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 S i ec a eigh f he i e. c) The S i a a c ce ai . d) The S i a de 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 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 eigh each i o b) The a e age molec la Sol ion eigh of he mi e. 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 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 i g a c ii : 93.5% b) Weigh f ac i S i e f CH4 i he i e. Ba i : 100 f LNG c) The LNG i hea ed 300 K a d 140 Pa, a d a ga i e de he e c di i . S i i e c e e . E i a e he de i f he 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 Le e e (a a pe cen ) = mola f ac ion of a e in he mi Ini ial e ima e F om he gi en a e eigh f ac ion (0.0617): A ga i ea a a e e f 150 Pa a d 295 K c ai 20% H2. I c i fa e i i e fa ia a d ai .5b. 40% O2. a d 12 / . a . Material balances around an ammonia gas absorber. ca c a e he a ia f ac i i he ga ea ab be . a d 40% H2O b e. A ga ea f a he a e f 10.1. S i : a e ci (defi ed a a f ae e i e be 1. S i : Ba i : 1 ec a be he f 2. The ab be e e i c i d ica i h a i e a dia e e a) Neg ec i g he e a a i f a e .6b.5 . e ec i e . Velocities and flu es in a gas mi ture. a d i a ai . i g he d A=a ia B = ai b) Ca c a e he e ga c .ec i a a ea). The ga e e h gh he b f a ac ed bed ga ab he e i f c ec e a ea f e i id a e ha ab b 90% f a f a ia. -2 / . The ab e e ci ie f each ecie a e -10 / .0 3/ a 300 K a d 102 Pa. a) Determine the mass average velocit . Solution: . and the molar average velocit . NO2. JO2. for the mi ture. nO2. v. Solution Molar average velocit . V Mass average velocit . vm Basis: 1 kmole of gas mi ture b) Evaluate the four flu es: jO2.in the direction of the -a is. V. . 3 Pa i f i g e ie f he ga i e: a) M e f ac i f ae a .3 c ai e a 340 K a d 101. P ope ie of ai a a ed i h a e apo . . a a d -b b e e a e f 30 C a d a h idi f 0. 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. The a e e e he c de e a 29 C a d ea e a 45 C. c) T a a c ai ed i he a .500 g/ f d ai . Ai . S C A i ide he ai ea i g he i ee ai f ae a : e a a ed a 38 C.016 g f a e / g f d ai . The c i g a e f ae he c de e f a big c a -fi ed e a i 8.1. I ea e he c i g e a a ed i h a e a a 38 C.970 g/ . b) A e age ec a eigh f he i e. ed i a 30. Wa e balance a o nd an ind ial cooling o e . F he c de e . d) Ma f ae a i he a .11).8c.7b. De e i e he F b) a a a ed i e c) d) 1. a) Ca c a e he a e e b e a a i i he c i g e. S a) A i i ee ai f ae a : a a ed i h a e a . The ai e e he c i g e a he a e f 6. a fe face . he ci c a i g a e i c ai e ha 2. S i W= M= i dage e = 500 a e ae aeB=b c = 2000 d ae I i ia e i a e Wa e ba a ce S id ba a ce: . a a a f he ci c a i g a e be de ibe a e di ca ded (b d ). The ef e. E i a e he a e . Thi a e a e c ai 500 g/L f di ed id . T a id f i g f he c de e hea . a 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 . Wi dage e f he e ae e i a ed a 0.2% f he eci c a i a e.a e e i e e .000 g/L f di ed id .E= ae b ea ai i he ai b) T acc f ae e i he c i g e . in kg/min.1. Solution . Solution c) Calculate the water-vapor partial pressure and relative humidit in the air leaving the dr er. The air enters the dr er at the rate of 30. and initial water-vapor partial pressure of 1.3 kPa. in kg of water/kg of dr air.9b. a) Calculate the moisture content of the entering air. Solution b) Calculate the flow rate of dr air. 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. The dr er operates at constant temperature and pressure. Water balance around a soap dr er.6 kPa.0 m3/min at 350 K. 101. Calc la e a e a e e a 350 K A i ee ai f ae a : . . A.10b. I a eff ed ce he e ec e f hi ga . P.500 3 a 298 K a d 101. 1987.): e he e W i i ib i he ca b e i ib i e e a ia e ad ii . a d * i be be ee 0.3% e e i ai .1. a d he a f e e ad bed b he ca b . i i ia f ee f e e. material balances. a d cc ie a e f 2. ed. a d g f e e/ g f ca b .i e.7 a d 345 Pa. NC. Re ea ch T ia g e Pa . A i g ha he ai d e ad b he ca b . E. S. The e i a ed each e i ib i a c a e ea e a d e e.. i Pa.3 Pa. i i e ed 100 g f ac i a ed ca b . Acti ated carbon adsorption. A a e ga c ai 0. he S i M= = a e f ca b f e e ad bed I i ia e i a e . 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 . 3 d. ca c a e he e i ib i c ce a i f e e i he ga e ha e. U. Solution .11b. Estimate how much activated carbon should be used if the s stem is allowed to reach equilibrium at constant temperature and pressure. Acti ated carbon adsorption.1.5% of the toluene originall present in the waste gas of Problem 1. It is desired to adsorb 99.10. material balances. O eg S a e U i e i . M. . E i a e he diff i i ai a 298 K a d 1 a a d e ed i a e a 0. Estimation of gas diffusivit b the Wilke-Lee equation. E. f S F A B i e di 1. La (MS he i . d. Estimation of gas diffusivit b the Wilke-Lee equation. d.093 c 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.12a.13a. 1964) ea ed he diff i i f ch 2/ .1. 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.087 cm2/ epo ed in Appendi A. Sol ion . and i c i ical ol me i 254 cm3/mol.a) E ima e he diff ii of naph halene (C10H8) in ai a 303 K and 1 ba . and i c i ical ol me i 413 cm3/mol. The no mal boiling poin of naph halene i 491.4 K.1 K.437 cm2/ epo ed in Appendi A. Compa e i i h he e pe imen al al e of 0. Compa e i i h he e pe imen al al e of 0. The no mal boiling poin of p idine i 388. E g. Compa e i i h he e pe imen al al e of 0. and i c i ical ol me i 274 cm3/mol..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.061 cm2/ (G illiland. E. The no mal boiling poin of aniline i 457. Sol ion E pe imen al al e . I d. R. 2 :681. Che . 1934)..6 K. but the collision integral is now given by mp = dipole moment..1.14d. P. TbB) Solution . Chem. Equation (1-49) is still used.162 m3)1/2] 10-25 (J- a) Modify the Mathcad routine of Figure 1. VA . Process Design De elop. Eng. MB. Brokaw (Ind. debyes [1 debye = 3. mB. 1969) has suggested an alternative method for this case. VB. Use the function name DABp(T. MA .3 to implement Brokaw's method. mA . a modified Lennard-Jones relation is often used. Diffusivit of polar gases If one or both components of a binary gas mixture are polar. 8:240. TbA . . 1 1.8 M 36..9 . 1987): ide S f di ide S i 1.1 373.b) E i a e he diff i 323 K.8 . K 249. 1987): Pa a e e H d ge ch Tb . e a .14) a e h be (Reid..15d. 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 .2 Vb .6 43.1 263. The da a e e B a ' e a i ( ee P b e 1. c 3/ 30. deb e 1.2 Vb . deb e 1. c 3/ 50.078 c eB a ' (Reid. a d c aei ea i a e h be Pa a e e Me h ch Tb .9 1.6 M 50.5 18 ide Wa e i ed . K 188.06 c efficie f a i e f e h ch ide a d f di ide a 1 ba a d 2/ .6 18. The da a e i ed he e e i e a a e f 0.5 64. e a . 16d.5 ba .S i 1.08 64.. 1987): S i .6 M 34.8 . K 189. c 3/ 35.06 di ide fide i f di ide a 298 K a d 1. The da a 1. e a .9 1.6 263.14) a e h be (Reid. Diffusivit of polar gases E a a e he diff i c efficie f h d ge e i ed e B a ' e a i ( eeP b e Pa a e e H d ge fide S f Tb . deb e 0.03 43.2 Vb . carbon dio ide (4) Calculate binar MS diffusivities from Wilke-Lee equation . carbon mono ide (3).1.ution Calculate mole fractions on a nitrogen (1)-free basis: o gen (2).d.5 bar. The mi ture composition is: O2 15 mole % CO 30% CO2 35% N2 20% Sol. Calculate the effective diffusivit of nitrogen through a stagnant gas mi ture at 373 K and 1.17a. Effective diffusivit in a multicomponent stagnant gas mi ture. An important parameter of the model is the effective diffusivit of mercuric chloride vapor traces in the flue gas. Meserole et al. If the flue gas is at 1. 12% CO2. and its composition (on a mercuric chloride-free basis) is 6% O2. One promising approach for removing mercur from fossil-fired flue gas involves the direct injection of activated carbon into the gas.22 cm2/s. A oc. estimate the effective diffusivit of mercuric chloride in the flue gas. Solution HgCl2 (1) O2 (2) CO2 (3) H2O (4) N2 (5) . 1999) describe a theoretical model for estimating mercur removal b the sorbent injection process.1.d.18a. Mercur is considered for possible regulation in the electric power industr under Title III of the 1990 Clean Air Act Amendments. (J.013 bar and 408 K. and 75% N2. et al. Ai & Wa e Manage. 49:694-704. 7% H2O. reported an effective diffusivit value of 0. Meserole.. Assume that onl the HgCl2 is adsorbed b the activated carbon. Mercur removal from flue gases b sorbent injection. . O e .9 c 3/ . The . C ae he e e i e a a e e ed b Reid. S. B. b) E i a e he diff i c efficie a 273 K f i c i f i id a e a 273 K i 1. he diff i c efficie i gi e b he Ne -Ha e e a i (Ha ed. F di e i f a i g e a . Wilke-Chang method for liquid diffusivit . "The Ph ica Che i f E ec ic S i . he diff i fa i ge a a be ea ed a ec a diff i .i a he ha ec e diff e. e a . Whe a a di cia e i i . a d B.1. I he ab e ce f a e ec ic e ia . H.. 1950): a) E i a e he diff S i i c efficie a 298 K f a e di e i f HC i ae.19a. 95. The i c i f i id e ha a 298 K i 1. 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. S i 1. The c i ica e f ca b e ach ide i 275.08 cP.20b. (1987) a 1. a e di e i f C SO4 i a e .79 cP. Diffusion in electrol te solutions.5 10-5 c 2/ ." ACS M g . The critical volume of oxygen is 73. O gen diffusion in ater: Ha duk and Minhas correlation.9 cP. 2nd ed..4 cm3/mol. the viscosity of water is 0.1 10 5 cm2/s (Cussler E. Estimate the diffusion coefficient of oxygen in liquid water at 298 K. At this temperature. UK. Solution .1. Cambridge. Cambridge University Press. L. 1997).21a. Use the Hayduk and Minhas correlation for solutes in aqueous solutions. The experimental value of this diffusivity was reported as 2. Diff sion. 7 10 5 cm2/s. The following data are available (Reid.. d. 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. Liquid diffusivit : Ha duk and Minhas correlation. 1987): Solution . Compare the estimate to the reported value of 3.22a. et al.1. et al. The viscosity of water at 288 K is 1. Based on this result. estimate the molar volume of allyl alcohol at its normal boiling point.15 cP. Solution Iniitial estimates From Table 2. and the Hayduk and Minhas correlation for aqueous solutions..2. E ima ing mola ol me f om li id diff ion da a.23b.1 . 1987). The diffusivity of allyl alcohol (C3H6O) in dilute aqueous solution at 288 K is 0.1.9 10 5 cm2/s (Reid. Compare it to the result obtained using the data on Table 1. .. 5 h ed. e a . 890.1. he ac i i c efficie f ace e i gi e b Wi e ai (S i h. he i fi i e di a 298 K i i diff ii f a e i e ha 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%.. S e . McG a -Hi C . Concentration dependence of binar liquid diffusivities. Ne Y . M. Trans. I c.. NY. 1996): S i E i a e he he d a ic fac . 1953): The e e i e a a e e c 2/ . B.42 10-5 S i E i a e he i fi i e di i f Ha d -Mi ha f a diff e ii i f e ha i a e a 298 K F A e di A. R. 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%.. J. Introduction to Chemical Engineering Thermod namics. a d R. F hi e a 298 K. 49. Farada Soc.. d. H.24b. U de he e c di i (Ha d. ed b Ha da dS e (1953) i 0. F A e di A E i a e he i fi i e di i f Ha d -Mi ha f a diff e ii i f ace ei a e a 298 K . The reaction at the surface ma be assumed to be instantaneous. gas-phase flu calculation.25b.The temperature of the gas film is 600 K. in kg/m2-min. d. ne t to the carbon surface. On the outside of the film. the thickness of this film is 1. and the pressure is 1 bar. Estimate the rate of combustion of the carbon. O2 (3) Calculate binar MS diffusivities from Wilke-Lee . there is virtuall no o gen. therefore. CO2 (2).0 mm.1. and 40% CO2. one-dimensional. Solution CO (1). 20% O2. the gas concentration is 40% CO. 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. Stead -state. 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) . . . : 100 H2O (36 ) (P ) .8 / 3. /100 C . S A = N 2SO4 B = H2O C )B A1 ( 288 K. Stead -state.240 / 3 (P 1973). T 288 K 999.T G ' 288 K 36 1. 1-20. T N 2SO4 288 K P 1. .A 0.085 .26b.A ( ) N 2SO4.153 P. one-dimensional.1. liquid-phase flu calculation. A G ' (N 2SO4 10H2O) E N 2SO4 . 77 10 5 cm2/s. Neglecting water evaporation.27c.42 mol percent and the concentration at the lower boundary of the water layer is esentially zero.The temperature of the system is 288 K and the total pressure is 1 atm. is being selectively removed from an air-NH3 mixture by absorption into water. The concentration of ammonia at the outer boundary of the gas layer is 3.Calculate diffusivity 1. in kg/m2-hr. NH3. ammonia is transferred by molecular diffusion through a stagnant gas layer 5 mm thick and then through a stagnant water layer 0. determine the rate of diffusion of ammonia. Molec la diff ion h o gh a ga -li id in e face. d. . Ammonia.1 mm thick. The diffusivity of ammonia in air under these conditions is 0. Assume that the gas and liquid are in equilibrium at the interface.215 cm2/s and in liquid water is 1. In this steady-state process. Sol ion: Ini ial e ima e : . .8 e ide f he fi a d 0. e a . B= ae hea f a i ai Ca c a e e ha Ca c a e a e hea f a i ai E i a e diff ii f Wi e-Lee . Stead -state molecular diffusion in gases. The a c h i a i ed a d a fe ed f he i id he a ha e.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. 1987) he e S i A = e ha i he ace ic fac .1 hic . The e e a e i 368 K a d he e e i 1 a .1. i g/ 2. 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. B h c e diff e h gh a ga fi 0.28c. A i e f e ha a d ae a i bei g ec ified i a adiaba ic di i a i c . . Le = 1.846 10-5 kg/m-s. Solution Estimate diffusivit from the Wilke-Lee equation . This observation. Compare our result with the value predicted b the WilkeLee equation. Based on the Lewis relation.01 kJ/kg-K.0 (Tre bal. and r = 1.0262 W/m-K.29a. Analog among molecular heat and mass transfer. called the Lewis relation. 1980). m = 1. has profound implications in humidification operations. It has been observed that for the s stem air-water vapor at near ambient conditions. For air at 300 K and 1 atm:Cp = 1. d.18 kg/m3. as will be seen later. k = 0. estimate the diffusivit of water vapor in air at 300 K and 1 atm.1. a i g ha c di i i he fi c a . The e e a e i 308 K a d he e ei 1 . d. Ca c a e he f e a d c ce a i fi e f he e a e h d ge (1). e a . 1987) S i fi 1. e i a e he d i he a e e e e da . If he e a i e h idi f he ai a he e edge f he fi i 20%.1.5 a e ei 1 e ai e i a ed f F A e di A 1.31b. i ge (2). Stead -state molecular diffusion in gases. 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. Wa e e a a i g f a d a 300 K d e b ec a diff i ac a ai hic . The a e e f ae a af ci f e e a e ca be acc a e he Wag e e a i (Reid.30b. a d he ba .. d. At one end of the diffusion path the concentration is 20 mole% H2. The MS diffusion coefficients are D12 = 83. D13 = 68. at the other end. The total molar flu is ero.8 mm2/s. the concentration is 50% H2. 20% N2. 40% CO2.0 Solution Appendi C-1: Solution of the Ma ell-Stefan equations for a multicomponent mi ture of ideal gases b orthogonal collocation ( C = 3). mm2/s.8 mm2/s.atm. 40% N2. = 0. 30% 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 . The diffusion path length is 86 mm. D23 = 16. Initial estimates of the flu es Initial estimates of the concentrations . . . S i b) Ca c a e he a .158. a) Ca c a e he a .02.a fe c efficie i he ga ha e a ha i i he e i e . S i 2.2. A a ce ai i i he ab be .ha e c ce a i i 0..62 g/ 2. hich b i a e a ia . he be e e e f ac i i he b f he i id ha e i 0.a fe c efficie i he i id ha e.an fe coefficien in a ga ab o be . / 3. ai a 347 K a d 1 a i b a high eed a a h ha e e (C10H8) he e. Ma . he dia e e . Ma . e e i g he d i i g f ce i e f e f ac i .a fe c efficie e e i g he d i i g f ce i e f S i i he ga ha e a ha i i a c ce a i . Whe he e e i e d a i ge begi . hi e he c e di g i e facia be e e i id.1a. I a ab a e e i e .an fe coefficien f om naph halene blima ion da a. hi e he c e di g i e facia be e e ga ha e c ce a i i 0. c) A he a e ace i he e i e . 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 . Ca c a e he a .125.0158. he e i e .2a. 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 . The be e e f a ha i i ea ed a 0. ba ed he a e age face a ea f he a ic e. f he d i i g f ce i e f a c ce ai .0 L f ace e e a a ed i 5 i . 1973). a e he a .32 i a e . S i he e i 2. i a b e ed ha 2. S i face f a The a i a . i g he d i i g f ce i e f a ia e e . E i a e a fe c efficie .D i g he a e i 27 . 2. ai a 300 K a d 1 a i b a high eed a a e he ec a g a ha a ha c ai i id ace e (C3H6O). I a ab a e e i e . i a e Pa (Pe a d Chi .0 c .85 c a) E i e e g/c 3. 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. The de i f id a h ha e e i 1. 1973). he dia e e f he he e i .145 i a e e a 347 K i 670 Pa (Pe a d Chi . A he e d f he e e i e . b) Ca c a e he S i a .79 g/c 3. ai ai i g a c a i id e e i he a e ei e a .a fe c efficie . hich e a a e a ia . The de i f i id ace e a 300 K i 0.3a. 1 g a d 50 c ide.f he 1.a fe c efficie . Mass-transfer coefficients from acetone e aporation data. 14. 0 g. Wa e a 308 K f d he i e a . Wi h he he f e a i (2-52).04 3/ i . D ai e e he b f he i e a he a e f 1. 50 i dia e e a d 1.c i fa ga i e.all e perimental data.a e e i e a e . I ea e he e ed ec i a 308 K a d i h a e a i e h idi f 34%. S i . e i a e he a e age a . Mass-transfer coefficients from etted.4b.2. ea ed a 308 K a d 1 a . i h he d i i g f ce i e f a f ac i .a fe c efficie . A e ed. 2.7c. a) I he i ba a e i e e hi S Ma an fe in an ann la pace. 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 e i e fa a i h a a h ha e e d. The a a c ed f a 51-OD 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 2- a 273 K a d 1 a , he i e iga f 12.2 g f ai / de e i ed ha he a ia 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 iga i , he Sch id be f he ga a 2.57, he i c i a 175 P, a d he a 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 e f c di i .A e ha e a i (2-52) a ie . i b) M ad a d Pe (T an . AIChE, 38, 593, 1942) hea - a fe c efficie i a a a ace: e e ed he f i gc ea i f he e d a d di a e he dia e e defi ed a ide a d i ide dia e e f he a , de i he e i ae Wied c efficie S i f he a a g he c di i e e i f f a a). C a a fe a d e i aeb h e . e i a e he a - a fe 2.8c. The Chilton-Colburn analog : flow across tube banks. W C (I d. E g. Che ., 40, 1087, 1948) .T .T 10 38-OD ( = 38 ) 57, : 310 K , 76 1 .T .T - G' -P . )R 310 K S P 310 K 1 (2-68) 0.074 2/ . , C D- / 2- , G / 2- .T 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 a fe c efficie i a agge ed be ba a a ge e i ia Che e : i gc ha e a i f he hea died b Wi di g a d U e he a - a fe e c efficie f a b). C S i e i a a g a e he e . e ai (2-69) e i a e he a - a fe 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 point 1: .Laminar flo At the bulk of the solution. point 2: At the interface. Vc = 209 c .11b.232 (Reid. he a e age ace e c ce a i i he ga fi i e a i e high. Zc = 0. 1987).2. Pc = 47.3.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 .. M = 58.1 K. e a . The f i g da a f ace e igh 3/ be eeded: Tc = 508. D i g he e e i e de c ibed i P b e 2. he ai e ci a ea ed a 6 / . e ie ch a de i a d i c i h d be e i a ed ca ef . S i e ie : A e age fi 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 .0 ba . a a e he ge ide f he a . Ma an fe f om a fla li id face. The ef e. d e he high a i i f ace e. E i a e he a . N ice ha . 9 for a drop of ater hich is originall 2 mm in diameter. E aporation of a drop of ater falling in air. Repeat E ample 2.12b. Solution .2. . Estimate the mass-transfer coefficient for the dissolution of sodium chloride from a cast sphere.02 10 6 m2/s. Assume that the kinematic viscosity at the average liquid film conditions is 1. Vc = 413 cm3/mol. if placed in a flowing water stream. Solution For air at 347 K and 1 atm: Estimate DAB from the Wilke-Lee equation Lennard-Jones parameters for naphthalene . Dissolution of a solid sphere into a flo ing liquid stream.0 m/s. 1973) .25 10 9 m2/s. 2.2. Solution From Prob.13b. The velocity of the 291 K water stream is 1.1 K. The solubility of NaCl in water at 291 K is 0.14b. During the experiment described in Problem 2.5 cm in diameter. the air velocity was measured at 10 m/s. Sublimation of a solid sphere into a gas stream. and the density of the saturated solution is 1.2.35 g/cm3. Estimate the mass-transfer coefficient predicted by equation (2-36) and compare it to the value measured experimentally.22 g/cm3 (Perry and Chilton.10: 2. and the mass diffusivity is 1. The following data for naphthalene might be needed: Tb = 491. 1. 26 is a sphere 2-cm in diameter.2. 1973). Solution . The densit of the cr stal is 1. The cr stal of Problem 1.15b. a) Estimate the cr stal's terminal velocit .464 kg/m3 (Perr and Chilton. It is falling at terminal velocit under the influence of gravit into a big tank of water at 288 K. Dissolution of a solid sphere into a flo ing liquid stream. 26: .b) Estimate the rate at which the cr stal dissolves and compare it to the answer obtained in Problem 1.26 From Prob.1. Solution From Prob.1.26. a) Show that a material balance on a length of pipe L leads to where v is the average fluid velocity. The pipe is 1-cm in diameter. and the mass diffusivity is 1.003 g/cm3. and c A * is the equilibrium solubility concentration. Water flows through a thin tube. Mass transfer inside a circular pipe. Water flows down the inside wall of a 25-mm ID wetted-wall tower of the design of Figure 2. The water flows slowly.2. Assume the air is everywhere at its average conditions of 309 K and 1 atm. at 298 K and 0. Dry air enters at the rate of 7 kg/m2-s. the water at 294 K.1 cm/s.16c.2. 1997). Solution 2.17b. b) What is the average concentration of benzoic acid in the water after 2 m of pipe. equation (2-63) applies. and the mass-transfer coefficient constant. the walls of which are lightly coated with benzoic acid (C7H6O2).all to er. Mass transfer in a etted. Under these conditions. Compute the average partial pressure of water in the air leaving if the tower is 1 . The solubility of benzoic acid in water at 298 K is 0.0 10 5 cm2/s (Cussler. while air flows upward through the core. 7. 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. . hi a ded b a 50-ID a h ha e e i e. Ai a 289 K a d 1 a f ed h gh he a a ace a a a e age e ci f 15 / . U e he e f P b e 2. a d a diff i i i ai f 0. A 289 K. e i a e he c efficie f he e face.06 c 2/ .a fe c efficie f he i e face.18c.2 Pa. a i e iga c c ed a 3. id a h ha e e d.g. a d e a i (2-47). a h ha e e ha a a e e f 5. The i e i e a ade f a 25-OD. i g he e i a e dia e e defi ed i 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.g a a d c . Ma an fe in an ann la pace. A ma e ial balance on a diffe en ial ol me elemen ield : Define: Then: Fo he in e io all: . i h a ea a2. he e ill be a mola fl f om he inne all. NA2. a1. i h pecific in e facial a ea. and a fl f om he o e all. NA1.Sol ion In hi i a ion. For the outer all: . . M = 78. Pc = 48. 2-45: .7 Pa. f 2 / . Zc = 0.9 ba . F be e e. The a e he e g h . 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 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 i id i a a e e a e f 315 K he e i e e a a e e f 26. S i e e c i de . Ben ene evaporation on the outside surface of a single c linder..271 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 . E i f he c i de .19c.2 K. Tc = 562. Vc = 259 c 3/ (Reid. e a . 1987).2. Wilke and Hougan (T an .20b. In one run.816 kg/m2-s temperature at the surface 311 K pressure 97. 41.7 kPa k G 4. and by evaporating this water under adiabatic conditions. calculate the gas-film mass-transfer coefficient using equation (2-55) and compare the result with the value reported by Wilke and Hougan. AIChE. the following data were reported: effective particle diameter 5.415 10 3 kmol/m2-s-atm With the assumption that the properties of the gas mixture are the same as those of air. 445. Solution From the Wilke-Lee equation . Ma an fe in a packed bed. Air was blown through a bed of porous celite pellets wetted with water. 1945) reported the mass transfer in beds of granular solids.2.71 mm gas stream mass velocity 0. they reported gas-film coefficients for packed beds. Solution From the Wilke-Lee equation: .2. The porosit of the bed is 40%.1 m long? The vapor pressure of iodine at 373 K is 6 kPa.21b. based on the empt cross section of the bed. Air at 373 K and 2 atm is passed through a bed 10-cm in diameter composed of iodine spheres 0.7-cm in diameter. Ma an fe and p e e d op in a packed bed. a) How much iodine will evaporate from a bed 0. The air flows at a rate of 2 m/s. b) E ima e he p e Sol ion e d op h o gh he bed. . Solution For dimensional consistency. 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 diffusivity of SO2 in water at 294 K is 1. 39.04 kg/m2-s. k c a.22b.7 10 9 m2/s. Sherwood and Holloway (Trans. when SI units are used exclusively. in many types of packing materials used in industrial towers is virtually impossible to measure. Both a and the mass-transfer coefficient depend on the physical geometry of the equipment and on the flow rates of the two contacting. 323.2. they are normally correlated together as the volumetric masstransfer coefficient. Compare the results. a) Consider the absorption of SO2 with water at 294 K in a tower packed with 25-mm Raschig rings. a. 36. inmiscible streams. The interfacial surface area per unit volume. estimate the liquid-film mass-transfer coefficient. Solution . 21. Volumetric mass-transfer coefficients in industrial to ers. add the constants: b) Whitney and Vivian (Chem. For the conditions described in part a). Progr. Eng. AIChE. 45. Empirical equations for the volumetric coefficients must be obtained experimentally for each type of mass-transfer operation. estimate the liquid-film mass-transfer coefficient using equation (2-72).. If the liquid mass velocity is L' = 2. 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. Accordingly. including fluidized beds. 45. through where vs is in cm/s.. to ferrocyanide. 75-cm high. 938. {Fe(CN)6} 3. They studied different arrangements of packed columns. {Fe(CN)6} 4. The bed was packed with 0. if the porosity of the bed is 60%. et al. with a particle density of 2. estimate the mass-transfer coefficient. The fluidized bed experiments were performed in a 5-cm-ID circular column.612 g/cm3.90 10 10 m2/s. in aqueous alkaline solutions. vs . The properties of the aqueous solutions were: density = 1.30 cP. Mass transfer in fluidi ed beds. Solution . (AIC E J. viscosity = 1.2. a) Using equation (2-56).083 kg/m3. They found that the porosity of the fluidized bed. k L.23b.534-mm spherical glass beads. 1999) studied the electrochemical reduction of ferrycianide ions. diffusivity = 5. Cavatorta. e. could be correlated with the superficial liquid velocity based on the empty tube. if he po o i of he bed i 60%.an fe 2. 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 hei fl idi ed bed e pe imen al n : .000 kg/h hile e e hing el e emain con an . Compa e o e l o ha of pa a). Mass transfer in a hollo -fiber boiler feed ater deaerator. e ima e he ma coefficien . Con ide he hollo -fibe BFW deae a o de c ibed in E ample 2-13. U ing hi co ela ion. L. calc la e he f ac ion of he en e ing di ol ed o gen ha can be emo ed.24b. Sol ion . Sol ion .an fe coefficien fo he e Re i ba ed on he emp be eloci .b) Ca a o a e al. 38 mg/L. a) Con ide he hollo -fibe BFW deae a o de c ibed in E ample 2-13.25b. calc la e he ga ol me flo a e and compo i ion a he l men o le . A ming ha onl o gen diff e ac o he memb ane. Mass transfer in a hollo -fiber boiler feed ater deaerator.2. hich mean a di ol ed o gen concen a ion of 8. Sol ion . The a e en e he hell ide a 298 K a a ed i h a mo phe ic o gen. a hic e f he fibe a S i fe c efficie a he a e age c di i i ide he he e i a i g he ga e ci i ide he e . Neg ec he 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 . e .b) Ca c a e he a . (From E ample 2.13) . Re ea E a e 3. Application of Raoult's law to a binar s stem.6 e e a e f 320 K.2b. Application of Raoult's law to a binar s stem. a) De e i e he c ii f he i id i e i ib i iha a be e e-40 e e ce e e if he e e i i a e e he e i ib i e e a e.1a.1.3. b f a i id c ce a i f 0. P edic I i ia e i a e . S i e f ac i f be e ea da 3. S i c ai i g 60 de 1 a e e e ce e. a) Wha d be he c ii f a he a e. f idea i . -C8H18. Henr 's la : saturation of ater ith o gen. P edic 3.27 10 e f ac i . A i ih ge di ed i a e c ai i g 0. Application of Raoult's la to a binar s stem.5 g O2/100 g f H2O i b gh i c i h a a ge e fa he ic ai a 283 K a d a a e e f 1 a .4a. S i i e i ib i i h a i id c ai i g 60 e e i i a e e de 1 a e e e ce e.3a.b) De e i e he c ii f he a be e e-40 e e ce e e if he 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 ac .a e e a 283 K e a 3. The He ' a 4a / c a f he ge . a he a e ha a a e e f 106 Pa a d a c a e f 47. A 373 K. N a he a e. a d a c a e.1 Pa. -C7H16.c a e i ha b i a 373 K de a 93 Pa e e? S i b) Wha (a)? S i d be he c ii f he a i e i ib i i h he i ha i de c ibed i 3. 5c.2 ca c a e: a) he a e e a e i ib i S i I i ia g e e .A e i ib i : Ba i : 1 L a e (1 g a e ) E I i ia c di i i ib i : c ce ai . c e = 11. The e i 10 ace e he i id. A i g ha he ga . Re ea E a e 3.3. 3.42 g ge /L The i gai ge . a d a e a e b gh i c ac i a 3 f ga c ed c ai e . b a i g ha he a ia. Material balances combined ith equilibrium relations. dif he Ma hcad ga i Fig e 3.ace e a d he e e a e e ai c a i e i ib i i achie ed. ai . 100 / . 0.0 .162 3.27 . A = 0.1 A/ 3 H ' KG = 0.an fe I . T . G. Ma .T / 2.6b.. = 60) A. .755 ) A : A = 0. ing ab o p ion.85 e i ance d A( .145. A.T / 3. ) S - .0 1. I 57% 2. .G = 0.L = 1.A : P = 1. k L.b) the liquid-film coefficient.i. A. Solution Initial estimates of interfacial concentrations: . Solution Basis: 1 m3 of aqueous solution c) the concentration on the liquid side of the interface. d) the mass flu of A. Solution Check this result b calculating the gas-phase flu : . 5.i i he e i ib i i id c ce a i i he a a a .0 e/ be e c e ed i a) he S i ea a .-a .L = 0. A e i ea c ai 9.i i he e i ib i i a f ac i . . he i id ea c ai 4. Mass-transfer resistances during absorption.10. b i h A. E a a e fe c efficie .G = 0.7b.5 e % a d he ga A. M dif he Ma hcad g a i Fig e 3. The i di id a ga -fi c efficie a 2. E e hi g e e e ai c a .3.a a ia e e i a a d A. 3. b) he S i a f f A.70 a d A.8d.6 e ea E a e 3. ( Ae = A *) c) The i S i id i e facia c ce ai f A. K . he he e A. F a e i hich c e A i a fe i g f e i ib i e a i i gi e b he ga ha e he i id ha e. Absorption of ammonia b water: use of F-t pe mass-transfer coefficients. Fif e ce f he e a e i a ce a a fe i he i id ha e.0 e% hi i i G = 3. The a e e i 1 a . Sol ion Ini ial g e e 3. Absorption of ammonia b water: use of F-t pe mass-transfer coefficients. b i h FL = 0. Sol ion Ini ial g e e .6 o epea E ample 3. Modif he Ma hcad p og am in Fig e 3.9d. E e hing el e emain con an .5..0050 kmol/m2. a fe c efficie : b) S i .3.10b.17 / 2-h -a . he i di id a fi c efficie e e e i a ed be L = 6. I he ab i fa ia i ae f a ai -a ia i e a 300 K a d 1 a . Ma .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 a) S i i e he f i g a .an fe e i ance d ing ab o p ion of ammonia. Solution 10 6 m (Prasad For oxygen in nitrogen at 298 K and 1 atm: . eM = membrane porosity. 34.c) Ky Solution d) Fraction of the total resistance to mass transfer that resides in the gas phase. and H is Henry's law constant. 177. across the membrane. the overall mass-transfer coefficient based on the liquid concentrations. 1910. Feb. Solution 3. For mass transfer across the hollow-fiber membrane contactors described in Example 2. Mass-transfer resistances in hollo -fiber membrane contactors.2. Nov. KL.. is given by (Yang and Cussler. and d = 25 and Sirkar. 32.13. eM = 0. For the membrane modules of Example 2. the gas equilibrium concentration divided by that in the liquid. tM = membrane tortuosity. and k c are the individual mass-transfer coefficients in the liquid. a) Calculate the corresponding value of k M.4. d = membrane thickness. 1988). 1988) where DAB = molecular diffusion coefficient in the gas filling the pores. respectively. AIC E J.tM = 2.13. and in the gas.11b. The mass-transfer coefficient across a hydrophobic membrane is from (Prasad and Sirkar. k M.. 1986) where k L. AIC E J. 2. 3. ca c a e KL.a fe c efficie i he i id ha e.and k-t pe coefficients: absorption of low-solubilit gases.i a) Sh e ai ha .12c.13. a F. E a f ac i f he a e i a ce a S F F i E a e 2. de he c di i de c ibed ab e. 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. D i g ab i f bi i ga e . he ga i e facia c ce ai a i fie he S I i he ga ha e: . P b.25: Vi a a f he e i a ce e ide i he i id ha e. a d e i a e ha a fe e ide i he i id fi .e c efficie be ed i he ga ha e. I ha ca e.i = . he i e facia c ce a i d i g ab i f bi i ga e a e e a ed h gh A. A. Si ce di e i id i a be He ' a .25.e a . a d P b e 2. Combined use of F. a h gh i i a ia e ea .b) U i g he e f a (a).13: e 2. E i a e he i e facia c ce a i a d he ca SO2 a f .5 129 2.a fe c efficie ee ca c a ed a FG = 0. .. Hg ( ) .0 0 0. The a . The e e a e a 303 K a d he a e e 1 a .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.5 42 1.0 85 1.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 0.002 303 K a e (Pe a d Chi g SO2/100 g a e Pa ia 0. The e i ib i SO2 bi i da a a e f SO2. 1973): e = 0.. Hg ( ) Defi e: = g SO2/100 g a e = Pa ia e e f SO2.160 / 2.0 176 2.5 224 S i / 2.2% SO2 b e. Ini ial g e : . and FL = 8. et al. The packing characteristics and flow rates at that point are such that FG = 1. the methanol content of the bulk of the gas phase is 76. Calculate the interfacial compositions and the local methanol flux.6 K Solution For methanol (A) For water (B) . To calculate the latent heats of vaporization at the new temperature..542 10 3 kmol/m2-s.13d.1 K. that of the bulk of the liquid phase is 60 mole %.2 mole %. Tc = 647. The temperature at that point in the tower is around 343 K. Tc = 512. At a different point in the packed distillation column of Example 3.650 10 3 kmol/m2-s. 1996): For water. modify the values given in Example 3.3.6.6 using Watson's method (Smith. for methanol. Distillation of a mi ture of methanol and ater in a packed to er: use of F-t pe mass-transfer coefficients. Pa ame e Ini ial e ima e . . Ni ge i ad bed. Ac i a ed ca b i ed ec e be e e f a i ge -be e e a i e.63 50 2.83 bed Pa ia e e be e e. Hg a) P a S i he e i ib i e e f1a da a a X' = g be . The e e a e a d a e e a e ai ai ed a 306 K a d 1 a .22 100 7. Y' = g be e e/ g i ge f a .95 40 1.55 25 0. e e/ g d ca b . Material balances: adsorption of ben ene vapor on activated carbon.26 80 4.88 90 6. 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 1.3.18 65 3.0 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 . 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 c 3 (STP)/g ca b 15 0. 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 .14b. ca e he i (X2.b) Ca c a e he i i e e i g ca b c ai S i f a e e i ed f he e e i g ac i a ed ca b e ad bed be e e). a he i e ec i f Y = Y1 i h he e i ib i c e. he i i e a i g i e i b ai ed b ca i g. X1 a . . ( e e be ha he O he XY diag a . 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 .Y2). 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 S i he c di i f a (c), ca c a e he be f idea age e i ed. 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 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 e e be e e, Pa 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: . 60 S i Ge e a e he XY diag a . 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 . I i de i ed d 10 g f a f 20% i eb eigh e ha 6% i eb 3 f ai a 350 K.56 4.06 a da ae-a a ia e e f 1.29 3. c ac i h h ai .76 2.96 7.3.79 6.63 8.40 9.83 6.58 19. 1 a .40 1.19 9. The e i a ed each e i ib i . Material balances in batch processes: dr ing of soap with air. The e a i aced i a e e c ai i g 8.6 Pa.33 12.17b.02 10.76 3.10 4. 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.42 15. 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 2.90 7. 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 .From the XY diagram. X = 0. 86 9.01 0.00422 .. 103 kg nicotine/kg kerosene 1. I d. E g.70 Solution From the XY diagram. The equilibrium data are as follo s (Claffe et al. 42.4 18.81 2.96 5.56 7. 166.0 kg of fresh. 1950): X'.02 4.. 103 kg nicotine/kg ater Y'. Nicotine in a ater solution containing 2% nicotine is to be e tracted ith kerosene at 293 K. 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 49.51 6.13 20.98 9. X = 0. Water and kerosene are essentiall insoluble.46 1. Che .ith kerosene. pure kerosene each. after 4 e tractions. . . a d i hi each age a a fe cc a if i c c e f . 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 . i ca be h ha (T e ba .19b. 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 S he i i g fac . 1980) he e S i age .27 i a che a ic diag a fac -f ca cade f idea age . Each age i e e e ed b a ci c e. a d N i he a be f e P be i 3.18. The d i g a d i id.18 i ge ai (3-60). e ec i e . bei g c ac ed i each age b a f e h V ha e. a e e a e faf c fig a i ca ed a c -f ca cade.3. Cross-flo cascade of ideal stages. Fig e 3.17 a d 3. a d c a e he e b ai ed b he eh d . VS/LS. c a f a age . 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 e i ed achie e a ea 95% e ac i efficie c . S i be f idea age 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 ab be . fP be 3.21, b i g a 2N e ha a i e i a 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 An ab o be i a ailable o ea he fl e ga of P oblem 3.3.5 e ilib i m age . Kremser equations: absorption of sulfur dio ide.23 hich i e i alen o 8. Calc la e al o he SO2 concen a ion in he a e lea ing he ab o be . Sol ion Ini ial e ima e .24b. 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. Le he a . 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 . The i i 1. The i e 3-he a c ai 0.037% (b eigh ) ace ic acid b e ac i i h 3-he a a 298 K. he e ha e i he L. F hi e .b) Wha i he e ce age e a f SO2 ha ca be achie ed i h hi ab a e ed i he a e ha a ca c a ed i P b e 3.828 W a i ace ic acid i ae S i e ha e be he V. Kremser equations: liquid e traction.23 (a)? S i be if he a e f I i ia e i a e 3.ha e. 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 . A a e ace ic acid i f a he a e f 1. e i ib i i gi e b W a i ace ic acid i e = 0.000 g/h .ha e.25b. I i de i ed ed ce he c ce a i f hi i 0.1% (b eigh ) ace ic acid.02% (b eigh ) ace ic acid. The butanol solution contains 4. A 1-butanol acid solution is to be e tracted ith pure ater.62 XAi. At 298 K. For practical purposes.10% (b eight) acid. 1-butanol and ater are inmiscible.26c. ho man equilibrium stages are required for a countercurrent cascade? Solution . here YAi is the eight ratio of acid in the aqueous phase and XAi is the eight ratio of acid in the organic phase. A total ater flo rate of 1005 kg/hr is used. Operation is at 298 K and 1 atm. a) If the outlet butanol stream is to contain 0. Co n e c en e c o -flo e ac ion.5% (b eight) of acetic acid and flo s at the rate of 400 kg/hr.Initial estimate: 3. the equilibrium data can be represented b YAi = 0. 242. The feed i f a he a e f 100 L/ i .19)? i ac -f ca cade.0 g/L. he e XAi i he g c e c ce a i i he e i . hi e he e i f a he a e f 250 L/ i .b) If he a e i i e a a ha i he e 1-b a c ce S i g he a e be f 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.27c. Ser.25 g f g c e/L.(g f g c e e i e f i ). 3. The c ce a i f he feed i i 15 g/L.961 XAi. 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 . Thei e i ib i e e i a 303 K i YAi = 1. Fi d he be f e i ib i age e i ed. S i .(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 . b a i ( ee P b e 3. The i e e i c ai 0. 81.. Glucose sorption on an ion e change resin. Chi g a d R h e (AIChE S mp. We a a e c ce a i f 1. 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. Solution . S i The ca i E a e 4. ca c a e he a ic i ac ed i h f he bed. ca c a e he di a ce f he a he fi fi e ch ca i . Void fraction near the alls of packed beds. a a di a ce f 100 A e 4. 4. C ide a c i d ica e e i h a dia e e f 305 f 50 . F I i ia e i a e f he ca be b ai ed f Fig.4 I i ia e i a e . 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. Beca e f he ci a a e f he id-f ac i adia a ia i f ac ed bed .1.1a. ea he a e a he a ic i he J0( d) = 0.1. F he bed de c ibed i E a e 4.4. id he e i h a dia e e S i A b) E i a e he id f ac i e f he a . Void fraction near the alls of packed beds.2b. a) F e a i (4-1). ca c a e he adia ca i f he fi fi e a i a. ca c a e he a i de f he id f ac i ci a i a h e i . Void fraction near the alls of packed beds. a d f fi e i i a. . (a) Sh ha he adia ca i f he a i a a d i i a f he f (4-1) a e he f he e a i ci de c ibed b E ai S b) F he fi S i i he ac ed bed f E a e 4.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 4.1.3c. mm ( *) Ampli de.Ini ial e ima e of he oo can be ob ained f om Fig.45) -16.85) 9.7 77.8 (4.0 (2. he follo ing e l a e ob ained: Ma ima: Minima .6 (1.3 (0. hi m happen a * be een 3.51) -27. mm ( *) Ampli de.4 Ini ial e ima e Ini ial e ima e Repea ing hi p oced e.1 95.2 67.32) -7.4 (4.7 (3.04) 37.39) -11.57) -56.3 58.39 and 3.98) 21.5 49.92) 13.2 (1.7 39.79) 6.3 (2.85 .4 86.6 c) Calc la e he di ance f om he all a hich he ab ol e al e of he po o i been dampened o le han 10% of he a mp o ic bed po o i . % .1 30.8 (1.2 11. % 20. 4. Sol ion F om he e fl c a ion ha l of pa (b).1 (3. The correlation is restricted to randomly packed beds in annular cylindrical containers of outside diameter Do. 2458-60. . It is well known that the wall in a packed bed affects the radial void fraction distribution. E..(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. such as in chemical reactors.1. 1999).4c. heat exchangers. estimate the average void fraction by numerical integration of equation (4-65) and estimate the ratio a / b . Annular packed beds (APBs) involving the flow of fluids are used in many technical and engineering applications. Solution 4. Void fraction near the alls of annular packed beds. AIChE J. and fusion reactor blankets. Since APBs have two walls that can simultaneously affect the radial void fraction distribution. inside diameter Di. G. Nov. A correlation for this purpose was recently formulated (Mueller. 45. equivalent diameter De = Do Di. Solution (b) Plot the void fraction. as predicted by Eq.consisting of equal-sized spheres of diameter dp. for r* from 0 to R*. inside diameter of 40 mm. 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. (a) Estimate the void fraction at a distance from the outer wall of 25 mm. . (4-66). Minimum liquid mass velocit for proper wetting of packing. L. Repea P oblem 4. .(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 Sol ion fo he APB de c ibed abo e. Sol ion Fo pla ic packing. Minimum liquid mass velocit for proper wetting of packing. in kg/ .5. Calc la e he minim m a e flo a e.5a. neded o en e p ope e ing of he packing face.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 . L. b ing ce amic in ead of pla ic In alo addle . A 1. F om S eam Table 4. 4. Sol ion Fo ce amic packing.6a.2 mm/ .min = 1.min = 0.15 mm/ . 3). S ecific i id h d a d id f ac i i fi -ge e a i a d ac i g. Ch = 0.2: .2.1: From Example 4.7b. Solution From Example 4.From Steam Tables 4. 1998).2: 4. Solution From Table 4. a = 200 m 1. For this packing.979.547 (Seader and Henley. Repeat Example 4. but using Montz metal B1-200 structured packing (very similar to the one shown in Figure 4.8b. S ecific i id h d a d id f ac i i c ed ac i g. but using 25-mm ceramic Berl saddles as packing material.2. Repeat Example 4. = 0. 9b.4. a d he h d a ic ecific a ea f he ac i g. A e ac ed i h 25ce 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. The i id a e ci i L' = 2. he id f ac i . E i a e he i id h d .0 cP a d i de i i 840 g/ 3.. The i c i f he i i 2.71 g/ 2. S ecific i id h d a d id f ac i i fi -ge e a i a d ac i g.1: . S i F Tab e 4. a. fraction Enter loading constant. Pressure drop in beds packed ith first-generation random packings. Pa-s Enter temperature. P. for this packing. Solution Packed Col mn Design Program This program calculates the diameter of a packed column to satisf a given pressure drop criterium. Enter data related to the gas and liquid streams Enter liquid flow rate. in kg/s Enter gas flow rate. T. in ft2/ft3 Enter specific area. Pa-s Enter gas viscosit . Ch Enter pressure drop constant.783. Assume that. in kg/s Enter liquid densit . mL. C = 1.10b. Repeat E ample 4. in Pa Enter data related to the packing Enter packing factor. in K Enter total pressure. X . and estimates the volumetric mass-transfer coefficients. Cp Enter allowed pressure drop. kg/m3 Enter liquid viscosit . in kg/m3 Enter gas densit . in Pa/m Calculate flow parameter.4. mG. m2/m3 Introduce a units conversion factor in Fp Enter porosit .3. Fp. but using 15-mm ceramic Raschig rings as packing material. in m . QG. D. in m3/s Calculate effective particle si e. in m3/s Calculate liquid volume flow rate. design for 70% of flooding Calculate gas volume flow rate.Calculate Y at flooding conditions Calculate gas velocit at flooding. QL. vGf As a first estimate of the column diameter. dp. in me e F ac ional app oach o flooding .I e a e o find he o e diame e fo he gi en p e e d op Col mn diame e . . P e e d op and app oach o flooding in c ed packing. X . in K Enter total pressure. Repeat Example 4. Cp Enter allowed pressure drop. For this packing. Pa-s Enter temperature. = 0. in Pa/m Calculate flow parameter.3.11b. Enter data related to the gas and liquid streams Enter liquid flow rate. kg/m3 Enter liquid viscosit . in ft2/ft3 Enter specific area. fraction Enter loading constant. a = 200 m 1. P.979. m2/m3 Introduce a units conversion factor in Fp Enter porosit . Ch Enter pressure drop constant. in Pa Enter data related to the packing Enter packing factor. in kg/s Enter liquid densit . in kg/s Enter gas flow rate. Pa-s Enter gas viscosit .4. but using Montz metal B1-200 structured packing (very similar to the one shown in Figure 4. in kg/m3 Enter gas densit . T. Fp = 22 ft2/ft3. mL. mG. Cp = 0. Solution Packed Col mn Design Program This program calculates the diameter of a packed column to satisf a given pressure drop criterium.355 (Seader and Henley.547. and estimates the volumetric mass-transfer coefficients. Ch = 0.3). 1998). a. Fp. Calculate Y at flooding conditions Calculate gas velocit at flooding. D. QG. dp. QL. in m3/s Calculate effective particle si e. in m . vGf As a first estimate of the column diameter. in m3/s Calculate liquid volume flow rate. design for 70% of flooding Calculate gas volume flow rate. all the ben ene is .12b. 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. containing 5 mole % ben ene.I e a e o find he o e diame e fo he gi en p e e d op Col mn diame e . Esentiall . Pressure drop in beds packed ith second-generation random packings. in me e F ac ional app oach o flooding 4. d. measured at 110 kPa and 298 K. The gas enters the to er at the rate of 1.5 m3/s. e dia e e be ed. The ac i g i be 50e a Pa i g . The b e 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. (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 D = 0. i iga ed ac i c bi a i Ca c a e he S i e e e. he i id de i i i i 2.e ed f 400 Pa/ f i iga ed ac i g. e e i ed b he ga h gh he ac i g.0 g/ . he i iga ed ac ed heigh i be 5 a d ha 1 f g i be aced e he i id i e ac a e ai e e a a . a d he e i g f ac i a a ach f di g. i c i be ch e (a) Ca c a e he (b) A e ha . a d he e dia e e d ce a ga .3 cP.ab bed b he 800 g/ 3. f he dia e e ch e .913 (b) Ca c a e he f = 0.825 e ed De ig P g a : h gh he d ac i g . 971.859 k Lah = 0.376 kmol/m3-s 4. d. but using Montz metal B1-200 structured packing (very similar to the one shown in Figure 4. d.0885 cm2/s From Table 4. Redesign the packed bed of Problem 4.14c. CL = 0.13b. Pressure drop in beds packed ith structured packings.12.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.00675 s 1. k y ah = 0. a = 200 m 1. A wastewater stream of 0. e = 0. DG = 0.355.00861 s 1. From the wilke-Lee equation.038 m3/s.0 10 10 m2/s.26 kmol/m3-s 4. CL = 1. k y ah = 0.2. Solution Using the Packed Column Design Program: D = 0. For this packing. Air stripping of aste ater in a packed column.3). containing 10 ppm (by weight) of benzene.85 m f = 0. Ch = 0.192. Cp = 0. is to be stripped .390 (Seader and Henley. 1998). CV = 0. Fp = 22 ft2/ft3. CV = 0.547.410 the Packed Column Design Program: Using k Lah = 0.979. Assume that DL = 5. 02 10 5 cm2/s (Cussler.096 cm2/s. The diffusivity of benzene vapor in air at 298 K and 1 atm is 0. 1998). Solution Calculate m.6 kPa-m3/mole (Davis and Cornwell. the slope of the equilibrium curve: For water at 298 K. the diffusivity of liquid benzene in water at infinite dilution at 298 K is 1. 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.005 ppm.with air in a packed column operating at 298 K and 2 atm to reduce the benzene concentration to 0. Calculate the tower diameter if the gas-pressure drop is not to exceed 500 Pa/m of packed height. 1997). Henry's law constant for benzene in water at this temperature is 0. The air flow rate to be used is 5 times the minimum. and y 2 (max) = y 2* = mx 2 . The packing specified is 50-mm plastic Pall rings. Estimate the corresponding mass-transfer coefficients. k y ah = 0.145 m f = 0.335 kmol/m3-s .Using the Packed Column Design Program: D = 1.032 s 1.766 k Lah = 0. Z . Stripping chloroform from water b sparging with air.4. Repeat E ample 4. Solution Initial estimate of the column height. but using an air flo rate that is t ice the minimum required.5.15b. Ini ial e ima e of ga hold p . Calculate po er required . 11 kPa-m3/mole (Perry and Chilton. each 3. A vessel 2. Repeat Example 4. Stripping chloroform from water b sparging with air. 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. 1973).22 kg/s at 298 K. The water will flow continuously downward at the rate of 7. and specifying a chloroform removal efficiency of 99%.4. estimate the chlorine removal efficiency achieved. The sparger is in the form of a ring. Henry's law constant for chlorine in water at this temperature is 0.16b.5. The diffusivity of chlorine at infinite dilution in water at 298 K is 1.5 kg/s with an initial chlorine concentration of 5 mg/L.0 mm in diameter.0 kW 4. (a) Assuming that all the resistance to mass transfer resides in the liquid phase.25 10 5 cm2/s (Cussler. containing 200 orifices. Airflow will be 0. Solution For 99% removal efficiency.17b.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. Solution . 25 cm in diameter. 1997).0 m in diameter and 2.15. xin/xout = 100 Z = 1. . Ini ial e ima e of ga hold p . d. A he b bb e i e. Batch aste ater aeration using spargers. I he ea e f a e a e .Ieai g i Z = 2.0 : (b) Ca c a e e e i ed 4.18c. 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. .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 . F he e ca e . each i g c e ed ai a a a e f 0. Oeff . a i g ha c A * e ai e e ia c a . O = a ge e ce i he ai ea i g he ae a i a . S i .F ba ch ae a i i ac a e a . B . Ind ial Wa e Poll ion Con ol.3 Pa c ai i g 20. each 3. J . The a ge e ce i he ai ea i g he ae a i a i e a ed efficie c .38 g/L (Da i a d C e . U de he e c di i . 3 d ed.0 g/L. P = ab e e e a he de h f ai e ea 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 . 1998). The a e e e a e i 298 K.a fe c efficie f he e c di i f e ai (4-23) a d (4-25). a d ca c a e he e i g ge a fe efficie c . Ma. 100 c i dia e e . . Each a ge i i he f f a i g. W. a he ic c di i a e 298 K a d 101. (a) E i a e he e ic a . c ai i g 20 ifice .0 i dia e e . McG a -Hi . he e ai i e ea ed a a i c ea ed i id de h. 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. if he echa ica efficie c f he c e i 60%.3 Pa. c = 8. 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 . 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. 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. (b) E i a e he i e e i ed ai e he di ed ge c ce a i f 0. 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 .5 g/L 6. W.9% ge ... P =a he ic e e.01 g/ . h gh he ge a fe he e C ide a 567 3 ae a i d ae a ed i h 15 a ge .0 a d c A. (c) E i a e he e e i ed e a e he 15 a ge . The a ge i be ca ed 5 be he face f he d. . Ini ial e ima e of ga hold p . Initial estimate of transfer efficienc (b) Calculate power required . . F he ae a i d fP be a a a e f Z = 3 .20c. da adc ec e he a e f .18. Ac i g e . Acc di g Ec e fe de (2000).4. hich i he ai e e a e. Batch aste ater aeration using spargers. C ide he i a i de c ibed i P b e 4. Hi : Re e be ha he a e f he d .ec i a a ea f he d cha ge a he a e de h U i g he 4. i fed i h a e a 316 2. Flooding conditions in a packed cooling to er.a fe c efficie i a ih i de h Z acc di g he e a i hi he e he e 4. The a e i c K a a a e f 25 g/ ac ed i h ai . i d f d he e.18. ac ed i h 75ce a ic Hif i g . e i a e he e ic a e f ai f . he f i g ec f e i ge e a ed e id e ha a a e ea 0. ca c a e f eg e i e ai c a cha ge .7 f e . f f b bb e-diff i ae a i e he e ic a .d. Neg ec i g e a a i f he a e a d cha ge 3/ . he ef e he c . 2 i dia e e . S i ga de e ed i P b 4. effect of liquid depth. 4 . 6 . S i LMV = i id a e ci GMV = ga a e ci . a 300 K a d 101. E i a e he c e di g a e f L a a i f he e . a d 7 .18.19c.3 Pa e e ia d .. 176 m = 0. and 4. he follo ing e D = 1. Design of a sieve-tra column for ethanol absorption Repea he calc la ion of E ample 4.8865 EMGE = 0.7. Sol ion F om he Sie e-Pla e De ign P og am.6.22 (no e ce i e eeping) E = 0.6 m P = 590 Pa/ a F o de No.0165 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.d.21c.878 .5 EOG = 0. = 1.8123 EMG = 0. 4.9 fo a col mn diame e co e ponding o 50% of flooding.8. 4. 90 Vapor: Rate = 5.d.0 kg/s Composition = 7. (a) Design a suitable cross-flow sieve-tray for such a tower.5 K Pressure = 100 kPa Liquid: Rate = 10.716 EMG = 0. tray spacing. but for a 45% approach to flooding.d. Solution EOG = 0.27 10 5 cm2/s (est.6 mole % aniline Density = 0. 1980). The composition of the liquid is calculated as 1.4. the average molecular weight = 19. Everything else remains the same as in Problem 4.0 kg/s Composition = 3.22. Solution From the Sieve-Plate Design Program.0636 (Treybal.116 cm2/s (est.1.22. Design of a sieve-tra column for aniline stripping. gas-pressure drop.93 m t = 0. Use a weir height of 40 mm. From the Sieve-Plate Design Program. Repeat Problem 4.045 Weir length = 1.45 . the following results are obtained for f = 0.6 m DP = 435 Pa/tray Froude No.22c. Design for a 75% approach to the flood velocity. The circumstances at the top of the tower.) Diffusivity = 0. the following results are obtained for f = 0.720 4.) The equilibrium data at this concentration indicates that = 0. Design of a sieve-tra column for aniline stripping. A sieve-tray tower is to be designed for stripping an aniline (C6H7N)-water solution with steam.5 mm on an equilateraltriangular pitch 12 mm between hole centers. Take do = 5.670 kg/m3 Viscosity = 118 mP (est.7.4 mole % aniline.00 mass % aniline Density = 961 kg/m3 Viscosity = 0. Report details respecting tower diameter. are: Temperature = 371. which are to be used to establish the design. punched in stainless steel sheet metal 2 mm thick. = 0.23c.75 D = 1. Check for excessive weeping.3 cP Surface tension = 58 dyne/cm Diffusivity = 4.722 EMGE = 0. The average molecular weight of the gas = 20. and entrainment in the gas.93 (no excessive weeping) E = 0. the molecular weight = 93. weir length.40 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a). Solution For aniline.) Foaming factor = 0. Use a weir height of 50 mm. = 0.D = 2.81 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).24c.1 kmol/s Composition = 18 mole % methanol Viscosity = 125 mP (est. Take do = 6.3 cP Surface tension = 40 dyn/cm Diffusivity = 5.0 mass % methanol Density = 961 kg/m3 Viscosity = 0.d.5 (Perry and Chilton.70 10 5 cm2/s (est. and entrainment in the gas.6 m DP = 399 Pa/tray Froude No. A dilute aqueous solution of methanol is to be stripped with steam in a sieve-tray tower. Solution From the Sieve-Plate Design Program. (a) Design a suitable cross-flow sieve-tray for such a tower.174 m t = 0. Design for 80% approach to the flood velocity. Solution . = 0. gas-pressure drop.660 EMG = 0.491 ( excessive weeping) E = 0.0 Vapor: Rate = 0. weir length.) The equilibrium data at this concentration indicates that = 2.666 EMGE = 0.6 m DP = 386 Pa/tray Froude No.80 D = 1. Design of a sieve-tra column for methanol stripping.853 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).0173 Weir length = 1.3 kPa Liquid: Rate = 0. The conditions chosen for design are: Temperature = 368 K Pressure = 101.) Diffusivity = 0.878 (no excessive weeping) E = 0.49 m t = 0. punched in stainless steel sheet metal 2 mm thick.) Foaming factor = 1.0 mm on an equilateraltriangular pitch 12 mm between hole centers. Check for excessive weeping. Report details respecting tower diameter.25 kmol/s Composition = 15.052 Weir length = 0.213 cm2/s (est. tray spacing. the following results are obtained for f = 0. Solution EOG = 0. 1973).665 4. 25c.035 cm2/s (est. Design for a 75% approach to the flood velocity. the following results are obtained for = 0. gas-pressure drop.561 (no excessive weeping) E = 0. but changing the perforation size to 4. 1987).14 10 5 cm2/s (est.26c. punched in stainless steel sheet metal 2 mm thick.) The system obeys Raoult's law.) Foaming factor = 0. the vapor pressure of n-butane at 310 K is 3. According to Bennett and Kovak (2000).24.6 cP Surface tension = 25 dyn/cm Diffusivity = 1. the optimal value of the ratio Ah/Aa is that which yields an orifice Froude number.EOG = 0.936 EMGE = 0.. Report details respecting tower diameter.853 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a). Fro = 0. Solution . Repeat Problem 4.6 m DP = 686 Pa/tray Froude No.50 kmol/s Average molecular weight = 150 Density = 850 kg/m3 Viscosity = 1. A gas containing methane. It is agreed to design a tray for the circumstances existing at the bottom of the tower . Design for the optimal value of do on an equilateral-triangular pitch 12 mm between hole centers. 12% C3H8.707 EMGE = 0.918 4. Solution From the Sieve-Plate Design Program. 2% C4H10 Viscosity = 113 mP (est.5.3 kmol/s Composition = 86% CH4. weir length.5 mm. effect of hole si e. = 1.d.d. Design of sieve-tra column for butane absorption.80 D = 1. Sieve-tra column for methanol stripping.700 4. and n-butane is to be scrubbed countercurrently in a sieve-tray tower with a hydrocarbon oil to absorb principally the butane. Solution EOG = 0. propane.) Diffusivity = 0. Use a weir height of 50 mm. (a) Design a suitable cross-flow sieve-tray for such a tower.793 EMG = 0. tray spacing.472 bar (Reid et al.9 Vapor: Rate = 0. where the conditions are: Temperature = 310 K Pressure = 350 kPa Liquid: Rate = 0. and entrainment in the gas.174 m t = 0.5987 EMG = 0.076 Weir length = 0. keeping everything else constant. d. NH3-H2O follows Henry's law. From the Sieve-Plate Design Program.7 kg/s Viscosity = 113 mP (est. Design of sieve-tra column for ammonia absorption.28 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a). The ammonia will be removed by scrubbing the gas countercurrently with pure liquid water in a sieve-tray tower.5 m DP = 624 Pa/tray . Solution From the Sieve-Plate Design Program.73 m t = 0. it was found that the orifice diameter that results in a Froude number = 0. tray spacing.42 10 5 cm2/s (est.230 cm2/s (est.27c. punched in stainless steel sheet metal 2 mm thick.5 (no excessive weeping) E = 0. the following results are obtained for = 0. and at 303 K the slope of the equilibrium curve is m = 0. Check for excessive weeping. By trial-and-error. Solution EOG = 0.0 kg/s Average molecular weight = 18 Density = 996 kg/m3 Viscosity = 0.75 D = 2. The gas will consist of H2 and N2 in the molar ratio 3:1.775 m t = 0. weir length.5 is do = 3.) Foaming factor = 1. Conditions at the bottom of the tower are: Temperature = 303 K Pressure = 200 kPa Liquid: Rate = 6.2.0.6 m DP = 811 Pa/tray Froude No.43 mm. and entrainment in the gas. Use a weir height of 40 mm. gas-pressure drop. (a) Design a suitable cross-flow sieve-tray for such a tower.0 Vapor: Rate = 0. Report details respecting tower diameter.75 mm on an equilateraltriangular pitch 12. 1980).034 4.038 EMGE = 1. containing 3% NH3 by volume. and residual uncracked ammonia is to be removed from the resulting gas.The slope of the equilibrium curve is approximately 1.013 Weir length = 2. A process for making small amounts of hydrogen by cracking ammonia is being considered. The average molecular weight of the gas is 20.5 mm between hole centers.) Diffusivity = 0. the following results are obtained for = 0. using the Sieve-Plate Design Program. Take do = 4.833 EMG = 1.80 D = 0.) For dilute solutions. Design for an 80% approach to the flood velocity.9 cP Surface tension = 68 dyn/cm Diffusivity = 2.85 (Treybal. = 0. (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).) Foaming factor = 0.60 D = 0. = 0.710 . Report details respecting tower diameter.8 mole/s Composition =48. Solution EOG = 0.0 mol % toluene Density = 726 kg/m3 Viscosity = 0.585 (no excessive weeping) E = 0.d.986 kg/m3 Viscosity = 337 mP (est.) The equilibrium data at this concentration indicates that m = 1.7 mm between hole centers.08 10 5 cm2/s (est.) Diffusivity = 0.28c.007 Weir length = 0. Use a weir height of 50 mm.5 m DP = 295 Pa/tray Froude No.611 EMG = 0.80 Vapor: Rate = 4. are: Temperature = 380 K Pressure = 98.152. A sieve-tray tower is to be designed for distillation of a mixture of toluene and methylcyclohexane. Sol ion a) From the Sieve-Plate Design Program. Solution EOG = 0.847 4.28 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).6 m t = 0.6 mole % toluene Density = 2. gas-pressure drop. Check for excessive weeping. Design of sieve-tra column for toluene-meth lc clohe ane distillation.54 mole/s Composition =44.22 cP Surface tension = 16.436 m (b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a). weir length. and entrainment in the gas.9 dyne/cm Diffusivity = 7.8 mm on an equilateraltriangular pitch 12.806 EMG = 0. Take do = 4.8 kPa Liquid: Rate = 4. Design for a 60% approach to the flood velocity. the following results are obtained for = 0.712 EMGE = 0.0386 cm2/s (est. The circumstances which are to be used to establish the design. tray spacing.033 (no excessive weeping) E = 0. = 1.85 EMGE = 0.048 Weir length = 2. (a) Design a suitable cross-flow sieve-tray for such a tower. punched in stainless steel sheet metal 2 mm thick.Froude No. A a e idi g i e i ib i age i ed f i i ga ia f a a e ae ea b ea fc e c e ai a 1 a a d 300 K. Ca c a e he c ce a i f a ia i he e i a e if he i e i id c ce a i i 0.873 a da d c bic e e f ai a e fed he e e i ga f a e a e . S i M h ee efficie c EMGE = 0.5.3a. C. i hi a ge f c ce a i a d 300 K. ca c a e he be f ea a . McG a -Hi . Ne Y . Ammonia stripping from a wastewater in a tra tower. 1971). he i e ai i f ee f a ia.i (Ki g.25.1 e%a ia. Sepa a ion P oce e . The e i ib i da a f hi e . ca be e e e ed b A. a d a a e age b) If he ab S i be e i e 5.414 A. U e8 a 5.75.1a. S i .34 e i ib i age . a d 1.i= 1. NY.. Overall tra efficienc C ide a a ab be i h a c a ab i fac A = 1. a) E i a e he e a a efficie c . J. 4a. Ammonia stripping from a wastewater in a tra tower.581. Solution Use 9 tra s 5.4 is reduced to 1.5 standard m3/kg of .Initial estimate 5. Ammonia stripping from a wastewater in a tra tower.3 and 5. The Murphree plate efficienc for the ammonia stripper of Problem 5. Estimate the number of real tra s required.5b. If the air flow rate to the absorber of Problems 5.3 is constant at 0. i. E i ib i f A i gi e a A. The hea i i he ec c ed bac he ce he e ce i bei g a i a ba i . Fi d he e a c efficie a e efficie c .0001 A.a c .0001 A.0 e /h Ga f a e = 2. a e ha he i id a d ga f ae ae gh c S i i ie f a Af he e A i i ed.i ea a 320 K i ed i a ab be e e di e a a ai ea . a d he M a .a e ie e. a e if 9 ea a ae ed a d he I i ia e i a e 5. Pi a da a a e a f : Li id f a e = 5.1%. A hea . h ee . ca c a e c ce a i fa M h ee efficie c e ai c a S i ia i he e i a 58.i ab be i a 16. Absorption of an air pollutant in a tra tower.5 e /h = 0.7 A.i = 0.04 A.i = 0.6b.i = 0. The c e c .a e . a d i f ai f ca e. The de i ed. Assuming that Henry s law applies.5. calculate the slope of the equilibrium line.2. An absorption column for laboratory use has been carefully constructed so that it has exactly 4 equilibrium stages.0027 mole fraction ammonia. and is being used to measure equilibrium data. Absorption of ammonia in a laborator -scale tra tower. Water is used as the solvent to absorb ammonia from air. Solution Initial estimate . The ratio of L/V = 1. inlet gas concentration is 0.01 mole fraction ammonia. The inlet water is pure distilled water.7b. The system operates isothermally at 300 K and 1 atm. the measured outlet gas concentration is 0. punched in sheet metal 2 mm thick. There is available a sieve-tray tower.8. 0.42 10 9 m2/s Gas viscosity = 1. The gas will consist of H2 and N2 in the molar ratio 3:1. at a pressure of 2 bars and a temperature of 303 K. 1980). Assume isothermal scrubbing with pure water at 303 K. containing 3% NH3 by volume. The water flow rate to be used should not exceed 50% of the maximum recommended for cross-flow sieve trays which is 0.5 m tray spacing.75 m diameter. A process for making small amounts of hydrogen by cracking ammonia is being considered and residual. b) Calculate the concentration of the gas leaving the absorber in part a).015 m3/s-m of tower diameter (Treybal.75 mm in diameter. The perforations are 4.5 mm centers. Absorption of ammonia in a seive-tra tower. The results are: mG = 0.0 Liquid diffusivity = 2. containing 6 cross-flow trays at 0.8c.d. arranged in triangular pitch on 12. The gas flow rate should not exceed 80% of the flooding value.657 kg/s Gas pressure drop = 615 Pa per tray .5. The weir height is 40 mm.122 10 5 kg/m-s Gas diffusivity = 0. uncracked ammonia is to be removed from the resulting gas.75 m at f = 0. = 0.23 cm2/s Slope of the equilibrium line. Data: Liquid density = 996 kg/m3 Surface tension = 0. a) Estimate the gas flow rate that can be processed in the column under the circumstances described above.85 Solution a) Run the sieve-tray design program of Appendix E with different values of gas flow rate until D = 0.068 N/m Foaming factor = 1. 841 F o = 1.048 b) Ini ial e ima e .EMGE = 0.042 E = 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. 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.2 a a d 298 K. The ga ea i g he c bbe i c ai 2% CO2.2 i e he i i i ecified. a) Ca c a e he i g a S i e e i g he e e c bic ee f e e i g ga . A i id. 6% O2.-ga a i f 1. a d 79% N2. i c ai 0. hich i ec c ed f a i e .5. F he i i i id f a e. A e i he a eai . The ga i be b i a b bb e-ca e c bbe a 1.i i he e f ac i f CO2 i f i he i id i . Absorption of carbon dio ide in a bubble-cap tra to er.9b. A a a fac i g d ice i b c e i ai d ce a f e ga hich. The c bbi g i id.A 298 K a d 1. he c ea ed a d c ed. 1 a i be i e i ib i i h N1 I i ia e i a e .058 e CO2/ i .2 a . i c ai 15% CO2. 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 . Solution . 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 . in cP 3 L = liquid densit . Estimate the overall tra efficienc for the absorber. the equilibrium-distribution curve is not a straight line.012 kg/m3. and the number of real tra s required.c) The monoethanolamine solution has a viscosit of 6. therefore m is not constant. in kg/m Hint: In this problem.0 cP and a densit of 1. Estimate thhe average value of m at liquid concentrations along the operating line and use theaverage in the correlation given. 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 for a gas velocity which is 70% of the flooding velocity.5 times the minimum.5. CS2. Assuming isothermal operation. The oil enters the absorber essentially pure at a rate 1. The partial pressure of CS2 in the original gas is 50 mm Hg.5%. is evaporated from the product in a dryer into an inert gas (essentially N2) in order to avoid an explosion hazard. Absorption of carbon disulfide in a sieve-tra tower. and the CS2 concentration in the outlet gas is not to exceed 0. The CS2-N2 mixture is to be scrubbed with an absorbent hydrocarbon oil.d. determine: a) Liquid flow rate.9 CS2 vapor pressure = 346 mm Hg Solution .030 N/m Foaming factor = 0.11c. Design a sieve-tray tower for this process. Carbon disulfide. and solutions of oil and CS2 follow Raoult s law. used as a solvent in a chemical plant. The gas will flow at the rate of 0. Ini ial e ima e . 5 m. 12 mm bet een centers Weir height = 50 mm.b) To er diameter and plate spacing Solution Plate design conditions: f = 0.705 m.5-mm holes in triangular pitch. Plate thickness = 2 mm From the Mathcad program in Appendi E: D = 0.712 E = 0. assuming straight equilibrium and operating lines .588 d) Number of real tra s required Solution Appro imation. t = 0.7 4.018 EMGE = 0. Pressure drop = 414 Pa/tra Fro = 0. Graphical solution: Generate operating line in diagram . Graphicall . the number of ideal stages is also slightl over 5 . Solution .Use 10 tra s e) Total gas-pressure drop. 2 CS2.1 CS2.91 kJ/mole. gas 46. Determine the number of equilibrium tra s for the absorber of Problem 5.9 the latent heat of vapori ation of CS2 at 297 K is 27. liquid CS2. The specific heats are: Substance J/mole-K Oil 362. Adiabatic absorption of carbon disulfide in a sieve-tra tower.12c. gaseous N2 at 297 K Assume T1 = 298. The vapor pressure of CS2 as a function of temperature is given b Solution Base for enthalpies: liquid oil.5.11.0 K .2 N2 29. liquid 76. assuming adiabatic operation. . . . . . . . Appro imatel 12 ideal stages Gas outlet temperature = 298 K . Equilibrium conditions at the operating temperature are approximated by Henry s law such that. 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. C6H6.07 kPa. The oil may be considered nonvolatile.5.13b. It enters the stripper containing 8 mole % benzene. Steam-stripping of ben ene in a sieve-plate column. a) How many theorethical stages are required? b) How many moles of steam are required per 100 moles of the oil-benzene mixture? c) If 85% of the benzene is to be recovered with the same steam and oil rates. how many theoretical stages are required? Solution a) . the equilibrium benzene partial pressure above the oil is 5. The steam leaving contains 3 mole % C6H6. when the oil phase contains 10 mole % benzene. 75% of which is to be removed. b) Ke e e ai c) (G a hica c ci ahead) . d. and total gas-pressure drop.15c. Solution From Prob. packed height. the conditions at the bottom of the absorber are: . Design a to er packed ith 50-mm ceramic Hiflo rings for the carbon disulfide scrubber of Problem 5. Assume isothermal operation and use a liquid rate of 1.0. Absorption of carbon disulfide in a random-packed to er. 5. Assume that Ch for the packing is 1.11.5 times the minimum and a gas-pressure drop not e ceeding 175 Pa/m of packing.Kremser equation 5.11. Calculate the to er diameter. 1 and 4.11: Calculate HtOG at the bottom of the to er . D Results: Estimate the packed height.From Tables 4. assume dilute solutions From Prob 5.2 Run the Packed To er Design Program of Ap. 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 . Solution For the driving force to be constant. using a packed tower that is 0.7 m: .7 m in diameter. The tower is packed with 35-mm plastic NORPAC rings.5. Flue gases usually contain less than 1 mole % of SO2. the operating line is parallel to the equilibrium line. The average gaspressure drop is 200 Pa/m. Assume that the properties of the liquid are similar to those of pure water..d. for a pressure drop of 200 Pa/m and D = 0.4x i. and the absorption factor A = 1. calculate the height of the packed section. 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. and that the properties of the flue gases are similar to those of air. Equilibrium is described by HenryÕs law with y i = 8. an air pollutant regulated by law. Absorption of sulfur dio ide in a random-packed to er. If the liquid flow is adjusted so that the driving force (y y*) is constant.0 It was found by trial and error with the Packed Column program that.16b. . Ben ene vapor recover s stem. Solution .0 mole % Operating pressure of absorber = 800 mm Hg Oil circulation rate = 2 m3/1.000 m3 of gas at STP entering the absorber.130 at 300 K NtOG = 5 transfer units Stripping: Pressure = 1 atm Steam at 1 atm.17c. 398 K Henry s law constant = 3.08 at 398 K Number of equilibrium stages = 5 a) In the winter.0 kg of steam is used in the stripper per 1.5. it is possible to cool the recycled oil to 293 K. = 0. Calculate the percent benzene recovery in the winter. Some data relative to the operation follow: Absorption: Benzene in entering gas = 1. at which temperature the absorber then operates. The resulting benzene-wash-oil solution is then heated to 398 K and stripped in a tray tower. using steam as the stripping medium. Benzene vapor in the gaseous effluent of an industrial process is scrubbed with a wash oil in a countercurrent packed absorber.095 at 293 K. The stripped wash oil is then cooled and recycled to the absorber.000 m3 of gas at STP Oil specific gravity = 0. Under these conditions 72.88 Molecular weight = 260 Henry s law constant = 0. 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 . a d ha N OG a d e i ib i age e ai he a e. ha e ec e f be e e ca be e ec ed? S i . i h he a e i a d ea a e . b he eam a e in he mme i inc ea ed b 50% o e he in e al e.c) If he oil a e canno be inc ea ed. ha mme eco e of ben ene can be e pec ed? . It is desired to absorb. et al. 99% of both GeCl4 and Cl2 in an existing 0. kg/m2-s For the two diffusing species.68 2.01774 m for 13-mm ceramic Raschig rings) = gas density.. The air also contains 540 kg/day of Cl2. 631. also will have no vapor pressure. However.. Thus. H. when dissolved. Ab o p ion of ge mani m e achlo ide ed fo op ical fibe . 17.850 kg/day of air containing 288 kg/day of GeCl4. Determine: a) Liquid flow rate. AIC E J. Both chlorides are oxidized at high temperature and converted to glasslike particles. L.75-m-diameter column that is packed to a height of 3. the dissolved GeCl4 has no vapor pressure and mass transfer is controlled by the gas phase.0 m with 13-mm ceramic Raschig rings. in kg/s.5.. DCl2 = 013 cm2/s.19c. at least. e = 0. Fp = 580 ft 1.Gy mass velocities. Because the solutions are very dilute. the equilibrium curve is a straight line of zero slope. At these conditions. it can be assumed that both gases are absorbed independenttly. take DGeCl4 = 0. The liquid rate should be set so that the column operates at 75% of flooding. kg/m3 Gx .0 kg/m2-s (Shulman.63. 1971): where: ds = equivalent packing diameter (0. 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. which. 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. Germanium tetrachloride (GeCl4) and silicon tetrachloride (SiCl4) are used in the production of optical fibers. The entering gas flows at the rate of 23. For the packing. Solution .06 cm2/s. . Ini ial e ima e b) The pe cen ab o p ion of GeCl4 and Cl2 ba ed on he a ailable 3.0 m of packing. Sol ion (dil e ol ion ) . . 93 Ch = 0. The characteristics of this packing are (Seader and Henley. Redesign the absorber of Problem 5. assume dilute solutions .15 using metal Montz B1-300 structured packing.d.482 Cp = 0.295 CL = 1.5.20c. Absorption of carbon disulfide in a structured-packed to er. 1998): Fp = 33 ft 1 a = 300 m 1 = 0.422 Solution From the Packed Tower Program in Appendix D: Estimate the packed height.165 CV = 0. 11: Calculate HtOG at the bottom of the to er .From Prob 5. 5.11.Calculate HtOG at the top of the tower From Prob. the conditions at the top of the absorber are: The mass-tranfer coefficients remain fairl constant along the tower . There is available a 0. Solution Calculate minimum water flow rate Try . Is the tower satisfactory.3-m-diameter tower packed with 25-mm ceramic Raschig rings to a depth of 3. what water rate should be used? At 300 K.5 m.5. and if so. with m = 1.0 to 0.d. ammonia-water solutions follow Henry s law up to 5 mole % ammonia in the liquid.04% by volume by water scrubbing.21c.414. Absorption of ammonia in a random-packed to er. It is desired to reduce the ammonia content of 0.05 m3/s of an ammonia-air mixture (300 K and 1 atm) from 5. From Appendi D Tr From Appendi D . Tr From Appendi D . Tr From Appendi D . ec i a a ea. A e ac ed i h e a M B1-300 c ed ac i g i be de ig ed f ai b f bed c ab b SO2 c bbi g i h a e . 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 ae he e a i (Seade a d He e .5 e % SO2. The e e i g ga .44 e / 2.f bed c . Absorption of sulfur dio ide in a structured-packed to er. c ai 20 e % f SO2.d. The e i i g ga i c ai 0.C bic i ei e ai 5. P e a e e e a a f a e f 1976 e / 2. 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.22b.ec i a a ea. a a SO2-f ee f a e f 37. 1998): a) De i e he f i g eai g i ee ai f he ab be : S i . Solution From Ap. and the gas-pressure drop per unit of packing height at the bottom of the absorber.b) If the absorber is to process 1. b trial-and-error: c) . calculate the water flow rate.0 m3/s (at 2 atm and 303 K) of the entering gas. D. the tower diameter. D. b trial-and-error: Average mass-transfer coefficients .At the top of the column: From Ap. A e age heigh Calc la e N G Ini ial e ima e . . 2a.0 7.534 1.334 toluene 591. 1987).110 2.3. Test your program with the data presented in Example 6. 6.2 48.3 37. Critical temperatures and pressures. A liquid mixture containing 50 mole % n-heptane (A).629 3.3 for ternary mixtures.381 2. Flash vapori ation of a ternar mi ture.8 41.3 7. at 303 K. at 303 K. et al.983 1. A liquid mixture containing 50 mole % n-heptane (A). 50 mole % n-octane (B).410 3.332 2. is to be continuously flash-vaporized at a pressure of 1 atm to vaporize 30 mole % of the feed. K Pc. toluene (B). is to be continuously flash-vaporized at a temperature of 350 K to vaporize 30 mole % of the feed.1a. 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 6.2 for a mixture of benzene (A). Flash vapori ation of a heptane-octane mi ture. 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). toluene. A = benzene.860 Solution Parameters of the Wagner equation.792 o-xylene 630. Flash vapori ation of a heptane-octane mi ture.9 6.286 1.3d. C = o-xylene: . Modify the Mathcad¨ program of Figure 6.6. 6. B.and the parameters of the Wagner equation for estimating vapor pressure (equation 6-5) are included in the following table (Reid. Component Tc.3. bar A B C D benzene 562. and o-xylene (C)..834 2. Ini ial e ima e . 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 e e f1a .3. C ide he e a i e fE a e 6. S i a .2 a d P b e 6. Flash vapori ation of a ternar mi ture. E i a e he e e a e.4d. 6. pressure.lene in the feed. Consider the ternar mi ture of E ample 6.2 and Problem 6. and to recover in the liquid 70% of the o. and the concentration of the liquid and gas phases Solution .3. Flash vapori ation of a ternar mi ture. fraction of the feed vapori ed. Calculate the temperature.5d. It is desired to recover in the vapor 75% of the ben ene in the feed. Batch distillation of a heptane-octane mi ture. Repeat the calculations of E ample 6.3. but for 80 mole % of the liquid distilled.6b. Solution Batch distillation of a mi ture of heptane and octane .6. 3. c e he c ii f he e id e af e 60 e% f he feed i ba ch-di i ed. fE a e 6. -he a e i h . 1980).8a.I i ia e i a e 6.3. S i I i ia e i a e: 6.7. F hi e . 1980).9b. F hi e .16 (T e ba . Binar batch distillation with constant relative volatilit . he a e age e a i e a i i i = 2.c a e a 1 = 2. he a e age e a i e a i i i he a e age c ce a i f he di i constant relative volatilit . Binar batch distillation with C ide he bi a ba ch di i a i a . C ide he bi a ba ch di i a i fE a e 6. U i g e a i (6-102) de i ed i .16 (T e ba .c a e a 1 a . U i g e a i (6-102) de i ed i P b e 6. The i e i be ba ch-di i ed i a e i 65 e % he a e. -he a e i h . Problem 6.10b. and 55 mole % n-he ane? The pressure is 1. compute the composition of the residue. Solution .7. 30 mole % n-pentane. Solution Initial estimates: 6.0 atm. Mi tures of light h drocarbons: m-value correlations. and the fraction of the feed that is distilled. What is the bubble point of a mi ture that is 15 mole % isopentane. 25 %. Mi tures of light h drocarbons: m-value correlations. a) Calculate the bubble point Solution .5 %.11c. propane. A solution has the follo ing composition. 0.4 to calculate equilibrium distribution coefficients.25 %. 56 %. the pressure is 10 bars. 18. e pressed as mole percent: ethane. n-butane. isobutane. In the follo ing. 0.25 %.Initial estimate 6. Use equation (6-103) and Table 6. isopentane. Initial estimate b) Calculate the de point Solution . Initial estimate c) The solution is flash-vapori ed to vapori e 40 mol % of the feed. Calculate the composition of the products. Solution Initial estimates . A product having an .12b. Binar batch distillation with constant relative volatilit . A 30 mole % feed of ben ene in toluene is to be distilled in a batch operation.Check! 6. S i .i e f ac i f i a .6 353.944 C ii f he a e ei = = 0. I i ia e i a e : 6.3 354. Batch distillation of a mi ture of isopropanol in ater. Ca c a e he a f e id e ef . a 1 a a e (Seade a d He e .5 a d F = 100 e .13b. 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. K 366 357 355.5 0.453 0.462 0.593 0. VLE da a f hi e .742 0.682 0.685. b i i g i f he a e e = 353.350 0.679 0.524 0.1 354.084 0. be d ced.2 353.3 353.198 0. Ca c a e he c ii f he i id e id e e ai i g i he i . 1998): T.220 0.012 0.769 0.916 0. a d he a e age c ii f he c ec ed di i a e.569 0.2 K.a e age c a i g ha S i ii f 45 e % be e ei = 2. Continuous distillation of a binar mi ture F c i di i a i f a bi a i e fc be ed e i a e he i i be f e i ib i U e Fe ee ai e i a e N i f di i a i A e ha . a e a i e a i i .I i ia e i a e 6. 6.4. Fe e e a i (6-58) ca age e i ed f he gi e e a a i . f hi e a 1 a .5. Continuous distillation of a binar mi ture of constant relative volatilit . 1980): ef a i ca (6-105) . aii i c a a = 2.15a. N i . 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 be de e i ed a a ica f he f i g e a i (T e ba . f he be e ee e i e fE a e 6.16a. he e a i e S i of constant relative volatilit . U e e a i (6-105) A e ha . 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. he e a i e f he be e eaii i c a e e i e fE a a = 2.5 e% i a .17b.5.13. VLE da a a e gi e i P b e 6. e 6.5 i e he i i i ed. Ni e -eigh e ce f he i a i he feed be ec e ed. f hi S i e i a e R i f di i a i e a 1 a . If a ef a i f 1.4. h a he e ica age i be e i ed: (a) If a a ia eb i e i ed? S i . I i ia e i a e 6. Continuous rectification of a ater-isopropanol mi ture. 6. a) Ca c a e he a f a e f he d c a d f he e id e. A distillation-membrane h brid for ethanol deh dration. The i i be fed a di i a i c eai ga a he ic e e.5 he i i . . i h a a ia eb i e a d a a c de e .24c. 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 d a e id e c ai i g 99 e% a e . 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. The ef ai i be 1. I a gi e a ica i . a a ge i ch. S i I i ia e i a e : b) Ca c a e he S i a f aea dc ii f he di i a e c i g f he c .99). The di i a e i e e a e b a e i h a a e e a = 70 a d = 0. 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 e b a e b he c ce a i ha he e ea e ea i he e ha .6. 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. 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 ea ea i e aii .ich d c ( P = 0. The mole fraction of the light ke in the distillate should be 0. and 35 mole % HK and is a saturated liquid.26b. Fenske-Under ood-Gilliland method. Use the FUG approach to estimate the number of ideal stages required and the optimal location of the feed-stage. The reflu ratio is 1.5 % recover of the light ke in the distillate.0 HK = 0. The feed is 10 mole % LNK.0 LK = 1.Initial estimates 6.75 Solution Kilomoles in distillate: . Equilibrium data: LNK = 4.75. 55 mole % LK. We desire 99.2 the minimum. A distillation column has a feed of 100 kmoles/h. q = 1 Initial estimate .Kilomoles in bottoms: Determine minimum reflu : Saturated liquid. 2 Rmin: Gilliland co ela ion Ini ial e ima e Ki kb ide e a ion: Ini ial e ima e: .De e mine n mbe of ideal age a R = 1. Feed a age 11 6. ca c a e he di i a e I i ia e i a e : .27b. O e h d ed e /h f a e a b bb e. fi e he e ica a db c i i b he Fe ee ai .4 5 B 0. S i age .2 3 C 0. a d a ef . Fenske-Under ood-Gilliland method.i f i gc ii : i e be e a a ed b di i a i ha he C e M e f ac i Re a i e a i i A 0.4 1 a) F a di i a e a e f 60 e /h. determine the minimum reflu ratio b the Under ood equation. Solution Initial estimate: .b) Using the separation in part a) for components A and C. determine the number of theoretical stages required. Solution . and the optimum feed location.2 times the minimum.Initial estimate: Initial estimate: c) For a reflu ratio of 1. C ce a De ig f Di i a i S McG a -Hi . .35c Flash Calculations: the Rachford-Rice Method for Ideal Mi tures. Ne Y . NY. 2001): a ed a e .I i ia e i a e Ki b ide e ai : I i ia e i a e: Feed a age 5 6. a) Sh ha he be f fa h a i a i fa ic e idea i e ca be ef gge ed b Rachf d a d Rice (D he a d Ma e. 2 using the Rachford-Rice method. Solution Initial estimate: .b) Solve E ample 6. 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 .