MASS TRANSFER Lesson 1 BY DR. ARI SEPPÄLÄ AALTO UNIVERSITY

Transcription

1 SS TRNSFER 2015 Lesson 1 BY DR. RI SEPPÄLÄ LTO UNIVERSITY

2 Structure of the course 6 lessons by r Seälä and Tuula Noonen 5 excercse lessons (one homework roblem n each excercse) by Votto Kotaho (1/6 onts) 3 large assgnements (1/3 onts) Exam (1/2 onts) To ass the course, at least 20% of max. onts from exam s needed ore detals at ycourses

3 Text books r Seälä & arkku Lamnen, neensrto-o (check the erratum n ycourses) dlbrs 18,90, Pkakrjakaua 5 untl (25 )! E. L. Cussler, Dffuson ass Transfer n Flud Systems dlbrs 55,80, mazon $ Humd ar thermodynamcs (n englsh) s avalable at ycourses Need to ask anythng? Please contact: ssstant (Votto Kotaho): emal (votto.kotaho@aalto.f), Tuesday or anytme he s at the offce (K4, room 404) Teachers (Tuula Noonen and r Seälä): emal (ar.seala@aalto.f or tuula.noonen@aalto.f) or after lectures

4 Course tocs xtures bascs Dffuson Dffuson coeffcents for gases and lquds dvecton + dffuson convecton Self-nduced convecton: Stefan flow, natural/free convecton nalogy between heat and mass transfer ass transfer correlatons and coeffcents Couled heat and mass transfer Thermal dffuson, thermohoress and other henomena ass transfer n orous meda Secal tocs: Evaoraton and condensng Enhanced coolng wth wet surfaces Dryng

5 Relaton to other scences Thermal conducton Forced and natural convecton Heat transfer coeffcent Heat transfer Lamnar and turbulent flow Reynolds number Flud mechancs Ideal mxtures (esecally deal gas) Thermodynamcs of humd ar xture chemstry Physcal chemstry ass transfer

6 Where are mass transfer henomena mortant? nywhere where movement of mxture comonents haens (bulk mxture may reman stagnant!) Esecally f the velocty of bulk mxture s not excessve comared to velocty of comonents Nature: Lvng cells, ollutant dsersal n atmoshere, ground and n natural waters, movement of ground water, fog formaton, hase transton n mxtures edcal scences - hysology Techncal scences: energy converson (e.g. combuston, fuel cells, fuel dryng, enhanced heat transfer ), chemcal ndustry, buldng hyscs, membrane rocesses (tycally searaton) etc. etc.

7 Ideal gas mxtures RT V m r r n n y B m m x B x r r B B B x - x x B + Volatle and non-volatle B hahtuva Comonent () of an deal gas olar fracton (bsolute) humdty x

8 Ideal gas mxtures of volatle comonent and non-volatle B h Enthaly H and secfc enthaly h H m H m h + m B h B, T (T) l ( lqud;t ) + c (T)dT h (T) ref Tref humd gas (k humd) h xh + h h k k B B B T T ref c B (T)dT ( l + c (T - T )) + c (T T ) h k x,ref ref B - H & m & B h k ref h Humd ar water vaor, B dry ar (N2, O2, ) Reference states for enthaly: - dry ar: dry ar at 0 o C - water/water vaor: lqud water at 0 o C k x Relatve vaor ressure (sometmes also referred as relatve humdty) j ( t) where t temerature ( '( T ) t, kj/kg o C) dry ar Relatve humdty f x x '( T )

9 Saturated ar Kyllänen lma axmum concentraton of vaor n ar Based on thermodynamcally stable state Note! Hgher concentratons are common n nature -> metastable states x x ' ' j 1(or 100%) f 1 DewontTD Kasteste Cool a surface that s n contact wth ar at a temerature of Tar Temerature TD where the condensng starts (saturated condton reached) s the dew ont -> (Tar) (TD) If total ressure s constant, for deal gas also: x(tar)x (TD)

10 Wet bulb temerature (tm) ärkälämötla Temerature of a wet surface n contact wth ar settles at a temerature lower than the temerature of the ar called the wet bulb temerature of the ar Coolng by evaoraton and heat transfer from ar to water are balanced Suffcent amount of unbounded (free) water s needed If thermal radton can be assumed to be neglgble, then t - t m rc RT Le 1-n l(t ) ln - ' - ( t ) Le D Br c l

11 Determnethe humdtyx of ar j 50%, T 20 o C, 1 bar From Table: (20 oc) 2337 Pa 0.5 * 2337 Pa 1169 Pa x kgh2o / kg 5 d. a ateral roertes of ar saturated wth water vaor (1 atm). Temerature C humdty kg H2O / kg dry ar Partal ressure of water vaor kpa Partal densty of water vaor kg H2O / m 3 Partal densty of dry ar Kg dry ar /m 3 Secfc enthaly of mxture kj/kg dry ar dry ar Secfc heat of mxture J/kgK

12 Ideal solutons - Raoult Law Equlbrum wth lqud mxture and t s comonents n gas hase Höyry, sekotus :ta ja B:tä (g) (l) B(g) B(l) Ideal-dlute soluton - Henry s Law x ' Luos, sekotus :ta ja B:tä x K artal ressure of comonent n gas hase vaor ressure of ure lqud comonent x molar fracton n a lqud hase x 0 1 Both solute and solvent obey Raoult law luennut ane luotn K constant that deends on tye of solvent, solute, temerature (and ressure) x molar fracton of solute for small x only Only solute obeys Henry s law Instead, solvent obeys Raoult law ax

13 Nature of Dffuson Fck s law of dffuson Slow rocess for gases! Very slow for lquds (arox tmes slower)!! Lmts the rogress of rocesses!!! Haens for mutually mscble/mxable artcles only j -D j c(z,t) z w n -rd z w -D r r c

14 olar (total) flux of comonent of a mxture J c v v velocty of comonent of a mxture (comared to a fxed reference frame) Total molar flux of mxture J J cv v molar average velocty of mxture c cv c v v v c x v Relatve molar flux,.e. molar dffuson flux of j ( v - v) J c ( c v + j ) v c + j J olar (total) flux of comonent c cv + j cv - D z convecton advecton + dffuson flux flux flux

15 Because J ( c v + j ) v c + j J cv c c j 0 Bnary dffuson (comonents and B) j -j B D B DB

16 Examle Determne molar flow (mol/s) of CO2 () and ar (B) through a hole n a buldng wall. Draw veloctes of CO2 and ar nsde the hole. Cross-sectonal area of the hole 1 cm 2 Hole length 50 mm CO2 molar fracton nsde 0.1 % outsde 0.01 %. Pressure nsde and outsde 1.05 bar Temerature 30 o C. D B (30 C,1.05 bar) o -5 m 2 /s Consder a one-dmensonal steady state condton.

17 CO2 ar E E-007 v (m/s) v B (m/s) -2.86E E E z (cm) -2.80E z (cm)

18 Dfferent defntons of mxture velocty olar averaged ass averaged v u c c v x v r v v r w IF constant -> uw cconstant -> vw Volume averaged Note v!!!! w c u V n v artal volume

19 Couled dffuson and advecton/convecton Smlfcatons bnary mxtures one-dmensonal steady state condton (J, J, JB constant) J c v - D B dc dz c dc v - J dz D B dc If» 0 then v» constant dz because J cv J v c 2 - c Ł 1- e 1 e vl D B vl D B ł J B v c B2 - c Ł 1- e B1 e vl D B vl D B ł Pe (Peclet number) vl D B