Tubular Flow with Laminar Flow (CHE 512) M.P. Dudukovic Chemical Reaction Engineering Laboratory (CREL), Washington University, St.

Transcription

1 Tubular Flw wth Lamnar Flw (CHE 5) M.P. Dudukvc Chemcal Reactn Engneerng Labratry (CREL), Washngtn Unversty, St. Lus, MO

2 4. TUBULAR REACTORS WITH LAMINAR FLOW Tubular reactrs n whch hmgeneus reactns are cnducted can be empty r packed cnduts f varus crss-sectnal shape. Ppes,.e tubular vessels f cylndrcal shape, dmnate. The flw can be turbulent r lamnar. The questns arse as t hw t nterpret the perfrmance f tubular reactrs and hw t measure ther departure frm plug flw behavr. We wll start by cnsderng a cylndrcal ppe wth fully develped lamnar flw. Fr a Newtnan flud the velcty prfle s gven by u = u r R () where u = u max s the mean velcty. By makng a balance n speces A, whch may be a reactant r a tracer, we arrve at the fllwng equatn: D C A z u C A z + D r r r C A r r A = C A t () Fr an exercse ths equatn shuld be derved by makng a balance n an annular cylndrcal regn f length z, nner radus r and uter radus r. We shuld render ths equatn dmensnless by defnng: = z L ; = r R ; = t t ;c = C A C A where L s the ppe length f nterest, R s the ppe radus, t = L/u s the mean resdence tme, C A s sme reference cncentratn. Let us assume an n-th rder rate frm. The abve equatn () nw reads: D c ( ) c u L + D L c kc n u R R A0 t c n = c (a) We defne: Pe a = u L D = L / D L / u = ( ) ( ) characterstc dffusn tme axal prcess tme characterstc cnvectn tme = axal Peclet number. (3)

3 Pe r = u R D = radal Peclet number (4) u R DL = R / D characterstc radal dffusn tme = = L / u characterstc cnvectn tme Pe r x R L = ( radal Peclet number) x ( ppe aspect rat) n Da n = kc A t = L/u kc A n = prcess tme characterstc reactn tme (5) In terms f the abve dmensnless grups whch nvlve: characterstc reactn tme, characterstc dffusn tme, characterstc cnvectn r prcess tme and the aspect rat, we can rewrte the abve equatns as: Pe a = u L D = u d t D L d t = Pe Pe r = u R D = u d t D R = Pe d t L d t (3a) (4a) Pe 64 7 a 48 d t Pe L Pe 64 7 a 48 d t Pe L c ( ) c + Pe c ( ) c L Pe r R L d t L Pe r R L Pe d t c Da n cn = c c Da n cn = c (b) where Pe = u d t D = Re Sc s the Peclet number fr the flw. In terms f Pe a and Pe r we have: Pe a c ( ) c + Pe r L R c Da n cn = c (c) The abve equatn can be smplfed f we deal wth a steady state reactr prblem fr whch c = 0, r f we deal wth a nnreactve tracer dynamc respnse, fr whch Da n = 0. In ether

4 event we need t slve a cumbersme partal dfferental equatn and need tw bundary cndtns n axal crdnate and tw n radal crdnate. 4. Segregated Flw Mdel The questn arses whether we really need t slve the abve equatn (c) numercally at all tmes r whether we can fnd smple slutns whch are vald under certan cndtns. Snce reactant r tracer dmensnless cncentratn, c, s a smth functn, based n physcal arguments, the value f the functn and ts dervatves s f smlar rder f magntude except perhaps at a fnte number f pnts. It can be shwn then that when Pe a >> and R Pe r >> the frst and thrd term f eq (c) can be neglected. Fr a steady state reactr L prblem ths results n the fllwng equatn: ( ) c Da n cn = 0 (6) = 0 ; c = (7) The ext cncentratn c ( =, ) s a functn f radal pstn,.e f the stream lne n whch the reactant travels. The verall average ext cncentratn, r mxng cup cncentratn, s btaned as c ex = C A ext C A = where u = u r R R ruc A R ru C A ( z = L,r) dr dr. Usng dmensnless varables we get: R 4 ru r R C A ( L,r) dr = R u C A ( ) c, c ex = 4 ( )d (8) Fr a frst rder reactn (n = ) we readly fnd frm eqs. (6) and (7) that c ( =, )= e Da ( ) (9) Usng eqn (9) n eqn (8) we btan the ext mxng cup cncentratn: 3

5 ( ) e Change varables t = u ; d = du t get Da c ex = 4 ( ) d (0) ( ) c ex = 4 where E n ( x) = ( ) 3 xu e u n e Da Da u u du = e Da du = E u 3 3 du s the n-th expnental ntegral. In cntrast, the crss-sectnal average cncentratn s: c ex = ( ) c, d = e Da u = e Da u du = E c Da ex = E = e Da Da Da ( ) d E Da (0a) () (a) Thus, f we measure by an nstrument the crss-sectnal average cncentratn, and try t nfer reactant cnversn frm t, ur results may be n errr snce cnversn s nly btanable frm mxng cup (flw averaged) cncentratn and clearly there s a dscrepancy between equatn (a) and eq (0a).. Yu shuld examne the devatn f eq (a) cmpared t eq (0a) and plt t as a functn f Da. The needed expnental ntegral are tabulated by Abramwtz and Stegun (Handbk f Mathematcal Functns, Dver Publ. 964). Usng the fllwng relatnshp amng expnental ntegrals [ ] () E n+ ()= z n ez ze n ( z) we get the fllwng expressn fr cnversn frm eqn (0a) x A = c ex = Da e Da + Da 4 E Da (0b) 4

6 xu e where E ( x) = du = e t u dt t x We shuld realze mmedately, upn reflectn, that by elmnatng the dffusn terms n eq (c) and n arrvng at eq (6-7) we deal wth the segregated flw system. Indeed, n absence f dffusnal effects there s n mxng amng varus stream lnes. The reactant that enters n a partcular stream lne exts n the same stream lne,.e at the same radal pstn, and hence s surrunded by elements f the same age as ts wn at all tmes durng ts jurney thrugh the reactr. Every stream lne has a dfferent resdence tme. The shrtest resdence tme s experenced by the elements n the center lne ( t / ), the mean resdence tme () t s experenced by the flud travelng n the stream lne that has the mean velcty,.e, at = / = 0.707, whle nfnte resdence tme s experenced by the elements n the stream lne at the wall ( = ). Hwever, snce the stream lne at the wall receves nfntesmal amunt f new flud the mean resdence tme fr the system exsts and s t. We recall that fr the segregated flw cndtn and frst rder reactn the ext cncentratn can be wrtten as: c ex = e Da E ( ) d = E ( s) s = Da (3) Ths means that we have fund the Laplace transfrm f the ext age densty functn fr fully develped lamnar flw f Newtnan flud n a ppe s L{ E ( )}= E ()= s E 3 = s e + s 4 E s s (4) Hwever, even the extensve transfrm pars f Bateman (Tables f Integral Transfrm Vl. ) d nt shw ths transfrm. We can, hwever, derve the RTD r the F functn fr lamnar flw readly based n physcal arguments. Let us magne that we have swtched frm whte t red flud at the nlet at t = 0. Red flud wll appear at the utlet at z = L nly startng at tme t /. The fractn f the utflw that s yunger than t s gven by the fractn f the flud whch s red. Ths s btaned by ntegratng the flw frm the center stream lne, where the red flud s present at the utlet frm tme t /, t the stream lne at pstn r at whch red flud just at the utlet at tme t and by dvdng ths flw rate by the ttal flw rate. 5

7 Recall that t = L u = L u r R By defntn the F curve s gven by: Ft ()= r r' udr' Q = r 4 r' u r' R R u dr' (5) Ft ()= 4 ' ( ' )d', fr t t (5a) where t = t ( ) t r = t /. Hence, Ft ()= = fr t t (6) t Ft ()= t t t = t t + t t = t 4t r 0 Ft ()= t 4t t < t t t (7) and Ft ()= t 4t H t t F( ) = H 4 (7a) (7b) The ext age densty functn s: 6

8 E(t) = df dt = t t 3 H t t (8) The dmensnless ext age densty functn s: E ( )= t E( t ) = 3 H (8a) Fr any reactn rder we can then wrte fr segregated flw lamnar flw c ex = c ( batch ) 3 H d = c batch( ) 3 d (9) Prevus studes have shwn that ths mdel s vald when Pe > 000 and Pe r that Pe a = Pe L dt,pe r = Pe. R > 85. Recall L 7

9 4.. Use f Segregated Flw Mdel n Lamnar Flw When (Anathakrshman et al. AIChE J.,, 063 (963),, 906 (966), 3, 939 (968) Pe r = u R D > 500 Pe r R L = u R D R L = R /D > 85 t cnvectn s the nly mprtant mde f transprt and the lamnar flw reactr behaves as n segregated flw snce dffusn effects are nt felt. Cnversn s then gven fr Newtnan flud n a cylndrcal ppe by Several pnts shuld be made: x A = C C batch A d = 3 c batch d (0) 3 ) Segregated flw s mst lkely t ccur n plymerc systems due t lw dffusvtes encuntered n such systems. Because such systems ften behave as nn-newtnan, new E( ) curves based n velcty prfles fr nn-newtnan fluds shuld be derved. Such expressns can be btaned fr pwer law fluds, Bngham fluds, etc., and actual devatns are left fr the exercses. / ) Snce the cndtns fr segregated flw t hld requre L dt < Pe r 70 ; Pe r > 500 r L dt < Pe ;Pe> and the cndtns fr lamnar flw are Re = u d t v <,00. Recall that Pe r = ( Sc) ( Re) and that we must ensure that the flw s truly fully develped befre t enters the reactr sectn. The entrance length, L e, s f the rder L e = d t Re ( L e d t Re als s used). Fr example f R e = 00 and S c = 000, Pe r = 50,000. Then Pe = 00,000 and L d t < 94 whle L e = 3.5 d t. If actual reactr length L = 50 d t the cndtns fr segregated flw are satsfed and the entry length represents nly.4% f the ttal length and can be neglected. Hwever, f Re =,000 and Sc =,000, Pe r = 500,000 and L/d t < 940 whle L e 35d t f 8

10 L = 50 d t segregated flw cndtns are satsfed but nw the entry length s 4% f the ttal length and mght nt be neglgble any mre n ts effect. Then the entry length prblem,.e the regn f develpng lamnar prfle needs full attentn. Hwever, the mdel s nw vald fr L = 500 d t and entry length effects are nw neglgble. ) Often even n lamnar flw t s mprtant t create a narrw ext age densty functn r a steeper RTD (F curve) s that devatns frm plug flw are mnmzed and plug flw perfrmance apprached. Narrwer RTD's are useful n certan type f cnsecutve reactns where ntermedates are the desred prduct, and when t s necessary t prevent undesred reactns f large reactn tmes t ccur n a flud wth resdence tmes much larger than the mean. Narrwer E curves r, equvalently, sharper F r W curves can be btaned by usng - parallel plates cnfguratns - annular flw - helcal cled ppes - statc mxers Fr example, the fully develped velcty prfle fr annular flw s: u( ) = u ( )l n ( 4 )ln + ( ) ln l n () Then = r R ; = R n R ut Ft ()= W()= t u d ; fr t t u mn Ft ()= ( )l n ( 4 )ln + ( ) 4 4 ln ln 4 () where, are gven by the slutn f the fllwng transcendental equatn 9

11 t ( 4 )ln + ( ) t = ( )ln l n l n (3) Thus, F (t ) must be evaluated numercally. The maxmum velcty ccurs at max = ln / ( ) / (4) and the earlest appearance tme s at t ( 4 )l n + ( ) t mn = ( )ln max l n l n max (5) The sketches f the washut curve fr a number f cases are gven n the attached fgure taken frm Nauman. Clearly, as s ncreased ne departs mre and mre frm the crcular tube W (t ) curve and appraches that fr flw amng parallel plates whch s clser t plug flw. Remarkably, addng even a thn wre n the center f the ppe such as R wre /R ppe = = makes the RTD much clser t that f plug flw! 000 Fr a helcal cl, the slutn fr F (t ) s lengthy, cmplex and numercal. Hwever, a gd apprxmatn s btaned by W()= t 0.00 t/t ( ).84 + t < 0.63 t ( t/t ) ; t > 0.63 t (6) A sngle screw extruder als gves a rather narrw E curve wth t mn = 0.75 t. Fr detals and references cnsult Nauman and Buffham (Mxng n Cntnuus Flw Systems, Wley 983). Statc mxers als create a narrwer E curve and apprxmate analyss has been presented by Nauman. v) The fnal pnt t be made s that n lamnar segregated flw n rder t prperly nterpret tracer nfrmatn the tracer must be njected at the njectn plane prprtnally t flw 0

12 at each pnt, and at the ext the mxng cup cncentratn must be measured (mean flw cncentratn). g Mathematcally, f ( r,t) cm s s the lcal flux densty f the tracer at pstn r f the njectn plane, and v (r ) s the nrmal velcty at pnt r f the njectn plane, prper flw taggng requres Fr a step nput ths means Mxng cup, r mean flw cncentratn s: r,t ( )= ct ()vr ( )fr all r. (7) s ( r,t) = C vr ( ) (7a) c = vr ( )cr,t ( )da / v ()da r (8) A where c (r,t ) s the cncentratn at the ext plane, v (r ) s the velcty nrmal t the ext plane at pnt r and A s the crss-sectnal area f the ext plane. If hwever ne ether uses crss sectnal area taggng whch fr a step nput s A r,t ( ) = () t (9) s ( r,t) = (9a) r mntrs crss-sectnal average cncentratn c = A cr,t ( )da A (30) errneus results n terms f nterpretng a step tracer respnse as an F curve are btaned as dscussed belw.

13 4.. Lmtatns n the Tracer Methd n Lamnar Flw Let us defne the fllwng quanttes: v( r,t) = velcty f the mvng flud perpendcular t the crss-sectnal area S,v( r,t) = v r s r Indcatr flux densty (.e flux) per unt area ( r,t) ( r,t )= c( r,t)v( r,t) (3) Amunt f ndcatr cllected at the utflw between t and t + T, I I = t + T t t + T ( r,t )dadt = c( r,t) v ( r,t )dadt (3) S t S The flw f ndcatr acrss S at tme t s Mean flw cncentratn c (): t c ( r,t) v( r,t)da = c ()Qt t () S c ()= t S c ( r,t)v( r,t)da S v ( r,t)da = S ( r,t)da Qt () (33) Nw I = t + T where Q()= t v ( r,t )da s the vlumetrc flw rate. S c ()Qt t ()dt (34) t

14 In the tme nterval (t, t + T ) the mean flw cncentratn s: c = t + T t c ()Qt t ()dt t + T t Qt ()dt (35) Mean flw s Q = T t + T Qt ()dt (36) t In steady state flw Q = Q = cnst. Mean Crss-sectnal Cncentratn c (): t c ()= t S c ( r,t)da S da (37) Tw Ways f Injectng Tracer nt Steady Flw: Flw taggng -durng a tme nterval the ndcatr s njected at (r flws thrugh) the crss sectn (z = 0) n such a way that fr any tme t n ths nterval ( r,t)= µ ()v t ( r ) (38) Fr all r n the crss sectn c ( r,t)= µ () t (38a) The abve njectn s prprtnal t flw, hence, the name flw taggng. Crss-sectnal taggng - the ndcatr s njected at a rate () t unfrm n the crss-sectn y = 0.e f fr very tme t n the nterval n questn ( r,t)= (). t (39) 3

15 Then fr any r at z = 0, c ( r,t)= () t v ( r ) (39a) In fully develped Newtnan lamnar flw the velcty prfle n a cylndrcal tube s vr ()= Q ( R r ) = u R r ( ) =u ( r R 4 R R ) = u r R 0 r R wth u = Q R Mean Flw Cncentratn at z = L c ()= t Q R rc ( r,t )vr ()dr (40) Mean Crss-sectnal Cncentratn at z = L R c ()= t R rc( r,t)dr (4) Fr a partcle wth radal crdnate r that at tme t ' was at z = 0 and at tme t at z = L the fllwng relatn hlds: ( t t' )vr ()= L = ( t t' ) Q ( R 4 R r Q )= t a R 4 t t' t a = L u L = vr ()t ( t' ) L = t a u vr () = t a u t t' = r R r = R t a t t' / t' t t a (4) Fr a fxed t ths defnes r as a functn f t'. Hence fr t' = 0 r = R t / a t (4a) 4

16 dr = R t a t / t a t (4b) Cnsder nw varus cmbnatns f taggng at the njectn plane and cncentratn mntrng at the ext plane. a) Flw taggng step nput, mean flw cncentratn at samplng ste, c( z= 0,t) = µ ()= t µh() t c ()= t Q R t a t / 0 0 < t < t a µ rv()dr r = µ t a t fr t t a (43) Et ()= d dt c () t µ = t a t 3 t t a (44) t a Et ()dt = t a t 3 t a t = t a t a = (45) t = t a te() t dt =t a t dt = t a t ta t a = t a (46) Indeed an E curve s btaned snce bth the mass balance,.e zerth mment, and the central vlume prncple,.e frst mment, are satsfed. b) Flw taggng step nput, mean crss-sectnal cncentratn at z = L, c( y= 0,t )= µ Ht () R t / a t c ()= t r µ dr R ()= µ t a t c t t t a (47) ( ) E ()= t d c µ dt = t a t fr t t a (48) 5

17 t a E ()dt t = t a t t a dt = t a t ta = (49) t = t a t E ()dt t = t a t dt = t a lnt ta t a = (50) The btaned mpulse respnse, E, clearly s nt an E curve snce the mean des nt exst. c) Crss-sectnal taggng step nput, mean flw cncentratn, ( z= 0,t)= H() t cr,t ( ) z = 0 = () r t > 0 (5) c ()= t Q R t / a t rc ( r,t )( r)dr = R Q t a t t t a (5) Nw F ()= t Qc () t R ; E ()= t df dt E () t s the same as E () t n b). It cannt be an E-curve snce t. d) Crss-sectnal taggng step nput, mean crss-sectnal cncentratn, ( z= 0,t)= H() t c ()= t R R t / a t r dr = () r R R t / a t r R dr QR r ( ) ( ) R c ()= t Q l nr r R t a t / = R Q ln t t a fr t t a (53) E ()= t t and Et ()dt = t dt = l nt t a t a t a = (54) Ths certanly cannt be an E-curve snce the area under the curve s nt fnte! 6

18 4. Taylr Dffusn and the Axal Dspersn Mdel When P e r > 500 but L/d t > Pe r /70 the cndtns fr the segregated flw mdel are vlated. Nw Pe a = Pe r L dt > Pe r >> s that axal dffusn can be neglected cmpared t radal 85 dffusn and cnvectn terms. Snce L s large, radal dffusn has suffcent tme t be felt d t and cannt be neglected snce R /D s nw less than 85,.e characterstc radal dffusn tme, t R /D, becmes mre cmparable t the characterstc cnvectn tme, t. Fr a reactr at steady state the fllwng prblem wuld have t be slved: Pe r L R c ( ) c Da n cn = 0 (55) whle the nert tracer respnse s descrbed by: Pe r L R c ( ) c = c (56) Bth eq (55) and (56) are subject t the apprprate bundary cndtns (B.C.). Agan a cmplex PDE needs t be slved and t wuld be helpful t fnd an apprxmate slutn. The dea f utlzng the B.C.'s n.e. = 0, c c = 0 and =, = 0 by defnng a crsssectnal mean cncentratn cd s nt mmedately frutful because f the ( ) term multplyng c. Sme tme ag G.I. Taylr made an experment by njectng a dye nt lamnar flw. He bserved that the slug f dye traveled tgether and spread ut as t mved dwnstream rather unfrmly acrss the tube dameter. It dd nt frm a parabld f dye as expected. Whle stream lnes clse t the center tend t mve the dye faster than thse clse t the walls, a cncentratn gradent develps n the lateral drectn, and the dye s transprted by dffusn frm the centrally lcated stream lnes t thers at the leadng edge f the frnt and frm the stream lnes clse t the wall t centrally lcated nes at the tralng edge. G.I. Taylr (Prc. Ryal Scety, Lndn, A 9, 86 (953); A 3, 446 (954), A 4, 473 (954)) descrbed ths behavr mathematcally by fully utlzng the expermental bservatns. 7

19 He ntced that the centrd f the dye slug mves at the mean velcty f flw. Hence, a transfrmatn f crdnates t a mvng crdnate system at mean flw velcty s useful. Ths requres: Thus ' =, = = ' = + ' + whch transfrms eq (56) fr the tracer respnse t: = ' ' = ' Pe r L c R ( ) c = c ' (57) Furthermre, expermental bservatns ndcated that the cncentratn at a pnt mvng at the mean flw velcty vares extremely slwly n tme, hence c 0. Fnally, G.I. Taylr ' assumed that the axal cncentratn gradent s ndependent f radal pstn, agan as supprted by expermental bservatns. Wth these assumptns eq (57) becmes: c = Pe R r ( 3) c L ( 58) Integrate frm = ; c = 0 t c = Pe R r L 3 c (59) Nw ntegrate between = 0, c = 0 and agan 8

20 R c c( = 0)= Pe r L c (60) The cncentratn at the center lne s unknwn, c = c ( = 0) and shuld be elmnated n terms f the mean crss sectnal area cncentratn c = c d = c d + Pe r R L 3 5 c d c = c + Pe r R L c = c + Pe r R L c (6) Elmnatng c n eq (60) usng eq (6) gves: c = c + Pe r R 4 L 4 c 3 (6) The dmensnless flux f tracer that crsses the plane that mves at the mean velcty f flw s: j td = c ( )d = 3 Pe r R L ( ) c d + ( 3 ) 4 c d 3 (63) j td = c ( 3 )d + Pe r R c L d = c 4 + Pe r R c L = 0 - Pe r R c L 4 = Pe r 48 R L c (64) Based n the prevus assumptns c = c snce c s ndependent f. The dmensnless flux densty f tracer acrss the plane mvng at the mean velcty f flw s: 9

21 j td = Pe r 48 R L c Ths yelds the fllwng expressn fr the dmensnal tracer flux densty, J t J t j td = u C = Pe r 48 R L c J t = u u R R 48 D L C t z = u R C t 48 D z (65) Equatn (65) fr the relatve tracer flux densty n the axal drectn wth respect t the plane mvng at the mean flw velcty has the frm f Fck's law: J t = D app C t z (66) The apparent dffusn ceffcent, called the axal dspersn ceffcent, D app, by cmparsn f eq (65) and eq (66) has the frm D app = u R 48 D (67) Ths s the famus frmula fr Taylr dffusvty fr lamnar flw f Newtnan flud n a crcular ppe. Snce the frmula depends n the velcty prfle, t s clear that dfferent D app s btaned fr dfferent gemetres (parallel planes, rectangular crss-sectn, annular flw) r fr Nn- Newtnan fluds. A reader wh s nt famlar wth the abve representatn f the dmensnless tracer flux densty wth respect t the mvng crdnate system shuld rederve eq (65) startng frm the begnnng. Ttal tracer that passes a plane n the stagnant crdnate system per unt tme s m t = R rxu r R Cr,L ( )dr where C = C c (68) The flux densty f tracer wth respect t stagnant crdnates s 0

22 N t = m t R = 4 u C Upn substtutn f eq (6) nt eq (69) and ntegratn ne gets: ( )c (,)d (69) N t = u C u Pe r 48 = u C u R 48D R C L C z = u C D app C z (70) The frst term s a cnvectve term and the secnd s the already establshed dspersn term. Thus, the flux densty wth respect t the stagnant crdnate system equals the flux wth respect t the mvng crdnate system plus the cnvectve flux. Later, Ars (Prc. Ry. Sc., A 35, 67 (956)) shwed that the apparent dffusvty r effectve dspersn ceffcent shuld take the fllwng frm: D app = D + u R 48D (7) Clearly the mlecular dffusn term s neglgble when the secnd term s much larger. The axal dspersn ceffcent, D app, cmbnes the effect f mlecular dffusn and f the velcty prfle. The net result s that the effects f the velcty prfle and f radal dffusn can be expressed by an equvalent axal dspersn term. Eq. (56) can nw be rewrtten as: Pe app c c Da n cn = c (7) whch represents the axal dspersn mdel wth Pe app = u L D app ; = D app Pe app u L (73) The apprprate bundary cndtns fr the mdel requre flux cntnuty at the entrance and at the ext. In addtn, cncentratn must be cntnuus at the ext. Snce nlet lnes are nrmally f much lesser dameter than the reactr, r cntan packng n rder t ntermx the feed, the D app fr these lnes s usually very small and the dspersn flux can be neglected cmpared t the cnvectve flux. The nlet bundary cndtn then s:

23 = 0 ; c c Pe app = c ( ) (74) At the ext The ntal cndtn s: = ; c = 0 (75) = 0 ; c= c ( ) (76) where c ( )= C nlet ( ), c C ( ) = C ntal ( ) C, Pe app = u L D app (77) Please nte that the new Peclet number, r the axal dspersn Peclet number, s defned nw n terms f the apparent r effectve dspersn ceffcent. D app = D + u R 48D (7) In crcular ppes fr Newtnan fluds n lamnar flw ths can be expressed as: 4.. Regn f Valdty 9 Re Sc Pe app = 9 + Re Sc L (78) The abve Taylr dffusn mdel wth the axal dspersn ceffcent gven by eq (67) s vald when ) The characterstc radal dffusn tme < characterstc cnvectn tme ) mlecular axal dffusvty << axal dspersn ceffcent Ths mples Re <,00 (t guarantee lamnar flw), d t L d t > 0.08 Pe r r L d t > 0.08 u R D (79)

24 and Pe = Pe r > 6.9 (80) accrdng t G.I. Taylr. Cmparsn wth numercal slutns and experments ndcates that the range f applcablty s mre lke Re <,000 ; L d t > Pe r > 50 (8) Agan fr a lamnar flw reactr ne shuld check whether the entry length L e s small cmpared t reactr L n rder fr the abve mdel t be applcable. Addtnal checks f the Taylr-Ars dffusn mdel aganst numercal slutns have been made. Wen and Fan summarze the fndngs n the enclsed graph whch shws the applcablty f varus lmtng cases. Presumably n the regn labeled dspersn mdel the Taylr-Ars expressn s nt vald but ther frms fr D app have t be ftted t emprcal data. Other than the regmes already mentned, there s a case f neglgble cnvectn and strct ne dmensnal dffusn whch s nt f great practcal sgnfcance. 3

25 4.. Addenda It s f nterest t pnt ut the fllwng facts.. G.I. Taylr based n hs expermental evdence reasned that "the tme necessary fr apprecable effects t appear wng t cnvectve transprt s lng cmpared wth the tme f decay durng whch radal varatns f cncentratn are reduced t a fractn f ther ntal value thrugh the actn f mlecular dffusn". The characterstc tme fr radal dffusn s btaned by slvng The slutn s: c = Pe r L R = 0, c = 0 =, c = 0 c c = A n e n J n n= L RPe r / (8) (8a) (8b) (83) where n are the egenvalues that satsfy the fllwng equatn: J n L RPe r / = 0 (84) If ne represents the abve seres slutn fr cncentratn by ts leadng term, n antcpatn f gd cnvergence, and assumes that the dye was ntally present nly at the center lne, then 4

26 The frst rt f J s 3.83 s that c = e J L RPe r L RPe r / / (85) = 3.83 (86) L = 3.83 RPe r / 4 L = 3.83 d t Pe / (87) By cnventn, the characterstc dffusn tme s taken as the ne when the cncentratn f unty at the center lne has drpped t e - f ts rgnal value.e. D = D = = R ( 3.83) L Pe r = R L Pe r = t D t = d t Pe 4 L (88) Therefre, the characterstc tme fr cnvectve change must be lng cmpared t the characterstc radal dffusn tme. t c = L = L u max u = t (89) c = (89a) Thus > R L Pe r (90) L d t > Pe r ; L d t > Pe (9) 4.7 L d t > Pe r ; 9.4 L d t > Pe (9) Practce shws that the requrement L d t > Pe r s suffcent ; ften 8 L d t > Pe r s als suffcent. 5

27 The ther cndtn arses frm the requrement that the mlecular dffusn be small cmpared t the Taylr dffusn effect u R 48 D > D u R D > 48 Pe r > 48 r Pe r > 48 = 6.9 Pe > 3.8 Practce and cmparsn wth numercal slutns shw that Pe r > 0 r preferably 50 are requred fr the perfect match between apprxmate Taylr slutn and data r the numercal slutn f the exact equatn. It s mprtant, hwever, t understand hw the abve crtera fr valdty f the Taylr slutn were establshed. The ther mprtant thng t realze s that Taylr apprach prvdes the expressn fr the behavr f the crss-sectnal average cncentratn c ' = Pe c (93) n terms f the mvng crdnate system, and nt f the mxng cup cncentratn. Thus, nterpretng the results n terms f the mxng cup cncentratn mght be n errr. Summary: Lamnar Flw n Tubular Reactrs Cnvectve mdel (Segregated Flw) s vald fr u d t D >000 and L < u d t d t 340 D Taylr Dffusvty Mdel fr Axal Dspersn s vald when Pe r > 48 = 6.9 Pe r = u R D L d t > Pe r 6

28 r u d t D > 3.8 and L > u d t d t D Need at least L/d t > 0 fr axal dspersn mdel. Pure dffusn u L D < r L d t < u d t D 7

29 THE AXIAL DISPERSION MODEL (CHE 5) M.P. Dudukvc Chemcal Reactn Engneerng Labratry (CREL), Washngtn Unversty, St. Lus, MO

30 5. THE AXIAL DISPERSION MODEL The axal dspersn mdel can readly be extended t apply t turbulent flw cndtns. Under such cndtns the velcty prfle s almst flat and averagng wth respect t radal pstn s pssble. The axal dspersn mdel fr turbulent flw has the frm as gven befre wth c Pe z ξ c ξ Da nc n = c θ (94) ξ = 0 ; c c Pe ξ = c ( θ ) (95) ξ = ; c ξ = 0 (96) θ = 0 ; c= c ξ ( ) (97) Pe z = u L E z (98) where E z s the axal dspersn ceffcent. Taylr suggested that the axal dspersn ceffcent n fully turbulent flw shuld be prprtnal t the frctn velcty u * E z d t u * (99) where the frctn factr f s gven by u * = f u (00) f = Re 0.5 (Blasus equatn) (0) Taylr suggested specfcally: E z = 3.57 u d t f (0) An emprcal crrelatn fr the axal dspersn ceffcent, E z, n ppes, whch ncludes the transtn and fully develped turbulent flw, s: wth E z = 3 x /8 (03) u d t Re Re

31 Re = u d t υ The axal dspersn ceffcent n ppes can be als btaned frm the enclsed graph (Fgure 6). The effect f beads, cnstrctns, etc., has been descrbed by Wen and Fan. The reactr steady state prblem nw can be descrbed as fllws: d c Pe dξ dc dξ Da nc n = 0 (04) ξ = 0, c Pe dc dξ = (05) ξ =, dc dξ = 0 (06) Cautn shuld be taken n that the Peclet number s nw prperly nterpreted as Pe = u L D app fr la mn ar flw r Pe = u L E z fr turbulent flw whle the crrelatns r graphs prduce u d t E z s that Pe = u d t E z L x d t Analytcal slutn f eqs (04-06) has been fund nly fr zerth and frst rder reactn. The slutn fr a frst rder reactn s: x A = c()= Da Pe 4 + 4Da Pe exp Pe + 4Da Pe + 4 Da exp Pe + 4 Da Pe Pe (07)

32 The lmt as Pe becmes the slutn fr the PFR lm c( )= c( ) Pe PFR = e Da (07a) The ther lmt f Pe 0 s nly f academc nterest, and, ndeed, t prperly cnverges t the CSTR behavr lm c()= c CSTR = (07b) Pe 0 + Da It shuld be remembered at all tmes that, due t the assumptns that led t the axal dspersn mdel, ths mdel nly makes physcal sense at large Peclet numbers, r small dspersn numbers N D < 0. (Pe > 0). At lw Pe numbers representatn f realty by the axal dspersn mdel s f dubtful accuracy and s ll funded. The reader shuld derve the slutn t the axal dspersn mdel n case f a zerth rder reactn. Fr ther reactn rders and fr cmplex reactn schemes eqs. (04) - (06) must be slved numercally. Several appraches can be chsen: - Shtng methd. Create tw st rder equatns equvalent t the rgnal secnd rder equatn (04) by chsng y = c, y = dc and ntegrate them backwards frm dξ ξ =, y = 0, y = y where y s the guessed value fr c( ξ = ). See f cndtn (05) at ξ = 0 s met y Pe y = and f t s nt, devse an algrthm fr crrectng y at ξ = untl the cndtn s met. - Quaslnearzatn. Lnearze the equatn arund an assumed slutn and btan the answer by teratvely slvng the lnear equatns. - Green's functn. Use the Green's functn t cnvert the prblem t an ntegral equatn whch s slved teratvely. - Fnte dfferences. Slve by standard fnte dfference schemes. Detals are left fr the appled mathematcs curse. 3

33 Apprxmate slutns that rely n the perturbatn thery are als qute useful n assessng dspersn effects and are descrbed n the Appendx. Snce the dspersn mdel wth B.C. (05) and (06) s a clsed system, then eq (07) represents the Laplace transfrm f E θ θ ( ) when s s substtuted fr Da. L{ E θ () θ }= c (, s)= + + 4s Pe 4 + 4s Pe exp Pe + 4s Pe + 4s exp Pe + 4s Pe Pe (08) where E θ ( θ ) s the slutn f the fllwng transent prblem: c Pe ξ c ξ = c θ (09) ξ = 0, c Pe c ξ = δ( θ) (0) ξ =, c ξ = 0 () θ = 0, c= 0 () Then c ( ξ =,θ )= E θ ( θ) (3) Inversn f eq (08) gves: E θ (θ ) = e Pe ω n snω n Pe [ + 4ω n ]exp Pe + 4ω n θ 4 Pe Pe Pe (4) + 4 Pe + 4 ω n n= [ ] where ω n are the pstve rts f: tan ω n = 4ω n Pe 4ω n Pe (5) Unfrtunately, expressn (4) s nt cnvenent fr calculatns. It cnverges very slwly at small θ and alternatve functnal frms are needed fr evaluatn f E θ (θ ) at small θ. It als requres extra cautn n calculatns fr Pe > 6 t prevent verflw f expnental terms. 4

34 An apprxmate expressn can be derved by replacng B.C. (0) wth ξ = 0, c= δ( θ ) (0a) It turns ut that, althugh ths makes the system pen, the dfferences n the respnse are small, and the result s: Pe E θ ( θ) 4 πθ exp Pe( θ) 4θ (6) Ths result s a gd apprxmatn at Pe > 6. Snce at Pe > 6 the E θ ( θ ) becmes qute narrw, the detals f mcrmxng shuld nt affect reactr perfrmance by much. Therefre, f ne wants t avd slvng a nnlnear bundary value prblem gven by eqs (04-06) fr reactn rders nt equal t ne, ne can btan an apprxmate slutn by usng the segregated flw cncept C ext = C b ( θ )E θ ()dθ θ (7) where eq (6) s used fr the E θ ( θ ). Please nta bene, that eq (7) des nt represent the physcal realty f the axal dspersn mdel but s based n the fact that fr narrw RTD's mcrmxng effects cannt be very large except at very hgh cnversns clsng n n. One shuld establsh, as an exercse, that the mments f E θ ( θ ) gven by eq (4) can readly be btaned frm ts Laplace transfrm,.e eq (08) and are: µ 0 = µ = (recall ths s θ scale θ =t/t ) σ D = Pe Pe e Pe ( ) (8) 5

35 Fr reasnable values f Pe (Pe > 0) the secnd term n the expressn fr the dmensnless varance s neglgble σ D (8a) Pe Thus, the dmensnless varance f the mpulse tracer respnse can be nterpreted n terms f the Peclet number f the axal dspersn mdel. Fr ther detals f tracer studes and ther nterpretatn see Levenspel r Wen and Fan. Recall nw the Tanks n Seres mdel fr whch the E-curve and varance are: E θ ( θ)= NN θ N ( N )! e Nθ σ D = N (9) By usng the equalty f varance fr the axal dspersn and the tanks n seres mdel the tw mdels can be related: σ = Pe = N N = Pe app (0) Ths allws us t extend the frm f the E-curve fr N-CSTRs t a case when N s a nn nteger.e fr an arbtrary value f the varance. where σ s the dmensnless varance. E ax.dsp θ ()= Pe Pe Pe θ Γ Pe = σ Γ e Pe θ σ σ θ e θ / σ σ () 6

36 5. Addenda: (Inversn f the Transfer Functn fr the Dspersn Mdel) It mght be nstructve t shw hw t btan the slutn t eq (09) and the effect f B.C. n that slutn. By takng the Laplace transfrm f eq (09) we get where We can rewrte ths as: Pe d c d ξ dc dξ sc = 0 c = L{}= c e sθ c()dθ θ d c dξ Pe d c dξ spe c = 0 (A) The characterstc equatn f the abve rdnary dfferental equatn s: m Pem spe= 0 and the rts are m, = Pe ± Pe + 4sPe = Pe 4s ± + Pe The slutn n the Laplace dman s: c = Ae Pe + Pe + 4 s ξ Pe + Be Pe Pe + 4 s ξ Pe (A) The cnstants A and B have t be fund by satsfyng the bundary cndtns. In case f cndtns (0) and () requred fr the clsed system ths means A + B A Pe Pe + Pe + 4 s B Pe Pe Pe Pe + 4s = Pe A Pe Pe + Pe + 4s e Pe Pe s Pe + B Pe Pe Pe + 4s e Pe Pe + 4 s Pe = 0 Slve fr A and B, substtute nt (A) and shw that when yu evaluate c at ξ = eq (08) s btaned.

37 E θ 4 + 4s Pe exp Pe + 4s Pe ()= s LE { θ () θ }= c (, s)= + + 4s + 4s exp Pe + 4s Pe Pe Pe (08) In cntrast f eq (0) s replaced by the cndtn ξ = 0,c= δ ( θ) the equatns t be slved fr A and B are A s e Pe Pe + + 4s Pe A + B = + B + 4s e Pe Pe + 4 s Pe = 0 Nw c ( ξ =, s) s c (,s)= + + 4s e Pe Pe + 4s Pe Pe e + 4s Pe + 4s e Pe Pe + 4 s Pe r Pe + 4 s + 4s Pe e Pe c (, s)= s Pe + 4s (A3) e Pe + 4 s / Pe Pe Hw d we nvert frms lke eq (08) r eq (A3) whch are unlkely t be fund n the pars f transfrms avalable n tables? Based n the physcal nature f the prblems we knw that there shuld nt be any branch cuts n the cmplex plane and that the resdue therem can be used. Then f

38 s the Laplace transfrm f f (t ).e f ()= s e st and s n are the ples f f () s.e the rts f Q (s ) ft ()dt Qs ( n ) = 0, n =,... then the nverse Laplace transfrm s btaned by: Let us apply ths t eq (08) f()= t Ps n Q' s n n= Ps ()= + 4 s Pe exp ( ) ( ) e s nt Pe + 4s Pe 3

39 Qs ()= s Pe + 4 s Pe e Pe + 4 s Pe = 0 Let + 4s Pe = z where z s a cmplex number z = x + y Then frm the equatn that dentfes the ples.e., Q( s) = 0 t fllws r + z z + z z = e = e Pe z Pe z y + x x e + x x Pe x x If z = x s real the abve equatn has nly ne rt at x = 0 as shwn n the sketch abve. (Fgure A). If z s cmplex, z = x + y, then + x + y x y = e Pe x e Pe y x y + y x ( ) + y = e Pe x e Pe y Ths requres ( x y )+ 4 y ( x) + y e y arctan x y = e Pe x e Pe y ( x y ) + 4 y ( x) + y = e Pe x 4

40 y arc tan x y = Pe y Ths s satsfed when x = 0 and y Pe = tan y y Let Pe y = ω y = ω. The abve equatn becmes: Pe Thus tan ω n = 4 Pe ω 4 Pe ω = tanω 4 4Peω n 4ω n Pe = whch s gven as egenvalue equatn (5) n the ntes. ω n Pe 4 ω n Pe (5) Thus at the rts f Q (s ) = s n Pe = Pe ω n and s the rts s n are Nw s n = Pe Pe ω n Q' ()= s s + Pe + 4s e Pe + 4 s Pe + Pe + 4s Pe Pe e Pe + 4 s Pe x 4 Pe + 4 s Pe 5

41 Q' ()= s Pe + 4s Pe + + 4s + e Pe Pe + 4s Pe + 4s + Pe Pe + 4 s Pe Q' ( s n )= Pe + Pe ω Pe ω n + e Pe n Pe ω n Pe ω n [ + Pe ω n] Q' ( s n )= + 4ω n ω n Pe + e ω n ω n Pe ( + Pe ω n) Ps ( n )= 4 Pe Pe ω n e e ω n ( ) ( ) = 8ω n Ps n Q' s n Pe + 4 ω n Pe e Pe e ω n + e ω n + Pe 4ω n Pe 4ω n + 4ω n Pe ( ) ( ) = 8ω n Ps n Q' s n Thus ne culd wrte the nversn frmula as: e Pe/ Pe 4 + Pe 4ω n Pe csω n 4ω + Pe snω n 8ω c(,θ)= n e Pe / e Pe Pe ω n θ Pe 4ω n + Pe snω n 4 + Pe 4ω n= n csω n Pe 8ω n exp Pe + 4ω n Pe θ = 4ω n ( Pe + )snω n Pe (A4) n= ( + 4 Pe 4ω n )csω n We can get ths nt the frm f eq (4) by usng varus algebrac manpulatns and trgnmetrc denttes as shwn belw and by nvkng eq (5). 4 ω n ( Pe + )snω n Pe ( + 4 Pe 4 ω n )csω n = csω n 4ω n ( Pe + )tan ω n Pe 4 Pe + 4ω n [ ] 6

42 [ ] cs ω = n 4 ω n Pe 6 Peω ( n Pe + ) ( 4ω n Pe )Pe ( + 4 Pe 4ω n ) snω = n snω n [ 4ω n Pe ] 6 Peω ( n Pe + ) ( 4ω Pe ) Pe [ ( + 4 Pe 4ω n n )] = tan ω n 6 Peω n ( Pe + ) ( 4ω n Pe )Pe [ ( + 4Pe 4ω n )] snω n [ + tan ω n ][ 4ω n Pe ] 4 Peω n 6 Peω n ( Pe + ) ( 4ω n Pe )Pe [ ( + 4Pe 4 ω n )] = snω n [ 4ω n Pe ] 6 Pe ω + n 4ω n Pe = 4 Peω n 6Peω n ( Pe + ) 4ω n ( Pe + 4 Pe)+6ω 4 n + Pe ( Pe + 4 Pe) 4ω n Pe [ ] snω n [ 4ω n + Pe ] [ ( )] sn ω n [ 4 ω n + Pe ] [ ( )+ Pe ( Pe + 4Pe) ] snω n [ 4ω n + Pe ] [ ( )+ Pe ( Pe + 4Pe) ] [ 4ω n + Pe ] snω n ( )+4ω n Pe [ ( + 4ω n )] 4Peω n Pe ( + 4Pe + 4ω n ) [ 4ω n + Pe ] = snω 4ω n ( n + Pe )sn ω = 4 Peω n 8 Pe ω n +6 Peω n + 6ω n 4 + Pe Pe + 4 Pe = 4 Peω n 4ω n Pe + 4 Pe + 4ω n = 4 Peω n 4ω n (Pe + 4 Pe) + 4ω n Pe + 4ω n = 4 Peω n (Pe + 4 Pe) 4 ω n + Pe Fnally, we have shwn that 4 ω n ( Pe + )snω n Pe ( + 4 Pe 4ω n )csω n ( ) = 4 Peω n Pe + 4 Pe + 4ω n ( 4ω n + Pe )snω n Then, substtutng the abve n eq (44) we get 7

43 c(,θ)= Pe e Pe ω n snω n ( 4ω n + Pe ω n Pe )e Pe Pe (A4a) + 4 Pe + 4ω n n= whch s equatn (4) n the ntes. [ ] Fr an exercse ne shuld be able t develp the respnse t a unt mpulse njectn fr a mdel whse transfer functn s gven by eq (A3). Ths mdel wuld be vald fr a farly lng reactr where the exactness f the nlet cndtn s nt that mprtant. θ 8

44 5. Mxng n a Ppelne When nstead f a reactr prblem we deal wth prblems f mxng f ne materal that flws after anther as bth are beng pumped thrugh the same ppe lne, we use the dspersn r Taylr dffusn mdel (dependng whether the flw s turbulent r lamnar) t descrbe the spreadng f the materal n the axal drectn. The gvernng equatns n the crdnate system that mves at the mean velcty f flw can be wrtten accrdng t the dffusn equatn: c θ ' = Pe c ζ () where Pe = u L D app and D app s ether the dspersn ceffcent, E z, r Taylr dffusvty. If we try t descrbe an mpulse respnse n a very lng ppe, then all the materal was cncentrated rgnally at the plane ζ = 0. At the same tme θ ' = 0; c = 0 except at ζ = 0 (3a) θ ' > 0; ζ c 0 (3b) n materal can reach axal pstn f nfnty at a fnte tme whch s expressed by the secnd cndtn abve. Fnally, snce the mass m t f the tracer materal njected at tme zer at the plane at zer axal pstn must be cnserved we have the last cndtn: m t c dζ = (4) AC L where A = π R s the crss-sectnal area f the system, L s the length wth respect t whch we have dmensnalzed the axal crdnate, C s a nrmalzng cncentratn. The slutn t the abve prblem can be btaned by ether a) takng the Laplace transfrm f the PDE, and B.C., slvng the resultng ODE and nvertng

45 the slutn, r b) by smlarty transfrm.e. by ntrducng new varables η = ζ θ' and u = c θ' The slutn s c ( ζ,θ')= m t π R LC πθ' Pe e ζ 4 θ ' Pe (5) If we cnsder c C = C actual cncentratn and turn back t the fxed crdnate system ξ = z L = ζ + θ ; θ = θ' = t t = u t L (6) C( ξ,θ)= m t π R L Pe 4 πθ e Pe( ξ θ) 4 θ (7) Cz,t ( )= m t πr L LPe 4πu t e Pe( z u t ) 4u Lt (8) r takng Pe = u L D app Cz,t ( )= m ( z u t) t πr 4πD app t e 4D app t (8a) The mpulse respnse at z = L, G (t) (nt necessarly the age densty functn because the system may be pen) s nw gven by and Gt ()= πr u C = m t G θ ()= θ t G()= t u 4πD app t L e 4πD app L u θ e ( L u t ) 4 D app t L ( θ ) L 4D app u θ (9) (30a)

46 G θ ()= θ Pe 4 πθ e Pe( θ ) 4θ (30) Pe = u L D app L Thse that had prbablty thery wll recall that the Gaussan densty functn s gven by: f ()= θ σ D π e ( θ µ ) σ D (3) It can readly be shwn that n lng beds G θ ( θ) gven by eq (30) s small fr all θ except thse n the vcnty f θ =. In the frst apprxmatn ne can represent the G θ ( θ) by a Gaussan densty functn ( θ ) G θ () θ σ D π e σ D (3) where σ D = Pe = D app (33) u L Ths shws that gven the flw velcty prfle u, then the mean velcty u and the apparent axal dspersn ceffcent u R s fxed, f Taylr dffusn mdel s applcable. 48 D The relatve spread f the curve arund the mean then s reduced as the length L between the njectn and mntrng statn s ncreased. The abslute spread, hwever, σ = σ D t = Pe L u = D appl u 3 (34) ncreases as the length f the cndut s ncreased. If the dspersn mdel hlds, the ncrease f the spread measured by σ = σ s prprtnal t L./. Step respnse s nw gven by: F s ()= θ σ D θ ( θ ) e σ D dθ π (35) 3

47 x = θ σ D, dθ = σ D dx, θ = + σ D x F s ()= θ π θ σ D e x dx = π e x dx + θ σ D e x dx = π e x dx + π θ σ D e x dx = + erf θ σ D = erf θ σ D = erfc θ wth σ σ D D = Pe whch s the nrmal dstrbutn. F s erf θ σ D frθ < ()= θ (36) + erf θ σ D frθ > Ths frmula can be used t a) determne after swtchng at θ = 0 at pstn z =0 frm flud A t flud B what fractn f flud B appears at the utflw and hw lng a tme t takes untl the utflw cntans 95% r mre f flud B. b) determne the length ver whch ne has a mxture between p % f flud A and p % f flud B. c) ther prblems f smlar type. 4

48 5.3 Determnatn f Mments Fnally, we shw hw t determne the mments f an mpulse respnse based n the example f the dspersn mdel. Fr the dspersn mdel we have that E θ (θ ) curve s gven by eq (4). Then the mments f the E θ curve are µ n = θ n E θ ( θ )dθ. Clearly, ths wuld requre sme lengthy ntegratns an seres manpulatns. Instead, we can use the Laplace transfrm f the E θ ( θ )curve E ( s) gven by eq (08) and recall that µ n = ( ) n d n E ds n s=o (37) Hwever, dfferentatng eq (08), althugh easer than ntegratn and summatn f equatn (4)), s als tedus. We can nstead recgnze that E ( s) can be expanded n Taylr seres abut s = 0. If we ntrduce the mments ths gves: E ()= s dn E s ds n s =O n=o ( ) s n n! (38) E ()= s ( ) n µ n s n = µ 0 µ s + µ n!! s µ 3 3! s3 (39) n= 0 If we expand eq (08) fr small s, and cmpare term by term wth the abve expansn gven by eq (39), we can readly dentfy all the mments. Really, we are nterested nly n the secnd mment. E ()= s 4 + 4s Pe Pe Pe e e + + 4s + 4s Pe Pe + 4s Pe e Pe + 4 s Pe (08)

49 Frst by Taylr seres: + 4s Pe = + Pe s 4 s Pe + ( ) Os3 ( ) + + 4s Pe = + Pe s 4 Pe s + Os3 ( ) + 4s Pe = Pe s + 4 s Pe + Os s = + Pe Pe s = Pe s + 4 Pe s 8 Pe s + Os 3 = Pe s 4 Pe s + Os ( 3 ) + 4s Pe = 4 Pe s + Os ( 3 ) 6 s Pe + ( ) Os3 ( ) e Pe = e Pe + 4s Pe = e Pe s + Pe s = e Pe s e s Pe e s + s + s = e Pe s + Pe + Pe = e Pe s + + s Pe + () Os3 e Pe + 4s Pe = e Pe s + Pe s = e Pe e s Pe s = e Pe s Pe s + 4s =e Pe s + Pe + s + 0 ( ) s3 s Cmbnng the abve E ()= s 4 + Pe s Pe s Pe s 4 Pe s 4 Pe e s + + s Pe Pe s e Pe s + Pe + e Pe s

50 Keepng nly the terms up t and ncludng s we get: E ()= s Pe s s + + Pe + Pe s Pe s ( + e Pe ) + Pe s E ()= s Expand the denmnatr by bnmal therem Pe s + Pe Pe + Pe s ( Pe + e Pe )s s s x = + x + x x = x + x... E ()= s Pe s + Pe Pe s Pe s+ ( Pe +e Pe )s + Pe s ( Pe + e Pe )s... E ()= s s + + Pe Pe ( ) Pe e s By cmparsn wth the E () s expansn n ts mments we dentfy: Ths s eq (8) n the ntes. µ 0 = µ = E ()= s µ 0 µ s + µ s µ = + Pe Pe e Pe ( ) σ D = µ µ = Pe Pe e Pe ( ) 3

51 5.4 Applcatn f Perturbatn Methds t the Dspersn Mdel fr Tubular Reactrs The axal dspersn mdel fr tubular reactrs at steady state can be descrbed by the fllwng equatns: d c Pe dz z = 0 dc dz R nc n = 0 () dc () dz z = = c Pe dc dz = 0 (3) where c = C A C A = s the dmensnless reactant cncentratn z = Z L = s the dmensnless dstance alng the reactr measured frm the entrance Pe = UL D = s the axal dspersn Peclet number L = ttal reactr length U = Q A = s the lnear mean velcty n the reactr (superfcal velcty n packed beds) Q = vlumetrc flw rate thrugh the reactr A = crss-sectnal area f the reactr D = dspersn ceffcent n the reactr (defned per ttal crss-sectnal area n packed beds). Nte: If D s defned per crss-sectnal area unccuped by slds n packed beds then Dε wll appear n the Pe nstead f D. ε s bed prsty. R n = k τ C n A - dmensnless reactn rate grup (Damkhler number). Nte: In packed beds ths mples that the rate was defned per unt reactr vlume. τ = L - reactr space tme (space tme based n superfcal velcty). U 4

52 Equatns (), () and (3) descrbe the behavr f tubular reactrs and catalytc packed bed reactrs under sthermal cndtns, at steady state fr an n-th rder rreversble, sngle reactn. Applcatn f these equatns t packed bed reactrs assumes that external and nternal mass transfer lmtatns (.e mass transfer frm flud t pellets and nsde the pellets) are nnexstent r have been prperly accunted fr n the verall rate expressn. Slutns t eqs (-3) fr an n-th rder reactn ((n 0,n ) can be btaned nly by numercal means. Hwever, the prblem s a dffcult nnlnear tw-pnt bundary value prblem. [See: P.H. McGnns, Chem. Engr. Prgr. Symp. Ser. N 55 Vl 6, p (968), Lee, E.S., AIChE J., 4(3), 490 (968), Lee, E.S., Quaslnearzatn (969)]. Snce n practcal applcatns Pe numbers are qute large ( Pe 5), and ften Pe = O (0 ), t s f nterest t develp apprxmate slutns t the dspersn mdel fr large Pe numbers. Remember, large Pe σ Pe means small varance and, hence, small varatn frm plug flw, and s exactly the cndtn under whch the dspersn mdel s applcable. Such apprxmate slutns can allw us t estmate well the departure frm plug flw perfrmance and wll save a lt f effrt whch s necessary fr numercal evaluatns f the mdel. We are nterested n large Pe, then Equatns (-3) can be wrtten as: = ε and ε s very small ε << Pe ( ). ε d c dz dc dz R nc n = 0 z = 0; = c ε dc dz (') (') z = ; dc dz = 0 (3') A. Outer Slutn Let us assume that a slutn f the fllwng frm exsts: c = Fz ()= ε n F n ()= z F 0 ()+ z ε F ()+ z ε F ()+ z... (4) n=0 If we can fnd such a slutn then, due t the fact that ε <<, we can hpe that the frst few 5

53 terms wll be adequate t descrbe ur slutn. [Strctly speakng ths wuld be true nly f the seres gven by Eq (4) cnverges fast. Remarkably, a gd apprxmatn t the slutn s btaned even fr dvergent seres! (Fr ths and further detals, see Nayfeh, A.H., Perturbatn Methds, Wley, 973) We d need n Eq (') an expressn fr c n. Hwever, as we are gng t keep nly the frst few terms f the seres (4) rasng t t an n-th pwer s nt that dffcult. c n = [ F + ε F + ε F + ε 3 F 3 +..] n = F n 0 + ε nf n 0 F + ε nn ( )F 0 n F + nf n F +ε 3 6 nn ( ) ( n )F 0 n 3 F 3 + nn ( )F n 0 F F +nf n 0 F 3 + O ε 4 ( ) (5) Dfferentate eq (4) nce and twce and substtute bth dervatves and eq (5) nt eq ('): ε d F 0 dz + ε d F dz + ε d F dz + ε 3 d F 3 dz +... df 0 dz + ε df dz + df ε dz R n F n 0 + εn F n 0 F + ε nn ( )F n 0 F +nf n 0 F +... = 0 (6) Substtute als eq (4) and ts dervatve nt eq ('): At z = 0 = F 0 + ε F + ε F +... ε df 0 dz + ε df dz + ε df dz +... (7) Grup nw the terms wth equal pwers f ε n eq (6) and eq (7) tgether, and requre that they?? t zer: ε 0 : df 0 dz + R nf 0 n = 0 (8) z = 0, F 0 = (8a) ε : df dz + R nnf 0 n F = d F 0 dz (9) 6

54 z = 0, F = df 0 dz (9a) ε : df dz + R nnf n 0 F = d F R n dz nn ( )F n 0 F (0) ε 3 : etc. z = 0, F = df dz (0a) Ntce at ths pnt that the dfferental equatns (D.E.'s) (eq 8-, etc.) are frst rder D.E.'s, whle the rgnal equatn (') was secnd rder. Eq (8) s frst rder fr F 0. Once eq (8) s slved and F 0 determned, the rght hand sde f eq (9) s knwn and the left hand sde s a frst rder D.E. fr F, etc. Thus, we can successvely determne all the F 's. Hwever, because all f these are frst rder equatns, we can nly satsfy ne f the rgnal bundary cndtns. If we tred t satsfy the bundary cndtn at the reactr ext, gven by eq (3'), we wuld have that df dz = 0 at z = fr all, mplyng that all F 's are cnstant due t the frm f D.E.'s ( eq 8-). Thus, we wuld nt be able t get any nfrmatn. Ths ndcates that we have t satsfy the cndtn at the reactr entrance gven by eq (') as ndcated by eq (7). Ths results n a set f cndtns gven by eqs(8a - 0a). The slutn f eq (8) wth I.C. (ntal cndtn) (8a) s readly btaned: [ ] F 0 ()= z + R n ( n )z n () Verfy that ths ndeed s a cncentratn prfle n a plug flw reactr fr an n-th rder reactn! Dfferentate eq () twce, substtute nt eq (9) and slve eq (9) wth I.C. eq (9a): [ ] n F ()= z R n + R n ( n )z n l n + R n ( n )z [ ] n n () Dfferentate eq () twce and substtute tgether wth eq () and eq () n the rght hand sde f eq (0). Slve eq (0) wth I.C. eq (0a): F ()= z R n n n n 3n u n + u n n n ( ) ln u nl nu + 7n 5 n (3) 7

55 where u = + R n (n - ) z (4) Nw we have fund F 0, F, F and we culd wrte ur slutn as F 0 + ε F + ε F. Hwever, we realze that such a slutn nly satsfes the B.C. eq (') at the reactr nlet and des nt satsfy cndtn eq (3') at the reactr utlet. Ultmately we are nterested nt n the cncentratn prfle alng the reactr but n ts value (cncentratn value) at the reactr ext. Snce the B.C. at the ext s nt satsfed we have a lt f reasns t dubt the valdty f ur slutn at the ext and must fnd ways f mprvng t. In perturbatn thery the slutn gven by eq (4) s called an "uter" slutn B. Inner Slutn c 0 ()= z F 0 ()+ z ε F ( z) + ε F ( z) + ε 3 F 3 ( z)+...c O uter slutn (5) We have seen befre that the reasn why the "uter" slutn cannt satsfy the B.C. at the ext (z= ) gven by eq (3') s that all the terms F 's f the "uter" slutn were btaned frm frst rder D.E.s. Nw we must expand ("stretch") ur crdnate system n such a way that we can take a "clser" lk at what happens n the very vcnty (n the "bundary layer") next t z =. Let η = z ε α α > 0 (6) Snce ε << eq (6) gves reasnably large values f η fr z very clse t. Thus, we are stretchng the crdnate system near z =. By chan rule we have: dc dz = dc dη dη dz = ε α dc dη (7) d c dz = d c ε α dη (8) Substtutng eq (7) and eq (8) nt eq (') we get: ε α d c dc + ε α dη dη R nc n = 0 (9) We requre nw that the dervatves shuld stand by equal pwers f ε.e α = α and plus that we wll get a nd rder D.E. Ths gves α =. Hence, 8

56 and becmes: d c ε dη d c dη η = z ε + dc ε dη R nc n = 0 (0) + dc dη ε R nc n = 0 () We assume nw that the slutn t eq () s the "nner" slutn c and that t can be represented by a pwer seres n ε : c = n= 0 ε n c n = c 0 + ε c + ε c + ε 3 c () Remember agan that we are develpng the nner slutn n the very vcnty f z =, s the "uter" slutn n that regn s s clse t that t can be represented by a Taylr seres expansn arund z =.e η = 0 c ()= z F 0 ' ()+ F 0 ()z + ε F + ε F ( )+ F '' 0 ( )+ F '' ' ()+ F ()z ' ()+ F ()z ( )+ F '' ( ) ( z ) +... ( z ) +... ( z ) O ε3 ( ) (3) where F ' = df dz Nw frm eq (0) z = εη ; ( z ) = ε η (4) Substtute eq (4) nt eq (3) and grup tgether the terms wth the same pwers f ε : c c c c ()= z c η ()= [ ] + ε F F 0 () + ε F () F 0 ()η ' () F ()η + F '' 0 ()η (5) 9

57 Ths s ften called an "nner-uter expansn", c,.e the expansn f the uter slutn n the nner (bundary) layer. Nw the nner slutn gven by the seres represented by eq () can be represented by the sum f tw seres: ne frm the nner-uter expansn and the ther frm the true nner slutn undetectable by the uter ne. c η c n ()= ε n n=o = ε n c n η n= O [ ()+ G n () η ] [ ()] ' = F 0 ()+ G 0 ()+ η ε F () F 0 ()η +G η + ε ' F () F ()η + F '' 0 ()η +G () η +O( ε3) (6) Dfferentate eq (6) nce and twce, rase t t the n-th pwer and substtute everythng nt eq () and requre that the sum f the terms wth equal pwers f ε be zer: d G 0 dη + dg 0 dη = 0 (7) d G dη + dg dη = F 0 ' ()+ R n F 0 ()+G 0 [ ] n (8) d G dη + dg dη = F 0 '' ' ()+ F [ ] n F '' () F 0 ()η + nr n F 0 ()+G 0 ' [ () F 0 ()η + G ] (9) Ths s a set f nd rder D.E.s fr the G 's. When the G 's are determned successvely the rght hand sde f the abve equatns s always knwn. Fr nd rder equatns we need B.C.s. One B.C. s btaned frm eq (3').e. whch leads t dc = 0 at η = 0 (30) dη dg 0 dη = 0 at η = 0 (7a) dg dη = F 0 ' () at η = 0 (8a) dg dη = F ' () at η = 0 (9a) 0

58 Remember that away frm the bundary at z = we d nt sense r detect the G 's as F 's ncely satsfy the B.C. at z = 0 and are prbably a gd representatn f the actual slutn fr >z 0 but nt at z =. Ths mples that as we mve away frm z =,.e as η ncreases, all G 's g t zer. Ths s the nd bundary cndtn: η G 0 fr = 0,,,3,... (7b,8b,9b) The slutn f eq (7) wth eq (7a) s G 0 ( η)= 0 (3) The slutn f eq (8) wth eq (8a) and eq (8b) s: [ ] G ( η)= R n + R n ( n ) n n e η (3) The slutn f eq (9) wth eq (9a) and eq (9b) s: [ ( )] G ()= R η n n + R n n n n 3+ η +ln + R n n [ ( )] If we substtute eq (0) nt eqs (3-33) we wll get G 's n terms f z. n n e η (33) The cncentratn prfle clse t the reactr ext s gven by equatn (6) (where η can be substtuted n terms f z). We are especally nterested n the utflw cncentratn at the ext,.e at z =, η= 0. Snce ε = /Pe we get: Let ρ n = [ + ( n )R n ] [ ( )] c ext = x A = F 0 ( z =)+ Pe F z = ( )+G η = 0 + [ F Pe ( z =)+G ( η = 0) ]+O Pe 3 (34) c ext = x A = ρ n n + Pe R n nρ n n ln ρ n n n + (35) 3n Pe n n n n + ρ n n n lnρ n + ρ n n + + O Pe 3 (36) After sme addtnal algebra: