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 i i f he i e i e f a f ac i . S i : Ba i : 100 e f he i e
b) The a e age ec a eigh f he i e.
S i
c) The a a c ce a i .
S i
d) The 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 eigh each i o be con in o l di illed a he a e of 4,000 kg/h De e mine:
a) The concen a ion of he mi e in e m of mole f ac ion . Sol ion
Ba i : 100 kg mi e
b) The a e age molec la eigh of he mi e. Sol ion
c) Calc la e he feed a e in kmol/h. Sol ion
1.3a. Concen a ion of li ified na al ga . A a e f i ified a a ga , LNG, f A a a ha he f i g a c i i : 93.5% CH4, 4.6% C2H6, 1.2% C3H8, a d 0.7% CO2. Ca c a e: a) A e age ec a eigh f he LNG i e. S i b) Weigh f ac i f CH4 i he i e. S i Ba i : 100 e f LNG
c) The LNG i hea ed 300 K a d 140 Pa, a d a i e c e e . E i a e he de i f he ga i e de he e c di i .
1.4b. Concen a ion of a fl e ga .
A fl e ga con i of ca bon dio ide, o gen, a e apo , and ni ogen. The mola f ac ion of CO2 and O2 in a ample of he ga a e 12% and 6%, e pec i el . The eigh f ac ion of H2O in he ga i 6.17%. E ima e he den i of hi ga a 500 K and 110 kPa.
Sol ion
Ba i : 100 kmole of ga mi e
Le = mola f ac ion of a e in he mi e (a a pe cen )
Ini ial e ima e
1.5b. Material balances around an ammonia gas absorber.
A ga ea f a he a e f 10.0 3/ a 300 K a d 102 Pa. I c i f a e i a i e f a ia a d ai . The ga e e h gh he b f a ac ed bed ga ab be he e i f c e c e a ea f e i id a e ha ab b 90% f a f he a ia, a d i a ai . The ab be e e i c i d ica i h a i e a dia e e f 2.5 . a) Neg ec i g he e a a i f a e , ca c a e he a ia f ac i i he ga ea i g he ab be . S i : Ba i : 1 ec d A = a ia B = ai b) Ca c a e he e ga a e ci (defi ed a a f a e e i e be c - ec i a a ea). S i :
1.6b. Velocities and flu es in a gas mi ture.
A ga i e a a a e e f 150 Pa a d 295 K c ai 20% H2, 40% O2, a d 40% H2O b e. The ab e e ci ie f each ecie a e -10 / , -2 / , a d 12 / , e ec i e , a
in the direction of the -a is.
a) Determine the mass average velocit , v, and the molar average velocit , V, for the mi ture. Solution
Molar average velocit , V
Mass average velocit , vm Basis: 1 kmole of gas mi ture
b) Evaluate the four flu es: jO2, nO2, JO2, NO2. Solution:
1.7b. P ope ie of ai a a ed i h a e apo . Ai , ed i a 30- 3 c ai e a 340 K a d 101.3 Pa i a a ed i h a e a . De e i e he f i g e ie f he ga i e: a) M e f ac i f a e a . b) A e age ec a eigh f he i e. c) T a a c ai ed i he a . d) Ma f a e a i he a . S i a) A i e e a i f a e a : F a a a ed i e b) c) d)
1.8c. Wa e balance a o nd an ind ial cooling o e .
The c i g a e f a e he c de e f a big c a -fi ed e a i 8,970 g/ . The a e e e he c de e a 29 C a d ea e a 45 C. F he c de e , he a e f a c i g e he e i i c ed d bac 29 C b c e c e c ac i h ai ( ee Fig e 1.11). The ai e e he c i g e a he a e f 6,500 g/ f d ai , a a d -b b e e a e f 30 C a d a h idi f 0.016 g f a e / g f d ai . I ea e he c i g e a a ed i h a e a a 38 C. a) Ca c a e he a e e b e a a i i he c i g e . S i C ide he ai ea i g he e a a ed a 38 C. A i e e a i f a e a :
E = a e b e a a i i he ai
b) T acc f a e e i he c i g e , a f he eff e f a ea b ici a a e a e ea e a i be ed a a e a e . Thi a e a e c ai 500 g/L f di ed id . T a id f i g f he c de e hea - a fe face , he ci c a i g a e i c ai e ha 2,000 g/L f di ed id . The ef e, a a a f he ci c a i g a e be de ibe a e di ca ded (b d ). Wi dage e f he e a e e i a ed a 0.2% f he eci c a i a e. E i a e he a e - a e e i e e . S i W = i dage e = 500 c = 2000 M = a e a e a e B = b d a e I i ia e i a e Wa e ba a ce S id ba a ce:
1.9b. Water balance around a soap dr er.
It is desired to dr 10 kg/min of soap continuousl from 17% moisture b weight to 4% moisture in a countercourrent stream of hot air. The air enters the dr er at the rate of 30.0 m3/min at 350 K, 101.3 kPa, and initial water-vapor partial pressure of 1.6 kPa. The dr er operates at constant temperature and pressure.
a) Calculate the moisture content of the entering air, in kg of water/kg of dr air. Solution
b) Calculate the flow rate of dr air, in kg/min. Solution
c) Calculate the water-vapor partial pressure and relative humidit in the air leaving the dr er. Solution
Calc la e a e a e e a 350 K
1.10b. Acti ated carbon adsorption; material balances. A a e ga c ai 0.3% e e i ai , a d cc ie a e f 2,500 3 a 298 K a d 101.3 Pa. I a eff ed ce he e e c e f hi ga , i i e ed 100 g f ac i a ed ca b , i i ia f ee f e e. The e i a ed each e i ib i a c a e e a e a d e e. A i g ha he ai d e ad b he ca b , ca c a e he e i ib i c ce a i f e e i he ga e ha e, a d he a f e e ad bed b he ca b . The ad i e i ib i f hi e i gi e b he F e d ich i he (EAB C C Ma a , 3 d. ed., U. S. E. P. A., Re ea ch T ia g e Pa , NC, 1987.): he e W i he ca b e i ib i ad i i , i g f e e/ g f ca b , a d * i he e i ib i e e a ia e e, i Pa, a d be be ee 0.7 a d 345 Pa. S i M = a f ca b = e f e e ad bed I i ia e i a e
1.11b. Acti ated carbon adsorption; material balances.
It is desired to adsorb 99.5% of the toluene originall present in the waste gas of Problem 1.10. Estimate how much activated carbon should be used if the s stem is allowed to reach equilibrium at constant temperature and pressure.
1.12a, d. Estimation of gas diffusivit b the Wilke-Lee equation. E. M. La (MS he i , O eg S a e U i e i , 1964) ea ed he diff i i f ch f i ai a 298 K a d 1 a a d e ed i a e a 0.093 c 2/ . E i a e he diff i c efficie b he Wi e-Lee e a i a d c a e i i h he e e i e a a e. S i F A e di B
a) E ima e he diff i i of naph halene (C10H8) in ai a 303 K and 1 ba . Compa e i i h he e pe imen al al e of 0.087 cm2/ epo ed in Appendi A. The no mal boiling poin of
naph halene i 491.1 K, and i c i ical ol me i 413 cm3/mol.
Sol ion
E pe imen al al e
b) E ima e he diff i i of p idine (C5H5N) in h d ogen a 318 K and 1 a m. Compa e i i h he e pe imen al al e of 0.437 cm2/ epo ed in Appendi A. The no mal boiling poin of p idine i 388.4 K, and i c i ical ol me i 254 cm3/mol.
E pe imen al al e
c) E ima e he diff i i of aniline (C6H7N) in ai a 273 K and 1 a m. Compa e i i h he e pe imen al al e of 0.061 cm2/ (G illiland, E. R., I d. E g. Che ., 2 :681, 1934). The no mal boiling poin of aniline i 457.6 K, and i c i ical ol me i 274 cm3/mol.
Sol ion
1.14d. Diffusivit of polar gases
If one or both components of a binary gas mixture are polar, a modified Lennard-Jones relation is often used. Brokaw (Ind. Eng. Chem. Process Design De elop., 8:240, 1969) has suggested an alternative method for this case. Equation (1-49) is still used, but the collision integral is now given by
mp = dipole moment, debyes [1 debye = 3.162 10-25 (J-m3)1/2]
a) Modify the Mathcad routine of Figure 1.3 to implement Brokaw's method. Use the function name
DABp(T, P, MA, MB, mA, mB, VA, VB, TbA, TbB)
b) E i a e he diff i c efficie f a i e f e h ch ide a d f di ide a 1 ba a d 323 K, a d c a e i he e e i e a a e f 0.078 c 2/ . The da a e i ed e B a ' e a i a e h be (Reid, e a ., 1987): Pa a e e Me h ch ide S f di ide Tb , K 249.1 263.2 Vb , c 3/ 50.6 43.8 , deb e 1.9 1.6 M 50.5 64.06 S i
1.15d. 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 . The da a e i ed e B a ' e a i ( ee P b e 1.14) a e h be (Reid, e a ., 1987): Pa a e e H d ge ch ide Wa e Tb , K 188.1 373.2 Vb , c 3/ 30.6 18.9 , deb e 1.1 1.8 M 36.5 18
S i
1.16d. Diffusivit of polar gases
E a a e he diff i c efficie f h d ge fide i f di ide a 298 K a d 1.5 ba . The da a e i ed e B a ' e a i ( eeP b e 1.14) a e h be (Reid, e a ., 1987): Pa a e e H d ge fide S f di ide Tb , K 189.6 263.2 Vb , c 3/ 35.03 43.8 , deb e 0.9 1.6 M 34.08 64.06 S i
1.17a,d. Effective diffusivit in a multicomponent stagnant gas mi ture.
Calculate the effective diffusivit of nitrogen through a stagnant gas mi ture at 373 K and 1.5 bar. The mi ture composition is:
O2 15 mole % CO 30% CO2 35% N2 20% Sol;ution
Calculate mole fractions on a nitrogen (1)-free basis: o gen (2); carbon mono ide (3); carbon dio ide (4)
1.18a,d. Mercur removal from flue gases b sorbent injection.
Mercur is considered for possible regulation in the electric power industr under Title III of the 1990 Clean Air Act Amendments. One promising approach for removing mercur from fossil-fired flue gas involves the direct injection of activated carbon into the gas. Meserole, et al. (J. Ai & Wa e
Manage. A oc., 49:694-704, 1999) describe a theoretical model for estimating mercur removal b
the sorbent injection process. An important parameter of the model is the effective diffusivit of mercuric chloride vapor traces in the flue gas. If the flue gas is at 1.013 bar and 408 K, and its composition (on a mercuric chloride-free basis) is 6% O2, 12% CO2, 7% H2O, and 75% N2, estimate the effective diffusivit of mercuric chloride in the flue gas. Assume that onl the HgCl2 is adsorbed b the activated carbon. Meserole et al. reported an effective diffusivit value of 0.22 cm2/s.
Solution
1.19a. Wilke-Chang method for liquid diffusivit .
E i a e he i id diff i i f ca b e ach ide i di e i i e ha a 298 K. C a e he e e i e a a e e ed b Reid, e a . (1987) a 1.5 10-5 c 2/ . The c i ica
e f ca b e ach ide i 275.9 c 3/ . The i c i f i id e ha a 298 K i 1.08 cP.
S i
1.20b. Diffusion in electrol te solutions.
Whe a a di cia e i i , i a he ha ec e diff e. I he ab e ce f a e ec ic e ia , he diff i f a i g e a a be ea ed a ec a diff i . F di e
i f a i g e a , he diff i c efficie i gi e b he Ne -Ha e e a i (Ha ed, H. S., a d B. B. O e , "The Ph ica Che i f E ec ic S i ," ACS M g . 95, 1950):
a) E i a e he diff i c efficie a 298 K f a e di e i f HC i a e .
S i
b) E i a e he diff i c efficie a 273 K f a e di e i f C SO4 i a e . The i c i f i id a e a 273 K i 1.79 cP.
1.21a. O gen diffusion in ater: Ha duk and Minhas correlation.
Estimate the diffusion coefficient of oxygen in liquid water at 298 K. Use the Hayduk and Minhas correlation for solutes in aqueous solutions. At this temperature, the viscosity of water is 0.9 cP. The critical volume of oxygen is 73.4 cm3/mol. The experimental value of this diffusivity was reported as 2.1 10 5 cm2/s (Cussler E. L., Diff sion, 2nd ed, Cambridge University Press, Cambridge, UK, 1997).
1.22a, d. Liquid diffusivit : Ha duk and Minhas correlation.
Estimate the diffusivity of carbon tetrachloride in a dilute solution in n-hexane at 298 K using the Hayduk and Minhas correlation for nonaqueous solutions. Compare the estimate to the reported value of 3.7 10 5 cm2/s. The following data are available (Reid, et al., 1987):
1.23b. E ima ing mola ol me f om li id diff ion da a.
The diffusivity of allyl alcohol (C3H6O) in dilute aqueous solution at 288 K is 0.9 10 5 cm2/s (Reid, et al., 1987). Based on this result, and the Hayduk and Minhas correlation for aqueous solutions, estimate the molar volume of allyl alcohol at its normal boiling point. Compare it to the result obtained using the data on Table 1.2. The viscosity of water at 288 K is 1.15 cP.
Solution
Iniitial estimates
1.24b, d. Concentration dependence of binar liquid diffusivities.
a) E i a e he diff i i f e ha i a e a 298 K he he f ac i f e ha i i i 40%. U de he e c di i (Ha d, B. R., a d R. H. S e , Trans. Farada Soc., 49, 890, 1953): The e e i e a a e e ed b Ha d a d S e (1953) i 0.42 10-5 c 2/ . S i E i a e he i fi i e di i diff i i f e ha i a e a 298 K f Ha d -Mi ha f a e i F A e di A, he i fi i e di i diff i i f a e i e ha a 298 K i
b) E i a e he diff i i f ace e i a e a 298 K he he f ac i f ace e i i i 35%. F hi e a 298 K, he ac i i c efficie f ace e i gi e b Wi e a i (S i h, J. M., e a ., Introduction to Chemical Engineering Thermod namics, 5 h ed, McG a -Hi C ., I c., Ne Y , NY, 1996):
S i
F A e di A
E i a e he i fi i e di i diff i i f ace e i a e a 298 K
1.25b, d. Stead -state, one-dimensional, gas-phase flu calculation.
A flat plate of solid carbon is being burned in the presence of pure o gen according to the reaction
Molecular diffusion of gaseous reactant and products takes place through a gas film adjacent to the carbon surface; the thickness of this film is 1.0 mm. On the outside of the film, the gas concentration is 40% CO, 20% O2, and 40% CO2. The reaction at the surface ma be assumed to be instantaneous, therefore, ne t to the carbon surface, there is virtuall no o gen.The temperature of the gas film is 600 K, and the pressure is 1 bar. Estimate the rate of combustion of the carbon, in kg/m2-min.
Solution
CO (1), CO2 (2), O2 (3)
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 length of the diffusion path is
The densit of the gas phase follows from the ideal gas law
Initial estimates of the flu es
Stoichiometric relations
1.26b. Stead -state, one-dimensional, liquid-phase flu calculation. A G ' (N 2SO4 10H2O) 288 K. E N 2SO4 . A 0.085 . A ( ) N 2SO4, . T G ' 288 K 36 /100 1,240 / 3 (P C , 1973). T N 2SO4 288 K P 1-20. T 288 K 999.8 / 3; 1.153 P. S A = N 2SO4 B = H2O C A1 ( )B : 100 H2O (36 ) (P )
Calculate diffusivity
1.27c, d. Molec la diff ion h o gh a ga -li id in e face.
Ammonia, NH3, is being selectively removed from an air-NH3 mixture by absorption into water. In this steady-state process, ammonia is transferred by molecular diffusion through a stagnant gas layer 5 mm thick and then through a stagnant water layer 0.1 mm thick. The concentration of ammonia at the outer boundary of the gas layer is 3.42 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. The diffusivity of ammonia in air under these conditions is 0.215 cm2/s and in liquid water is 1.77 10 5 cm2/s. Neglecting water evaporation, determine the rate of diffusion of ammonia, in kg/m2-hr. Assume that the gas and liquid are in equilibrium at the interface.
Sol ion:
1.28c. Stead -state molecular diffusion in gases.
A i e f e ha a d a e a i bei g ec ified i a adiaba ic di i a i c . The a c h i a i ed a d a fe ed f he i id he a ha e. Wa e a c de e (e gh he a e hea f a i a i eeded b he a c h bei g e a a ed) a d i a fe ed f he a he i id ha e. B h c e diff e h gh a ga fi 0.1 hic . The e e a e i 368 K a d he e e i 1 a . The e f ac i f e ha i 0.8 e ide f he fi a d 0.2 he he ide f he fi . Ca c a e he a e f diff i f e ha a d f a e , i g/ 2- . The a e hea f a i a i f he a c h a d a e a 368 K ca be e i a ed b he Pi e ace ic fac c e a i (Reid, e a ., 1987)
he e i he ace ic fac . S i A = e ha B = a e Ca c a e e ha hea f a i a i Ca c a e a e hea f a i a i E i a e diff i i f Wi e-Lee
1.29a, d. Analog among molecular heat and mass transfer.
It has been observed that for the s stem air-water vapor at near ambient conditions, Le = 1.0 (Tre bal, 1980). This observation, called the Lewis relation, has profound implications in
humidification operations, as will be seen later. Based on the Lewis relation, estimate the diffusivit of water vapor in air at 300 K and 1 atm. Compare our result with the value predicted b the Wilke-Lee equation. For air at 300 K and 1 atm:Cp = 1.01 kJ/kg-K, k = 0.0262 W/m-K, m = 1.846 10-5 kg/m-s, and r = 1.18 kg/m3.
Solution
1.30b, d. Stead -state molecular diffusion in gases.
Wa e e a a i g f a d a 300 K d e b ec a diff i ac a ai fi 1.5
hic . If he e a i e h idi f he ai a he e edge f he fi i 20%, a d he a e e i 1 ba , e i a e he d i he a e e e e da , a i g ha c di i i he fi e ai
c a . The a e e f a e a a f c i f e e a e ca be acc a e e i a ed f he Wag e e a i (Reid, e a ., 1987)
S i
F A e di A
1.31b, d. Stead -state molecular diffusion in a ternar gas s stem.
Ca c a e he f e a d c ce a i fi e f he e a e h d ge (1), i ge (2), a d ca b di ide (3) de he f i g c di i . The e e a e i 308 K a d he e e i 1
atm. The diffusion path length is 86 mm. At one end of the diffusion path the concentration is 20 mole% H2, 40% N2, 40% CO2; at the other end, the concentration is 50% H2, 20% N2, 30% CO2. The total molar flu is ero, = 0. The MS diffusion coefficients are D12 = 83.8 mm2/s, D13 = 68.0 mm2/s, D23 = 16.8 mm2/s.
Solution
Appendi C-1: Solution of the Ma
ell-Stefan equations for a
multicomponent mi ture of ideal gases b orthogonal collocation ( C =
3).
Orthogonal collocation matrices
The p e e and empe a e in he apo pha e a e
The Ma ell-S efan diff ion coefficicien a e
The leng h of he diff ion pa h i
2.1a. Ma - an fe coefficien in a ga ab o be . A ga ab be i ed e e be e e (C6H6) a f ai b c bbi g he ga i e i h a a i e i a 300 K a d 1 a . A a ce ai i i he ab be , he be e e e f ac i i he b f he ga ha e i 0.02, hi e he c e di g i e facia be e e ga -ha e c ce a i i 0.0158. The be e e f a ha i i ea ed a 0.62 g/ 2- . a) Ca c a e he a - a fe c efficie i he ga ha e a ha i i he e i e , e e i g he d i i g f ce i e f e f ac i . S i b) Ca c a e he a - a fe c efficie i he ga ha e a ha i i he e i e , e e i g he d i i g f ce i e f a c ce a i , / 3. S i c) A he a e ace i he e i e , he be e e e f ac i i he b f he i id ha e i 0.125, hi e he c e di g i e facia be e e i id- ha e c ce a i i 0.158. Ca c a e he a - a fe c efficie i he i id ha e, e e i g he d i i g f ce i e f e f ac i . S i
2.2a. Ma - an fe coefficien f om naph halene blima ion da a.
I a ab a e e i e , ai a 347 K a d 1 a i b a high eed a d a i g e
f he he e i 2.0 c . A he e d f he e e i e , 14.32 i a e , he dia e e f he he e i 1.85 c .
a) E i a e he a - a fe c efficie , ba ed he a e age face a ea f he a ic e, e e i g he d i i g f ce i e f a ia e e . The de i f id a h ha e e i 1.145 g/c 3, i a e e a 347 K i 670 Pa (Pe a d Chi , 1973).
S i
b) Ca c a e he a - a fe c efficie , f he d i i g f ce i e f a c ce a i .
S i
2.3a. Mass-transfer coefficients from acetone e aporation data.
I a ab a e e i e , ai a 300 K a d 1 a i b a high eed a a e he face f a ec a g a ha a ha c ai i id ace e (C3H6O), hich e a a e a ia . The a i 1 g a d 50 c ide. 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, ai ai i g a c a i id e e i he a . D i g a e e i e a , i a b e ed ha 2.0 L f ace e e a a ed i 5 i . E i a e he a
-a fe c efficie . The de i f i id -ace e a 300 K i 0.79 g/c 3; i a e e i 27 Pa (Pe a d Chi , 1973).
2.4b. Mass-transfer coefficients from etted- all e perimental data.
A e ed- a e e i e a e - c i f a g a i e, 50 i dia e e a d 1.0 g. Wa e a 308 K f d he i e a . D ai e e he b f he i e a he a e f 1.04
3/ i , ea ed a 308 K a d 1 a . I ea e he e ed ec i a 308 K a d i h a e a i e h idi f 34%. Wi h he he f e a i (2-52), e i a e he a e age a - a fe c efficie ,
i h he d i i g f ce i e f a f ac i .
2.7c. Ma an fe in an ann la pace.
a) I d i g a e f diff i f a h ha e e i ai , a i e iga e aced a 30.5-c ec i f he i e i e f a a i h a a h ha e e d. The a a c ed f a 51- -OD b a i e i e ded b a 76- -ID b a i e. Whi e e a i g a a a e ci i hi he a f 12.2 g f ai / 2- a 273 K a d 1 a , he i e iga de e i ed ha he a ia e e f a h ha e e i he e i i g ga ea a 0.041 Pa. U de he c di i f he i e iga i , he Sch id be f he ga a 2.57, he i c i a 175 P, a d he a e e f a h ha e e a 1.03 Pa. E i a e he a - a fe c efficie f he i e a f hi e f c di i . A e ha e a i (2-52) a ie . S i b) M ad a d Pe (T an . AIChE, 38, 593, 1942) e e ed he f i g c e a i f hea - a fe c efficie i a a a ace:
he e d a d di a e he ide a d i ide dia e e f he a , de i he e i a e dia e e defi ed a
W i e d he a a g e e i f a a fe a d e i e i a e he a - a fe c efficie f he c di i f a a). C a e b h e .
2.8c. The Chilton-Colburn analog : flow across tube banks. W C (I d. E g. Che ., 40, 1087, 1948) 310 K 1 . T , . T 10 38- -OD ( = 38 ) 57- , 76 . T . T : G' , / 2- , G / 2 --P . ) R (2-68) C D - . T 310 K 1 0.074 2/ . S P 310 K 1 : D :
b) E i a e he a - a fe c efficie be e ec ed f e a a i f - a c 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 a c h i ca b di ide a 300 K a d 1 a :
c) Za a a (Ad . Heat Transfer, , 93, 1972) ed he f i g c e a i f he hea -a fe c efficie i -a -agge ed be b-a -a -a ge e i i -a h-a died b Wi di g a d Che e :
U e he a - a fe e e i a a g e a i (2-69) e i a e he a - a fe c efficie f a b). C a e he e .
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) .
Laminar flo
At the bulk of the solution, point 2:
2.11b. Ma an fe f om a fla li id face.
D i g he e e i e de c ibed i P b e 2.3, he ai e ci a ea ed a 6 / , a a e he ge ide f he a . E i a e he a - a fe c efficie edic ed b e a i (2-28) (2-29) a d c a e i he a e ea ed e e i e a . N ice ha , d e he high a i i f ace e, he a e age ace e c ce a i i he ga fi i e a i e high. The ef e, e ie
ch a de i a d i c i h d be e i a ed ca ef . The f i g da a f ace e igh be eeded: Tc = 508.1 K, Pc = 47.0 ba , M = 58, Vc = 209 c 3/ , Zc = 0.232 (Reid, e a ., 1987).
S i
A e age fi e ie :
E i a e he i c i f he i e f L ca Me h d
2.12b. E aporation of a drop of ater falling in air.
Repeat E ample 2.9 for a drop of ater hich is originall 2 mm in diameter. Solution
2.13b. Dissolution of a solid sphere into a flo ing liquid stream.
Estimate the mass-transfer coefficient for the dissolution of sodium chloride from a cast sphere, 1.5 cm in diameter, if placed in a flowing water stream. The velocity of the 291 K water stream is 1.0 m/s.
Assume that the kinematic viscosity at the average liquid film conditions is 1.02 10 6 m2/s, and the mass 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
From Prob. 2.10:
2.14b. Sublimation of a solid sphere into a gas stream.
During the experiment described in Problem 2.2, the air velocity was measured at 10 m/s. Estimate the mass-transfer coefficient predicted by equation (2-36) and compare it to the value measured experimentally. The following data for naphthalene might be needed: Tb = 491.1 K, Vc = 413 cm3/mol.
Solution
For air at 347 K and 1 atm:
Estimate DAB from the Wilke-Lee equation
2.15b. Dissolution of a solid sphere into a flo ing liquid stream.
The cr stal of Problem 1.26 is a sphere 2-cm in diameter. It is falling at terminal velocit under the influence of gravit into a big tank of water at 288 K. The densit of the cr stal is 1,464 kg/m3 (Perr and Chilton, 1973).
a) Estimate the cr stal's terminal velocit . Solution
b) Estimate the rate at which the cr stal dissolves and compare it to the answer obtained in Problem 1.26.
Solution
From Prob.1.26
2.16c. Mass transfer inside a circular pipe.
Water flows through a thin tube, the walls of which are lightly coated with benzoic acid (C7H6O2). The water flows slowly, at 298 K and 0.1 cm/s. The pipe is 1-cm in diameter. Under these
conditions, equation (2-63) applies.
a) Show that a material balance on a length of pipe L leads to
where v is the average fluid velocity, and cA* is the equilibrium solubility concentration.
b) What is the average concentration of benzoic acid in the water after 2 m of pipe. The solubility of benzoic acid in water at 298 K is 0.003 g/cm3, and the mass diffusivity is 1.0 10 5 cm2/s
(Cussler, 1997).
Solution
2.17b. Mass transfer in a etted- all to er.
Water flows down the inside wall of a 25-mm ID wetted-wall tower of the design of Figure 2.2, while air flows upward through the core. Dry air enters at the rate of 7 kg/m2-s. Assume the air is
everywhere at its average conditions of 309 K and 1 atm, the water at 294 K, and the mass-transfer coefficient constant. Compute the average partial pressure of water in the air leaving if the tower is 1
g.
S i
F a e a 294 K
2.18c. Ma an fe in an ann la pace.
I d i g he b i a i f a h ha e e i a ai ea , a i e iga c c ed a 3- - g a a d c . The i e i e a ade f a 25- -OD, id a h ha e e d; hi a
ded b a 50- -ID a h ha e e i e.
Ai a 289 K a d 1 a f ed h gh he a a ace a a a e age e ci f 15 / . E i a e he a ia e e f a h ha e e i he ai ea e i i g f he be. A 289 K, a h ha e e ha a a e e f 5.2 Pa, a d a diff i i i ai f 0.06 c 2/ . U e he e f P b e 2.7
e i a e he a - a fe c efficie f he i e face; a d e a i (2-47), i g he e i a e dia e e defi ed i P b e 2.7, e i a e he c efficie f he e face.
Sol ion
In hi i a ion, he e ill be a mola fl f om he inne all, NA1, i h pecific in e facial a ea, a1, and a fl f om he o e all, NA2, i h a ea a2. A ma e ial balance on a diffe en ial ol me
elemen ield :
Define:
Then:
2.19c. Ben ene evaporation on the outside surface of a single c linder. Be e e i e a a i g a he a e f 20 g/h e he face f a 10-c -dia e e c i de . D ai a 325 K a d 1 a f a igh a g e he a i f he c i de a a e ci f 2 / . The i id i a a e e a e f 315 K he e i e e a a e e f 26.7 Pa. E i a e he e g h f he c i de . F be e e, Tc = 562.2 K, Pc = 48.9 ba , M = 78, Vc = 259 c 3/ , Zc = 0.271 (Reid, e a ., 1987). S i Ca c a e he a e age e ie f he fi F he Wi e-Lee e a i F he L ca Me h d F E . 2-45:
2.20b. Ma an fe in a packed bed.
Wilke and Hougan (T an . AIChE, 41, 445, 1945) reported the mass transfer in beds of granular solids. Air was blown through a bed of porous celite pellets wetted with water, and by evaporating this water under adiabatic conditions, they reported gas-film coefficients for packed beds. In one run, the following data were reported:
effective particle diameter 5.71 mm gas stream mass velocity 0.816 kg/m2-s temperature at the surface 311 K
pressure 97.7 kPa
kG 4.415 10 3 kmol/m2-s-atm
With the assumption that the properties of the gas mixture are the same as those of air, calculate the gas-film mass-transfer coefficient using equation (2-55) and compare the result with the value reported by Wilke and Hougan.
Solution
2.21b. Ma an fe and p e e d op in a packed bed.
Air at 373 K and 2 atm is passed through a bed 10-cm in diameter composed of iodine spheres 0.7-cm in diameter. The air flows at a rate of 2 m/s, based on the empt cross section of the bed. The porosit of the bed is 40%.
a) How much iodine will evaporate from a bed 0.1 m long? The vapor pressure of iodine at 373 K is 6 kPa.
Solution
b) E ima e he p e e d op h o gh he bed. Sol ion
2.22b. Volumetric mass-transfer coefficients in industrial to ers.
The interfacial surface area per unit volume, a, in many types of packing materials used 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, inmiscible streams. Accordingly, they are normally correlated together as the volumetric mass-transfer coefficient, kca.
Empirical equations for the volumetric coefficients must be obtained experimentally for each type of mass-transfer operation. Sherwood and Holloway (Trans. AIChE, 36, 21, 39, 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, when SI units are used exclusively.
a) Consider the absorption of SO2 with water at 294 K in a tower packed with 25-mm Raschig rings. If the liquid mass velocity is L' = 2.04 kg/m2-s, estimate the liquid-film mass-transfer coefficient. The diffusivity of SO2 in water at 294 K is 1.7 10 9 m2/s.
Solution
For dimensional consistency, add the constants:
b) Whitney and Vivian (Chem. Eng. Progr., 45, 323, 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. For the conditions described in part a), estimate the liquid-film mass-transfer coefficient using equation (2-72). Compare the results.
2.23b. Mass transfer in fluidi ed beds.
Cavatorta, et al. (AIC E J., 45, 938, 1999) studied the electrochemical reduction of ferrycianide ions, {Fe(CN)6} 3, to ferrocyanide, {Fe(CN)6} 4, in aqueous alkaline solutions. They studied different arrangements of packed columns, including fluidized beds. The fluidized bed experiments were performed in a 5-cm-ID circular column, 75-cm high. The bed was packed with 0.534-mm spherical glass beads, with a particle density of 2.612 g/cm3. The properties of the aqueous solutions were: density = 1,083 kg/m3, viscosity = 1.30 cP, diffusivity = 5.90 10 10 m2/s. They found that the porosity of the fluidized bed, e, could be correlated with the superficial liquid velocity based on the empty tube, vs, through
where vs is in cm/s.
a) Using equation (2-56), estimate the mass-transfer coefficient, kL, if the porosity of the bed is 60%.
b) Ca a o a e al. p opo ed he follo ing co ela ion o e ima e he ma - an fe coefficien fo hei fl idi ed bed e pe imen al n :
he e Re i ba ed on he emp be eloci . U ing hi co ela ion, e ima e he ma - an fe coefficien , L, if he po o i of he bed i 60%. Compa e o e l o ha of pa a).
Sol ion
2.24b. Mass transfer in a hollo -fiber boiler feed ater deaerator.
Con ide he hollo -fibe BFW deae a o de c ibed in E ample 2-13. If he a e flo a e inc ea e o 60,000 kg/h hile e e hing el e emain con an , calc la e he f ac ion of he en e ing di ol ed o gen ha can be emo ed.
2.25b. Mass transfer in a hollo -fiber boiler feed ater deaerator.
a) Con ide he hollo -fibe BFW deae a o de c ibed in E ample 2-13. A ming ha onl o gen diff e ac o he memb ane, calc la e he ga ol me flo a e and compo i ion a he l men o le . The a e en e he hell ide a 298 K a a ed i h a mo phe ic o gen, hich mean a di ol ed o gen concen a ion of 8.38 mg/L.
b) Ca c a e he a - a fe c efficie a he a e age c di i i ide he e . Neg ec he hic e f he fibe a he e i a i g he ga e ci i ide he e .
S i
Ca c a e he a e age f c di i i ide he fibe
Ca c a e he a e age ge a f ac i i he ga
F L ca e h d f e N2
3.1a. Application of Raoult's law to a binar s stem.
Re ea E a e 3.1, b f a i id c ce a i f 0.6 e f ac i f be e e a d a e e a e f 320 K.
S i
3.2b. Application of Raoult's law to a binar s stem.
a) De e i e he c i i f he i id i e i ib i i h a a c ai i g 60 e e ce be e e-40 e e ce e e if he e e i i a e e de 1 a e e. P edic
he e i ib i e e a e.
S i
b) De e i e he c i i f he a i e i ib i i h a i id c ai i g 60 e e ce be e e-40 e e ce e e if he e e i i a e e de 1 a e e. P edic
he e i ib i e e a e.
S i
3.3a. Application of Raoult's la to a binar s stem.
N a he a e, -C7H16, a d a c a e, -C8H18, f idea i . A 373 K, a he a e ha a a e e f 106 Pa a d a c a e f 47.1 Pa. a) Wha d be he c i i f a he a e- c a e i ha b i a 373 K de a 93 Pa e e? S i b) Wha d be he c i i f he a i e i ib i i h he i ha i de c ibed i (a)? S i
3.4a. Henr 's la : saturation of ater ith o gen.
A i i h ge di ed i a e c ai i g 0.5 g O2/100 g f H2O i b gh i c ac i h a a ge e f a he ic ai a 283 K a d a a e e f 1 a . The He ' a c a f he ge - a e e a 283 K e a 3.27 104 a / e f ac i . a) Wi he i gai e ge ? b) Wha i be he c ce a i f ge i he fi a e i ib i i ? S i
A e i ib i :
Ba i : 1 L a e (1 g a e )
E i ib i c ce a i , ce = 11.42 g ge /L I i ia c di i :
The i gai ge .
3.5c. Material balances combined ith equilibrium relations.
Re ea E a e 3.3, b a i g ha he a ia, ai , a d a e a e b gh i c ac i a c ed c ai e . The e i 10 3 f ga ace e he i id. A i g ha he ga - ace e a d he e e a e e ai c a i e i ib i i achie ed, dif he Ma hcad g a i Fig e 3.2 ca c a e:
a) he a e e a e i ib i
S i
A : P = 1.755
) .
A : A = 0.145; A= 0.162
3.6b. Ma - an fe e i ance d ing ab o p ion.
I A ( = 60) , A,G = 0.1 , A,L = 1.0 A/ 3 . T 2.0 ; 1,100 / 3. T H ' 0.85 / . T KG = 0.27 / 2- - . I 57% , ) - , G; S
b) the liquid-film coefficient, kL; Solution
Basis: 1 m3 of aqueous solution
c) the concentration on the liquid side of the interface, A,i; Solution
d) the mass flu of A. Solution
3.7b. Mass-transfer resistances during absorption.
F a e i hich c e A i a fe i g f he ga ha e he i id ha e, he e i ib i e a i i gi e b
he e A,i i he e i ib i a ia e e i a a d A,i i he e i ib i i id c ce a i i a f ac i . A e i i he a a a , he i id ea c ai 4.5 e % a d he ga
ea c ai 9.0 e % A. The a e e i 1 a . The i di id a ga -fi c efficie a hi i i G = 3.0 e/ 2- -a . Fif e ce f he e a e i a ce a a fe i be e c e ed i he i id ha e. E a a e a) he e a a - a fe c efficie , K ; S i b) he a f f A; S i ( Ae = A*) c) The i id i e facia c ce a i f A. S i
3.8d. Absorption of ammonia b water: use of F-t pe mass-transfer coefficients.
M dif he Ma hcad g a i Fig e 3.6 e ea E a e 3.5, b i h A,G = 0.70 a d A,L = 0.10. E e hi g e e e ai c a .
Sol ion
Ini ial g e e
3.9d. Absorption of ammonia b water: use of F-t pe mass-transfer coefficients.
Modif he Ma hcad p og am in Fig e 3.6 o epea E ample 3.5, b i h FL = 0.0050 kmol/m2- . E e hing el e emain con an .
Sol ion
3.10b. Ma - an fe e i ance d ing ab o p ion of ammonia. I he ab i f a ia i a e f a ai -a ia i e a 300 K a d 1 a , he i di id a fi c efficie e e e i a ed be L = 6.3 c /h a d G = 1.17 / 2-h -a . The e i ib i e a i hi f e di e i f a ia i a e a 300 K a d 1 a i De e i e he f i g a - a fe c efficie : a) S i b) S i
c) Ky Solution
d) Fraction of the total resistance to mass transfer that resides in the gas phase. Solution
3.11b. Mass-transfer resistances in hollo -fiber membrane contactors.
For mass transfer across the hollow-fiber membrane contactors described in Example 2.13, the overall mass-transfer coefficient based on the liquid concentrations, KL, is given by (Yang and Cussler, AIC E J., 32, 1910, Nov. 1986)
where kL, kM, and kc are the individual mass-transfer coefficients in the liquid, across the membrane, and in the gas, respectively; and H is Henry's law constant, the gas equilibrium concentration divided by that in the liquid. The mass-transfer coefficient across a hydrophobic membrane is from (Prasad and Sirkar, AIC E J., 34, 177, Feb. 1988)
where DAB = molecular diffusion coefficient in the gas filling the pores, eM = membrane porosity,
tM = membrane tortuosity, d = membrane thickness.
For the membrane modules of Example 2.13, eM = 0.4,tM = 2.2, and d = 25 10 6 m (Prasad and Sirkar, 1988).
a) Calculate the corresponding value of kM. Solution
b) U i g he e f a (a), E a e 2.13, a d P b e 2.25, ca c a e KL, a d e i a e ha f ac i f he a e i a ce a a fe e ide i he i id fi . S i F E a e 2.13: F P b. 2.25: Vi a a f he e i a ce e ide i he i id ha e.
3.12c. Combined use of F- and k-t pe coefficients: absorption of low-solubilit gases. D i g ab i f - bi i ga e , a a fe f a high c ce a ed ga i e a e di e i id i f e e a e ace. I ha ca e, a h gh i i a ia e e a
- e a - a fe c efficie i he i id ha e, a F- e c efficie be ed i he ga ha e. Si ce di e i id i a be He ' a , he i e facia c ce a i d i g ab i f - bi i ga e a e e a ed h gh A,i = A,i.
a) Sh ha , de he c di i de c ibed ab e, he ga i e facia c ce a i a i fie he e a i
S i
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 a e , 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 0.2% SO2 b e. The e e a e a 303 K a d he a e e 1 a . E i a e he i e facia c ce a i a d he ca SO2 a f . The a - a fe c efficie e e ca c a ed a FG = 0.002 / 2- , = 0.160 / 2- . The e i ib i SO2 bi i da a a 303 K a e (Pe a d Chi , 1973): g SO2/100 g a e Pa ia e e f SO2, Hg ( ) 0.0 0 0.5 42 1.0 85 1.5 129 2.0 176 2.5 224 S i Defi e: = g SO2/100 g a e = Pa ia e e f SO2, Hg ( )
3.13d. Distillation of a mi ture of methanol and ater in a packed to er: use of F-t pe mass-transfer coefficients.
At a different point in the packed distillation column of Example 3.6, the methanol content of the bulk of the gas phase is 76.2 mole %; that of the bulk of the liquid phase is 60 mole %. The temperature at that point in the tower is around 343 K. The packing characteristics and flow rates at that point are such that FG = 1.542 10 3 kmol/m2-s, and FL = 8.650 10 3 kmol/m2-s. Calculate the interfacial compositions and the local methanol flux. To calculate the latent heats of vaporization at the new temperature, modify the values given in Example 3.6 using Watson's method (Smith, et al., 1996):
For water, Tc = 647.1 K; for methanol, Tc = 512.6 K
Solution
For methanol (A)
Pa ame e
3.14b. Material balances: adsorption of ben ene vapor on activated carbon. Ac i a ed ca b i ed ec e be e e f a i ge -be e e a i e. A i ge -be e e i e a 306 K a d 1 a c ai i g 1% be e e b e i be a ed c e c e a he a e f 1.0 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 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 . The e e a e a d a e e a e ai ai ed a 306 K a d 1 a . Ni ge i ad bed. The e i ib i ad i f be e e hi ac i a ed ca b a 306 K i e ed a f : Be e e a ad bed Pa ia e e be e e, Hg c 3 (STP)/g ca b 15 0.55 25 0.95 40 1.63 50 2.18 65 3.26 80 4.88 90 6.22 100 7.83 a) P he e i ib i da a a X' = g be e e/ g d ca b , Y' = g be e e/ g i ge f a a e e f 1 a . S i
b) Ca c a e he i i f a e e i ed f he e e i g ac i a ed ca b ( e e be ha he e e i g ca b c ai e ad bed be e e). S i O he XY diag a , ca e he i (X2,Y2). 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 , he i i e a i g i e i b ai ed b ca i g, a he i e ec i f Y = Y1 i h he e i ib i c e, X1 a .
c) If the carbon flow rate is 20% above the minimum, what will be the concentration of ben ene adsorbed on the carbon leaving?
d) F he c di i f a (c), ca c a e he be f idea age e i ed.
S i
See e i e c c i 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 a e , 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 a e a f : Be e e a ad bed Pa ia e e be e e, Pa g be e e/100 g ca b 2.9 1.0 5.5 2.0 12.0 5.0 17.1 8.0 20.0 10.0 25.7 15.0 30.0 20.0 a) Ca c a e he i i ea f a e 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.
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:
3.17b. Material balances in batch processes: dr ing of soap with air.
I i de i ed d 10 g f a f 20% i e b eigh e ha 6% i e b c ac i h h ai . The e a i aced i a e e c ai i g 8.06 3 f ai a 350 K, 1 a , a d a a e - a a ia e e f 1.6 Pa. The e i a ed each e i ib i , a d
he he ai i he e e i e i e e aced b f e h ai f he igi a i e c e a d e e a e. H a i e he ce be e ea ed i de each he ecified a i e c e f e ha 6%? Whe hi a i e ed ai a 350 K a d 1 a , he e i ib i di ib i f i e be ee he ai a d he a i a f : W % i e i a Pa ia e e f a e , Pa 2.40 1.29 3.76 2.56 4.76 3.79 6.10 4.96 7.83 6.19 9.90 7.33 12.63 8.42 15.40 9.58 19.02 10.60 S i Ge e a e he XY diag a
From the XY diagram, at the e it of the fifth equilibrium stage, X = 0.06 and
ith kerosene.
Nicotine in a ater solution containing 2% nicotine is to be e tracted ith kerosene at 293 K. Water and kerosene are essentiall insoluble. Determine the percentage e traction of nicotine if 100 kg of the feed solution is e tracted in a sequence of four batch ideal e tractions using 49.0 kg of fresh, pure kerosene each. The equilibrium data are as follo s (Claffe et al., I d. E g. Che ., 42, 166, 1950):
X', 103 kg nicotine/kg ater Y', 103 kg nicotine/kg kerosene 1.01 0.81 2.46 1.96 5.02 4.56 7.51 6.86 9.98 9.13 20.4 18.70 Solution
3.19b. Cross-flo cascade of ideal stages.
The d i g a d i id- i id e ac i e a i de c ibed i P b e 3.17 a d 3.18,
e ec i e , a e e a e f a f c fig a i ca ed a c -f ca cade. Fig e 3.27 i a che a ic diag a f a c -f ca cade f idea age . Each age i e e e ed b a ci c e, a d i hi each age a a fe cc a if i c c e f . The L ha e f f e
age he e , bei g c ac ed i each age b a f e h V ha e. If he e i ib i -di ib 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 , 1980) he e S i he i i g fac , VS/LS, c a f a age , a d N i he a be f age . S e P b e 3.18 i g e a i (3-60), a d c a e he e b ai ed b he e h d . S i
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 f P b e 3.18 a d 3.19. Ca c a e he be f idea age e i ed achie e a ea 95% e ac i efficie c .
S i
U e 8 idea age
3.21b. Kremser equations: absorption of h drogen sulfide.
A che e f he e a f H2S f a f 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 i i 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 .
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 a e i
ac ica i ce i e i e a ge a f a e d e he bi i f H2S i a e . If a 2N i f e ha a i e (MEA) i a e 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 f P b e 3.21, b i g a 2N e ha a i e i a ab be .
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.
3.24b. Kremser equations: absorption of sulfur dio ide.
An ab o be i a ailable o ea he fl e ga of P oblem 3.23 hich i e i alen o 8.5 e ilib i m age .
a) Calc la e he a e flo a e o be ed in hi ab o be if 90% of he SO2 i o be emo ed. Calc la e al o he SO2 concen a ion in he a e lea ing he ab o be .
Sol ion
b) Wha i he e ce age e a f SO2 ha ca be achie ed i h hi ab be if he a e f a e ed i he a e ha a ca c a ed i P b e 3.23 (a)?
S i
I i ia e i a e
3.25b. Kremser equations: liquid e traction.
A a e ace ic acid i f a he a e f 1,000 g/h . The i i 1.1% (b eigh ) ace ic acid. I i de i ed ed ce he c ce a i f hi i 0.037% (b eigh ) ace ic acid b e ac i i h 3-he a a 298 K. The i e 3-he a c ai 0.02% (b eigh ) ace ic acid. A e ac i c i a ai ab e hich i e i a e a c e c e ca cade f 15 e i ib i age . Wha e f a e i e i ed? Ca c a e he c i i f he e
ha e ea i g he c . F hi e , e i ib i i gi e b W a i ace ic acid i e = 0.828 W a i ace ic acid i a e
S i
Initial estimate:
3.26c. Co n e c en e c o -flo e ac ion.
A 1-butanol acid solution is to be e tracted ith pure ater. The butanol solution contains 4.5% (b eight) of acetic acid and flo s at the rate of 400 kg/hr. A total ater flo rate of 1005 kg/hr is used. Operation is at 298 K and 1 atm. For practical purposes, 1-butanol and
ater are inmiscible. At 298 K, the equilibrium data can be represented b YAi = 0.62 XAi, here YAi is the eight ratio of acid in the aqueous phase and XAi is the eight ratio of acid in the organic phase.
a) If the outlet butanol stream is to contain 0.10% (b eight) acid, ho man equilibrium stages are required for a countercurrent cascade?
b) If he a e i i e a a g he a e be f age , b i a c -f ca cade, ha i he e 1-b a c ce a i ( ee P b e 3.19)?
S i
3.27c. Glucose sorption on an ion e change resin.
Chi g a d R h e (AIChE S mp. Ser., 81, . 242, 1985) f d ha he e i ib i f g 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. Thei e i ib i e e i a 303 K i YAi = 1.961 XAi, he e XAi i he g c e c ce a i i he e i .(g f g c e e i e f e i ) a d YAi i he g c e c ce a i i i .(g f g c e e i e f
i ).
a) We i h b g c e hi i e cha ge e i a 303 K i a c e c e ca cade f idea age . The c ce a i f he feed i i 15 g/L. We a a e c ce a i f 1.0 g/L. The i e e i c ai 0.25 g f g c e/L. The feed i f a he a e f 100 L/ i ,
hi e he e i f a he a e f 250 L/ i . Fi d he be f e i ib i age e i ed.
b) If 5 equilibrium stages are added to the cascade of part a), calculate the resin flo required to maintain the same degree of glucose sorption.
4.1a. Void fraction near the alls of packed beds.
C ide a c i d ica e e i h a dia e e f 305 ac ed i h id he e i h a dia e e f 50 . a) F e a i (4-1), ca c a e he a ic i f he bed. S i A e b) E i a e he id f ac i a a di a ce f 100 f he a . A e
4.2b. Void fraction near the alls of packed beds.
Beca e f he ci a a e f he id-f ac i adia a ia i f ac ed bed , he e a e a be f ca i c e he a he e he ca id f ac i i e ac e a he a ic a e f he bed. F he bed de c ibed i E a e 4.1, ca c a e he di a ce f he a he fi fi e ch ca i . S i The ca i ea he a e a he a ic i he J0( d) = 0. F E a e 4.1, I i ia e i a e f he ca be b ai ed f Fig. 4.4 I i ia e i a e
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
4.3c. Void fraction near the alls of packed beds.
(a) Sh ha he adia ca i f he a i a a d i i a f he f c i de c ibed b E a i (4-1) a e he f he e a i
S i
b) F he ac ed bed f E a e 4.1, ca c a e he adia ca i f he fi fi e a i a, a d f he fi fi e i i a; ca c a e he a i de f he id f ac i ci a i a h e i .
Ini ial e ima e of he oo can be ob ained f om Fig. 4.4 Ini ial e ima e
Ini ial e ima e
Repea ing hi p oced e, he follo ing e l a e ob ained: Ma ima: Minima
, mm ( *) Ampli de, % ,mm ( *) Ampli de, % 20.8 (1.04) 37.2 11.3 (0.57) -56.7
39.6 (1.98) 21.1 30.2 (1.51) -27.3 58.3 (2.92) 13.5 49.0 (2.45) -16.7 77.1 (3.85) 9.2 67.7 (3.39) -11.1 95.8 (4.79) 6.4 86.4 (4.32) -7.6
c) Calc la e he di ance f om he all a hich he ab ol e al e of he po o i fl c a ion ha been dampened o le han 10% of he a mp o ic bed po o i .
Sol ion
(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 4.1, estimate the average void fraction by numerical integration of equation (4-65) and estimate the ratio a / b.
Solution
4.4c. Void fraction near the alls of annular packed beds.
Annular packed beds (APBs) involving the flow of fluids are used in many technical and engineering applications, such as in chemical reactors, heat exchangers, and fusion reactor blankets. It is well known that the wall in a packed bed affects the radial void fraction distribution. Since APBs have two walls that can simultaneously affect the radial void fraction distribution, it is essential to include this variation in transport models. A correlation for this purpose was recently formulated (Mueller, G. E.,
AIChE J., 45, 2458-60, Nov. 1999). The correlation is restricted to randomly packed beds in annular
consisting of equal-sized spheres of diameter dp, with diameter aspect ratios of 4 De/dp 20. The correlation is
Consider an APB with outside diameter of 140 mm, inside diameter of 40 mm, packed with identical 10-mm diameter spheres.
(a) Estimate the void fraction at a distance from the outer wall of 25 mm. Solution
(c) Sho ha he a e age po o i fo an APB i gi en b
(d) E ima e he a e age po o i fo he APB de c ibed abo e. Sol ion
4.5a. Minimum liquid mass velocit for proper wetting of packing.
A 1.0-m diame e bed ed fo ab o p ion of ammonia i h p e a e a 298 K i packed i h 25-mm pla ic In alo addle . Calc la e he minim m a e flo a e, in kg/ , neded o en e p ope e ing of he packing face.
Sol ion
Fo pla ic packing, L,min = 1.2 mm/ .
F om S eam Table
4.6a. Minimum liquid mass velocit for proper wetting of packing. Repea P oblem 4.5, b ing ce amic in ead of pla ic In alo addle . Sol ion
From Steam Tables
4.7b. S ecific i id h d a d id f ac i i fi -ge e a i a d ac i g.
Repeat Example 4.2, but using 25-mm ceramic Berl saddles as packing material. Solution
From Table 4.1:
From Example 4.2:
4.8b. S ecific i id h d a d id f ac i i c ed ac i g.
Repeat Example 4.2, but using Montz metal B1-200 structured packing (very similar to the one shown in Figure 4.3). For this packing, a = 200 m 1, = 0.979, Ch = 0.547 (Seader and Henley, 1998).
Solution
4.9b. S ecific i id h d a d id f ac i i fi -ge e a i a d ac i g. A e ac ed i h 25- ce a ic Ra chig i g i be ed f ab bi g be e e a f a di e i e i h a i e ga i g a a h i a 300 K. The i c i f he i i 2.0 cP a d i de i i 840 g/ 3. The i id a e ci i L' = 2.71 g/ 2- . E i a e he i id h d , he id f ac i , a d he h d a ic ecific a ea f he ac i g. S i F Tab e 4.1:
4.10b. Pressure drop in beds packed ith first-generation random packings.
Repeat E ample 4.3, but using 15-mm ceramic Raschig rings as packing material. Assume that, for this packing, C = 1.783.
Solution
Packed Col mn Design Program
This program calculates the diameter of a packed column to satisf a given pressure drop criterium, and estimates the volumetric mass-transfer coefficients. Enter data related to the gas and liquid streams
Enter liquid flow rate, mL, in kg/s Enter gas flow rate, mG, in kg/s
Enter liquid densit , in kg/m3 Enter gas densit , kg/m3
Enter liquid viscosit , Pa-s Enter gas viscosit , Pa-s
Enter temperature, T, in K Enter total pressure, P, in Pa
Enter data related to the packing
Enter packing factor, Fp, in ft2/ft3 Enter specific area, a, m2/m3
Introduce a units conversion factor in Fp
Enter porosit , fraction Enter loading constant, Ch
Enter pressure drop constant, Cp Enter allowed pressure drop, in Pa/m
Calculate Y at flooding conditions
Calculate gas velocit at flooding, vGf
As a first estimate of the column diameter, D, design for 70% of flooding
Calculate gas volume flow rate, QG, in m3/s
Calculate liquid volume flow rate, QL, in m3/s
I e a e o find he o e diame e fo he gi en p e e d op
Col mn diame e , in me e
4.11b. P e e d op and app oach o flooding in c ed packing.
Repeat Example 4.3, but using Montz metal B1-200 structured packing (very similar to the one shown in Figure 4.3). For this packing, Fp = 22 ft2/ft3, a = 200 m 1, = 0.979, Ch = 0.547, Cp = 0.355 (Seader and Henley, 1998).
Solution
Packed Col mn Design Program
This program calculates the diameter of a packed column to satisf a given pressure drop criterium, and estimates the volumetric mass-transfer coefficients. Enter data related to the gas and liquid streams
Enter liquid flow rate, mL, in kg/s Enter gas flow rate, mG, in kg/s
Enter liquid densit , in kg/m3 Enter gas densit , kg/m3
Enter liquid viscosit , Pa-s Enter gas viscosit , Pa-s
Enter temperature, T, in K Enter total pressure, P, in Pa
Enter data related to the packing
Enter packing factor, Fp, in ft2/ft3 Enter specific area, a, m2/m3
Introduce a units conversion factor in Fp
Enter porosit , fraction Enter loading constant, Ch
Enter pressure drop constant, Cp Enter allowed pressure drop, in Pa/m
Calculate Y at flooding conditions
Calculate gas velocit at flooding, vGf
As a first estimate of the column diameter, D, design for 70% of flooding
Calculate gas volume flow rate, QG, in m3/s
Calculate liquid volume flow rate, QL, in m3/s
I e a e o find he o e diame e fo he gi en p e e d op
Col mn diame e , in me e
F ac ional app oach o flooding
4.12b, d. Pressure drop in beds packed ith second-generation random packings.
A packed to er is to be designed for the countercurrent contact of a ben ene-nitrogen gas mi ture ith kerosene to ash out the ben ene from the gas. The gas enters the to er at the rate of 1.5 m3/s, measured at 110 kPa and 298 K, containing 5 mole % ben ene. Esentiall , all the ben ene is
ab bed b he e e e. The i id e e he e a he a e f 4.0 g/ ; he i id de i i 800 g/ 3, i c i i 2.3 cP. The ac i g i be 50- e a Pa i g , a d he e dia e e
i be ch e d ce a ga - e e d f 400 Pa/ f i iga ed ac i g.
(a) Ca c a e he e dia e e be ed, a d he e i g f ac i a a ach f di g. (b) A e ha , f he dia e e ch e , he i iga ed ac ed heigh i be 5 a d ha 1 f
i iga ed ac i g i be aced e he i id i e ac a e ai e e a a . The b e c bi a i be ed a he ga i e i ha e a e a echa ica efficie c f 60%. Ca c a e he e e i ed b he ga h gh he ac i g. S i (a) De ig f c di i a he b f he e he e he a i f f ga a d i id cc Be e e e e i g i h he ga : A i g ha a f he be e e i ab bed: F he L ca e h d f i e f ga e : U i g he Pac ed C De ig P g a : D = 0.913 f = 0.825 (b) Ca c a e he e e d h gh he d ac i g
From the Lucas method for mixtures of gases:
From equation (4-11):
(c) Estimate the volumetric mass-transfer coefficients for the gas and liquid phases. Assume that DL = 5.0 10 10 m2/s.
From the wilke-Lee equation, DG = 0.0885 cm2/s From Table 4.2, CL = 1.192, CV =
0.410Using the Packed Column Design Program: kLah = 0.00675 s 1; kyah = 0.26 kmol/m3-s
4.13b, d. Pressure drop in beds packed ith structured packings.
Redesign the packed bed of Problem 4.12, but using Montz metal B1-200 structured packing (very similar to the one shown in Figure 4.3). For this packing, Fp = 22 ft2/ft3, a = 200 m 1, e = 0.979, Ch = 0.547, Cp =0.355, CL = 0.971, CV = 0.390 (Seader and Henley, 1998).
Solution
Using the Packed Column Design Program: D = 0.85 m f = 0.859
kLah = 0.00861 s 1; kyah = 0.376 kmol/m3-s
4.14c, d. Air stripping of aste ater in a packed column.
with air in a packed column operating at 298 K and 2 atm to reduce the benzene concentration to 0.005 ppm. The packing specified is 50-mm plastic Pall rings. The air flow rate to be used is 5 times the minimum. Henry's law constant for benzene in water at this temperature is 0.6 kPa-m3/mole (Davis and Cornwell, 1998). Calculate the tower diameter if the gas-pressure drop is not to exceed 500 Pa/m of packed height. Estimate the corresponding mass-transfer coefficients. The diffusivity of benzene vapor in air at 298 K and 1 atm is 0.096 cm2/s; the diffusivity of liquid benzene in water at infinite dilution at 298 K is 1.02 10 5 cm2/s (Cussler, 1997).
Solution
Calculate m, the slope of the equilibrium curve: For water at 298 K,
Calculate the minimum air flow rate:
Convert liquid concentrations from ppm to mole fractions
At these low concentrations the equilibrium and operating lines are straight, and y2 (max) = y2* = mx2
