Mass Transfer (Stoffaustausch) Fall 2012

Transcription

1 Mass Transfer (Stoffaustaush) Fall Examination 9. Januar Name: Legi-Nr.: Edition Diffusion by E. L. Cussler: none nd rd Test Duration: minutes The following materials are not permitted at your table and have to be deposited in front or bak of the examination room during the examination: bags and jakets exerises of the mass transfer leture (also handwritten on summary sheet or textbook) notebooks, mobile phones, devies with wireless ommuniation ability The following materials are permitted at your table: alulator opy of the book Diffusion ( nd or rd edition) by E. L. Cussler printout of the leture sript sheet ( pages) summary in format DIN A4 or equivalent Please read these points: write your name and Legi-Nr. on eah sheet of your solution begin eah problem on a new sheet write only on the front side of eah sheet /6

2 Problem (5 points) In a big room with pure air suddenly liquid benzene is spilled on the floor. The spill shape is a ylinder with a diameter of.5 m and a thikness of.5 m. The liquid is volatile and evaporates into the air. Assume the entire system to be at onstant 7 C. During evaporation the ylinder diameter remains onstant and the diffusion takes plae only in vertial diretion. The pressure in the room is atm. a) Make a detailed sketh of the problem and the onentration profile. (4 points) b) Calulate the diffusion oeffiient of benzene in air at 7 C using the Chapman- Enskog equation. (9 points) ) Estimate the onentration of the liquid m above the liquid level after minutes. (7 points) d) How muh will the benzene thikness derease after 5 hours? (5 points) Data: Benzene density:.876 g/m Moleular weight benzene: 78. g/mol Moleular weight air: 9 g/mol Vapour pressure benzene in air at 7 C:.4 atm Universal gas onstant: R=.8 atm L/(mol K) erf x = x π e t dt x erf(x) x erf(x) x erf(x) x erf(x) x erf(x) x erf(x) /6

3 Solution a) b) ( ) (5.46.7) / kb ( / kb)( / kb) kt B / /8.67 = T.86 M M D.89 m / s p ) Semi-infinite slab approah erf z m.4 4Dt m s s erf (.4).5695 C C C C sat.5695 C C C sat psat 6 mol sat 5.7 RT m mol m d) D j z sat t dm dv dh A AjM dt dt dt /6

4 dh dt M j dh D sat M dt t Integration: h M D t dh dt t sat M D h sat t Prof. Dr. Sotiris E. Pratsinis g m mol 5.7 mol s 5 6 6s m g.876 m.m 4/6

5 Problem (5 points) A tubular perfume diffuser is plaed in a room. The perfume diffuses a dog repellent smell radially. The tube is 4 m long with diameter of.44 m. The saturated onentration on the surfae of the tube is.48 g/m. 6 m away from the enter of the tube the perfume onentration is. μg/m. Assume that the perfume onentration in the room is at steady state. a) Draw a detailed sketh of the problem, inluding a onentration profile. (5 points) b) Calulate first the onentration profile and then determine at what distane, from the diffuser enter, the dog an smell the onentration of perfume of 6.8 μg/m? Start with the omplete generalized mass balane and state your assumptions to simplify it. ( points) ) A tube ontains. g of perfume. How often the tubes needs to be replaed during one month (= days)? (8 points) Data: Diffusion oeffiient perfume in air:.59 m /s in German: abstossend 5/6

6 Solution a) Sketh C sat Tubular dog repellent diffusor L = 4 m r i C d r d r b) The generalized mass balane for ylindrial oordinates is given by: v D r v r vz r t r r r r z r r z Assumptions: steady state: d/dt = symmetri: d/d = no reation: r = sine we assume that the system is dilute, onvetion an be negleted: v v v r z The mass balane then redues to: D r r r r The first integration: 6/6

7 K r r Prof. Dr. Sotiris E. Pratsinis The seond integration: = Klnr K The boundary onditions are: r = r i r = r d = sat = d Using these boundary onditions: ( - ) K and K K ln r sat d d d ln ri ln rd The onentration profile as a funtion of radius (r): ( - ) ( - ) (r) ln r + ln r ln r ln r ln r ln r sat d sat d d i d i d d ln r ln r d (r) - sat d d ln ri ln rd ln rt ln r d t (r t ) d sat - d ln ri ln rd ( (r ) ) ln(r / r ) ln r ln r t t d t d i d sat - d (6.8. )g / m m / m. t d 6 6 ln(r / r ) ln( ).48-. g / m m / m 6 ln(r / r ).9 t d rt exp(.9) 6 m rt m The dog will stay m away from the enter of the repellent tube. ) To alulate the life-time of a stik we need to alulate the rate of mass transfer. For 7/6

8 that we will start with Fik s first law: Prof. Dr. Sotiris E. Pratsinis d jd dr We an use the same onentration profile derived earlier with new onentration tolerane and the distane, i.e. with new BC, r = r t = t ln r ln r t (r) - sat t t ln ri ln rt ln r ln r t d{ sat - t t} ln ri ln rt j D dr sat t j D ln r i ln r t r - Flux on the tube surfae at r = r i sat - t sat - t jrr D D i ln r ln r r ln(r / r ) r i t i i t i Plugging all the values we get, -.59 m / s g / m m / m 6 6 sat t jrr D i ln(r i / r t ) r i ln(. / 594.5). m jrr 7.4 g / m s i The mass flowrate from radial diretion: J j A j r L rri rri i J 7.4 g / (m s). m 4 m 8 7 J.8 g / s 8.9 g / min The aroma mass that one stik an produe is. g, 8/6

9 The stik last for following time, Prof. Dr. Sotiris E. Pratsinis 7 time. g / 8. g / min 4479 min days Therefore for a month we need about three stiks of the repellents. 9/6

10 Problem (5 points) In laundries perhloroethylene (per) is used as a solvent and must be removed for eologial reasons from the waste water. The per ontaining water is purified using a paked bed of porous ativated arbon ( Aktivkohle ). The paked bed has a diameter of m and a height of 45 m and removes 99.7% of the per. In the paked bed, per will be adsorbed immediately by the ativated arbon. The water flow rate through the paked bed is 75 L/h. You an assume steady state and onstant properties of the paked bed. a) Draw a sketh of the paked bed showing the important problem parameters. (5 points) b) What is the diameter of the ativated arbon partiles? Assume all partiles to be the same and spherial. ( points) Data: kinemati visosity of water:. m /s diffusion oeffiient of per in water: m /s speifi surfae area of the ativated arbon per bed volume: 9 m /m /6

11 Solution a) wastewater Q=75 L/h L BED =45 m z purified water d BED = m b) Mass balane of per in water over a differential bed length (height) Δz: (aumulation) = (flow in minus flow out) - (amount adsorbed by partiles per time) ( Arossv ) ( Arossv ) kainterf ( i) z z z () where A ross : the ross setion towards the flow, v : the superfiial veloity of the liquid, : the onentration of per in water, k: the loal mass transfer oeffiient for the adsorption of per by the ativated arbon partiles, A interf : the interfaial area for the adsorption and i : the onentration of per at the liquid-partile interfae. For dilute systems the superfiial veloity eq. () beomes: v is onstant, while A interf A ross z. Thus, A ) ( ) k A z( ) rossv ( z z z ross i Rearranging eq. () leads to: () ( ) v zz z ( ) z k ( i ) () /6

12 or: v k z i The adsorption on the surfae is instantaneous: i. Thus: (4) v k z d k dz v (5) The boundary onditions are: z =, z = L,. If we integrate eq.(5) using the B.C.s, we have:. d k v H BED dz (6) ln k v. z H BED (7). ln k v H BED (8) v ln. k H BED (9) The superfiial veloity v of the paked bed is the veloity if the tube would be empty: v Q 4Q A d BED L m / L 475 h 6s / h m m.84 s () Therefore, from eq. (9): m.84 ln. s m k.64 m s 9 45m m The mass transfer orrelation for paked beds is (Table 8.-/Cussler, rd ed): /6

13 k d.4 / v D v.7 Prof. Dr. Sotiris E. Pratsinis () where k is the loal mass transfer oeffiient, v the superfiial veloity, ν the kinemati visosity of the fluid (assume of water sine the solution is dilute), D the diffusion oeffiient of the material being transferred and d the partile (arbon spheres) diameter. Solving eq. () for d, we get: k d v.7v D / /.4.m / s.64 m / s.m / s d 6.84m / s.7.84m / s 8. m / s d.59m / /.4 /6

14 Problem 4 (5 points) Air ontaining a reative ompound A at a onentration of mol% is transported through a laboratory gas pipe of 5 m long and 5 m in diameter. The pipe is oated on the inside with a layer of polymer (B). The polymer layer (B) reats slowly with the reative ompound A aording to the reation A + B C. This reation is irreversible and first order with respet to A, and has the reation rate onstant m/s. The total gas flow rate at the pipe inlet is ml/min (at 5 C and bar). At these onditions the flow through the tube is laminar. a) Draw a sketh of the problem inluding the expeted onentration profile of A over the reator length. (5 points) b) Selet the appropriate mass transfer orrelation and alulate the overall mass transfer oeffiient. (7 points) ) At steady state, find an equation for the onentration of A in the axial diretion. (8 points) d) Calulate the onentration of speies A [mol/m ] at the reator outlet at steady state. How muh (%) of A has reated with the polymer layer? (5 points) Data: Visosity of air (5 C):.8-5 kg/m s Density of air (5 C):.5 kg/m Diffusion oeffiient of A in air (5 C):.6 m /s Gas onstant R: 8.4 J/mol K 4/6

15 Solution 4 a) A A z AL A 5 C, bar d z= z=l B A+C L b) The gas veloity an be found from the gas feed rate: ml m min min 6 Q ml 6 se u.5 m / s A ross.5 m 4 For an irreversible, first order heterogeneous reation the overall MTC an be found: K k Where k is the resistane to mass transfer and κ is the reation rate onstant. For k we an find the appropriate mass transfer orrelation, whih is laminar flow through a irular tube: / d u D k.6 LD d Entering the given values gives: / /.6 / s / m m s m 4 m m s.5m k.5.6 / Entering the values in the overall MTC: m s 5/6

16 K 4 5 k.5 m / s 7. m / s / m s Prof. Dr. Sotiris E. Pratsinis ) 8 points The mass balane over a setion of pipe is ( ) ( ), A u z z z K A 4 ross A A ir A A wall d u A( z) A( z z) K dz A A, wall ( ) ( ) 4, d u z z z K z A A A A wall d 4 K A A A, wall dt d u The boundary onditions are: z all z A A, wall A After integration we obtain the onentration profile of A as funtion of z: 4K A A exp z du d) 5 points The onentration at the reator inlet A, an be determined by using the ideal gas law 5 Pa p mol A XA..8 RT 8.4 J / mol K 88.5 K m The onentration at z=l: 5 m s 4K / AL A exp L.8 mol / m exp.5m.47 mol / m d u.5 m.5 m / s To find the onversion of A: A A AL.8.47 % % 58.4%.8 6/6