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 i c) The a a c ce ai . S i d) The a de i . S i 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 o be 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 ion Ba i : 100 kg mi e b) The a e age molec la eigh of he mi e. Sol ion c) Calc la e he feed a e in kmol/h. Sol ion 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 LNG c) 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 he ga i e de he e c di i . S i 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): -2 / . 1. Velocities and flu es in a gas mi ture.0 3/ a 300 K a d 102 Pa.5b. Material balances around an ammonia gas absorber. 40% O2. S i : Ba i : 1 ec d A=a ia B = ai b) Ca c a e he e ga a e ci (defi ed a a f ae e i e be c . A ga ea f a he a e f 10. 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. S i : 1. a d 12 / .6b. I c i fa e i a i e fa ia a d ai .ec i a a ea). a) Neg ec i g he e a a i f a e . The ab e e ci ie f each ecie a e -10 / . The ab be e e i c i d ica i h a i e a dia e e f 2. a d i a ai . ca c a e he a ia f ac i i he ga ea i g he ab be . a . a d 40% H2O b e. e ec i e .5 . A ga i ea a a e e f 150 Pa a d 295 K c ai 20% H2. V Mass average velocit . a) Determine the mass average velocit . and the molar average velocit . for the mi ture. NO2. nO2. JO2. v. Solution: . vm Basis: 1 kmole of gas mi ture b) Evaluate the four flu es: jO2. V.in the direction of the -a is. Solution Molar average velocit . . 3 c ai e a 340 K a d 101.3 Pa i a a ed i h a e a . c) T a a c ai ed i he a . a a d -b b e e a e f 30 C a d a h idi f 0. a) Ca c a e he a e e b e a a i i he c i g e.500 g/ f d ai .8c. The ai e e he c i g e a he a e f 6. Ai . d) Ma f ae a i he a . A i ee ai f ae a : . I ea e he c i g e a a ed i h a e a a 38 C. 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 1.7b. The a e e e he c de e a 29 C a d ea e a 45 C. De e i e he f i g e ie f he ga i e: a) M e f ac i f ae a .970 g/ . P ope ie of ai a a ed i h a e apo . The c i g a e f ae he c de e f a big c a -fi ed e a i 8. ed i a 30. S i C ide he ai ea i g he e a a ed a 38 C. F he c de e . b) A e age ec a eigh f he i e.11).016 g f a e / g f d ai . S i a) A i ee ai f ae a : F a a a ed i e b) c) d) 1.1. Wa e balance a o nd an ind ial cooling o e . The ef e. T a id f i g f he c de e hea . E i a e he a 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 . 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: .a e e i e e .2% f he eci c a i a e. a a a f he ci c a i g a e be de ibe a e di ca ded (b d ). E= ae b ea ai i he ai b) T acc f ae e i he c i g e . Thi a e a e c ai 500 g/L f di ed id . Wi dage e f he e ae e i a ed a 0.a fe face .000 g/L f di ed id . he ci c a i g a e i c ai e ha 2. 6 kPa. The dr er operates at constant temperature and pressure. in kg/min. and initial water-vapor partial pressure of 1.0 m3/min at 350 K. in kg of water/kg of dr air. Solution . 101. It is desired to dr 10 kg/min of soap continuousl from 17% moisture b weight to 4% moisture in a countercourrent stream of hot air.1. a) Calculate the moisture content of the entering air.9b. Solution c) Calculate the water-vapor partial pressure and relative humidit in the air leaving the dr er. Water balance around a soap dr er. Solution b) Calculate the flow rate of dr air.3 kPa. The air enters the dr er at the rate of 30. Calc la e a e a e e a 350 K A i ee ai f ae a : . .3% e e i ai . 1987. A i g ha he ai d e ad b he ca b .1.7 a d 345 Pa. E. a d he a f e e ad bed b he ca b . S. 3 d. ca c a e he e i ib i c ce a i f e e i he ga e ha e. a d cc ie a e f 2. a d be be ee 0. U. The e i a ed each e i ib i a c a e ea e a d e e.10b. S i M= a f ca b = e f e e ad bed I i ia e i a e .500 3 a 298 K a d 101. ed. a d * i he e i ib i e e a ia e e. P. I a eff ed ce he e ec e f hi ga . i i ia f ee f e e.i g f e e/ g f ca b . i Pa.3 Pa. Re ea ch T ia g e Pa .): he e W i he ca b e i ib i ad ii . A a e ga c ai 0. Acti ated carbon adsorption.. A. material balances. The ad i e i ib i f hi e i gi e b he F e d ich i he (EAB C C Ma a . i i e ed 100 g f ac i a ed ca b . NC. 11b.10.1. Acti ated carbon adsorption. It is desired to adsorb 99. Solution . Estimate how much activated carbon should be used if the s stem is allowed to reach equilibrium at constant temperature and pressure. material balances.5% of the toluene originall present in the waste gas of Problem 1. 13a. E. E i a e he diff i c 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. M.12a. S i F A e di B 1. Estimation of gas diffusivit b the Wilke-Lee equation. . O eg S a e U i e i . La (MS he i . Estimation of gas diffusivit b the Wilke-Lee equation. 1964) ea ed he diff i i f ch f i ai a 298 K a d 1 a a d e ed i 2 a e a 0. d.093 c / .1. d. 1 K. and i c i ical ol me i 413 cm3/mol. Sol ion E pe imen al al e b) 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 he e pe imen al al e of 0. Sol ion . and i c i ical ol me i 254 cm3/mol.087 cm2/ epo ed in Appendi A.437 cm2/ epo ed in Appendi A. The no mal boiling poin of p idine i 388. 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 0.4 K. The no mal boiling poin of naph halene i 491. . Compa e i i h he e pe imen al al e of 0. Che . E pe imen al al e c) E ima e he diff ii of aniline (C6H7N) in ai a 273 K and 1 a m. R.6 K. The no mal boiling poin of aniline i 457. 2 :681. I d. E. E g. and i c i ical ol me i 274 cm3/mol. Sol ion E pe imen al al e .061 cm2/ (G illiland. 1934).. mB. VA . Equation (1-49) is still used.14d.162 10-25 (J- m3)1/2] a) Modify the Mathcad routine of Figure 1. Diffusivit of polar gases If one or both components of a binary gas mixture are polar. P. MA . VB. Eng. 8:240. a modified Lennard-Jones relation is often used. Process Design De elop. 1969) has suggested an alternative method for this case. Use the function name DABp(T. Chem. TbA .1. Brokaw (Ind.. TbB) Solution . MB. but the collision integral is now given by mp = dipole moment. mA .3 to implement Brokaw's method. debyes [1 debye = 3. . 1 1.14) a e h be (Reid.2 Vb ..15d.9 .9 1.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 d 323 K.2 Vb . deb e 1. K 249.8 M 36..8 . c 3/ 30.06 S i 1. The da a e i ed eB a ' ea i a e h be (Reid.1 373.6 43. c 3/ 50. K 188.5 18 . The da a e i ed e B a ' e a i ( ee P b e 1. 1987): Pa a e e H d ge ch ide Wa e Tb . Diffusivit of polar gases E a a e he diff i c efficie f h d ge ch ide i a e a 373 K a d 1 ba . 1987): Pa a e e Me h ch ide S f di ide Tb .078 c / .6 18.6 M 50. e a .1 263. deb e 1. a d c aei 2 he e e i e a a e f 0.5 64. e a . 5 ba .14) a e h be (Reid. e a .2 Vb . Diffusivit of polar gases E a a e he diff i c efficie f h d ge fide i f di ide a 298 K a d 1. S i 1. K 189. 1987): Pa a e e H d ge fide S f di ide Tb ..16d. The da a e i ed e B a ' e a i ( eeP b e 1.03 43. c 3/ 35. deb e 0.08 64.9 1.6 263.06 S i .8 .6 M 34. 5 bar. Calculate the effective diffusivit of nitrogen through a stagnant gas mi ture at 373 K and 1.d. carbon dio ide (4) Calculate binar MS diffusivities from Wilke-Lee equation .ution Calculate mole fractions on a nitrogen (1)-free basis: o gen (2). Effective diffusivit in a multicomponent stagnant gas mi ture.17a.1. carbon mono ide (3). The mi ture composition is: O2 15 mole % CO 30% CO2 35% N2 20% Sol. reported an effective diffusivit value of 0.013 bar and 408 K. Assume that onl the HgCl2 is adsorbed b the activated carbon. Mercur is considered for possible regulation in the electric power industr under Title III of the 1990 Clean Air Act Amendments. 1999) describe a theoretical model for estimating mercur removal b the sorbent injection process.22 cm2/s. 49:694-704. Mercur removal from flue gases b sorbent injection. Meserole. One promising approach for removing mercur from fossil-fired flue gas involves the direct injection of activated carbon into the gas. 7% H2O. and its composition (on a mercuric chloride-free basis) is 6% O2. A oc.d. If the flue gas is at 1. Meserole et al. An important parameter of the model is the effective diffusivit of mercuric chloride vapor traces in the flue gas. 12% CO2. (J. et al. Ai & Wa e Manage. and 75% N2. estimate the effective diffusivit of mercuric chloride in the flue gas.1.18a.. Solution HgCl2 (1) O2 (2) CO2 (3) H2O (4) N2 (5) . . S.08 cP. (1987) a 1. "The Ph ica Che i f E ec ic S i .79 cP. The c i ica e f ca b e ach ide i 275. F di e i f a i g e a . The i c i f i id e ha a 298 K i 1.20b.. a d B. he diff i fa i ge a a be ea ed a ec a diff i . I he ab e ce f a e ec ic e ia . B. S i 1.5 10-5 c 2/ ." ACS M g .1. 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 . e a .9 c 3/ . O e . 95.19a. Whe a a di cia e i i .i a he ha ec e diff e. he diff i c efficie i gi e b he Ne -Ha e e a i (Ha ed. The i c i f i id a e a 273 K i 1. 1950): a) E i a e he diff i c efficie a 298 K f a e di e i f HC i ae. Diffusion in electrol te solutions. Wilke-Chang method for liquid diffusivit . C ae he e e i e a a e e ed b Reid. 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. H. . Diff sion. Use the Hayduk and Minhas correlation for solutes in aqueous solutions. Cambridge University Press. The critical volume of oxygen is 73. At this temperature.1 10 5 cm2/s (Cussler E. 2nd ed. L. 1997).9 cP.4 cm3/mol.21a. the viscosity of water is 0. Estimate the diffusion coefficient of oxygen in liquid water at 298 K. Solution . O gen diffusion in ater: Ha duk and Minhas correlation. UK.1. Cambridge. The experimental value of this diffusivity was reported as 2.. 1987): Solution . The following data are available (Reid. Liquid diffusivit : Ha duk and Minhas correlation..22a. et al. Estimate the diffusivity of carbon tetrachloride in a dilute solution in n-hexane at 298 K using the Hayduk and Minhas correlation for nonaqueous solutions.7 10 5 cm2/s. Compare the estimate to the reported value of 3.1. d. Solution Iniitial estimates From Table 2. estimate the molar volume of allyl alcohol at its normal boiling point.. and the Hayduk and Minhas correlation for aqueous solutions. et al.15 cP.23b. Based on this result. Compare it to the result obtained using the data on Table 1. The viscosity of water at 288 K is 1. The diffusivity of allyl alcohol (C3H6O) in dilute aqueous solution at 288 K is 0. E ima ing mola ol me f om li id diff ion da a.1 . 1987).2.1.9 10 5 cm2/s (Reid. Ne Y . he ac i i c efficie f ace e i gi e b Wi e ai (S i h. I c. R. 49. S e . F hi e a 298 K.42 10-5 c 2/ . McG a -Hi C . e a . B. 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%. Trans. 5 h ed. 1953): The e e i e a a e e ed b Ha da dS e (1953) i 0. NY.. H. M.. 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. 890. a d R.. U de he e c di i (Ha d.. 1. Concentration dependence of binar liquid diffusivities.24b.. Introduction to Chemical Engineering Thermod namics. Farada Soc. d. J. 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%.. 1996): S i E i a e he he d a ic fac . 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 . Estimate the rate of combustion of the carbon. the gas concentration is 40% CO. d.25b. A flat plate of solid carbon is being burned in the presence of pure o gen according to the reaction Molecular diffusion of gaseous reactant and products takes place through a gas film adjacent to the carbon surface. O2 (3) Calculate binar MS diffusivities from Wilke-Lee .The temperature of the gas film is 600 K. gas-phase flu calculation. Stead -state. and the pressure is 1 bar. The reaction at the surface ma be assumed to be instantaneous. On the outside of the film. 20% O2. one-dimensional. ne t to the carbon surface.1. the thickness of this film is 1. therefore. CO2 (2).0 mm. and 40% CO2. there is virtuall no o gen. in kg/m2-min. Solution CO (1). Appendi C-2: 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 length of the diffusion path is The densit of the gas phase follows from the ideal gas law Initial estimates of the flu es Initial estimates of the concentrations . Stoichiometric relations (No o gen) . . . one-dimensional.1. Stead -state. T 288 K 999.T G ' 288 K 36 /100 1.8 / 3. . 1973).A 0. E N 2SO4 . S A = N 2SO4 B = H2O C A1 ( )B : 100 H2O (36 ) (P ) . 1.240 / 3 (P C . A G ' (N 2SO4 10H2O) 288 K.085 . liquid-phase flu calculation.A ( ) N 2SO4. T N 2SO4 288 K P 1-20.153 P.26b. in kg/m2-hr.215 cm2/s and in liquid water is 1.42 mol percent and the concentration at the lower boundary of the water layer is esentially zero. . The diffusivity of ammonia in air under these conditions is 0. Neglecting water evaporation. In this steady-state process. is being selectively removed from an air-NH3 mixture by absorption into water. Molec la diff ion h o gh a ga -li id in e face. Calculate diffusivity 1. Assume that the gas and liquid are in equilibrium at the interface. ammonia is transferred by molecular diffusion through a stagnant gas layer 5 mm thick and then through a stagnant water layer 0.The temperature of the system is 288 K and the total pressure is 1 atm. determine the rate of diffusion of ammonia.77 10 5 cm2/s.1 mm thick. NH3. The concentration of ammonia at the outer boundary of the gas layer is 3. Ammonia. d.27c. Sol ion: Ini ial e ima e : . 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 . The e e a e i 368 K a d he e e i 1 a . i g/ 2. B h c e diff e h gh a ga fi 0. Stead -state molecular diffusion in gases.28c.. 1987) he e i he ace ic fac ..2 he he ide f he fi . The a e hea f a i a i f he a c h a d a e a 368 K ca be e i a ed b he Pi e ace ic fac c e a i (Reid.1. A i e f e ha a d ae a i bei g ec ified i a adiaba ic di i a i c . The a c h i a i ed a d a fe ed f he i id he a ha e.8 e ide f he fi a d 0. e a .1 hic . 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. Ca c a e he a e f diff i f e ha a d f a e . The e f ac i f e ha i 0. . Solution Estimate diffusivit from the Wilke-Lee equation . called the Lewis relation. has profound implications in humidification operations. For air at 300 K and 1 atm:Cp = 1.1. d. as will be seen later. estimate the diffusivit of water vapor in air at 300 K and 1 atm. k = 0. Analog among molecular heat and mass transfer.0262 W/m-K. Based on the Lewis relation. Compare our result with the value predicted b the Wilke- Lee equation. This observation. 1980).846 10-5 kg/m-s. m = 1.01 kJ/kg-K. It has been observed that for the s stem air-water vapor at near ambient conditions.29a.0 (Tre bal. Le = 1. and r = 1.18 kg/m3. If he e a i e h idi f he ai a he e edge f he fi i 20%. 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. a d he a e ei 1 ba . e i a e he d i he a e e e e da .5 hic .30b. d. i ge (2). 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. Ca c a e he f e a d c ce a i fi e f he e a e h d ge (1). Stead -state molecular diffusion in gases.1. d. 1987) S i F A e di A 1.31b. e a . a d ca b di ide (3) de he f i g c di i . Stead -state molecular diffusion in a ternar gas s stem. The e e a e i 308 K a d he e ei 1 . a i g ha c di i i he fi e ai c a .. The MS diffusion coefficients are D12 = 83. D13 = 68.8 mm2/s. 40% CO2. 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 . at the other end.0 mm2/s. At one end of the diffusion path the concentration is 20 mole% H2. = 0. 30% CO2.atm. 40% N2.8 mm2/s. The total molar flu is ero. Solution Appendi C-1: Solution of the Ma ell-Stefan equations for a multicomponent mi ture of ideal gases b orthogonal collocation ( C = 3). The diffusion path length is 86 mm. 20% N2. the concentration is 50% H2. D23 = 16. Initial estimates of the flu es Initial estimates of the concentrations . . . 125.2a.158. A a ce ai i i he ab be .a fe c efficie i he i id ha e. hi e he c e di g i e facia be e e ga - ha e c ce a i i 0.62 g/ 2.0158. / 3. The be e e f a ha i i ea ed a 0. e e i g he d i i g f ce i e f a c ce a i .ha e c ce a i i 0. he dia e e .a fe c efficie i he ga ha e a ha i i he e i e . hi e he c e di g i e facia be e e i id. he be e e e f ac i i he b f he i id ha e i 0. Ca c a e he a . Ma .02.1a. 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 . S i 2. Whe he e e i e begi . a) Ca c a e he a . 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. 2.an fe coefficien f om naph halene blima ion da a. he be e e e f ac i i he b f he ga ha e i 0.. e e i g he d i i g f ce i e f e f ac i .a fe c efficie i he ga ha e a ha i i he e i e .an fe coefficien in a ga ab o be . Ma . hich b i a e a ia . e e i g he d i i g f ce i e f e f ac i . S i c) A he a e ace i he e i e . I a ab a e e i e . S i b) Ca c a e he a . 1973). Mass-transfer coefficients from acetone e aporation data.0 c . I a ab a e e i e . I i c ec ed a e e i c ai i g i id ace e hich a a ica e ace he ace e e a a ed. f he d i i g f ce i e f a c ce ai . i a e e a 347 K i 670 Pa (Pe a d Chi .79 g/c 3. The de i f id a h ha e e i 1. E i a e he a - a fe c efficie .3a. A he e d f he e e i e . 1973). i a b e ed ha 2. a) E i a e he a .32 i a e . hich e a a e a ia . f he he e i 2.85 c . S i 2. e e i g he d i i g f ce i e f a ia e e .0 L f ace e e a a ed i 5 i .a fe c efficie . ba ed he a e age face a ea f he a ic e. The a i 1 g a d 50 c ide. S i b) Ca c a e he a . i a e e i 27 Pa (Pe a d Chi . The de i f i id ace e a 300 K i 0.D i g a e ei e a .145 g/c 3. 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).a fe c efficie . 14. S i . he dia e e f he he e i 1. ai ai i g a c a i id e e i he a . Mass-transfer coefficients from etted. Wi h he he f e a i (2-52). S i . 50 i dia e e a d 1. e i a e he a e age a .a e e i e a e .c i fa ga i e.all e perimental data. Wa e a 308 K f d he i e a . A e ed. I ea e he e ed ec i a 308 K a d i h a e a i e h idi f 34%.2.04 3/ i .a fe c efficie . D ai e e he b f he i e a he a e f 1.0 g. i h he d i i g f ce i e f a f ac i . ea ed a 308 K a d 1 a .4b. 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- -OD ba 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 he a 2 f 12.2 g f ai / - 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 he i 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 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 10 38- -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 : b) E i a e he a - a fe c efficie be e ec ed f e a a i f - ac h i ca 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 d Che e : U e he a - a fe e e i a a g e ai (2-69) e i a e he a - a fe c efficie f a b). C a e he e . S i 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 20 m/s. The air is at 310 K and 101.3 kPa. The naphthalene remains at 290 K; at this temperature the vapor pressure of naphthalene is 26 Pa. Estimate the moles of naphthalene lost from the plate per hour, if the end effects can be ignored. Solution 2.10b. 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 a velocity of 0.6 m/s. The 291 K seawater has a salt concentration of 0.0309 g/cm3. Estimate the rate at which the salt goes into solution if the edge effects can be ignored. Assume that the kinematic viscosity at the average liquid film conditions is 1.02 10 6 m2/s, and the diffusivity is 1.25 10 9 m2/s. The solubility of NaCl in water at 291 K is 0.35 g/cm3, and the density of the saturated solution is 1.22 g/cm3 (Perry and Chilton, 1973) . Solution Laminar flo At the bulk of the solution. point 1: . point 2: At the interface. 2.3.0 ba .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 . Vc = 209 c / . e a .232 (Reid. D i g he e e i e de c ibed i P b e 2. d e he high a i i f ace e. The f i g da a f ace e igh 3 be eeded: Tc = 508. The ef e. Zc = 0. e ie ch a de i a d i c i h d be e i a ed ca ef . a a e he ge ide f he a . he ai e ci a ea ed a 6 / . M = 58. he a e age ace e c ce a i i he ga fi i e a i e high. 1987). N ice ha .1 K. S i A 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 .11b.. Pc = 47. Ma an fe f om a fla li id face. E i a e he a . 2.9 for a drop of ater hich is originall 2 mm in diameter.12b. E aporation of a drop of ater falling in air. Repeat E ample 2. Solution . . 2.5 cm in diameter. 1.25 10 9 m2/s.10: 2. During the experiment described in Problem 2. 2. Sublimation of a solid sphere into a gas stream. The following data for naphthalene might be needed: Tb = 491. Dissolution of a solid sphere into a flo ing liquid stream. and the density of the saturated solution is 1.02 10 6 m2/s.14b. 1973) . Solution For air at 347 K and 1 atm: Estimate DAB from the Wilke-Lee equation Lennard-Jones parameters for naphthalene . The velocity of the 291 K water stream is 1. The solubility of NaCl in water at 291 K is 0.22 g/cm3 (Perry and Chilton.2.35 g/cm3. Assume that the kinematic viscosity at the average liquid film conditions is 1. Estimate the mass-transfer coefficient for the dissolution of sodium chloride from a cast sphere. the air velocity was measured at 10 m/s.1 K. and the mass diffusivity is 1.13b. Estimate the mass-transfer coefficient predicted by equation (2-36) and compare it to the value measured experimentally.0 m/s. Solution From Prob. Vc = 413 cm3/mol. if placed in a flowing water stream. It is falling at terminal velocit under the influence of gravit into a big tank of water at 288 K.15b. The cr stal of Problem 1.464 kg/m3 (Perr and Chilton. The densit of the cr stal is 1. Solution . 1973).2. a) Estimate the cr stal's terminal velocit .26 is a sphere 2-cm in diameter. Dissolution of a solid sphere into a flo ing liquid stream. 1.26 From Prob.b) Estimate the rate at which the cr stal dissolves and compare it to the answer obtained in Problem 1.1.26. Solution From Prob.26: . Dry air enters at the rate of 7 kg/m2-s. and the mass diffusivity is 1.16c. and the mass-transfer coefficient constant. Assume the air is everywhere at its average conditions of 309 K and 1 atm. Mass transfer in a etted. The pipe is 1-cm in diameter. b) What is the average concentration of benzoic acid in the water after 2 m of pipe.003 g/cm3. 1997).all to er.2. while air flows upward through the core.1 cm/s. the walls of which are lightly coated with benzoic acid (C7H6O2). Solution 2. Water flows through a thin tube. at 298 K and 0. 2. Under these conditions. Compute the average partial pressure of water in the air leaving if the tower is 1 . a) Show that a material balance on a length of pipe L leads to where v is the average fluid velocity. Water flows down the inside wall of a 25-mm ID wetted-wall tower of the design of Figure 2.0 10 5 cm2/s (Cussler. the water at 294 K. Mass transfer inside a circular pipe. and c A * is the equilibrium solubility concentration. The water flows slowly. equation (2-63) applies. The solubility of benzoic acid in water at 298 K is 0.17b. U e he e f P b e 2.7 e i a e he a . I d i g he b i a i f a h ha e e i a ai ea . S i F a e a 294 K 2. -ID a h ha e e i e.a fe c efficie f he i e face. a i e iga c c ed a 3. -OD. 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.2 Pa. The i e i e a ade f a 25. i g he e i a e dia e e defi ed i P b e 2. . g. a d e a i (2-47). Ma an fe in an ann la pace.06 c 2/ . a h ha e e ha a a e e f 5. hi a ded b a 50. e i a e he c efficie f he e face. id a h ha e e d.18c.g a a d c . 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 / .7. a d a diff i i i ai f 0. A 289 K.. Sol ion In hi i a ion. NA1. a1. i h pecific in e facial a ea. i h a ea a2. and a fl f om he o e all. A ma e ial balance on a diffe en ial ol me elemen ield : Define: Then: Fo he in e io all: . he e ill be a mola fl f om he inne all. NA2. For the outer all: . . M = 78. F be e e. 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 .2 K.19c.2.. 1987). Pc = 48. Vc = 259 c 3/ . e a . E i a e he e g h f he c i de . Tc = 562. Zc = 0.271 (Reid. 2-45: .9 ba . 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 / .7 Pa. Ben ene evaporation on the outside surface of a single c linder. 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 . The i id i a a e e a e f 315 K he e i e e a a e e f 26. calculate the gas-film mass-transfer coefficient using equation (2-55) and compare the result with the value reported by Wilke and Hougan.415 10 3 kmol/m2-s-atm With the assumption that the properties of the gas mixture are the same as those of air. Wilke and Hougan (T an .71 mm gas stream mass velocity 0.816 kg/m2-s temperature at the surface 311 K pressure 97. 1945) reported the mass transfer in beds of granular solids. the following data were reported: effective particle diameter 5.7 kPa k G 4. Solution From the Wilke-Lee equation . 445.20b. Ma an fe in a packed bed. they reported gas-film coefficients for packed beds. and by evaporating this water under adiabatic conditions. In one run. AIChE. Air was blown through a bed of porous celite pellets wetted with water. 41.2. The porosit of the bed is 40%. based on the empt cross section of the bed. Ma an fe and p e e d op in a packed bed. Solution From the Wilke-Lee equation: . a) How much iodine will evaporate from a bed 0.2. The air flows at a rate of 2 m/s.21b.7-cm in diameter. Air at 373 K and 2 atm is passed through a bed 10-cm in diameter composed of iodine spheres 0.1 m long? The vapor pressure of iodine at 373 K is 6 kPa. b) E ima e he p e e d op h o gh he bed. Sol ion . Compare the results. when SI units are used exclusively. For the conditions described in part a). 1949) measured rates of absorption of SO2 in water and found the following expression for 25-mm Raschig rings at 294 K where k a is in kmole/m2-s. inmiscible streams. Accordingly. k c a. Solution For dimensional consistency. Progr. 1940) obtained the following correlation for the liquid-film mass-transfer coefficient in packed absorption towers The values of a and n to be used in equation (2-71) for various industrial packings are listed in the following table.. The interfacial surface area per unit volume. estimate the liquid-film mass-transfer coefficient. Volumetric mass-transfer coefficients in industrial to ers. 45.22b.7 10 9 m2/s. Sherwood and Holloway (Trans. Eng. 323. 39. AIChE. If the liquid mass velocity is L' = 2. 21. 36. a. Solution .2. Both a and the mass-transfer coefficient depend on the physical geometry of the equipment and on the flow rates of the two contacting. Empirical equations for the volumetric coefficients must be obtained experimentally for each type of mass-transfer operation. a) Consider the absorption of SO2 with water at 294 K in a tower packed with 25-mm Raschig rings. estimate the liquid-film mass-transfer coefficient using equation (2-72).04 kg/m2-s. in many types of packing materials used in industrial towers is virtually impossible to measure. add the constants: b) Whitney and Vivian (Chem. they are normally correlated together as the volumetric mass- transfer coefficient. The diffusivity of SO2 in water at 294 K is 1. 083 kg/m3. They found that the porosity of the fluidized bed. through where vs is in cm/s. a) Using equation (2-56).534-mm spherical glass beads. Solution . in aqueous alkaline solutions. k L. if the porosity of the bed is 60%. 75-cm high. The properties of the aqueous solutions were: density = 1. et al. including fluidized beds. could be correlated with the superficial liquid velocity based on the empty tube. 938.30 cP. (AIC E J. viscosity = 1. estimate the mass-transfer coefficient. 45. The fluidized bed experiments were performed in a 5-cm-ID circular column. 1999) studied the electrochemical reduction of ferrycianide ions. diffusivity = 5. Mass transfer in fluidi ed beds. to ferrocyanide..23b. with a particle density of 2. {Fe(CN)6} 3.90 10 10 m2/s.612 g/cm3.2. They studied different arrangements of packed columns. Cavatorta. The bed was packed with 0. vs . e. {Fe(CN)6} 4. If he a e flo a e inc ea e o 60. p opo ed he follo ing co ela ion o e ima e he ma . U ing hi co ela ion. Mass transfer in a hollo -fiber boiler feed ater deaerator. Compa e o e l o ha of pa a). Sol ion .b) Ca a o a e al. calc la e he f ac ion of he en e ing di ol ed o gen ha can be emo ed. e ima e he ma . L.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 .24b. Con ide he hollo -fibe BFW deae a o de c ibed in E ample 2-13. if he po o i of he bed i 60%. Sol ion 2.000 kg/h hile e e hing el e emain con an .an fe coefficien . 2. hich mean a di ol ed o gen concen a ion of 8.38 mg/L.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 gen diff e ac o he memb ane. Sol ion . calc la e he ga ol me flo a e and compo i ion a he l men o le . The a e en e he hell ide a 298 K a a ed i h a mo phe ic o gen. a fe c efficie a he a e age c di i i ide he e .b) Ca c a e he a . 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 . (From E ample 2.13) . 6 e f ac i f be e ea da e e a e f 320 K. S i 3. Re ea E a e 3.1a. S i I i ia e i a e . P edic he e i ib i e e a e. 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 ce be e e-40 e e ce e e if he e e i i a e e de 1 a e e.1.2b. 3. Application of Raoult's law to a binar s stem. Application of Raoult's law to a binar s stem. b f a i id c ce a i f 0. P edic he e i ib i e e a e. 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 . S i 3.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 i 3. Henr 's la : saturation of ater ith o gen. f idea i .a e e a 283 K e a 3. A i ih ge di ed i a e c ai i g 0. a d a c a e. a) Wha d be he c ii f a he a e.1 Pa. A 373 K.27 10 a / 4 e f ac i . 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. a he a e ha a a e e f 106 Pa a d a c a e f 47. Application of Raoult's la to a binar s stem.3a. -C8H18.4a. The He ' a c a f he ge .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 . N a he a e. -C7H16. The e i 10 3 f ga ace e he i id. b a i g ha he a ia. A i g ha he ga . a d a e a e b gh i c ac i a c ed c ai e . Re ea E a e 3. 3.2 ca c a e: a) he a e e a e i ib i S i I i ia g e e .42 g ge /L I i ia c di i : The i gai ge . ai . A e i ib i : Ba i : 1 L a e (1 g a e ) E i ib i c ce ai . c e = 11.3.ace e a d he e e a e e ai c a i e i ib i i achie ed. dif he Ma hcad ga i Fig e 3. Material balances combined ith equilibrium relations.5c. S .145. Ma .an fe e i ance d ing ab o p ion. A : P = 1.755 ) . T H ' 0. G.162 3. A.T 2..6b.0 A/ 3 .T KG = 0.L = 1.85 / . I A( = 60) . 1. A = 0. ) .1 .100 / 3. A : A = 0.27 / 2. A.0 . . . I 57% .G = 0. i. A.b) the liquid-film coefficient. Solution Initial estimates of interfacial concentrations: . Solution Basis: 1 m3 of aqueous solution c) the concentration on the liquid side of the interface. k L. d) the mass flu of A. Solution Check this result b calculating the gas-phase flu : . S i ( Ae = A *) c) The i id i e facia c ce ai f A. Absorption of ammonia b water: use of F-t pe mass-transfer coefficients.L = 0. he i id ea c ai 4.5 e % a d he ga ea c ai 9.8d. S i b) he a f f A.70 a d A.a fe c efficie . The a e e i 1 a .i i he e i ib i a ia e e i a a d A.5.10.0 e/ 2. M dif he Ma hcad g a i Fig e 3. Mass-transfer resistances during absorption.3. b i h A. F a e i hich c e A i a fe i g f he ga ha e he i id ha e. K . S i 3.G = 0. 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. . he e i ib i e a i i gi e b he e A.i i he e i ib i i id c ce a i i a f ac i . E e hi g e e e ai c a .6 e ea E a e 3.0 e% A. The i di id a ga -fi c efficie a hi i i G = 3.-a . E a a e a) he ea a . A e i i he a a a .7b. 0050 kmol/m2. E e hing el e emain con an . Sol ion Ini ial g e e . Sol ion Ini ial g e e 3. Absorption of ammonia b water: use of F-t pe mass-transfer coefficients..9d. Modif he Ma hcad p og am in Fig e 3.5.6 o epea E ample 3. b i h FL = 0. he i di id a fi c efficie e e e i a ed be L = 6.3.3 c /h a d G = 1. The e i ib i ea i hi f e di e i fa ia i a e a 300 K a d 1 a i De e i e he f i g a .a fe c efficie : a) S i b) S i .10b. I he ab i fa ia i ae f a ai -a ia i e a 300 K a d 1 a .17 / 2-h -a .an fe e i ance d ing ab o p ion of ammonia. Ma . 34. AIC E J.tM = 2. across the membrane.. The mass-transfer coefficient across a hydrophobic membrane is from (Prasad and Sirkar. 32. Solution 3. eM = membrane porosity. tM = membrane tortuosity. AIC E J.. eM = 0. 1988) where DAB = molecular diffusion coefficient in the gas filling the pores. KL. is given by (Yang and Cussler.13. Feb. the gas equilibrium concentration divided by that in the liquid.13. and d = 25 10 6 m (Prasad and Sirkar.2. Solution For oxygen in nitrogen at 298 K and 1 atm: . 1988). 1910. and H is Henry's law constant. a) Calculate the corresponding value of k M. the overall mass-transfer coefficient based on the liquid concentrations. Mass-transfer resistances in hollo -fiber membrane contactors. For the membrane modules of Example 2. respectively. d = membrane thickness.4.11b. c) Ky Solution d) Fraction of the total resistance to mass transfer that resides in the gas phase. For mass transfer across the hollow-fiber membrane contactors described in Example 2. 177. 1986) where k L. and in the gas. and k c are the individual mass-transfer coefficients in the liquid. k M. Nov. i.13: F P b.a fe c efficie i he i id ha e. Si ce di e i id i a be He ' a .25: Vi a a f he e i a ce e ide i he i id ha e. ca c a e KL. bi i ga e . S i F E a e 2.25. de he c di i de c ibed ab e.12c. E a e 2. a h gh i i a ia e ea . he ga i e facia c ce ai a i fie he e ai S i I he ga ha e: .b) U i g he e f a (a). 3. D i g ab i f .and k-t pe coefficients: absorption of low-solubilit gases. Combined use of F.e c efficie be ed i he ga ha e.e a . a a fe f a high c ce a ed ga i e a e di e i id i fe e a e ace.i = A. I ha ca e. 2.13. he i e facia c ce a i d i g ab i f . a) Sh ha . a d P b e 2. bi i ga e a e e a ed h gh A. a F. a d e i a e ha f ac i f he a e i a ce a a fe e ide i he i id fi . Hg ( ) .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 g 0.160 / 2.0 0 0. Hg ( ) 0. The e e a e a 303 K a d he a e e 1 a .a fe c efficie ee ca c a ed a FG = 0. E i a e he i e facia c ce a i a d he ca SO2 a f .5 224 S i Defi e: = g SO2/100 g a e = Pa ia e e f SO2. 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.. = 0.0 85 1.002 / 2.0 176 2. 1973): g SO2/100 g a e Pa ia e e f SO2.. The a .5 129 2. The e i ib i SO2 bi i da a a 303 K a e (Pe a d Chi .5 42 1.2% SO2 b e. Ini ial g e : . 2 mole %. The packing characteristics and flow rates at that point are such that FG = 1. Calculate the interfacial compositions and the local methanol flux.542 10 3 kmol/m2-s.6 K Solution For methanol (A) For water (B) .3.6. 1996): For water. for methanol. The temperature at that point in the tower is around 343 K.13d.. that of the bulk of the liquid phase is 60 mole %.650 10 3 kmol/m2-s.1 K. Tc = 512. et al. Tc = 647. At a different point in the packed distillation column of Example 3. modify the values given in Example 3. Distillation of a mi ture of methanol and ater in a packed to er: use of F-t pe mass-transfer coefficients.6 using Watson's method (Smith. To calculate the latent heats of vaporization at the new temperature. the methanol content of the bulk of the gas phase is 76. and FL = 8. Pa ame e Ini ial e ima e . . 14b.63 50 2.95 40 1.88 90 6.55 25 0.18 65 3. The e e i g ac i a ed ca b c ai 15 c 3 be e e a (a STP) ad bed e g a f he ca b . Material balances: adsorption of ben ene vapor on activated carbon. The e e a e a d a e e a e ai ai ed a 306 K a d 1 a .83 a) P he e i ib i da a a X' = g be e e/ g d ca b .26 80 4.22 100 7.0 3/ a i g ea f ac i a ed ca b a e e 85% f he be e e f he ga i a c i ce . 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. Y' = g be e e/ g i ge f a a e e f1a .3. Ac i a ed ca b i ed ec e be e e f a i ge -be e e a i e. Hg c 3 (STP)/g ca b 15 0. 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 ed c ec e a he a e f 1. S i . Si ce he e a i g i e i ab e he e i ib i c e a d he e i ib i c e i c ca e a d . a he i e ec i f Y = Y1 i h he e i ib i c e. ca e he i (X2. .Y2).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 he e e i g ca b c ai e ad bed be e e). S i O he XY diag a . he i i e a i g i e i b ai ed b ca i g. X1 a . c) If the carbon flow rate is 20% above the minimum, what will be the concentration of ben ene adsorbed on the carbon leaving? Solution 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 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 gas mixture leaving the desorber, and the number of ideal stages required. Solution 3.16b. Material balances: adsorption of ben ene vapor on activated carbon; cocurrent operation. If the adsorption process described in Problem 3.14 took place cocurrentl , calculate the minimum flow rate of activated carbon required. Solution Fom Problem 3.14: From the XY diagram: . The e i a ed each e i ib i .40 9. 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 .02 10.10 4. Material balances in batch processes: dr ing of soap with air.79 6.60 S i Ge e a e he XY diag a .06 3 f ai a 350 K.40 1.58 19.29 3. I i de i ed d 10 g f a f 20% i eb eigh e ha 6% i eb c ac i h h ai .90 7. he e 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 . Pa 2.76 2. a da ae-a a ia e e f 1.42 15.76 3.83 6.6 Pa.19 9. The e a i aced i a e e c ai i g 8. 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.33 12. 1 a .17b.56 4.96 7.63 8.3. X = 0. From the XY diagram.18b. at the e it of the fifth equilibrium stage.06 and 3. Material balances in batch processes: e traction of an aqueous nicotine solution . 0 kg of fresh. pure kerosene each. ith kerosene.56 7. 42. 103 kg nicotine/kg kerosene 1. I d.02 4. Determine the percentage e traction of nicotine if 100 kg of the feed solution is e tracted in a sequence of four batch ideal e tractions using 49.01 0. Water and kerosene are essentiall insoluble. 1950): X'.96 5.46 1. X = 0.81 2. Che . after 4 e tractions.. Nicotine in a ater solution containing 2% nicotine is to be e tracted ith kerosene at 293 K.70 Solution From the XY diagram. 103 kg nicotine/kg ater Y'. 166.86 9. The equilibrium data are as follo s (Claffe et al. E g.13 20..00422 .4 18.98 9.51 6. . c a f a age .27 i a che a ic diag a fac -f ca cade f idea age . S e P be 3. 1980) he e S i he i i g fac .3. a d c a e he e b ai ed b he eh d . bei g c ac ed i each age b a f e h V ha e. Each age i e e e ed b a ci c e. VS/LS. If he e i ib i -di ib i c e f he c -f ca cade i e e he e aigh a d f e .17 a d 3. i ca be h ha (T e ba . a e e a e faf c fig a i ca ed a c -f ca cade.18 i ge ai (3-60). e ec i e . Cross-flo cascade of ideal stages. The L ha e f f e age he e .i id e ac i eai de c ibed i P b e 3. S i . a d N i he a be f age . The d i g a d i id.19b. a d i hi each age a a fe cc a if i c c e f .18. Fig e 3. I i ia e i a e 3.20a. 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 age 3.21b. 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 i f He ' a , acc di g Yi = 48.3Xi, he e Xi = e H2S/ e f a e ; Yi = e H2S/ 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 minimum flo rate is used. Solution at SC b) Determine the composition of the e iting liquid. Solution c) Calculate the number of ideal stages required. Solution 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 ed i id f a e i ed ced d a a ica beca e he MEA eac i h he ab bed H2S i he i 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 ed b (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 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. The design water flow rate will be 50% higher than the minimum. Under these conditions, the equilibrium 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 5 e ilib i m age .24b. Calc la e al o he SO2 concen a ion in he a e lea ing he ab o be . Kremser equations: absorption of sulfur dio ide. Sol ion Ini ial e ima e .3. 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.23 hich i e i alen o 8. An ab o be i a ailable o ea he fl e ga of P oblem 3. e i ib i i gi e b W a i ace ic acid i e = 0.1% (b eigh ) ace ic acid. A e ac i c i a ai ab e hich i e i a e ac e c e ca cade f 15 e i ib i age .23 (a)? S i I i ia e i a e 3.000 g/h . 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 3.037% (b eigh ) ace ic acid b e ac i i h 3-he a a 298 K. .ha e.02% (b eigh ) ace ic acid. he e ha e i he L. Kremser equations: liquid e traction. F hi e .ha e. The i i 1. 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 . I i de i ed ed ce he c ce a i f hi i 0. A a e ace ic acid i f a he a e f 1. The i e 3-he a c ai 0.25b.828 W a i ace ic acid i ae S i Le he a e ha e be he V. A 1-butanol acid solution is to be e tracted ith pure ater.62 XAi. here YAi is the eight ratio of acid in the aqueous phase and XAi is the eight ratio of acid in the organic phase. ho man equilibrium stages are required for a countercurrent cascade? Solution . A total ater flo rate of 1005 kg/hr is used. Co n e c en e c o -flo e ac ion. the equilibrium data can be represented b YAi = 0. 1-butanol and ater are inmiscible.10% (b eight) acid. At 298 K. Operation is at 298 K and 1 atm.26c.5% (b eight) of acetic acid and flo s at the rate of 400 kg/hr. The butanol solution contains 4. a) If the outlet butanol stream is to contain 0. Initial estimate: 3. For practical purposes. Glucose sorption on an ion e change resin. b i ac -f ca cade.(g f g c e e i e f e i ) a d YAi i he g c e c ce a i i 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 f idea age . 1985) f d ha he e i ib i fg c e a i e cha ge e i i he ca ci f a i ea f c ce a i be 50 g/L. ha i he e 1-b a c ce a i ( ee P b e 3. The feed i f a he a e f 100 L/ i . he e XAi i he g c e c ce a i i he e i .25 g f g c e/L.(g f g c e e i e f i ).. hi e he e i f a he a e f 250 L/ i . S i . The c ce a i f he feed i i 15 g/L. The i e e i c ai 0. Thei e i ib i e e i a 303 K i YAi = 1. 242.27c. Ser.19)? S i 3. Chi g a d R h e (AIChE S mp. b) If he a e i i e a a g he a e be f age . We a a e c ce a i f 1. 81. .961 XAi. Fi d he be f e i ib i age e i ed.0 g/L. Solution .b) If 5 equilibrium stages are added to the cascade of part a). calculate the resin flo required to maintain the same degree of glucose sorption. 4. ca c a e he di a ce f he a he fi fi e ch ca i . Beca e f he ci a a e f he id-f ac i adia a ia i f ac ed bed . a) F e a i (4-1). I i ia e i a e f he ca be b ai ed f Fig.4 I i ia e i a e . 4. A e 4. F E a e 4. F he bed de c ibed i E a e 4. ca c a e he a ic i f he 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.1a.2b. Void fraction near the alls of packed beds.1. S i A e b) E i a e he id f ac i a a di a ce f 100 f he a . 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 . Void fraction near the alls of packed beds.1. S i The ca i ea he a e a he a ic i he J0( d) = 0. (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 i b) F he ac ed bed f E a e 4. a d f he fi fi e i i a.3c. 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 e 4.1. ca c a e he adia ca i f he fi fi e a i a. S i . Void fraction near the alls of packed beds. ca c a e he a i de f he id f ac i ci a i a h e i . 4 Ini ial e ima e Ini ial e ima e Repea ing hi p oced e.3 58.6 (1.1 30.1 95.7 77.04) 37.45) -16.8 (1. Sol ion F om he e l of pa (b).3 (2.85) 9.79) 6.92) 13. % .4 (4.85 .1 (3.32) -7.51) -27.3 (0.7 (3. he follo ing e l a e ob ained: Ma ima: Minima .2 67.mm ( *) Ampli de. hi m happen a * be een 3. 4.0 (2.6 c) 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 ha been dampened o le han 10% of he a mp o ic bed po o i .8 (4.57) -56.4 86.7 39. Ini ial e ima e of he oo can be ob ained f om Fig. % 20. mm ( *) Ampli de.39) -11.39 and 3.5 49.2 (1.98) 21.2 11. Annular packed beds (APBs) involving the flow of fluids are used in many technical and engineering applications.1. E. A correlation for this purpose was recently formulated (Mueller. G. Solution 4. The correlation is restricted to randomly packed beds in annular cylindrical containers of outside diameter Do. heat exchangers. 2458-60. Since APBs have two walls that can simultaneously affect the radial void fraction distribution. and fusion reactor blankets.. such as in chemical reactors. It is well known that the wall in a packed bed affects the radial void fraction distribution. 45.4c. Nov.(d) What fraction of the cross-sectional area of the packed bed is characterized by porosity fluctuations which are within 10% of the asymptotic bed porosity? Solution From part (c) (f) For the packed bed of Example 4.. it is essential to include this variation in transport models. estimate the average void fraction by numerical integration of equation (4-65) and estimate the ratio a / b . Void fraction near the alls of annular packed beds. inside diameter Di. AIChE J. equivalent diameter De = Do Di. . 1999). consisting of equal-sized spheres of diameter dp. The correlation is Consider an APB with outside diameter of 140 mm. with diameter aspect ratios of 4 De/dp 20. packed with identical 10-mm diameter spheres. (4-66). inside diameter of 40 mm. as predicted by Eq. for r* from 0 to R*. Solution (b) Plot the void fraction. . (a) Estimate the void fraction at a distance from the outer wall of 25 mm. min = 1.min = 0. Minimum liquid mass velocit for proper wetting of packing. A 1.2 mm/ . Calc la e he minim m a e flo a e. neded o en e p ope e ing of he packing face.6a.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 h 25-mm pla ic In alo addle . Sol ion 4. Sol ion Fo pla ic packing. Repea P oblem 4.5a.15 mm/ . (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. b ing ce amic in ead of pla ic In alo addle . Minimum liquid mass velocit for proper wetting of packing. F om S eam Table 4. L. L. in kg/ .5. Sol ion Fo ce amic packing. . Ch = 0.547 (Seader and Henley.2: .1: From Example 4.3). From Steam Tables 4. S ecific i id h d a d id f ac i i c ed ac i g.2: 4. Repeat Example 4.7b.2. Repeat Example 4. For this packing. but using 25-mm ceramic Berl saddles as packing material. S ecific i id h d a d id f ac i i fi -ge e a i a d ac i g.8b.2. 1998).979. Solution From Example 4. = 0. but using Montz metal B1-200 structured packing (very similar to the one shown in Figure 4. Solution From Table 4. a = 200 m 1. 9b. a d he h d a ic ecific a ea f he ac i g. The i id a e ci i L' = 2. S i F Tab e 4. S ecific i id h d a d id f ac i i fi -ge e a i a d ac i g.0 cP a d i de i i 840 g/ 3. A e ac ed i h 25.4.. The i c i f he i i 2. ce a ic Ra chig i g i be ed f ab bi g be e e a f a di e i e i h a i e ga i g a a h i a 300 K. he id f ac i .1: .71 g/ 2. E i a e he i id h d . X . Fp. Solution Packed Col mn Design Program This program calculates the diameter of a packed column to satisf a given pressure drop criterium. P. Ch Enter pressure drop constant. but using 15-mm ceramic Raschig rings as packing material. mL. in kg/m3 Enter gas densit . kg/m3 Enter liquid viscosit . Assume that.10b. Repeat E ample 4. C = 1. Cp Enter allowed pressure drop. T. in Pa/m Calculate flow parameter. and estimates the volumetric mass-transfer coefficients. m2/m3 Introduce a units conversion factor in Fp Enter porosit . in K Enter total pressure. Pa-s Enter temperature. fraction Enter loading constant. in Pa Enter data related to the packing Enter packing factor. Enter data related to the gas and liquid streams Enter liquid flow rate. a. in kg/s Enter gas flow rate. in ft2/ft3 Enter specific area.3. Pa-s Enter gas viscosit . Pressure drop in beds packed ith first-generation random packings. for this packing. in kg/s Enter liquid densit .4.783. mG. in m . vGf As a first estimate of the column diameter. dp. Calculate Y at flooding conditions Calculate gas velocit at flooding. QL. design for 70% of flooding Calculate gas volume flow rate. in m3/s Calculate effective particle si e. in m3/s Calculate liquid volume flow rate. QG. D. 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 . . X . and estimates the volumetric mass-transfer coefficients. Pa-s Enter temperature. Pa-s Enter gas viscosit . Solution Packed Col mn Design Program This program calculates the diameter of a packed column to satisf a given pressure drop criterium. fraction Enter loading constant. in Pa Enter data related to the packing Enter packing factor. Cp = 0.979. in kg/s Enter gas flow rate. For this packing. in K Enter total pressure. but using Montz metal B1-200 structured packing (very similar to the one shown in Figure 4. kg/m3 Enter liquid viscosit . in kg/m3 Enter gas densit . Fp = 22 ft2/ft3.547. 1998). T.3.3). mG. Fp. Ch Enter pressure drop constant.4. Repeat Example 4. m2/m3 Introduce a units conversion factor in Fp Enter porosit . in kg/s Enter liquid densit . a. Ch = 0. P e e d op and app oach o flooding in c ed packing. a = 200 m 1. Enter data related to the gas and liquid streams Enter liquid flow rate. P. in ft2/ft3 Enter specific area. mL. in Pa/m Calculate flow parameter.355 (Seader and Henley.11b. Cp Enter allowed pressure drop. = 0. design for 70% of flooding Calculate gas volume flow rate. in m . QL. dp. Calculate Y at flooding conditions Calculate gas velocit at flooding. D. in m3/s Calculate liquid volume flow rate. in m3/s Calculate effective particle si e. QG. vGf As a first estimate of the column diameter. all the ben ene is . 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.5 m3/s. containing 5 mole % ben ene. in me e F ac ional app oach o flooding 4. Pressure drop in beds packed ith second-generation random packings. measured at 110 kPa and 298 K. Esentiall . The gas enters the to er at the rate of 1. I e a e o find he o e diame e fo he gi en p e e d op Col mn diame e .12b. d. The ac i g i be 50. e a Pa i g .ab bed b he e e e. (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. 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%. The i id e e he e a he a e f 4.0 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. (b) A e ha .e ed f 400 Pa/ f i iga ed ac i g. i c i i 2. a d he e dia e e i be ch e d ce a ga .913 f = 0. he i id de i i 800 g/ 3.3 cP.825 (b) Ca c a e he e ed h gh he d ac i g . Ca c a e he e e i ed b he ga h gh he ac i g. Solution Using the Packed Column Design Program: D = 0.14c. d. k y ah = 0.390 (Seader and Henley. Cp = 0.26 kmol/m3-s 4.971.547. d. CV = 0.0885 cm2/s From Table 4.85 m f = 0. Fp = 22 ft2/ft3. CL = 0. is to be stripped . but using Montz metal B1-200 structured packing (very similar to the one shown in Figure 4.410 Using the Packed Column Design Program: k Lah = 0. From the wilke-Lee equation.355. 1998). 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. For this packing.3). Pressure drop in beds packed ith structured packings. containing 10 ppm (by weight) of benzene. Redesign the packed bed of Problem 4.376 kmol/m3-s 4. Air stripping of aste ater in a packed column. a = 200 m 1. DG = 0.038 m3/s.12.859 k Lah = 0. CV = 0. e = 0.00675 s 1. CL = 1. k y ah = 0.0 10 10 m2/s.13b.979.00861 s 1. Ch = 0. Assume that DL = 5.192.2. A wastewater stream of 0. Solution Calculate m. and y 2 (max) = y 2* = mx 2 .096 cm2/s.02 10 5 cm2/s (Cussler. The packing specified is 50-mm plastic Pall rings.with air in a packed column operating at 298 K and 2 atm to reduce the benzene concentration to 0. Estimate the corresponding mass-transfer coefficients. The diffusivity of benzene vapor in air at 298 K and 1 atm is 0. 1998). Henry's law constant for benzene in water at this temperature is 0. 1997).6 kPa-m3/mole (Davis and Cornwell.005 ppm. The air flow rate to be used is 5 times the minimum. 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. the diffusivity of liquid benzene in water at infinite dilution at 298 K is 1. the slope of the equilibrium curve: For water at 298 K. Calculate the tower diameter if the gas-pressure drop is not to exceed 500 Pa/m of packed height. 032 s 1.145 m f = 0. k y ah = 0.766 k Lah = 0.335 kmol/m3-s .Using the Packed Column Design Program: D = 1. Z .15b.5. Solution Initial estimate of the column height.4. Repeat E ample 4. Stripping chloroform from water b sparging with air. but using an air flo rate that is t ice the minimum required. Ini ial e ima e of ga hold p . Calculate po er required . Stripping chloroform from water b sparging with air. containing 200 orifices. 1973).15.11 kPa-m3/mole (Perry and Chilton. and specifying a chloroform removal efficiency of 99%.5.0 kW 4. Airflow will be 0. Solution For 99% removal efficiency. 1997). but using the same air flow rate used in Problem 4.30 m W T(Z) = 5. Stripping chlorine from water b sparging with air.5 kg/s with an initial chlorine concentration of 5 mg/L. Henry's law constant for chlorine in water at this temperature is 0. xin/xout = 100 Z = 1. each 3. The diffusivity of chlorine at infinite dilution in water at 298 K is 1. The water will flow continuously downward at the rate of 7.0 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.4. Repeat Example 4.0 mm in diameter. 25 cm in diameter. A vessel 2. (a) Assuming that all the resistance to mass transfer resides in the liquid phase.17b.25 10 5 cm2/s (Cussler. estimate the chlorine removal efficiency achieved.0 m in diameter and 2. Solution . The sparger is in the form of a ring.16b.22 kg/s at 298 K. . Ini ial e ima e of ga hold p . Ieai g i Z = 2. I he ea e f a e a e . A he b bb e i e.d. e ca be a fe ed f he ga he i id f he i id he ga de e di g he c ce a i d i i g f ce. de i ab e ga e a e f e e i ed de bed f he ae.18c.a d ge i ad bed i he a e he b bb e f ai a e di e ed ea he b f ae a i a d . Batch aste ater aeration using spargers.0 : (b) Ca c a e e e i ed 4. . 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.38 g/L (Da i a d C e . (b) E i a e he i e e i ed ai e he di ed ge c ce a i f 0. each i g c e ed ai a a a e f 0. .9% ge . P = ab e e e a he de h f ai e ea e. c ai i g 20 ifice . J . McG a -Hi . if he echa ica efficie c f he c e i 60%. S i . F he e ca e .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 . he e ai i e ea ed a a i c ea ed i id de h. 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 . B . P =a he ic e e.0 i dia e e .F ba ch ae a i i ac a e a . 1998). 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 .. Each a ge i i he f f a i g.01 g/ . a he ic c di i a e 298 K a d 101. O = a ge e ce i he ai ea i g he ae a i a . h gh he e C ide a 567 3 ae a i d ae a ed i h 15 a ge . 100 c i dia e e .3 Pa.5 g/L 6. (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). a i g ha c A * e ai e e ia c a . W. each 3. U de he e c di i . The a e e e a e i 298 K. 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 . Oeff . 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 . c = 8. W. 3 d ed.3 Pa c ai i g 20. The a ge i be ca ed 5 be he face f he d. Ma.0 a d c A. 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. Ind ial Wa e Poll ion Con ol. 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.0 g/L. . Ini ial e ima e of ga hold p . Initial estimate of transfer efficienc (b) Calculate power required . . he ef e he c . C ide he i a i de c ibed i P b e 4.d.ec i a a ea f he d cha ge a he a e de h cha ge . F he ae a i d fP be 4. ac ed i h 75.. i fed i h a e a 316 2 K a a a e f 25 g/ . he f i g ec f e i ge e a ed 4. 2 i dia e e . hich d f d he e. E i a e he c e di g a e f f eg e i a a i f he e . a d 7 . S i U i g he ga de e ed i P b 4. Ac i g e . da adc ec e he a e f . effect of liquid depth.18. Hi : Re e be ha he a e f he d e ai c a . The a e i c ac ed i h ai .18.7 f e . 4 .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. S i LMV = i id a e ci GMV = ga a e ci . f e f b bb e-diff i ae a i e he e ic a . a 300 K a d 101. Acc di g Ec e fe de (2000).18. i 3/ . 6 . e i a e he e ic a e f ai f . 4. Flooding conditions in a packed cooling to er.20c. Neg ec i g e a a i f he a e a d cha ge i he ai e e a e. ce a ic Hif i g . Batch aste ater aeration using spargers.3 Pa e e ia d .19c. ca c a e La a a e f Z = 3 . 878 .6. Design of a sieve-tra column for ethanol absorption Repea he calc la ion of E ample 4.6 m P = 590 Pa/ a F o de No.0165 EOG = 0. Sol ion F om he Sie e-Pla e De ign P og am. and 4.22 (no e ce i e eeping) E = 0.8123 EMG = 0.8865 EMGE = 0.21c.5 D = 1. 4.d.7.176 m = 0.8. he follo ing e l a e ob ained fo = 0. In od ce a ni con e ion fac o in Fp Ini ial e ima e of GMV 4.9 fo a col mn diame e co e ponding o 50% of flooding. = 1. 4. 6 m DP = 435 Pa/tray Froude No. Everything else remains the same as in Problem 4.116 cm2/s (est. The composition of the liquid is calculated as 1. Design for a 75% approach to the flood velocity.716 EMG = 0.93 m t = 0. the molecular weight = 93. tray spacing.45 .) Foaming factor = 0.720 4. Solution From the Sieve-Plate Design Program.23c. punched in stainless steel sheet metal 2 mm thick. 1980).75 D = 1.7. but for a 45% approach to flooding. weir length. Use a weir height of 40 mm.670 kg/m3 Viscosity = 118 mP (est.d. and entrainment in the gas.d. the following results are obtained for f = 0. gas-pressure drop.90 Vapor: Rate = 5. (a) Design a suitable cross-flow sieve-tray for such a tower.40 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).5 K Pressure = 100 kPa Liquid: Rate = 10. the average molecular weight = 19. are: Temperature = 371. Take do = 5.1.00 mass % aniline Density = 961 kg/m3 Viscosity = 0.722 EMGE = 0. 4.0 kg/s Composition = 7.27 10 5 cm2/s (est. From the Sieve-Plate Design Program. the following results are obtained for f = 0. Repeat Problem 4.) The equilibrium data at this concentration indicates that = 0.22c.5 mm on an equilateral- triangular pitch 12 mm between hole centers. Solution For aniline. The average molecular weight of the gas = 20. Check for excessive weeping. Report details respecting tower diameter.22. Design of a sieve-tra column for aniline stripping.0636 (Treybal. Solution EOG = 0. which are to be used to establish the design.3 cP Surface tension = 58 dyne/cm Diffusivity = 4.) Diffusivity = 0.045 Weir length = 1. A sieve-tray tower is to be designed for stripping an aniline (C6H7N)-water solution with steam.0 kg/s Composition = 3.22. The circumstances at the top of the tower.6 mole % aniline Density = 0. Design of a sieve-tra column for aniline stripping.4 mole % aniline.93 (no excessive weeping) E = 0. = 0. tray spacing.70 10 5 cm2/s (est.0 mm on an equilateral- triangular pitch 12 mm between hole centers. Take do = 6.174 m t = 0.24c.80 D = 1.6 m DP = 399 Pa/tray Froude No. (a) Design a suitable cross-flow sieve-tray for such a tower. = 0. and entrainment in the gas. Use a weir height of 50 mm.1 kmol/s Composition = 18 mole % methanol Viscosity = 125 mP (est.25 kmol/s Composition = 15.666 EMGE = 0.491 ( excessive weeping) E = 0. D = 2. Solution EOG = 0. Solution .) Diffusivity = 0.) Foaming factor = 1.660 EMG = 0.d.3 kPa Liquid: Rate = 0. Check for excessive weeping. gas-pressure drop. weir length.665 4. = 0. 1973).0173 Weir length = 1. A dilute aqueous solution of methanol is to be stripped with steam in a sieve-tray tower.052 Weir length = 0.3 cP Surface tension = 40 dyn/cm Diffusivity = 5.0 Vapor: Rate = 0.6 m DP = 386 Pa/tray Froude No. Design for 80% approach to the flood velocity. Report details respecting tower diameter. The conditions chosen for design are: Temperature = 368 K Pressure = 101.) The equilibrium data at this concentration indicates that = 2.0 mass % methanol Density = 961 kg/m3 Viscosity = 0.853 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).213 cm2/s (est. punched in stainless steel sheet metal 2 mm thick.5 (Perry and Chilton. Design of a sieve-tra column for methanol stripping.81 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).878 (no excessive weeping) E = 0. Solution From the Sieve-Plate Design Program. the following results are obtained for f = 0.49 m t = 0. 174 m t = 0.80 D = 1. the vapor pressure of n-butane at 310 K is 3. Repeat Problem 4. tray spacing.3 kmol/s Composition = 86% CH4.5.076 Weir length = 0. Design for a 75% approach to the flood velocity.) Foaming factor = 0.24.9 Vapor: Rate = 0. EOG = 0. Design of sieve-tra column for butane absorption. Sieve-tra column for methanol stripping.5 mm.14 10 5 cm2/s (est. punched in stainless steel sheet metal 2 mm thick.50 kmol/s Average molecular weight = 150 Density = 850 kg/m3 Viscosity = 1.936 EMGE = 0. but changing the perforation size to 4.6 cP Surface tension = 25 dyn/cm Diffusivity = 1.5987 EMG = 0. effect of hole si e. Use a weir height of 50 mm. keeping everything else constant.6 m DP = 686 Pa/tray Froude No.d.700 4.793 EMG = 0.d. A gas containing methane.26c.853 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a). = 1. Design for the optimal value of do on an equilateral-triangular pitch 12 mm between hole centers. Solution EOG = 0. 2% C4H10 Viscosity = 113 mP (est.561 (no excessive weeping) E = 0..035 cm2/s (est. Fro = 0. It is agreed to design a tray for the circumstances existing at the bottom of the tower .25c. gas-pressure drop. (a) Design a suitable cross-flow sieve-tray for such a tower. weir length.) Diffusivity = 0. the following results are obtained for = 0. Report details respecting tower diameter. 1987). and entrainment in the gas. and n-butane is to be scrubbed countercurrently in a sieve-tray tower with a hydrocarbon oil to absorb principally the butane.707 EMGE = 0. the optimal value of the ratio Ah/Aa is that which yields an orifice Froude number. According to Bennett and Kovak (2000).472 bar (Reid et al.) The system obeys Raoult's law. Solution . propane.918 4. Solution From the Sieve-Plate Design Program. where the conditions are: Temperature = 310 K Pressure = 350 kPa Liquid: Rate = 0. 12% C3H8. 5 is do = 3. Report details respecting tower diameter. Solution EOG = 0.) Foaming factor = 1. (a) Design a suitable cross-flow sieve-tray for such a tower. it was found that the orifice diameter that results in a Froude number = 0.85 (Treybal.75 mm on an equilateral- triangular pitch 12.0 Vapor: Rate = 0.775 m t = 0. Solution From the Sieve-Plate Design Program.5 (no excessive weeping) E = 0. A process for making small amounts of hydrogen by cracking ammonia is being considered.80 D = 0. Use a weir height of 40 mm. Check for excessive weeping.5 mm between hole centers. and at 303 K the slope of the equilibrium curve is m = 0.73 m t = 0.) For dilute solutions. The slope of the equilibrium curve is approximately 1.) Diffusivity = 0. = 0.42 10 5 cm2/s (est. Design of sieve-tra column for ammonia absorption.034 4. gas-pressure drop.6 m DP = 811 Pa/tray Froude No.7 kg/s Viscosity = 113 mP (est.9 cP Surface tension = 68 dyn/cm Diffusivity = 2.833 EMG = 1.0 kg/s Average molecular weight = 18 Density = 996 kg/m3 Viscosity = 0. 1980). containing 3% NH3 by volume. Conditions at the bottom of the tower are: Temperature = 303 K Pressure = 200 kPa Liquid: Rate = 6.0.75 D = 2.2. punched in stainless steel sheet metal 2 mm thick.43 mm. and residual uncracked ammonia is to be removed from the resulting gas.28 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a). the following results are obtained for = 0. and entrainment in the gas. The gas will consist of H2 and N2 in the molar ratio 3:1.038 EMGE = 1. tray spacing. NH3-H2O follows Henry's law. By trial-and-error.5 m DP = 624 Pa/tray . using the Sieve-Plate Design Program. Take do = 4. The average molecular weight of the gas is 20. weir length.d.013 Weir length = 2.230 cm2/s (est. Design for an 80% approach to the flood velocity. From the Sieve-Plate Design Program. The ammonia will be removed by scrubbing the gas countercurrently with pure liquid water in a sieve-tray tower.27c. the following results are obtained for = 0. 08 10 5 cm2/s (est. = 1.585 (no excessive weeping) E = 0. Design for a 60% approach to the flood velocity.8 mole/s Composition =48.28c.28 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).986 kg/m3 Viscosity = 337 mP (est.85 EMGE = 0.) The equilibrium data at this concentration indicates that m = 1. Use a weir height of 50 mm.60 D = 0.033 (no excessive weeping) E = 0. gas-pressure drop. Check for excessive weeping. Take do = 4. (a) Design a suitable cross-flow sieve-tray for such a tower.007 Weir length = 0.611 EMG = 0.7 mm between hole centers. = 0.8 kPa Liquid: Rate = 4. A sieve-tray tower is to be designed for distillation of a mixture of toluene and methylcyclohexane. Solution EOG = 0. Design of sieve-tra column for toluene-meth lc clohe ane distillation.8 mm on an equilateral- triangular pitch 12. Solution EOG = 0.5 m DP = 295 Pa/tray Froude No. Report details respecting tower diameter.0 mol % toluene Density = 726 kg/m3 Viscosity = 0.710 .847 4.d.) Diffusivity = 0. (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).9 dyne/cm Diffusivity = 7. and entrainment in the gas. weir length. Sol ion a) From the Sieve-Plate Design Program.22 cP Surface tension = 16. punched in stainless steel sheet metal 2 mm thick.6 m t = 0. The circumstances which are to be used to establish the design.80 Vapor: Rate = 4.152.048 Weir length = 2.712 EMGE = 0.6 mole % toluene Density = 2.) Foaming factor = 0. the following results are obtained for = 0.0386 cm2/s (est.436 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a). Froude No. tray spacing.806 EMG = 0. are: Temperature = 380 K Pressure = 98.54 mole/s Composition =44. J. McG a -Hi .1 e%a ia.1a.i= 1.3a.. S i . i hi a ge f c ce a i a d 300 K. 5.i (Ki g. 1971).34 e i ib i age . ca c a e he be f ea a . S i b) If he ab be e i e 5. Sepa a ion P oce e . NY.25.873 a da d c bic e e f ai a e fed he e e i ga f a e a e .414 A. a) E i a e he e a a efficie c . C. Ne Y . 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. he i e ai i f ee f a ia.75. Overall tra efficienc C ide a a ab be i h a c a M h ee efficie c EMGE = 0. a d 1. ca be e e e ed b A. Ca c a e he c ce a i f a ia i he e i a e if he i e i id c ce a i i 0. The e i ib i da a f hi e . Ammonia stripping from a wastewater in a tra tower. S i U e8 a 5. a d a a e age ab i fac A = 1. 581. Estimate the number of real tra s required. Initial estimate 5.4a. Solution Use 9 tra s 5.5b.3 and 5.3 is constant at 0. If the air flow rate to the absorber of Problems 5.5 standard m3/kg of . Ammonia stripping from a wastewater in a tra tower. Ammonia stripping from a wastewater in a tra tower. The Murphree plate efficienc for the ammonia stripper of Problem 5.4 is reduced to 1. = 0. Pi a da a a e a f : Li id f a e = 5.i = 0. A hea .0001 A.1%. a e .04 A.0 e /h Ga f a e = 2. The ce i bei g a i a ba i . The hea i i he ec c ed bac he ce he e he e A i i ed.i = 0. a e ha he i id a d ga f ae ae gh c a . a d i f ai f ca e.a e ie e.i de i ed. Fi d he e a c efficie c . The c e ab be i a 16.a c .7 A. 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.i = 0.i ea a 320 K i ed i a ab be e e di e a i ie f a Af a ai ea . a d he M h ee a e efficie c .5 e /h A. S i .i. S i I i ia e i a e 5. Absorption of an air pollutant in a tra tower.0001 E i ib i f A i gi e a A.6b. 0027 mole fraction ammonia. The system operates isothermally at 300 K and 1 atm. Solution Initial estimate .2.7b. Assuming that Henry s law applies.01 mole fraction ammonia. calculate the slope of the equilibrium line. the measured outlet gas concentration is 0. The ratio of L/V = 1. Water is used as the solvent to absorb ammonia from air.5. inlet gas concentration is 0. Absorption of ammonia in a laborator -scale tra tower. The inlet water is pure distilled water. An absorption column for laboratory use has been carefully constructed so that it has exactly 4 equilibrium stages. and is being used to measure equilibrium data. punched in sheet metal 2 mm thick. Assume isothermal scrubbing with pure water at 303 K.5 mm centers. Absorption of ammonia in a seive-tra tower. arranged in triangular pitch on 12. The gas will consist of H2 and N2 in the molar ratio 3:1.5 m tray spacing.015 m3/s-m of tower diameter (Treybal. = 0. The perforations are 4. The results are: mG = 0. at a pressure of 2 bars and a temperature of 303 K.8c.23 cm2/s Slope of the equilibrium line.42 10 9 m2/s Gas viscosity = 1. b) Calculate the concentration of the gas leaving the absorber in part a).d.0 Liquid diffusivity = 2. The weir height is 40 mm.122 10 5 kg/m-s Gas diffusivity = 0.657 kg/s Gas pressure drop = 615 Pa per tray . The gas flow rate should not exceed 80% of the flooding value.85 Solution a) Run the sieve-tray design program of Appendix E with different values of gas flow rate until D = 0.75 m at f = 0. containing 6 cross-flow trays at 0.8. A process for making small amounts of hydrogen by cracking ammonia is being considered and residual. a) Estimate the gas flow rate that can be processed in the column under the circumstances described above.068 N/m Foaming factor = 1.5. The water flow rate to be used should not exceed 50% of the maximum recommended for cross-flow sieve trays which is 0.75 mm in diameter. containing 3% NH3 by volume. 0. uncracked ammonia is to be removed from the resulting gas. 1980).75 m diameter. Data: Liquid density = 996 kg/m3 Surface tension = 0. There is available a sieve-tray tower. EMGE = 0.841 F o = 1.048 b) Ini ial e ima e .042 E = 0. A i id. he c ea ed a d c ed.i i he e f ac i f CO2 i he i id i . 1 a i be i e i ib i i h N1 I i ia e i a e . A e i he a eai . The ga i be b i a b bb e-ca e c bbe a 1. A a a fac i g d ice i b c e i ai d ce a f e ga hich. 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 .-ga a i f 1. 6% O2. Absorption of carbon dio ide in a bubble-cap tra to er. S i F he i i i id f a e. i c ai 15% CO2.2 i e he i i i ecified.9b.058 e CO2/ i . The c bbi g i id. a d 79% N2.5. i c ai 0. 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. hich i ec c ed f a i e . 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. The ga ea i g he c bbe i c ai 2% CO2.A 298 K a d 1.2 a a d 298 K.2 a . Calculate molecular weight of MEA solution Basis: 100 gm of 30% MEA in water b) Determine the number of theoretical tra s required for part a) Solution Generate XY diagram . 012 kg/m3. therefore m is not constant. and the number of real tra s required. the equilibrium-distribution curve is not a straight line. Seader and Henle (1998) proposed the following empirical correlation to estimate the overall efficienc of absorbers and strippers using bubble-cap tra s (it has also been used to obtain rough estimates for sieve-tra towers): where: EO = overall fractional efficienc m = slope of equilibrium curve mL = liquid viscosit . Estimate the overall tra efficienc for the absorber. Estimate thhe average value of m at liquid concentrations along the operating line and use theaverage in the correlation given. in cP 3 L = liquid densit . in kg/m Hint: In this problem.c) The monoethanolamine solution has a viscosit of 6. Solution .0 cP and a densit of 1. 9 CS2 vapor pressure = 346 mm Hg Solution . Carbon disulfide. The oil enters the absorber essentially pure at a rate 1. used as a solvent in a chemical plant. kg/s Data (at 297 K): Oil average molecular weight = 254 Oil viscosity = 4 cP Oil density = 810 kg/m3 Surface tension = 0.4 m3/s at 297 K and 1 atm. Design a sieve-tray tower for this process.11c. Absorption of carbon disulfide in a sieve-tra tower. is evaporated from the product in a dryer into an inert gas (essentially N2) in order to avoid an explosion hazard. Assuming isothermal operation.5 times the minimum. Design for a gas velocity which is 70% of the flooding velocity.5. The partial pressure of CS2 in the original gas is 50 mm Hg. determine: a) Liquid flow rate. The gas will flow at the rate of 0.030 N/m Foaming factor = 0. The CS2-N2 mixture is to be scrubbed with an absorbent hydrocarbon oil. CS2.d.5%. and solutions of oil and CS2 follow Raoult s law. and the CS2 concentration in the outlet gas is not to exceed 0. Ini ial e ima e . t = 0. 12 mm bet een centers Weir height = 50 mm. Pressure drop = 414 Pa/tra Fro = 0.705 m.712 E = 0.018 EMGE = 0.5 m. Plate thickness = 2 mm From the Mathcad program in Appendi E: D = 0.7 4. assuming straight equilibrium and operating lines .5-mm holes in triangular pitch.b) To er diameter and plate spacing Solution Plate design conditions: f = 0.588 d) Number of real tra s required Solution Appro imation. Graphical solution: Generate operating line in diagram . the number of ideal stages is also slightl over 5 .Graphicall . Solution . Use 10 tra s e) Total gas-pressure drop. gaseous N2 at 297 K Assume T1 = 298. liquid 76. Determine the number of equilibrium tra s for the absorber of Problem 5. liquid CS2. The vapor pressure of CS2 as a function of temperature is given b Solution Base for enthalpies: liquid oil. Adiabatic absorption of carbon disulfide in a sieve-tra tower.12c. The specific heats are: Substance J/mole-K Oil 362.5.1 CS2.9 the latent heat of vapori ation of CS2 at 297 K is 27.11.2 CS2. assuming adiabatic operation.2 N2 29.0 K .91 kJ/mole. gas 46. . . . . . . . Appro imatel 12 ideal stages Gas outlet temperature = 298 K . It enters the stripper containing 8 mole % benzene.5. A straw oil used to absorb benzene from coke-oven gas is to be steam-stripped in a sieve-plate column at atmospheric pressure to recover the disolved benzene.13b. 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. Equilibrium conditions at the operating temperature are approximated by Henry s law such that. The oil may be considered nonvolatile. The steam leaving contains 3 mole % C6H6. Steam-stripping of ben ene in a sieve-plate column. when the oil phase contains 10 mole % benzene.07 kPa. 75% of which is to be removed. C6H6. the equilibrium benzene partial pressure above the oil is 5. how many theoretical stages are required? Solution a) . b) Ke e e ai c) (G a hica c ci ahead) . Solution From Prob.11.d.15c.5 times the minimum and a gas-pressure drop not e ceeding 175 Pa/m of packing. Absorption of carbon disulfide in a random-packed to er. 5. Assume that Ch for the packing is 1.11. Design a to er packed ith 50-mm ceramic Hiflo rings for the carbon disulfide scrubber of Problem 5.0. packed height. Assume isothermal operation and use a liquid rate of 1. and total gas-pressure drop. the conditions at the bottom of the absorber are: . Kremser equation 5. Calculate the to er diameter. 2 Run the Packed To er Design Program of Ap.11: Calculate HtOG at the bottom of the to er .From Tables 4. D Results: Estimate the packed height.1 and 4. assume dilute solutions From Prob 5. 5.Calculate HtOG at the top of the tower From Prob.11. the conditions at the top of the absorber are: The mass-tranfer coefficients remain fairl constant along the tower . an air pollutant regulated by law. the operating line is parallel to the equilibrium line. Flue gases usually contain less than 1 mole % of SO2.0 It was found by trial and error with the Packed Column program that. using a packed tower that is 0.16b. The average gas- pressure drop is 200 Pa/m..d. and that the properties of the flue gases are similar to those of air.5. Solution For the driving force to be constant. It is desired to remove 90% of the sulfur dioxide in a flue gas stream at 298 K and 1 atm by countercurrent absorption with pure water at the same temperature. The tower is packed with 35-mm plastic NORPAC rings. Equilibrium is described by HenryÕs law with y i = 8. for a pressure drop of 200 Pa/m and D = 0.4x i. If the liquid flow is adjusted so that the driving force (y y*) is constant. calculate the height of the packed section. Absorption of sulfur dio ide in a random-packed to er. Assume that the properties of the liquid are similar to those of pure water.7 m: .7 m in diameter. and the absorption factor A = 1. . 0 mole % Operating pressure of absorber = 800 mm Hg Oil circulation rate = 2 m3/1.08 at 398 K Number of equilibrium stages = 5 a) In the winter. it is possible to cool the recycled oil to 293 K. The resulting benzene-wash-oil solution is then heated to 398 K and stripped in a tray tower. using steam as the stripping medium. Ben ene vapor recover s stem. Solution .0 kg of steam is used in the stripper per 1.88 Molecular weight = 260 Henry s law constant = 0. 398 K Henry s law constant = 3. = 0.130 at 300 K NtOG = 5 transfer units Stripping: Pressure = 1 atm Steam at 1 atm. at which temperature the absorber then operates. The stripped wash oil is then cooled and recycled to the absorber. Calculate the percent benzene recovery in the winter. Under these conditions 72.000 m3 of gas at STP Oil specific gravity = 0.095 at 293 K. Benzene vapor in the gaseous effluent of an industrial process is scrubbed with a wash oil in a countercurrent packed absorber.5.17c. Some data relative to the operation follow: Absorption: Benzene in entering gas = 1.000 m3 of gas at STP entering the absorber. ha e ec e f be e e ca be e ec ed? S i . A i g ha he ab be he e a e a 300 K. 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 he a ai ab e c i g a e . i h he a e i a d ea a e . a d ha N OG a d e i ib i age e ai he a e. c) If he oil a e canno be inc ea ed. ha mme eco e of ben ene can be e pec ed? . b he eam a e in he mme i inc ea ed b 50% o e he in e al e. 5.. Gas-phase mass-transfer coefficients for GeCl4 and Cl2 can be estimated from the following empirical equations developed from experimental studies with 13-mm Raschig rings for liquid mass velocities between 0..Gy mass velocities. For the packing. Fp = 580 ft 1. The entering gas flows at the rate of 23. 1971): where: ds = equivalent packing diameter (0. it can be assumed that both gases are absorbed independenttly. at least.19c. e = 0. Because the solutions are very dilute. 17. It is desired to absorb. 631. take DGeCl4 = 0. kg/m2-s For the two diffusing species.0 kg/m2-s (Shulman.06 cm2/s. which. Germanium tetrachloride (GeCl4) and silicon tetrachloride (SiCl4) are used in the production of optical fibers. 99% of both GeCl4 and Cl2 in an existing 0. H. et al. The liquid rate should be set so that the column operates at 75% of flooding. Thus. At these conditions.63. when dissolved. The air also contains 540 kg/day of Cl2. the GeCl4 oxidation is quite incomplete and it is necessary to scrub the unreacted GeCl4 from its air carrier in a packed column operating at 298 K and 1 atm with a dilute caustic solution. Solution .75-m-diameter column that is packed to a height of 3. AIC E J. also will have no vapor pressure. L. DCl2 = 013 cm2/s. the dissolved GeCl4 has no vapor pressure and mass transfer is controlled by the gas phase.68 2. Both chlorides are oxidized at high temperature and converted to glasslike particles. Ab o p ion of ge mani m e achlo ide ed fo op ical fibe . Determine: a) Liquid flow rate. However. the equilibrium curve is a straight line of zero slope. in kg/s..01774 m for 13-mm ceramic Raschig rings) = gas density.850 kg/day of air containing 288 kg/day of GeCl4.0 m with 13-mm ceramic Raschig rings. kg/m3 Gx . . 0 m of packing. Sol ion (dil e ol ion ) . Ini ial e ima e b) The pe cen ab o p ion of GeCl4 and Cl2 ba ed on he a ailable 3. . 165 CV = 0. 1998): Fp = 33 ft 1 a = 300 m 1 = 0.20c. assume dilute solutions . Redesign the absorber of Problem 5. The characteristics of this packing are (Seader and Henley.422 Solution From the Packed Tower Program in Appendix D: Estimate the packed height.482 Cp = 0. Absorption of carbon disulfide in a structured-packed to er.93 Ch = 0.d.295 CL = 1.15 using metal Montz B1-300 structured packing. 5. From Prob 5.11: Calculate HtOG at the bottom of the to er . the conditions at the top of the absorber are: The mass-tranfer coefficients remain fairl constant along the tower . 5.11.Calculate HtOG at the top of the tower From Prob. and if so. There is available a 0.5.04% by volume by water scrubbing. Is the tower satisfactory.3-m-diameter tower packed with 25-mm ceramic Raschig rings to a depth of 3. ammonia-water solutions follow Henry s law up to 5 mole % ammonia in the liquid.414.05 m3/s of an ammonia-air mixture (300 K and 1 atm) from 5.0 to 0.21c. Solution Calculate minimum water flow rate Try .5 m. It is desired to reduce the ammonia content of 0. with m = 1. Absorption of ammonia in a random-packed to er. what water rate should be used? At 300 K.d. From Appendi D Tr From Appendi D . Tr From Appendi D . Tr From Appendi D . ec i a a ea. C bic i ei e ai 5. P e a e e e a a f a e f 1976 e / 2. c ai 20 e % f SO2. a a SO2-f ee f a e f 37.f bed c . The e i i g ga i c ai 0.44 e / 2- f bed c . Absorption of sulfur dio ide in a structured-packed to er. E i ib i da a f bi i f SO2 i a e a 303 a d 1 a ha e bee fi ed b ea .d.22b. 1998): a) De i e he f i g eai g i ee ai f he ab be : S i . A e ha ei he ai a e i a fe be ee he ha e a d ha he e eae i he a a 2 a a d 303 K. A e ac ed i h e a M B1-300 c ed ac i g i be de ig ed ab b SO2 f ai b c bbi g i h a e .5 e % SO2. The e e i g ga .ec i a a ea. ae he e a i (Seade a d He e . calculate the water flow rate.0 m3/s (at 2 atm and 303 K) of the entering gas. b trial-and-error: c) . Solution From Ap. b) If the absorber is to process 1. the tower diameter. D. and the gas-pressure drop per unit of packing height at the bottom of the absorber. b trial-and-error: Average mass-transfer coefficients . D. At the top of the column: From Ap. A e age heigh Calc la e N G Ini ial e ima e . . 3 37.3 for ternary mixtures. 6.and the parameters of the Wagner equation for estimating vapor pressure (equation 6-5) are included in the following table (Reid. Test your program with the data presented in Example 6.3.286 1.. B.792 o-xylene 630.0 7. and o-xylene (C).534 1. Modify the Mathcad¨ program of Figure 6.2a.9 6. 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. 50 mole % n-octane (B).3. bar A B C D benzene 562. at 303 K. toluene. at 303 K. toluene (B). Flash vapori ation of a heptane-octane mi ture.332 2.3d. 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. 50 mole % n-octane (B).334 toluene 591. 1987).410 3.8 41. A liquid mixture containing 50 mole % n-heptane (A). Component Tc. is to be continuously flash-vaporized at a temperature of 350 K to vaporize 30 mole % of the feed.2 48.2 for a mixture of benzene (A). Flash vapori ation of a heptane-octane mi ture.3 7.110 2. is to be continuously flash-vaporized at a pressure of 1 atm to vaporize 30 mole % of the feed.860 Solution Parameters of the Wagner equation. K Pc. Critical temperatures and pressures.983 1.834 2.1a. C = o-xylene: . A = benzene.381 2. 6. Flash vapori ation of a ternar mi ture. et al. 6. A liquid mixture containing 50 mole % n-heptane (A).629 3. Ini ial e ima e . 3. E i a e he e e a e. S i .2 a d P b e 6. C ide he e a i e fE a e 6.6. a d c 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 .4d. Flash vapori ation of a ternar mi ture. 5d. Calculate the temperature. fraction of the feed vapori ed.lene in the feed. and to recover in the liquid 70% of the o.2 and Problem 6.3. pressure. and the concentration of the liquid and gas phases Solution .6. Consider the ternar mi ture of E ample 6. Flash vapori ation of a ternar mi ture. It is desired to recover in the vapor 75% of the ben ene in the feed. Batch distillation of a heptane-octane mi ture.6.6b. Solution Batch distillation of a mi ture of heptane and octane .3. but for 80 mole % of the liquid distilled. Repeat the calculations of E ample 6. c a e a 1 a . 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.9b. -he a e i h . he a e age e a i e a i i i = 2.c a e a 1 a .8a.3. U i g e a i (6-102) de i ed i P b e 6.16 (T e ba . he a e age e a i e a i i i = 2. 1980).16 (T e ba . F hi e .3. Binar batch distillation with constant relative volatilit . -he a e i h .7. c e he c ii f he e id e af e 60 e% f he feed i ba ch-di i ed. C ide he bi a ba ch di i a i fE a e 6. 1980). C ide he bi a ba ch di i a i fE a e 6. Binar batch distillation with constant relative volatilit . U i g e a i (6-102) de i ed i . I i ia e i a e 6. F hi e . S i I i ia e i a e: 6. Solution .0 atm.10b. Solution Initial estimates: 6. compute the composition of the residue. Mi tures of light h drocarbons: m-value correlations. and the fraction of the feed that is distilled.7. What is the bubble point of a mi ture that is 15 mole % isopentane. 30 mole % n-pentane. Problem 6. and 55 mole % n-he ane? The pressure is 1. isobutane. e pressed as mole percent: ethane. 0. 18. 25 %. 56 %. 0. Mi tures of light h drocarbons: m-value correlations. isopentane.4 to calculate equilibrium distribution coefficients.25 %. a) Calculate the bubble point Solution .11c. Initial estimate 6. Use equation (6-103) and Table 6. In the follo ing. the pressure is 10 bars.5 %. A solution has the follo ing composition. n-butane. propane.25 %. Initial estimate b) Calculate the de point Solution . Solution Initial estimates . Initial estimate c) The solution is flash-vapori ed to vapori e 40 mol % of the feed. Calculate the composition of the products. Binar batch distillation with constant relative volatilit . A product having an . A 30 mole % feed of ben ene in toluene is to be distilled in a batch operation.12b. Check! 6. 1 354.944 C ii f he a e ei = = 0. S i I i ia e i a e : 6.685.5 0.524 0.198 0.453 0.742 0.2 K. Batch distillation of a mi ture of isopropanol in ater.5 a d F = 100 e .569 0.084 0. a 1 a a e (Seade a d He e . VLE da a f hi e .593 0. K 366 357 355.3 354. a i g ha = 2.6 353. Ca c a e he a f e id e ef . S i . b i i g i f he a e e = 353. A i e f 40 e%i a i ae i be ba ch-di i ed a 1 a i 70 e % f he cha ge ha bee a i ed. a d he a e age c ii f he c ec ed di i a e.012 0.i e f ac i f i a . Ca c a e he c ii f he i id e id e e ai i g i he i .220 0.13b. 1998): T. a e age c ii f 45 e % be e ei be d ced.3 353.2 353.679 0.462 0.916 0.682 0.769 0.350 0. Fe e e a i (6-58) ca be 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 . A e ha .5. I i ia e i a e 6. U e Fe ee ai e i a e N i f di i a i f he be e e.16a. 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 ca be de e i ed a a ica f he f i g e a i (T e ba . he e a i e aii i c a a = 2. 1980): (6-105) . Continuous distillation of a binar mi ture of constant relative volatilit . e e i e fE a e 6.4. S i 6. N i . 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 . f hi e a 1 a .15a. 13.5. he e a i e aii i c a a = 2. S i I i ia e i a e 6. VLE da a a e gi e i P b e 6.5 i e he i i i ed. Continuous rectification of a ater-isopropanol mi ture.4. A e ha .17b.U e e a i (6-105) e i a e R i f di i a i f he be e e. f hi e a 1 a . Ni e -eigh e ce f he i a i he feed be ec e ed.5 e% i a . If a ef a i f 1. h a he e ica age i be e i ed: (a) If a a ia eb i e i ed? S i . e e i e fE a e 6. 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. i h a a ia eb i e a d a a c de e . I a gi e a ica i . 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. The di i a e i e e a e b a e i h a a e e a = 70 a d = 0. The e e a e ea i e ed a a a a ed i id he c he a a he ea e i id c ce a i . a a ge i ch. The i i be fed a di i a i c eai ga a he ic e e. S i .6.24c. Ma i d ia i a i id e a e diffic i ib e e aaeb i ec i di i a i beca e he ha e beha i c ai a a e e. a d a e id e c ai i g 99 e% a e . 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. The e b a e b he c ce a i ha he e ea e ea i he e ha . a ea ea i e aii .99). 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 . 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) Ca c a e he a f a e f he d c a d f he e id e. The ef ai i be 1. A distillation-membrane h brid for ethanol deh dration.ich d c ( P = 0.6.5 he i i . The mole fraction of the light ke in the distillate should be 0. A distillation column has a feed of 100 kmoles/h.2 the minimum. Equilibrium data: LNK = 4.26b. The feed is 10 mole % LNK. Initial estimates 6. Fenske-Under ood-Gilliland method. The reflu ratio is 1. We desire 99.5 % recover of the light ke in the distillate.0 HK = 0. 55 mole % LK. and 35 mole % HK and is a saturated liquid.0 LK = 1.75. Use the FUG approach to estimate the number of ideal stages required and the optimal location of the feed-stage.75 Solution Kilomoles in distillate: . q = 1 Initial estimate .Kilomoles in bottoms: Determine minimum reflu : Saturated liquid. 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: . Fenske-Under ood-Gilliland method.4 1 a) F a di i a e a e f 60 e /h.i i e be e a a ed b di i a i ha he f i gc ii : C e M e f ac i Re a i e a i i A 0.4 5 B 0. a d a ef . S i I i ia e i a e : . Feed a age 11 6. fi e he e ica age .27b.2 3 C 0. ca c a e he di i a e a db c i i b he Fe ee ai . O e h d ed e /h f a e a b bb e. Solution Initial estimate: . determine the minimum reflu ratio b the Under ood equation.b) Using the separation in part a) for components A and C. and the optimum feed location. Solution . Initial estimate: Initial estimate: c) For a reflu ratio of 1.2 times the minimum. determine the number of theoretical stages required. NY. McG a -Hi . Ne Y . I i ia e i a e Ki b ide e ai : I i ia e i a e: Feed a age 5 6. C ce a De ig f Di i a i S e .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. 2001): . b) Solve E ample 6. Solution Initial estimate: .2 using the Rachford-Rice method. 2 Single e traction.7. Appendi G-1: Single-Stage E traction Ini ial e ima e : . 7.3. Single-stage e traction: insoluble liquids 7.4. Single-stage e traction: insoluble liquids 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 ae 7.6 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 Ini ial e ima e : Ini ial g e e : Ini ial g e e : Composited e tract Check material balances Acetone Chloroform Water .
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