Mass Transfer Solution Manual
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Mass Transfer Operations by Treybal Solution Manual

Mass Transfer Operations by Treybal Solution Manual

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Mass Transfer Solution Manual Document Transcript

  • 1. 1.1a. Concentration of a gas mi ture. A i e f b e ga e [he i (1), a g (2), (3), a d e (4)] i a a a e e f 200 Pa a d a e e a e f 400 K. If he i e ha e a e f ac i f each f he ga e , de e i e: a) The c ii f he i ei e f a f ac i . S i : Ba i : 100 e f he i e b) The a e age ec a eigh f he i e. S ic) The a a c ce ai .S i d) The a de i . S i
  • 2. 1.2a. Concen a ion of a li id ol ion fed o a di illa ion col mn.A ol ion of ca bon e achlo ide (1) and ca bon di lfide (2) con aining 50% b eigh each i obe con in o l di illed a he a e of 4,000 kg/h De e mine:a) The concen a ion of he mi e in e m of mole f ac ion .Sol ionBa i : 100 kg mi eb) The a e age molec la eigh of he mi e.Sol ionc) Calc la e he feed a e in kmol/h.Sol ion
  • 3. 1.3a. Concen a ion of li ified na al ga .A a e f i ified a a ga , LNG, f A a a ha he f i g a c ii : 93.5%CH4, 4.6% C2H6, 1.2% C3H8, a d 0.7% CO2. Ca c a e:a) A e age ec a eigh f he LNG i e.S i b) Weigh f ac i f CH4 i he i e. S i Ba i : 100 e f LNGc) The LNG i hea ed 300 K a d 140 Pa, a d a i e c e e . E i a e he de i f hega i e de he e c di i .S i
  • 4. 1.4b. Concen a ion of a fl e ga . A fl e ga con i of ca bon dio ide, o gen, a e apo , and ni ogen. The mola f ac ion of CO2 and O2 in a ample of he ga a e 12% and 6%, e pec i el . The eigh f ac ion of H2O in he ga i 6.17%. E ima e he den i of hi ga a 500 K and 110 kPa. Sol ion Ba i : 100 kmole of ga mi e Le = mola f ac ion of a e in he mi e (a a pe cen )Ini ial e ima e F om he gi en a e eigh f ac ion (0.0617):
  • 5. 1.5b. Material balances around an ammonia gas absorber. A ga ea f a he a e f 10.0 3/ a 300 K a d 102 Pa. I c i fa e i a i e fa ia a d ai . The ga e e h gh he b f a ac ed bed ga ab be he e i f c ec e a ea f e i id a e ha ab b 90% f a f he a ia, a d i a ai . The ab be e e i c i d ica i h a i e a dia e e f 2.5 . a) Neg ec i g he e a a i f a e , ca c a e he a ia f ac i i he ga ea i g he ab be . S i : Ba i : 1 ec d A=a ia B = aib) Ca c a e he e ga a e ci (defi ed a a f ae e i e bec - ec i a a ea).S i :1.6b. Velocities and flu es in a gas mi ture.A ga i ea a a e e f 150 Pa a d 295 K c ai 20% H2, 40% O2, a d 40% H2Ob e. The ab e e ci ie f each ecie a e -10 / , -2 / , a d 12 / , e ec i e , a
  • 6. in the direction of the -a is.a) Determine the mass average velocit , v, and the molar average velocit , V, for the mi ture.Solution Molar average velocit , V Mass average velocit , vm Basis: 1 kmole of gas mi ture b) Evaluate the four flu es: jO2, nO2, JO2, NO2. Solution:
  • 7. 1.7b. P ope ie of ai a a ed i h a e apo .Ai , ed i a 30- 3 c ai e a 340 K a d 101.3 Pa i a a ed i h a e a . De e i e hef i g e ie f he ga i e:a) M e f ac i f ae a .b) A e age ec a eigh f he i e.c) T a a c ai ed i he a .d) Ma f ae a i he a .S ia) A i ee ai f ae a : F a a a ed i e b) c) d) 1.8c. Wa e balance a o nd an ind ial cooling o e . The c i g a e f ae he c de e f a big c a -fi ed e a i 8,970 g/ . The a e e e he c de e a 29 C a d ea e a 45 C. F he c de e , he a e f ac i g e he e i i c ed d bac 29 C b c ec e c ac i h ai ( ee Fig e 1.11). The ai e e he c i g e a he a e f 6,500 g/ f d ai , a a d -b b e e a e f 30 C a d a h idi f 0.016 g f a e / g f d ai . I ea e he c i g e a a ed i h a e a a 38 C. a) Ca c a e he a e e b e a a i i he c i g e. S i C ide he ai ea i g he e a a ed a 38 C. A i ee ai f ae a :
  • 8. E= ae b ea ai i he aib) T acc f ae e i he c i g e , a f he eff e f a ea b ici a a e a e ea e a i be ed a a e a e . Thi a e a e c ai 500 g/L f di ed id . T a id f i g f he c de e hea - a fe face , he ci c a i g a ei c ai e ha 2,000 g/L f di ed id . The ef e, a a a f heci c a i g a e be de ibe a e di ca ded (b d ). Wi dage e f he e aee i a ed a 0.2% f he eci c a i a e. E i a e he a e - a e e i e e .S i W= i dage e = 500 c = 2000 M= a e ae aeB=b d ae I i ia e i a e Wa e ba a ce S id ba a ce:
  • 9. 1.9b. Water balance around a soap dr er.It is desired to dr 10 kg/min of soap continuousl from 17% moisture b weight to 4% moisture ina countercourrent stream of hot air. The air enters the dr er at the rate of 30.0 m3/min at 350 K,101.3 kPa, and initial water-vapor partial pressure of 1.6 kPa. The dr er operates at constanttemperature and pressure.a) Calculate the moisture content of the entering air, in kg of water/kg of dr air.Solution b) Calculate the flow rate of dr air, in kg/min. Solution c) Calculate the water-vapor partial pressure and relative humidit in the air leaving the dr er. Solution
  • 10. Calc la e a e a e e a 350 K A i ee ai f ae a :
  • 11. 1.10b. Acti ated carbon adsorption; material balances.A a e ga c ai 0.3% e e i ai , a d cc ie a e f 2,500 3 a 298 K a d 101.3 Pa. I a eff ed ce he e ec e f hi ga , i i e ed 100 g f ac i a edca b , i i ia f ee f e e. The e i a ed each e i ib i a c a e ea ea d e e. A i g ha he ai d e ad b he ca b , ca c a e he e i ib ic ce a i f e e i he ga e ha e, a d he a f e e ad bed b he ca b .The ad i e i ib i f hi e i gi e b he F e d ich i he (EAB C CMa a , 3 d. ed., U. S. E. P. A., Re ea ch T ia g e Pa , NC, 1987.): he e W i he ca b e i ib i ad ii ,i g f e e/ g f ca b , a d * i hee i ib i e e a ia e e, i Pa, a d be be ee 0.7 a d 345 Pa. S i M= a f ca b = e f e e ad bed I i ia e i a e
  • 12. 1.11b. Acti ated carbon adsorption; material balances.It is desired to adsorb 99.5% of the toluene originall present in the waste gas of Problem 1.10.Estimate how much activated carbon should be used if the s stem is allowed to reach equilibriumat constant temperature and pressure.Solution
  • 13. 1.12a, d. Estimation of gas diffusivit b the Wilke-Lee equation.E. M. La (MS he i , O eg S a e U i e i , 1964) ea ed he diff i i f ch fi ai a 298 K a d 1 a a d e ed i a e a 0.093 c 2/ . E i a e he diff ic efficie b he Wi e-Lee e a i a d c a e i i h he e e i e a a e.S iFA e diB1.13a, d. Estimation of gas diffusivit b the Wilke-Lee equation.
  • 14. a) E ima e he diff ii of naph halene (C10H8) in ai a 303 K and 1 ba . Compa e i i h he e pe imen al al e of 0.087 cm2/ epo ed in Appendi A. The no mal boiling poin of naph halene i 491.1 K, and i c i ical ol me i 413 cm3/mol. Sol ion E pe imen al al eb) E ima e he diff ii of p idine (C5H5N) in h d ogen a 318 K and 1 a m. Compa e i i h hee pe imen al al e of 0.437 cm2/ epo ed in Appendi A. The no mal boiling poin of p idine i388.4 K, and i c i ical ol me i 254 cm3/mol. Sol ion
  • 15. E pe imen al al ec) E ima e he diff ii of aniline (C6H7N) in ai a 273 K and 1 a m. Compa e i i h hee pe imen al al e of 0.061 cm2/ (G illiland, E. R., I d. E g. Che ., 2 :681, 1934). Theno mal boiling poin of aniline i 457.6 K, and i c i ical ol me i 274 cm3/mol. Sol ion E pe imen al al e
  • 16. 1.14d. Diffusivit of polar gasesIf one or both components of a binary gas mixture are polar, a modified Lennard-Jones relation isoften used. Brokaw (Ind. Eng. Chem. Process Design De elop., 8:240, 1969) has suggested analternative method for this case. Equation (1-49) is still used, but the collision integral is now given by mp = dipole moment, debyes [1 debye = 3.162 10-25 (J- m3)1/2] a) Modify the Mathcad routine of Figure 1.3 to implement Brokaws method. Use the function name DABp(T, P, MA , MB, mA , mB, VA , VB, TbA , TbB) Solution
  • 17. b) E i a e he diff i c efficie f a i e f e h ch ide a d f di ide a 1 ba a d323 K, a d c aei he e e i e a a e f 0.078 c 2/ . The da a e i ed eB a ea i a e h be (Reid, e a ., 1987):Pa a e e Me h ch ide S f di ideTb , K 249.1 263.2Vb , c 3/ 50.6 43.8 , deb e 1.9 1.6M 50.5 64.06 S i1.15d. Diffusivit of polar gasesE a a e he diff i c efficie f h d ge ch ide i a e a 373 K a d 1 ba . The da a e i ed e B a e a i ( ee P b e 1.14) a e h be (Reid, e a ., 1987):Pa a e e H d ge ch ide Wa eTb , K 188.1 373.2Vb , c 3/ 30.6 18.9 , deb e 1.1 1.8M 36.5 18
  • 18. S i1.16d. Diffusivit of polar gasesE a a e he diff i c efficie f h d ge fide i f di ide a 298 K a d 1.5 ba . The da a e i ed e B a e a i ( eeP b e 1.14) a e h be (Reid, e a ., 1987):Pa a e e H d ge fide S f di ideTb , K 189.6 263.2Vb , c 3/ 35.03 43.8 , deb e 0.9 1.6M 34.08 64.06 S i
  • 19. 1.17a,d. Effective diffusivit in a multicomponent stagnant gas mi ture.Calculate the effective diffusivit of nitrogen through a stagnant gas mi ture at 373 K and 1.5 bar.The mi ture composition is:O2 15 mole %CO 30%CO2 35%N2 20%Sol;ution Calculate mole fractions on a nitrogen (1)-free basis: o gen (2); carbon mono ide (3); carbon dio ide (4) Calculate binar MS diffusivities from Wilke-Lee equation
  • 20. 1.18a,d. Mercur removal from flue gases b sorbent injection.Mercur is considered for possible regulation in the electric power industr under Title III of the 1990Clean Air Act Amendments. One promising approach for removing mercur from fossil-fired flue gasinvolves the direct injection of activated carbon into the gas. Meserole, et al. (J. Ai & Wa eManage. A oc., 49:694-704, 1999) describe a theoretical model for estimating mercur removal bthe sorbent injection process. An important parameter of the model is the effective diffusivit ofmercuric chloride vapor traces in the flue gas. If the flue gas is at 1.013 bar and 408 K, and itscomposition (on a mercuric chloride-free basis) is 6% O2, 12% CO2, 7% H2O, and 75% N2,estimate the effective diffusivit of mercuric chloride in the flue gas. Assume that onl the HgCl2 isadsorbed b the activated carbon. Meserole et al. reported an effective diffusivit value of 0.22cm2/s.Solution HgCl2 (1) O2 (2) CO2 (3) H2O (4) N2 (5)
  • 21. 1.19a. Wilke-Chang method for liquid diffusivit .E i a e he i id diff i i f ca b e ach ide i di e i i e ha a 298 K.C ae he e e i e a a e e ed b Reid, e a . (1987) a 1.5 10-5 c 2/ . The c i ica e f ca b e ach ide i 275.9 c 3/ . The i c i f i id e ha a 298 K i 1.08 cP.S i1.20b. Diffusion in electrol te solutions.Whe a a di cia e i i ,i a he ha ec e diff e. I he ab e ce f ae ec ic e ia , he diff i fa i ge a a be ea ed a ec a diff i . F di e i f a i g e a , he diff i c efficie i gi e b he Ne -Ha e e a i (Ha ed,H. S., a d B. B. O e , "The Ph ica Che i f E ec ic S i ," ACS M g . 95, 1950): a) E i a e he diff i c efficie a 298 K f a e di e i f HC i ae. S i b) E i a e he diff i c efficie a 273 K f a e di e i f C SO4 i a e . The i c i f i id a e a 273 K i 1.79 cP.
  • 22. 1.21a. O gen diffusion in ater: Ha duk and Minhas correlation.Estimate the diffusion coefficient of oxygen in liquid water at 298 K. Use the Hayduk andMinhas correlation for solutes in aqueous solutions. At this temperature, the viscosity of wateris 0.9 cP. The critical volume of oxygen is 73.4 cm3/mol. The experimental value of thisdiffusivity was reported as 2.1 10 5 cm2/s (Cussler E. L., Diff sion, 2nd ed, CambridgeUniversity Press, Cambridge, UK, 1997).Solution
  • 23. 1.22a, d. Liquid diffusivit : Ha duk and Minhas correlation.Estimate the diffusivity of carbon tetrachloride in a dilute solution in n-hexane at 298 K using theHayduk and Minhas correlation for nonaqueous solutions. Compare the estimate to the reportedvalue of 3.7 10 5 cm2/s. The following data are available (Reid, et al., 1987): Solution
  • 24. 1.23b. E ima ing mola ol me f om li id diff ion da a.The diffusivity of allyl alcohol (C3H6O) in dilute aqueous solution at 288 K is 0.9 10 5 cm2/s(Reid, et al., 1987). Based on this result, and the Hayduk and Minhas correlation for aqueoussolutions, estimate the molar volume of allyl alcohol at its normal boiling point. Compare it to theresult obtained using the data on Table 1.2. The viscosity of water at 288 K is 1.15 cP.Solution Iniitial estimates From Table 2.1
  • 25. 1.24b, d. Concentration dependence of binar liquid diffusivities. a) E i a e he diff i i f e ha i a e a 298 K he he f ac i f e ha i i i 40%. U de he e c di i (Ha d, B. R., a d R. H. S e , Trans. Farada Soc., 49, 890, 1953):The e e i e a a e e ed b Ha da dS e (1953) i 0.42 10-5c 2/ . S i E i a e he i fi i e di i diff ii f e ha i a e a 298 K f Ha d -Mi ha f a e i F A e di A, he i fi i e di i diff ii f a e i e ha a 298 K i b) E i a e he diff i i f ace e i a e a 298 K he he f ac i f ace e i i i 35%. F hi e a 298 K, he ac i i c efficie f ace e i gi e b Wi e ai (S i h, J. M., e a ., Introduction to Chemical Engineering Thermod namics, 5 h ed, McG a -Hi C ., I c., Ne Y , NY, 1996): S i E i a e he he d a ic fac
  • 26. F A e di A E i a e he i fi i e di i diff ii f ace ei a e a 298 K f Ha d -Mi ha f a e i
  • 27. 1.25b, d. Stead -state, one-dimensional, gas-phase flu calculation.A flat plate of solid carbon is being burned in the presence of pure o gen according to thereaction Molecular diffusion of gaseous reactant and products takes place through a gas film adjacent to the carbon surface; the thickness of this film is 1.0 mm. On the outside of the film, the gas concentration is 40% CO, 20% O2, and 40% CO2. The reaction at the surface ma be assumed to be instantaneous, therefore, ne t to the carbon surface, there is virtuall no o gen.The temperature of the gas film is 600 K, and the pressure is 1 bar. Estimate the rate of combustion of the carbon, in kg/m2-min. Solution CO (1), CO2 (2), O2 (3) Calculate binar MS diffusivities from Wilke-Lee
  • 28. Appendi C-2: Solution of the Ma ell-Stefan equations for amulticomponent mi ture of ideal gases b orthogonal collocation ( C =3). Orthogonal collocation matricesThe p e e and empe a e in he apo pha e a eThe Ma ell-S efan diff ion coefficicien ae
  • 29. The length of the diffusion path isThe densit of the gas phase follows from the ideal gas law Initial estimates of the flu es Initial estimates of the concentrations
  • 30. Stoichiometric relations (No o gen)
  • 31. 1.26b. Stead -state, one-dimensional, liquid-phase flu calculation.A G (N 2SO4 10H2O) 288 K.E N 2SO4 .A0.085 .A ( ) N 2SO4, .T G 288 K 36 /100 1,240 / 3 (P C ,1973). T N 2SO4 288 K P 1-20. T 288 K 999.8 / 3; 1.153 P.SA = N 2SO4 B = H2O C A1 ( )B : 100 H2O (36 ) (P )
  • 32. Calculate diffusivity1.27c, d. Molec la diff ion h o gh a ga -li id in e face.Ammonia, NH3, is being selectively removed from an air-NH3 mixture by absorption into water. Inthis steady-state process, ammonia is transferred by molecular diffusion through a stagnant gaslayer 5 mm thick and then through a stagnant water layer 0.1 mm thick. The concentration ofammonia at the outer boundary of the gas layer is 3.42 mol percent and the concentration at thelower boundary of the water layer is esentially zero.The temperature of the system is 288 K andthe total pressure is 1 atm. The diffusivity of ammonia in air under these conditions is 0.215 cm2/sand in liquid water is 1.77 10 5 cm2/s. Neglecting water evaporation, determine the rate ofdiffusion of ammonia, in kg/m2-hr. Assume that the gas and liquid are in equilibrium at theinterface.
  • 33. Sol ion: Ini ial e ima e :
  • 34. 1.28c. Stead -state molecular diffusion in gases.A i e f e ha a d ae a i bei g ec ified i a adiaba ic di i a i c . Thea c h i a i ed a d a fe ed f he i id he a ha e. Wa e a c de e(e gh he a e hea f a i a i eeded b he a c h bei g e a a ed) a d i a fe ed f he a he i id ha e. B h c e diff e h gh a ga fi 0.1 hic . The e e a e i 368 K a d he e e i 1 a . The e f ac i f e ha i 0.8 e ide f he fi a d 0.2 he he ide f he fi . Ca c a e he a e f diff i f e haa d f a e , i g/ 2- . The a e hea f a i a i f he a c h a d a e a 368 K ca bee i a ed b he Pi e ace ic fac c e a i (Reid, e a ., 1987) he e i he ace ic fac . S i A = e ha B= ae Ca c a e e ha hea f a i ai Ca c a e a e hea f a i ai E i a e diff ii f Wi e-Lee
  • 35. 1.29a, d. Analog among molecular heat and mass transfer.It has been observed that for the s stem air-water vapor at near ambient conditions, Le = 1.0(Tre bal, 1980). This observation, called the Lewis relation, has profound implications inhumidification operations, as will be seen later. Based on the Lewis relation, estimate the diffusivitof water vapor in air at 300 K and 1 atm. Compare our result with the value predicted b the Wilke-Lee equation. For air at 300 K and 1 atm:Cp = 1.01 kJ/kg-K, k = 0.0262 W/m-K, m = 1.846 10-5kg/m-s, and r = 1.18 kg/m3. Solution Estimate diffusivit from the Wilke-Lee equation
  • 36. 1.30b, d. Stead -state molecular diffusion in gases.Wa e e a a i g f a d a 300 K d e b ec a diff i ac a ai fi 1.5 hic . If he e a i e h idi f he ai a he e edge f he fi i 20%, a d he a e ei 1ba , e i a e he d i he a e e e e da , a i g ha c di i i he fi e aic a . The a e e f ae a af ci f e e a e ca be acc a e e i a ed f he Wag e e a i (Reid, e a ., 1987) S i F A e di A 1.31b, d. Stead -state molecular diffusion in a ternar gas s stem. Ca c a e he f e a d c ce a i fi e f he e a e h d ge (1), i ge (2), a d ca b di ide (3) de he f i g c di i . The e e a e i 308 K a d he e ei 1
  • 37. atm. The diffusion path length is 86 mm. At one end of the diffusion path the concentration is 20mole% H2, 40% N2, 40% CO2; at the other end, the concentration is 50% H2, 20% N2, 30% CO2.The total molar flu is ero, = 0. The MS diffusion coefficients are D12 = 83.8 mm2/s, D13 = 68.0mm2/s, D23 = 16.8 mm2/s.Solution Appendi C-1: Solution of the Ma ell-Stefan equations for a multicomponent mi ture of ideal gases b orthogonal collocation ( C = 3). Orthogonal collocation matrices The p e e and empe a e in he apo pha e a e The Ma ell-S efan diff ion coefficicien ae The leng h of he diff ion pa h i The den i of he ga pha e follo f om he ideal ga la
  • 38. Initial estimates of the flu es Initial estimates of the concentrations
  • 39. 2.1a. Ma - an fe coefficien in a ga ab o be . A ga ab be i ed e e be e e (C6H6) a f ai b c bbi g he ga i e iha a i e i a 300 K a d 1 a . A a ce ai i i he ab be , he be e e e f ac i i he b f he ga ha e i 0.02, hi e he c e di g i e facia be e e ga - ha e c ce a i i 0.0158. The be e e f a ha i i ea ed a 0.62 g/ 2- . a) Ca c a e he a - a fe c efficie i he ga ha e a ha i i he e i e , e e i g he d i i g f ce i e f e f ac i . S i b) Ca c a e he a - a fe c efficie i he ga ha e a ha i i he e i e , e e i g he d i i g f ce i e f a c ce a i , / 3. S i c) A he a e ace i he e i e , he be e e e f ac i i he b f he i id ha e i 0.125, hi e he c e di g i e facia be e e i id- ha e c ce a i i 0.158. Ca c a e he a - a fe c efficie i he i id ha e, e e i g he d i i g f ce i e f e f ac i . S i2.2a. Ma - an fe coefficien f om naph halene blima ion da a.I a ab a e e i e , ai a 347 K a d 1 a i b a high eed a d a i ge a h ha e e (C10H8) he e, hich b i a e a ia . Whe he e e i e begi , he dia e e
  • 40. f he he e i 2.0 c . A he e d f he e e i e , 14.32 i a e , he dia e e f he he e i1.85 c .a) E i a e he a - a fe c efficie , ba ed he a e age face a ea f he a ic e,e e i g he d i i g f ce i e f a ia e e . The de i f id a h ha e e i 1.145g/c 3, i a e e a 347 K i 670 Pa (Pe a d Chi , 1973).S i b) Ca c a e he a - a fe c efficie , f he d i i g f ce i e f a c ce ai . S i2.3a. Mass-transfer coefficients from acetone e aporation data.I a ab a e e i e , ai a 300 K a d 1 a i b a high eed a a e he face f a ec a g a ha a ha c ai i id ace e (C3H6O), hich e a a e a ia . The a i1 g a d 50 c ide. I i c ec ed a e e i c ai i g i id ace e hicha a ica e ace he ace e e a a ed, ai ai i g a c a i id e e i he a .D i ga e ei e a , i a b e ed ha 2.0 L f ace e e a a ed i 5 i . E i a e he a - a fe c efficie . The de i f i id ace e a 300 K i 0.79 g/c 3; i a e e i 27 Pa (Pe a d Chi , 1973).S i
  • 41. 2.4b. Mass-transfer coefficients from etted- all e perimental data.A e ed- a e e i e a e - c i fa ga i e, 50 i dia e e a d 1.0 g.Wa e a 308 K f d he i e a . D ai e e he b f he i e a he a e f 1.04 3/ i , ea ed a 308 K a d 1 a . I ea e he e ed ec i a 308 K a d i h a e a i eh idi f 34%. Wi h he he f e a i (2-52), e i a e he a e age a - a fe c efficie , i h he d i i g f ce i e f a f ac i .S i
  • 42. 2.7c. Ma an fe in an ann la pace.a) I d i g a e f diff i f a h ha e e i ai , a i e iga e aced a 30.5-c ec i f he i e i e fa a i h a a h ha e e d. The a a c ed f a 51- -ODba i e i e ded b a 76- -ID b a i e. Whi e e a i g a a a e ci i hi hea f 12.2 g f ai / 2- a 273 K a d 1 a , he i e iga de e i ed ha he a ia e e f a h ha e e i he e i i g ga ea a 0.041 Pa. U de he c di i f hei e iga i , he Sch id be f he ga a 2.57, he i c i a 175 P, a d he a e e f a h ha e e a 1.03 Pa. E i a e he a - a fe c efficie f he i e a f hi e f c di i .A e ha e a i (2-52) a ie .S i b) M ad a d Pe (T an . AIChE, 38, 593, 1942) e e ed he f i gc ea i f hea - a fe c efficie i a a a ace: he e d a d di a e he ide a d i ide dia e e f he a , de i he e i ae dia e e defi ed a Wied he a a g e e i f a a fe a d e i e i a e he a - a fe c efficie f he c di i f a a). C aeb h e . S i
  • 43. 2.8c. The Chilton-Colburn analog : flow across tube banks.W C (I d. E g. Che ., 40, 1087, 1948) 310 K 1 .T , .T 1038- -OD ( = 38 ) 57- , 76 .T - .T : G , / 2- , G / 2- -P . )R (2-68) C D- .T 310 K 1 0.074 2/ . S P 310 K 1 : D :
  • 44. b) E i a e he a - a fe c efficie be e ec ed f e a a i f - ac h ica b di ide f he a e ge e ica a a ge e he he ca b di ide f a a a i e ci f 10 / a 300 K a d 1 a . The a e e f - a c h a 300 K i 2.7 Pa.S i P e ie f di e i e f ac h i ca b di ide a 300 K a d 1 a :c) Za a a (Ad . Heat Transfer, , 93, 1972) ed he f i gc e a i f he hea - a fe c efficie i a agge ed be ba a a ge e i ia ha died b Wi di g a dChe e :U e he a - a fe e e i a a g e ai (2-69) e i a e he a - a fec efficie f a b). C a e he e .S i
  • 45. 2.9b. Ma an fe f om a fla pla e.A 1-m square thin plate of solid naphthalene is oriented parallel to a stream of air flowing at 20m/s. The air is at 310 K and 101.3 kPa. The naphthalene remains at 290 K; at this temperaturethe vapor pressure of naphthalene is 26 Pa. Estimate the moles of naphthalene lost from the plateper hour, if the end effects can be ignored.Solution2.10b. Ma an fe f om a fla pla e.A thin plate of solid salt, NaCl, measuring 15 by 15 cm, is to be dragged through seawater at avelocity of 0.6 m/s. The 291 K seawater has a salt concentration of 0.0309 g/cm3. Estimate therate at which the salt goes into solution if the edge effects can be ignored. Assume that thekinematic viscosity at the average liquid film conditions is 1.02 10 6 m2/s, and the diffusivity is1.25 10 9 m2/s. The solubility of NaCl in water at 291 K is 0.35 g/cm3, and the density of thesaturated solution is 1.22 g/cm3 (Perry and Chilton, 1973) .Solution
  • 46. Laminar floAt the bulk of the solution, point 2:At the interface, point 1:
  • 47. 2.11b. Ma an fe f om a fla li id face.D i g he e e i e de c ibed i P b e 2.3, he ai e ci a ea ed a 6 / , a a e he ge ide f he a . E i a e he a - a fe c efficie edic ed b e a i (2-28)(2-29) a d c aei he a e ea ed e e i e a . N ice ha , d e he high a i i face e, he a e age ace e c ce a i i he ga fi i e a i e high. The ef e, e ie ch a de i a d i c i h d be e i a ed ca ef . The f i g da a f ace e ighbe eeded: Tc = 508.1 K, Pc = 47.0 ba , M = 58, Vc = 209 c 3/ , Zc = 0.232 (Reid, e a ., 1987).S iA e age fi e ie : E i a e he i c i f he i ef L ca Me h d E i a e he diff ii f he Wi e-Lee e ai
  • 48. 2.12b. E aporation of a drop of ater falling in air.Repeat E ample 2.9 for a drop of ater hich is originall 2 mm in diameter.Solution
  • 49. 2.13b. Dissolution of a solid sphere into a flo ing liquid stream.Estimate the mass-transfer coefficient for the dissolution of sodium chloride from a cast sphere, 1.5cm in diameter, if placed in a flowing water stream. The velocity of the 291 K water stream is 1.0m/s.Assume that the kinematic viscosity at the average liquid film conditions is 1.02 10 6 m2/s, andthe mass diffusivity is 1.25 10 9 m2/s. The solubility of NaCl in water at 291 K is 0.35 g/cm3, andthe density of the saturated solution is 1.22 g/cm3 (Perry and Chilton, 1973) .Solution From Prob. 2.10:2.14b. Sublimation of a solid sphere into a gas stream.During the experiment described in Problem 2.2, the air velocity was measured at 10 m/s. Estimatethe mass-transfer coefficient predicted by equation (2-36) and compare it to the value measuredexperimentally. The following data for naphthalene might be needed: Tb = 491.1 K, Vc = 413cm3/mol.SolutionFor air at 347 K and 1 atm: Estimate DAB from the Wilke-Lee equation Lennard-Jones parameters for naphthalene
  • 50. 2.15b. Dissolution of a solid sphere into a flo ing liquid stream.The cr stal of Problem 1.26 is a sphere 2-cm in diameter. It is falling at terminal velocit under theinfluence of gravit into a big tank of water at 288 K. The densit of the cr stal is 1,464 kg/m3 (Perrand Chilton, 1973).a) Estimate the cr stals terminal velocit .Solution
  • 51. b) Estimate the rate at which the cr stal dissolves and compare it to the answer obtained in Problem1.26.Solution From Prob.1.26 From Prob.1.26:
  • 52. 2.16c. Mass transfer inside a circular pipe. Water flows through a thin tube, the walls of which are lightly coated with benzoic acid (C7H6O2). The water flows slowly, at 298 K and 0.1 cm/s. The pipe is 1-cm in diameter. Under these conditions, equation (2-63) applies. a) Show that a material balance on a length of pipe L leads to where v is the average fluid velocity, and c A * is the equilibrium solubility concentration. b) What is the average concentration of benzoic acid in the water after 2 m of pipe. The solubility of benzoic acid in water at 298 K is 0.003 g/cm3, and the mass diffusivity is 1.0 10 5 cm2/s (Cussler, 1997). Solution2.17b. Mass transfer in a etted- all to er.Water flows down the inside wall of a 25-mm ID wetted-wall tower of the design of Figure 2.2, whileair flows upward through the core. Dry air enters at the rate of 7 kg/m2-s. Assume the air iseverywhere at its average conditions of 309 K and 1 atm, the water at 294 K, and the mass-transfercoefficient constant. Compute the average partial pressure of water in the air leaving if the tower is 1
  • 53. g.S i F a e a 294 K2.18c. Ma an fe in an ann la pace.I d i g he b i a i f a h ha e e i a ai ea , a i e iga c c ed a 3- - ga a d c . The i e i e a ade f a 25- -OD, id a h ha e e d; hi a ded b a 50- -ID a h ha e e i e.Ai a 289 K a d 1 a f ed h gh he a a ace a a a e age e ci f 15 / . E i a e he a ia e e f a h ha e e i he ai ea e i i g f he be. A 289 K, a h ha e eha a a e e f 5.2 Pa, a d a diff i i i ai f 0.06 c 2/ . U e he e f P b e 2.7 e i a e he a - a fe c efficie f he i e face; a d e a i (2-47), i g hee i a e dia e e defi ed i P b e 2.7, e i a e he c efficie f he e face.
  • 54. Sol ionIn hi i a ion, he e ill be a mola fl f om he inne all, NA1, i h pecific in e facial a ea, a1,and a fl f om he o e all, NA2, i h a ea a2. A ma e ial balance on a diffe en ial ol meelemen ield :Define: Then: Fo he in e io all:
  • 55. For the outer all:
  • 56. 2.19c. Ben ene evaporation on the outside surface of a single c linder.Be e e i e a a i g a he a e f 20 g/h e he face f a 10-c -dia e e c i de .D ai a 325 K a d 1 a f a igh a g e he a i f he c i de a a e ci f 2 / . The i id i a a e e a e f 315 K he e i e e a a e e f 26.7 Pa. E i a e he e g h f he c i de . F be e e, Tc = 562.2 K, Pc = 48.9 ba , M = 78, Vc = 259 c 3/ , Zc = 0.271(Reid, e a ., 1987).S i Ca c a e he a e age e ie f he fi F he Wi e-Lee e ai F he L ca Me h d F E . 2-45:
  • 57. 2.20b. Ma an fe in a packed bed.Wilke and Hougan (T an . AIChE, 41, 445, 1945) reported the mass transfer in beds of granularsolids. Air was blown through a bed of porous celite pellets wetted with water, and by evaporatingthis water under adiabatic conditions, they reported gas-film coefficients for packed beds. In one run,the following data were reported:effective particle diameter 5.71 mmgas stream mass velocity 0.816 kg/m2-stemperature at the surface 311 Kpressure 97.7 kPak G 4.415 10 3 kmol/m2-s-atmWith the assumption that the properties of the gas mixture are the same as those of air, calculatethe gas-film mass-transfer coefficient using equation (2-55) and compare the result with the valuereported by Wilke and Hougan.Solution From the Wilke-Lee equation
  • 58. 2.21b. Ma an fe and p e e d op in a packed bed.Air at 373 K and 2 atm is passed through a bed 10-cm in diameter composed of iodine spheres0.7-cm in diameter. The air flows at a rate of 2 m/s, based on the empt cross section of the bed.The porosit of the bed is 40%.a) How much iodine will evaporate from a bed 0.1 m long? The vapor pressure of iodine at 373 K is6 kPa.Solution From the Wilke-Lee equation:
  • 59. b) E ima e he p e e d op h o gh he bed.Sol ion
  • 60. 2.22b. Volumetric mass-transfer coefficients in industrial to ers.The interfacial surface area per unit volume, a, in many types of packing materials used inindustrial towers is virtually impossible to measure. Both a and the mass-transfer coefficientdepend on the physical geometry of the equipment and on the flow rates of the two contacting,inmiscible streams. Accordingly, they are normally correlated together as the volumetric mass-transfer coefficient, k c a.Empirical equations for the volumetric coefficients must be obtained experimentally for each type ofmass-transfer operation. Sherwood and Holloway (Trans. AIChE, 36, 21, 39, 1940) obtained thefollowing correlation for the liquid-film mass-transfer coefficient in packed absorption towersThe values of a and n to be used in equation (2-71) for various industrial packings are listed in thefollowing table, when SI units are used exclusively.a) Consider the absorption of SO2 with water at 294 K in a tower packed with 25-mm Raschigrings. If the liquid mass velocity is L = 2.04 kg/m2-s, estimate the liquid-film mass-transfercoefficient. The diffusivity of SO2 in water at 294 K is 1.7 10 9 m2/s.SolutionFor dimensional consistency, add the constants:b) Whitney and Vivian (Chem. Eng. Progr., 45, 323, 1949) measured rates of absorption of SO2 inwater and found the following expression for 25-mm Raschig rings at 294 Kwhere k a is in kmole/m2-s. For the conditions described in part a), estimate the liquid-filmmass-transfer coefficient using equation (2-72). Compare the results.Solution
  • 61. 2.23b. Mass transfer in fluidi ed beds.Cavatorta, et al. (AIC E J., 45, 938, 1999) studied the electrochemical reduction of ferrycianideions, {Fe(CN)6} 3, to ferrocyanide, {Fe(CN)6} 4, in aqueous alkaline solutions. They studieddifferent arrangements of packed columns, including fluidized beds. The fluidized bed experimentswere performed in a 5-cm-ID circular column, 75-cm high. The bed was packed with 0.534-mmspherical glass beads, with a particle density of 2.612 g/cm3. The properties of the aqueoussolutions were: density = 1,083 kg/m3, viscosity = 1.30 cP, diffusivity = 5.90 10 10 m2/s. Theyfound that the porosity of the fluidized bed, e, could be correlated with the superficial liquid velocitybased on the empty tube, vs , throughwhere vs is in cm/s.a) Using equation (2-56), estimate the mass-transfer coefficient, k L, if the porosity of the bed is60%.Solution
  • 62. b) Ca a o a e al. p opo ed he follo ing co ela ion o e ima e he ma - an fe coefficien fo hei fl idi ed bed e pe imen al n : he e Re i ba ed on he emp be eloci . U ing hi co ela ion, e ima e he ma - an fe coefficien , L, if he po o i of he bed i 60%. Compa e o e l o ha of pa a). Sol ion 2.24b. Mass transfer in a hollo -fiber boiler feed ater deaerator. Con ide he hollo -fibe BFW deae a o de c ibed in E ample 2-13. If he a e flo a e inc ea e o 60,000 kg/h hile e e hing el e emain con an , calc la e he f ac ion of he en e ing di ol ed o gen ha can be emo ed. Sol ion
  • 63. 2.25b. Mass transfer in a hollo -fiber boiler feed ater deaerator.a) Con ide he hollo -fibe BFW deae a o de c ibed in E ample 2-13. A ming ha onl o gendiff e ac o he memb ane, calc la e he ga ol me flo a e and compo i ion a he l meno le . The a e en e he hell ide a 298 K a a ed i h a mo phe ic o gen, hich mean adi ol ed o gen concen a ion of 8.38 mg/L.Sol ion
  • 64. b) Ca c a e he a - a fe c efficie a he a e age c di i i ide he e . Neg ec he hic e f he fibe a he e i a i g he ga e ci i ide he e .S i Ca c a e he a e age f c di i i ide he fibe Ca c a e he a e age ge a f ac i i he ga F L ca eh df e N2 F he Wi e-Lee e ai
  • 65. (From E ample 2.13)
  • 66. 3.1a. Application of Raoults law to a binar s stem. Re ea E a e 3.1, b f a i id c ce a i f 0.6 e f ac i f be e ea da e e a e f 320 K. S i3.2b. Application of Raoults law to a binar s stem.a) De e i e he c ii f he i id i e i ib i iha a c ai i g 60 e e cebe e e-40 e e ce e e if he e e i i a e e de 1 a e e. P edic he e i ib i e e a e.S i I i ia e i a e
  • 67. b) De e i e he c ii f he a i e i ib i i h a i id c ai i g 60 e e ce be e e-40 e e ce e e if he e e i i a e e de 1 a e e. P edic he e i ib i e e a e. S i 3.3a. Application of Raoults la to a binar s stem. N a he a e, -C7H16, a d a c a e, -C8H18, f idea i . A 373 K, a he a e ha a a e e f 106 Pa a d a c a e f 47.1 Pa. a) Wha d be he c ii f a he a e- c a e i ha b i a 373 K de a 93 Pa e e? S i b) Wha d be he c ii f he a i e i ib i i h he i ha i de c ibed i (a)? S i3.4a. Henr s la : saturation of ater ith o gen.A i ih ge di ed i a e c ai i g 0.5 g O2/100 g f H2O i b gh i c ac i h a a ge e fa he ic ai a 283 K a d a a e e f 1 a . The He ac a f he ge - a e e a 283 K e a 3.27 10 4a / e f ac i .a) Wi he i gai e ge ?b) Wha i be he c ce a i f ge i he fi a e i ib i i ?S i
  • 68. A e i ib i : Ba i : 1 L a e (1 g a e ) E i ib i c ce ai , c e = 11.42 g ge /L I i ia c di i : The i gai ge .3.5c. Material balances combined ith equilibrium relations.Re ea E a e 3.3, b a i g ha he a ia, ai , a d a e a e b gh i c ac i ac ed c ai e . The e i 10 3 f ga ace e he i id. A i g ha he ga - ace ea d he e e a e e ai c a i e i ib i i achie ed, dif he Ma hcad ga iFig e 3.2 ca c a e:a) he a e e a e i ib iS i I i ia g e e
  • 69. A : P = 1.755 ) . A : A = 0.145; A = 0.1623.6b. Ma - an fe e i ance d ing ab o p ion.I A( = 60) , A,G = 0.1 , A,L = 1.0 A/ 3 .T 2.0 ; 1,100 / 3. T H 0.85 / .T KG = 0.27 / 2- - . I 57% , ) - , G;S
  • 70. b) the liquid-film coefficient, k L;Solution Basis: 1 m3 of aqueous solution c) the concentration on the liquid side of the interface, A,i; Solution Initial estimates of interfacial concentrations:
  • 71. d) the mass flu of A.Solution Check this result b calculating the gas-phase flu :
  • 72. 3.7b. Mass-transfer resistances during absorption.F a e i hich c e A i a fe i g f he ga ha e he i id ha e, hee i ib i e a i i gi e b he e A,i i he e i ib i a ia e e i a a d A,i i he e i ib i i id c ce a i i a f ac i . A e i i he a a a , he i id ea c ai 4.5 e % a d he ga ea c ai 9.0 e% A. The a e e i 1 a . The i di id a ga -fi c efficie a hi i i G = 3.0 e/ 2- -a . Fif e ce f he e a e i a ce a a fe i be e c e ed i he i id ha e. E a a e a) he ea a - a fe c efficie , K ; S i b) he a f f A; S i ( Ae = A *) c) The i id i e facia c ce ai f A. S i3.8d. Absorption of ammonia b water: use of F-t pe mass-transfer coefficients.M dif he Ma hcad g a i Fig e 3.6 e ea E a e 3.5, b i h A,G = 0.70 a d A,L =0.10. E e hi g e e e ai c a .
  • 73. Sol ion Ini ial g e e 3.9d. Absorption of ammonia b water: use of F-t pe mass-transfer coefficients. Modif he Ma hcad p og am in Fig e 3.6 o epea E ample 3.5, b i h FL = 0.0050 kmol/m2- . E e hing el e emain con an . Sol ionIni ial g e e
  • 74. 3.10b. Ma - an fe e i ance d ing ab o p ion of ammonia.I he ab i fa ia i ae f a ai -a ia i e a 300 K a d 1 a , hei di id a fi c efficie e e e i a ed be L = 6.3 c /h a d G = 1.17 / 2-h -a . Thee i ib i ea i hi f e di e i fa ia i a e a 300 K a d 1 a iDe e i e he f i g a - a fe c efficie :a)S i b) S i
  • 75. c) Ky Solution d) Fraction of the total resistance to mass transfer that resides in the gas phase. Solution3.11b. Mass-transfer resistances in hollo -fiber membrane contactors.For mass transfer across the hollow-fiber membrane contactors described in Example 2.13, theoverall mass-transfer coefficient based on the liquid concentrations, KL, is given by (Yang andCussler, AIC E J., 32, 1910, Nov. 1986) where k L, k M, and k c are the individual mass-transfer coefficients in the liquid, across the membrane, and in the gas, respectively; and H is Henrys law constant, the gas equilibrium concentration divided by that in the liquid. The mass-transfer coefficient across a hydrophobic membrane is from (Prasad and Sirkar, AIC E J., 34, 177, Feb. 1988) where DAB = molecular diffusion coefficient in the gas filling the pores, eM = membrane porosity, tM = membrane tortuosity, d = membrane thickness. For the membrane modules of Example 2.13, eM = 0.4,tM = 2.2, and d = 25 10 6 m (Prasad and Sirkar, 1988). a) Calculate the corresponding value of k M. Solution For oxygen in nitrogen at 298 K and 1 atm:
  • 76. b) U i g he e f a (a), E a e 2.13, a d P b e 2.25, ca c a e KL, a d e i a e haf ac i f he a e i a ce a a fe e ide i he i id fi .S iF E a e 2.13:F P b. 2.25: Vi a a f he e i a ce e ide i he i id ha e.3.12c. Combined use of F- and k-t pe coefficients: absorption of low-solubilit gases.D i g ab i f - bi i ga e , a a fe f a high c ce a ed ga i ea e di e i id i fe e a e ace. I ha ca e, a h gh i i a ia e ea - e a - a fe c efficie i he i id ha e, a F- e c efficie be ed i he ga ha e. Si ce di e i id i a be He a , he i e facia c ce a i d i gab i f - bi i ga e a e e a ed h gh A,i = A,i.a) Sh ha , de he c di i de c ibed ab e, he ga i e facia c ce ai a i fie hee ai S i I he ga ha e:
  • 77. I he i id ha e: F He La : The : Rea a gi g:b) I a ce ai a a a ed f he ab i f SO2 f ai b ea f ae,a e i i he e i e he ga c ai ed 30% SO2 b e a d a i c ac i h a i id c ai i g0.2% SO2 b e. The e e a e a 303 K a d he a e e 1 a . E i a e hei e facia c ce a i a d he ca SO2 a f . The a - a fe c efficie eeca c a ed a FG = 0.002 / 2- , = 0.160 / 2- . The e i ib i SO2 bi i da a a303 K a e (Pe a d Chi , 1973): g SO2/100 g a e Pa ia e e f SO2, Hg ( )0.0 00.5 421.0 851.5 1292.0 1762.5 224S i Defi e: = g SO2/100 g a e = Pa ia e e f SO2, Hg ( )
  • 78. Ini ial g e :
  • 79. 3.13d. Distillation of a mi ture of methanol and ater in a packed to er: use of F-t pemass-transfer coefficients.At a different point in the packed distillation column of Example 3.6, the methanol content of thebulk of the gas phase is 76.2 mole %; that of the bulk of the liquid phase is 60 mole %. Thetemperature at that point in the tower is around 343 K. The packing characteristics and flow ratesat that point are such that FG = 1.542 10 3 kmol/m2-s, and FL = 8.650 10 3 kmol/m2-s.Calculate the interfacial compositions and the local methanol flux. To calculate the latent heats ofvaporization at the new temperature, modify the values given in Example 3.6 using Watsonsmethod (Smith, et al., 1996): For water, Tc = 647.1 K; for methanol, Tc = 512.6 K Solution For methanol (A) For water (B)
  • 80. Pa ame eIni ial e ima e
  • 81. 3.14b. Material balances: adsorption of ben ene vapor on activated carbon.Ac i a ed ca b i ed ec e be e e f a i ge -be e e a i e. A i ge -be e e i e a 306 K a d 1 a c ai i g 1% be e e b ei be a edc ec e a he a e f 1.0 3/ a i g ea f ac i a ed ca b a e e85% f he be e e f he ga i a c i ce . The e e i g ac i a ed ca b c ai15 c 3 be e e a (a STP) ad bed e g a f he ca b . The e e a e a d a e e a e ai ai ed a 306 K a d 1 a . Ni ge i ad bed. The e i ib i ad i f be e e hi ac i a ed ca b a 306 K i e ed a f :Be e e a ad bed Pa ia e e be e e, Hgc 3 (STP)/g ca b15 0.5525 0.9540 1.6350 2.1865 3.2680 4.8890 6.22100 7.83 a) P he e i ib i da a a X = g be e e/ g d ca b , Y = g be e e/ g i ge f a a e e f1a . S i
  • 82. b) Ca c a e he i i f a e e i ed f he e e i g ac i a ed ca b ( e e be ha hee e i g ca b c ai e ad bed be e e).S iO he XY diag a , ca e he i (X2,Y2). Si ce he e a i g i e i ab e he e i ib ic e a d he e i ib i c e i c ca e a d , he i i e a i g i e i b ai ed b ca i g, a he i e ec i f Y = Y1 i h he e i ib i c e, X1 a .
  • 83. c) If the carbon flow rate is 20% above the minimum, what will be the concentration of ben eneadsorbed on the carbon leaving?Solution
  • 84. d) F he c di i f a (c), ca c a e he be f idea age e i ed.S i See e i ec ci he XY g a h 3.15b. Material balances: desorption of ben ene vapor from activated carbon. The ac i a ed ca b ea i g he ad be f P b e 3.14 i ege e a ed b c e c e c ac i h ea a 380 K a d 1 a . The ege e a ed ca b i e ed he ad be , hi e he i e f ea a d de bed be e e a i c de ed. The c de a e e a a e i a ga ic a d a a e ha e a d he ha e a e e a a ed b deca a i . D e he bi i f be e e i ae, f he be e e i be c ce a ed i he ga ic ha e, hi e he a e ha e i c ai ace f be e e. The e i ib i ad i da a a 380 K aea f : Be e e a ad bed Pa ia e e be e e, Pa g be e e/100 g ca b 2.9 1.0 5.5 2.0 12.0 5.0 17.1 8.0 20.0 10.0 25.7 15.0 30.0 20.0 a) Ca c a e he i i ea f ae e i ed. S i E i ib i c e
  • 85. From Problem 3.14 From the XY diagram:b) For a steam flow rate of twice the minimum, calculate the ben ene concentration in the gasmixture leaving the desorber, and the number of ideal stages required. Solution
  • 86. 3.16b. Material balances: adsorption of ben ene vapor on activated carbon; cocurrentoperation.If the adsorption process described in Problem 3.14 took place cocurrentl , calculate theminimum flow rate of activated carbon required.SolutionFom Problem 3.14: From the XY diagram:
  • 87. 3.17b. Material balances in batch processes: dr ing of soap with air.I i de i ed d 10 g f a f 20% i eb eigh e ha 6% i ebc ac i h h ai . The e a i aced i a e e c ai i g 8.06 3 f ai a 350 K, 1 a ,a da ae-a a ia e e f 1.6 Pa. The e i a ed each e i ib i , a d he he ai i he e e i e i e e aced b f e h ai f he igi a i ec e a d e e a e. H a i e he ce be e ea ed i de each he ecified a i ec e f e ha 6%? Whe hi a i e ed ai a 350 K a d 1 a , hee i ib i di ib i f i e be ee he ai a d he a i a f :W % i ei a Pa ia e e f a e , Pa2.40 1.293.76 2.564.76 3.796.10 4.967.83 6.199.90 7.3312.63 8.4215.40 9.5819.02 10.60S iGe e a e he XY diag a
  • 88. From the XY diagram, at the e it of the fifth equilibrium stage, X = 0.06 and3.18b. Material balances in batch processes: e traction of an aqueous nicotine solution
  • 89. ith kerosene.Nicotine in a ater solution containing 2% nicotine is to be e tracted ith kerosene at 293 K.Water and kerosene are essentiall insoluble. Determine the percentage e traction of nicotine if100 kg of the feed solution is e tracted in a sequence of four batch ideal e tractions using 49.0 kgof fresh, pure kerosene each. The equilibrium data are as follo s (Claffe et al., I d. E g. Che .,42, 166, 1950):X, 103 kg nicotine/kg ater Y, 103 kg nicotine/kg kerosene1.01 0.812.46 1.965.02 4.567.51 6.869.98 9.1320.4 18.70Solution From the XY diagram, after 4 e tractions, X = 0.00422
  • 90. 3.19b. Cross-flo cascade of ideal stages.The d i g a d i id- i id e ac i eai de c ibed i P b e 3.17 a d 3.18, e ec i e , a e e a e faf c fig a i ca ed a c -f ca cade. Fig e 3.27 i a che a ic diag a fac -f ca cade f idea age . Each age i e e e ed b a ci c e,a d i hi each age a a fe cc a if i c c e f . The L ha e f f e age he e , bei g c ac ed i each age b a f e h V ha e. If he e i ib i -di ib ic e f he c -f ca cade i e e he e aigh a d f e , i ca be h ha(T e ba , 1980) he e S i he i i g fac , VS/LS, c a f a age , a d N i he a be f age . S e P be 3.18 i ge ai (3-60), a d c a e he e b ai ed b he eh d . S i
  • 91. I i ia e i a e3.20a. Cross-flo cascade of ideal stages: nicotine e traction.C ide he ic i e e ac i fP be 3.18 a d 3.19. Ca c a e he be f idea age e i ed achie e a ea 95% e ac i efficie c .S i U e 8 idea age3.21b. Kremser equations: absorption of h drogen sulfide.A che e f he e a f H2S f af f 1.0 d 3/ f a a ga b c bbi g i h a e a 298 K a d 10 a i bei g c ide ed. The i i ia c ii f he feed ga i 2.5 e e ce H2S. A fi a ga ea c ai i g 0.1 e e ce H2S i de i ed. The ab bi g a e i e e he e f ee f H2S. A he gi e e e a e a d e e, he e if He a , acc di g Yi = 48.3Xi, he e Xi = e H2S/ e f a e ; Yi = eH2S/ e f ai .a) F a c e c e ab be , de e i e he f a e f a e ha i e i ed if 1.5 i e he
  • 92. minimum flo rate is used. Solution at SCb) Determine the composition of the e iting liquid.Solutionc) Calculate the number of ideal stages required.Solution
  • 93. 3.22b. Absorption ith chemical reaction: H2S scrubbing ith MEA.A h i P b e 3-21, c bbi g f h d ge fide f a a ga i g ae i ac ica i ce i e i e a ge a f ae d e he bi i f H2S i a e . If a 2N i f e ha a i e (MEA) i ae i ed a he ab be , h e e , he e i edi id f a e i ed ced d a a ica beca e he MEA eac i h he ab bed H2S i hei id ha e, effec i e i c ea i g i bi i .F hi i e g h a d a e e a e f 298 K, he bi i f H2S ca be a i a edb (de Ne e , N., Air Poll ion Con rol Engineering, 2 d ed., McG a -Hi , B , MA, 2000): Re ea he ca c a i fP be 3.21, b i g a 2N e ha a i e i a ab be . S i
  • 94. 3.23b. Kremser equations: absorption of sulfur dio ide.A flue gas flows at the rate of 10 kmol/s at 298 K and 1 atm with a SO2 content of 0.15 mole %.Ninety percent of the sulfur dioxide is to be removed by absorption with pure water at 298 K. Thedesign water flow rate will be 50% higher than the minimum. Under these conditions, theequilibrium line is (Ben tez, J., P oce Enginee ing and De ign fo Ai Poll ion Con ol,Prentice Hall, Englewood Cliffs, NJ, 1993): where Xi = moles SO2/mole of water; Yi = moles SO2/mole of air. a) Calculate the water flow rate and the SO2 concentration in the water leaving the absorber. Solution b) Calculate the number of ideal stages required for the specified flow rates and percentage SO2 removal. Solution
  • 95. 3.24b. Kremser equations: absorption of sulfur dio ide.An ab o be i a ailable o ea he fl e ga of P oblem 3.23 hich i e i alen o 8.5e ilib i m age .a) Calc la e he a e flo a e o be ed in hi ab o be if 90% of he SO2 i o be emo ed.Calc la e al o he SO2 concen a ion in he a e lea ing he ab o be .Sol ion Ini ial e ima e
  • 96. b) Wha i he e ce age e a f SO2 ha ca be achie ed i h hi ab be if he a e f a e ed i he a e ha a ca c a ed i P b e 3.23 (a)? S i I i ia e i a e3.25b. Kremser equations: liquid e traction.A a e ace ic acid i f a he a e f 1,000 g/h . The i i 1.1% (b eigh )ace ic acid. I i de i ed ed ce he c ce a i f hi i 0.037% (b eigh ) ace icacid b e ac i i h 3-he a a 298 K. The i e 3-he a c ai 0.02% (b eigh ) ace icacid. A e ac i c i a ai ab e hich i e i a e ac e c e ca cade f 15e i ib i age . Wha e f a e i e i ed? Ca c a e he c ii f he e ha e ea i g he c . F hi e , e i ib i i gi e bW a i ace ic acid i e = 0.828 W a i ace ic acid i aeS iLe he a e ha e be he V- ha e; he e ha e i he L- ha e.
  • 97. Initial estimate:3.26c. Co n e c en e c o -flo e ac ion.A 1-butanol acid solution is to be e tracted ith pure ater. The butanol solution contains4.5% (b eight) of acetic acid and flo s at the rate of 400 kg/hr. A total ater flo rate of1005 kg/hr is used. Operation is at 298 K and 1 atm. For practical purposes, 1-butanol and ater are inmiscible. At 298 K, the equilibrium data can be represented b YAi = 0.62 XAi, here YAi is the eight ratio of acid in the aqueous phase and XAi is the eight ratio of acid inthe organic phase.a) If the outlet butanol stream is to contain 0.10% (b eight) acid, ho man equilibriumstages are required for a countercurrent cascade?Solution
  • 98. b) If he a e i i e a a g he a e be f age , b i ac -f ca cade, ha i he e 1-b a c ce a i ( ee P b e 3.19)? S i3.27c. Glucose sorption on an ion e change resin.Chi g a d R h e (AIChE S mp. Ser., 81, . 242, 1985) f d ha he e i ib i fg c e ai e cha ge e i i he ca ci f a i ea f c ce a i be 50 g/L. Thei e i ib ie e i a 303 K i YAi = 1.961 XAi, he e XAi i he g c e c ce a i i he e i .(g fg c e e i e f e i ) a d YAi i he g c e c ce a i i i .(g f g c e e i e f i ).a) We i h bg c e hi i e cha ge e i a 303 K i a c e c e ca cade fidea age . The c ce a i f he feed i i 15 g/L. We a a e c ce a i f 1.0g/L. The i e e i c ai 0.25 g f g c e/L. The feed i f a he a e f 100 L/ i , hi e he e i f a he a e f 250 L/ i . Fi d he be f e i ib i age e i ed.S i
  • 99. b) If 5 equilibrium stages are added to the cascade of part a), calculate the resin flo required tomaintain the same degree of glucose sorption.Solution
  • 100. 4.1a. Void fraction near the alls of packed beds.C ide a c i d ica e e i h a dia e e f 305 ac ed i h id he e i h a dia e e f 50 .a) F e a i (4-1), ca c a e he a ic i f he bed. S i A eb) E i a e he id f ac i a a di a ce f 100 f he a . A e4.2b. Void fraction near the alls of packed beds.Beca e f he ci a a e f he id-f ac i adia a ia i f ac ed bed , he e a e a be f ca i c e he a he e he ca id f ac i i e ac e a he a ic a e f he bed. F he bed de c ibed i E a e 4.1, ca c a e he di a ce f he a hefi fi e ch ca i . S i The ca i ea he a e a he a ic i he J0( d) = 0. F E a e 4.1, I i ia e i a e f he ca be b ai ed f Fig. 4.4 I i ia e i a e
  • 101. I i ia e i a e I i ia e i a e I i ia e i a e I i ia e i a e4.3c. Void fraction near the alls of packed beds.(a) Sh ha he adia ca i f he a i a a d i i a f he f ci de c ibed b E ai(4-1) a e he f he e a i S ib) F he ac ed bed f E a e 4.1, ca c a e he adia ca i f he fi fi e a i a, a d f he fi fi e i i a; ca c a e he a i de f he id f ac i ci a i a h e i .S i
  • 102. Ini ial e ima e of he oo can be ob ained f om Fig. 4.4 Ini ial e ima e Ini ial e ima e Repea ing hi p oced e, he follo ing e l a e ob ained: Ma ima: Minima , mm ( *) Ampli de, % ,mm ( *) Ampli de, % 20.8 (1.04) 37.2 11.3 (0.57) -56.7 39.6 (1.98) 21.1 30.2 (1.51) -27.3 58.3 (2.92) 13.5 49.0 (2.45) -16.7 77.1 (3.85) 9.2 67.7 (3.39) -11.1 95.8 (4.79) 6.4 86.4 (4.32) -7.6c) Calc la e he di ance f om he all a hich he ab ol e al e of he po o i fl c a ion habeen dampened o le han 10% of he a mp o ic bed po o i .Sol ionF om he e l of pa (b), hi m happen a * be een 3.39 and 3.85
  • 103. (d) What fraction of the cross-sectional area of the packed bed is characterized by porosityfluctuations which are within 10% of the asymptotic bed porosity?SolutionFrom part (c)(f) For the packed bed of Example 4.1, estimate the average void fraction by numerical integration ofequation (4-65) and estimate the ratio a / b .Solution4.4c. Void fraction near the alls of annular packed beds.Annular packed beds (APBs) involving the flow of fluids are used in many technical and engineeringapplications, such as in chemical reactors, heat exchangers, and fusion reactor blankets. It is wellknown that the wall in a packed bed affects the radial void fraction distribution. Since APBs have twowalls that can simultaneously affect the radial void fraction distribution, it is essential to include thisvariation in transport models. A correlation for this purpose was recently formulated (Mueller, G. E.,AIChE J., 45, 2458-60, Nov. 1999). The correlation is restricted to randomly packed beds in annularcylindrical containers of outside diameter Do, inside diameter Di, equivalent diameter De = Do Di,
  • 104. consisting of equal-sized spheres of diameter dp, with diameter aspect ratios of 4 De/dp 20. Thecorrelation isConsider an APB with outside diameter of 140 mm, inside diameter of 40 mm, packed with identical10-mm diameter spheres.(a) Estimate the void fraction at a distance from the outer wall of 25 mm.Solution (b) Plot the void fraction, as predicted by Eq. (4-66), for r* from 0 to R*.
  • 105. (c) Sho ha he a e age po o i fo an APB i gi en b (d) E ima e he a e age po o i fo he APB de c ibed abo e. Sol ion4.5a. Minimum liquid mass velocit for proper wetting of packing.A 1.0-m diame e bed ed fo ab o p ion of ammonia i h p e a e a 298 K i packed i h25-mm pla ic In alo addle . Calc la e he minim m a e flo a e, in kg/ , neded o en ep ope e ing of he packing face.Sol ion Fo pla ic packing, L,min = 1.2 mm/ . F om S eam Table4.6a. Minimum liquid mass velocit for proper wetting of packing.Repea P oblem 4.5, b ing ce amic in ead of pla ic In alo addle .Sol ion Fo ce amic packing, L,min = 0.15 mm/ .
  • 106. From Steam Tables4.7b. S ecific i id h d a d id f ac i i fi -ge e a i a d ac i g.Repeat Example 4.2, but using 25-mm ceramic Berl saddles as packing material.Solution From Table 4.1: From Example 4.2:4.8b. S ecific i id h d a d id f ac i i c ed ac i g.Repeat Example 4.2, but using Montz metal B1-200 structured packing (very similar to the oneshown in Figure 4.3). For this packing, a = 200 m 1, = 0.979, Ch = 0.547 (Seader and Henley,1998).Solution From Example 4.2:
  • 107. 4.9b. S ecific i id h d a d id f ac i i fi -ge e a i a d ac i g.A e ac ed i h 25- ce a ic Ra chig i g i be ed f ab bi g be e e a f adi e i e i h a i e ga i g a a h i a 300 K. The i c i f he i i 2.0 cP a d ide i i 840 g/ 3. The i id a e ci i L = 2.71 g/ 2- . E i a e he i id h d , he id f ac i , a d he h d a ic ecific a ea f he ac i g.S i F Tab e 4.1:
  • 108. 4.10b. Pressure drop in beds packed ith first-generation random packings.Repeat E ample 4.3, but using 15-mm ceramic Raschig rings as packing material. Assume that, forthis packing, C = 1.783.SolutionPacked Col mn Design Program This program calculates the diameter of a packed column to satisf a given pressure drop criterium, and estimates the volumetric mass-transfer coefficients. Enter data related to the gas and liquid streams Enter liquid flow rate, mL, in kg/s Enter gas flow rate, mG, in kg/s Enter liquid densit , in kg/m3 Enter gas densit , kg/m3 Enter liquid viscosit , Pa-s Enter gas viscosit , Pa-s Enter temperature, T, in K Enter total pressure, P, in Pa Enter data related to the packing Enter packing factor, Fp, in ft2/ft3 Enter specific area, a, m2/m3 Introduce a units conversion factor in Fp Enter porosit , fraction Enter loading constant, Ch Enter pressure drop constant, Cp Enter allowed pressure drop, in Pa/m Calculate flow parameter, X
  • 109. Calculate Y at flooding conditions Calculate gas velocit at flooding, vGfAs a first estimate of the column diameter, D, design for 70% of flooding Calculate gas volume flow rate, QG, in m3/s Calculate liquid volume flow rate, QL, in m3/s Calculate effective particle si e, dp, in m
  • 110. I e a e o find he o e diame e fo he gi en p e e d op Col mn diame e , in me e F ac ional app oach o flooding
  • 111. 4.11b. P e e d op and app oach o flooding in c ed packing.Repeat Example 4.3, but using Montz metal B1-200 structured packing (very similar to the oneshown in Figure 4.3). For this packing, Fp = 22 ft2/ft3, a = 200 m 1, = 0.979, Ch = 0.547, Cp =0.355 (Seader and Henley, 1998).SolutionPacked Col mn Design Program This program calculates the diameter of a packed column to satisf a given pressure drop criterium, and estimates the volumetric mass-transfer coefficients. Enter data related to the gas and liquid streams Enter liquid flow rate, mL, in kg/s Enter gas flow rate, mG, in kg/s Enter liquid densit , in kg/m3 Enter gas densit , kg/m3 Enter liquid viscosit , Pa-s Enter gas viscosit , Pa-s Enter temperature, T, in K Enter total pressure, P, in Pa Enter data related to the packing Enter packing factor, Fp, in ft2/ft3 Enter specific area, a, m2/m3 Introduce a units conversion factor in Fp Enter porosit , fraction Enter loading constant, Ch Enter pressure drop constant, Cp Enter allowed pressure drop, in Pa/m Calculate flow parameter, X
  • 112. Calculate Y at flooding conditions Calculate gas velocit at flooding, vGfAs a first estimate of the column diameter, D, design for 70% of flooding Calculate gas volume flow rate, QG, in m3/s Calculate liquid volume flow rate, QL, in m3/s Calculate effective particle si e, dp, in m
  • 113. I e a e o find he o e diame e fo he gi en p e e d op Col mn diame e , in me e F ac ional app oach o flooding4.12b, d. Pressure drop in beds packed ith second-generation random packings.A packed to er is to be designed for the countercurrent contact of a ben ene-nitrogen gas mi ture ith kerosene to ash out the ben ene from the gas. The gas enters the to er at the rate of 1.5m3/s, measured at 110 kPa and 298 K, containing 5 mole % ben ene. Esentiall , all the ben ene is
  • 114. ab bed b he e e e. The i id e e he e a he a e f 4.0 g/ ; he i id de i i800 g/ 3, i c i i 2.3 cP. The ac i g i be 50- e a Pa i g , a d he e dia e e i be ch e d ce a ga - e ed f 400 Pa/ f i iga ed ac i g.(a) Ca c a e he e dia e e be ed, a d he e i g f ac i a a ach f di g.(b) A e ha , f he dia e e ch e , he i iga ed ac ed heigh i be 5 a d ha 1 f i iga ed ac i g i be aced e he i id i e ac a e ai e e a a . The b e - c bi a i be ed a he ga i e i ha e a ea echa ica efficie c f 60%.Ca c a e he e e i ed b he ga h gh he ac i g.S i(a) De ig f c di i a he b f he e he e he a i f f ga a d i id cc Be e e e e i g i h he ga : A i g ha a f he be e e i ab bed: F he L ca eh df i e f ga e : U i g he Pac ed C De ig P g a : D = 0.913 f = 0.825 (b) Ca c a e he e ed h gh he d ac i g
  • 115. From the Lucas method for mixtures of gases: From equation (4-11):(c) Estimate the volumetric mass-transfer coefficients for the gas and liquid phases. Assume that DL= 5.0 10 10 m2/s. From the wilke-Lee equation, DG = 0.0885 cm2/s From Table 4.2, CL = 1.192, CV = 0.410 the Packed Column Design Program: Using k Lah = 0.00675 s 1; k y ah = 0.26 kmol/m3-s4.13b, d. Pressure drop in beds packed ith structured packings.Redesign the packed bed of Problem 4.12, but using Montz metal B1-200 structured packing (verysimilar to the one shown in Figure 4.3). For this packing, Fp = 22 ft2/ft3, a = 200 m 1, e = 0.979, Ch= 0.547, Cp = 0.355, CL = 0.971, CV = 0.390 (Seader and Henley, 1998).Solution Using the Packed Column Design Program: D = 0.85 m f = 0.859 k Lah = 0.00861 s 1; k y ah = 0.376 kmol/m3-s4.14c, d. Air stripping of aste ater in a packed column.A wastewater stream of 0.038 m3/s, containing 10 ppm (by weight) of benzene, is to be stripped
  • 116. with air in a packed column operating at 298 K and 2 atm to reduce the benzene concentration to0.005 ppm. The packing specified is 50-mm plastic Pall rings. The air flow rate to be used is 5 timesthe minimum. Henrys law constant for benzene in water at this temperature is 0.6 kPa-m3/mole(Davis and Cornwell, 1998). Calculate the tower diameter if the gas-pressure drop is not to exceed500 Pa/m of packed height. Estimate the corresponding mass-transfer coefficients. The diffusivity ofbenzene vapor in air at 298 K and 1 atm is 0.096 cm2/s; the diffusivity of liquid benzene in water atinfinite dilution at 298 K is 1.02 10 5 cm2/s (Cussler, 1997). SolutionCalculate m, the slope of the equilibrium curve:For water at 298 K, Calculate the minimum air flow rate: Convert liquid concentrations from ppm to mole fractions At these low concentrations the equilibrium and operating lines are straight, and y 2 (max) = y 2* = mx 2
  • 117. Using the Packed Column Design Program:D = 1.145 m f = 0.766 k Lah = 0.032 s 1; k y ah = 0.335 kmol/m3-s
  • 118. 4.15b. Stripping chloroform from water b sparging with air.Repeat E ample 4.5, but using an air flo rate that is t ice the minimum required.Solution Initial estimate of the column height, Z
  • 119. Ini ial e ima e of ga hold p
  • 120. Calculate po er required
  • 121. 4.16b. Stripping chloroform from water b sparging with air.Repeat Example 4.5, but using the same air flow rate used in Problem 4.15, and specifying achloroform removal efficiency of 99%.Solution For 99% removal efficiency, xin/xout = 100 Z = 1.30 m W T(Z) = 5.0 kW 4.17b. Stripping chlorine from water b sparging with air. A vessel 2.0 m in diameter and 2.0 m deep (measured from the gas sparger at the bottom to liquid overflow at the top) is to be used for stripping chlorine from water by sparging with air. The water will flow continuously downward at the rate of 7.5 kg/s with an initial chlorine concentration of 5 mg/L. Airflow will be 0.22 kg/s at 298 K. The sparger is in the form of a ring, 25 cm in diameter, containing 200 orifices, each 3.0 mm in diameter. Henrys law constant for chlorine in water at this temperature is 0.11 kPa-m3/mole (Perry and Chilton, 1973). The diffusivity of chlorine at infinite dilution in water at 298 K is 1.25 10 5 cm2/s (Cussler, 1997). (a) Assuming that all the resistance to mass transfer resides in the liquid phase, estimate the chlorine removal efficiency achieved. Solution
  • 122. Ini ial e ima e of ga hold p
  • 123. Ieai g i Z = 2.0 : (b) Ca c a e e e i ed4.18c,d. Batch aste ater aeration using spargers.I he ea e f a e a e , de i ab e ga e a e f e e i ed de bed f he ae,a d ge i ad bed i he a e he b bb e f ai a e di e ed ea he b fae a i a d . A he b bb e i e, e ca be a fe ed f he ga he i idf he i id he ga de e di g he c ce a i d i i g f ce.
  • 124. F ba ch ae a i i ac a e a ,a ge a -ba a ce ca be i e a he e c A * i he ge a a i c ce a i . I eg a i g be ee he i e i i e a d a d he c e di g di ed ge c ce a i i i c A,0 a d c A, ; a i g ha c A * e ai e e ia c a , a d ha a he e i a ce a a fe e ide i he i id ha e: I ae a i a , he e ai i e ea ed a a i c ea ed i id de h, he bi i f ge i i f e ced b h b he i c ea i g e e f he ai e e i g he ae a i a a d b he dec ea i g ge a ia e e i he ai b bb e a ge i ab bed. F he e ca e , he e f a ea a a i a ec e di g he ae a i a idde h i gge ed (Ec e fe de , W. W. J ., Ind ial Wa e Poll ion Con ol, 3 d ed., McG a -Hi , B , Ma, 2000): he e c = a a i di ed ge c ce a i i f e h a e e ed a he ic ai a 101.3 Pa c ai i g 20.9% ge , P = ab e e e a he de h f ai e ea e, P =a he ic e e, O = a ge e ce i he ai ea i g he ae a i a . The a ge e ce i he ai ea i g he ae a i a i e a ed he ge a fe efficie c , Oeff , h gh he e C ide a 567 3 ae a i d ae a ed i h 15 a ge , each i g c e ed ai a a a e f 0.01 g/ . Each a ge i i he f f a i g, 100 c i dia e e , c ai i g 20 ifice , each 3.0 i dia e e . The a ge i be ca ed 5 be he face f he d. The a e e e a e i 298 K; a he ic c di i a e 298 K a d 101.3 Pa. U de he e c di i , c = 8.38 g/L (Da i a d C e , 1998). (a) E i a e he e ic a - a fe c efficie f he e c di i f e ai (4-23) a d (4-25). (b) E i a e he i e e i ed ai e he di ed ge c ce a i f 0.5 g/L 6.0 g/L, a d ca c a e he e i g ge a fe efficie c . (c) E i a e he e e i ed e a e he 15 a ge , if he echa ica efficie c f he c e i 60%. S i
  • 125. Ini ial e ima e of ga hold p
  • 126. Initial estimate of transfer efficienc(b) Calculate power required
  • 127. 4.19c,d. Batch aste ater aeration using spargers; effect of liquid depth. C ide he i a i de c ibed i P b e 4.18. Acc di g Ec e fe de (2000), f e f b bb e-diff i ae a i e he e ic a - a fe c efficie i a ih i id de h Z acc di g he e a i hi he e he e e ha a a e ea 0.7 f e . F he ae a i d fP be 4.18, ca c a e La a a e f Z = 3 , 4 , 6 , a d 7 . E i a e he c e di g a e f f eg e i a a i f he e . Hi : Re e be ha he a e f he d e ai c a , he ef e he c - ec i a a ea f he d cha ge a he a e de h cha ge . S i U i g he ga de e ed i P b 4.18, he f i g ec f e i ge e a ed4.20c. Flooding conditions in a packed cooling to er.Ac i g e , 2 i dia e e , ac ed i h 75- ce a ic Hif i g , i fed i h a e a 316K a a a e f 25 g/ 2- . The a e i c ac ed i h ai , a 300 K a d 101.3 Pa e e ia d ,da adc ec e he a e f . Neg ec i g e a a i f he a e a d cha gei he ai e e a e, e i a e he e ic a e f ai f , i 3/ , hich d f d he e.S i LMV = i id a e ci GMV = ga a e ci
  • 128. In od ce a ni con e ion fac o in FpIni ial e ima e of GMV 4.21c,d. Design of a sieve-tra column for ethanol absorption Repea he calc la ion of E ample 4.6, 4.7, 4.8, and 4.9 fo a col mn diame e co e ponding o 50% of flooding. Sol ion F om he Sie e-Pla e De ign P og am, he follo ing e l a e ob ained fo = 0.5 D = 1.176 m = 0.6 m P = 590 Pa/ a F o de No. = 1.22 (no e ce i e eeping) E = 0.0165 EOG = 0.8123 EMG = 0.8865 EMGE = 0.878
  • 129. 4.22c,d. Design of a sieve-tra column for aniline stripping. A sieve-tray tower is to be designed for stripping an aniline (C6H7N)-water solution with steam. The circumstances at the top of the tower, which are to be used to establish the design, are: Temperature = 371.5 K Pressure = 100 kPa Liquid: Rate = 10.0 kg/s Composition = 7.00 mass % aniline Density = 961 kg/m3 Viscosity = 0.3 cP Surface tension = 58 dyne/cm Diffusivity = 4.27 10 5 cm2/s (est.) Foaming factor = 0.90 Vapor: Rate = 5.0 kg/s Composition = 3.6 mole % aniline Density = 0.670 kg/m3 Viscosity = 118 mP (est.) Diffusivity = 0.116 cm2/s (est.) The equilibrium data at this concentration indicates that = 0.0636 (Treybal, 1980). (a) Design a suitable cross-flow sieve-tray for such a tower. Take do = 5.5 mm on an equilateral- triangular pitch 12 mm between hole centers, punched in stainless steel sheet metal 2 mm thick. Use a weir height of 40 mm. Design for a 75% approach to the flood velocity. Report details respecting tower diameter, tray spacing, weir length, gas-pressure drop, and entrainment in the gas. Check for excessive weeping. Solution For aniline, the molecular weight = 93. The average molecular weight of the gas = 20.7. The composition of the liquid is calculated as 1.4 mole % aniline, the average molecular weight = 19.1. From the Sieve-Plate Design Program, the following results are obtained for f = 0.75 D = 1.93 m t = 0.6 m DP = 435 Pa/tray Froude No. = 0.93 (no excessive weeping) E = 0.045 Weir length = 1.40 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a). Solution EOG = 0.716 EMG = 0.722 EMGE = 0.7204.23c,d. Design of a sieve-tra column for aniline stripping.Repeat Problem 4.22, but for a 45% approach to flooding. Everything else remains the same as inProblem 4.22.Solution From the Sieve-Plate Design Program, the following results are obtained for f = 0.45
  • 130. D = 2.49 m t = 0.6 m DP = 399 Pa/tray Froude No. = 0.491 ( excessive weeping) E = 0.0173 Weir length = 1.81 m(b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).Solution EOG = 0.660 EMG = 0.666 EMGE = 0.6654.24c,d. Design of a sieve-tra column for methanol stripping.A dilute aqueous solution of methanol is to be stripped with steam in a sieve-tray tower. Theconditions chosen for design are:Temperature = 368 K Pressure = 101.3 kPaLiquid:Rate = 0.25 kmol/s Composition = 15.0 mass % methanolDensity = 961 kg/m3 Viscosity = 0.3 cPSurface tension = 40 dyn/cmDiffusivity = 5.70 10 5 cm2/s (est.) Foaming factor = 1.0Vapor:Rate = 0.1 kmol/s Composition = 18 mole % methanolViscosity = 125 mP (est.) Diffusivity = 0.213 cm2/s (est.)The equilibrium data at this concentration indicates that = 2.5 (Perry and Chilton, 1973).(a) Design a suitable cross-flow sieve-tray for such a tower. Take do = 6.0 mm on an equilateral-triangular pitch 12 mm between hole centers, punched in stainless steel sheet metal 2 mm thick.Use a weir height of 50 mm. Design for 80% approach to the flood velocity. Report detailsrespecting tower diameter, tray spacing, weir length, gas-pressure drop, and entrainment in thegas. Check for excessive weeping.Solution From the Sieve-Plate Design Program, the following results are obtained for f = 0.80 D = 1.174 m t = 0.6 m DP = 386 Pa/tray Froude No. = 0.878 (no excessive weeping) E = 0.052 Weir length = 0.853 m(b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).Solution
  • 131. EOG = 0.5987 EMG = 0.707 EMGE = 0.700 4.25c,d. Sieve-tra column for methanol stripping; effect of hole si e. Repeat Problem 4.24, but changing the perforation size to 4.5 mm, keeping everything else constant. Solution From the Sieve-Plate Design Program, the following results are obtained for = 0.80 D = 1.174 m t = 0.6 m DP = 686 Pa/tray Froude No. = 1.561 (no excessive weeping) E = 0.076 Weir length = 0.853 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a). Solution EOG = 0.793 EMG = 0.936 EMGE = 0.9184.26c,d. Design of sieve-tra column for butane absorption.A gas containing methane, propane, and n-butane is to be scrubbed countercurrently in a sieve-traytower with a hydrocarbon oil to absorb principally the butane. It is agreed to design a tray for thecircumstances existing at the bottom of the tower , where the conditions are:Temperature = 310 K Pressure = 350 kPaLiquid:Rate = 0.50 kmol/s Average molecular weight = 150Density = 850 kg/m3 Viscosity = 1.6 cPSurface tension = 25 dyn/cmDiffusivity = 1.14 10 5 cm2/s (est.) Foaming factor = 0.9Vapor:Rate = 0.3 kmol/s Composition = 86% CH4, 12% C3H8, 2% C4H10Viscosity = 113 mP (est.) Diffusivity = 0.035 cm2/s (est.)The system obeys Raoults law; the vapor pressure of n-butane at 310 K is 3.472 bar (Reid et al.,1987).(a) Design a suitable cross-flow sieve-tray for such a tower. According to Bennett and Kovak (2000),the optimal value of the ratio Ah/Aa is that which yields an orifice Froude number, Fro = 0.5. Designfor the optimal value of do on an equilateral-triangular pitch 12 mm between hole centers, punched instainless steel sheet metal 2 mm thick. Use a weir height of 50 mm. Design for a 75% approach tothe flood velocity. Report details respecting tower diameter, tray spacing, weir length, gas-pressuredrop, and entrainment in the gas.Solution
  • 132. The slope of the equilibrium curve is approximately 1.0. The average molecular weight of the gas is 20.2. By trial-and-error, using the Sieve-Plate Design Program, it was found that the orifice diameter that results in a Froude number = 0.5 is do = 3.43 mm. From the Sieve-Plate Design Program, the following results are obtained for = 0.75 D = 2.73 m t = 0.6 m DP = 811 Pa/tray Froude No. = 0.5 (no excessive weeping) E = 0.013 Weir length = 2.28 m(b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).Solution EOG = 0.833 EMG = 1.038 EMGE = 1.034 4.27c,d. Design of sieve-tra column for ammonia absorption. A process for making small amounts of hydrogen by cracking ammonia is being considered, and residual uncracked ammonia is to be removed from the resulting gas. The gas will consist of H2 and N2 in the molar ratio 3:1, containing 3% NH3 by volume. The ammonia will be removed by scrubbing the gas countercurrently with pure liquid water in a sieve-tray tower. Conditions at the bottom of the tower are: Temperature = 303 K Pressure = 200 kPa Liquid: Rate = 6.0 kg/s Average molecular weight = 18 Density = 996 kg/m3 Viscosity = 0.9 cP Surface tension = 68 dyn/cm Diffusivity = 2.42 10 5 cm2/s (est.) Foaming factor = 1.0 Vapor: Rate = 0.7 kg/s Viscosity = 113 mP (est.) Diffusivity = 0.230 cm2/s (est.) For dilute solutions, NH3-H2O follows Henrys law, and at 303 K the slope of the equilibrium curve is m = 0.85 (Treybal, 1980). (a) Design a suitable cross-flow sieve-tray for such a tower. Take do = 4.75 mm on an equilateral- triangular pitch 12.5 mm between hole centers, punched in stainless steel sheet metal 2 mm thick. Use a weir height of 40 mm. Design for an 80% approach to the flood velocity. Report details respecting tower diameter, tray spacing, weir length, gas-pressure drop, and entrainment in the gas. Check for excessive weeping. Solution From the Sieve-Plate Design Program, the following results are obtained for = 0.80 D = 0.775 m t = 0.5 m DP = 624 Pa/tray
  • 133. Froude No. = 1.033 (no excessive weeping) E = 0.048 Weir length = 2.28 m(b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).Solution EOG = 0.806 EMG = 0.85 EMGE = 0.847 4.28c,d. Design of sieve-tra column for toluene-meth lc clohe ane distillation. A sieve-tray tower is to be designed for distillation of a mixture of toluene and methylcyclohexane. The circumstances which are to be used to establish the design, are: Temperature = 380 K Pressure = 98.8 kPa Liquid: Rate = 4.8 mole/s Composition =48.0 mol % toluene Density = 726 kg/m3 Viscosity = 0.22 cP Surface tension = 16.9 dyne/cm Diffusivity = 7.08 10 5 cm2/s (est.) Foaming factor = 0.80 Vapor: Rate = 4.54 mole/s Composition =44.6 mole % toluene Density = 2.986 kg/m3 Viscosity = 337 mP (est.) Diffusivity = 0.0386 cm2/s (est.) The equilibrium data at this concentration indicates that m = 1.152. (a) Design a suitable cross-flow sieve-tray for such a tower. Take do = 4.8 mm on an equilateral- triangular pitch 12.7 mm between hole centers, punched in stainless steel sheet metal 2 mm thick. Use a weir height of 50 mm. Design for a 60% approach to the flood velocity. Report details respecting tower diameter, tray spacing, weir length, gas-pressure drop, and entrainment in the gas. Check for excessive weeping. (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a). Sol ion a) From the Sieve-Plate Design Program, the following results are obtained for = 0.60 D = 0.6 m t = 0.5 m DP = 295 Pa/tray Froude No. = 0.585 (no excessive weeping) E = 0.007 Weir length = 0.436 m(b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).Solution EOG = 0.611 EMG = 0.712 EMGE = 0.710
  • 134. 5.1a. Overall tra efficienc C ide a a ab be i h a c a M h ee efficie c EMGE = 0.75, a d a a e age ab i fac A = 1.25. a) E i a e he e a a efficie c . S i b) If he ab be e i e 5.34 e i ib i age , ca c a e he be f ea a . S i U e8 a5.3a. Ammonia stripping from a wastewater in a tra tower.A a e idi g i e i ib i age i ed f i i ga ia f a a e ae ea b ea fc e c e ai a 1 a a d 300 K. Ca c a e he c ce a i fa ia i he e i a e if he i e i id c ce a i i 0.1 e%a ia, he i e ai if ee f a ia, a d 1.873 a da d c bic e e f ai a e fed he e e i ga f a e a e . The e i ib i da a f hi e , i hi a ge f c ce a i a d 300 K, cabe e e e ed b A,i= 1.414 A,i (Ki g, C. J., Sepa a ion P oce e , McG a -Hi , Ne Y ,NY, 1971).S i
  • 135. Initial estimate5.4a. Ammonia stripping from a wastewater in a tra tower.The Murphree plate efficienc for the ammonia stripper of Problem 5.3 is constant at 0.581.Estimate the number of real tra s required.Solution Use 9 tra s 5.5b. Ammonia stripping from a wastewater in a tra tower. If the air flow rate to the absorber of Problems 5.3 and 5.4 is reduced to 1.5 standard m3/kg of
  • 136. a e , ca c a e c ce a i fa ia i he e i a e if 9 ea a ae ed a d he M h ee efficie c e ai c a a 58.1%. S i I i ia e i a e5.6b. Absorption of an air pollutant in a tra tower.A hea - i ea a 320 K i ed i a ab be e e di e a i ie f a Afa ai ea . The hea i i he ec c ed bac he ce he e he e A i i ed. The ce i bei g a i a ba i , a d i f ai f ca e- i de i ed. The c eab be i a 16- a e ie e- a c . Pi a da a a e a f :Li id f a e = 5.0 e /hGa f a e = 2.5 e /h A,i = 0.04 A, = 0.0001 A,i = 0.0001E i ib i f A i gi e a A,i = 0.7 A,i. Fi d he e a c efficie c , a d he M h ee a e efficie c . a e ha he i id a d ga f ae ae gh c a .S i
  • 137. 5.7b. Absorption of ammonia in a laborator -scale tra tower.An absorption column for laboratory use has been carefully constructed so that it has exactly 4equilibrium stages, and is being used to measure equilibrium data. Water is used as the solventto absorb ammonia from air. The system operates isothermally at 300 K and 1 atm. The inletwater is pure distilled water. The ratio of L/V = 1.2, inlet gas concentration is 0.01 mole fractionammonia; the measured outlet gas concentration is 0.0027 mole fraction ammonia. Assumingthat Henry s law applies, calculate the slope of the equilibrium line.Solution Initial estimate
  • 138. 5.8c,d. Absorption of ammonia in a seive-tra tower.A process for making small amounts of hydrogen by cracking ammonia is being consideredand residual, uncracked ammonia is to be removed from the resulting gas. The gas will consistof H2 and N2 in the molar ratio 3:1, containing 3% NH3 by volume, at a pressure of 2 bars anda temperature of 303 K.There is available a sieve-tray tower, 0.75 m diameter, containing 6 cross-flow trays at 0.5 mtray spacing. The perforations are 4.75 mm in diameter, arranged in triangular pitch on 12.5mm centers, punched in sheet metal 2 mm thick. The weir height is 40 mm. Assumeisothermal scrubbing with pure water at 303 K. The water flow rate to be used should notexceed 50% of the maximum recommended for cross-flow sieve trays which is 0.015 m3/s-mof tower diameter (Treybal, 1980). The gas flow rate should not exceed 80% of the floodingvalue.a) Estimate the gas flow rate that can be processed in the column under the circumstancesdescribed above.b) Calculate the concentration of the gas leaving the absorber in part a).Data:Liquid density = 996 kg/m3 Surface tension = 0.068 N/mFoaming factor = 1.0 Liquid diffusivity = 2.42 10 9 m2/sGas viscosity = 1.122 10 5 kg/m-s Gas diffusivity = 0.23 cm2/sSlope of the equilibrium line, = 0.85Solutiona) Run the sieve-tray design program of Appendix E with different values of gas flow rate until D = 0.75 m at f = 0.8. The results are: mG = 0.657 kg/s Gas pressure drop = 615 Pa per tray
  • 139. EMGE = 0.841 F o = 1.042 E = 0.048b) Ini ial e ima e
  • 140. 5.9b. Absorption of carbon dio ide in a bubble-cap tra to er.A a a fac i g d ice i b c e i ai d ce a f e ga hich, he c ea ed a dc ed, i c ai 15% CO2, 6% O2, a d 79% N2. The ga i be b i a b bb e-ca e c bbe a 1.2 a a d 298 K, be c bbed c ec e i h a 30 % e ha a i e(C2H7ON) a e i e e i g a 298 K. The c bbi g i id, hich i ec c ed f a i e , i c ai 0.058 e CO2/ i . The ga ea i g he c bbe i c ai 2%CO2. A i id- -ga a i f 1.2 i e he i i i ecified. A e i he a eai .A298 K a d 1.2 a , he e i ib i e f ac i f ca b di ide e a e i f e ha a i e (30 %) i gi e b he e A,i i he e f ac i f CO2 i he i id i . a) Ca c a e he i g a f i e e i g he e e c bic ee f e e i g ga . S i F he i i i id f a e, 1 a i be i e i ib i i h N1 I i ia e i a e
  • 141. Calculate molecular weight of MEA solution Basis: 100 gm of 30% MEA in waterb) Determine the number of theoretical tra s required for part a)Solution Generate XY diagram
  • 142. c) The monoethanolamine solution has a viscosit of 6.0 cP and a densit of 1,012 kg/m3.Estimate the overall tra efficienc for the absorber, and the number of real tra s required.Seader and Henle (1998) proposed the following empirical correlation to estimate the overallefficienc of absorbers and strippers using bubble-cap tra s (it has also been used to obtainrough estimates for sieve-tra towers):where: EO = overall fractional efficiencm = slope of equilibrium curvemL = liquid viscosit , in cP 3 L = liquid densit , in kg/mHint: In this problem, the equilibrium-distribution curve is not a straight line, therefore m is notconstant. Estimate thhe average value of m at liquid concentrations along the operating lineand use theaverage in the correlation given.Solution
  • 143. 5.11c,d. Absorption of carbon disulfide in a sieve-tra tower.Carbon disulfide, CS2, used as a solvent in a chemical plant, is evaporated from the product in adryer into an inert gas (essentially N2) in order to avoid an explosion hazard. The CS2-N2 mixtureis to be scrubbed with an absorbent hydrocarbon oil. The gas will flow at the rate of 0.4 m3/s at297 K and 1 atm. The partial pressure of CS2 in the original gas is 50 mm Hg, and the CS2concentration in the outlet gas is not to exceed 0.5%. The oil enters the absorber essentially pureat a rate 1.5 times the minimum, and solutions of oil and CS2 follow Raoult s law. Design asieve-tray tower for this process. Design for a gas velocity which is 70% of the flooding velocity.Assuming isothermal operation, determine:a) Liquid flow rate, kg/sData (at 297 K):Oil average molecular weight = 254 Oil viscosity = 4 cPOil density = 810 kg/m3 Surface tension = 0.030 N/mFoaming factor = 0.9 CS2 vapor pressure = 346 mm HgSolution
  • 144. Ini ial e ima e
  • 145. b) To er diameter and plate spacingSolution Plate design conditions: f = 0.7 4.5-mm holes in triangular pitch; 12 mm bet een centers Weir height = 50 mm; Plate thickness = 2 mm From the Mathcad program in Appendi E: D = 0.705 m; t = 0.5 m; Pressure drop = 414 Pa/tra Fro = 0.712 E = 0.018 EMGE = 0.588 d) Number of real tra s required Solution Appro imation, assuming straight equilibrium and operating lines
  • 146. Graphical solution: Generate operating line in diagram
  • 147. Graphicall , the number of ideal stages is also slightl over 5
  • 148. Use 10 tra se) Total gas-pressure drop.Solution
  • 149. 5.12c. Adiabatic absorption of carbon disulfide in a sieve-tra tower.Determine the number of equilibrium tra s for the absorber of Problem 5.11, assuming adiabaticoperation. The specific heats are:Substance J/mole-KOil 362.2CS2, liquid 76.2N2 29.1CS2, gas 46.9the latent heat of vapori ation of CS2 at 297 K is 27.91 kJ/mole. The vapor pressure of CS2 as afunction of temperature is given b Solution Base for enthalpies: liquid oil, liquid CS2, gaseous N2 at 297 K Assume T1 = 298.0 K
  • 150. Appro imatel 12 ideal stagesGas outlet temperature = 298 K
  • 151. 5.13b. Steam-stripping of ben ene in a sieve-plate column.A straw oil used to absorb benzene from coke-oven gas is to be steam-stripped in a sieve-platecolumn at atmospheric pressure to recover the disolved benzene, C6H6. Equilibrium conditions atthe operating temperature are approximated by Henry s law such that, when the oil phase contains10 mole % benzene, the equilibrium benzene partial pressure above the oil is 5.07 kPa. The oil maybe considered nonvolatile. It enters the stripper containing 8 mole % benzene, 75% of which is to beremoved. The steam leaving contains 3 mole % C6H6.a) How many theorethical stages are required?b) How many moles of steam are required per 100 moles of the oil-benzene mixture?c) If 85% of the benzene is to be recovered with the same steam and oil rates, how many theoreticalstages are required? Solution a)
  • 152. b) Ke e e aic) (G a hica c ci ahead)
  • 153. Kremser equation5.15c,d. Absorption of carbon disulfide in a random-packed to er.Design a to er packed ith 50-mm ceramic Hiflo rings for the carbon disulfide scrubber ofProblem 5.11. Assume isothermal operation and use a liquid rate of 1.5 times the minimum anda gas-pressure drop not e ceeding 175 Pa/m of packing. Calculate the to er diameter, packedheight, and total gas-pressure drop. Assume that Ch for the packing is 1.0.SolutionFrom Prob. 5.11, the conditions at the bottom of the absorber are:
  • 154. From Tables 4.1 and 4.2Run the Packed To er Design Program of Ap. DResults:Estimate the packed height; assume dilute solutionsFrom Prob 5.11: Calculate HtOG at the bottom of the to er
  • 155. Calculate HtOG at the top of the tower From Prob. 5.11, the conditions at the top of the absorber are: The mass-tranfer coefficients remain fairl constant along the tower
  • 156. 5.16b,d. Absorption of sulfur dio ide in a random-packed to er.It is desired to remove 90% of the sulfur dioxide in a flue gas stream at 298 K and 1 atm bycountercurrent absorption with pure water at the same temperature, using a packed tower thatis 0.7 m in diameter. The tower is packed with 35-mm plastic NORPAC rings. The average gas-pressure drop is 200 Pa/m. Equilibrium is described by HenryÕs law with y i = 8.4x i.. If theliquid flow is adjusted so that the driving force (y y*) is constant, calculate the height of thepacked section. Flue gases usually contain less than 1 mole % of SO2, an air pollutantregulated by law. Assume that the properties of the liquid are similar to those of pure water,and that the properties of the flue gases are similar to those of air.Solution For the driving force to be constant, the operating line is parallel to the equilibrium line, and the absorption factor A = 1.0 It was found by trial and error with the Packed Column program that, for a pressure drop of 200 Pa/m and D = 0.7 m:
  • 157. 5.17c. Ben ene vapor recover s stem.Benzene vapor in the gaseous effluent of an industrial process is scrubbed with a wash oil in acountercurrent packed absorber. The resulting benzene-wash-oil solution is then heated to 398K and stripped in a tray tower, using steam as the stripping medium. The stripped wash oil isthen cooled and recycled to the absorber. Some data relative to the operation follow:Absorption:Benzene in entering gas = 1.0 mole %Operating pressure of absorber = 800 mm HgOil circulation rate = 2 m3/1,000 m3 of gas at STPOil specific gravity = 0.88 Molecular weight = 260Henry s law constant = 0.095 at 293 K;= 0.130 at 300 KNtOG = 5 transfer unitsStripping:Pressure = 1 atm Steam at 1 atm, 398 KHenry s law constant = 3.08 at 398 KNumber of equilibrium stages = 5a) In the winter, it is possible to cool the recycled oil to 293 K, at which temperature theabsorber then operates. Under these conditions 72.0 kg of steam is used in the stripper per1,000 m3 of gas at STP entering the absorber. Calculate the percent benzene recovery in thewinter. Solution
  • 158. I i ia e i a e :b) I he e i i i ib e c he ec c ed a h i e ha 300 K i h hea ai ab e c i g a e . A i g ha he ab be he e a e a 300 K, i h he a e ia d ea a e , a d ha N OG a d e i ib i age e ai he a e, ha e ec e f be e e ca be e ec ed?S i
  • 159. c) If he oil a e canno be inc ea ed, b he eam a e in he mme i inc ea ed b 50%o e he in e al e, ha mme eco e of ben ene can be e pec ed?
  • 160. 5.19c. Ab o p ion of ge mani m e achlo ide ed fo op ical fibe .Germanium tetrachloride (GeCl4) and silicon tetrachloride (SiCl4) are used in the production ofoptical fibers. Both chlorides are oxidized at high temperature and converted to glasslike particles.However, the GeCl4 oxidation is quite incomplete and it is necessary to scrub the unreacted GeCl4from its air carrier in a packed column operating at 298 K and 1 atm with a dilute caustic solution.At these conditions, the dissolved GeCl4 has no vapor pressure and mass transfer is controlled bythe gas phase. Thus, the equilibrium curve is a straight line of zero slope. The entering gas flows atthe rate of 23,850 kg/day of air containing 288 kg/day of GeCl4. The air also contains 540 kg/day ofCl2, which, when dissolved, also will have no vapor pressure.It is desired to absorb, at least, 99% of both GeCl4 and Cl2 in an existing 0.75-m-diameter columnthat is packed to a height of 3.0 m with 13-mm ceramic Raschig rings. The liquid rate should beset so that the column operates at 75% of flooding. Because the solutions are very dilute, it can beassumed that both gases are absorbed independenttly.Gas-phase mass-transfer coefficients for GeCl4 and Cl2 can be estimated from the followingempirical equations developed from experimental studies with 13-mm Raschig rings for liquid massvelocities between 0.68 2.0 kg/m2-s (Shulman, H. L., et al., AIC E J., 17, 631, 1971): where: ds = equivalent packing diameter (0.01774 m for 13-mm ceramic Raschig rings) = gas density, kg/m3 Gx ,Gy mass velocities, kg/m2-s For the two diffusing species, take DGeCl4 = 0.06 cm2/s; DCl2 = 013 cm2/s. For the packing, Fp = 580 ft 1, e = 0.63. Determine: a) Liquid flow rate, in kg/s. Solution
  • 161. Ini ial e ima eb) The pe cen ab o p ion of GeCl4 and Cl2 ba ed on he a ailable 3.0 m of packing.Sol ion (dil e ol ion )
  • 162. 5.20c,d. Absorption of carbon disulfide in a structured-packed to er. Redesign the absorber of Problem 5.15 using metal Montz B1-300 structured packing. The characteristics of this packing are (Seader and Henley, 1998): Fp = 33 ft 1 a = 300 m 1 = 0.93 Ch = 0.482 Cp = 0.295 CL = 1.165 CV = 0.422 Solution From the Packed Tower Program in Appendix D:Estimate the packed height; assume dilute solutions
  • 163. From Prob 5.11: Calculate HtOG at the bottom of the to er
  • 164. Calculate HtOG at the top of the tower From Prob. 5.11, the conditions at the top of the absorber are: The mass-tranfer coefficients remain fairl constant along the tower
  • 165. 5.21c,d. Absorption of ammonia in a random-packed to er.It is desired to reduce the ammonia content of 0.05 m3/s of an ammonia-air mixture (300 K and 1atm) from 5.0 to 0.04% by volume by water scrubbing. There is available a 0.3-m-diameter towerpacked with 25-mm ceramic Raschig rings to a depth of 3.5 m. Is the tower satisfactory, and if so,what water rate should be used? At 300 K, ammonia-water solutions follow Henry s law up to 5mole % ammonia in the liquid, with m = 1.414.Solution Calculate minimum water flow rate Try
  • 166. From Appendi DTr From Appendi D
  • 167. Tr From Appendi D
  • 168. Tr From Appendi D
  • 169. C bic i ei e ai5.22b,d. Absorption of sulfur dio ide in a structured-packed to er.A e ac ed i h e a M B1-300 c ed ac i g i be de ig ed ab b SO2f ai b c bbi g i h a e . The e e i g ga , a a SO2-f ee f a e f 37.44 e / 2- f bed c - ec i a a ea, c ai 20 e % f SO2. P e a e e e a a f a e f 1976 e / 2- f bed c - ec i a a ea. The e i i g ga i c ai 0.5 e % SO2.A e ha ei he ai a e i a fe be ee he ha e a d ha he e eaei he a a 2 a a d 303 K. E i ib i da a f bi i f SO2 i a e a 303 a d 1a ha e bee fi ed b ea - ae he e a i (Seade a d He e , 1998): a) De i e he f i g eai g i ee ai f he ab be : S i
  • 170. b) If the absorber is to process 1.0 m3/s (at 2 atm and 303 K) of the entering gas, calculate the water flow rate, the tower diameter, and the gas-pressure drop per unit of packing height at the bottom of the absorber. Solution From Ap. D, b trial-and-error:c)
  • 171. At the top of the column:From Ap. D, b trial-and-error: Average mass-transfer coefficients
  • 172. A e age heigh Calc la e N G Ini ial e ima e
  • 173. 6.1a. Flash vapori ation of a heptane-octane mi ture. A liquid mixture containing 50 mole % n-heptane (A), 50 mole % n-octane (B), at 303 K, is to be continuously flash-vaporized at a pressure of 1 atm to vaporize 30 mole % of the feed. What will be the composition of the vapor and liquid and the temperature in the separator if it behaves as an ideal stage? Solution From the program in Figure 6.3, 6.2a. Flash vapori ation of a heptane-octane mi ture. A liquid mixture containing 50 mole % n-heptane (A), 50 mole % n-octane (B), at 303 K, is to be continuously flash-vaporized at a temperature of 350 K to vaporize 30 mole % of the feed. What will be the composition of the vapor and liquid and the pressure in the separator if it behaves as an ideal stage? Solution From the program in Figure 6.3,6.3d. Flash vapori ation of a ternar mi ture.Modify the Mathcad¨ program of Figure 6.3 for ternary mixtures. Test your program with the datapresented in Example 6.2 for a mixture of benzene (A), toluene (B), and o-xylene (C). Criticaltemperatures and pressures,and the parameters of the Wagner equation for estimating vaporpressure (equation 6-5) are included in the following table (Reid, et al., 1987).Component Tc, K Pc, bar A B C Dbenzene 562.2 48.9 6.983 1.332 2.629 3.334toluene 591.8 41.0 7.286 1.381 2.834 2.792o-xylene 630.3 37.3 7.534 1.410 3.110 2.860SolutionParameters of the Wagner equation, A = benzene, B, toluene, C = o-xylene:
  • 174. Ini ial e ima e
  • 175. 6.4d. Flash vapori ation of a ternar mi ture.C ide he e a i e fE a e 6.2 a d P b e 6.3. E i a e he e e a e, a dc ii f he i id a d a ha e he 60% f he i e ha bee a i ed a a c a e e f1a .S i
  • 176. 6.5d. Flash vapori ation of a ternar mi ture.Consider the ternar mi ture of E ample 6.2 and Problem 6.3. It is desired to recover in the vapor 75%of the ben ene in the feed, and to recover in the liquid 70% of the o- lene in the feed. Calculate thetemperature, pressure, fraction of the feed vapori ed, and the concentration of theliquid and gas phasesSolution
  • 177. 6.6b.Batch distillation of a heptane-octane mi ture.Repeat the calculations of E ample 6.3, but for 80 mole % of the liquid distilled.SolutionBatch distillation of a mi ture of heptane and octane
  • 178. I i ia e i a e6.8a. Binar batch distillation with constant relative volatilit .C ide he bi a ba ch di i a i fE a e 6.3. F hi e , -he a e i h - c a e a 1a , he a e age e a i e a i i i = 2.16 (T e ba , 1980). U i g e a i (6-102) de i ed i P b e6.7, c e he c ii f he e id e af e 60 e% f he feed i ba ch-di i ed.S i I i ia e i a e: 6.9b. Binar batch distillation with constant relative volatilit . C ide he bi a ba ch di i a i fE a e 6.3. F hi e , -he a e i h - c a e a 1 a , he a e age e a i e a i i i = 2.16 (T e ba , 1980). The i e i be ba ch-di i ed i he a e age c ce a i f he di i a e i 65 e % he a e. U i g e a i (6-102) de i ed i
  • 179. Problem 6.7, compute the composition of the residue, and the fraction of the feed that is distilled. Solution Initial estimates:6.10b. Mi tures of light h drocarbons: m-value correlations.What is the bubble point of a mi ture that is 15 mole % isopentane, 30 mole % n-pentane, and 55mole % n-he ane? The pressure is 1.0 atm.Solution
  • 180. Initial estimate6.11c. Mi tures of light h drocarbons: m-value correlations.A solution has the follo ing composition, e pressed as mole percent: ethane, 0.25 %; propane, 25 %;isobutane, 18.5 %; n-butane, 56 %; isopentane, 0.25 %. In the follo ing, the pressure is 10 bars. Useequation (6-103) and Table 6.4 to calculate equilibrium distribution coefficients.a) Calculate the bubble pointSolution
  • 181. Initial estimateb) Calculate the de pointSolution
  • 182. Initial estimatec) The solution is flash-vapori ed to vapori e 40 mol % of the feed. Calculate the composition of theproducts.Solution Initial estimates
  • 183. Check!6.12b. Binar batch distillation with constant relative volatilit .A 30 mole % feed of ben ene in toluene is to be distilled in a batch operation. A product having an
  • 184. a e age c ii f 45 e % be e ei be d ced. Ca c a e he a f e id e ef , a i g ha = 2.5 a d F = 100 e . S i I i ia e i a e :6.13b. Batch distillation of a mi ture of isopropanol in ater.A i e f 40 e%i a i ae i be ba ch-di i ed a 1 a i 70 e % f hecha ge ha bee a i ed. Ca c a e he c ii f he i id e id e e ai i g i he i ,a d he a e age c ii f he c ec ed di i a e. VLE da a f hi e ,i e f ac i fi a , a 1 a a e (Seade a d He e , 1998):T, K 366 357 355.1 354.3 353.6 353.2 353.3 354.5 0.220 0.462 0.524 0.569 0.593 0.682 0.742 0.916 0.012 0.084 0.198 0.350 0.453 0.679 0.769 0.944C ii f he a e ei = = 0.685; b i i g i f he a e e = 353.2 K.S i
  • 185. I i ia e i a e6.15a. Continuous distillation of a binar mi ture of constant relative volatilit .F c i di i a i f a bi a i e fc a e a i e a i i , Fe e e a i (6-58) cabe ed e i a e he i i be f e i ib i age e i ed f he gi e e a a i , N i .U e Fe ee ai e i a e N i f di i a i f he be e e- e e i e fE a e 6.4.A e ha , f hi e a 1 a , he e a i e aii i c a a = 2.5.S i6.16a. Continuous distillation of a binar mi ture of constant relative volatilit .F c i di i a i f a bi a i e fc a e a i e a i i , he i i ef a i cabe de e i ed a a ica f he f i g e a i (T e ba , 1980):(6-105)
  • 186. U e e a i (6-105) e i a e R i f di i a i f he be e e- e e i e fE a e 6.4.A e ha , f hi e a 1 a , he e a i e aii i c a a = 2.5.S i I i ia e i a e 6.17b. Continuous rectification of a ater-isopropanol mi ture. A a e -i a i e a i b bb e i c ai i g 10 e%i a i be c i ec ified a a he ic e e d ce a di i a e c ai i g 67.5 e% i a . Ni e -eigh e ce f he i a i he feed be ec e ed. VLE da a a e gi e i P b e 6.13. If a ef a i f 1.5 i e he i i i ed, h a he e ica age i be e i ed: (a) If a a ia eb i e i ed? S i
  • 187. 6.24c. A distillation-membrane h brid for ethanol deh dration.Ma i d ia i a i id e a e diffic i ib e e aaeb i ec idi i a i beca e he ha e beha i c ai a a e e, a a ge i ch, a ea ea i e aii . O e i i c bi e di i a i ih e ec e e a e aai ech gie f a h b id. A e a e f ch a c bi a i i he deh d a i f e ha i g a di i a i - e b a e h b id.I a gi e a ica i , 100 e / f a a a ed i id c ai i g 37 e % e ha a d 63 e% ae be e a a ed ie d a d c hich i 99 e % e ha , a d a e id e c ai i g 99 e% a e . The i i be fed a di i a i c eai ga a he ic e e, i h a a ia eb i e a d a a c de e . The ef ai i be 1.5 he i i . The di i a e i e e a e b a e i h a a e e a = 70 a d = 0.6. The e b a e b he c ce a i ha he e ea e ea i he e ha - ich d c ( P = 0.99). The e e a e ea i e ed a a a a ed i id hec he a a he ea e i id c ce a i .a) Ca c a e he a f a e f he d c a d f he e id e.S i I i ia e i a e : b) Ca c a e he a f aea dc ii f he di i a e c i g f he c . S i
  • 188. Initial estimates6.26b. Fenske-Under ood-Gilliland method.A distillation column has a feed of 100 kmoles/h. The feed is 10 mole % LNK, 55 mole % LK, and 35mole % HK and is a saturated liquid. The reflu ratio is 1.2 the minimum. We desire 99.5 % recover ofthe light ke in the distillate. The mole fraction of the light ke in the distillate should be 0.75. Use theFUG approach to estimate the number of ideal stages required and the optimal location of the feed-stage.Equilibrium data: LNK = 4.0 LK = 1.0 HK = 0.75Solution Kilomoles in distillate:
  • 189. Kilomoles in bottoms: Determine minimum reflu : Saturated liquid, q = 1 Initial estimate
  • 190. De e mine n mbe of ideal age a R = 1.2 Rmin: Gilliland co ela ion Ini ial e ima e Ki kb ide e a ion: Ini ial e ima e:
  • 191. Feed a age 116.27b. Fenske-Under ood-Gilliland method.O e h d ed e /h f a e a b bb e- i i e be e a a ed b di i a i ha hef i gc ii :C e M e f ac i Re a i e a i iA 0.4 5B 0.2 3C 0.4 1a) F a di i a e a e f 60 e /h, fi e he e ica age , a d a ef , ca c a e he di i a ea db c i i b he Fe ee ai .S i I i ia e i a e :
  • 192. b) Using the separation in part a) for components A and C, determine the minimum reflu ratio b theUnder ood equation.Solution Initial estimate:
  • 193. Initial estimate:Initial estimate: c) For a reflu ratio of 1.2 times the minimum, determine the number of theoretical stages required, and the optimum feed location. Solution
  • 194. I i ia e i a e Ki b ide e ai : I i ia e i a e: Feed a age 56.35c Flash Calculations: the Rachford-Rice Method for Ideal Mi tures.a) Sh ha he be f fa h a i a i fa ic e idea i e ca be ef a ed a gge ed b Rachf d a d Rice (D he a d Ma e, C ce a De ig f Di i a i S e ,McG a -Hi , Ne Y , NY, 2001):
  • 195. b) Solve E ample 6.2 using the Rachford-Rice method.Solution Initial estimate:
  • 196. 7.2 Single e traction. Appendi G-1: Single-Stage E traction Ini ial e ima e :
  • 197. 7.3. Single-stage e traction: insoluble liquids 7.4. Single-stage e traction: insoluble liquids
  • 198. 7.5 Multistage crosscurrent e traction: insoluble liquids. Rea a gi g e ai (3-60) S le fl ae e age T al le fl ae7.6 Multistage crosscurrent e traction. Appendi G-2: Multistage Crosscurrent E traction Data presented are from E ample 7-3: acetone- ater-chloroform s stem at 298 K
  • 199. Ini ial e ima e :
  • 200. Ini ial g e e :Ini ial g e e :
  • 201. Composited e tract Check material balances Acetone Chloroform Water