Mass Transfer (Stoffaustausch)

Transcription

1 Mass Tansfe (Stoffaustaush) Examination 3. August 3 Name: Legi-N.: Edition Diffusion by E. L. Cussle: none nd 3 d Test Duation: minutes The following mateials ae not pemitted at you table and have to be deposited in font o bak of the examination oom duing the examination: bags and jakets exeises of the mass tansfe letue (also handwitten on summay sheet o textbook) notebooks, mobile phones, devies with wieless ommuniation ability The following mateials ae pemitted at you table: alulato opy of the book Diffusion ( nd o 3 d edition) by E. L. Cussle pintout of the letue sipt sheet ( pages) summay in fomat DIN A4 o equivalent Please ead these points: wite you name and Legi-N. on eah sheet of you solution begin eah poblem on a new sheet wite only on the font side of eah sheet /7

2 Poblem (5 points) A iula pond of 5 m diamete and 5 m depth is filled with wate of 3 C. The ai above the pond is stagnant and initially wate wate-fee. Desibe the evapoation of wate by assuming diffusion only in vetial dietion. a) Daw a detailed sketh of the poblem and inlude the onentation pofiles. (4 points) b) How does the wate onentation in the ai hange as a funtion of time and distane fom the wate sufae. Please stat with the diffeential equations and indiate the bounday and initial onditions. (6 points) ) Calulate the diffusion oeffiient of wate in ai at 3 C using the Chapman- Enskog equation. (9 points) d) How muh has the wate level dopped afte hous? (6 points) Data: Moleula weight of wate: 8 g/mol Density of wate: g/m 3 Satuation pessue of wate: ln(p sat [Pa]) = /T [K] Pond = Teih (Geman) /7

3 Solution a) b) To desibe the evolution of the wate onentation the semi-infinite slab appoah should be onsideed: C x The initial and bounday onditions ae Fo t =, all z, C A =. C t D Fo t >, z = (wate-salt intefae), C = C o Fo t >, z The solution is: (fa away fom the wate level), C = C =. C C x x ef C / C ef / C C D t D t ( ) ( ) 3.76 ) / kb ( / kb)( / kb) kt B / 33/5.. = T M M 8 9 D. m / s p ln psat d) 3/7

4 psat 4.33kPa Pof. D. Sotiis E. Patsinis sat psat 4.33kPa 6 mol RT m kpa m K mol K D j z sat t dm dv dh A AjM dt dt dt HO dh dt j M HO dh D sat dt t Integation: M HO M D dh dt t h HO t sat M HO D h sat t g m 8. 6 mol.7 mol s 36s 3 m g 3 m.3m 4/7

5 Poblem (5 points) A fisheman ties a new at epellent aoma to keep the stay ats away fom his daily ath. The aoma diffuses adially out of a spheial apsule of 5. m 3 volume. The aoma satuation onentation at the diffuse sufae is.5 g/m 3. The onentation pofile has eahed steady state. Cats an toleate only an aoma onentation up to 9.8g/m 3 and feel unomfotable if they ae lose to the diffuse. a) Daw a detailed sketh of the poblem. (5 points) b) Assume that the initial distane between the ente of the diffuse and the at is. m and the aoma onentation at this point is. g/m 3. How fa does the at need to move (in adial dietion) until it an toleate the onentation? Stat with the omplete genealized mass balane and state you assumptions to simplify it. ( points) ) Calulate the flux at the sufae of the diffuse. How many times a day the diffuse needs to be eplaed to keep ats away if eah diffuse ontains.45 g of epellent aoma? (8 points) Data: Diffusion oeffiient of epellent aoma in ai.59 m /s 5/7

6 Solution Pof. D. Sotiis E. Patsinis a) Sketh sat i C d C t Spheial at epellent diffuse i d C t b) The genealized mass balane fo spheial oodinates is given by: v v D (sin ) v t sin sin sin Assumptions: steady state: d/dt = symmeti: d/d, d/d = no eation: = sine we assume that the system is dilute, onvetion an be negleted: v v v z The mass balane then edues to: D The fist integation: K 6/7

7 The seond integation: -K = K The bounday onditions ae: = i = d = sat = d Using these bounday onditions: ( - ) K K and K sat d d d d i We need to alulate the initial adius of the sphee fom the given volume: 4 V 3 3 i 3 i i 3V 35 m m 3 3 (.5 -.)g/m K = -.59 g / m m m.6 m m 4 K g / m +. = -3.9 g / m m m m 5 3 The onentation pofile as a funtion of adius (): () K K K t ( ) K t 7/7

8 4 (.594 g / m ) t m = 34 m 9.8 g / m ( 3.94 g / m ) Pof. D. Sotiis E. Patsinis Theefoe the at has to move ~4 m away fom the initial position. ) Flux on the sufae of the sphee is alulated using Fik s fist law: d jd d K () K j D K d d{ K } j D K Flux on the tube sufae at = i j D K i i Plugging all the values we get, j.59 m / s.59 g / m m / m (.6 m) 4 4 i 9 j 3.67 g / m s i The mass flowate fom adial dietion: J j A j 4 i i i J 3.67 g / (m s) 4 (.6 m) J 5.8 g / s 3. g / min The aoma mass that one aoma sphee an podue is.45 g, 8/7

9 The diffuso last fo following time, Pof. D. Sotiis E. Patsinis 6 time.45 g / 3. g / min 446 min 4 hs Theefoe fo we need only one diffuso a day. 9/7

10 Poblem 3 (5 points) The off-gas of a eato is flowing upwads though a vetial wetted-wall olumn ( subbe ). Theeby a wate-soluble hemial is emoved fom the off-gas. The inne wall of the ylindial olumn is wetted by initially pue wate that ounte-flows fom the top to the bottom at m/s foming a film of onstant film thikness. The olumn length is 8.5 m and the film thikness is muh smalle than the olumn diamete. The diffusion oeffiient of the hemial in wate is. -5 m /s. At the bottom of the olumn the onentation of hemial in wate is 5% of the satuation onentation. Assume that the hemial in the off-gas is always highe than the satuation onentation in wate. a) Daw a sketh of the wetted-wall olumn fo gas subbing, showing the impotant poblem paametes. (5 points) b) Calulate the film thikness. ( points) /7

11 Solution 3 (See Example 8.3-) a) puified off-gas pue wate Pof. D. Sotiis E. Patsinis ɭ L=7.5 m d [] off-gas [] wate with dissolved vapo b) Mass balane of wate-soluble vapo in the falling wate-film ove a diffeential olumn length Δz: (aumulation) = (flow in minus flow out) + (amount absobed by wate pe time) ( Aossv ) ( Aossv ) kaintef ( ( sat) ) z zz () whee A oss : the oss setion towads the flow, v : the aveage film veloity, : the onentation of the wate-soluble vapo in wate, k: the loal mass tansfe oeffiient fo the absoption of the vapo by the wate, A intef : the intefaial aea fo the absoption and sat : the satuation onentation of the vapo at the gas-liquid intefae. The oss setion of the annula wate film is given by: A ( ) ( )( ) ( ( )) ( d ) d, sine d oss out in out in out in out out while the intefaial aea fo the absoption is: A dz intef. Thus, eq. () beomes: ) ( ) kdz ( ) d v () ( z zz, sat ( ) o: zz z ( ) z k v (, sat ) (3) /7

12 d k dz v ( ), sat (4) The mass tansfe oelation fo falling films is (Table 8.3-/Cussle, 3 d ed): kz zv.69 D D.5 (5) o:.5 k.69 Dv z (6) whee k: the loal mass tansfe oeffiient, z: the position along the film, v the supefiial veloity, D the diffusion oeffiient of the mateial being tansfeed and v : the aveage film veloity. Substituting k in eq. (4) by eq.(6), we have: ( ( d.69 Dv ) v, sat d.69 ), sat z.5 D / v dz z.5 The bounday onditions ae: z =, z = L, L. 5, sat dz (7) If we integate eq.(7) using the B.C.s, we have:.85, sat d( (, sat ).69 D ), sat /, sat v L z.5 dz (8).69 D / v.5 L ln( z (9).85, sat, sat ), sat.38 D / v.5 ln(.85 ) ln( ) L (), sat, sat /7

13 .38 LD/ v ln( ) m. m. s ln( ).85 5 m s ().8m 3/7

14 Poblem 4 (5 points) The ievesible gas phase eation A + B C takes plae on the sufae of spheial atalyst patiles (diamete d p =.5 m) that ae distibuted in a paked bed eato (diamete d eato =.4 m). The gas at the inlet onsists of A with exess B, and has a flow ate of 4 L/min. The eation is fist ode with espet to A and has a eation ate onstant of 83 m/s. Assume that the atalyst bed is poous and does not affet the gas veloity. a) Daw a sketh of the poblem and indiate all impotant poesses, dimensions and onentations. (5 points) b) Selet the appopiate mass tansfe oelation and alulate the oveall mass tansfe oeffiient. Is the eation mass tansfe-limited o eation-limited? (8 points) ) Deive an equation fo the onentation of A as funtion of the eato length. Stat with a mass balane ove a diffeential eato volume dv. (8 points) d) How long should the eato be to eat 99.5% of A? (4 points) Data: Aveage gas visosity:.48-5 m /s Diffusion oeffiient of A:. m /s Speifi sufae aea of the atalyst: 4 m atalyst / m 3 eato 4/7

15 Solution 4 Chemial Reations a) A+B+C z=l L d p k κ A+B C d bed z= A+B v b) The gas veloity an be found fom the gas feed ate: v 3 L m min 4' min 6 Q ml 6 se 5.35 m / s A oss.4 m 4 Fo an ievesible, fist ode heteogeneous eation the oveall MTC an be found: K k Whee k is the esistane to mass tansfe and κ is the eation ate onstant. Fo k we an find the appopiate mass tansfe oelation, whih is fo a paked bed:.4 /3 dv p D k.7 v Enteing the given values gives: m m s.4 / /. m / s k m / s.4 m / s m / s.48 m / s 5/7

16 Enteing the values in the oveall MTC: K.7 m / s k.4 m / s.83 m / s Pof. D. Sotiis E. Patsinis This eation is neithe mass tansfe limited no eation limited sine both play a moe o less equal ole (same ode of magnitude). ) The mass balane ove eato volume dv is ( ) ( ) A v z z z K A oss A A exhange A Ai whee A dv A dz exhange bed oss ( ) ( ) A v z z z K A dz oss A A oss A Ai A( z) A( z z) K A dz v d dz A K A v The bounday onditions ae: z A z L all z A Ai, A AL Ai Ai Afte integation we obtain the onentation pofile of A at the eato outlet as funtion of L: K A AL A exp L v d) Fo 99.5% onvesion: A AL A A AL.5 % 99.5% 6/7

17 AL ln v A L K Filling in the numbes: m s 3 m s m m ln / L.8m.7 / 4 at / eato Pof. D. Sotiis E. Patsinis 7/7