Exercise 10: Theory of mass transfer coefficient at boundary

Transcription

1 Partle Tehnology Laboratory Prof. Sotrs E. Pratsns Sonneggstrasse, ML F, ETH Zentrum Tel.: U Stoffaustaush HS 7 Exerse : Theory of mass transfer oeffent at boundary Chapter, Problem Fnd the dssoluton rate of a holesterol gallstone n dameter mmersed n a soluton of ble salts. The solublty of holesterol n ths soluton s about.5 - g/. The densty dfferene between the ble saturated wth holesterol and that ontanng no holesterol s about - g/ ; the nemat vsosty of ths soluton s about.6 /se; the dffuson oeffent of holesterol s.8-6 /se. A shemat of the problem an be seen n Fgure. r r Solute flux away Fgure. Steady dssoluton of a sphere The dssoluton rate of the gall stone an be defned le: dm M N A A( sat ) () dt Where the surfae of the sphere s A d and the onentraton far away s whereas at the surfae of the sphere, the onentraton s the saturaton onentraton, whh s the solublty of holesterol n ble soluton and s gven. Beause of the densty dfferene between saturated ble soluton and pure ble soluton, we get a free onveton around the sphere. For ths ase, Cussler (p.5) gves us the mass transfer orrelaton: ble = Galle

2 d d g..6 D D We an solve ths for the mass transfer oeffent and obtan: D d g D..6 d D d For the densty of ble soluton, we an tae the densty of water: g All other quanttes are gven n the text or are well-nown onstants. So we get for the mass transfer oeffent: 6.8 s ().8 s Fnally, we an alulate the dssoluton rate: S d sat g s s day g.8 (There s a typo n the text;. g per month =>. g per day.) day () (a)

3 Chapter 8, Problem 8 Calulate the fraton of the resstane to SO transport n the gas and lqud membrane phases for an SO srubber operatng at C. The membrane lqud, largely ethylene glyol, s 5 - th. In t, the SO has a dffuson oeffent of about.85-5 /s and a solublty of.6 (molso/l)/(mmhg of SO). In the sta gas, there s an unstrred flm adjaent to the membrane. th, and the SO has a dffuson oeffent of. /s. Hnt: Resstane, R MTCoverall H R () p L R p G D. p s 5.59 l l atm mmhg prt K K mol atm s l mmhg p mol 5.59 mol s l mmhg mol s l mmhg () () H mol.6 l mmhg l mmhg 8.6 mol () L 5.85 D s.7 L ll 5 s (5) R L l mmhg 8.6 H mol.6 L.7 s l mmhg s mol (6) R L 9.7% R (7) R G 7.% R (8)

4 Chapter, Problem You need to estmate an overall mass transfer oeffent for solute adsorpton from an aqueous soluton of densty. g/ nto hydrogel beads. n dameter. The oeffent sought Ky s defned by N = K y (y y ) Where N has unts of (g/ se), and the y s have unts of solute mass fraton n the water. The mass transfer oeffent s n the soluton s - /se; that wthn the beads s gven by B = 6D d Where d s partle dameter and D the dffuson oeffent equal to 6 /se. Beause the beads are of hydrogel, the partton oeffent s one. Estmate Ky n the unts gven. From Chapter 8.5., we now that the total mass transfer oeffent has the form: K y = [ + H ] s B Beause the flux has unts (g/ se), we need to orret by the densty K y = ρ K y = ρ [ + H ] s B Insertng for the partton oeffent H (gven n Text) and the gven equaton for B, the overall MTC beomes: K y = [ ρ s + = [ ρ s + d 6D Insertng the nown Data nto equaton results n: B ] () ] () Thus gvng an overall MTC: = [. g. + se 6( 6 se) ] (5) K y =.9 g se (6)

5 Chapter, Problem Garda, a mrobal dsease, nfets most large anmals n the western Unted States. When you anoe n the northern Mnnesota, you rs the beaver fever f you drn lae water nfeted wth garda ysts. A ommeral deve to remove these ysts onssts of a small bed of on exhange beads loaded wth I -. When you su water through the bed nto your mouth, you quly ll the ysts. a) Wrte a dfferental equaton gvng the perent of ysts lled versus flow. b) Imagne the bed s of onstant volume and operates at a gven flow. Explan why a tall snny bed s better than a short fat bed. ) In fat, the short fat bed s used. Explan why. a) The mass balane s A( v ) ( v V z ) z z () For dlute system, v = onstant Av V z z z () Sne V Az, rearrange the equaton z z z v () z v () z Sne there s a rapd reaton on the surfae, Therefore, v (5) z, v and a reman onstant throughout the bed, and the boundary ondtons are (B.C.) at z=, (6) (B.C.) at z=l, If we ntegrate eq.(5) wth B.C.s, l d dz (7) v l ln z v (8) ln l v (9) exp l v () b) Constant flow,.e. volume of water per tme Sne V Av onstant, v when A. Also, volume of bed ( V bed Al )s onstant. Therefore, 5

6 V Vl v A V bed Vbed exp l exp l exp v Vl / Vbed V Therefore, l s not nfluenng the onentraton dretly. However, we have to onsder the nfluene of the length on veloty at a gven bed volume and a gven flow rate, see eq. (). Ths veloty may nfluene the mass transfer oeffent. For paed bed, ths s gven by (Cussler, p. 5) u or, u. du D.7 (u ).58 v Vl V bed Þ /.58 (l) Therefore, as the bed beomes longer and thnner, the veloty nreases. Thus, the mass transfer oeffent nreases and the onentraton at the end of the bed dereases. ) Pressure drop for a paed bed (Perry s Handboo, p. 7-, 7th edton) P L, a = f(olumn and gas haratersts) L P Wth our mouth, we an only supply a lmted pressure and therefore we have also a lmt n the beds length. () 6