold in an absorption column. How wiél the following change? (Circle appropriate response) ,5 V , i h e 44“" i i; ii i. back pressure? DECREASE No CHANGE i ii. exiting liquid concentration x0? INCREASE o CHANGE a 1 a“ i rte rm“ 3 5:? Emma“? entering gas concentration yo? lNCREASE DECREASE i i " it}? 4 I, i ‘1’ 5, big” Mfiffifi infifitgvina% 2m, fitie‘i? at" i, - ~ Mr W Ea" r t 4.1 E if is My IV. liqwd Side mass transfer coeffiCIent RFQECREAS flit/R @SE r O CHANGE 5 fl " Q J :5?" a N has" AWmea i i/ x in r 76% . Aggy/Lewes“ v, driving force at the top of the coiumn? ENCREASE@ NO CHANGE ‘ ‘3 “WWW [gigfiir‘k 5&9 first Short problem 5. Write an overall mas ansfer coefficient in terms/(4f the individual mass transfer coefficients for =m5 hollow fiber membrane tube used in blood dialysis and shown below. Which layer sistance dominates the overali resistance? (5 pts) i /’r'\ i 64;- Lruelc A»; {a ’1} __/_' 3, .TL i 4' f1]- r’A. h /' JLL , K fit») D'E/ /bjfp% if} 53’ng fiteexiraea if) W t“ .i “it * ‘ a it,” {35% ‘ PO 3’0 k3 M a—f aflojer M412 dirde 'm r “ ma“ " :- . if? iflw‘éfi'fla‘gé; 4°09 fart/Ti. tfiflfifiq‘fiw f‘g‘w‘jmfifiéflfiyfi «arm we i a, agitate mfig i?“ ital ", r“ em, 6. You want to absorb 99% of the carbon dioxide in 4.2 kg/ sec of nearly pure air by washing it with pure water in a 3 m tall column. The equilibrium line under the operating conditions is y*= 16x. a. b. .0 Sketch the operating and equilibrium lines in the limit of minimum flow points on the operating lineal/(Wig??- Sketch the two lines for 1.5 times the minimum flow. Wptfiifgj \ How much water (kg! sec) will be needed and what is the exitifigliquid concentration (retative to yo)? (5 pts) (label terminal 00130 T r) 7’. Ajoint venture has been underway for severai years to deveiop a membrane process to remove (302 Wfrom high~pressure sour natural gas. Typicai feed and product conditions are: Retentate, 60 atm 98% CH4 ‘- 3 CW- $0.3 um 2% C02 fl' 500?. Feed, 60 atm 70% CH4 30% CO .2 Hollow fiber membrane material targets are: Seiectivity or: COTCH4 : 50 C302 permeability (=DH): 10 (mciimzvsecymletm Membrane thickness: 0.3 em = 3 x 10-7 m C. o( a m 91%! 33sz 3 (Xil'yfiKleyz) x; = permeate mote fraction y; = retentate mole fraction .[LW _ iv_@ 21. Using 0:, find the permeate concentrations of CO; and CH4. {5 pte) Eek-55%; } 1. Lorelei “LC 3 b. Using a, find the permeebéiity of CH4. (5 ms) m__ “a. barrage-3": m‘imefi .3 Write an equation for the flux of 002 across the membrane reiative to the flux of methane and find a number forthis ratio. (NOTE: Use the average of the feed and \ r W.“ “‘1: retentate concentrations for the upstream driving force; no unit conversions required fig 93% use correct concentration units; no fiim resistances.) (19 pts) 1‘, 71‘ airtight/‘th m ‘12} L0¢41\\i\fiw ,1 err M”) were e v, U v g 1 :r Xm/ m— m; «03* t “e \r e @r'“i‘*j‘wb . = 50 z j K 09W"; red I , ’ if a, ' (kW/a rm) 0 we -’ .949) X54 ‘ 1‘ M2 '1. , 0 50 L SO’SGXQL 5 1:23: .79 #602. 2% ) WM 0 9. ‘ : 0.5! ' c? 8 166202 3 3,0 g C“ XC‘H /(” 0H4“) : C i“ Koo 0K 0.0.2.) fi) 50 "gram. , {Eff—it- “ o, 3 0. 02. Xwe/ Liaiin xwé/ ,7; gt '79 vac-(«1+ “Km”; 049 T2765: l0 l0 [3 MM 0.5K (P35#éoa+nfll“[{dm'o‘49 [OJ (memtrafi I Pvt-01W- Eguations which may be usefu§ N1 = kLAcL = kGAch kxAx = kyAy = kpApi : (DHe/fl) A concentration N1 = cm =j1 + CW" : -D(dc;/ dz) + 011/" 01?»!!! dt 3 AN; E=HTU*NTU m gm m 1 .in Mlflfll) Kya 1—(mGIL) (yowmxo)} Operating line: y = {ya — (L/G)xi] + (L/G)x Equilibrium Sine: y* = mx Of. :- D1H1/ DgHg 5 (X1Iy1)!{X2/y2) xi 3 permeate mole fraction; yg : reten‘iate mole fraction R (gas constant) = 8.31 J! (moi—K) = 82.05 aim-cm? (moi—K) = 1.98 ea” moi-K Seiect correlations for k: ...
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