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HW5_1and2 - 3 Water containing 0.1 M benzoic acid flows at...

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Unformatted text preview: 3. Water containing 0.1 M benzoic acid flows at 0.1 cm/sec through a rigid tube made of celiuiose acetate with a 1 cm diameter. The walls of the tube are porous, 0.01 cm thick and are permeable to email electrolytes; solutes diffuse through these wails as they would through water. The tube is immersed in a well—stirred water bath. Under these conditions, the fiux of benzoic acid from the bulk to the inner tube wall can be described by the correlation below. Remember, there is more than one layer of resistance. '51‘ 21.62ng ’3 D DL 3. Write a mass balance over a slice of membrane with thickness Az; include convection and mass transfer. Solve this differential equation to get an expression for the average concentration in the tube at a particular position 2. (Similar to HW 3 except we are now considering a holiow fiber membrane) b. After 50 cm oftube, what fraction of a 0.1 M benzoic acid solution has been removed? (i.e., c finai I c initial) 0. What would your answer be if the benzoic acid solution in the tube is dissolved in benzene, but the membrane and surrounding liquid are still aqueous? benzoic acid flows in tube ® Si‘wiim i1. clissiilw‘j mass transfer of ' Fodinj’; Sew . benzoic acid PM FinaL mac, if ”HM (59 1111:5003».- AM mMu—M ] ~— . tilde-$50 a“) J C(i‘ifi can.) ”(has Xfiul 0W fit. an. 1M4”. gmgg 4i rigid & M‘Emiomm will i '5 ‘9’ 1 a V3 L” {9% C"”“"‘*“""\) ‘ W ' WA .‘ . [c -- ELL-k:- . .m to %L gomj £1.10 T: Q‘4 XEOWSCM/KC Kama/u : E —; I'ixio”5CM1/W X OtQt C4,“. $153 Let/W. '1. -L- flu-v "I K (inf limit) ”w- 3155 xio'g Wage. ugx'om in I?“ df'rfiMLibPLt W gm ‘lm‘ a) fig M45 oil'fil ' ' Q Flam.) MASS sumac on Q“ S tat-6’2 oilfiLWSch . ri=o= i—C‘Brs'c; 2E {V3 A1 0‘ J wLSdrE-ME TT n“ L Q T "U’CJ “- :17.— 'yU-I*€’:,1( P“ Waifig 5K9 (C! __ $3») % %‘té% i*+{vf‘fl(;fi9 W @) arm Fw afn'u; FMC ‘m M L “-3 Q w) W XE; Q1 pcaacgw‘lwegm in M: alva M I Flow :22le “immJ Cm ‘- me. im W'Ewh“M:fiM Mm avnMJ’x‘ukt (“’0 ’2 K XFoL Rem +wb& ”kw?“ mww .4, WTCEEJ+ j(clfio>=o M,» C i ’3 $1M ~o BC/ 2:30) (it"C'o :Q( m CH, C 4 “any: 3:9 mm. ”—1- : (3)633 #53- K3) ”J— : CKP(" :WJhfi" "84' I"? ”flag; té 0{ Clo l [9“) W“M EMEJ— _ “17% Hmmj, 91 mi ‘ @— =Cfifi+ad¥ “Mg ”WM d4 wag-mac Oaflfimi'w? IU— vmm+ 'a-F was.“ wave. V; as ”wife—F? ‘Uevjlec‘i’flwi 3% {6‘3”- 4; WSQS‘MMQQ W'E‘Sffitiw «we pmcm Catfiw. lain mu cffer‘WTS “M wM-tr" : G . 10) 01,5410 054 <3." {WE‘LW ' Par+="[am Mm : C5 K " f 3 J Wm K , '3“ + ML"— ‘ fi la »ibz(i33""j(iww K2 6 km H (4—14.) m I {agxio'fi ($0) 3 L53? xgo'4-6m/m I f "E 5'" . (L93: %I, {mars (Wm/3:41;. ._ {QMEGMJEL K “WV-“q .OM m‘fi) ~ :2 £93 Kl; ngm a m 3 Q1 - $5.3 .3 W (, 401W?" Zia) aw? @jr» 5% (:10 a if? (Upgx @{ 1. You are using a 1000 A thick silicon rubber membrane to separate oxygen from air (see diagram below). Find the permeate (=downstream) 02 concentration and membrane area required in the foilowing two cases. Q = flowrate in ft2l min. a. The selectivity (02! N2) = 3.2; permeabitity (DH/RT) of 02 = 500 Barrer; b. The selectivity (02! N2) is 10; permeability 02 = 1 Barrer c. HINT: use the selectivity to get the downstream concentrations; then relate volumetric flowrate Q to a flux equation (ANq). d. News: 1 Barrer : 10"” cm3 (STP)—cmr (cmZ—sec—cm Hg) where cm3 (srpy (ch-sec) is the volumetric transmembrane flux of the diffusing species at 0 C and 1 atm; cm refers to the membrane thickness and cmHg refers to the transmembrane partial pressure driving force for the diffusing species. 1 barrer = 3.348 x 10'19 (kmoI-m)/ (mz-sec-Pa) These are, unfortunately, the standard units of permeability. (see p. 439 text re: units) {gamma TQ— Silicone rubbe;r membrane PQJQ M141 Ti; 9—1! .J Air 02 rich P: 1 atm P: 0 atm T = 25 C T 2 25 C Q = 100 scfm Luau) ragga 0} Dav. Cam lot/0M & =2 SCFWFC iésiuo (rum: WWI/Leg‘ wfimbw Wt Um M +23» 05a 51'). ”Fan-{n+6 t 0’) Select-Surly a: 2 it; : eel/:jjr/Vreéw-iafi 1 4:11/ 1 2, a. “XL/j: (l-xJ/(hal) .w to: sew, «x w — {-o C ”ma 5'3 z a. [-0 0.8-l °< (0.41 —t 0.9.: ‘9 ’55:: r— @NOTfi: if (figttf‘ anew-x attutate and; __. item we i'iis" at image/w sfi't- a M or w/ W3 31“.”:322/ M ‘ b) Hmflha W: +ma. 5f MES Raul/2713A =HMUTHKM, M 1+ is WW ff m: we. fit Uoium+dc wafi P if"! PtrnmLz-ll}5 , Ba-V‘V'Cr‘s 7}? '3 a __.,_ / : (1m 2 Q ‘3 )1 (P02,¢‘,,Pm> \Pm <0~1=><rw am 2‘ J; w watt. W73014=<oaw<oma=0a+m M5? fiMflQh/Fufihntl Gym-Mil a, ' + .. (DO-Fm. 144an W WPO 1 (Oq3)(0am) MW F’Mhfl’q PWL/ “ 9*; “WW/Q 454 J .22 3 “ “H W . ,‘7‘0 __ é W3 61 500%.)». 35c? 9.3 fl— - ”26m .. $2.83 X10 ; QSTP \ j (O‘cfimlw 5 b H WA: M ewe/«m ? F—‘u’Llll " PO (‘3, _ ‘ ‘ a... M“ f) 1 (6594“) 8' CMP >(0'lt +4. $écd‘}) Cwl- WCKJC‘L3 “H.“ 3 ‘3.‘i><tos ml “‘2" a C 1040-3- 5%"; =,4 m/ sowwm ...
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